Samacheer Kalvi 11th Chemistry Solutions Chapter 14 Haloalkanes and Haloarenes

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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 14 Haloalkanes and Haloarenes

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Samacheer Kalvi 11th Chemistry Haloalkanes and Haloarenes Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Haloalkanes and Haloarenes Multiple Choice Questions

Question 1.
The IUPAC name of Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes  is ………….
(a) 2-Bromopent – 3 – ene
(b) 4-Bromopent – 2 – ene
(c) 2-Bromopent – 4 – ene
(d) 4-Bromopent – 1 – ene
Answer:
(b) 4 – Bromopent – 2 – ene

Question 2.
Of the following compounds. which has the highest boiling point’?
(a) n-Butyl chloride
(b) Isobutyl chloride
(c) t-Butyl chloride
(d) n-propyl chloride
Answer:
(a) n-Butyl chloride

Question 3.
Arrange the following compounds in increasing order of their density.
(A) CCl4
(B) CHCl3
(C) CH2Cl2
(D) CH3Cl
(a) D<C<B<A
(b) C<B<A<D
(c) A<B<C<D
(d) C<A<B<D
Answer:
(a) D<C<B<A

Question 4.
With respect to the position of – Cl in the compound CH3 – CH = CH – CH2 – Cl, it is classified as ………..
(a) Vinyl
(b) Allyl
(c) Secondary
(d) Aralkyl
Answer:
(b) Allyl

Question 5.
What should be the correct IUPAC name of diethyl chioromethane?
(a) 3-Chioropentane
(b) 1-Chloropentane
(c) 1-Chloro- 1, 1, diethylmethanc
(d) 1-Chloro- 1 -ethylpropane
Answer:
(a) 3-Chioropentane

Question 6.
C-X bond is strongest in …………
(a) Chioromeihane
(b) lodomethane
(c) Bromomethane
(d) Fluoromethanc
Answer:
(d) Fluoromethane

Question 7.
In the reaction Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes  X + N2, X is …………
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 8.
Which of the following compounds will give racemic mixture on nucleophilic substitution by OH ion?
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
(a) (i)
(b) (ii) and (iii)
(c) (iii)
(d) (i) and (ii)
Answer:
(c) (iii)

Question 9.
The treatment of ethyl formate with excess of RMgX gives –
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Answer:
(c) R—CHO

Question 10.
Benzene reacts with Cl2 in the presence of FeCl2 and in absence of sunlight to form –
(a) Chiorobenzene
(b) Benzyl chloride
(c) Benzal chloride
(d) Benzene hexachloride
Answer:
(a) Chiorobenzene

Question 11.
The name of C2F4Cl2 is –
(a) Freon – 112
(b) Freon – 113
(c) Freon – 114
(d) Freon – 115
Answer:
(c) Freon – 114

Question 12.
Which of the following reagent is helpful to differentiate ethylene dichloride and ethylidene chloride?
(a) Zn / methanol
(b) KOH / ethanol
(c) Aqueous KOH
(d) ZnCl2 / Cone. HCl
Answer:
(e) Aqueous KOH

Question 13.
Match the compounds given in Column I with suitable items given in Column II.
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes -249
Code:
(a) A→2,B→4,C→1,D→3
(b) A→3,B→2,C→4,D→1
(c) A→1,B→2,C→3,D→4
(d) A→3,B→1,C→4,D→2
Answer:
(d) A→3,B→1,C→4,D→2

Question 14.
Assertion : in monohaloarenes, electrophilic substitution occurs at oriho and para positions.
Reason : Halogen atom is a ring deactivator.
(a) If both assertion and reason arc true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) If assertion is true but reason is false.
(d) If both assertion and reason are false.
Answer:
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.

Question 15.
Consider the reaction, CH3CH2CH2Br + NaCN → CH3CH2CH2CN + NaBr This reaction will be the fastest in …………
(a) ethanol
(b) methanol
(c) DMF (N, N’ – dimethyl forrnamide)
(d) water
Answer:
(c) DMF (N, N’ – dimethyl fonuarnide)

Question 16.
Freon-12 is manufactured from tetrachioromethane by …………
(a) Wurtz reaction
(b) Swarts reaction
(c) DMF (N, N’ – dimethyl tòrmamide)
(d) water
Answer:
(c) DMF (N, N’ – dimethyl formarnide)

Question 16.
Frcon-12 is manufactured from tetrachioromethane by …………
(a) Wurtz reaction
(b) Swans reaction
(c) Haloform reaction
(d) Gattennann reaction
Answer:
(b) Swarts reaction

Question 17.
The most easily hydrolysed molecule under SN1 condition is …………
(a) allyl chloride
(b) ethyl chloride
(c) isopropyl chloride
(d) benzyl chloride
Answer:
(a) benzyl chloride

Question 18.
The carbocation formed in SN1 reaction of alkyl halide in the slow step is …………
(a) sp3 hybridised
(b) sp2 hybridised
(c) sp hybridised
(d) none of these
Answer:
(b) sp2 hybridised

Question 19.
The major products obtained when chiorobenzene is nitrated with HNO3 and cone. H2SO4
(a) 1-chloro-4-nitrobenzenc
(b) 1-chloro-2-n itrobenzene
(c) 1-chloro-3-n itroberizene
(d) 1-chloro- 1 -nitrobenzene
Answer:
(a) 1 -chloro-4-nitrobenzene

Question 20.
Which one of the following is most reactive towards nucleophilic substitution reaction?
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 21.
Ethylidene chloride on treatment with aqueous KOH gives …………
(a) acetaldehyde
(b) ehtylene glycol
(c) formaldehyde
(d) glyoxal
Answer:
(a) acetaldehyde

Question 22.
The raw material for Rasching process is …………
(a) chiorobenzene
(b) phenol
(c) benzene
(d) anisole
Answer:
(c) benzene

Question 23.
Chloroform reacts with nitric acid to produce …………
(a) nitro-toluene
(b) nitro-glycerine
(c) chloropicrin
(d) chioropicric acid
Answer:
(c) chioropicrin

Question 24
Acetone Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes X, X is …………
(a) 2-propanol
(b) 2-methyl-2-propanol
(c) 1 -propanol
(d) acetonol
Answer:
(b) 2-methyl-2-propanol

Question 25.
Silver propionate when refluxed with Bromine in carbon tetrachioride gives …………
(a) propionic acid
(b) chioroethane
(c) bromoethane
(d) chioropropane
Answer:
(c) bromoethane

Samacheer Kalvi 11th Chemistry Haloalkanes and Haloarenes Short Answer Questions

Question 26.
Classify the following compounds in the form of alkyl. allylic, vinyl, benzylic halides:
(a) CH3-CH = CH-Cl
(b) C6H5CH2I
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
(d) CH2 = CH – Cl
Answer:
(a) CH3-CH = CH-Cl – Allylic halide
(b) C6H5CH2 – Benzylic halide
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
(d) CH2 = CHCl – Vinyl halide

Question 27.
Why chlorination of methane is not possible in dark?
Answer:

  • Chlorination of methane is a free radical substitution reaction.
  • Before chlorine reacts with methane, the Cl-Cl single bond must break to form free radicals and this can only be done in the presence of ultraviolet light.
  • In dark, chlorine free radicals formation is not possible and so chlorination of methane is not possible in dark.
  • The ultraviolet light is a source of energy and it being used to break of Cl-Cl and
    produce Cl free radical Free radical’s which can attack methane. in dark this is not possible.

Question 28.
How will you prepare n-propyl iodide from n-propyl bromide?
Answer:
Finkeistein reaction:
n-propyl bromide on heating with a concentrated solution of sodium iodide in dry acetone gives n-propyl iodide. This SN2 reaction is called Finkelstein reaction.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 29.
Which alkyl halide from the following pair is –
1. chiral
2. undergoes faster SN2 reaction?
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Answer:
1.Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes Here the C*is chiral carbon atom and it is surrounded by four different groups. Br
2. Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes  i.e., 1-chiorohutane undergoes faster SN2 reaction. CH3-CH2-CH2Cl is a primary alkyl halide and so it undergoes faster SN2 reaction.

Question 30.
How does chiorobenzene react with sodium in the presence of ether? What is the name of the reaction?
Answer:
Chiorobenzene when heated with sodium in ether solution will forrn biphenyl as the product. This reaction is called fitting reaction.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 31.
Give reasons for polarity of C-X bond in haloalkancs.
Answer:

  • Carbon halogen bond is a polar bond as halogens are more electronegative than carbon.
  • The carbon atom exhibits a partial positive change (δ+) and halogen atom acquires a partial negative change. (δ)
  • Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 32.
Why is it necessary to avoid even traces of moisture during the use of Grignard reagent?
Answer:
Grignard reagents are highly reactive substances. They react with any source of proton to form hydrocarbons. Even water is sufficiently acidic to convert it into the corresponding hydrocarbon. So it is necessary to avoid even traces of moisture with the Grignard reagent as they arc highly reactive.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 33.
What happens when acetyl chloride is treated with excess of CH3 MgI?
Answer:
When acetyl chloride is treated with excess of Grignard reagent, the product formed is tertiary butyl alcohol.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 34.
Arrange the following alkyl halideïniiicreasing order of bond enthalpy of RX: CH3Br, CH3F, CH3Cl, CH3I
Answer:
increasing order of bond enthalpy of RX is …………
CH3 I bond enthalpy 234 kJ mol-1
CH3 Br bond enthalpy 293 kJ mol-1
CH3 Cl bond enthalpy 351 kJ mol-1
CH3 F bond enthalpy 452 IcI mol-1
CH3 F > CH3 Cl> CH3Br < CH3 I

Question 35.
What happens when chloroform reacts with oxygen in the presence of sunlight?
Answer:
Chloroform undergoes oxidation in the presence of sunlight and air to form phosgene (carboxyl chloride)
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Since phosgene is very poisonous, therefore its presence makes chloroform unfit for use as anaesthetic.

Question 36.
Write down the possible isomers of C5H11Br and give their IUPAC and common names.
Answer:
C5H11Br : 8 isomers are possible.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 37.
Mention any three methods of preparation of haloalkanes from alcohols.
Answer:
Haloalkanes are prepared from alcohols by treating with –

  1. HCl
  2. PCl5
  3. SO2Cl2

Answer:
1. Reaction of alcohol with hydrogen halide:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

2. Reaction of alcohol with phosphorous halide:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

3. Reaction of alcohol with thionyl chloride:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 38.
Compare SN1 and SN2 reaction mechanisms.
Answer:
SN1 Mechanism:
1. It is a unimolecular nucleophilic substitution reaction of first order.
2. It takes place in two steps.
3. It leads to racemisation.
4. It mostly take place in tertiary alkyl halides.
5. The rate of the reaction depends only on the concentration one of substrate and so it is a first order reaction.
Step – I
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Step – II
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

SN2 Mechanism:
1. It is a bimolecular nucleophilic substitution reaction of second order.
2. It takes place in one step.
3. It leads to invertion of configuration.
4. It mostly take place in primary alkyl halides.
5. The rate of the reaction depends on the concentration of both the substrate as well as the nucleophile and so it is a second order reaction.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 39.
Compare SN1 and SN2 reaction mechanisms.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 40.
Discuss the aromatic nucicophilic substitution reactions of chiorobenzene.
Answer:
Aromatic nucleophilic substitution reactions:
Dow’s process:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 41.
Account for the following –
1. t-butyl chloride reacts with aqueous KOH by SN1 mechanism while n-butyl chloride reacts with SN2 mechanism.
2. p-dichlorobenzene has higher melting point than those of o-and m-dichlorobenzene.
Answer:
1. In t-butyl chloride, there is niore steric hindrance and it involves formation of a stable tertiary carbocation. Therefore it reacts with KOH by SN1 mechanism rather than SN2 mechanism because SN1 mechanism is faourable in case of steric crowding and is directly proportional to partial positive charge on carbon atom. In n-butyl chloride, there is least steric hindrance and involves formation of less stable primary carbocation. Thus it takes place in one step and is favoured by SN2 mechanism.

2. Melting point of p – dichiorobenzene is higher than that of ortho and meta-dichiorobenzene. This is due to the fact that is has a symmetrical structure and therefore, its molecules can easily pack closely in the crystal lattice. p-dichlorobcnzene being more symmetrical fits closely in the crystal lattice and has stronger intermolecular attraction than o & m isomers. So p-isomer has high melting point than the corresponding o & m-isomers.

Question 42.
In an experiment ethyliodide in ether is allowed to stand over magnesium pieces. Magnesium dissolves and product is formed
(a) Name the product and write the equation for the reaction.
(b) Why all the reagents used in the reaction should be dry? Explain.
(c) How is acetone prepared from the product obtained in the experiment?
Answer:
(a) When ethyl iodide in ether is allowed to stand over magnesium pieces. the product formed will be Ethyl magnesium iodide.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

(b) Gngnard reagent are highly reactive compounds and react with any source of proton to give hydrocarbons. Even when water is there, it is sufficiently acidic to convert it into the corresponding hydrocarbon. So it is necessary to avoid even traces of moisture with the grignard reagent as they are highly reactive. That is why the all reagents used in the reaction should be dry.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Methyl magnesium iodide reacts with acetyl chloride to form acetone.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 43.
Write a chemical reaction useful to prepare the following:
1. Freon- 12 from carbon tetrachioride.
2. Carbon tetrachioride from carbon disulphide.
Answer:
1. Freon-12 from carbon tetrachioride:
Freon- 12 is prepared by the action of hydrogen fluoride on carbon tetrachioride in the presence of catalytic amount of antimony pentachloride. This is called “swartz reaction”.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

2. Carbon tetrachioride from carbon disuiphide:
Carbon disuiphide reacts with chlorine gas in the presence ofanhydrous AlCl3 as catalyst to give carbon tetrachloridc.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 44.
What are Freons? Discuss their uses and environmental effects.
Answer:
Freons are the chiorofluoro derivatives of methane and ethane.
Freon is represented as Freon – cba
Where, c = number of carbon atoms, b = number of hydrogen atoms, a = total number of fluorine atoms.
CF2Cl2 = 0
c = 1 – 1 = 0
H = 0 +1 = 1
F = 2
So Freon – 12 is CF2Cl2

Uses of Freons:

  • Freons are used as refrigerants in refrigerators and air conditioners.
  • it is used as a propellant for aerosols and foams.
  • it is used as propellant for foams to spray out deodorants, shaving creams and insecticides.

Environmental effects of Freons:

1. Freon gas is a very powerful greenhouse gas which means that it traps the heat normally radiated from the earth out into the space. This causes the earth’s temperature to increase, resulting in rising sea levels, droughts. stronger storms, flash floods and a host of other very unpleasant effect.

2. As freon moves throughout the air, its chemical ingredients causes depletion of ozone layer. Depletion of ozone increases the amount of ultraviolet radiations that reaches the earths surface, resulting in serious risk to human health. High levels of ozone, in turn, causes resoiratory problems and can also kill olants.

Question 45.
Predict the products when bromoethane is treated with the fòllowing:
1. KNO2
2. AgNO2
Answer:
1. When bromoethane is treated with KNO2 the product formed is Ethyl nitrite.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
2. When bromoethane is treated with AgNO2 nitroethane will be formed as the product.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 46.
Explain the mechanism of SN1 reaction by highlighting the stereochemistry behind it.
Answer:
1. SN1 stands for unimolecular nucleophilic substitution reaction of first order reaction.
2. The rate of the following SN1 reaction depends upon the concentration of alkyl halide
(RX) and it is independent of the concentration of the nucicophile (OH ) used.
3. R-Cl + OH R – OH + Cl
4. This SN1 reaction follows first order kinetics and it occurs in two steps:
5. SN1 reaction mechanism takes place in tertiary butyl bromide with aqueous KOH a follows:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
6. the 2 steps of the reaction are:
Step 1:
Formation of carbocation:
The polar C-Br bond breaks first forming a carbocation and bromide ion. This step is slow and hence it is the rate determining step.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
The carbocation has two equivalent lobes of the vaccant 2p orbital, So it can react equally fast form either face.

Step 2:
The nucleophile immediately reacts with the carbocation. This step is fast and hence does not affect the rate of the reaction.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
The nucleophile OH can attack carbocation from both the sides.

Question 47.
Write short notes on the the following:
1. Raschig process
2. Dows Process
3. Darzens process
Answer:
1. Raschig process:
Chiorobenzene is commercially prepared by passing a mixture of benzene vapour in air and HUt over heated cupric chloride.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

2. Dow’s process:
Chiorobenzene is boiled with Sodium hydroxide to get Phenol. This reaction is called Dow’s process.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

3. Darzens process:
Ethanol reacts with SOCl, in the presence of pyridine to form chloroethane. This reaction is called Darzens process.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 48.
Starting from CH3MgI. How will you prepare the following’?

  1. Acetic acid
  2. Acetone
  3. Ethyl acetate
  4. Isopropyl alcohol
  5. Methyl cyanide

Answer:
1. Acetic acid from grignard reagent:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

2. Acetone from Gngnard reagent:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

3. Ethyl acetate from Grignard reagent:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

4. Isopropyl alcohol from Grignard reagent:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

5. Methyl cyanide form Grignard reagent:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 49.
Complete the following reactions:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 50.
Explain the preparation of the following compounds:

  1. DDT
  2. Chloroform
  3. Biphenyl
  4. Chioropicrin
  5. Freon-12

Answer:
1. DUT
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

2. Chloroform:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

3. Biphenyl:
Fittig reaction:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

4. Chioropicrin.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

5. Freon – 12.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 51.
An organic compound (A) with molecular formula C2H5Cl reacts with KOH gives compounds (B) and with alcoholic KOH gives compound (C). Identify (A), (B) and (C)
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
1. An organic compound (A) of molecular formula C2H5Cl is identified as Chioroethane with molecular formula CH2-CH2Cl from the formula.

2. Chioroethane reacts with aqueous KOH to give ethanol, i.e., Aqueous C2H5OH as (B) by substitution reaction.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

3. ChIooethane reacts with alcoholic KOH to give ethene C2H4 as (C) by elimination reaction.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes -250
Question 52.
Simplest alkene (A) reacts with HCl to form compound (B).Compound (B) reacts with ammonia to form compound (C) of molecular formula C2H2N. Compound (C) undergoes carbylarnine test. Identify (A), (B), and (C).
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
1. The simplest alkene (A) is CH2 = CH2, ethene.

2. Ethene reacts with HCl to give Chioroethane CH2-CH2Cl as (B) by addition reaction.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

3. Chioroethane reacts with ammonia to give Ethylamine CH3-CH2 NH2 as (C). It is a primary amine and Carbylamine test is the characteristic test for 1° amine.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 53.
A hydrocarbon C3H6 (A) reacts with HBr to form compound (B). Compound (B) reacts with aqueous potassium hydroxide to give (C) of molecular formula C3H8O. What are (A) (B) and (C). Explain the reactions.
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
1. The hydrocarbon with molecular formula C3H6 (A) is identified as propene,
CH3-CH = CH2

2. Propene reacts with HBr to form brornopropane CH3—CH2—CH2Br as (B).
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

3. 1 -bromopropanc react with aqueous potassium hydroxide to give 1 -propanol
CH3 – CH2 CH2OH as (C).

4. 2-bromo propane reacts with aqueous KOH to give 2-propanol as (C)
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 54.
Two isomers (A) and (B) have the same molecular formula C2H4Cl2. Compound (A) reacts with aqueous KOK, gives compound (C) of molecular formula C2H4O. Compound (B) reacts with aqueous KOH, gives compound (D) of molecular formula C2H6O2. Identif’ (A), (B), (C) and (D).
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
1. The compounds (A) and (B) with the molecular formula C2H4Cl2 are Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes and Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes respectively with IUPAC names: 1, 1-dichloroethane (A) and 1.

2. 1, 1-dichloroethane reacts with aqueous KOH tp give CH3CHO Acetaldehyde as (C).

3. 1 ,2-dichloroethane reacts with aqueous KOH to give ethylene glycol Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes  as (D).
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

[Evaluate Yourself ]

Question 1.
Write the IUPAC name of the following:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 2.
Write the structure of the following compounds:

  1. 1-Bromo-4-cthylcyclo hexane
  2. 1 4-Dichlorobut-2-ene
  3. 2 Chloro-3-methyl pentane

Answer:
1. 1-Bromo-4 ethylcyclo hexane
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

2. 1. 4-Dichioro but-2-ene
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

3. 2-Çhloro-3-methyl pentane
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 3.
Write all possible chain isomers with molecular formula C5H11Cl
Answer:
C5H11Cl:
8 isomers arc possible. These are –
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 4.
neo-pentyl bromide undergoes nucleophilic substitution reactions very slowly. Justify.
Answer:
1.Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes   neo-pentyl bromide undergoes nucleophilic substitution reactions very slowly due to steric hindrance of many alkyl groups. When bromide is attached to neo-pentyl carbon atom, the heterolytic cleavage of C-Br takes place very slowly and substitution is also a very slow reaction.

2. Due to bulky neo-pentyl group, it becomes difficult for a nucleophile to attack from the back side of carbon having C-Br bond.

3. Splitting of C-Br gives a primary carbocation which is very less stable. So neo-pentyl bromide undergoes nucleophilic substitution reactions very slowly.

Question 5.
Why Grignard reagent should be prepared in anhydrous condition’?
Answer:
Grignard reagent are highly reactive species and react with any source of proton to give hydrocarbons. Even when water is there, it is sufficiently acidic to convert it into the corresponding hydrocarbon. So it is necessary to avoid even traces of moisture while preparing grignard reagent as they are highly reactive and that is why gngnard reagent should be prepared in anhydrous conditions.

Question 6.
Haloalkanes undergo nucleophilic substitution reaction whereas haloarenes undergo electrophilic substitution reaction. Comment.
Answer:
Haloalkanes undergo nucleophilic substitution reactions due to high clectronegativity of the halogen atom. The C-X bond in haloalkanes is slightly polar, thereby the C atom acquires a slight positive charge in C-X bond. [lence C atom is a good target for attack for nucleophiles. Therefore C-X atom of haloalkanes is replaced by a nucleophile easily.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

On the other hand in haloarenes, the halogen atom releases electron to the benzene nucleus relatively electron rich with respect to the halogen atom. As a result the electrophilic attack occurs at ortho and para positions. Hence haloarenes undergo electrophilic substitution reactions.

Question 7.
Chloroform is kept with a little ethyl alcohol in a dark coloured bottle. Why?
Answer:
1. Chlorofòrm is slowly oxidised by air in the presence of light to an extremely poisonous gas, carboxyl chloride (phosgene). it is therefore stored in closed dark coloured bottles completely filled so that air is kept out.

2. With the use of 1% ethanol we can stabilise chloroform, because ethanol can convert the poisonous COCl2 gas into non poisonous diethyl carbonate.
COCl2 + 2C2H5OH CO(OC2H5)2 + 2HCl

Question 8.
What is the IUPAC name of the insecticide DDT? Why is their use banned in most of the countries?
Answer:
1. The IUPAC name of the insecticide DDT isp, p-dichloro-diphenyl trichioroethane.
2. Even DDT is an effective insecticide. Now-adays it is banned because of its long term toxic effects.
3. DDT is very persistent in the environment and it has a high affinity for fatty tissues. As a result, DDT gets accumulated in animal tissue fat, in particular that of birds of prey with subsequent thinning of their eggs shells and impacting their rate of reproduction. That is why DDT is banned in most of the countries.

Samacheer Kalvi 11th Chemistry Haloalkanes and Haloarenes Additional Questions Solved

I. Choose The Correct Answer.

Question 1.
Which of the following is an example for polyhalo compounds?
(a) Vnyl iodide
(b) Chiorobenzene
(c) Allyl chloride
(d) Chloroform
Answer:
(d) Chloroform

Question 2.
Which of the following is a secondary haloalkane?
(a) Bromoethane
(b) 2-Chioropropane
(c) 2-Jodo-2-methylpropane
(d) 1-Chloropropane
Answer:
(b) 2-Chioropropane

Question 3.
How many isomers are possible for the formula C4H9Cl?
(a) 3
(b) 2
(c) 4
(d) 5
Answer:
(c) 4

Question 4.
How many isomers are possible for the formula C5H11Br?
(a) 11
(b) 8
(c) 4
(d) 5
Answer:
(b) 8

Question 5.
Which of the following is called Lucas reagent?
(a) Conc. H2SO4 + Anhydrous CuSO4
(b) Conc.HCl + Anhydrous ZnCl2
(c) Dil.HCl + AlCl3
(d) Conc.HCl + ConcHNO2
Answer:
(b) Conc.HCl + Anhydrous ZnCl2

Question 6.
Which of the following mechanism is followed in the halogenation of alkanes in the presence of U-V light?
(a) Nucleophilic substitution
(b) Electrophilic addition
(c) Free radical substitution
(d) Elimination reaction
Answer:
(c) Free radical substitution

Question 7.
The reactivity of alcohols with haloacid is –
(a) 3° > 2° > 1°
(b) 1° > 2° > 3°
(c) 2° > 3° > 1°
(d) 3° > l° > 2°
Answer:
(a) 3° > 2°> 1°

Question 8.
Which of the following reagent is not used to convert alcohol to haloalkane?
(a) H-X
(b) PX5
(c) CCl4
(d) SOCl2
Answer:
(c) CCl4

Question 9.
What is the name of the reaction in which bromoethane is converted to iodoethane by reacting with NaI in acetone?
(a) Hunsdicker reaction
(b) Dow’s process
(c) Finkelstein reaction
(d) Swarts reaction
Answer:
(c) Finkeistein reaction

Question 10.
Identify the correct order of boiling point of haloalkanes?
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 11.
Which of the following pair functional groups represents ambident nucleophiles?
(a)-SH-&-OH
(b)-CN-&-NO2
(c)-Br-&-Cl
(d)-O-&-CHO
Answer:
(b)-CN & NO2

Question 12.
Which one following mechanism will be followed when Tartiary butyl chloride is treated with alcoholic KOH?
(a) SN1 mechanism
(b) E1 mechanism
(c) SN2 mechanism
(d) E2 mechanism
Answer:
(b) E1 mechanism

Question 13.
Which one of the following is used for producing pesticìdes?
(a) CHI3
(b) CHCl3
(c) CCl3 NO2
(d) CCl4
Answer:
(b) CHCl3

Question 14.
Which one of the following react with gringnard reagent followed by hydrolysis will yield primary alcohol?
(a) CH3 CHO
(b) HCHO
(c) CH3 COCH3
(d) CO2
Answer:
(b) HCHO

Question 15.
Which one of the following reacts with CH3MgI followed by hydrolysis and gives isopropyl alcohol?
(a) CH3 COCH3
(b) CH3 CHO
(c) HCHO
(d) CNCl
Answer:
(b) CH3 CHO

Question 16.
Which one of the following reacts with CH3 Mg I followed by hydrolysis to yield tert. butyl alchol ?
(a) CH3CHO
(b) HCHO
(c) CH3COOC2H5
(d) CH3COCH3
Answer:
(d) CH3COCH3

Question 17.
Which one of the following reacts with CH3 Mg I followed by acid hydrolysis to yield acetic acid ?
(a) CNCl
(b) CH3COOC2H5
(c) HCOOC2H5
(d) CO2
Answer:
(d) CO2

Question 18.
Which one of the following reagent react with methyl magnesium iodide followed by acid hydrolysis to give ethyl acetate?
(a) Chiorodimethyl ether
(b) Ethyl chloroformate
(c) Ethyl formate
(d) Acetaldehyde
Answer:
(b) Ethyl chioroformate

Question 19.
Which one of the following is used as fibre-swelling agent in textile processing?
(a) Chiorohenzene
(b) Chloroform
(c) Chlorai
(d) Chloroethane
Answer:
(a) Chiorobenzene

Question 20.
Which one of the fillowing is a gemdihalide?
(a) CH3CHCl2
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
(c) CH3-CH2Cl
(d) C6H4Cl2
Answer:
(a) CH3CHCl2

Question 21.
Which of the following reagent is used to distinguish gem-dihalides and vicinal dihalides?
(a) Alcoholic KOH
(b) Aqueous KOH
(c) FeCl3 / Cl2
(d) Ethanol
Answer:
(b) Aqueous KOH

Question 22.
Which one of the following is used in the conversion of ethyliden dichioride to Acetylene?
(a) Zn + Methanol
(b) KOH + Ethanol
(c) Aqueous NaOH
(d) Alcoholic KOH
Answer:
(b) KOH + Ethanol

Question 23.
Which one of the following is used as a metal cleaning solvent?
(a) Isopropylidene chloride
(b) Methylene chloride
(c) Chloroform
(d) lodoform
Answer:
(b) Methylene chloride

Question 24.
Which one of the following is used as an insecticide and as a soil sterilising agent?
(a) Chloroform
(b) Chlorai
(c) Chioropicrin
(d) Tetrachioromethane
Answer;
(c) Chloropicrin

Question 25.
Which one of the following is used to test primary amincs?
(a) Schiff’s test
(b) Carbylarnine test
(c) Dye test
(d) Silver mirror test
Answer:
(b) Carbyianiine test

Question 26.
Which one of the following is used as propellant for aerosols and foams?
(a) Freons
(b) Methylidene chloride
(c) Chlorai
(d) Chloroform
Answer:
(a) Freons

Question 27.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
the product X is –

Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 28.
Which one of the following will undergo SN1 reaction faster?
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 29.
Which one of the following compounds does not undergo nucleophilic substitution reactions at all?
(a) Ethyl bromide
(b) Vinyl chloride
(c) Benzyl chloride
(d) isopropyl chloride
Answer:
(b) Vinyl chloride

II. Match the following.

Question 1.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 2.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 3.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 4.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 5.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes “17” />

Question 6.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

III. Fill in the blanks.

Question 1.
is used in the treatment of typhoid ……….
Answer:
chloramphenicol

Question 2.
………. is used in the treatment of malaria
Answer:
Chioroquine

Question 3.
………. is used as an anesthetic ……….
Answer:
Halothane

Question 4.
………. is used for cleaning electronic equipments ……….
Answer:
Trichioroethylene

Question 5.
The IUPAC name of CH2=CH-CH2Cl is ……….
Answer:
3-Chioro-prop- 1-ene

Question 6.
The structure of Vinyl iodide is ……….
Answer:
CH2=CHI

Question 7.
2-iodo-2-methylpropane belongs to type ……….
Answer:
3°haloalkanes

Question 8.
The structure of 2-Iodo-2-methylpropanc is ………
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 9.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
the IUPAC name of this compound is ……..
Answer:
1-Bromo-2, 2-dimethyipropane

Question 10.
The IUPAC name of CH2=CHCl is ………
Answer:
Chioroethene

Question 11.
The IUPAC name of Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes is ………
Answer:
2-Bromo-3-chloro-2, 4-dimethylpentane.

Question 12.
The IUPAC name of Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes is ………
Answer:
3-Chloro-2-methyl-prop- 1 -ene

Question 13.
The reactivity of haloacids (HCl , HBr, HI) with alcohol is in the order ………
Answer:
HI > HBr > HCl

Question 14.
The decreasing order of bond length among alkyl halides (CH2I , CH2 Br , CH3F , CH3Cl ) is in the order ………
Answer:
CH3F < CH3Cl < CH3Br < CH3I

Question 15.
The bond strength of C-X for the C-Cl , C-Br, C-I , C-F decreases in the order is ………
Answer:
C-F < C-Cl<C-Br < C-I

Question 16.
The catalyst used in Darzen halogenation of alcohol is ………
Answer:
Pyridine

Question 17.
In Finkeistein reaction, the mechanism followed is
Answer:
SN2

Question 18.
Silver salt of fatty acid is converted to bromo alkanc by ………
Answer:
Hunsdicker reaction

Question 19.
In Swarts reaction, chioroalkane is converted to ………
Answer:
Fluoroalkane.

Question 20.
The conversion of bromoalkane to fluroalkane by heating with AgF is called ………
Answer:
Swartz reaction

Question 21.
The decreasing order of boiling point of haloalkanes CH3 Br , CH3Cl , CH3F CH3I is ………
Answer:
CH3I > CH3Br > CH3CI > CH3F

Question 22.
The correct increasing order of boiling point of haloalkanes CH3Cl , CHCl3 , CH2Cl2 , CCl4 is ………
Answer:
CCl4 > CHCl3 > CH2Cl2 > CH3Cl

Question 23.
Haloalkanes reacts with aqueous solution of KOH to form ………
Answer:
Alcohol

Question 24.
Haloalkane reacts with alcoholic solution of KOH to form ………
Answer:
Alkene

Question 25.
Ethyl bromide reacts with alcoholic AgCN to form ………
Answer:
CH3CH2NC

Question 26.
Ethyl bromide reacts with alcoholic KNO2 to form ………
Answer:
Ethyl nitrite

Question 27.
The reaction in which sodium alkoxide react with haloalkane to form ether in called ………
Answer:
Williamson’s synthesis

Question 28.
Primary alkyl halide react with aqueous NaOH follows ………
Answer:
SN2

Question 29.
Tertiary butyl bromide reacts with aqueous KOH follows ………
Answer:
SN2 mechanism

Question 30.
Ethyl bromide reacts with alcoholic KOH following mechanism. ………
Answer:
E1

Question 31.
The product formed when tertiary butyl chloride is treated with alcoholic KOH is ………
Answer:
Iso-butylene

Question 32.
When 2-bromobutane react with alcoholic KOH, the products formed are ………
Answer;
1-butene & 2-butene

Question 33.
The product formed when iodoethane is treated with HI in the presence of red phosphorous is ………
Answer:
CH3-CH3

Question 34.
……… is used as an antiseptic
Answer:
Iodoform

Question 35.
……… is used for extinguishing the fire caused by oil or petrol under the commercial name pyrene
Answer:
Tetrachioromethane

Question 36.
……… Ethyl formate reacts with methyl magnesium iodide followed by acid hydrolysis to yield
Answer:
Acetaldehyde

Question 37.
……… Ethyl-methyl ether ethene is obtained by the action of methyl magnesium iodide with
Answer:
Chioro-aimethyl-ether

Question 38.
The hybridised state of carbon in haloarenes is ………
Answer:
sp2

Question 39.
The catalyst used in the preparation of chiorobenzene from benzene is ………
Answer:
FeCl3

Question 40.
In the Gattermann reaction of preparation of chiorobenzene from benzene, the catalyst used is ………
Answer:
Cu / HCl

Question 41.
The conversion of benzene diazoniurn chloride to chiorobenzene in the presence of Cu2Cl2 + HCl is named as ………
Answer:
Sandmeyer reaction

Question 42.
Fluorobenzene is prepared from benzene diazonium chloride by ………
Answer:
BaIz-Scheimann reaction

Question 43.
Conversion of benzene to chiorobenzene in the presence of CuCl2 / HCl is named as ………..
Answer:
Gattermann reaction

Question 44.
The conversion of chiorobenzene to phenol by the action of NaOH is called ………..
Answer:
Dow’s process

Question 45.
In Wurtz fittig reaction, chiorohenzene is converted to by reacting it with ethyl chloride.
Answer:
Ethylbenzene

Question 46.
The product obtained in fittig reaction of chiorobenzene is ………..
Answer:
Biphenyl

Question 47.
The reagent used in the conversion of Chiorobenzene to Benzene is ………..
Answer:
Ni-Al/NaOH

Question 48.
The catalyst used in the preparation of Phenyl magnesium chloride from chiorobenzene is ………..
Answer:
THF

Question 49.
Iso-propylidene chloride is an example of ………..
Answer:
vicinal dihalide

Question 50.
The IUPAC name of Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes is ………..
Answer:
1, 2-dibromo-2-methylpropane

Question 51.
The reagent used in the conversion of ethylene dichioride is ………..
Answer:
Zn + CH3OH

Question 52.
Chloroform is converted to methylene-chioride by the action of ………..
Answer:
Zn + HCl and H2 / Ni

Question 53.
The reagents used in the preparation of chloroform are ………..
Answer:
Ethanol + Bleaching powder

Question 54.
The formula of Chioropicrin is ………..
Answer:
CCl3NO2

Question 55.
The product formed when methylamine react with chloroform and alkali is ………..
Answer:
CH3NC

Question 56.
The product formed when CCI4 reacts with hot water vapours is
Answer:
COCl2

Question 57.
The formula of Freon -11 is ………..
Answer:
CFCl3

Question 58.
The formula of Freon – 12 is ………..
Answer:
CF2Cl2

Question 59.
The catalyst used in the preparation of CCl2F2 from CCl4 and HF is ………..
Answer:
SbCl5

Question 60.
The reagents used to prepare DDT are ………..
Answer:
Chlorai and Chiorobenzene

Question 61.
……….. is used to kill various insects like housefly and mosquitoes.
Answer:
(c) DDT

Question 62.
The name of CFCl3 is
Answer:
Freon-11

Question 63.
The treatment of acetone with excess of RMgX gives:
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 64.
The most easily hydrolysed molecule under SN2 reaction is………..
Answer:
Ten. butyl chloride

Question 65.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes. The product ‘X’ is ………..
Answer:
Ethanol

Question 66.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes The product X is ………..
Answer:
CH4

Question 67.
On heatrng CHCl3 with aqueous NaOH solution, the product formed is
Answer:
HCOONa

Question 68.
Depletion of ozone layer is caused by
Answer:
Freons

Question 69.
Chloropicrin is used as
Answer:
soil sterilizing agent

Question 70.
lodoform can be used as
Answer:
Antiseptic.

Question 71.
in oil fire extinguisher, the compound used pyrene is chemically
Answer:
CCl4

Question 72.
Reaction of ethyl chloride with sodium metal leads to the formation of
Answer:
n-Butane

Question 73.
When chloroform is treated with primary amine and KOH, we get ………..
Answer:
Offensive odours

IV. Choose the odd one out.

Question 1.
(a) CH3Br
(b) CH3-CH2Br
(c) (CH3)3C-Br
(d) CH3-CH2-CH2Br
Answer:
(c) (CH3)3C-Br. It is a haloalkane whereas others are 1° haloalkanes.

Question 2.
(a) Finkeistein reaction
(b) Wurtz reaction
(c) Swarts reaction
(d) Friedel crafts alkylation
Answer:
(d) Friedel crafts alkylation. It is used to prepare aromatic hydrocarbon whereas others are used to prepare alkanes.

Question 3.
(a) PCl5
(b) SOCl2
(c) HCl
(d) HF
Answer:
(cl) HF It is not used to prepare directly fluoro alkane whereas others are used to prepare directly chioro alkanes.

Question 4.
(a) Aerosol spray propellant
(b) Metal cleaning agent
(c) Anaesthetic agent
(d) Solvent in paint remover
Answer:
(c) Anaesthetic agent. It is not the use of methylene chloride whereas others are uses of methylene chloride.

V. Choose the correct pair.

Question 1.
(a) Chiorobenzene + chloral : DDT
(b) Chloroform + HNO3 : Phosgene
(c) Chloroform + Zn / HCl : Methyl isocyanide
(d) Methane + 4Cl2 : Carbon tetra chloride
Answer:
(c) Chlorobenzene + chloral : DDT

Question 2.
(a) Chloroform : Analgesic
(b) Freon : Propellant
(c) Chioropicrin : Antiseptic
(d) DDT : Soil sterilizing agent
Answer:
(b) Freon : Propellant

Question 3.
(a) Freon : Refrigerant
(b) DDT : Antiseptic
(c) Methylene : Soil sterilizing agent
(d) Iodotorni : Anaesthetic
Answer:
(a) Freon : Refrigerant

Question 4.
(a) HCOH + CH3MgI : Secondary alcohol
(b) CH3CHO + CH3MgI : Tertiary alcohol
(c) CH3COCH3 + CH3MgI : Primary alcohol
(d) CO2 + CH3MgI : Acetic acid
Answer:
(d) CO2 + CH3MgI : Acetic acid

Question 5.
(a) (C3H)3C-Cl + alcoholic KOH : SN1 reaction
(b) (CH3)3C-Cl + alcoholic KOH : E1 reaction
(c) (CH3)3C-Cl + aqueous KOH : SN2 reaction
(cl) CH3-Cl + aqueous KOH : SN1 reaction
Answer:
(b) (CH3)3C-Cl + alcoholic KOH : E1 reaction

VI. Choose the incorrect pair.

Question 1.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 2.
(a) CH3I > CH3Br > CH3Cl > CH3F : decreasing order of boiling point
(b) CCl4 > CHCl3> CH2Cl2 > CH3Cl : increasing order of boiling point
(c) CH3-CH2-CH2Cl <CH3 CH2 Cl <CH3Cl : increasing order of boiling point
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Answer:
(c) CH3-CH2-CH2Cl <CH3 CH2Cl <CH3Cl : increasing order of boiling point

Question 3.
(a) (CH3)3 C-Br + aqueous KOH : SN1 reaction
(b) (CH3)3 C-Br + alcoholic KOH: E1 reaction
(c) CH3Br – aqueous KOH : E2 reaction
(d) CH3Br + aqueous KOH : SN2 reaction
Answer:
(c) CH3Br + aqueous KOH : E2 reaction

Question 4.
(a) 1-chioro propane + Alcoholic KOH : Propene
(b) Tert.butyl bromide + Alcoholic KOH : Isobutylenc
(c) CH3-CH2I + HI + Red p : Ethanc
(d) CH3CHO + CH3 Mg I : Ten. Butyl alcohol
Answer:
(d) CH3CHO + CH3 Mg I : Ten. Butyl alcohol

Question 5.
(a) HCHO + CH3MgI : Primary alcohol
(b) CH3CHO + CH3MgI : Secondary alcohol
(c) CH3COCH3 + CH3MgI : Aromatic alcohol
(d)CO2 + CH3MgI : Acetic acid
Answer:
(c) CH3COCH3 + CH3MgI : Aromatic alcohol

VII. Assertion & Reason.

Question 1.
Assertion (A): The C-I in CH3X is weak.
Reason (R): Larger the size, greater is the bond length and weaker is the bond formed.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(a) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 2.
Assertion (A): Haloalkanes have higher boiling point and melting point than the parent alkanes having the same number of carbon.
Reason (R): The intermolecular forces of attraction and dipole-dipole interactions are
stronger in haloalkanes.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 3.
Assertion (A) : Among isomenc halides, the boiling point decreases with increase in branching in alkyl group.
Reason (R) : With the increase in branching, the molecule attains spherical shape with less
surface area and less forces of interaction.
(a) Both (A) and (R) are correct and (R) is not the correct explanation of (A).
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 4.
Assertion (A): The melting point of para halóbenzene is higher than that of ortho and meta isomers.
Reason (R): The higher melting point of p-isomer is due to its symmetry which leads to more close packing of its molecules in the crystal and subsequently p-isomer have strong intermolecular attractive forces.
(a) both (A) and (R) are correct and (R) is the correct explanation of(A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) both (A) and (R) are correct and (R) is the correct explanation of(A).

Question 5.
Assertion (A) : Haloarenes are insoluble in water.
Reason (R) : Haloarenes are able to form hydrogen bonds with water.
(a) Both (A) and (R) are correct but (R) is the correct explanation of (A).
(b) both (A) and (R) are correct and (R) is not the correct explanation of (A).
(e) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct
Answer:
(c) (A) is correct but (R) is wrong.

Question 6.
Assertion (A) : Haloarenes do not undergo nucleophilic substitution reactions readily.
Reason (R) : The C-X bond in aryl halides is short and stronger and also the aromatic ring is a center of high electron density.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(h) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 7.
Assertion (A): Chloroform vapours can be used as an anaesthetic.
Reason (R): Chloroform vapours depresses the central nervous system and cause unconsciousness.
(a) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 8.
Assertion (A): Nowadays chloroform is not used as an anaesthetic.
Reason (R): Chloroform undergoes oxidation in the presence of light and air to form highly poisonous phosgene.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(cl) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 9.
Assertion (A): DDT is banned now-a-days.
Reason (R): DDT has a long term toxic effect.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Assertion is false but reason is true.
Answer:
(a) Both assertion and reason are true and reason is the correct explanation of assertion.

VIII. Choose the correct statement.

Question 1.
(a) Halo alkanes have higher boiling point than the parent alkane with same number of carbons because of strong inter molecular forces of attraction.
(b) The boiling point of halo alkanes decreases with the increase of halogen atoms.
(c) The boiling point of mono halo alkanes decreases with the increase in the number of carbon atoms.
(d) Halo alkanes are soluble in water.
Answer:
(a) Halo alkanes have higher boiling point than the parent alkane with same number of carbons because of strong inter molecular forces of attraction.

Question 2.
(a) Halo alkanes are soluble in water.
(b) The boiling point of halo alkanes increase with the increase in the number of halogen atoms.
(c) The melting point of mono halo alkane decrease with the increase in the number of carbon atoms.
(d) The density of alkyl halides are lesser than those of hydrocarbons of comparable molecular weight.
Answer:
(b) The boiling point of halo alkanes increase with the increase in the number of halogen atoms.

Question 3.
(a) Williamson’s synthesis of ether is an example of nucleophilic substitution reaction.
(b) Reaction of methyl bromide with aqueous potassium hydroxide is an example ofelimination reaction.
(c) Reaction of Tertiary butyl bromide with alcoholic KOH is an example of SN2 reaction.
(d) Reaction of Tertiary butyl bromide with alcoholic KOH is an example of E, reaction.
Answer:
(a) Williamson’s synthesis of ether is an example of nucleophilic substitution reaction.

Samacheer Kalvi 11th Chemistry Haloalkanes and Haloarenes 2 Marks Questions and Answers

Question 1.
Write the IUPAC names of –

  1. CH2=CHCI
  2. CH2=CH-CH2Br.

Answer:

  1. CH2=CHCl : Chloroethene
  2. CH2=CH-CH2Br : 3-Bromo-prop- 1-ene

Question 2.
Write the structural formula of the following compounds:

  1. 2-Chloro-2-Methylpropane
  2. 1 -Bromo-2, 2-Dimethylpropane

Answer:
1. 2-Chloro-2-Methylpropane
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

2. 1 -Brorno-2, 2-dimethyipropane
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 3.
How many isomers are possible for the formula C3H7F? Give their structures and names.
Answer:
C3H7F – 2 isomers are possible.
1. CH3-CH2-CH2F – 1- fiuoropropene
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 4.
Write the isomeric structures and names for the formula C2H4Cl2.
Answer:

Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes Ethylene dichioride (or) 1, 2-dichioroethane.

Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes Ethylidene chloride (or) 2-dichioroethane.

Question 5.
Draw the structures of –

  1. 1-brorno-2, 3-dichiorobutane
  2. 2-bromo-3-chloro-2, 4-dimethyl pentane

Answer:
1. 1 -bromo-2, 3-dichiorobutane
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

2. 2-bromo-3-chloro-2, 4-dimethy 1 pentane
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 6.
What happens when HI reacts with ten. butyl alcohol?
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 7.
Explain the action of
(i) PCl5
(ii) PCl3 with ethanol.
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 8.
How does HBr react with propene?
Answer:
Propene react with HBr and follows Markovnikov’s rule.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 9.
How methane reacts with Cl2 in the presence of light?
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 10.
Explain-Finkelstein reaction.
Answer:
Chioro (or) bromoalkane on heating with sodium iodide in dry acetone gives jodo alkane. This reaction is called Finkeistein reaction.
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 11.
Explain Swans reaction.
Answer:
Chioro (or) bromoalkanes on heating with AgF give fluoroalkanes. This reaction is called Swans reaction.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 12.
What happens when silver propionate reacts with Br, in CCl4?
Answer:
Silver salt of fatty acids (CH3CH2COOAg), e.g., silver propionate treated with Br2 / CCl4 gives bromoalkane. This reaction is called Hunsdicker reaction.

Question 13.
Haloalkanes have higher boiling point and melting point than the parent alkane. Justify this statement.
Answer:
Haloalkanes have higher boiling point than the parent alkane having the same number of carbon atoms because the intermolecular forces of attraction and dipole-dipole interactions are comparatively stronger in haloalkanes.

Question 14.
CCl4 > CHCl3 > CH2Cl2 > CH3Cl is the decreasing order of boiling point in haloalkanes. Give reason.
Answer:
The boiling point of chioro, bromo and iodoalkanes increases with increase in the number of halogen atoms. So the correct decreasing order of boiling point of haloalkanes is:
CCl4> CHCl3 > CH2Cl2> CH3Cl.

Question 15.
Arrange the following in increasing order of boiling point. Give reason.
(CH3CH2CH2Cl, CH3CH2Cl, CH3Cl)
Answer:
The correct order of boiling point of haloalkanes is:
CH3CH2CH2Cl > CH3CH2Cl > CH3Cl.
The boiling point of monoalkanes increase with the increase in the number of carbon atoms.

Question 16.
Why haloalkanes arc insoluble in water but soluble in organic solvents?
Answer:
Haloalkanes are polar covalent compounds, soluble in organic solvents, but insoluble in water because they are unable to form hydrogen bond with water molecules.

Question 17.
What happens when bromoethane is treated with moist silver oxide?
Answer:
When bromoethane is treated with moist silver oxide, ethanol will be formed as product:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 18.
Complete the following reactions:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 19.
Explain the action of sodium hydrogen suiphide with bromoethane?
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 20.
Explain Williamson’s synthesis.
Answer:
Williamson’s synthesis:
Haloalkanes when boiled with sodium alkoxide gives the corresponding ether.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes -251
Question 21.
Explain the action of alcoholic potash with bromoethane.
Answer:
Elimination reaction takes place when alcoholic potash reacts with bromoethane:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 22.
State-Saytzeff’s rule.
Answer:
Some haloalkanes when treated with alcoholic KOH yield a mixture of olefins with different amounts. It is explained by Saytzeff’s rule which states that in a dehydrohalogenation reaction, the preferreed product is that alkene which has more number ofalkyl group attached to the doubly bonded carbon atom.

Question 23.
How will you convert 1-chioropropanc to propene?
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 24.
What is Grignard reagent? How is it prepared from ethyl bromide?
Answer:
When a solution of haloalkane in either is treated with magnesium, we will get alkyl magnesium halide known as Grignard reagent, ethyl magnesium bromide is prepared from ethyl bromide as:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 25.
How will you prepare ethyl lithium’?
Answer:
When bromoethane is treated with an active metal like lithium in the presence of dry ether, then ethyl lithium will be formed.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 26.
What is tetraethyl Lead? How is it prepared?
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 27.
Convert bromoethane to ethane.
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 28.
How is iodoethane converted to ethane?
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 29.
Starting from CH3MgI, how will you prepare acetaldehyde?
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 30.
How will you get acetone from methyl manesium iodide?
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 31.
Explain the action of Ethyl chioroformate with Methyl magnesium iodide.
Answer:
Ethyl chioroformate reacts with Grignard reagent to form esters as follows:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 32.
Write the IUPAC names of:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes 1 – bromo-4-fluoro-2-iodobenzene
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes 1 – bromo-2-fluoro-4-iodobenzene

Question 33.
How is benzene directly converted to chiorobenzene?
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 34.
Explain Sandmeyer’s reaction.
Answer:
Sandmeyer’s reaction:
The reaction in which benzene-diazonium chloride is converted to chiorobenzene on heating it with Cu2Cl2/HCl is known as sand Meyer’s reaction.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 35.
Explain Gattcrrnann reaction.
Answer:
Gattermann reaction:
The reaction in which benzene diazonium chloride react with copper in the presence of HCl, chlorobenzene is formed, called Gatterrnann reaction.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 36.
How will you prepare iodobenzene?
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 37.
Explain-Balz-Schiernann reaction.
Answer:
Fluorobenzene is prepared by treating benzene diazonium chloride with fluoro boric acid. This reaction produces diazonium fluoroborate which on heating produce fluorobenzene. this reaction is called Balz-Schiernann reaction.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 38.
p-dichlorobenzcnc has higher melting point than ortho and meta dichloro benzene. Why?
Answer:
The melting point of p-dichloro benzene is higher than the melting point of the corresponding ortho and mcta isomers. The higher melting which leads to more close packing of its molcules in the crystal lattice and consequently strong intermolecular attractive forces which requires more energy for melting.
p-dichlorobenzene > o-dichlorobenzene > m-dichlorobenzene

Question 39.
Explain Wurtz-fìtting reaction.
Answer:
Wurtz-fitting reaction:
Chlorobenzene and chioroethane are heated with sodium in ether solution to form ethylbenzene. This reaction is called Wurtz-fìtting reaction.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 40.
How will you get benzene from chlorobenzene?
Answer:
Chiorobenzene is reduced with Ni-Al Alloy in the presence of NaOH to give benzene.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 41.
Explain about the preparation of phenyl magnesium chloride.
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 42.
How will you prepare ethylidcne dichioride from acetylene?
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 43.
What is gern-dihalide? Give one example and explain its preparation.
Answer:
1. If two halogen atoms are attached to one carbon atom of alkyl halide, it is named as gem dihalide. e.g., CH3-CHCl2 (ethylidene dichloride.)

2. Preparationof CH3-CHCl2:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 44.
Explain the action of zinc and HCl on chloroform.
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 45.
How does nickel react with chloroform?
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 46.
Convert methane to methylene chloride.
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 47.
Explain Carbylamine reaction. (or) Give the characteristic test for primary amine.
Answer:
Chlorofonn reacts with aliphatic or aromatic primary amines and alcoholic caustic potash to give foul smelling alkyl isocyanide (carbylamine).
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 48.
How will you prepare carbon tetrachioride?
Answer:
The reaction of methane with excess of chlorine in the presence of sunlight give carbon tetrachioride as major product.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 49.
How will you convert carbon tetrachioride to chloroform?
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 50.
What are freons? How are they named? Give two examples.
Answer:
(i) The chioro-fluoro derivatives of methane and ethane are called freons.
(ii) Freon is represented as freon-cba
c = number of C atoms –
b = number of H atoms + 1
a = total number ofF atoms
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes -252
Question 51.
What are ambident nucleophiles’? Explain with an example.
Answer:
Nucleophiles which can attack through two different sites are called ambident nucicophiles. For example, cyanide group is a resonance hybrid of two contributing structures and therefore it can act as a nucleophile in two different ways:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
It can attack through carbon to form cyanides and through nitrogen to form isocyanides or carbylamines.

Question 52.
Which compound in each of the following pairs will react faster in SN2 reaction with OH ion?
1. CH3Br or CH3I
2. (CH3)3CCl or CH3Cl
Answer:
1. Since I ion a better leaving group than Br ion, therefore CH3I will react faster than CH3Br in SN2 reaction with OH ion.
2. On steric grounds 1° alkyl halides are more reactive than tert alkyl halide in SN2 reactions. Therefore, CH3Cl will react at a faster rate than (CH3)3CCl in a SN2 reaction with OH ion.

Question 53.
The treatment ofalkyl chlorides with aqueous KOH solution leads to the formation of alcohols but in the presence of alcoholic KOH solution, alkenes are the major product. Explain.
Answer:
In aqueous solution, KOH is almost completely ioniscd to give OH ions which being a strong nucleophile brings about a substitution reaction of alkyl halides to form alcohols. In aqueous solution, OH ions are highly hydrated. This solvation reduces the basic character of OH ions which therefore, abstract fails to abstract a hydrogen atom from the n-carbon of the alkyl chloride to form an alkene. In contrast, an alcoholic solution of KOH contains alkoxide (RO) ions which being a much stronger base than OH ions preferentially eliminates a molecule of HCl from an alkyl chloride to form alkenes.

Question 54.
Give one example of each of the following reactions:
1. Wurtz Reaction
2. Wurtz – Fittig reaction
Answer:
1. Wurtz Reaction:
It involves conversion of alkyl halides into alkane.
Example:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

2. Wurtz-Fittig reaction:
It involves the reaction of an aryl halide and alkyl halide to form the corresponding hydrocarbon.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 55.
How will you distinguish between the following pair of compounds:
1. Chloroform and carbon tetrachioride,
2. Benzyl alcohol and chiorobenzene.
Answer:
1. On heating chloroform and carbon tetrachloride with aniline with ethanoic acid and potassium hydroxide separately, chloroform forms a pungent smelling isocyanide compound but carbon tetrachloride does not form this compound.

2. On adding sodium hydroxide and silver nitrate to both the compounds, benzyl chloride forms a white precipitate but chiorobenzene does not form any white precipitate.

Samacheer Kalvi 11th Chemistry Haloalkanes and Haloarenes 3 Marks Questions And Answers

Question 1.
Give one example with structure and name for each of the following compounds.
(a) Primary haloalkane
(b) Secondary haloalkane
(c) Tertiary haloalkane
Answer:
(a) Primary haloalkane
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

(b) Secondary haloalkane
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

(c) Tertiary haloalkane
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 2.
Write the possible isomers for the formula C4H9Cl with structures and names.
Answer:
C4H9Cl (4 isomers):
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 3.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes -253
Answer:
For a compound with molecular formula C4H9Cl – 3 isomers (1°, 2°, 3°) arc possible. Among the isomeric alkyl halides, the boiling point decreases with the increase in branching in the alkyl chain. This is because with increase in branching, the molecule attains spherical shape with less surface area. As a result, the intermolecular forces become weak, resulting in lower boiling point. Therefore the boiling point decreases in the order:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Hence, (CH3)3 C-Cl has the lowest boiling point.

Question 4.
Explain ammonolysis of haloalkanes. (or) How excess of haloalkanc react with alcoholic ammonia?
Answer:
With excess of haloalkanes, ammonia react to give primary, secondary, tertiary amines along with quarternary ammonium salt:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 5.
Explain how bromoethane reacts with –
1. alcoholic KCN
2. alcoholic AgCN
Answer:
1. Bromoethanc reacts with alcoholic KCN to form ethyl cyanide.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

2. Bromoethane reacts with alcoholic AgCN to form alkyl isocyanide.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 6.
Explain the hydrolysis of 2-bromobutane with aqueous KOB.
Answer:

  • 2-bromobutane is optically active and it undergoes SN1 reaction with aqueous KOH.
  • The product obtained will be an optically inactive racemic mixture.
  • As nucleophilic reagent OH ion can attack the carbocation from both sides to form equal proportions of dextro and levo rotatory optically active isomers, it results in the formation of optically inactive racemic mixture.

Question 7.
Explain the action of alcoholic KOH with 2-bromobutanc.
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
According to Saytzcff’s rule, when 2-bromobutanc reacts with alcoholic KOK, yields a mixture of olefins in different amounts.

(ii) In a dehydrohalogenation reaction, the preferred product is that alkene which has more number of alkyl groups attached to the doubly bonded carbon alkene.

Question 8.
Mention the uses of chloroform.
Answer:

  • Chloroform is used as a solvent in pharmaceutical industry.
  • It is used for producing pesticides and drugs.
  • It is used as an anaesthetic.

Question 9.
What are the uses of carbon tetrachioride?
Answer:

  • Carbon tetrachioride in used as a dry cleaning agent.
  • It is used as a solvent for oils, fats and waxes.
  • As the vapours of CCl4 is non-combustible, it is used under the name pyrene for extinguishing the fire caused by oil (or) petrol.

Question 10.
What are ogranometallic compounds? Give one example. Explain the nature of carbon-metal bond.
Answer:

  • Organometallic compounds are organic compounds in which there ¡s a direct carbon-metal bond.
  • Example :  CH3MgI. methyl magnesium iodide.
  • The carbon-magnesium bond in Grignard reagent is covalent but highly polar. The carbon atom is more electronegative than magnesium. Hence, the carbon atom has partial negative charge and magnesium atom has a partial positive charge.
    Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 11.
How would you prepare acetic acid from methyl magnesium iodide?
Answer:
Solid carbon dioxide reacts with grignard reagent to form addition product which on hydrolysis yields acetic acid.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 12.
Explain the nature of C-X bond in haloarenes and its resonance structure.
Answer:

  1. In haloarenes, the carbon atom is sp2 hybridised. The sp2 hybridised orbitais arc shorter and holds the electron pair bond more.
  2. Halogen atom contains p-orbital with lone pair of electrons which interacts with π – orbitaIs of berizene ring to form extended conjugated system of π – orbitals.
  3. The delocalisation of these electrons give double bond character to C-X bond. The resonance structure of halobenzene are given as:
    Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 13.
Compare the bond length C-X in haloarenes and C-X in haloalkanes.
Answer:
1. Due to this double bond character of C-X bond in haloarenes, the C-X bond length is shorter length and stronger than in haloalkanes.
2. Example:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 14.
What are the uses of chlorobenzene?
Answer:

  • Chiorobenzene is used in the manufacture of pesticides like DDT.
  • It is used as high boiling solvent in organic synthesis.
  • It is used as a fibre – swelling agent in textile processing.

Question 15.
What are polyhalogen compounds? Give its types with example.
Answer:

  1. Carbon compounds containing more than one halogen atom are called polyhalogen compounds.
  2. They are classified as
    • gem dihalides
    • Vicinal dihalides.

(a) Gem dihalide:
In this compound, two halogen atoms are attached to one carbon atom.
e.g, CH3CHCl2 – ethylidene chloride.

(b) Vicinal dihalide:
In this compound, two halogen atoms are attached to two adjacent carbon atoms.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarene 1,2-dichloroethane

Question 16.
Give two examples for –
1.  gem dihalide
2. vicinal dihalide.
Answer:
1. Gem dihalides:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
2. Vicinal dihalides:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 17.
Explain two methods of preparation of ethylene diehioride.
Answer:
1. Addition of Cl2 to Ethylene:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
2. Action of PCl5 on Ethylene glycol:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 18.
How would you distinguish gern-dihalides and iinal dihalides?
Answer:
(i) Gem-dihalides on hydrolysis with aqueous KOH gives an aldehyde or a ketone whereas vicinal dihalides on hydrolysis with aqueous KOH give glycols.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
The above reaction can be used to distinguish between gem-dihalides and vicinal dihalides.

Question 19.
Explain the action of metallic zinc with –
(i) Ethylidene dichioride
(ii) Ethylene dichloridc.
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 20.
What happens when alcoholic KOH is treated with
(i) Ethylidene dichioride
(ii) Ethylene dich I onde?
Answer:
(i) Gem-dihalides and vicinal dihalides both on treatment with alcoholic KOH give alkynes.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 21.
What are the uses of methylene chloride?
Answer:
Methylene chloride is used as:

  • Aerosol spray propellant.
  • Solvent in paint remover.
  • Process solvent in the manufacture of drugs.
  • A metal cleaning agent.

Question 22.
Explain the laboratory preparation of chloroform.
Answer:
Haloform reaction:
Chloroform is prepared in the laboratory by the reaction between ethyl alcohol with bleaching powder followed by the distillation of the final product chloroform. Bleaching powder acts as a source of chlorine and calcium hydroxide. The reaction take place in 3 steps.
Step 1:
Oxidation-
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Step 2:
Chlorination-
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Step 3:
Hydrolysis-
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 23.
What is chloropicrin? How is it obtained? Mention its uses.
Answer:
1. Chioropicrin is CCl3NO2.

2. Chloroform reacts with nitric acid to form chioropicrin (trichioro-nitromethane):
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

3. Chioropicrin is used as an insecticide and soil sterilising agent.

Question 24.
What are the uses of freons?
Answer:

  • Freons are used as refrigerants in refrigerators and air conditioners.
  • It is used as a propellant for foams and aerosols.
  • It is used as a propellant for foams to spray out deodorants, shaving creams and insecticides.

Question 25.
What is DDT? How is it prepared’?
Answer:
(i) DDT is p, p’-dichloro-diphenyi-trichloroethane
(ii) DDT can be prepared by heating a mixture of chloro benzene with chlorai in the presence of conc. H2SO4
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 26.
Mention the uses of DDT.
Answer:

  • DDT is used lo control certain insects which carries diseases like malaria and yellow fever.
  • It is used in farms to control some agricultural pests.
  • It is used in building construction as pest control agent.
  • It is used to kill certain kind of insects like housefly and mosquitoes due to its high and specific toxicity.

Question 27.
Write the equations for the preparation of I-iodobutane from:
(i) 1 -butanol
(ii) 1 -chiorobutane
(iii) but- 1-ene
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 28.
Explain why:

  1. the dipole moment of chiorobenzene is lower than that of cyclohexyl chloride?
  2. alkyl halides though polar, are immiscible with water?
  3. Gringard reagents should be prepared under anhydrous conditions?

Answer:

  1. Chlorobenzene is stabilised by resonance and there is negative charge on ‘Cl’ in 3 out of 5 resonating structures, therefore it has a lower dipole moment than cyclohexyl chloride in which there is no such negative charge.
  2. Alkyl halides cannot form H-bond with water and cannot break H-bonds between water molecules, therefore they are insoluble in water.
  3. Grignard reagents react with H2O to form alkanes, therefore they are prepared under anhydrous conditions.

Question 29.
Explain as to why haloarenes arc much less reactive than haloalkanes towards nucleophilic substitution reactions?
Answer:
Haloarenes are much less reactive than haloalkanes towards nucleophilic substitution reactions due to the following reasons:

1. Resonance effect:
In haloarenes the electron pair on the halogen atom is in conjugation with the π – electrons of the ring and the following resonating structures are possible. C-Cl bond acquires a partial double bond character due to resonance. As a result, the bond cleavage in haloarenes is difficult than in case of haloalkanes and therefore they are less reactive towards nucleophilic substitution reactions.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

2. The C-Cl bond length in haloalkanes is 177 pm while in haloarenes it is 169 pm. Since it is difficult to break shorter bond than a longer bond. Therefore, haloarenes are less reactive than haloalkanes towards nucleophilic substitution reactions.

Question 30.
Do the following conversions:
(i) Methyl bromide to acetone
(ii) Benzyl alcohol to 2-phenylacetic acid
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarene

Question 31.
Give reasons for the following:

  1. Ethyl iodide undergoes SN2 reactions faster than ethyl bromide.
  2. (±) 2-Butanol is optically inactive.
  3. C-X bond length in halobenzene is smaller than C-X bond length in CH3-X.

Answer:

  1. Because in ethyl iodide, iodide being the best leaving group among all the halide ions. Rate of SN2 reaction ability of leaving group.
  2. (±) 2-hutanol is a racemic mixture which is optically inactive due to the external compensation.
  3. Due to resonance in halobenzene, it has a smaller bond length value as compared to CH3-X.

Question 32.
Write the structure of diphenyl. How is it prepared from chlorobcnzene?
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Samacheer Kalvi 11th Chemistry Haloalkanes and Haloarenes 5 Marks Question And Answers

Question 1.
Explain the classification of organic compounds with example.
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarene

Question 2.
Explain SN2 mechanism with suitable examples.
Answer:

  • SN2 reaction means bimolecular nucleophilic substitution reaction of second order.
  • The rate of SN2 reaction depends upon the concentration of both alkyl halides and the nucleophile.
  • This reaction involves the formation of a transition state in which both the reactant molecules are partially bonded to each other. The attack of nucleophile occurs from the back side.
  • The carbon at which substitution occurs has inverted configuration during the course of reaction just as an umbrella has the tendency to invert in a wind-storm. This inversion of configuration is called walden inversion.
    Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarene

Question 3.
SN2 reaction of an optically active haloalkane is accompanied by inversion of configuration at the asymmetric centre. Prove it.
Answer:
1. SN2 reaction of an optically active haloalkane (e.g., 2-bromo-octane) is accompanied by inversion of configuration at the asymmetric center.

2. When 2-bromooctane is heated with sodium hydroxide, 2-octanol is formed with inversion of configuration. (-) 2-bromo-octane on heating with sodium hydroxide gives (+) 2-octanol is formed in which -OH group occupies a position opposite to what bromine had occupied in the optically active haloalkane.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarene

Question 4.
Explain E2 reaction mechanism with a suitable example.
Answer:

  • E2 reaction stands bimolecular elimination reaction of second order. The rate of E2 reaction depends on the concentration of alkyl halide and the base.
  • Primary alkyl halide undergoes this reaction in the presence of alcoholic KOH.
  • It is a one step process in which the abstraction of the proton from the 3 carbon atom and expulsion of halide from the x carbon atom occur simultaneously.
  • The mechanism that follows is shown below:
    Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarene

Question 5.
Describe E1 reaction mechanism with a suitable example.
Answer:
1. E1 stands for unimolecular elimination reaction of first order.
2. Tertiary alkyl halides undergoes elimination reaction in the presence of alcoholic KOH.
3. It takes place in two steps.
4.  Step 1.
Heterolytic fission to yield a carbocation:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

5. Step 2.
Elimination of a proton from the fl-carbon to produce an alkene:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 6.
Starting from methyl magnesium iodide how would you prepare –

  1. Ethanol
  2. 2-propanol
  3. Tert-butyl alcohol.

Answer:
1. Ethanol:
Formaldehyde reacts with CH3MgI to give an addition product which on hydrolysis yields ethanol.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

2. 2-Propanol :
Acetaldehyde react with CH3MgI to give an addition product which on hydrolysis yields 2-propanol.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 7.
Starting from methyl magnesium iodide, how would you prepare –

  1. Ethyl methyl ether
  2. methyl cyanide
  3. methane.

Answer:
1. Ethyl methyl ether:
Lower halogenated ether reacts with grignard reagent to form higher ether.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

2. Methyl cyanide:
Grignard reagent reacts 4rith cyanogen chloride to form alkyl cyanide.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

3.  Methane:
Grignard reagent reacts with waler to give methane as product.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 8.
Describe electrophilic substitution reaction of chiorohenzene with equations.
Answer:
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 9.
An organic compound Ⓐ of molecular formula C3H6 react with HBr in the presence of peroxide to give C3H7Br as Ⓑ Ⓑ on reaction with aqueous KOH gives © with molecular formula C3H8O. Identify Ⓐ Ⓑ and ©
Answer:
1. An organic compound Ⓐ of molecular formula C3H6 is CH3-CH=CH2, Propene

2. Propene Ⓐ reacts ith HRr in the presence of peroxide to give 1-bromopropane
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

3. 1-Bromopropane on reaction with aqueous KOH undergoes hydrolysis to give  Propan- 1-ol. CH3-CH2-CH2OH as ©
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 10.
An organic compound Ⓐ of molecular formula C2H6O reacts with thionyl chloride in the presence of pyridine gives Ⓑ C2H5Cl. Ⓑ on reaction with alcoholic KOH gives Ⓒ, C2H4. Ⓒ on treatmeni with Cl2 gives C2H4Cl2 as Ⓓ. Identify Ⓐ Ⓑ Ⓒ Ⓓ and explain the reaction.
Answer:
1. An organic compound Ⓐ of molecular formula C2H6O is Ethanol CH3-CH2OH.

2. Ethanol reacts with thionyl chloride in the presence of pyridine to give CH3-CH2Cl Ⓑ as product.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

3. Ethyl chloride on treatment with alcoholic KOH, undergoes dehydrohalogenation to give C2H4, ethylene Ⓒ as the product.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

4. Ethylene on reaction with Cl2 yield ethylene dichioride as Ⓓ as the product.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 11.
The simplest aromatic hydrocarbon C6H6 reacts Ⓑ on treatment with sodium hydroxide will (C6H5OH), Phenol, Ⓒ as the product. Also Cl2 to give Ⓐ which on reaction with sodium hydroxide gives Ⓑ.Ⓑ of molecular formula C6H6O. Ⓑ on treatment with ammonia will give C6H7N as D. Identify Ⓐ, Ⓑ,Ⓒ, and explain the reactions involved.
Answer:
1. The aromatic hydrocarbon Ⓑ is Benzene, C6H6.

2. Benzene reacts with Cl, to give chiorobenzene (C6H5Cl), as Ⓑ as the product.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

3. Chiorobenzene Ⓑ reacts with sodium hydroxide to give (C6H5OH) phenol, Ⓒ as the product.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

4. Chiorobenzene reacts with ammonia to give Aniline, C6H5NH2 Ⓓ as the product.Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Question 12.
An organic compound Ⓐ of molecular formula C2H2 reacts with HCl to give C2H4Cl2 as Ⓑ. Ⓑ on reaction with aqueous KOH will give C2H4O as Ⓒ Identify Ⓐ, Ⓑ,Ⓒ and explain the reactions involved.
Answer:
1. An organic compound Ⓐ of molecular formula C2H2 is CH ≡ CH (Acetylene)

2. Acetylene Ⓐ reacts with HCl to give ethylene dichioride Ⓑ as the product.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

3. Ethylidene dichioride on reaction with aqueous KOH will give acetaldehyde, © as the product.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Question 13.
Two isomers of formula C4H9Br are Ⓐ and B. Ⓐ on reaction with alcoholic KOH gives of molecular formula C8H8 by E1 reaction. Ⓑ on reaction with alcoholic KOH gives Ⓓ and Ⓔ as products by Saytzefì’s rule. Idcnti’ A, B, C, D, E.
Answer:
1. C4H9Br: Two isomers may be there: Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes Tert butyl bromide and Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

2. Ⓐ on reaction with alcoholic KOH gives iso butylene © as the product.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

3. 2-bromobutane Ⓑ on reaction with alcoholic KOH follows Saytzeff s rule to give a mixture of olefins in different amounts.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 14.
A simple aromatic hydrocarbon Ⓐ reacts with Cl2 to give Ⓑ of molecular formula C6H5Cl. Ⓑ on reaction with ethyl chloride along with sodium metal gives © of formula C8H10. © alone reacts with Na metal in the presence of ether to give Ⓓ C12H10. Identify Ⓐ Ⓑ Ⓒ & Ⓓ
Answer:
1. The simple aromatic hydrocarbon is C6H6. Benzene (A)

2. Benzene reacts with Cl2 to give chiorobenzene.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

3. Chiorobenzene reacts with ethyl chloride to form ethyl benzene, © as the product
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

4. Chiorobenzene on reaction Na metal gives Biphenyl compound © as the product.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

Question 15.
An organic compound Ⓐ of molecular formula CH2O reacts with methyl magnesium iodide followed by acid hydrolysis to give Ⓑ of moLecular formula C2H6O. Ⓑ on reaction with PCl5 gives Ⓑ. © on reaction with alcoholic KOK gives Ⓓ an alkene as the product. Identif’ Ⓐ Ⓑ Ⓒ Ⓓ and explain the reactions involved.
Answer:
1. Ⓐ of molecular formula CH2O is identified as HCHO, formaldehyde.

2. Formaldehyde reacts with CH3MgI followed by hydrolysis to give ethanol, CH3-CH2OH B as the product.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

3. Ethanol Ⓑ reacts with PCl5 to give C2H5Cl, Ethyl chloride © as the product.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

4. CH3-CH2Cl © on reaction with alcoholic KOH undergoes dehydrohalogenation to give ethylene CH2=CH3 Ⓓ as the product.
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes
Samacheer Kalvi 11th Chemistry Solution Chapter 14 Haloalkanes and Haloarenes

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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

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Samacheer Kalvi 11th Chemistry Hydrocarbons Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Hydrocarbons Multiple Choice Questions.

Question 1.
The correct statement regarding the comparison of staggered and eclipsed conformations of ethane is ……….. [NEET]
(a) the eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain.
(b) the staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.
(c) the staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has torsional strain.
(d) the staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has no torsional strain.
Answer:
(b) the staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.

Question 2.
The above reaction is an example of which of the following?
(a) Reirner Tiemann reaction
(b) Wurtz reaction
(c) Aldol condensation
(d) Hoffmann reaction
Answer:
(b) Wurtz reaction

Question 3.
An alkyl bromide (A) reacts with sodium in ether to form 4, 5-diethyloctane, the compound (A) is ……….
(a) CH3(CH2)3Br
(b) CH3(CH2)5Br
(c) CH3(CH2)3CH(Br)CH3
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons


Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons - 256

Question 4.
The C-H bond and C-C bond in ethane are formed by which of the following types of overlap ………..
(a) sp3 – s and sp3 – sp3
(b) sp2 – s and sp3 – sp3
(c) sp – sp and sp – sp
(d) p – s and p – p
Answer:
(a) sp3 – s and sp3 – sp3

Question 5.
In the following reaction Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
the major product obtained is ………
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 6.
Which of the following is optically active?
(a) 2 – Methylpentane
(b) Citric acid
(c) Glycerol
(d) none of these
Answer:
(a) 2 – Methylpentane

Question 7.
The compounds formed at anode in the electrolysis of an aqueous solution of potassium acetate are ……….
(a) CH4 and H2
(b) CH4 and CO2
(c) C2 H6 and CO2
(d) C2 H6 and Cl2
Answer:
(c) C2 H6 and CO2

Question 8.
The general formula for cycloalkanes is …………
(a) CnHn
(b) CnH2n
(c) CnH2n-2
(d) CnH2n+2
Answer:
(b) CnH2n

Question 9.
The compound that will react most readily with gaseous bromine has the formula ………….[NEET]
(a) C3H6
(b) C2H2
(c) C4H10
(d) C2H4
Answer:
(a) C3H6

Question 10.
Which of the following compounds shall not produce propene by reaction with HBr followed by elimination (or) only direct elimination reaction? [NEET]
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
(b) CH3 – CH2 – CH2 – OH
(c) H2C – C = O
(d) CH3 – CH2 – CH2Br
Answer:
(c) H2C = C = O

Question 11.
Which among the following alkenes on reductive ozonolysis produces only propanone?
(a) 2 – Methylpropene
(b) 2 – Methylbut – 2 – ene
(c) 2, 3 – Dimethylbut – 1 – ene
(d) 2, 3 – Dimethylbut – 2 – ene
Answer:
(d) 2, 3 – Dimethylbut – 2 – ene

Question 12.
The major product formed when 2 bromo – 2 – methylbutane is refluxed with ethanolic KOH is ……..
(a) 2 – methylbut – 2 – ene
(b) 2 – methylbutan – 1 – ol
(c) 2 – methyl but – 1 – ene
(d) 2 – methylbutan -2- ol
Answer:
(a) 2 – methylbut – 2 – ene

Question 13.
Major product of the below mentioned reaction is ……….
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
(a) 2 – chloro – 1 – iodo – 2 – methylpropane
(b) 1 – chloro – 2 – iodo – 2 – methylpropane
(c) 1 ,2 – dichioro – 2 – methylpropane
(d) 1, 2 diiodo 2 – methylpropane
Answer:
(a) 2 – chioro – 1 – lodo – 2 – methylpropane

Question 14.
The IUPAC name of the following compound is ………
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
(a) trans – 2 – chloro-3 iodo – 2 – pentane
(b) cis – 3 – iodo – 4 – chloro – 3 – pentane
(c) trans – 3 – iodo – 4 – chloro – 3 – pentene
(d) cis – 2 chloro – 3 – lodo -2 – pdntene
Answer:
(a) trans -2 – chloro -3 – iodo – 2 – pentane

Question 15.
cis – 2 – butene and trans – 2 – butene are ……….
(a) conformational isomers
(b) structural isomers
(c) configurational isomers
(d) optical isomers
Answer:
(c) configurational isomers

Question 16.
Identify the compound (A) in the following reaction.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 17.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons where a is ………
(a) Zn
(b) Conc. H2SO4
(c) Alc. KOH
(d) Dil. H2SO4
Answer:
(c) Alc. KOH

Question 18.
Consider the nitration of benzene using mixed conc. FeSO4 and HNO3, if a large quantity of KHSO4 is added to the mixture, the rate of nitration will be ………
(a) unchanged
(b) doubled
(c) faster
(d) slower
Answer:
(d) slower

Question 19.
In which of the following molecules, all atoms are co-planar?
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
(d) both (a) and (b)

Question 20.
Propyne on passing through red hot iron tube gives ………
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 21.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 22.
Which one of the following is non-aromatic?
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 23.
Which of the following compounds will not undergo Friedal – crafts reaction easily? [NEET]
(a) Nitrobenzene
(b) Toluene
(c) Cumene
(d) Xyiene
Answer:
(a) Nitrobenzene

Question 24.
Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating?
(a) – COOH
(b) – NO2
(c) – C N
(d) – SO3H
Answer:
(b) – NO2

Question 25.
Which of the following can be used as the halide component for friedal – crafts reaction?
(a) Chiorobenzene
(b) Bromobenzene
(c) Chloroethene
(d) Isopropyl chloride
Answer:
(d) Isopropyl chloride

Question 26.
An alkane is obtained by decarboxylation of sodium propionate. Same alkane can be prepared by ……..
(a) Catalytic hydrogenation of propene
(b) action of sodium metal on iodomethane
(c) reduction of 1 – chloropropane
(d) reduction of bromomethane
Answer:
(b) action of sodium metal on iodomethane

Question 27.
Which of the following is aliphatic saturated hydrocarbon?
(a) C8H18
(b) C9H18
(c) C8H14
(d) All of these
Answer:
(a) C8H18

Question 28.
Identify the compound ‘Z’ in the following reaction.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
(a) Formaldehyde
(b) Acetaldehyde
(c) Formic acid
(d) None of these
Answer:
(a) Formaldehyde

Question 29.
Peroxide effect (Kharasch effect) can be studied in case of ………
(a) Oct – 4 – ene
(b) Hex – 3 – ene
(c) Pent – 1 – ene
(d) But – 2 – ene
Answer:
(a) Pent – 1 – ene

Question 30.
2 – butyne on chlorination gives ………
(a) 1 – chlorobutane
(b) 1, 2 – dichlorobutane
(c) 1, 1, 2, 2 – tetrachlorobutane
(d) 2, 2, 3, 3 – tetrachlorobutane
Answer:
(d) 2, 2, 3, 3 tetra chiorobutane

Samacheer Kalvi 11th Chemistry Hydrocarbons Short Answer Questions

Question 31.
Give IUPAC names for the fllowing compounds …………
(i) CH3 CH = CH – CH= CH – C ≡C – CH3
(ii) Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

(iii) (CH3)3 C-C≡C-CH (CH3)2
(iv) Ethyl – isopropyl – acetytene
(v) CH≡C – C = C – C≡CH
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 32.
Identify the compound A. B, C and D in the following series of reactions.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 33.
Write a short note on ortho – para directors aromat&c e1ropiJic substitution reactions.
Answer:
The group which increases the eleiron deity at oìo and para positions of the ring are known as ortho-para directors.
Example:
-OH, -NH2 -NHR -CH3, -OCH3 etc.
Let us consider the directive influences of phenolic( -OH) group. Phenol is the resonace hybrid of following structure.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
In these resonance structures the negative charge residue is present on ortho and para posrtions of the ring structure. Therefore the electron density at ortho and para positions increases as compared to the metci position, thus phenolic group activities the benzene ring for electrophilic attack at ortho and para positions and hetice – OH group is an ortho-para direction or and an activator.

Question 34.
How is propyne prepared from an alkyene dihalide?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 35.
An alkyl halide with molecular formula C6H13Br on dehydrohalogenation gave two isomeric alkenes X and Y with molecular formula C6H12. On reductive ozonolysis, X and Y gave four compounds CH3COCH3. CH3CHO, CH3CH2CHO and (CH3)2 CHCHO. Find the alkyl halide.
Answer:
1. C6H13Br is 3 – Bromo – 4 methylpentanc.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
2. 3 – Bromo -4 methylpentane on dehydrogenation give two isomers X and Y as follows:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
There fore C6H13 Br is 3 – Bromo – 4 – methy ipentane.

Question 36.
Describe the mechanism of Nitration of benzene.
Answer:
Step-1 :
Generation of \(\overset { \oplus }{ { NO }_{ 2 } } \) electrophile.
HNO3 + H2SO4 → \(\overset { \oplus }{ { NO }_{ 2 } } \) + H\(\overset { \oplus }{ { SO }_{ 4 } } \) + H2O

Step-2:
Attack of the electrophile on benzene ring to form arenium ion.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Step-3:
Rearomatisation of arenium ion.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Overall Reaction:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 37.
How does Huckel rule help to decide the aromatic character of a compound?
Answer:
A compound is said to be aromatic, if it obeys the following rules:

  • The molecule must be cyclic.
  • The molecule must be co-planar.
  • Complete delocalisation of it-electrons in the ring.
  • Presence of (4n + 2) π electrons in the ring where n is an integer (n = 0,1,2 …)

This is known as Huckel’s rule.
Example:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
1. It is cyclic one.
2. It is a co-planar molecule.
3. It has six delocalised ir electrons.
4. 4n + 2 = 6
4n = 6 – 2
4n = 4
n = 1
It obey Huckel’s rule, with n = 1, hence benzene is aromatic in nature.

Question 38.
Suggest the route for the preparation of the following from benzene.
1. 3 – chioro – nitrobenzene
2. 4- chlorotoluene
3. Bromobenzene
4. in – dinitrobenzene
Answer:
1. Preparation of 3 – chloronitro – benzene from benzene:
Benzene undergoes nitration and followed by chlorination and it leads to the formation of 3- chloronitrobenzene.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

2. Preparation 4-chiorotoluene from benzene:
Benzene undergoes Fnedel crafi’s alkylation followed by chlorination and it leads to the formation of 4-chiorotoluene.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

3. Prepar2tion of Bromobenzene from benzene:
Bezene undergo bromination to give bromobenzene.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

4. Preparation of m-dinitrobenzene from benzene:
Benzene undergo twice the time nitration to give m-dinitrobenzene.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 39.
Suggest a simple chemical test to distinguish propane and propene.
Answer:
Chemical test to distinguish between propane and propene:
1. Bromine water test:
Propene contains double bond, therefore when we pour the bromine water to propene sample, it decolounses the bromine water whereas propane which is a saturated hydrocarbon does not decolourise the bromine water.

2. Baeyer’s test:
When propene reacts with Bayer’s reagent it gives 1,2 dihydroxypropene. Propane does not react with Baeyer’s reagent.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 40.
What happens when isobutylene is treated with acidified potassium permanganate?
Answer:
Isobutylene is treated with acidified KMnO4 to give acetone.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 41.
how will you convert ethyl chloride in to –
1. ethane
2. n – butane
Answer:
1. Conversion of ethyl chloride into ethane:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

2. Conversion of ethyl chloride into n-butane:
Wurtz reaction:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 42.
Describe the conformers of n-butane.
Answer:
n-butane may be considered as a derivative of ethane as one hydrogen on each carbon atom is replaced by a methyl group.

Edipsed conformation:
In this conformation, the distance between the two methyl groups is minimum so there is maximum repulsion between them and it is the least stable conformer.

Anti or staggered form:
In this conformation, the distance between the two methyl groups is maximum and so there is minimum repulsion between them. It is the most stable conformer. The following potentially energy diagram shows the relative stability of various conformers of n-butane.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 43.
Write the chemical equations for combustion of propane.
Answer:
Chemical equations for combustion of propane:
The general combustion reaction for any alkane is:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 44.
Explain Markovnikoffs rule with suitable example.
Answer:
Markovnikoff’s rule: When an unsymmetrical alkene reacts with hydrogen halide, the hydrogen adds to the carbon atom that has more number of hydrogen and halogen adds to the carbon atom having fewer hydrogen atoms.
Example:
Addition of HBr to Propene:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 45.
What happens when ethylene is passed through cold dilute alkaline potassium permanganate.
Answer:
Ethylene reacts with cold dilute alkaline KMnO4 solution to give ethylene glycol:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 46.
Write the structures of following alkanes.
1. 2, 3 – Dimethyl – 6 – (2 – methylpropyl) decane
2. 5 – (2 – Ethylbutyl) – 3, 3 – dimethyldecane
3. 5 (1,2 – Dimethyipropyl) – 2 – methylnonane
Answer:
1. 2, 3 – Dimethyl – 6 – (2 – rnethylpropyl) decane
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
2. 5 – (2 – Ethylbutyl) – 3, 3- dimethyldecane
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
3. 5 – (1,2 – Dimethylpropyl) – 2- methylnonane
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 47.
How will you prepare propane from a sodium salt of fatty acid?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Sodium salt of butyric acid on heating with sodalime gives propane.

Question 48.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons - 257
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 49.
How will you distinguish 1 – butyne and 2 – butyne?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 50.
How will you distinguish 1 – butyene and 2 – butyne?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

In 1-butyne, terminal carbon atom contains atom one acidic hydrogen, therefore it will react with silver nitrate in the presence of ammonium hydroxide to give silver butynide. Whereas 2-butyne does not undergo such type of the reaction, because of the absence of acidic hydrogen.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Samacheer Kalvi 11th Chemistry Hydrocarbons Evaluate Your self

Question 1.
Write the structural formula and carbon skeleton formula for all possible chain isomers of C6H14 (Hexane).
Answer:
C6H14 has five possible isomeric structures:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 2.
Give the IUPAC name for the following alkane.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 3.
Draw the structural formula for 4,5 -diethyl -3,4,5-trimethyloiane
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 4.
Water destroys Grignard reagents. Why?
Answer:
CH3MgX + HOH → CH4 – Mg(OH)X.
Water would protonate the grignard reagent and destroy the gngnard reagent, because the grignard carbon atom is highly nucleophilic. This would form a hydrocarbon. Therefore to make a grignard solution, only ether is the best solvent and water or alcohol are not used for that purpose.

Question 5.
Is it possible to prepare methane by Kolbe’s electrolytic method.
Answer:
Kolbe’s electrolytic method is suitable for the preparation of symmetrical alkanes, that is alkanes containing even number of carbon atoms. Methane has only one carbon, hence it cannot be prepared by Kolbe’s electrolytic method.

Question 6.
Write down the combustion reaction of propane whose AH° = -2220 kJ
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 7.
Why ethane is produced in chlorination of methane?
Answer:
Chlorination of methane involves free radical mechanism. During the propagation step methyl free radical is produced, which is involved in the termination step, the two methyl free radical then combines to form ethane.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 8.
How toluene can be prepared by this method?
(i) From n-heptane,
(ii) From 2-methyihexane
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

(ii)Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 9.
Write the IUPAC names for the following alkenes.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 10.
Draw the structures for the following alkenes.

  1. 6 – Bromo – 2,3 – dimethyl 2 – hexene
  2. 5 – Bromo – 4 – chloro 1 – heptene
  3. 2,5 – Dimethyl 4 – octene
  4. 4 – Methyl – 2 pentene

Answer:
1. 6 – Bromo – 2,3 – dirnethyl – 2 – hexene:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

2. 5 Bromo -4 – Chloro – 1 – heptene:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

3. 2.5 – dimethyl – 4 – Octene:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

4. 4 – methyl 2 pentene:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 11.
Draw the structure and write down the IUPAC name for the isomerism exhibited by the molecular formulae:
1. C5H10 – Pentene (3 isomers)
2. C6H12 – Hexene (5 isomers)
Answer:
1. C5H10 – Pentene (3 isomers)
(a) CH3 – CH2 – CH2 – CH = CH2 → 1- penlene (or) pent- 1-ene
(b) CH3 – CH2 – CH = CH – CH3 → 2- pentene (or) pent-2-ene
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

2. C6H12 – Hexene (5 isomers)
(a) CH3 – CH2 – CH2 – CH2 – CH = CH2 → Hex-I-ene
(b) CH3 – CH2 – CH2 – CH = CH – CH3 → Hex – 2 – ene
(c) CH3 – CH2 – CH = CH – CH2 – CH3 → Hex – 3 – ene
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
These two compounds exhibits constitutional isomerism.

Question 12.
Determine whether each of the following alkenes can exist as cis-trans isomers?
(a) 1 – Chloropropene
(b) 2 – Chloropropene
Answer:
(a) 1 – Chloropropene:
CH3 – CH = CHCl
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Therefore – 1 – Chloropropene has cis-trans isomers.

(b) 2 – Chloropropene:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

In 2-Chloropropene, it deviates from the rule, that is “At least one same group is present on the two doubly bonded carbon atom”. Therefore-2-chloropropenc cannot exist as cis – trans isomers.

Question 13.
Draw cis-trans isomers for the following compounds
(a) 2- chloro-2-butene
(b) CH3 – CH= CH – CH2 – CH3
Answer:
(a) 2-Chloro-2-butene:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

(b) CH3 – CH = CH – CH – CH3
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 14.
How propene is prepared form 1, 2-dichloropropane?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 15.
How ozone reacts with 2-methyl propene?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 16.
An organic compound (A) on ozonolysis gives only acetaldehyde. (A) reacts with Br2 /CCl4 to give compound (B). Identify the compounds (A) and (B). Write the IUPAC name of (A) and (B). Give the geometrical isomers of (A).
Answer:
2-Rutene undergo ozonolysis to give acetaldehyde only.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Geometrical isomers of 2 – Butene (A):
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 17.
An organic compund (A) C2H4 decolourises bromine water. (A) on reaction with chlorine gives (Br). A reacts with HBr to give (C). Identify (A),(B),(C). Explain the reactions.
Answer:
(i) C2H4 (A), decolourises the bromine water. Therefore it contains double bond. Hence (A) is ethylene.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 18.
Prepare propyne from its corresponding alkene.
Answer:
Preparation of propyne from propene:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 19.
Write the products A & B for the following reaction.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 20.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 21.
Calculate the number of rings present in C18H12.
Answer:
Double bond equivalent formula = ( C – \(\frac {1}{2}\) + \(\frac {1}{2}\))
Where C = no. of carbon atoms. H = no. of hydrogen and halogen atoms and N = no. of nitrogen atoms.
So. in C18H12 Double bond equkalent = 18 – \(\frac {12}{2}\) + 0 + 1 = 18 – 6 + 1 = 13
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
One ring is equal to one double bond equivalent.
∴ here four rings are there, four double bond equivalent arc used. So remaining, 13 – 4 = 9.
nine double bonds are present in the ring. Hence, C18H12 contain four aromatic rings.

Question 22.
write all possible isomers for an aromatic benzenoid compound having the molecular formula C8H10.
Answer:
Possible isomers for C8H10:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 23.
Write all possible isomers Iòr a monosubstituted aromatic benzenoid compound having the molecular formula C9H12
Answer:
Possible isomers for C9H12:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 24.
how benzene can be prepared by Grignard Reagent?
Answer:
Phenyl magnesium bromde reagent reacts with water molecule to give benzene:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 25.
Why benzene undergoes electrophilic substitution reaction whereas alkenes undergoes addition reaction?
Answer:

  • Benzene possess an unhybridised p-orbital containing one electron. The lateral overlap of theirp-orbitals produces 3 it bond.
  • The six electron of the p-orbitais cover all the six carbon atoms and arc said to be delocalised.
  • Due to delocalisation, strong it-bond is formed which makes the molecule stable. Therefore benzene undergoes electrophilic substitution reaction, whereas alkenes undergoes addition reaction.

Question 26.
Convert Ethyne to Benzene and name the process.
Answer:
Conversion of Ethyne into Benzene:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

This process is one of the cyclic polymerisation process.

Question 27.
Toluene undergoes nitration easily than benzene.Why?
Answer:
1. Toluene has a methyl group on the benzene ring which is electron releasing group and hence activate the benzene ring by pushing the electrons on the benzene ring.
2. CH3 group is ortho – para director and ring activator. Therefore in toluene, ortho and para positions are the most reactive towards an electrophile, thus promoting electrophilic substitution reaction.
3. The methyl group hence makes it around 25 times more reactive than benzene. Therefore it undergoes nitration easily than benzene.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Samacheer Kalvi 11th Chemistry Hydrocarbons Additional Questions Solved

Choose the correct statement.

Question 1.
Statement-I : Methane. ethane, propène and butane are alkane group compounds.
Statement-II : They are obeying C2H2n+2 + formula and each member differs from it proceeding member by a CH2 group.
(a) Statement-I and II are correct and Statement-II is correct explanation of statement-I.
(b) Statement-I and II are correct but statement-II is not correct explanation of statement-I
(e) Statëment-I is correct but statement-II is wrong.
(d) Statement-I is wrong but statement-II is correct.
Answer:
(a) Statement-I and II are correct and Statement-II is correct explanation of statement-I.

Question 2.
Statement – I : n – butane and iso – butane are isomers.
Statement – II : Because they are having same molecular formula but differs only in the structural formula.
(a) Statement -I and II are correct and statement – II is correct explanation of statement – I
(h) Statement – I and II are correct but statement – II is not correct explanation of statement -I
(c) Statement – I is correct but statement – II s wrong.
(d) Statement – I is wrong but statement – II is correct.
Answer:
(a) Statement -I and II are correct and statement – II is correct explanation of statement-I

Question 3.
Which one of the following shows three possible isomeric structures’?
(a) C4H10
(b) C5H12
(c) C6H12
(d) C3H4
Answer:
(b) C5H12

Question 4.
Find out the branched hydrocarbon from the following compounds.
(a) 1 – propane
(b) n – propane
(c) iso – butane
(d) n – butane –
Answer:
(c) iso – butane

Question 5.
Which of the following compound cannot be prepared by Kolbe’s electrolytic method’?
(a) CH3-CH3
(b) CH4
(c) CH2 = CH2
(d) CH = CH
Answer:
(b) CH4

Question 6.
Statement – I : Boiling point of methane is lower than that of butane.
Statement – II : The boiling point of continuous chain alkanes increases with increase in length of carbon chain.
(a) Statement – I and II are correct and statement – II is correct explanation of statement – I
(b) Statement – I and II are correct but statement – II is not correct explanation of statement -I
(c) Statement – I is correct but statement – II is wrong.
(d) Statement – I is wrong but statement – I is correct.
Answer:
(a) Statement – I and II are correct and statement – II is correct explanation of statement -I.

Question 7.
Statement – I : Alkenes shows both structural and geometrical isomerism.
Statement – II : Because of the presence of double bond.
(a) Statement – I and II are correct and statement-II is correct explanation of statement – I.
(b) Statement – I and II are correct but statement-II is not correct explanation of statement – I
(c) Statement – I is correct but statement – II is wrong.
(d) Statement – II is wrong but statement – II is correct.
Answer:
(a) Statement – I and II are correct and statement – II is correct explanation of statement – I.

Question 8.
Consider the following statements.
(i) The process of reduction using sodium in liquid ammonia is called as Birch reduction.
(ii) Birch reduction is stereospecific in reaction.
(iii) Aleynes can be reduced to cis – alkenes using Birch reduction. Which of the above statement is/are correct’?
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) only
(d) (ii) only
Answer:
(a) (i) and (ii)

Question 9.
Statement – I : Alkenes are more reactive than alkanes.
Statement – II : Because of the presence of a double bond.
(a) Statement – I and II are correct and statement – II is correct explanation of statement – I
(b) Statement – I and Il are correct but statement – II is not correct explanation of statement -I
(c) Statement – I is correct but statement – II is wrong.
(d) Statement – I is wrong but statement – II is correct.
Answer:
(a) Statement – I and II are correct and statement – II is correct explanation of statement – I

Question 10.
Peroxide effect is not observed in ………..
(a) HCl
(b) HI
(c) HBr
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 11.
Which one of the following has garlie odour?
(a) Ethane
(b) Ethene
(c) Ethyne
(d) Ethanol
Answer:
(c) Ethyne

Question 12.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons Identify the X.
(a) Propane
(b) Acetone
(c) Acetaldehyde
(d) Formaldehyde
Answer:
(b) Acetone

Question 13.
Which one of the following is not a monocyclic aromatic hydrocarbon?
(a) Benzene
(b) Phenol
(c) Toluene
(d) Naphthalene
Answer:
(d) Naphthalene

Question 14.
Which one of the following is a polynuclear aromatic hydrocarbon?
(a) Anthracene
(b) Phenol
(c) Benzene
(d) Toluene
Answer:
(a) Anthracene

Question 15.
Which one of the following is an aromatic compounds?
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 16.
Molecular formula of benzene is ………….
(a) C6H6
(b) C6H5
(c) C7H8
(d) CH4
Answer:
(a) C6H6

Question 17.
Statements-I : Unlike alkenes and alkynes benzene undergoes substitution reactions rather than addition reaction under normal conditions.
Statements-II : Because of the delocalisation of electrons a strong t bond is formed which makes the molecule stable.
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.
(b) Statement-I and II are correct but statement-II is not correct explanation of statement-I
(c) Statement-I ¡s correct but statement-II is wrong.
(d) Statement-I is wrong but statement-II is correct.
Answer:
(a) Statement-I and II are correct and statement-II ls correct explanation of statemen-I.

Question 18.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons A → B + H2O. Identify A and B.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 19.
Benzene undergoes Birch reduction to form .
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 20.
Which one of the following is not an ortho-para director?
(a) – NO2
(b) – CH3
(c) – OH
(d) – C2H5
Answer:
(a) – NO2

Question 21.
Which one of the following is not a meta director’?
(a) – NH2
(b) – NO2
(c) – COOR
(d) – SO3H
Answer:
(a) – NH2

Question 22.
Which one of the following benzene ring deactivator?
(a) – CHO
(b) – OH
(c) – CH3
(d) – OCH3
Answer:
(a) – CHO

Question 23.
Fine the odd one out:
(a) Benzene
(b) Ethane
(c) Ethene
(d) Propyne
Answer:
(a) Benzene

Question 24.
Which among these is not associated with atiphatic compounds?
(a) They contain (4n+2) π electrons
(b) They contain straight chain compounds.
(c) They contain branched chain compounds.
(d) They have appropriate number of H-atoms and functional groups.
Answer:
(a) They contain (4n+2) π electron.

Question 25.
Which of the following compounds will exhibit cis-trans isomerism
(a) 2 – Butene
(b) 2 – Butyne
(c) 1 – Butene
(d) 2 – Butanol
Answer:
(a) 2 – Butene

Question 26.
Which conformation of ethane has the lowest potential energy?
(a) Eclipsed
(b) Staggered
(c) Skew
(d) All will have equal potential energy.
Answer:
(b) Staggered

Question 27.
Which of the following is less reactive than benzene towards electrophilic substitution reactions?
(a) Nitrobenzene
(b) Aniline
(c) Bromobenzene
(d) Chlorobenzene
Answer:
(a) Nitrobenzene

II. Match the following
Question 1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 3.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 4.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 5.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 6.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 7.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 8.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

III. Fill in the blanks.

Question 1.
Liquefied petroleum gas consist of a mixture of ………..
Answer:
Propane + butane

Question 2.
Mangoes contain ………….
Answer:
Cyclo hexane

Question 3.
Methane gas is also called as ………….
Answer:
Marsh gas

Question 4.
The IUPAC name of the following compound is:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
3-Ethyl, 2-methylpentane

Question 5.
Sodalime is the mixture of ……………
Answer;
NaOH + CaO

Question 6.
Wurtz reaction used in the preparation of …………
Answer:
higher alkanes

Question 7.
The major reagent present in Corey-House reaction is ………..
Answer:
Lithium dimethyl cuparate

Question 8.
General formula for grignard reagents is ………..
Answer:
R – MgX

Question 9.
The rotation of C – C single bond leads to different isomenc structure called as …………
Answer:
conformers

Question 10.
The least stable conformer of ethane is form ……….
Answer:
eclipsed

Question 11.
The potential energy difference between the staggered and eclipsed conformation of ethane is …………
Answer:
12.5 kJ /mol

Question 12.
The most stable conformer of butane is ………..
Answer:
Staggered

Question 13.
Paraffin is the older name for the group family of compounds.
Answer:
alkane

Question 14.
Paraffin means ……….
Answer:
Little activity

Question 15.
Preparation of methyl chloride is followed by mechanism.
Answer:
Free radical

Question 16.
n – hexane passed over chromic oxide supported on alumina at 873 K will give ……….
Answer:
Bcnzene

Question 17.
The number of possible isomers of C6H12 is …………..
Answer:
5

Question 18.
When ethanol is heated at 440 K with excess of concentrated H2SO4, it will give …………
Answer:
Ethene

Question 19.
Alkynes undergoes reduction using Lindlar’s catalyst to give ………….
Answer:
cix – alkenes

Question 20.
Alkynes undergoes reduction using sodium in liquid ammonia to give …………
Answer:
trans – alkenes

Question 21.
Aqueous solution of potassium succinate is electrolysed to give ………….
Answer:
Ethane

Question 22.
The order of reactivity of different hydrogen halides (HCl, HI , HBr) is ………..
Answer:
HI >HBr >HCl

Question 23.
Addition of hydrohalides to alkene is an example for …………
Answer:
Electrophilic additton

Question 24.
Ethane reacts with HBr to form …………
Answer:
Bromo ethane

Question 25.
Homolytic fission of benzoyl peroxide will give ………..
Answer:
C6H5

Question 26.
Propene reacts with HBr in the presence of peroxide to form …………
Answer:
1 – Bromopropane

Question 27.
Baeyer’s reagent is …………
Answer:
alkaline KMNO4

Question 28.
Ozonolysis of ethene produces ………….
Answer:
HCHO

Question 29.
Electrolysis of potassium maleate yields ………
Answer:
Ethyne

Question 30.
Ozonolysis of acetylene gives ………….
Answer:
HCOOH

Question 31.
Three molecules of acetylene undergoes polymerisation to give …………
Answer:
benzene

Question 32.
Benzene reacts with bromine in the presence of AlCl3 to form bromobenzene and it is an example of reaction
Answer:
Electrophilic substitution

Question 33.
The six carbon atoms of benzene are hybridised.
Answer:
sp2

Question 34.
Bond angle in benzene is …………
Answer:
120º

Question 35.
Benzene contains a bonds and it bonds …………
Answer:
12,3

Question 36.
Wurtz-fittig reaction helps to prepare compounds …………
Answer:
aromatic

Question 37.
When phenol reacts with Zn-dust under dry distillation conditions it gives ………..
Answer:
Benzene

Question 38.
Benzene is insoluble in …………
Answer:
water

Question 39.
Benzene reacts with hydrogen in the presence of Pt to yield ……….
Answer:
Cyclohexane

Question 40.
Benzene reacts with Cl2 in the presence of sunlight to give ………..
Answer:
BHC (Ben.zene hexachioride)

Question 41.
The step in which Cl-Cl bond homolysis occurs is called ……….
Answer:
Initiation step

Question 42.
Dienes are the name given to compounds with ……….
Answer:
Exactly two double bonds

Question 43.
The hybridisation state of a carbocation is …………
Answer:
sp2

Question 44.
The peroxide effect in anti-markovnikoff addition involves a ………. mechanism.
Answer:
free radical.

IV Choose the odd one out
Question 1.
(a) Ethane
(b) Benzene
(c) Ethene
(d) Ethyne
Answer:
(b) Benzene. it is an aromatic hydrocarbon whereas others are aliphatic hydrocarbons.

Question 2.
(a) Zn + HCl
(b) Zn + CH3COOH
(c) LiAlH4
(d) Acidified K2Cr2O7
Answer:
Acidified K2Cr2O7. It is an oxidising agent whereas other three reagents are reducing

Question 3.
(a) Soft drink bottle
(b) Jars
(c) Vegetable oil bottle
(d) Straws
Answer:
(d) Straws. It is made up of polypropylene whereas others are made of PET (Polyethylene terephthalate).

Question 4.
(a) Straws
(b) Foam cups
(c) Diapers
(d) Toys
Answer:
(b) Foam cups. It is made up of polystyrene whereas others are made up of polypropylene.

Question 5.
(a) Orlon
(b) Neoprene rubber
(c) PVC
(d) PET
Answer:
(d) PET. It is prepared by the polymerisation of glycol and terephthalic acid, whereas others are prepared from acetylene.

Choose the correct pair.

Question 1.
(a) Propene + O3 : HCHO + CH3 CH0
(b) Ethene + O3 : CH3 CHO + H2O2
(c) But-2-ene + O3 : HCHO + H2 O2
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
(a) Propene + O3 : HCHO + CH3CHO

Question 2.
(a) PET : shampoo bottles
(b) PS : disposable utensils
(c) PP : grocery bags
(d) HDPE : plastic pipes
Answer:
(b) PS : disposable utensils

Question 3.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 4.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

VI. Choose the incorrect pair.

Question 1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 2.
(a) CH2 = CH2 + O3 : 2HCOH
(b) CH3 – CH = CH2 + O3 : CH3CHO + HCHO
(c) CH3 – CH = CH – CH3 + O3 : 2CH3COCH3
(d) CH3 – CH = CH – CH3 + O3 : 2CH3CHO
Answer:
(c) CH3 – CH = CH – CH3 + O3 : 2CH3COCH3

Question 3.
(a) PET : Jars
(b) HDPE : Juice containers
(c) PS : Disposable utensils
(d) PVC : Grocery bags
Answer:
(d) PVC : Grocery bags

VII. Assertion & Reason.
Question 1.
Assertion (A) : Methane is called marsh gas.
Reason (R) : Decomposition of plant and animal matter in am oxygen deficient environment like swamps, mashes and bogs produce methane gas.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 2.
Assertion (A) : Water destroys grignard reagent and so it is not used as solvent for RMgX.
Reason (R) : Water decomposes grignard reagent (RMgX) to give alkane.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 3.
Assertion (A) : The boiling point of straight chain isomers have higher boiling point as compared to branched chain isomers.
Reason (R) : The boiling point decreases with increase in branching as the molecule becomes compact and the area of contact decreases.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A)is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 4.
Assertion (A) : The eclipsed conformation ofethane is less stable than staggered conformation of ethane.
Reason (R) : In eclipsed conformation, the distance between the two methyl group is minimum and so there is maximum repulsion between them and it is the least stable conformer.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A)and (R) are correct hut (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Samacheer Kalvi 11th Chemistry Hydrocarbons 2 Mark Questions and Answers

Question 1.
What are unsaturated hydrocarbons?
Answer:
hydrocarbons having localised carbon – carbon multiple bonds are called unsaturated hydrocarbons.
Example :
alkenes and alkynes.

Question 2.
What is marsh gas?
Answer:
Answer:

  • Methane is the major component of the atmosphere of jupiter, Saturn, Uranus and Neptune but only minor component of earth atmosphere.
  • Decomposition of plant and animal matter in an oxygen deficient environment like swamps, marshes bogs and the sediments of lakes produces methane gas. It is otherwise known as marsh gas.

Question 3.
Write a note on methane clathrates.
Answer:

  • Frozen mixture of water and methane gas is chemically known as methane clathrate.
  • The methane molecule which is produced by biological process under the deep -ocean (at 4°C and 50 atm) does not simply reach the surface instead each molecule is trapped inside the clusters of 6 to 18 water molecules forming methane clathratcs.

Question 4.
What is isomerism? Mention the types of isomerism?
Answer:
The phenomenon in which the same molecular formula may exhibit different structural arrangement is called isomerism.
There are two types of isomerism. namely.

  • Structural isomerism
  • Stereoisomerism

Question 5.
Draw and name the possible structural formula for C4H10
Answer:
C4H10 has two possible structural formula, they are,
(i) CH3-CH2-CH2-CH3 n-Butane
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 6.
Give the IUPAC name of the following compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 7.
What is Sabatier- Sendersens reaction?
Answer:
The process of addition of H2 to unsaturated compounds (alkenes or alkynes) in known as hydrogenation. The above process can be catalysed by nickel at 298 K. This reaction is known as Sabalier-Sendersens reaction.
For example:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 8.
What are decarboxylation reactions? Given an example.
Answer:
When a mixture of sodium salt of carboxylic acid and sodalime is heated an alkane is formed. During this process CO2 molecule is eliminated process and this process is known as decarboxylation reaction.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 9.
Write a note on Kolbe’s electrolytic method?
Answer:
When potassium or sodium salt of carboxylic acid is electrolysed, a higher alkane is formed.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 10.
How will you prepare propane from Chloropropane?
Answer:
Nascent hydrogen reacts with chloropropane to give propane.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 11.
What is Wurtz reaction?
Answer:
When a solution of haloalkanc in dry ether is treated with sodium metal, higher alkanes are produced. This reaction is known as Wurtz reaction. For example:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 12.
Write shnrt flotes on Corey-House reaction?
Answer:
An alkyl halide and lithium dialkyl cupratc are reacted to give higher alkanes. This reaction is known as Corey-House reaction.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 13.
How will you prepare methane from grignard reagent’?
Answer:
Methyl magnesium chloride reacts with water to give methane.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 14.
What are conformers?
Answer:
The rotation about C-C single bond axis yielding several arrangements of a hydrocarbon called conformers.

Question 15.
Draw the conformations of ethane using Newman projection formula method?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 16.
What are combustion reactions?
Answer:
A combustion reaction is a chemical reaction between a substance and oxygen with evolution of heat and light. In the presence of sufficient oxygen alkanes undergoes combustion when ignited and produces carbon dioxide and water.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 17.
What is arornatisation?
Answer:
Alkanes with six to ten carbon atoms are converted into homologous of benzene at higher temperature and in the presence of a catalyst. This process is known as aromatisation. For example, n-Hexane passed over Cr2O3 supported on alumina at 873 K gives benzene.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 18.
Write notes on isomerisation.
Answer:
1. Isomerisation is a chemical process by which a compound is transformed into any of its isomeric form.
2. Normal alkanes can be converted into branched alkanes in the presence ofAlCl3 and HCl at 298 K.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
3.  This process is of great industrial importance, the quantity of gasoline is improved by isomerising its components.

Question 19.
Mention the uses of alkanes.
Answer:

  • Alkanes are extensively used as fuels.
  • Methane present in natural gas is used in home heating.
  • A mixture of propane and butane is known as LPG gas which is used for domestic cooking purpose.
  • Gasoline is a complex mixture of many hydrocarbons used as a fuel for internal combustion engines.

Question 20.
Draw and name the structural formula for C4H8’?
Answer:
(i) CH3 – CH = CH – CH3 -w 1 – Butene
(ii) CH2 = CH – CH2 – CH2 – 2 Butene
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 21.
cis-isomers are less stable than rans-isomers?
Answer:
Consider:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Among cis and trans isomers, cis isomer is less stable than trans isomer. In the cis isomer, similar groups arc very near to each other. vander Waal’s repulsion and steric hindrance make the molecule much more unstable. But in trans isomer, similar groups are diagonally opposite to each other and there is no such steric hindrance. Due to more steric interaction cis isomers is less stable than Irons isomer.

Question 22.
How will you prepare ethene from ethanol?
Answrer:
When ethanol is heated at 430 K – 440 K with excess cone. H2SO4. a molecule of water from alcohol is removed and ethene or ethylene is formed.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 23.
How will you convert 1-brornopropane into popene?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
1-bromopropane reacts with alcoholic KOH and eliminate hydrogen bromide resulting in the formation of propene.

Question 24.
How will you prepare ethene by Kolbe’s electrolytic method?
Answer:
When an aqueous solution of potassium succinate is electrolysed between two platinum electrodes, ethene is produced at the anode.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 25.
Write any two test for alkenes.
Answer:

  • Rapid decolourisation of bromine in CCl4 without evolution of hydrogen bromide.
  • Decolourisation of cold dilute aqueous solution of KMnO4

Question 26.
State Markovnikoff’s rule.
Answer:
When an unsymmetrical alkene reacts with hydrogen halide, the hydrogen atom adds to the carbon atom that has more number of hydrogen atoms and halogen add to the carbon atom having fewer hydrogen atoms.

Question 27.
What is peroxide effect?
Answer:
The addition of HBr to an alkene in the presence of organic peroxide gives the antì-Markovnikofl’s product. This effect is called as peroxide effect.

Question 28.
Identify the products A and B from the following reaction.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 29.
What happens when propcne reacts with concentrated H,S04?
Answer:
Propene reacts rjth cone. H2S04 to form 2 – propyl hydrogen sulphate in accordance with Markovnikoffs rule Further hydrolysis yields 2-propanol.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 30.
Complete the following reaction and identify A, B and C.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 31.
Mention the uses of alkenes.
Answer:

  • Alkenes are used as starting material in the synthesis of alcohols, plastics, detergents and fuels.
  • Ethenc is the most important organic fcedstock in the polymer industry. Examples are, PVC, Sarans and Polythene. These polymers are used in the manufacture of floor tiles, shoe-soles, synthetic fibres, raincoats, pipes etc.

Question 32.
What are gem dihalides? How will you prepare propyne from gem dihalides?
Answer:
Compounds containing two halogen atoms on the same carbon atom are called gem dihalides. On heating 1,1 -dichloropropane with alcoholic KOH, it will give propyne.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 33.
How will you prepare acetylene from potassium maleate?
Answer:
Electrolysis of potassium maleate yields acetylene. This is one of Kolbe’s electrolytic reaction.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 34.
How will you prepare acetylene from calcium carbide?
Answer:
Acetylene can be manufactured in large scale by action of calcium carbide with water.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 35.
How will you convert ethyne into ethanol?
Answer:
Ethyne undergo hydration on warming with mercuric sulphate and dil H2SO4 at 333 K to form ethanol (Acetaldehyde).
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 36.
Mention the uses of Acetylene.
Answer:

  • Acetylene is used in oxy acetylene torch used for welding and cutting metals.
  • It is used for manufacture of PVC, polyvinyl acetate, Polyvinyl ether, orlon and neoprene rubbers.

Question 37.
What are all the conditions for aromaticity?
Answer:
Huckel proposed that aromaticity is a function of electronic structure of an organic compound.
A compound may be aromatic, if it obey the following rules:

  • The molecule must have a cyclic structure.
  • The molecule must be co-planar.
  • Complete delocalisation of it electrons in the ring.
  • Presence of (4n + 2) it electrons in the ring where n is an integer (n = 0,1,2….), this is known as Huckel’s rule.

Question 38.
Classify the following compounds by using aromaticity concepts:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

  • It is co-planar molecule.
  •  It has six delocalised ir electrons.
  •  4n + 26
    4n = 6 – 2
    4,n = 4
    n = 1, obeys Huckel’s rule with n = 1

(b) Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons Cyclo-octatetraene.

1. Molecule is non-planar.
Hence it is a non-aromatic compound.

(c) Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons Cyclopropyl cation.

  • It has a co-planar structure.
  • It has two delocalised it electrons.
    4n + 2 = 2
    4n = 0
    n = 0

Hence is an aromatic compound.

Question 39.
Write notes on Resonance of benzene?
Answer:
1. The phenomenon in which two or more structures can be written for a substance which has identical position of atoms is called resonance.
2. The actual structure of the molecule is said to be a resonance hybrid of the various possible alternative structures.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
3.  In benzene, Kekule’s structures I and II represented the resonance structures, and structure III is the resonance hybrid of structures I and II.
4. The structures I and II exist only in theory. The actual structure of benzene is the hybrid of two hypothetical resonance structures.

Question 40.
How will you convert phenol into benzene?
Answer:
When phenol vapours are passed over zinc dust then it is reduced to benzene:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 41.
What is Wurtz-fitting reaction?
Answer:
When a solution of bromobenzene and iodomethane in dry ether is treated with metallic sodium, toluene is formed.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 42.
What are activating and deactivating groups?
Answer:

  • When mono substituted benzene undergoes an electrophilic substitution reaction, the rate of the reaction and the site of attack of the incoming electrophile depends on the functional group already attached to it.
  •  Some groups increases the reactivity of benzene ring and are known as activating groups.
  • Some groups decreases the reactivity of benzene ring and are known as de-activating groups.
  • Example:
    Activating groups: – NH2 – OH. – CH3 etc.
    Deactivating groups: – NO3. CN, – CHO etc,

Question 43.
Why does benzcnc undergo clectrophilic substitution reactions easily and nucleophilic substitution with difficulty?
Answer:
Due to the presence of an electron cloud containing 6π – electrons above and below the plane of the ring, benzene is a rich source of electrons, Consequently, it attracts the electrophilic reagents towards it and repels the nucleophilic reagents. As a result, benzene undergoes electrophilic substitution reactions easily and nucleophilic substitution with difficulty.

Question 44.
Out of benzene, m – dinitrobenzene and toluene which will undergo nitration most easily and why?
Answer:
CH3 group is an electron-donating group, while -NO2 group is electron withdrawing group. Therefore, maximum electron density will be there in toluene, followed by benzene and it is least in pn-dinitrobenzene. Therefore, the ease of nitration decreases in the order:
Toluene > benzene > m – dinitrobenzene.

Question 45.
Why are alkanes called paraffins?
Answer:
Paraffins means little affinity. Alkanes due to strong C-C and C-H bonds are relatively chemically inert. They are thus called as paraflins.

Samacheer Kalvi 11th Chemistry Hydrocarbons 3 Mark Questions and Answers

Question 1.
Explain the classification of hydrocarbons.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 2.
How to write the possible isomers of C5H12?
Answer:
1. To begin draw the carbon backbone of the straight chain isomer.
C-C-C-C-C

2. To determine the carbon backbone structure of the other isomers, arrange the carbon atoms in the other way:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

3. Fill in ail the hydrogen atoms so that each carbon forms four bonds,
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 3.
Explain how to draw the structural formula for 3-ethyl, 2, 3-dimethylpentane.
Answer:
3-ethyl, 2, 3-dimethylpentane.
Step 1:
The parent hydrocarbon is pentane. Draw the chain of five carbon atoms and number
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Step 2:
Complete the carbon skeleton by attaching the alkyl group as they are specified in the name. An ethyl group is attached to carbon atom 3 and two methyl groups are attached to carbon atoms 2 and 3.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Step 3:
Add hydrogen atoms to the carbon skeleton so that each carbon atoms have four bonds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 4.
In alkane compounds with same number of carbon atoms, straight chain isomers have higher boiling point as compared to branched chain isomers-justify statement.
Answer:

  • The boiling point of continuous chain alkanes increases with increase in the length of carbon chain roughly about 30°C for every added carbon atom to the chain.
  • Alkanes are non-polar compounds and having weak Vander Wal’s force which depends upon molecular surface area and hence increases with increase in molecular size.
  • The boiling point decreases with increase in branching on the molecule i.e. as it becomes compact and the area of the contract decreases. Hence in alkanes with same number of carbon atoms, straight chain isomers have higher boiling point as compared to branched chain isomers.

Question 5.
Why oil spills in aqueous environment spread so quickly?
Answer:

  • Water molecules are polar and akanes are non-polar. The insolubility of alkanes in water makes them a good water repellent for metals which protects the metal surface from corrosion.
  • Because of their lower density than water they tòrm two layers and occupies the top layer. The density difference between alkanes and water explains why oil spills in aqueous environment spread so quickly.

Question 6.
Explain pyrolysis method.
Answer:
1. Pyrolysis is defined as the thermal decomposition of an organic compound into smaller fragments in the absence of air through the application of heat. Pyrolysis of alkanes is also named as cracking.

2. In the absence of air, whai alkane are vapours passed through red-hot metal it breaks into simpler hydrocarbons.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

3. The product depends upon the nature of alkane, temperature, pressure and presence or absence of the catalyst.

Question 7.
Write notes on Geometrical isomerism or cis-trans isomerism.
Answer:
1. It is a type of stereoisomerism and it is also called cis-trans isomerism. Such type of isomerism results due to the restricted rotation of doubly bounded carbon atoms.

2. If the similar groups lie on the same side then the geometrical isomers are called as cis-is omers.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

3. If the similar groups lie on the opposite side then the geometrical isomers are called as trans-isomers.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 8.
Explain how 2-butyne reacts with (a) Lindlar’s catalyst and (b) Sodium in liquid ammonia?
Answer:
(a) 2-butyne reacts with Lindlar’s catalyst:
2-butyne can be reduced to cis-2 butene using CaCO3 supported in Pd -metal partially deactivated with sulphur.
This reaction is stereo specific giving only the cis-2-butene.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

(b) 2-butyne reacts with sodium in liquid ammonia:
2-butyne can also be reduced to trans-2-butene using sodium in liquid ammonia. This reaction is stereospecific giving only the trans-2-butene.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 9.
What are vicinal dihalides? How will you prepane alkene from vicinal dihaldes?
Answer:
1. The compounds in which two halogen atoms are attached to adjacent carbon-atoms are called as vicinal dihalides.

2. When vicinal dihalides are warmed with granulated zinc in methanol they lose a molecule
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 10.
Why alkenes are more reactive than alkanes?
Answer:
1. Alkenes are more reactive than alkanes due to the presence of a double bond.

2. The a-bond is strong but the it-bond is weak. The typical reactions of alkenes involve the addition of an electrophile across the double bond proceeding through ionic mechanism. However addition reactions proceed through free-radical mechanism also. Ozonolysis and nolvmerisation are some of the characteristic reaction of alkenes.

Question 11.
Explain the mechanism of addition of HBr to propenc.
Answer:
Step 1:
Formation of electrophile:
In HBr, br is more electronegative than H. When bonded electron move towards Br, polarity is deve[oped and it creates electrophile H+ which attacks to the double bond to form a carbocation.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Step 2 :
Secondary carbocation is more stable than primary carhocation and it predominates over the primary carbocation.

Step 3 :
The Br-1 ion attack the 2°- carbocation to form 2-Bromo propane as the major product.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 12.
Explain the mechanism of addition of FlBr to 3-methyl-1-butene.
Answer:
Consider the addition of HBr to 3-methyl-1-butene. Here the expected product according to Markovnikoff’s rule is 2-bromo-3-methylbutane but the actual major product is 2-bromo- 2-methylbutane. This is because, the secondary carbocation formed during the reaction is rearranged to give the more stable tertiary carbocation. Attack of Br-1 on this tertiary carbocation gives the major product 2-bromo-2-methylbutane
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 13.
Why peroxide effect is not observed in HCl and HI?
Answer:
The H-Cl bond is stronger (430.5 k.J mol-1) than H-Hr bond (363.7 kJ mol-1), thus H-Cl is not cleaved by the free radical. The H-I bond is further weaker (296.8 kJ mol-3) than H-Cl bond. Thus H-I bond breaks easily hut iodine free radicals combine to form iodine molecules instead of adding to the double bond and hence peroxide effect is not observed in case of HCl and HI.

Question 14.
Explain the ozonolysis of (a) Ethene and (b) propene
Answer:
Ozonolysis is a method ofoxidative cleavage of alkenes using ozone and it form two carbonyl compounds. Alkenes react with ozone to form ozonide and it is cleaved by Zn/H2O to form smaller molecules.
This reaction is often used to identify the structure of unknown alkene by detecting the position of double or triple bond.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 15.
What is polymerisation? Explain with suitable example.
Answer:
A polymer is a large molecule formed by the combination of large number of small molecules (monomers). This process is known as polymerisation. A few examples are:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 16.
Complete the following reactions:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 17.
Explain the ozonolysis of (a) Acetylene (b) Propyne
Answer:
Ozone adds to carbon-carbon triple bond of alkynes to form ozonides. The ozonides are hydrolysed by water to form carbonyl compounds. The hydrogen peroxide formed in the reaction may oxidise the carbonyl compound to carboxylic acid.
(a) Acetylene:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

(b) propyne:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 18.
Explain the polyrnerisation of acetylene molecules.
Answer:
Acetylene undergoes two types of polymerisation reactions, they are:

  1. Linear polymerisation
  2. Cyclic polymerisation

1. Linear polymerisation:
Acetylene forms linear polymer, when passed into a solution of cuprous chloride and ammonium chloride.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

2. Cyclic polymerisation:
Acetylene undergoes cyclic polymerisation on passing through red hot iron tube. Three molecules of acetylene polymerises to form benzene.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 19.
Discuss the Kekule’s structure of benzene.
Answer:
Kekule suggested that benzene consists of a cyclic planar structure of six carbon with alternate single and double bonds.
There were two objections:
1. Bnzene forms only one ortho disubstituted products whereas the Kekule’s structure predicts two ortho disubstituted products as shown:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

2. Kekule’s structure failed to explain why benzene with three double bonds did not give addition reaction like alkenes. To overcome this objection, Kekule suggested that henzene was a resonance hybrid of two forms(1 and 2) which are in rapid equilibrium.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 20.
Why benzene undergoes substitution reaction rather than addition reactions under normal conditions?
Answer:
1. Each carbon atom in henzene possess an unhybridised p-orbital containing one electron. The lateral overlap of their p-orbitais produces 3π- bond, six electrons of the p-orbitais over all the six carbon atoms and are said lo be delocalised.

2. Due to delocalisation, strong it-bond is formed which makes the molecule stable. Hence unlike alkenes and alkynes, benzene undergoes substitution reactions rather than addition reactions under normal conditions.

Question 21.
Explain the industrial preparation of benzene from coal tar.
Answer:
Coal tar is a viscous liquid obtained by the pyrolysis of coal. During fractional distillation, coal tar is heated and distills away its volatile compounds, namely, benzene, toluene, xylene in the temperature range of 350 K to 443 K. These vapours are collected at the upper part of the fractionating column.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 22.
Explain the suiphonation of benzene.
Answer:
Benzcne reacts with fuming sulphuric acid and give benzene suiphonic acid. Although SO3 molecule it does not have positive charge, yet it is a strong electrophile. This is because the octet of electrons around the sulphur atom is not reacted. This reached is reversible nd desulphonation occus readily in aqueous medium.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 23.
What is BHC? How will you prepare BHC? Mention its uses.
Answer:
1. BHC is I3enzene hexachioride.
2. Benzene reacts with three molecule of Cl2in the presence of sunlight or UV light to yield BHC.
This is also called as gammaxane or Lindane.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
3. BHC is a powerful insecticide.

Question 24.
In aryl halides, halogen group is a oriho -para director and a deactivator towards electrophilic substitution reactions, why?
Answer:
1.  In aryl halides, the strong -I effect of the halogens decreases the electron density of benzene ring, thereby deactivating it for electrophilic attack.
2. The presence of lone pair on halogens is involved in resonance with π-electrons of the benzene ring and it increases the electron density at ortho and para position. Hence the halogen group is an ortho-para director and a deactivator.

Question 25.
Explain the carcinogenity and toxicity of aromatic hydrocarbons.
Answer:
Benzene and polycyclic aromatic hydrocarbons (PAH) are ubiquitous environmental pollutants generated during incomplete combustion of coal, oil, petrol and wood. Some (PAH) originate from open burning, natural seepage of petroleum and coal deposits and volcanic activities. They are toxic, mutagenic and carcinogenic. It has hematological, immunological and neutrological effect on humans They are radioactive and prolonged exposure leads to genetic damage. Some of the examples of PAH arc “L” shaped polynuclear hydrocarbons, which are much more toxic and carcinogenic.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
It is found in cigarette smoke, in tobacco and and charcoal boiled food.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
It is found in gasoline exhaust and barbecued food.

Question 26.
Arrange benzene, n-hexane and ethyne in decreasing order of acidic behavior. Also give reason for this behavior?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Since s-electrons arc closer to the nucleus, therefore as the s-character of the orbital making the C-H bond increases the electrons of C-H bond lies closer and closer to the carbon atom. In other words, the partial +ve charge on the H-atom and hence the acidic character increases as the s-character of the orbital increases. Thus, the acidic character decreases in the order
Ethyne > Benzene > Hexane

Samacheer Kalvi 11th Chemistry Hydrocarbons 5 Mark Questions

Question 1.
Explain the conformational analysis of ethane.
Answer:
The two tetrahedral methyl groups can rotate about the carbon-carbon bond axis yielding several arrangements called conformers. The extreme conformations are staggered and eclipsed conformations. There can be number of other arrangements between staggered and eclipsed forms and their arrangements are known as skew forms.

Eclipsed conformation:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

In this conformation, the hydrogen of one carbon atom is directly behind those of the other. The repulsion between the atoms is maximum and it is the least stable conformer.

Staggered conformation:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
In this conformation, the hydrogen’s of both the carbon atoms are tir apart from each other. The repulsion between the atoms is minimum and it is the most stable conformer.

Skew formation:
The infinite numbers of possible intermediate conformations between the two extreme conformations are referred as skew conformations. The stabilities of various conformations of ethane are
Staggered > Skew> Eclipsed
The potential energy difference between the staggered and eclipsed conformation of ethanc is around 12.5 kJ mol-1HBr The various conformations can be represented by Newman projection formula.

Newman Projection formula for Ethane:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 2.
Explain the structure of benzene.
Answer:
1. Molecular formula:
Elemental analysis and molecular weight determination have proved that the molecular formula of benzene is C6H6. This indicates that benzene is a highly unsaturated compound.

2. Straight chain structure is not possible:
Benzene could be constructed as a straight chain but it not feasible since it does not show the properties of alkenes or alkynes. For example, it does not decolorise the bromine water in CCl4.

3. Evidence of cyclic structure:
(1) In the presence of Nickel, benzene reacts with hydrogen to give cyclohexane, a six membered ring. This proves that benzene is a hexagonal
molecule with three double bonds.

(2) Benzene reacts with bromine in the presence of iron to give substituted C6H5Br. No isomers of C6H5Br was identified. On further reaction with bromine three isomeric disubstituted products C6H4Br2 are formed. On this basis Kekule proposed that benzene consists of ring of carbon atoms with alternate single and double bonds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

4. Resonance description of benzene:
The phenomenon in which two or more structures can be written for a substance which has identical position ofatomsis called resonance. The actual structure of the molecule is said to be a resonance hybrid of various possible alternative structures. In benzene. Kekule’s structure (I) and (II) represented the resonance structures and structure (III) is the resonance hybrid of structure I and II.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

5. Spectroscopic measurements:
X-ray and electron dîflìaction studies indicated that all carbon-carbon bonds are of equal length which is in between that of a single bond (1 .45Å ) and that of a double bond (1.34Å ).

6. Molecular orbital structure :
(1). Benzene is a flat hexagonal molecule with all carbons and hydrogen lying in the same plane with a bond angle of 120º. Each carbon atom has sp2 hybrid orbitais of carbon, overlap with each other and with s-orbitals of six hydrogen atoms forming six sigma (σ) C-H bonds and six sigma (σ) C – C bonds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

(2). All the σ-bonds in benzene lies in one plane with bond angle of 120º. Each C-atom in benzene possess an unhybridised p-orbital containing one electron. The lateral overlap of their p-orbitais produces 3π-bond, the six electrons of the p-orbitais cover all the six C-atoms and are said to be delocalised. Due to this delocalisation, strong π-bond is formed which makes the molecule stable.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

7. Representation of benzene:
Hence, there are three ways is which benzene can be represented.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 3.
Explain the mechanism of the reaction between methane and chlorine.
Answer:
Methane reacts with chlorine in the presence of light which proceeds through the free radical chain mechanism. This mechanism is characterised by three steps: initiation, propagation and termination.

1. Chain initiation:
The chain is initiated by UV light, leading to homolytic fission of chlorine molecules into free radicals (chlorine atoms).
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Here we choose Cl-Cl bond for fission because C- C and C-H bonds are stronger than Cl-Cl bond.

2. Chain propagation:
It proceeds as follows –
(a) Chlorine free radiais attack the methane molecule and breaks the C-H bond resulting in the generation of methyl free radicals.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

(b) The methyl free radicals thus obtained attacks the second molecule of chlorine to give chloromethane (CH3Cl) and a chlorine free radical as follows –
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

(c) This chlorine free radical then cycles back to step (a) and both steps (a) and (b) are repeated many times and thus a chain of reaction is set up.

3. Chain termination:
Aller sometime, the reaction stops due to the consumption of reactant and the chain is terminated by the combination of free radicals.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 4.
Write the mechanism for following reaction.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
The reaction proceeds through freee radical mechanism:

Step 1.
The weak O-O single bond linkage of peroxide undergoes homolytic cleavage to generate free radical.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Step 2.
The radicals abstracts a hydrogen atom from HBr thus generating bromine free radical.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Step 3.
The Bromine free radical adds lo the double bond in the way so as to form the more stable alkyl free radical,
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Step 4.
Addition of HBr to secondary free radical.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 5.
Explain the acidic nature of alkynes.
Answer:
An alkyiic shows acidic nature only if it contains terminal hydrogen atom. This can be explained by considering the Sp hybrid orbitais of carbon atom in alkynes. The percentage of s-character of sp hybrid orbital (50%) is more than sp2 hybrid orbital ofalkenes (33%) and sp3 hybrid orbital of alkanes (25%). Because of this, carbon atom becomes more electronegative, thus facilitating donation of H ions to bases. So hydrogen attached to triply bonded carbon atoms is acidic in nature.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 6.
Write the mechanism of chlorination of benzene.
Answer:
Step 1:
Generation of Cl electrophile.
AlCl + Cl – Cl + AlClΘ

Step 2:
Attack of the electrophile on the benzene ring to form arenium ion:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Step 3:
Rearomatisation of arenium ion:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 7.
Describe the mechanism of suiphonation of benzene.
Answer:
Step 1:
Generation of SO3 electrophile:
2H2SO4 → H3O + SO3 + HSO4Θ

Step 2:
Attack of the electrophile on benzene ring to form arenium ion:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Step 3:
Rearomatisation of Areniurn ion:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 8.
Describe the mechanism of Freidel craft’s alkylation.
Answer:
Step 1:
Generation of CH3 electrophile:
AlCl3 + CH3Cl → CH3 + AlClΘ

Step 2 :
Attack of the electrophile on benzene ring to form areniurn ion:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Step 3:
Rearornatisation of arenium ion:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 9.
Write the mechanism of Freidel craft’s acylation.
Answer:
Step 1:
Generation of CH3CO electrophile:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Step 2 :
Attack of the electrophile on benzene ring to form arenium ion.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Step 3 :
Rearomatisation of arenium ion:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 10.
An organic compound (A) of a molecular formula C6H6 which is a simple aromatic hydrocarbon. A undergoes hydrogenation to give a cyclic compound (B). A reacts with chlorine in the presence of UV-light to give C which is used as insecticide. Identify A, B and C. Explain the reactions with equation.
Answer:
1. Simple aromatic hydrocarbon, C2H6 is benzene.
2. Benzene (A) reacts with H2 in the presence of Pt to give cyclohexane (B).
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
3. Benzene (A) reacts with Cl2 in presence of UV-light to give benzene hexachioride (C).
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 11.
What are ortho-para directors? Explain why -OH group is an ortho-para director and activator.
Answer;
The group which increases the electron density at ortho and para positions are called as ortho-para directors,
Example:
-OH. -NH2 -NHR. -OCH3, -CH3, -C2H5 etc.

Let us consider the directive influence of phenolic group. Phenol is the resonance hybrid of the following structures:
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
In these resonance structures, the negative charge residue in the present at orilio and para positions of the ring structure. h is quite evident that the lone pair of electrons on the atom which is attached lo the ring is involved in resonance and it makes the ring more electron rich than benzene.

The electron density at oriho and para position increases as compared to the meta position. Therefore phenolic group activates the benzene ring for electrophilic attack at oriho and para positions. Hence 01-1 group is an orilio para director and activator.

Question 12.
An organic compound (A) of a molecular formula C6H6 is a simple aromatic hydrocarbon. A reacts with O2 in the presence of VO5 at 773 K to give B. A is further treated with sodium and liquid ammonia to give C which is a dienc compound. Identify A, B, and C and explain the reactions.
Answer:
1. A is benzene (C6H6), a simple aromatic hydrocarbon

2. Benzene (A) reacts with O2 is the presence of V2O5 at 773K to give maleic anhydride (B).
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
3. Benzene (A) is treated with sodium and liquid ammonia to give 1.4 – cyclohexadiene (C)
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 13.
What are inea drectors’? Explain with suitable example.
Answer:
The group which increases the electron density at mcta position are called as rneta directors.
For example:
-NO2. -CN, -CHO, -COOH. -SO3H etc.
Let us consider the directive influence of aldehydic (-CHO) group. Benzaldehyde is a resonence hybrid of the following structures.
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
In these resonance structures, the positive charge residue is present on the ring structure. It is quite evident that resonance deLocalises the positive charge on the atoms of the ring, making the ring less electron rich than henzene. Here overall electron density of benzene ring decreases due to -I effect of’ -CHO group, thereby deactivating the henzene for electrophilic atlack. However resonating structures shows that electron density is more at meto position as compared to ortho and para positions. Flence -CHO group is a ineta -director and a deactivator.

Question 14.
An organic compound (A) of molecular formula C2H4 which is a simple alkene reacts with Baeyer’s reagent to give B of molecular formula C2H6O2 A again reacts with ozone followed by hydrolysis in the presence of zinc to form C of molecular formula CH2O. Identify A, B and C. Explain with reactions.
Answer:
(i). A is CH2 = CH2 (Ethylene)

(ii). A (ethylene) reacts with Baeyer’s reagent (cold alkaline KMnO4) to give ethylene glycol (ethane 1 ,2 diol).
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Question 15.
An organic compound 1,1-dichioropropane reacts with aicholic KOH to give A of molecular formula C3H4. A reacts with mercuric sulphate and dil. H2SO4 at 333 K to give B. A on passing through red hot iron tube at 873 K will give C which is a cyclic compound. Identify A, B and C. Explain the reactions.
Answer:
1. 1,1-dichioropropane reacts with aicholic KOI-I to give propyne (A).
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

2. Propyne (A) reacts with mercuric sulphate and dil. H2SO4 at 333 K to give acetone (B).
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

3. Propyne (A) on passing through red hot iron tube at 873 K will give mesitylene (C).
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons
Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

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Samacheer Kalvi 11th Bio Zoology Solutions Chapter 12 Trends in Economic Zoology

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Tamilnadu Samacheer Kalvi 11th Bio Zoology Solutions Chapter 12 Trends in Economic Zoology

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Samacheer Kalvi 11th Bio Zoology Trends in Economic Zoology Text Book Back Questions and Answers

Textbook Evaluation Solved
Choose The Correct Answer
Question 1.
Which one of the following is not related to vermiculture?
(i) Maintains soil fertility
(ii) Breakdown of inorganic matter
(iii) Gives porosity, aeration and moisture holding capacity
(iv) Degradation of non biodegradable solid waste
(a) (i) and (ii) is correct
(b) (iii) and (iv) is correct
(c) (ii) and (iv) is not correct
(d) (i) and (iii) is not correct
Answer:
(c) (ii) and (iv) is not correct

Question 2.
Which one of the following is not an endemic species of earthworm?
(a) Perionyx
(b) Lampito
(c) Eudrillus
(d) Octochaetona
Answer:
(c) Eudrillus

Question 3.
Match the following
(1) Bombyx mori – a) Champa – (1) Muga
(2) Antheraea assamensis – b) Mulberry – (II) Eri
(3) Antheraea mylitta – c) Arjun – (III)Tassar
(4) Attacus ricini – d) Castor – (IV) Mulberry Select the correct one.
(a) 1 – b – IV
(b) 2 – a – I
(c) 3 – c – III
(d) 4 – d – II
Answer:
(c) 3 – c – III

Question 4.
Silk is obtained from
(a) Laccifer lacca
(b) Nosema bombycis
(c) Attacus ricini
(d) Attacus mylitta
Answer:
(c) Attacus ricini

Question 5.
Assertion: Nuptial flight is a unique flight taken by the queen bee followed by several drones.
Reason: The queen bee produces a chemical substance called pheromone. The drones in that area are attracted to the pheromone and then mating takes place.
(a) Assertion and reason is correct but not related
(b) Assertion and reason is incorrect but related
(c) Assertion and reason is correct but related
(d) Assertion and reason is incorrect but not related
Answer:
(c) Assertion and reason is correct but related

Question 6.
Rearing of honey bee is called …………………..
(a) Sericulture
(b) Lac culture
(c) Vermiculture
(d) Apiculture
Answer:
(d) Apiculture

Question 7.
Which of the statement regarding Lac insect is TRUE?
(a) A microscopic, resinous crawling scale insect
(b) Inserts its proboscis into plant tissue suck juices and grows
(c) Secretes lac from the hind end of body
(d) The male lac insect is responsible for large scale production of lac
Answer:
(c) Secretes lac from the hind end of body

Question 8.
Aquaponics is a technique which is ……………….
(a) A combination of aquaculture and fish culture
(b) A combination of aquaculture and hydroponics
(c) A combination of vermiculture and hydroponics
(d) A combination of aquaculture and prawn culture
Answer:
(b) A combination of aquaculture and hydroponics

Question 9.
Prawn belongs to the class …………………
(a) Crustacea
(b) Annelida
(c) Coelenterata
(d) Echinodermata
Answer:
(a) Crustacea

Question 10.
Pearl oyster belongs to the Class …………………
(a) Gastropoda
(b) Cephalopoda
(c) Scaphapoda
(d) Pelecypoda
Answer:
(d) Pelecypoda

Question 11.
Inland fisheries are ………………..
(a) Deep sea fishing
(b) Capturing fishes from sea coast
(c) Raising and capturing fishes in fresh water
(d) Oil extraction from fish
Answer:
(c) Raising and capturing fishes in fresh water

Question 12.
Induced breeding technique is used in …………………
(a) Marine fishery
(b) Capture fishery
(c) Culture fishery
(d) Inland fishery
Answer:
(b) Capture fishery

Question 13.
Isinglass is used in …………………..
(a) Preparation
(b) Clearing of wines
(c) Distillation of wines
(d) Preservation of wines
Answer:
(b) Clearing of wines

Question 14.
Animal husbandry is the science of rearing, feeding and caring, breeding and disease control of animals. It ensures supply of proper nutrition to our growing population through activities like increased production and improvement of animal products like milk, eggs, meat, honey, etc.
(a) Poultry production depends upon the photoperiod. Discuss.
(b) Polyculture of fishes is of great importance.
Answer:

(a) Light is an important aspect in the poultry production. Light stimulates the secretion of FSH and LH. The wavelength between 400 and 700 nm is required. The decrease in the photoperiod will affect the egg production.

(b) A few selected fishes belonging to different species are stocked together in proper proportion in a pond. This mixed forming is termed as composite fish farming or polyculture. It is of great importance because

  1. All available riches are fully utilized
  2. Compatible species do not harm each other
  3. There is no competition among different species

 

Question 15.
Assertion: The best quality of pearl is known as lingha pearl and obtained from marine oysters.
Reason: Nacre is secreted continuously by the epithelial layer of the mantle and deposited around the foreign particle.
(a) Assertion is true, Reason is false
(b) Assertion and Reason are false
(c) Assertion is false But Reason is true
(d) Assertion and Reason are true
Answer:
(c) Assertion is false But Reason is true

Question 16.
Choose the correctly matched pair.
(a) Egg layers – Brahma
(b) Broiler types – Leghorn
(c) Dual purpose – White Plymouth rock
(d) Ornamental breeds – Silkie
Answer:
(d) Ornamental breeds – Silkie

Question 17.
Write the advantages of vermicomposting?
Answer:

  1. Vermicomposting provides excellent organic manure for sustainable agro-practices.
  2. Marketing of vermicompost can provide a supplementary income.
  3. Vermicompost is rich in essential plant nutrients.
  4. It improves soil structure, texture, aeration, and water holding capacity and prevents soil erosion.
  5. It is rich in nutrients and an eco-friendly amendment to soil for farming and terrace gardening,
  6. It enhances seed germination and ensures good plant growth.

Question 18.
Name the three castes in a honey bee colony?
Answer:
The Queen, Drones and Workers.

Question 19.
Name the following?

  1. The largest bee in the colony:
  2. The kind of flight which the new virgin queen takes along with the drones out of the hive:

Answer:

  1. The queen.
  2. Nuptial flight.

 

Question 20.
What are the main duties of a worker bee?
Answer:
Each worker has to perform different types of work in her life time. During the first half of her life, she becomes a nurse bee attending to indoor duties such as secretion of royal jelly, prepares bee- bread to feed the larvae, feeds the queen, takes care of the queen and drones, secretes bees wax, builds combs, cleans and fans the bee hive. Then she becomes a soldier and guards the bee hive. In the second half her life lasting for three weeks, she searches and gathers the pollen, nectar,, propolis and water.

Samacheer Kalvi 11th Bio Zoology Solutions Chapter 12 Trends in Economic Zoology img 1

Question 21.
What happens to the drones after mating flight?
Answer:
They die after copulation.

Question 22.
Give the economic importance of Silkworm?
Answer:

  1. Rearing of silkworm on a commercial scale is called sericulture.
  2. It is an agro-based industry comprising of
    1. – Cultivation of food plants for the silkworms.
    2. – Rearing of silkworms.
    3. – Reeling and spinning of silk.
  3. Silk fibres are utilized in preparing silk clothes.
  4. Silk is used in industries and for military purposes.
  5. Silk is used in the manufacture of fishing fibres, parachutes, cartridge bags, insulation coils for telephone, wireless receivers, tyres of racing cars, filter fibres, in medical dressings and as suture materials.

Question 23.
What are the Nutritive values of fishes?
Answer:
Economic importance of fish:-
Fishes form a rich source of protein food and provide a good staple food to tide over the nutritional needs of man. Fish species such as sardines, mackerel, tuna, herrings have high amino acids concentration particularly histidine which is responsible for the meaty flavor of the flesh. It is rich in fat such as omega 3 fatty acids. Minerals such as calcium, magnesium, phosphorus, potassium, iron, manganese, iodine and copper.

Some of the fish by-products are: Fish oil is the most important fish by-product. It is derived from fish liver and from the fish body.

Fish liver oil is derived from the liver which is rich in vitamin A and D, whereas fish body oil has high content of iodine, not suitable for human consumption, but is used in the manufacture of laundry soaps, paints and cosmetics. Fish meal is prepared from fish waste after extracting oil from the fish.

The dried wastes are used to prepare food for pig, poultry and cattle. The wastes obtained during the preparation of fish meal are widely used as manure.

Isinglass is a high-grade collagen produced from dried air bladder or swim bladder of certain fishes viz. catfish and carps. The processed bladder which is dissolved in hot water forms a gelatin having adhesive property. It is primarily used for clarification of wine, beer and vinegar.

Question 24.
Give the economic importance of prawn fishery?
Answer:
The flesh of prawn is palatable and rich in glycogen, protein with low fat content.

Question 25.
Give the economic importance of lac insect?
Answer:
Economic importance of Lac:-

  1. Lac is largely used as a sealing wax and adhesive for optical instruments. It is used in electric industry, as it is a good insulator.
  2. It is used in preparations of shoe and leather polishes and as a protective coating of wood.
  3. It is used in laminating paper board, photographs, engraved materials and plastic moulded articles.
  4. Used as a filling material for gold ornaments.

Question 26.
List any three common uses of shellac?
Answer:

  1. Shellac with denatured alcohol is used to remove dust on the walls.
  2. Coating of metals with shellac prevents rusting.
  3. Shellac coating on citrus fruits increases their shelf life.

Question 27.
Name any two trees on which lac insect grows?
Answer:
Acacia catechu, Acacia nilotica.

Question 28.
What is seed lac?
Answer:
Lac cut from the host plant is called ‘stick lac’. The lac present on the twig is scraped and collected. After grinding, the unnecessary materials like dusts and fine particles are removed. The resultant lac is called ‘seed lac’.

Question 29.
Define cross breeding?
Answer:
Breeding between a superior male of one breed with a superior female of another breed is known as cross breeding.

Question 30.
What are the advantages of artificial insemination?
Answer:
Advantages of artificial insemination:

  1. It increases the rate of conception
  2. It avoids genital diseases
  3. Semen can be collected from injured bulls which have desirable traits
  4. Superior animals located apart can be bred successfully

Question 31.
Discuss the various techniques adopted in cattle breeding?
Answer:
There are two methods of animal breeding, namely inbreeding and outbreeding.
1. Inbreeding:
Breeding between animals of the same breed for 4-6 generations is called inbreeding.

2. Outbreeding:
The breeding between unrelated animals is called outbreeding. It is done in three ways;

  • Out crossing: It is the breeding between unrelated animals of the same breed but having no common ancestry. The offspring of such a cross is called outcross.
  • Cross breeding: Breeding between a superior male of one breed with a superior female of another breed. The cross bred progeny has superior traits (hybrid vigour or heterosis).
  • Interspecific hybridization: In this method of breeding mating is between male and female of two different species.

 

Question 32.
Mention the advantages of MOET?
Answer:
Multiple Ovulation Embryo Transfer Technology (MOET) is a method of propagation of animals with desirable traits. This technology is used to produce high milk yielding females and high quality meat yielding bulls in a short time.

Question 33.
Write the peculiar characters of duck?
Answer:
Peculiarity of ducks:
The ducks body is fully covered with oily feathers. They have a layer of fat under their skin which prevents it from getting wet. They lay eggs at night or in the morning. The ducks feed on rice bran, kitchen wastes, waste fish and snails.

In – Questions Solved

Question 1.
My vennicompost manufacturing unit is plagued by a number of red ants. Are there any bio-friendly measures to tackle the menace as I do not want to use any chemicals?
Answer:
There are few bio-friendly measures to tackle the menace. For eg. Sprinkling of cinnamon powder or cloves can repel ants.

Question 2.
India has the distinction of producing all the four types of silk i.e.
(a) Mulberry silk (9-1.7%)
(b) Tasar silk (1.4%)
(c) Eri silk (6.4%) and
(d) Muga silk (0.5%) which are produced by different species of silkworms. Name the species that produces large amount and least amount of silk in India?
Answer:
Bombyx mori produces large amount of silk in India and Antheraea assamensis produces least amount of silk in India.

Question 3.
(a) Growing fish or other aquatic animals and plants together in an integrated system. The fish wastes provides nutrients for the plants and the plants filter the water. Additionally bacteria break down by-products such as ammonia?
(b) Growing plants in a nutrient solution instead of soil. Fish kept in the water provide the required nutrients. Write the appropriate scientific terms for above (a) and (b) and differentiate between them?
Answer:
(a) Aquaponics
(b) Hydroponics

Aquaponics

Hydroponics

1. It is a technique of combination of growing fish (aquaculture) and growing plants in non-soil media and nutrient-laden water (hydroponics)1. It is the method of growing plants in non-soil media and nutrient-laden water.
2. It prevents toxic water run off.2. It provides fresh nutrients to the plants.
3. It also maintains ecosystem balance by recycling the waste and excretory products by the fish.3. It provides scope for alternate method of plant cultivation.

 

Question 4.
Why are fish so efficient at converting feed to flesh?
Answer:
Fish is more efficient in converting a greater proportion of feed they receive into flesh because as fish live in water they expend less energy to maintain internal body temperature and also by excreting a more simple form of urine (ammonia).

Samacheer Kalvi 11th Bio Zoology Trends in Economic Zoology Additional Questions & Answers

I. Choose The Correct Answer

Question 1.
During the process of vermiculture which of the following does not happen?
(a) Decomposition of organic food waste
(b) Supplying nutrients to the soil
(c) Use of earthworms in the process
(d) Synthesis of organic substances
Answer:
(d) Synthesis of organic substances

Question 2.
The breakdown of organic matter by the activity of the earthworms and its elimination from its body is called …………….
(a) Vermitech
(b) Vermicast
(c) Vermicompost
(d) Vermiculture
Answer:
(b) Vermicast

Question 3.
The technology of composting and bioremediation of soils and other activities by application of earthworm is called? …………………
(a) Vermitech
(b) Vermicast
(c) Vermicompost
(d) Vermiculture
Answer:
(a) Vermitech

Question 4.
Which is not related to humus formers?
(a) They are darker in colour
(b) They feed on organic matter
(c) They make the soil porous
(d) They are surface dwellers
Answer:
(c) They make the soil porous

Question 5.
Which of the following is the exotic species used for vermicomposting?
(a) Eudrilus eugeniae
(b) Lampito mauritii
(c) Periyonyx excavatus
(d) Octochaetona serrata
Answer:
(a) Eudrilus eugeniae

Question 6.
Antheraea mylitta feeds on ……………
(a) Mulberry
(b) Arjun
(c) Castor
(d) Champa
Answer:
(b) Arjun

Question 7.
Attacus ricini produces ……………. silk?
(a) Muga
(b) Tassar
(c) Mulberry
(d) Eri
Answer:
(d) Eri

Question 8.
The process of killing the cocoons is called ……………..
(a) Reeling
(b) Stifling
(c) Rearing
(d) Culturing
Answer:
(b) Stifling

Question 9.
Febrine is a dangerous disease to silkworms caused by ………………
(a) Streptococceus
(b) Nosema bombycis
(c) BmNPV
(d) Beauveria bassiana
Answer:
(b) Nosema bombycis

Question 10.
Muscardine is a- disease of silkworms caused by a …………………..
(a) Bacterium
(b) Virus
(c) Protozoan
(d) Fungus
Answer:
(d) Fungus

Question 11.
Among the honey bees, workers are ………………..
(a) Fertile females
(b) Fertile males
(c) Sterile females
(d) Sterile males
Answer:
(c) Sterile females

Question 12.
Honey is used as ………………..
(1) An antiseptic
(2) A laxative
(3) A sedative
(4) A substitute for sugar
(a) 1, 2, 3 only
(b) 2, 3, 4 only
(c) 2,4 only
(d) 1, 2, 3, & 4
Answer:
(d) 1, 2, 3, & 4

Question 13.
Lac is secreted by lac insect for ………………
(a) Growth
(b) Development
(c) Protection
(d) Moulting
Answer:
(c) Protection

Question 14.
Growing plants in non-soil media and nutrient-laden water is called ……………….
(a) Aquaponics
(b) Hydroponics
(c) Aquaculture
(d) Apiculture
Answer:
(b) Hydroponics

Question 15.
Raft based method of aquaponic gardening is known as ………………..
(a) Deep water culture
(b) Aqua vertica
(c) Media based method
(d) Nutrient film technique
Answer:
(a) Deep water culture

Question 16.
Culturing of animals in the water having salinity range 0.5-30 ppt are called as ………………
(a) Freshwater aquaculture
(b) Marine water aquaculture
(c) Brackish water aquaculture
(d) Metahaline culture
Answer:
(c) Brackish water aquaculture

Question 17.
Culturing of animals in the water salinity ranges from 30-35% is called ……………
(a) Brackish water aquaculture
(b) Mariculture
(c) Metahalive culture
(d) Fresh water aquaculture
Answer:
(b) Mariculture

Question 18.
Labeo and catla belong to ………………. fishes?
(a) Indigenous fresh water
(b) Salt water
(c) Exotic
(d) Brackish water
Answer:
(a) Indigenous fresh water

Question 19.
In composite fish farming …………………. are stocked together in proper proportion in a pond?
(a) Different species of fishes
(b) Fishes of the same species
(c) All animals
(d) All plants and animals
Answer:
(a) Different species of fishes

Question 20.
Fish liver oil is rich in vitamins ……………
(a) A & B
(b) A & C
(c) A & D
(d) A & E
Answer:
(c) A & D

Question 21.
The best quality of pearl is obtained from …………… oysters?
(a) Land
(b) Brackish water
(c) Fresh water
(d) Marine
Answer:
(d) Marine

Question 22.
Breeding between animals of the same breed for 4-6 generation is called ……………
(a) Outbreeding
(b) Inbreeding
(c) Cross breeding
(d) Outcrossing
Answer:
(A) Inbreeding

Question 23.
The breeding between unrelated animals is called ……………
(a) Inbreeding
(b) Cross breeding
(c) Outbreeding
(d) Outcrossing
Answer:
(c) Outbreeding

Question 24.
Breeding between a superior male of one breed with a superior female of another breed is called ……………
(a) Inbreeding
(b) Outbreeding
(c) Outcrossing
(d) Cross breeding
Answer:
(d) Cross breeding

Question 25.
Jersy is a breed?
(a) Milch
(b) Drought purpose
(c) Dual purpose
(d) Common
Answer:
(a) Milch

Question 26.
Leghorns are preferred in commercial farms because ……………
(a) They give good quality flesh
(b) They mature late
(c) They mature early and begin to lay eggs at the age of 5 months
(d) They look beautiful
Answer:
(c) They mature early and begin to lay eggs at the age of 5 months

Question 27.
…………… is a dual purpose chicken breed?
(a) Aseel
(b) Silkie
(c) Brahma
(d) White plymouth rock
Answer:
(c) Brahma

Question 28.
The incubation period of chick embryo development is …………… days?
(a) 10-12
(b) 21-22
(c) 25-30
(d) 15-17
Answer:
(b) 21-22

Question 29.
Ranikhet, Coccidiosis and Fowl pox are ……………
(a) Chicken breeds
(b) Types of poultry farming
(c) Poultry diseases
(d) Types of breeding methods
Answer:
(c) Poultry diseases

Question 30.
Which of the following is the native breed of duck?
(a) Pekin
(b) Campbell
(c) Muscori
(d) Syhlet meta
Answer:
(d) Syhlet meta

II. Fill in the Blanks

Question 1.
…………… are called as ‘farmer’s friends’.
Answer:
(Earthworms)

Question 2.
The breakdown of organic matter by the activity of the earthworms and its elimination from its body is called ……………
Answer:
(Vermicast)

Question 3.
Humus feeders are …………… worms.
Answer:
(Burrowing)

Question 4.
…………… are the earthworms used for vermicomposting.
Answer:
(Humus feeders)

Question 5.
The exotic species of earthworms are ……………
Answer:
(Eisenia fetida/Eudrilus eugeniae)

Question 6.
…………… is a liquid collected after the passage of water through a column of vermibed.
Answer:
(Vermiwash)

Question 7.
Vermiwash is obtained from formed by earthworms.
Answer:
(Burrows/drilospheres)

Question 8.
Production of silk from the silkworm by rearing practices on a commercial scale is called ……………
Answer:
(Sericulture)

Question 9.
The eggs of silkworms after ten days of incubation hatch into larva called as ……………
Answer:
(Caterpillar)

Question 10.
The cultivation of mulberry is called as ……………
Answer:
(Moriculture)

Question 11.
The process of killing the cocoons is called ……………
Answer:
(Stifling)

Question 12.
The process of removing the threads of silk from the killed cocoon is called ……………
Answer:
(Reeling)

Question 13.
Nosema bombycis causes …………… to silkworms,
Answer:
(Pobrine)

Question 14.
…………… is the viral disease of Bombyx mori caused by BmNPV.
Answer:
(Grasserie)

Question 15.
White muscardine is caused by a fungus
Answer:
(Beauveria bassiana)

Question 16.
Apiculture is the care and management of on a commercial scale.
Answer:
(honeybees)

Question 17.
…………… is the Indian honey bee,
Answer:
(Apis indica)

Question 18.
…………… is the rock bee.
Answer:
(Apis dorsata)

Question 19.
…………… is the little bee.
Answer:
(Apisflorea)

Question 20.
Apis mellifera is the …………… bee.
Answer:
(European)

Question 21.
……………. is the functional female bee present in each bee hive.
Answer:
(Queen)

Question 22.
The Queen bee produces …………… to attract drones.
Answer:
(Pheromone)

Question 23.
Among the honey bees, are sterile females.
Answer:
(Workers)

Question 24.
………………. is the functional male member of the honey bee colony.
Answer:
(Drone)

Question 25.
A unique flight by the queen bee followed by several drones is called ……………
Answer:
(Nuptial flight)

Question 26.
The process of leaving the colony by the queen with a large group of worker bees to form a new colony is called ……………
Answer:
(Swarming)

Question 27.
Honey is used as an antiseptic, laxative and as a ……………
Answer:
(Sedative)

Question 28.
The resinous chemical substance present in the bee wax is called ……………
Answer:
(Propolis)

Question 29.
Lac is produced by the lac insect
Answer:
(Laccifer lacca/Tachardia lacca)

Question 30.
Karanagalli is the host plant for ……………
Answer:
(Lac insect)

Question 31.
The mass emergence of larvae from the eggs of lac in search of a host plant is called ……………
Answer:
(Swarming)

Question 32.
A condition in which a secondary parasite develops within a previously existing parasite is called ……………
Answer:
(Hyper-parasitism)

Question 33.
The collection of lac from the host plant is known as ……………
Answer:
(Harvesting)

Question 34.
Lac cut from the host plant is called ……………
Answer:
(Stick lac)

Question 35.
The seed lac is sundried and then melted to produce ……………
Answer:
(Shellac)

Question 36.
Growing fish is known as ……………
Answer:
(Aquaculture)

Question 37.
Chanos chanos are cultured in ……………
Answer:
(Brackish water)

Question 38.
Culturing of animals in the salinity ranges from 36-40% is called …………… culture.
Answer:
(Metahaline)

Question 39.
…………… is commonly known as the brine shrimp.
Answer:
(Artemia)

Question 40.
Catla, Labeo are fresh water fishes.
Answer:
(Native/indigenous)

Question 41.
Fish oil is derived from the …………… of fish.
Answer:
(Liver)

Question 42.
……………… is prepared from fish waste after extracting oil from the fish.
Answer:
(Fish meal)

Question 43.
………………. is a high grade collagen produced from dried air bladder of catfish.
Answer:
(Isinglass)

Question 44.
Prawn is rich in …………… protein with low fat content.
Answer:
(Glycogen)

Question 45.
Most of the marine prawns caught along the Indian coast belongs to the family ……………
Answer:
(Penaeidae)

Question 46.
Metapenaeus dobsoni is a species of ……………
Answer:
(Prawn)

Question 47.
Fresh water bivalve …………… is used in artificial pearl culture.
Answer:
(Lamellidens)

Question 48.
The piece of tissue which is inserted inside the mantle of the oyster is called as …………… tissue.
Answer:
(Graft)

Question 49.
90% of pearl is ……………
Answer:
(Calcium carbonate)

Question 50.
The best quality of pearl is obtained from …………… oysters.
Answer:
(Marine)

Question 51.
The practice of breeding and raising livestock is called as ……………
Answer:
(Animal husbandry)

Question 52.
……………… increases homozygosity and exposes harmful recessive genes.
Answer:
(Inbreeding)

Question 53.
…………… helps to produce new and favourable traits, new breeds and hybrids-with superior qualities
Answer:
(Outbreeding)

Question 54.
……………….. is a technique in which the semen collected from the male is injected to the reproductive tract of the selected female.
Answer:
(Artificial insemination)

Question 55.
…………… is the production and marketing of milk and its products.
Answer:
(Dairying)

Question 56.
Sindhi, Gir, Jersy are …………… breeds.
Answer:
(Dairy/Milch)

Question 57.
…………… is the rearing and propagation of avian species.
Answer:
(Poultry)

Question 58.
…………… is the type of chicken breeds well known for fast growth and soft quality meat.
Answer:
(Broiler)

Question 59.
Caring and management of young chicks for 4-6 weeks immediately after hatching is called ……………
Answer:
(Brooding)

Question 60.
Ranikhet is a …………… disease.
Answer:
(Poultry)

Question 61.
Domesticated ducks have been derived from the wild duck named ……………
Answer:
(Anas Aoscav/Mallard)

Question 62.
Droppings of poultry can be used as in …………… fields.
Answer:
(Manure)

Question 63.
The droppings of poultry are rich in nitrogen, potash and ……………
Answer:
(Phosphates)

Question 64.
The eggs and poultry meat are the richest sources of …………… and vitamins.
Answer:
(Proteins)

III. Answer The Following Questions

Question 1.
What is Economic Zoology?
Answer:
Economic zoology is a branch of Science that deals with economically useful animals. It involves the study of application of animals for human welfare.

Question 2.
How are animals classified on the basis of economic importance?
Answer:

  1. Animals for food and food products
  2. Economically beneficial animals
  3. Animals of aesthetic importance
  4. Animals for scientific research

Question 3.
What is Vermiculture?
Answer:
Vermiculture is the process of using earthworms to decompose organic food waste, into a nutrient-rich material capable of supplying necessary nutrients which helps to sustain plant growth.

Question 4.
What is Vermitech?
Answer:
Applications of earthworm in technology of composting and bioremediation of soils and other activities is called Vermitech.

Question 5.
Why are earthworms called as ‘friends of farmers’?
Answer:
Earthworms play a vital role in maintaining soil fertility. Hence, they are called as ‘friends of farmers’.

Question 6.
What is Vermicast?
Answer:
The breakdown of organic matter by the activity of the earthworms and its elimination from its body is called vermicast.

Question 7.
What are the different groups of earthworms?
Answer:
There are two major groups of earthworms. The humus formers dwell on the surface and feed on organic matter. They are generally darker in colour. They are used for vermicomposting. The humus feeders are burrowing worms that are useful in making the soil porous and mixing and distributing humus throughout the soil.

Question 8.
What are endemic species and exotic species of earthworms?
Answer:
The native species of earthworms cultured in India for vermicomposting such as Periyonyx excavatus, Lampito mauritii, Octochaetona serrata are endemic species. Some earthworm species introduced from other countries such as Eisenia fetida, Eudrilus eugeniae are exotic species.

Question 9.
What is Vermicompost?
Answer:
Vermicompost is. the compost produced by the action of earthworms in association with all other organisms in the compost unit.

Question 10.
Write a note on Vermiwash?
Answer:
Vermiwash is a liquid collected after the passage of water through a column of vermibed. It is useful as a foliar spray to enhance plant growth and yield. It is obtained from the burrows or drilospheres formed by earthworms. Nutrients, plant growth promoter substances and some useful microorganisms are present in vermiwash.

Question 11.
What are the pests of earthworms?
Answer:
Ants, springtails, centipedes, slugs, mites, certain beetle larvae, birds, rats, snakes, mice, toads and other insects or animals which feed on worms.

Question 12.
What, are the internal parasites of earthworms?
Answer:
Protozoans, some nematodes and the larvae of certain flies.

Question 13.
What are the advantages of vermicompost?
Answer:

  1. Vermicompost is rich in essential plant nutrients.
  2. It improves soil structure texture, aeration, and water holding capacity and prevents soil erosion.
  3. Vermicompost is rich in nutrients and an eco-friendly amendment to soil for farming and terrace gardening.
  4. It enhances seed germination and ensures good plant growth.

Question 14.
What is sericulture?
Answer:
Sericulture is an agro-based industry which denotes commercial production of silk through silkworm rearing.

Question 15.
What are the components of sericulture?
Answer:

  1. Cultivation of food plants for the silkworms,
  2. Rearing of silkworms, and
  3. Reeling and spinning of silk.

The first two are agricultural and the last one is an industrial component. Only felt species of silkworms are used in the sericulture industry.

Question 16.
Tabulate the different types of silkworm?
Answer:
Samacheer Kalvi 11th Bio Zoology Solutions Chapter 12 Trends in Economic Zoology img 2

Question 17.
What are the various races of Bombyx mori?
Answer:
On the basis of the moults they undergo in the larval stage, B. mori is divided into three races. They are:

  1. Tri moulters – 3 moults
  2. Tetra moulters – 4 moults
  3. Penta moulters – 5 moults
    • On the basis of voltinism (the number of broods raised per year), three kinds of races are recognised. They are:
    • Univoltines – one brood only
    • Bivoltines – two broods only
    • Multivoltines – more than two broods

 

Question 18.
What is Moriculture?
Answer:
The cultivation of mulberry is called as Moriculture.

Question 19.
What are the improved mulberry varieties?
Answer:
Victory 1, S3 6, G2 and G4

Question 20.
What is the favourable season for cultivating mulberry plants?
Answer:
June, July, November and December

Question 21.
What is stifling?
Answer:
The process of killing the cocoons is called stifling,

Question 22.
What is reeling?
Answer:
The process of removing the threads from the killed cocoon is called reeling.

Question 23.
What is cooking?
Answer:
The process of soaking cocoons in hot water (95° – 97°) for 10 – 15 minutes to soften the gum that binds the silk threads together is called as cooking.

Question 24.
What are the uses of silk?
Answer:
Uses of Silk:-
1. Silk fibers are utilized in preparing silk clothes. Silk fibers are now combined with other natural or synthetic fibers to manufacture clothes like Teri-Silk, Cot-Silk etc. Silk is dyed and printed to prepare ornamented fabrics. They are generally made from Eri-silk or spun silk.

2. Silk is used in industries and for military purposes.

3. It is used in the manufacture of fishing fibers, parachutes, cartridge bags, insulation coils for telephone, wireless receivers, tyres of racing cars, filter fibres, in medical dressings and as suture materials.

Question 25.
Write a note on the diseases and pests of silkworms? Diseases and Pests of Silkworm?
Answer:
The profitable silk industry is threatened by various diseases caused by the virus, fungal, bacterial and protozoan infections but also by insect predators, birds and other higher animals. Ants, crows, kites, rats, feed upon silkworms thereby causing a great loss to silk industry.

Pebrine, is a dangerous disease to silkworms and the causative organism is Nosema bombycis, a protozoan. This silkworm disease is transmitted through the egg of the mother silkworm and also through ingestion of contaminated food.

Flacherie generally occurs in the mature larvae and is caused mainly by bacteria like Streptococcus and Staphylococcus. Grasserie is a most dominant and serious viral disease.

It is caused by Bombyx mori nuclear polyhedrosis virus (BmNPV) a Baculovinis, which belongs to sub group ‘A’ of the Baculoviridae. Among the fungal diseases, white muscardine is common. This disease is caused by fungus Beauveria bassiana.

Question 26.
What is Apiculture?
Answer:
Care and management of honey bees on a commercial scale for the production of honey is called Apiculture or Bee keeping.

Question 27.
What are the types of bees?
Answer:

  1. Apis dorsata – Rock bee
  2. Apisflorea – Little bee
  3. Apisindica – Indian bee
  4. Apis mellifera – European bee
  5. Apis adamsoni – African bee

Question 28.
Comment on the social organization of honey bees?
Answer:
In honey bees, a highly organized division of labour is found. A well developed honey bee colony consists of the Queen, Drones and Workers. All the three types depend on each other for their existence. There is normally one queen, 10,000 to 30,000 workers and few hundred drones (male bees) in a colony.

Question 29.
Write a note on Queen, Worker and Drones?
Answer:
Queen:

  1. It is a functional female present in each hive.
  2. It feeds on royal jelly.
  3. Its function is to lay eggs throughout its life span.
  4. It lays about 15 lakh eggs.

Workers:

  1. Workers are sterile females and the smallest of all the bees.
  2. Each worker bee secretes royal jelly, prepares bee-bread to feed the larvae, feeds the queen, takes care of the queen and drones, secretes bee wax, builds combs, cleans and fans the bee hive.
  3. It guards the bee hive, gathers pollen, nectar, propolis and water.

Drones:

  1. Drones are the fuctional male members of the colony.
  2. The sole duty of the drone is to fertilize the virgin queen and hence called ‘king of the colony’.

Question 30.
What is nuptial flight?
Answer:
During the breeding season in winter, a unique flight taken by the queen bee followed by , several drones is called nuptial flight.

Question 31.
Explain the structure of a bee hive?
Answer:
Structure of a Bee Hive. The house of honey bee is termed as bee hive or comb. The hive consists of hexagonal cells made up of wax secreted by the abdomen of worker bees arranged in opposite rows on a common base. These hives are found hanging vertically from the rocks, building or branches of trees . The young stages of honey bees accommodate the lower and central cells of the hive called the brood cells.

In Apis dorsata, the brood cells are similar in size and shape but in other species, brood cells are of three types viz., queen cell for queens, worker cell for workers and drone cells for drones. The cells are intended for storage of honey and pollen in the upper portion of the comb whereas the lower portions are for brood rearing.

Question 32.
Explain the parts of the Langstroth bee hive?
Answer:
The Langstroth bee hive is made up of wood and consists of six parts.
(1) Stand is the basal part of the hive on which the hive is constructed. The stands are adjusted to make a slope for rain water to drain

(2) Bottom board is situated above the stand and forms the proper base for the hive. It has two gates, one gate functions as an entrance while the other acts as an exit.

(3) Brood chamber is the most important part of the hive. It is provided with 5 to 10 frames arranged one above the other through which the workers can easily pass. The frame is composed of wax sheet which is held in vertical position up by a couple of wires. Every sheet of wax is known as Comb Foundation. The comb foundation helps in obtaining a regular strong worker brood cell comb which can be used repeatedly.

(4) Super is also a chamber without cover and base. It is provided with many frames containing comb foundation to provide additional space for expansion of the hive. 5) Inner cover is a wooden piece used for covering the super with many holes for proper ventilation. 6) Top cover is meant for protecting the colonies from rains. It is covered with a sheet which is plain and sloping.

Question 33.
What is comb foundation?
Answer:
Comb foundation is a sheet of bee wax, on both sides of which the exact shape of different cells of the comb is made in advance.

Question 34.
What is the use of Bee gloves?
Answer:
Bee gloves are used by bee keepers for protecting their hands while inspecting the hives.

Question 35.
What is Bee veil?
Answer:
Bee veil is a device made of fine nettings to protect the bee-keeper from bee sting.

Question 36.
What is the use of Smoker?
Answer:
Smoker is used to scare the bees during hive maintenance and honey collection by releasing smoke.

Question 37.
What is hive tool?
Answer:
Hive Tool is a flat, narrow and long piece of iron which helps in scraping excess propolis or wax from hive parts. . .

Question 38.
What is the use of:
(a) Uncapping knife
(b) Bee brush
(c) Queen introducing cage
(d) Feeders
(e) Honey Extractor
(f) Hive Entrance Guard
Answer:
(a) Uncapping knife:
Uncapping knife is a long knife which helps in removing the cap from the combs as a first step in honey extraction.

(b) Bee Brush:
Bee brush is a large brush often employed to brush off bees from honey combs particularly at the time of extraction.

(c) Queen introducing cage:
Queen introducing cage is a pipe made of wire nets used for keeping the queen for about 24 hours for acquaintance with the hive and worker bees.

(d) Feeder:
Feeder is a basin with sugar syrup covered by grass to feed the bees during drought season. The grass prevents the bees from sinking into the syrup.

(e) Honey Extractor:
Honey Extractor is a stainless-steel device which spins the combs rapidly to extract honey.

(f) Hive Entrance Guard:
Hive Entrance Guard is a device similar to queen excluder in front of the hive entrance which prevents the escape of queen during warming season.

Question 39.
What are the products of bee keeping?
Answer:
Products of bee keeping and their economic importance. The chief products of bee keeping industry are honey and bee wax. Honey is the healthier substitute for sugar.

The major constituents of honey are: levulose, dextrose, maltose, other sugars, enzymes, pigments, ash and water. It is an aromatic sweet material derived from nectar of plants. It is a natural food, the smell and taste depends upon the pollen taken by the honey bee. It is used as an antiseptic, laxative and as a sedative.

It is generally used in Ayurvedic and Unani systems of medicine. It is also used in the preparation of cakes, breads and biscuits. Bee wax is secreted by the abdomen of the worker bees at the age of two weeks. The wax is masticated and mixed with the secretions of the cephalic glands to convert it into a plastic resinous substance.

The resinous chemical substance present in the wax is called propolis which is derived from pollen grains. The pure wax is white in colour and the yellow colour is due to the presence of carotenoid pigments. It is used for making candles, water proofing materials, polishes for floors, furniture, appliances, leather and taps. It is also used for the production of comb foundation sheets in bee keeping and used in pharmaceutical industries.

Question 40.
What is Lac culture?
Answer:
The culture of lac insect using techniques for the procurement of lac on large scale is known as Lac culture.

Question 41.
Name the insect that produces Lac?
Answer:
Tachardia lacca.

Question 42.
What is ‘swarming’?
Answer:
The mass emergence of larvae from the egg in search of a host plant is called ‘swarming’.

Question 43.
What is hyper-parasitism?
Answer:
Hyper-parasitism – A condition in which a secondary parasite develops within a previously existing parasite.

Question 44.
What is harvesting? Write on the types of lac?
Answer:
The collection of lac from the host plant is known as harvesting. Harvesting may be done before swarming (immature) or after swarming (mature). Immature harvesting produces ‘Ari lac’ whereas mature harvesting produces the mature lac. Lac cut from the host plant is called ‘Stick lac’.

The lac present on the twig is scraped and collected. After grinding, the unnecessary materials like dusts and fine particles are removed. The resultant lac is called ‘seed lac’. The seed lac is sun dried and then melted to produce ‘shellac’.

Question 45.
What is Aquaponics?
Answer:
Aquaponics is a technique which is a combination of aquaculture (growing fish)’ and hydroponics (growing plants in non-soil media and nutrient-laden water).

Question 46.
What are the methods of aquaponic gardening?
Answer:
(i) Deep water culture is otherwise known as raft based method. In this method a raft floats in water. Plants are kept in the holes of raft and the roots float in water. This method is applicable for larger commercial scale system. By this, method fast growing plants are cultivated.

(ii) Media based method involves growing plants in inert planting media like clay pellets or shales. This method is applicable for home and hobby scale system. Larger number of fruiting plants, leafy green plants, herbs and other varieties of plants can be cultivated.

(iii) Nutrient Film technique involves the passage of nutrient rich water through a narrow trough or PVC pipe. Plants are kept in the holes of the pipe to allow the roots to be in free contact with in the water stream.

(iv) Aqua vertica is otherwise known as vertical aquaponics. Plants are stacked on the top of each other in tower systems. Water flows in through the top of the tower. This method is suitable for growing leafy greens, strawberries and other crops that do not need supporting solid substratum to grow.

Samacheer Kalvi 11th Bio Zoology Solutions Chapter 12 Trends in Economic Zoology img 3

Question 47.
What are the advantages of Aquaponic gardening?
Answer:
Advantages of Aquaponic gardening.
Water conservation:
No need of water discharge and recharge as the water is maintained by recycling process.

Soil:
Bottom soil may be loaded with freshwater. Microbes in water can convert the waste materials into usable forms like ammonia into nitrates which are used by the plants. Thus the soil fertility is maintained.

Pesticides:
In this system use of pesticides is avoided and hence it is eco-friendly.

Weeds:
Since the plants are cultured in confined conditions, growth of weeds is completely absent. The utilization of nutrient by plants is high in this method.

Artificial food for fishes:
In this system plant waste and decays are utilized by fishes as food. So, the need for the use of supplementary feed can be minimized.

Fertilizer usage:
Artificial or chemical fertilizers is not required for this system since the plants in the aquaponics utilize the nutrients from the fish wastes dissolved in water.

Question 48.
What are the fauna and flora cultured and cultivated in aquaponics?
Answer:
Cultivable fishes like tilapia, trout, koi, gold fish, bass etc., are cultured in aquaponics. Common cultivable plants like tomato, pepper, lettuce, cucumber, and rose are co-cultivated in this method.

Question 49.
What is aquaculture?
Answer:
Aquaculture is a branch of science that deals with the farming of aquatic organisms such as fish, molluscs, crustaceans and aquatic plants.

Question 50.
What are the three types of aquaculture on the basis of the source?
Answer:
On the basis of source, aquaculture can be classified into three categories. They are

  1. Freshwater aquaculture
  2. Brackish water aquaculture
  3. Marine water aquaculture.

 

Question 51.
What is pisciculture?
Answer:
Culturing of fishes is called fish culture or pisciculture.

Question 52.
What should be the pH and salinity of freshwater used for aquaculture?
Answer:
pH of the freshwater should be around neutral and salinity below 5ppt.

Question 53.
Write a short note on Brackish water fishes?
Answer:
Brackish water fishes spend most of its life in river mouths (estuaries) back waters, mangrove swamps and coastal lagoons. Estuarine fish are more common in Bengal and Kerala. Culturing of animals in the water having salinity range 0.5 – 30 ppt are called as brackish water culture. Fishes cultured in brackish water are Milk fish (Chanos Chanos), Sea bass (‘Koduva’), Grey mullet (‘Madavai’), Pearl spots (‘Kari’meen) etc.

Question 54.
What are the varieties of marine fisheries?
Answer:
Mackerels, sardines, sharks, catfish.

Question 55.
What is Mariculture?
Answer:
Culturing of animals in the water salinity ranges from 30 – 35% is called Mariculture. Some fishes like Chanos sp, Mugil cephalus are cultured here.

Question 56.
What is Metahaline culture?
Answer:
Culturing of animals in the water salinity ranges from 36 – 40% is called Metahaline culture. Eg, Brine shrimp (Artemia salina).

Question 57.
What is Artemia?
Answer:
Artemia is commonly known as the brine shrimp. It is a crustacean and lives in high saline waters because of its high osmoregulatory capacity.

Question 58.
What are the characteristics of cultivable fishes?
Answer:
Characteristics of cultivable fishes.
The special characteristic features of cultivable fishes are:

  1. Fishes should have high growth rate in short period of culture.
  2. They should accept supplementary diet. ‘
  3. They should be hardy enough to resist some common diseases and infection of parasites.
  4. Fishes proposed for poly culture should be able to live together without interfering or attacking other fishes.
  5. They should have high conversion efficiency so that they can effectively utilize the food.

Question 59.
What are the types of cultivable fishes?
Answer:
Types of cultivable fish:
Cultivable fish are of 3 types:-

  1. Indigenous or native fresh water fishes (Major carps, Catla, Labeo, Clarias).
  2. Salt water fishes acclimatized for fresh water (Chanos, Mullet).
  3. Exotic fishes are imported from other counties (Common carps).

 

Question 60.
What are the reasons for culturing carps in India?
Answer:
Major carps have proved to be best suited for culture in India, because the carps

  1. Feed on zooplanktons and phytoplanktons, decaying weeds, debris and other aquatic plants.
  2. They can survive in turbid water with slightly higher temperature.
  3. Can tolerate O2 variations in water.
  4. Can be transported from one place to other easily.
  5. They are highly nutritive and palatable.

Question 61.
What are the external factors affecting fish culture?
Answer:
External factors affecting fish culture. The factors that affect fish culture are temperature, light rain, water, flood, water current, turbidity of water, pH hardness, salinity and dissolved O2. Light and temperature also play an important role in fish breeding.

Question 62.
What is Natural breeding of fishes?
Answer:
Natural breeding (Bundh breeding):
These are special types of ponds where natural riverine conditions or any natural water resources are managed for breeding of culturable fishes. These bundhs are constructed in large low-lying areas that can accommodate large quantity of rain water. The shallow area of such bundhs is used as spawning ground.

Question 63.
What are the advantages of composite fish farming?
Answer:
The advantages include:-

  1. All available niches are fully utilized.
  2. Compatible species do not harm each other.
  3. No competition among different species is found.
  4. Catla catla, Labeo rohita and Cirrhinus mrigala (surface feeder) are the commonly used fish species for composite fish farming.

Question 64.
What are exotic fishes?
Answer:
Exotic fishes. The fishes imported into a country for fish culture are called exotic fishes and such fish culture is known as exotic fish culture. Examples of such exotic fishes introduced in India are Cyprinus carpio and Oreochromis mossambicus.

Question 65.
What are the types of Prawn fishery?
Answer:
Types of prawn fishery

  1. Shallow water prawn fishery – located on the west coast restricted to shallow waters.
  2. Estuaries and back waters or saline lake prawn fishery – The area of production of prawns are the back waters seen along the Western coast, Ennur, Pulicat, Chilka lake and Estuaries of Ganga and Brahmaputra rivers.
  3. Freshwater prawn fishery – Prawns are caught from the rivers and lakes throughout India.
  4. Marine prawn fishery – Most of the marine prawns are caught along the Indian coast belonging to the .family Penaeidae.

 

Question 66.
What are species of Prawn distributed in water bodies?
Answer:
Species of prawn:
A number of species of prawn are distributed in water resources such as Penaeus indicus, Penaeus monodon, Metapenaeus dobsoni and Macrobrachium rosenbergii.

Question 67.
Name the genus that produces high quality pearls?
Answer:
Pinctada.

Question 68.
How is pearl formed?
Answer:
Pearl Formation:
When a foreign particle accidently enters into the space between mantle and shell of the oyster, it adheres to the mantle. The mantle epithelium encloses it like a sac and starts to secrete concentric layers of nacre around it as a defensive mechanism.

Nacre is secreted continuously by the epithelial layer of the mantle and is deposited around the foreign particle and over a period time the formation of repeated layers of calcium carbonate makes the hard and glossy pearl. When the pearl enlarges the oyster dies. The shell is then carefully opened and the pearls are manually separated and graded.

Question 69.
What are the steps of insertion of nucleus into oyster?
Answer:
Following steps are taken for the insertion of nucleus:-

a. Fitness of oyster for operation:
The selected oysters for the insertion of nucleus should be healthy and strong enough to overcome the stress during operation.

b. Preparation of graft tissues:
The piece of tissue which is inserted inside the mantle is called as ‘GRAFT’ tissue. The outer edges of these graft squares must be known because nacre secreting cells are found only on the outer surface of the mantle so it is essential to keep the outer surface in contact with the inserted nucleus.

c. Preparation of nucleus:
Any small particle may function as nucleus to initiate the pearl formation but it is reported that calcareous nucleus is the best because the deposition of nacre was found to be more on calcarious nucleus.

d. Insertion of nucleus:
For the insertion of nucleus, oysters are fixed in a desk clamp in the position of right valve facing upward. Mantle folds are smoothly touched to expose the foot and the main body mass, followed by an incision into the epithelium of the foot and a slender channel into the main mass one graft tissue which functions as a bed for the nucleus.

e. Post operation care:
Nucleated oysters are placed into cages and suspended into sea water and attached with floating rafts to a depth of 2 to 3 metres for about 6 to 7 days to recover from the shocks due to operation. This period of 6 to 7 days is known as ‘Recovery period’. About 3000 to 3600 nucleated oysters are kept in different cages suspended in sea water at 2 to 3 meters depth for 3 to 6 years and undisturbed except at the time of clearing and inspection.

f. Harvesting of pearl:
Pearls are harvested in the month of December to February which may slightly vary according to climatic conditions. After the completion of 3 years of the insertion of nucleus, pearl oysters are harvested from the sea and the pearls are taken out from the shell.

g. Clearing of pearls:
After taking out the pearls from the oysters shell they are washed properly, cleared with the soap solution.

Question 70.
What is the composition of pearl?
Answer:
Composition of pearl:
Pearl comprises of water, organic matter, calcium carbonate and the residue.

  1. Water: 2-4%
  2. Organic matter: 3.5-5.9%
  3. Calcium carbonate: 90% and
  4. Residue: 0.1-0.8% carbonate: 90%

Question 71.
What is Animal husbandry?
Answer:
Animal husbandry is the practice of breeding and raising livestock cattles like cows, buffaloes, and goats and birds etc. that are useful to human beings.

Question 72.
What are the objectives of animal breeding?
Answer:
Objectives of Animal breeding

  1. To improve growth rate.
  2. Enhancing the production of milk, meat, eggs etc.,
  3. Increasing the quality of the animal products.
  4. Improved resistance to diseases.
  5. Increased reproductive rate.

 

Question 73.
What are the methods of animal breeding?
Answer:
Methods of Animal breeding:
There are two methods of animal breeding, namely inbreeding and outbreeding:

1. Inbreeding:
Breeding between animals of the same breed for 4-6 generations is called inbreeding. Inbreeding increases homozygosity and exposes the harmful recessive genes.

Continuous inbreeding reduces fertility and even productivity, resulting in “inbreeding depression”. This can be avoided by breeding selected animals of the breeding population and they should be mated with superior animals of the same breed but unrelated to the breeding population. It helps to restore fertility and yield.

2. Outbreeding:
The breeding between unrelated animals is called outbreeding. Individuals produced do not have common ancestors for 4-6 generations.

Outbreeding helps to produce new and favourable traits, to produce hybrids, with superior qualities and helps to create new breeds. New and favourable genes can be introduced into a population through outbreeding.

Question 74.
What is Out crossing?
Answer:
Out crossing:
It is the breeding between unrelated animals of the same breed but having no common ancestry. The offspring of such a cross is called outcross. This method is suitable for breeding animals below average in productivity.

Question 75.
What is Cross breeding?
Answer:
Cross breeding:
Breeding between a superior male of one breed with a superior female of another breed. The cross bred progeny has superior traits (hybrid vigour or heterosis.).

Question 76.
What is Interspecific hybridization?
Answer:
Interspecific hybridization. In this method of breeding mating is between male and female of two different species. The progeny obtained from such crosses are different from their parents, and may possess the desirable traits of the parents.

Question 77.
What is Artificial insemination?
Answer:
Artificial insemination. Artificial insemination is a technique in which the semen collected from the male is injected to the reproductive tract of the selected female.

Question 78.
Explain the Multiple Ovulation Embryo Transfer (MOET)?
Answer:
Multiple ovulation embryo transfer technology (MOET). It is another method of propagation of animals with desirable traits. This method is applied when the success rate of crossing is low even after artificial insemination. In this method Follicle stimulating hormone (FSH) is administered to cows for inducing follicular maturation and super ovulation. Instead of one egg per cycle, 6-8 eggs can be produced by this technology.

The eggs are carefully recovered non-surgically from the genetic mother and fertilized artificially. The embryos at 8-32 celled stages are recovered and transferred to a surrogate mother. For another round of ovulation, the same genetic mother is utilized. This technology can be applied to cattle, sheep and buffaloes. Advantage of this technology is to produce high milk yielding females and high-quality meat yielding bulls in a short time.

Question 79.
What is Dairying and Dairy Operation?
Answer:
Dairying is the production and marketing of milk and its products. Dairy operation consists of proper maintenance of cattle, the collection and processing of milk and its by-products.

Question 80.
What are the groups of cattle?
Answer:
(I) Dairy breeds or Milch breeds:
They are high milk yielders with extended lactation. Eg., Sindhi, Gir, Sahiwal, Jersy, Brown Swiss, Holstein cattle.

(II) Draught purpose breeds:
Bullocks are good for draught purpose. Eg. Kangayam, Malvi.

(III) Dual Purpose breeds:
Cows are meant for yielding more milk and bullocks are used for better drought purpose Eg. Ongole, Hariana.

Question 81.
Distinguish between a healthy cattle and ill cattle?
Answer:

S.No

Healthy Cattle

Ill Cattle

1.A healthy cattle eats, drinks and sleeps well regularly.An ill cattle does not eat, drink and sleep properly.
2.It appears bright, alert and active in their movement with a shiny coat.An ill cattle appears dull, has a restless and changes posture frequently with drop in milk yield.

Question 82.
Name the main diseases of dairy cattle?
Answer:
Rinderpest, foot and mouth disease, cow pox, hemorrhagic fever, anthrax.

Question 83.
Write a note on milk products?
Answer:
Milk products:
Milk is produced by dairy animals which is an emulsion of fat and lactose. Milk also contains enzymes which are destroyed during pasteurization.

Milk is a rich source of vitamin A, B„ Bp and deficient in Vitamin C. Due to its high nutrition value, it serves as a . complete food for infants. Dairy products such as yoghurt, cheese, butter, ice cream, condensed milk, curd, and milk powder processed from milk make dairy, a highly farming attraction.

Question 84.
Write on the nutritive value of meat?
Answer:
Meat:
Meat is rich in protein and also contains many minerals like iron, zinc, vitamins and selenium. It also contains vitamins needed for human diet.

Question 85.
What is Poultry Farming?
Answer:
Poultry Farming refers to the rearing and propagation of avian species such as chicken, ducks, turkeys, geese, quail and guinea fowls.

Question 86.
Write on the chicken breeds of egg layers?
Answer:
Egg layers:
These are farmed mainly for the production of egg.

Leghorn:
This is the most popular commercial breed in India and originated from Italy. They are small, compact with a single comb and wattles with white, brown or black colour. They mature early and begin to lay eggs at the age of 5 or 6 months. Hence these are preferred in commercial farms. They can also thrive well in dry areas.

Chittagong:
It is the breed chiefly found in West Bengal. They are golden or light yellow coloured. The beak is long and yellow in colour. Ear lobes and wattles are small and red in colour. They are good egg layers and are delicious.

Question 87.
W-ite on the Broiler breed?
Answer:
Bioiler type:
These are well known for fast growth and soft quality meat.

White Plymouth rock:
They have white plumage throughout the body. It is commonly used in troiler production. This is an American breed. It is a fast growing breed and well suitable for growing intensively in confined farms.

Question 88.
What are Dual purpose chicken breeds?
Answer:
Dual purpose breeds:
These are for both meat and egg production purpose.

Brahma:
It is a breed popularly known for its massive body having heavy bones, well feathered and proportionate body. Pea comb is one of the important breed characters. It has two common varieties namely, Light Brahma and Dark Brahma.

Question 89.
Write a short note on Game breeds?
Answer:
(I) Game breeds:
Since ancient times, special breed of roosters have been used for the sport of cockfighting.

(II) Aseel:
This breed is white or black in colour. The hens are not good egg layers but are good in incubation of eggs. It is found in all states of India. Aseel is noted for its pugnacity, high stamina, and majestic gait and dogged fighting qualities. Although poor in productivity, this breed is well-known for their meat qualities.

Question 90.
Write a short note on Ornamental chicken breeds?
Answer:
Ornamental breeds:
Ornamental chicken are reared as pets in addition to their use for egg production and meat.

Silkie:
It is a breed of chicken having a typical fluffy plumage, which is said to feel like silk and satin. The breed has numerous additional special characters, such as black skin and bones, blue earlobes, and five toes on each foot, while the majority chickens only have four. They are exhibited in poultry .shows, and come out in various colours. Silkies are well recognized for their calm, friendly temperament. Silkie chicken is especially simple to maintain as pets.

Question 91.
What are the types of Poultry farming?
Answer:
Types of Poultry farming:
There are different methods used to rear both broiler and layer chicken. The types of poultry farming are Free range farming, Organic method, Yarding method, Battery cage method and Furnished cage method. Among these, Battery cage method is widely used in large scale poultry farms. The Free range, Organic and Yarding methods are eco-friendly and the eggs produced by such farming practices are preferred in the market.

Question 92.
What are the stages involved in rearing of chicken?
Answer:
Stages involved in rearing:
There are some steps involved in rearing of chicken:-

1. Selection of the best layer:
An active intelligent looking bird, with a bright comb, not obese should be selected.

2. Selection of eggs for hatching: Eggs should be selected very carefully. Eggs should be fertile, medium sized, dark brown shelled and freshly laid eggs are preferred for rearing. Eggs should be washed, cleaned and dried.

3. Incubation and hatching:
The maintenance of newly laid eggs in optimum condition till hatching is called incubation. The fully developed chick emerges out of egg after an incubation period of 21 – 22 days.

There are two types of incubation namely natural incubation and artificial incubation. In the natural incubation method, only a limited number of eggs can be incubated by a mother hen. In artificial incubation, more number of eggs can be incubated in a chamber (Incubator).

3. Brooding:
Caring and management of young chicks for 4 – 6 weeks immediately after hatching is called brooding. It can also be categorized into two types namely natural and artificial brooding.

4. Housing of Poultry:
To protect the poultry from sun, rain and predators, it is necessary to provide housing to poultry. Poultry house should be moisture-proof, rat proof and it should be easily cleanable and durable.

5. Poultry feeding:
The diet of chicks should contain adequate amount of water, carbohydrates, proteins, fats, vitamins and minerals.

Question 93.
What are poultry products?
Answer:
Eggs and meat.

Question 94.
What are poultry byproducts?
Answer:
A number of poultry byproducts like blood-meal, feather meal, poultry byproduct meal and hatchery by-product meal are used as good sources of nutrients for meat producing animals and poultry. These byproducts supply proteins, fats, vitamins and good amount of minerals.

Question 95.
Name Poultry diseases?
Answer:
Poultry diseases:
Ranikhet, Coccidiosis and Fowl pox are some common poultry diseases.

Question 96.
What are the benefits of Poultry farming?
Answer:
Benefits of Poultry farming are:

  1. It does not require high capital for construction and maintenance of the poultry farming.
  2. It does not require a big space.
  3. It ensures high return of investment within a very short period of time.
  4. It provides fresh and nutritious food and has a huge global demand.
  5. It provides employment opportunities for the people.

 

Question 97.
What are the native and exotic duck breeds?
Answer:

  1. Native Breeds – Indian Runner, Syhlet meta.
  2. Exotic Breeds – Muscori, Pekin, Aylesbury, Campbell.

Question 98.
What are the types of duck breeds?
Answer:
There are three types of ducks depending on the purpose for which it is formed. They are meat productive duck breeds, egg productive duck breeds, and breeds for both meat and egg production.

Question 99.
What are the advantages of duck farming?
Answer:
Advantages of duck farming:
They can be reared in small backyards where water is available and needs less care and management as they are very hardy. They can adapt themselves to all types of environmental conditions and are bred for feed efficiency, growth rate and resistance to diseases.

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Samacheer Kalvi 11th Bio Zoology Chemical Coordination and Integration Text Book Back Questions and Answers

Textbook Evaluation Solved
Choose The Correct Answer
Question 1.
The maintenance of constant internal environment is referred as .
(a) Regulation
(b) Homeostasis
(c) Co-ordination
(d) Hormonal control
Answer:
(b) Homeostasis

Question 2.
Which of the following are exclusive endocrine glands?
(a) Thymus and testis
(b) Adrenal and ovary
(c) Parathyroid and adrenal
(d) Pancreas and parathyroid
Answer:
(c) Parathyroid and adrenal

Question 3.
Which of the following hormone is not secreted under the influence of pituitary gland?
(a) Thyroxine
(b) Insulin
(c) Oestrogen
(d) Glucocorticoids
Answer:
(A) Insulin

Question 4.
Spermatogenesis in mammalian testes is controlled by?
(a) Luteinising hormone
(b) Follicle stimulating hormone
(c) FSH and prolactin
(d) GH and prolactin
Answer:
(A) Follicle stimulating hormone

Question 5.
Serum calcium level is regulated by?
(a) Thyroxine
(b) FSH
(c) Pancreas Assertion is true, but Reason is false
(d) Thyroid and parathyroid
Answer:
(d) Thyroid and parathyroid

Question 6.
Iodised salt is essential to prevent?
(a) Rickets
(b) Scurvy
(c) Goitre
(d) Acromegaly
Answer:
(c) Goitre

Question 7.
Which of the following gland is related with immunity?
(a) Pineal gland
(b) Adrenal gland
(c) Thymus
(d) Parathyroid gland
Answer:
(c) Thymus

Question 8.
Which of the following statement about sex hormones is correct?
(a) Testosterone is produced by Leydig cells under the influence of luteinizing hormone
(b) Progesterone is secreted by corpus luteum and softens pelvic ligaments during child birth
(c) Oestrogen is secreted by both sertoli cells and corpus luteum
(d) Progesterone produced by corpus luteum is biologically different from the one produced by placenta.
Answer:
(a) Testosterone is produced by Leydig cells under the influence of luteinizing hormone

Question 9.
Hypersecretion of GH in children leads to
(a) Cretinism
(b) Gigantism
(c) Graves disease
(d) Tetany
Answer:
(b) Gigantism

Question 10.
A pregnant female delivers a baby who suffers from stunted growth, mental retardation, low intelligence quotient and abnormal skin.This is the result of …………….
(a) Low secretion of growth hormone
(b) Cancer of the thyroid gland
(c) Over secretion of pars distalis
(d) Deficiency of iodine in diet.
Answer:
(b) Cancer of the thyroid gland

Question 11.
The structure which connects the hypothalamus with anterior lobe of pituitary gland is the
(a) Dendrites of neuro hypophysis
(b) Axons of neurohypophysis
(c) Bands of white fibers from cerebellar region
(d) Hypophysial portal system
Answer:
(b) Axons of neurohypophysis

Question 12.
Comment on homeostasis?
Answer:
Homeostasis: Maintenance of constant internal environment of the body by the different coordinating system. ‘

Question 13.
Which one of the following statement is correct
(a) Calcitonin and thymosin are thyroid hormones
(b) Pepsin and prolactin are selected in stomach
(c) Secretin and rhodopsin are polypeptide hormones
(d) Cortisol and aldosterone are steroid hormones
Answer:
(d) Cortisol and aldosterone are steroid hormones .

Question 14.
Which of the given option shows all wrong statement for thyroid gland Statements
(i) It inhibits process of RBC formation
(ii) It hepls in maintenance of water and electrolytes
(iii) Its more secretion can reduce blood pressure
(iv) It stimulates osteoblast
(a) (i) and (ii)
(b) (iii) and (iv)
(c) (i) and (iv)
(d) (i) and (iii)
Answer:
(c) (i) and (iv)

Question 15.
Hormones are known as chemical messenger? Justify?
Answer:
Hormones are released into the blood stream and circulated as chemical signals. These act specifically on certain organs or tissues called target organs or tissues. These speed up or slow down or alter the activity of target tissues or organs. Hence they are known as chemaical messengers.

Question 16.
Write the role of oestrogen in ovulation?
Answer:
Oestrogen is the ovarian hormone secreted during the proliferation phase of menstrual cycle from the 6th day to 14th day of the cycle. On 14th day of the cycle, under the influence of uteinizing hormone, ovum is released from the graffian follicles. This process is known as ovulation.

Question 17.
Comment on Acini of thyroid gland?
Answer:
Thyroid gland is a bilobed endocrine gland. Each lobe is made up of many lobules. The lobules consist of follicles called acini. Each acinus is lined with glandular, cuboidal or squamous epithelial cells. The lumen of acinus is filled with colloid, a thick glycoprotein mixture consisting of thyroglobulin molecules.

Question 18.
Write the causes for diabetes mellitus and diabetes insipids?
Answer:
Diabetes mellitus is caused due to reduced secretion of insulin. As a result, blood glucose level is elevated. Diabetes insipidus is caused due to under secretion of ADH or vasopression. As a result, reabsorption of water gets affected and hende large amount of urine is produced.

Question 19.
Specify the symptoms of acromegaly?
Answer:
Acromegaly is caused due to excessive secretion of growth hormone in adults. The symptoms of acromegaly are overgrowth of hand bones, feet bones, jaw bones malfunctioning of gonads. enlargement of viscera, tongue, lungs, heart, liver, splean and endocrine glands like thyroid, or adrenal glands.

Question 20.
Write the symptoms of cretinism?
Answer:
Cretinism is caused due to hypothyroidism in infants. A cretin child shows the following symptoms:-

  1. Retarded skeletal growth.
  2. Absence of sexual maturity
  3. Retarded mental ability
  4. Thick and short limbs
  5. Thick wrinkled skin
  6. Bloated face
  7. Protruded enlarged tongue
  8. Low BMR, slow pulse rate,subnormal body temperature and elevated blood cholesterol levels

 

Question 21.
Briefly explain the structure of thyroid gland?
Answer:
The thyroid gland is butterfly shaped, bilobed situated below the larynx on each side of the upper trachea. The two lobes are connected by a median tissue mass called isthmus. Each lobe is made up of many lobules. The lobules consist of follicles called acini. Each acinus is lined with glandular, cuboidal or squamous epithelial cells.

The lumen of acini is filled with colloid, a thick glycoprotein mixture consisting of thyroglobulin molecules. The thyroid gland secretes Tri-iodothyronine (T,) and tetra-iodothyronine (T4) or thyroxine hormones. These are concerned with metabolism.

Samacheer Kalvi 11th Bio Zoology Solutions Chapter 11 Chemical Coordination and Integration img 1

Question 22.
Name the layers of adrenal cortex and mention their secretions?
Answer:
A pair of adrenal glands are located at the anterior end of the kidneys. Elence, they are called suprarenal glands.The outer region is called cortex and the inner region is medulla. The adrenal cortex has three distinct zones,

  1. Zona Glomerulosa
  2. Zona fasciculata
  3. Zona reticularis

1. Zona Glomerulosa:

  • It is the outer thin layer. It constitutes about 15% of the cortex.
  • It secretes mineralocorticoids.

2. Zona fasciculata:

  • It is the middle wide layer constituting about 15% of adrenal cortex.
  • It secretes glucocorticoids such as cortisol, corticosterone and trace amounts of adrenal androgen and oestrogen.

3. Zona reticularis:

  • It is the inner zone of adrenal cortex.
  • It constitutes about 10% of adrenal cortex.
  • It constitutes about 10% of adrenal cortex.
  • It secretes the adrenal androgen, trace amount of oestrogen and glucocorticoids.

Question 23.
Differentiate hyperglycemia from hypoglycemia?
Answer:

Hyperglycemia

Hypoglycemia

1. Elevation in the blood sugar level is called hyperglycemia1. Decrease in the blood sugar level is called hypoglycemia.
2. This happens due to reduced secretion of insulin.2. This happens due to increased secretion of insulin.

 

Question 24.
Write the functions of CCK?
Answer:
CholecystokininCholecystokinin is secreted by duodenum in response to the presence of fat and acid in the diet. It acts on the gall bladder to release bile into duodenum and stimulates the secretion of pancreatic enzymes and its discharge.

Question 25.
Growth hormone is important for normal growth. Justify the statement?
Answer:
Growth hormone promotes growth of all the tissues and metabolic process of the body. It influences the metabolism of carbohydrate, proteins and lipids. It increases the rate of protein biosynthesis in the cells.

It stimulates chondrogenesis (cartilage formation), osteogenesis (bone formation) and helps in the retention of minerals like nitrogen, potassium, phosphorus, sodium etc in the body. It increases the release of fatty acid from adipose tissue and decreases the rate of glucose utilization for energy by the cells. The hyposecretion of growth hormones causes dwarfism in children.

Question 26.
Pineal gland is an endocrine gland, write its role?
Answer:
The pineal gland or epiphysis cerebri or conarium is located behind the third ventricle of brain. It is formed of paranchymal cells and interstitial cells. It secretes the hormone, melatonin.

It plays a central role in the regulation of circadian rhythm of our body and maintains normal sleep wake cycle. It also regulates the timing of sexual maturation of gonads. It also influences metabolism, pigmentation, menstrual cycle and defence mechanism of our body.

Question 27.
Comment on the functions of adrenalin?
Answer:
Adrenalin increases liver glycogen breakdown into glucose and increases the release of fatty acids from fat cells. During emergency, it increases heart beat rate and blood pressure.lt stimulates the smooth muscles of cutaneous and visceral arteries to decrease blood flow. It increases blood flow to the skeletal muscles and nervous tissue.

Question 28.
Predict the effects of removal of pancreas from the human body?
Answer:
Pancreas is both exocrine and endocrine gland. It is located just below the stomach as a leaf like structure. It secretes digestive enzymes and hormones like insulin and glucogon.

The digestive enzymes digest carbohydrates, proteins and fats. Insulin and glucose regulate blood sugar level. If pancreas is removed from the body, digestion and main’taince of blood sugar level gets afftected.

Question 29.
Enumerate the role of kidney as an endocrine gland?
Answer:
Kidney has endocrine tissues which act as partial endocrine gland. It secretes renin, erythropoietin and calcitripl. Renin is secreted by juxta glomerular cells. It increases blood pressure when angiotensin is formed in blood.

Erythropoietin is also secreted by the juxta glomerular cells of the kidney and stimulates erythropoieis in bone marrow. Calcitriol is secreted by proximal tubes of nephrons. It is an active form of vitamin D3 which promotes calcium and phosphorus absorption from intestine and accelerates bone formation.

Question 30.
Write a detailed account of gastro intestinal tract hormones?
Answer:
Group of specialized endocrine cells present in gastro-intestial tract secretes hormones such as gastrin, cholecystokinin (CCK), secretin and gastric inhibitory peptides (GIP). Gastrin acts on the gastric glands and stimulates the secretion of HC1 and pepsinogen.

Cholecystokinin (CCK) is secreted by duodenum in response to the presence of fat and acid in the diet. It acts on the gall bladder to release bile into duodenum and stimulates the secretion of pancreatic enzymes and its discharge. Secretin acts on acini cells of pancreas to secrete bicarbonate ions and water to neutralize the acidity. Gastric inhibitory peptide (GIP) inhibits gastric secretion and motility.

In Text Questions Solved

Question 1.
Pituitary gland is located in a depression in the sphenoid bone of skull below the brain, so is also called hypothalamus cerebri. Discuss the following:
(a) Pituitary gland is commonly called “master gland” of the body?
(b) Discuss the role of hypothalamus and pituitary as a coordinated unit in maintaining physiological processes?
(c) How does the posterior lobe of pituitary help in osmoregulation?
Answer:
(a) Pituitary gland is commonly called “master gland” of the body because it regulates the activity of other endocrine glands such as adrenal gland, thyroid gland, testis and ovary.

(b) The hypothalamus sends signals to the pituitary gland to release or inhibit pituitary hormone releasing factor production thus maintaining Homeostasis of the body.

(c) The posterior lobe of pituitary helps in osmoregulation by secreting the hormones namely vasopressin and oxytocin which increases tubular reabsorption of water in the nephrons.

Question 2.
Endocrine glands control and coordinate the body functions through secreting certain chemical messengers called hormones. Due to certain physiological reasons, the blood glucose level of an otherwise normal person?
(a) Give the possible cause for the increases in blood glucose level?
(b) What is the chemical nature of this hormone? Discuss its role in the body?
(c) How can this condition be reversed?
Answer:
(a) Stress and anxiety could have been the possible causes for the sudden increase in blood glucose level.

(b) Insulin is a polypeptide hormone. Insulin is secreted by the Islets of Langerhans of pancreas. It helps in converting glucose into glycogen or fat and allowing-oxidation of glucose inside the cells. Thus, insulin helps maintain the blood sugar level.

(c) This condition can be reversed by relaxing and taking a walk.

Question 3.
Identify the peaks of FSH, LH, Oestrogen and Progesterone hormones through out the menstrual cycle?
Answer:

Samacheer Kalvi 11th Bio Zoology Solutions Chapter 11 Chemical Coordination and Integration img 2

Oestrogen, FSH are in the peak during the follicular phase (6 -13 days) of menstruation. LH is at the peak during ovulation that is on the 14th day of menstrual cycle. Progesterone is at the peak in the luteal phase of menstruation (14 – 28 days).

Samacheer Kalvi 11th Bio Zoology Chemical Coordination and Integration Additional Questions & Answers

I. Choose The Correct Answer

Question 1.
Which of the following is not related to hormones?
(a) They are secreted by ductless glands
(b) They act on the target
(c) They act as chemical messengers
(d) They play an important role in disgestion
Answer:
(d) They play an important role in disgestion

Question 2.
Which of the following is a partial endocrine?
(a) Thymus gland
(b) Pineal gland
(c) Gonads
(d) Parathyroid gland
Answer:
(c) Gonads

Question 3.
Which of the following is the function of Growth Hormone?
(a) It increases blood pressure
(b) It influences the metabolism of carbohydrates, proteins and lipids
(c) It stimulates melanin synthesis in melanocytes
(d) It promotes reabsorption of water by nephrons
Answer:
(b) It influences the metabolism of carbohydrates, proteins and lipids

Question 4.
Which of the following stimulates milk secretion after the child birth in females?
(a) Luteotropic hormone
(b) Luteinizing hormone
(c) Follicle stimulating hormone
(d) Somatotropin
Answer:
(a) Luteotropic hormone

Question 5.
Which of the following is the reason for calling vasopressin as the anti Jiuretic hormone?
(a) It promotes dilution of urine
(b) It reduces dilution of urine
(c) It increases blood pressure
(d) It constricts blood vessels
Answer:
(b) It reduces dilution of urine

Question 6.
A person is unable to sleep normally. He may not get proper secretion of?
(a) Thyroxine
(b) Vasopression
(c) Melatonin
(d) Oxytocin
Answer:
(c) Melatonin

Question 7.
Why is parathormone called hypercalcemic hormone?
(a) It decreases blood calcium level
(b) It increases blood potassium level
(c) It increases blood calcium level
(d) It decreases blood potassium level
Answer:
(b) It increases blood potassium level

Question 8.
Which of the following is not the function of cortisol?
(a) It produces anti inflammatory reactions
(b) It stimulates RBC production
(c) It supresses the immune response
(d) It regulates water and electrolyte balance of the body
Answer:
(d) It regulates water and electrolyte balance of the body

Question 9.
Which adrenal hormone is concerned with maintaining electrolytes, osmotic pressure and blood pressure?
(a) Cortisol
(b) Glucocorticoids
(c) Aldosterone
(d) Adrenalin
Answer:
(c) Aldosterone

Question 10.
Which of the following hormones prepares the body for meeting emergency situations?
(a) Aderenalin and noradrenalin
(b) Glucocorticoids and mineralocorticoids
(c) Cortisol and Aldosterone
(d) Glucocorticoids and corticosterone
Answer:
(a) Aderenalin and noradrenalin

Question 11.
Which of the following is the function of insulin?
(a) It promotes Glycogenolysis
(b) It increases the uptake of glucose into the body cells
(c) It promotes gluconeogenesis
(d) It reduces the cellular uptake and utilization of glucose
Answer:
(b) It increases the uptake of glucose into the body cells

Question 12.
Which of the following is the functiion of Glucogon?
(a) It increases the cellular utilization of glycogen into glucose
(b) It promotes the breakdown of glycogen into glucose
(c) It inhibits the breakdown of glycogen into glucose
(d) It inhibits the production of glucose from non-carbohydrate sources
Answer:
(b) It promotes the breakdown of glycogen into glucose

Question 13.
Which of the following promotes the appearance of secondary sexual characters in male and female respectively?
(a) Testosterone and progesterone
(b) Progesterone and testosterone
(c) Testosterone and oestrogen
(d) Oestrogen and testosterone
Answer:
(c) Testosterone and oestrogen

Question 14.
Which of the following is not the function of progesterone?
(a) Implantation of zygote in the uterus
(b) Decreasing the contraction of uterus
(c) Formation of placenta
(d) Maturation of reproductive organs
Answer:
(d) Maturation of reproductive organs

Question 15.
Which of the following hormone is secreted by proximal tubules of nephron?
(a) Renin
(b) Calcitriol
(c) Erythropoietin
(d) Cholecystokinin
Answer:
(b) Calcitriol

Question 16.
A person attains a maximum of 4 feet height. What could be the reason for this?
(a) Hypersecretion of somatotropic hormone
(b) Hypersecretion of ACTH
(c) Hyposecretion of TSH
(d) Hyposecretion of somatotropic hormone
Answer:
(d) Hyposecretion of somatotropic hormone

Question 17.
An excessive secretion of growth hormone in adults lead to ……………….
(a) Gigantism
(b) Dwarfism
(c) Acromegaly
(d) Cretinism
Answer:
(c) Acromegaly

Question 18.
Why does cretenism occur?
(a) Hypothyroidism in adults
(b) Hypothyroidism in infants
(c) Hyperthroidism in adults
(d) lower level of iodine in blood
Answer:
(b) Hypothyroidism in infants

Question 19.
Increased heartbeat, high BP and protrusion of eyeball are the characteristics of …………….
(a) Endemic goitre
(b) Tetany
(c) Myxodema
(d) Grave’s disease
Answer:
(d) Grave’s disease

Question 20.
Hypocalcemia occures due to?
(a) Hypersecretion of parathyroid hormone
(b) Hyposecretion of parathyroid hormone
(c) Hypersecretion of cortisol
(d) Hyposecretion of cortisol
Answer:
(b) Hyposecretion of parathyroid hormone

Question 21.
Demineralization occurs due to?
(a) Hypoparathyroidism
(b) Hypothyroidism
(c) Hyperthyroidism
(d) Hyperparathyroidism
Answer:
(d) Hyperparathyroidism

Question 22.
Hyper pigmentation of skin and low metabolic rate are the symptoms of …………………..
(a) Cushing’s syndrome
(b) Addison’s disease
(c) Grave’s disease
(d) Gull’s disease
Answer:
(b) Addison’s disease

Question 23.
Excess secretion of cortisol leads to
(a) Addison’s disease
(b) Grave’s disease
(c) Cushing’s syndrome
(d) Gull’s diesease
Answer:
(c) Cushing’s syndrome

Question 24.
In Type I diabetes ………………
(a) Insulin is produced in sufficient quantity
(b) Insulin resistance occurs
(c) Insulin is produced more
(d) Insulin is not produced
Answer:
(d) Insulin is not produced

Question 25.
Excessive intake of food in diabetes is called as …………………..
(a) Polydipsia
(b) Polyurea
(c) Polyphagia
(d) Ketosis
Answer:
(c) Polyphagia

Question 26.
The hyposecretion of vasopressin results in …………….. ?
(a) Diabetes mellitus
(b) Diabetes insipidus
(c) Ketosis
(d) Hyperglycemia
Answer:
(b) Diabetes insipidus

Question 27.
What does peptide hormone do as it cannot cross the phospholipid cell membrane?
(a) It combines with cAMP
(b) It binds to the receptors on the exterior cell surface
(c) It combines with steroids
(d) It combines with adenylate cyclases
Answer:
(a) It binds to the receptors on the exterior cell surface

Question 28.
In case of steroids hormones, dimer of receptor-hormone complex is formed. What does it do then?
(a) It binds with golgi complex
(b) It binds with tRNA
(c) It binds with mRNA
(d) It binds with DNA
Answer:
(d) It binds with DNA

II. Fill in the Blanks

Question 1.
The hormones are converted into inactive compounds by the …………………….. and excreted by the ……………………..
Answer:
(liver, kidneys )

Question 2.
Hormones act as …………………….. to perform specific functions in the target organs.
Answer:
( organic catalysts / coenzymes )

Question 3.
The pituitary gland is controlled by the ……………………..
Answer:
(hypothalamus)

Question 4.
Hypothalamus produces …………………….. which regulate the secretions of the pituitary gland.
Answer:
(neurotransmitters)

Question 5.
The pituitary gland is located in the bony cavity of the sphenoid bone called ……………………..
Answer:
( sella turcica)

Question 6.
The pituitary gland is connected to the hypothalamus by a stalk called ……………………..
Answer:
(infundibulum)

Question 7.
The anterior lobe of the pituitary gland originates from the embriyonic invagination of pharyngeal epithelium called ……………………..
Answer:
(Rathke’s pouch)

Question 8.
…………………….. decreases the rate of glucose
Answer:
(Growth Hormone)

Question 9.
ACT4 stimulates melanin synthesis in ……………………..
Answer:
(Melanocytes)

Question 10.
LH is also known as ……………………..
Answer:
(Interstitial cell stimulating Hormone)

Question 11.
FSH and LH are collectively referred as ……………………..
Answer:
(gonadotropins)

Question 12.
High prolactin secretion during lactation supresses LH secretion and ……………………..
Answer:
(ovulation)

Question 13.
…………………….. promotes reabsorption of water and electrolytes by distal tubules of nephron.
Answer:
(ADH / vasopressin )

Question 14.
…………………….. stimulates ejection of milk from the mammary glands.
Answer:
(oxytocin)

Question 15.
The pineal gland secretes …………………….. which plays a central role in the regulation of circadian rhythm of our body.
Answer:
(melatonin)

Question 16.
The two lateral lobes of thyroid gland are connectd by a median tissue mass called ……………………..
Answer:
(isthmus)

Question 17.
…………………….. gland produces ‘T’ lymphocytes which provides cell mediated immunity.
Answer:
(Thymus)

Question 18.
…………………….. produces anti inflammatory reactions and suppresses the immune response.
Answer:
(cortisol)

Question 19.
…………………….. increases the breakdown of liver glycogen into glucose
Answer:
(Adrenalin)

Question 20.
Both adrenalin and nor adrenalin are ……………………..
Answer:
(catecholamines)

Question 21.
…………………….. stimulates the reabsorption of sodium and water.
Answer:
(Aldosterone)

Question 22.
The beta cells of islets of langerhans secrete ……………………..
Answer:
(Insulin)

Question 23.
The alpha cells of islets of langerhans secrete ……………………..
Answer:
(glucagon)

Question 24.
The delta cells of islets of langerhans secrete ……………………..
Answer:
(somatostatin)

Question 25.
Insulin inhibits the breakdown of …………………….. into glucose.
Answer:
(glycogen)

Question 26.
…………………….. is called a hypoglycemic hormone.
Answer:
(Insulin)

Question 27.
…………………….. promotes the breakdown of glycogen into glucose.
Answer:
(Glucogon)

Question 28.
The breakdown of glucose is called ……………………..
Answer:
(glycogenolysis)

Question 29.
The synthesis of glucose from non – carbohydrate molecules is called ……………………..
Answer:
(gluconeolysis)

Question 30.
…………………….. reduces the cellular update and utilization of glucose.
Answer:
(Glugocon)

Question 31.
…………………….. is called a hyperglycemic hormone.
Answer:
(Glucogon)

Question 32.
The …………………….. cells of testes secrete male sex hormone.
Answer:
(Leydig)

Question 33.
…………………….. is reposible for development of secondary sexual charactes of female.
Answer:
(Oestrogen)

Question 34.
…………………….. prepares the uterus for implantation of the fertilized ovum.
Answer:
(Progesterone)

Question 35.
…………………….. on the atrial wall secretes atrial natriuretic factor to reduce the blood pressure .
Answer:
(cardiocytes)

Question 36.
…………………….. increases blood pressure when angiotension is formed in blood.
Answer:
(Renin)

Question 37.
stimulates the erythyropoiesis in bone marrow.
Answer:
(Erythropoietin)

Question 38.
…………………….. is the hormone secreted by proximal tubules of nephron which promotes calcium and phosphorus absorption from intestine.
Answer:
(Calcitriol)

Question 39.
…………………….. stimulates the secretion of HCl and pepsinogen.
Answer:
(Gastrin)

Question 40.
…………………….. stimulates the release of bile into duodenum and secretion of panereatic enzymes.
Answer:
(Cholecystokinin)

Question 41.
…………………….. acts on acini cells of pancreas to secrete bicarbonate ions and water to neutralise the acidity.
Answer:
(secretin)

Question 42.
…………………….. is caused due to hyposecretion of growth hormone in children.
Answer:
(dwarfism)

Question 43.
…………………….. is caused due to hypersecretion of growth hormone in childre.
Answer:
(Gigantism)

Question 44.
…………………….. is caused due to excessive secretion of growth hormone in adults.
Answer:
(Acromegaly)

Question 45.
The hypothyroidism in children causes ……………………..
Answer:
(cretinism)

Question 46.
The hyposecretion of thyriod in adults causes ……………………..
Answer:
(Myxodema)

Question 47.
Myxodema is otherwise called ……………………..
Answer:
(Gull’s disease)

Question 48.
The hypersecretion of thyriod gland causes as ……………………..
Answer:
(exopthalmic goitre)

Question 49.
The hyposecretion of parathyroid hormone causes ……………………..
Answer:
(Tetany)

Question 50.
The exophthalmic goitre is called as ……………………..
Answer:
(Grave’s disease / thyrotoxicosis)

Question 51.
Due to hyposecretion of parathyroid hormone serum …………………….. level decreases.
Answer:
(calcium)

Question 52.
Demineralisation is caused due to ……………………..
Answer:
(hyperparathyroidism)

Question 53.
…………………….. is caused due to hyposecretion of glucocorticoids and mineralocorticoids from the adrenal cortex.
Answer:
(Addison’s disease)

Question 54.
The hypersecretion of cortisol causes ……………………..
Answer:
(Cushing’s syndrome)

Question 55.
Type II diabetes is caused due to ……………………..
Answer:
(insulin resistance)

Question 56.
Type I diabetes is ……………………..
Answer:
(insulin dependent)

III. Answer The Following Questions

Question 1.
Distinguish between Exocrine glands and Endocrine glands?
Answer:

Exocrine

Endocrine glands

1. Exocrine glands have ducts to carry their secretions.1. Endocrine gland do not have ducts to carry their secretions. They are directly released into the blood. Hence these are called ductless glands.
2. These are concerned with digestion and other functions.2. These are concerned with growth, puberty, metabalosim and maintenance of homeostasis.
3. eg. liver, teargland, mammary gland.3. eg. pituitary gland, thyroid gland.

 

Question 2.
Why is hypothalamus considered as neuro endocrine gland?
Answer:
The hypothalamus alone with its neural function produces hormones. Hence it is considered as neuro-endocrine gland.

Question 3.
What are the functions of hypothalamus?
Answer:
Hypothalamus maintains homeostasis, blood pressure, body temperature, cardio and fluid electrolyte balance of the body. It influences various emotional responses.

Question 4.
Write on the origin of pituitary gland?
Answer:
The anterior lobe of pituitary gland originates from the embryonic invagination of pharyngeal epithelium called Rathke’s pouch. The posterior lobe originates from the base of the brain as an outgrowth of hypothalamus.

Question 5.
What are the lobes of adenohypophysis?
Answer:
Pars distalis, Pars intermedia and Pars tuberalis.

Question 6.
What is Pars nervosa?
Answer:
The neurohypophysis of the pituitary gland is known as pars nervosa.

Question 7.
Name the hormones secreted by the pituitary gland?
Answer:
The anterior lobe of pituitary secretes six tropic hormones such as growth hormone (GH), thyroid stimulating hormone (TSH), adreno corticotropic hormone (ACTH), follicle stimulaiton hormone (FSH), luteinizing hormone (LH), luteotropic hormone (LTH) and melanocyte stimulating hormone (MSH) (in lower animals only). The posterior lobe of pituitary secretes the hormones namely vasopressin and oxytocin.

Question 8.
Write a paragraph on the Growth Hormone?
Answer:
Growth hormone (GH):
It is also known as somatotropic hormone (STH) or Somatotropin. It is a peptide hormone Growth hormone promotes growth of all the tissues and metabolic process of the body. It influences the metabolism of carbohydrates, proteins and lipids and increases the rate of protein biosynthesis in the cells.

It stimulates chondrogenesis (cartilage formation), osteogensis (bone formation) and helps in the retention of minerals like nitrogen, potassium, phosphorus, sodium etc., in the body. GH increases the release of fatty acid from adipose tissue and decreases the rate of glucose utilization for energy by the cells. Thus it conserves glucose for glucose dependent tissues, such as the brain.

Question 9.
Write on the role of Thyroid stimulating hormone and Adreno cortico tropic hormone? Thyroid stimulating hormone (TSH) or thyrotropin?
Answer:
TSH is a glycoprotein hormone, which stimulates the thyroid gland to secrete Triiodothyronine (T3) and thyroxine (T4). TSH secretion is regulated by negative feedback machanism. It’s release from the anterior pitutary is induced by the thyrotropin releasing hormone (TRH). When thyroxine level in the blood increases, TRH acts on both the pituitary and hypothalamus to inhibit TSH secretion.

Adreno cortico tropic hormone (ACTH):
ACTH is a peptide hormone that stimulates the adrenal cortex to secrete glucocorticoids and mineralocorticoids. It stimulates melanin synthesis in melanocytes, induces the release of fatty acids from adipose tissues and stimulates insulin secretion. ACTH secretion is regulated by negative feedback mechanism.

Question 10.
What are the functions of follicle stimulation hormone?
Answer:
Follicle stimulating hormone (FSH):
FSH is a glycoprotein hormone which regulates the functions of the gonads (ovary and testis). In males, FSH along with androgens acts on the germinal epithelium of seminiferous tubules and stimulates the production and release of sperms (spermatogenesis). In females, FSH acts on the ovaries and brings about the development and maturation of gratfran follicles.

Question 11.
What is the role of Luteinizing hormone (LH)?
Answer:
Luteinizing Hormone is a glycoprotein hormone which is also known as interstitial cell stimulating hormone (ICSH). In males, ICSH acts on the interstitial cells of testis to produce the male sex hormone, testosterone. In females, LH along with FSH matures the ovarian follicles.

LH independently induces ovulation, maintains the corpus luteum and promotes synthesis and release of ovarian hormones. FSH and LH are collectively referred as gonadotropins. FSH and LH are not produced during childhood. The secretion of FSH and LH starts only during prepubertal period.

Question 12.
What is the significance of Luteotropic hormone (LTH)?
Answer:
Luteo Tropic Hormone is also called luteotropin or lactogenic hormone or prolactin or mammotropin. It is a protein hormone which stimulates milk secretion after the child birth in females. High prolactin secretion during lactation suppresses LH secretion and ovulation since it induces the corpus luteum hence named as luteo tropic hormone.

Question 13.
What are the hormones of neurohypophysis?
Answer:
(i) Vasopressin or antidiuretic hormone (ADH):
ADH is a peptide hormone which promotes reabsorption of water and electrolytes by distal tubules of nephron and thereby reduces loss of water through urine. Hence it is called as anti diuretic hormone. It also causes constriction of blood vessels when released in large amount and increases blood pressure. ADH deficiency causes Diabetes insipidus which induces the production of large amount of urine.

(ii) Oxytocin (means quick birth):
It is a peptide hormone which stimulates vigorous contraction of the smooth muscles of uterus during child birth and ejection of milk from the mammary glands.

Question 14.
What are the functions of pineal gland?
Answer:
In human, the pineal gland or epiphysis cerebri or conarium is located behind the third ventricle of brain and is formed of parenchymal cells and interstitial cells.

It secretes the hormone, melatonin, which plays a central role in the regulation of circadian rhythm of our body and maintains the normal sleep wake cycle. It also regulates the timing of sexual maturation of gonads. In addition melatonin also influences metabolism, pigmentation, menstrual cycle and defence mechanism of our body.

Question 15.
Write on the role of parathyroid gland?
Answer:
In human, four tiny parathyroid glands are found in the posterior wall of the thyroid glands. This galnd is composed of two types of cells, the chief cells and oxyphil cells. The chief cells secrete parathyroid hormone (PTH) and the functions of oxyphil cells are not known. Parathyroid hormone or parathormone (PTH).

PTH is a hypercalcemic hormone. It is a peptide hormone involved in controlling the calcium and phosphate homeostasis. The secretion of PTH is controlled by calcium level in the blood. It increases the blood calcium level by stimulating osteoclasts to dissolve the bone matrix.

As a result calcium and phosphate are released into the blood. PTH enhances the reabsorption of calcium and excretion of phosphates by the renal tubules and promotes activation by vitamin D to increase calcium absorption by intestinal muscosal cells.

Question 16.
Write a paragraph on Thymus gland?
Answer:
Thymus gland is partially an endocrine and partially a lymphoid organ. It is a bilobed structure located just above the heart and aorta, behind the sternum. It is covered by fibrous capsule and anatomically it is divisble into an outer cortex and an inner medulla.

It secretes four hormones such as thymulin, thymosin, thymopoietin and thymic humoral factor (THF). The primary function of thymus is the production of immuno competent ‘T’ lymphocytes which provides cell mediated immunity.

Question 17.
Explain the hormones secreted by Adrenal gland?
Answer:
A pair of adrenal glands are located at the anterior end of the kidneys, hence also called suprarenal glands. Anatomically the outer region is the cortex and the inner region is the medulla.

Histologically the adrenal cortex has three distinct zones, zona glomerulosa, zona fasciculata and zona reticularis. Zona glomerulosa an outer thin layer constitutes about 15% of adrenal cortex, and secretes mineralocorticoids.

Zona fasciculata, the middle widest layer constitutes about 75% of adrenal cortex and secretes glucocorticoids such as cortisol, corticosterone and trace amounts of adrenal androgen and oestrogen. Zona reticularis, an inner zone of adrenal cortex constitute about 10% of adrenal cortex and secretes the adrenal androgen, trace amount of oestrogen and glucocorticoids.

Adrenal medulla:
It is the central part of adrenal gland and is composed of ovoid and columnar cells, which are found around the network of blood capillaries. Adrenalin (epinephrine) and nor adrenalin (nor epinephrine) are the two hormones secreted by the adrenal medulla. Both adrenalin and nor adrenalin are catecholamines.

Function of adrenal hormones:
Glucocorticoids stimulate gluconeogenesis, lipolysis and proteolysis (the life saving avtivity). Cortisol is a glucocorticoid invloved in maintaining cardio vascular and kidney functions.

It produces anti-inflammatory reactions and suppresses the immune response. It stimulates the RBC production. It is also known as stress combat hormone. Mineralocorticoids regulates water and electrolyte balance of our body.

Aldosterone stimulates the reabsorption of sodium and water and eliminates potassium and phosphate ions through excretion,, thus it helps in maintainting electrolytes, osmotic pressure and blood pressure. Adrenal androgen plays a role in hair growth in the axial region, pubis and face during puberty.

The adrenal medulla secretes the hormones adrenalin and noradrenalin and are referred as “3F hormone” (fight, flight and fright hormone). Adrenalin increases liver glycogen breakdown into glucose and increases the release of fatty acids from fat cells.

During emergency it increases heart beat rate and blood pressure. It stimulates the smooth muscles of cutaneous and visceral arteries to decrease blood flow. It increases blood flow to the skeletal muscles therby increases the metabolic rate of skeletal muscles, cardiac muscles and nervous tissue.

Question 18.
What are the hormones responsible for maintaining blood glucose level?
Answer:
Insulin and glucogon are the hormones secreted by Islets of langerhans of pancreas:-
Insulin:
Insulin is a peptide hormone and plays an important role in glucose homeostasis. It’s main effect is to lower blood glucose levels by increasing the uptake of glucose into the body cells, especially muscle and fat cells. Insulin also inhibits the breakdown of glycogen to glucose, the conversion of amino acids or fats to glucose, so insulin is rightly called a hypoglycemic hormone.

Glucagon:
Glucagon is a polypeptide hormone. It is a potent hyperglycaemic hormone that acts on the liver and promotes the breakdown of glycogen to glucose (Glycogenolysis), synthesis of glucose from lactic acid and from non-carbohydrate molecules (gluconeogenesis).

Releases glucose from the liver cells, increasing the blood glucose levels. Since glucagon reduces the cellular uptake and utilisation of glucose it is called a hyperglycemic hormone. Prolonged hyperglycemia leads to the disorder called diabetes mellitus.

Question 19.
Explain the endorenal function of gonads?
Answer:
Testis:
A pair of testis is present in the scrotal sac of males. The testis functions as a sex organ and also as an endocrine gland. The testis is composed of seminiferous tubules and interstitial cells or Leyding cells. The Leyding cells secrete several male sex hormones, collectively called androgens, mainly testosterone.

Functions of testosterone:
Under the influence of FSH and LH testosterone initiates maturation of male reproductive organs, and the appearance of secondary sexual characters, muscular growth, growth of facial and axillary hair, masculine voice and male sexual behaviour. It enhances the total bone matrix and plays a stimulating role in the process of spermatogenesis.

Ovary:
Females have a pair of ovaries located in the pelvic region of the abdomen. The ovary is composed of ovarian follicles and stromal tissues. Besides producing the eggs or ova, the ovaries secrete the steroid hormones oestrogen and progesterone. Oestrogen is responsible for the maturation of reproductive organs and the development of secondary sexual characters at puberty. Along with progesterone, oestrogens promotes breast development and initiate the cyclic changes during menstrual cycle.

Progesterone prepares the uterus for implantaion of the fertilized ovum. It decreases the uterine contraction during pregnancy and stimulates mammary glands and milk secretion. It is responsible for premenstrual changes in the uterus and is essential for the formation of placenta.

Question 20.
Write the disorders associated with growth hormone?
Answer:
Dwarfism is due to hyposecretion of growth hormone (GH) in children, skeletal growth and sexual maturity is arrested. They attain a maximum height of 4 feet only. Gigantism is due to hypersecreion of growth hormone (GH) in children. Overgrowth of skeletal structure occurs (up to 8 feet) and the visceral growth is not appropriate with that of limbs.

Acromegaly is due to excessive secretion of growth hormone in adults. Over growth of hand bones, feet bones, jaw bones, malfunctioning of gonads, enlargement of viscera, tongue, lungs, heart, liver, spleen and endocrine gland like thyroid, adrenal etc., are the symptoms of acromegaly.

Question 21.
Write on Hyperthyroidism and Hypothyroidism?
Answer:
In infants, hypothyroidiam causes cretinism. A cretins shows retarded skeletal growth, absence of sexual maturity, retarded mental ability, thick wrinkled skin, protruded enlarged tongue, bloated face, thick and short limbs occurs. The other symptoms are low BMR, slow pulse rate, subnormal body temperature and elevated blood cholesterol level.

Hyposecretion of thyroid in adults causes myxodema. It is otherwise called Gull’s disease. This disease is characterised by decreased mental activity, memory loss, slowness of movement, speech, and general weakness of body, dry coarse skin, scarce hair, puffy appearence, disturbed sexual function, low BMR, poor appetite, and subnormal body temperature.

Grave’s disease also called as thyrotoxicosis or exophthalmic goitre. This disease is caused due to hyper secretion of thyriod. It is characterised by enlargement of thyroid gland, increases BMR (50% -100%), elevated respiratory and excretory rates, increased heart beat, high BP, increases body temperature, protrusion of eyeball and weakness of eye muscles and weight loss.

Simple goitre is also known as Endemic goitre. It is caused due to hyposecretion of thyroxine. The symptoms includes enlargement of thyroid gland, fall in serum thyroxine level, increases TSH secretion.

Question 22.
Write on the disorders associated with parathryroid gland?
Answer:
Tetany is caused due to the hyposecretion of parathyroid hormone (PTH). Due to hyposecretion of PTH serum calcium level decreases (Hypocalcemia), as a result serum phosphate level increases. Calcium and phosphate excretion level decreses. Generalized convulsion, locking of jaws increased heart beat rate, increases body temperature, muscular spasm are the major symptoms of tetany. Hyperparathyroidism is caused due to excess PTH in blood. Demineralisation of bone, loss of muscle tone, general weakness, renal disorders are the symptoms of hyperparathyroidism.

Question 23.
Write a paragraph on Addison’s disease?
Answer:
Addison’s disease is caused due to hyposecretion of glucocorticoids and mineralocorticoids from the adrenal cortex. Muscular weakness, low BP, loss of appetite, vomiting, hyper pigmentation of skin, low metabolic rate, subnormal temperature, reduced blood volume – weight loss are the symptoms that occur in Addison’s disease. Reduces aldosterone secretion increases urinary excretion of Na Cl. and decreases potassium excretion leading to dehydration.

Question 24.
Write a paragraph on Cushing’s syndrome?
Answer:
Cushing’s syndrome is caused due to excess secretion of cortisol. Obesity of the face and trunk, redness of face, hand, feet, thin skin, excessive hair growth, loss of minerals from bone (osteoporoses) systolic hypertension are features of Cushing’s syndrome. Suppression of sexual function like atrophy of gonads are the other symptoms of Cushing’s syndrome.

Question 25.
Write on Hypoglycaemia?
Answer:
Hypoglycaemia is due to increased secretion of insulin thereby blood glucose level decreases. In this disorder blood glucose level lowers than normal fasting index. Increased heartbeat, weakness, nervousness, headache, confusion, lack of co-ordination, slurred speech, serious brain defects like epilepsy and coma occurs.

Question 26.
Write a paragraph on different types of diabetes mellitus?
Answer:
Hyperglycaemia is otherwise known as Diabetes mellitus. Is is caused due to reduced secretion of insulin. As the result, blood glucose level is elevated. Diabetes mellitus is of two types, Type! Diabetes and Type II Diabetes. Type I diabetes is also known Insulin dependent diabetes, caused by the lack of insulin secretion due to illness or viral infections. Type II diabetes is also known as Non-Insulin dependent diabetes, caused due to reduced sensitivity to insulin, often called as insulin resistance.

Symptoms of diabetes includes, polyurea (excessive urination), polyphagia (excessive intake of food), polydipsia (excessive consumption of liquids due to thirst), ketosis (breakdown of fat into glucose results in accumulation of ketone bodies) in blood. Gluconeogenesis (Conversion of non- carbohydrate) also occur in diabetes.

Question 27.
Explain the mechanism of hormone action?
Answer:
Peptide hormones cannot cross the phospolipid cell membrane and bind to the receptors on the exterior cell surface. They are transported to the golgi, which is the site of modification. It acts as a first messenger in the cell. Hormones on binding to their receptors do not enter target cell but generate the production of second messengers such as cyclic AMP (cAMP), which in turn regulates cellular metabolism.

This is catalyzed by the enzyme adenylate cyclase. The interaction between the hormone at the surface and the effect brought out by cAMP within the cell is known as signaling cascade. At each step there is a possibility of amplification.

  1. One hormone molecule may bind to multiple receptor molecule before it is degraded.
  2. Each receptor may activate several adenylate cyclases each of which make much CAMP.
  3. Thus there is more signal after each step.

The actions of CAMP are terminated by phosphodiesterases. The effect of peptide hormones like insulin, glucagon, somatotropin are uauslly short lived because they work through second messenger system.

Steroid hormones can easily cross the cell membrane, and bind to their receptors, which are intracellular or intranuclear. Upon binding to the receptors, they pair up with another receptor – hormone complex (dimerize). This dimer can then bind to DNA and alter its transcription.

The effect of steroid hormones such as aldosterone, FSH are long lived, as they alter the amount of mRNA and protein in a cell.

Amino acid derived hormones are derived from one or two aminoacid with a few additonal modifications. Thyroid hormone is synthesised from thyrosine and includes the addtion of several iodine atoms. Epinephrine an amino acid derivative may function through second messenger system like peptide hormone or they may actually enter the cell and function like steroid hormones.

Question 28.
Tabulate the major hypothalamic hormones and their functions?
Answer:

S.No

Hormones

Functions

1.Thyrotropin releasing hormone (TRH)Stimulate the secretion of TSH
2.Gonadotropin releasing hormone (GnRH)Stimulates the secretion of FSH
3.Corticotropin releasing hormone (CRH)Stimulates the secretion of ACTH
4.Growth hormone releasing hormone (GHRH)Stimulates the secretion of GH
5.Prolactin releasing hormone (PRH)Stimulates the secretion of Prolactin
6.Luteinizing hormone releasing hormone (LHRH)Stimulates the secretion of LH
7.MSH releasing hormoneStimulates the secretion of MSH
8.Growth hormone-inhibiting hormone (GHIH)Inhibits the secretion of GH
9.Prolactin inhibiting hormone (PIH)Inhibits the secretion of Prolactin
10.MSH inhibiting hormoneInhibits the secretion of MSH

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Samacheer Kalvi 11th Bio Zoology Solutions Chapter 10 Neural Control and Coordination

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Tamilnadu Samacheer Kalvi 11th Bio Zoology Solutions Chapter 10 Neural Control and Coordination

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Samacheer Kalvi 11th Bio Zoology Neural Control and Coordination Text Book Back Questions and Answers

Textbook Evaluation Solved
Choose The Correct Answer
Question 1.
Which structure in the ear converts pressure waves to action potentials?
(a) Tympanic membrane
(b) Organ of Corti
(c) Oval window
(d) Semicircular canal
Answer:
(b) Organ of Corti

Question 2.
Which of the following pairings is correct?
(a) Sensory nerve – afferent
(b) Motor nerve – afferent
(c) Sensory nerve – ventral
(d) Motor nerve – dorsal
Answer:
(a) Sensory nerve – afferent

Question 3.
During synaptic transmission of nerve impulse, neurotransmitter (P) is released from synaptic vesicles by the action of ions (Q)? Choose the correct P and Q?
(a) P = Acetylcholine, Q = Ca+
(b) P = Acetylcholine, Q = Na+
(c) P = GABA, Q = Na+
(d) P = Cholinesterase, Q = Ca++
Answer:
(a) P = Acetylcholine, Q = Ca+

Question 4.
Examine the diagram of the two cell types A and B given below and select the correct option?
Samacheer Kalvi 11th Bio Zoology Solutions Chapter 10 Neural Control and Coordination img 1
(a) Cell-A is the rod cell found evenly all over retina
(b) Cell-A is the cone cell more concentrated in the fovea centralis
(c) Cell-B is concerned with colour vision in bright light
(d) Cell-A is sensitive to bright light intensities
Answer:
(c) Cell-B is concerned with colour vision in bright light

Question 5.
Assertion: The imbalance in concentration of Na+, K+ and proteins generates action potential.
Reason: To maintain the unequal distribution of Na+ and K+, the neurons use electrical energy.
(a) Both Assertion and Reason are true and Reason is the correct explanation of the Assertion.
(b) Both Assertion and Reason are true but the Reason is not the correct explanations of Assertion.
(c) Assertion is true, but Reason is false.
(d) Both Assertion and Reason are false.
Answer:
(a) Both Assertion and Reason are true and Reason ¡s the correct explanation of the Assertion.

Question 6.
Which part of the human brain is concerned with the regulation of body temperature?
(a) Cerebellum
(b) Cerebrum
(c) Medulla oblongata
(d) Hypothalamus
Answer:
(a) Cerebellum

Question 7.
The respiratory centre is present in the …………………..
(a) Medulla oblongata
(b) Hypothalamus
(c) CerebeLlum
(d) Thalamus
Answer:
(a) Medulla oblongata

Question 8.
Match the following human spinal nerves in column-I with their respective number in column-II and choose the correct option?

Column -1

Column – II

PCervical nerves(i)5 pairs
QThoracic nerve(ii)1 pair
RLumbar nerve(iii)12 pairs
SCoccygeal nerve(iv)8 pairs

(P – iv), (Q – iii), (R – i), (S – ii)
(b) (P – iii), (Q – i), ( R – ii), (S – iv)
(c) (P – iv), (Q – i), (R – ii), ( S – iii)
(d) (P – ii), (Q – iv), (R – i), (S – iii)
Answer:
(a) (P -iv)
(b) (Q – iii)
(c) (R – i)
(d) (S – ii)

Question 9.
Which of the following cranial nerve controls the movement of eye ball ?
(a) Trochlear nerve
(b) Optic nerve
(c) Olfactory nerve
(d) Vagus nerve
Answer:
(a) Trochlear nerve

Question 10.
The abundant intracellular cation is ………………….
(a) H+
(b) K+
(c) Na+
(d) Ca++
Answer:
(b) K+

Question 11.
Which of the following statement is wrong regarding conduction of nerve impulse?
(a) In a resting neuron, the axonal membrane is more permeable to K+ ions and nearly impermeable to Na+ ions.
(b) Fluid outside the axon has a high concentration of Na+ ions and low concentration of K+, in a resting neuron.
(c) Ionic gradients are maintained by Na-K pumps across the resting membrane, which transport 3Na+ ions outwards for 2K+ into the cell.
(d) A neuron is polarized only when the outer surface of the axonal membrane possess a negative charge and its inner surface is positively charged.
Answer:
(d) A neuron is polarized only when the outer surface of the axonal membrane possess a negative charge and its inner surface is positively charged.

Question 12.
All of the following are associated with the myelin sheath except ………………….
(a) Faster conduction of nerve impulses
(b) Nodes of Ranvier forming gaps along the axon
(c) Increased energy output for nerve impulse conduction
(d) Saltatory conduction of action potential
Answer:
(c) Increased energy output for nerve impulse conduction

Question 13.
Several statements are given here in reference to cone cells. Which of the following option indicates all correct statements for cone cells?
Statements:
(i) Cone cells are less sensitive in bright light than Rod cells
(ii) They are responsible for colour vision
(iii) Erythropsin is a photo pigment which is sensitive to red colour light
(iv) They are present in fovea of retina
(a) (iii), (ii) and (i)
(b) (ii), (iii) and (iv)
(c) (i), (iii) and (iv)
(d) (i), (ii) and (iv)
Answer:
(b) (ii), (iii) and (iv)

Question 14.
Which of the following statement concerning the somatic division of the peripheral neural system is incorrect?
(a) Its pathways innervate skeletal muscles
(b) Its pathways are usually voluntary
(c) Some of its pathways are referred to as reflex arcs
(d) Its pathways always involve four neurons
Answer:
(d) Its pathways always involve four neurons

Question 15.
When the potential across the axon membrane is more negative than the normal resting potential, the neuron is said to be in a state of …………………
(a) Depolarization
(b) Hyperpolarization
(c) Repolarization
(d) Hypopolarization
Answer:
(b) Hyperpolarization

Question 16.
Why is the blind spot called so?
Answer:
Slightly below the posterior pole of the eye, the optic nerve and the retinal blood vessels enter the eye. This region is devoid of rods and cones. Hence, this region is called blind spot.

Question 17.
Sam’s optometrist tells him that his intraocular pressure is high. What is this condition called and which fluid does it involve?
Answer:
The aqueous humour present in between iris and lens and the cornea and iris is produced and drained at the same rate, maintaining a constant intra ocular pressure of about 16 mm Hg. Any block in the canal of schlemm increases the intra ocular pressure of aqueous humour. This condition is called ‘Glaucoma’. Due to pressure, optic nerve and the retina are compressed. This leads to blindness.

Question 18.
Why are we getting running nose while crying?
Answer:
When we cry, the tears come out of the tear glands under the eyelids and drain through the tear duct that empty into the nose. It mixes with mucus there and the nose runs.

Question 19.
The action potential occurs in response to a threshold stimulus; but not at subthreshold stimuli. What is the name of the principle involved?
Answer:
All or none principle.

Question 20.
Pleasant smell of food urges Ravi to rush into the kitchen. Name the parts of the brain involved in the identification of food and emotional responses to odour?
Answer:
Olfactory nerve carries the sense of smell to the sensory strip present in the cerebrum. The sensory areas are present in the parietal lobe of the cerebrum. The stimuli of smell reaches the mammillary bodies present in the hypothalamus. This produces olfactory reflexes and emotional responses to odour.

Question 21.
Cornea transplant in humans is almost never rejected. State the reason?
Answer:
Cornea does not have blood vessels. Hence there is no possibility of rejection when cornea is transplanted from one person to another person.

Question 22.
At the end of repolarization, the nerve membrane gets hyperpolarized. Why?
Answer:
At the end of repolarization, the membrane potential inside the axolemma becomes negative due to the efflux of K+ ions. When it becomes more negative than the resting potential -70 mV to about – 90mV, it becomes hyperpolarised.

Question 23.
Label the parts of the neuron?
Answer:
Samacheer Kalvi 11th Bio Zoology Solutions Chapter 10 Neural Control and Coordination img 2

  1. Nucleolus
  2. Node of Ranvier
  3. Dendrite
  4. Myelin sheath
  5. Axon
  6. Nucleus

 

Question 24.
The choroid plexus secretes cerebrospinal fluid. List the function of it?
Answer:
Cerebro spinal fluid provides buoyancy to the central nervous system.

  1. It acts as a shock absorber for the brain and spinal cord.
  2. It nourishes the brain cells by transporting food and oxygen.
  3. It carries harmful metabolic wastes from the brain to the blood.
  4. It maintains a constant pressure inside the cranial vessels.

Question 25.
What is the ANS controlling centre? Name the parts that are supplied by the ANS?
Answer:
Hypothalamus is the ANS controlling centre. The Autonomic neural system innervates smooth muscles, glands and cardiac muscle.

Question 26.
Why the limbic system is called the emotional brain? Name the parts of it?
Answer:
The limbic system is a set of components located on both side of the thalamus present in the inner part of the cerebral hemisphere. It includes the olfactory bulbs, cingulate gyrus, mammillary body, amygdala, hippocampus and hypothalamus. The limbic system plays a primary role in the regulation of pleasure, pain, anger, fear, sexual feeling, affection and memory. Hence it is called emotional brain.

Question 27.
Classify receptors based on type of stimuli?
Answer:

Receptors

Stimulus

Effector organs

Mechano receptorsPressure and vibrationMechano receptors are present in the cochlea of the inner ear and the semi circular canal and utriculus
ChemoreceptorsChemicalsTaste buds in the tongue and nasal epithelium
ThermoreceptorsTemperatureSkin
PhotoreceptorsLightRod and cone cells of the retina in the eye

Samacheer Kalvi 11th Bio Zoology Solutions Chapter 10 Neural Control and Coordination

Question 28.
Name the first five cranial nerves, their nature and their functions?
Answer:

Cranial nerves

Nature of nerve

Function

1.Olfactory nerveSensorySense of smell
2.Optic nervesSensorySense of sight
3.Oculomotor nervesMotorMovement of the eye
4.Trochlear nerveMotorRotation of the eye ball
5.Trigeminal nerveSensory and motor (mixed)Functioning of facial parts

Question 29.
The sense of taste is considered to be the most pleasurable of all senses? Describe the structure of the receptor involved with a diagram?
Answer:
Gustatory receptor: The sense of taste is considered to be the most pleasurable of all senses. The tongue is provided with many small projections called papillae which give the tongue an abrasive feel. Taste buds are located mainly on the papillae which are scattered over the entire tongue surface.

Most taste buds are seen on the tongue few are scattered on the soft palate, inner surface of the cheeks, pharynx and epiglottis of the larynx. Taste buds are flask-shaped and consist of 50 – 100 epithelial cells of two major types.

Gustatory epithelial cells (taste cells) and Basal epithelial cells (Repairing cells). Long microvilli called gustatory hairs project from the tip of the gustatory cells and extends through a taste pore to the surface of the epithelium where they are bathed by saliva.

Gustatory hairs are the sensitive portion of the gustatory cells and they have sensory dendrites which send the signal to the brain. The basal cells that act as stem cells, divide and differentiate into new gustatory cells.
Samacheer Kalvi 11th Bio Zoology Solutions Chapter 10 Neural Control and Coordination img 2

Question 30.
Describe the structures of olfactory receptors?
Answer:
The smell receptors are excited by air borne chemicals that dissolve in fluids. The yellow coloured patches of olfactory epithelium form the olfactory organs that are located on the roof of the nasal cavity.

The olfactory epithelium is covered by a thin coat of mucus layer below and olfactory glands bounded connective tissues, above. It contains three types of cells: supporting cells, Basal cells and millions of pin shaped olfactory receptor cells (which are unusual bipolar cells).

The olfactory glands and the supporting cells secrete the mucus. The unmyelinated axons of the olfactory receptor cells are gathered to form the filaments of olfactory nerve [cranial nerve-I] which synapse with cells of olfactory bulb.

The impulse, through the olfactory nerves, is transmitted to the frontal lobe of the brain for identification of smell and the limbic system for the emotional responses to odour.

In-Text Questions Solved

Question 1.
Can you state why some areas of the brain and spinal cord are grey and some are white?
Answer:
Some areas of the brain and spinal cord are grey due to the presence of non-myelinated nerve cells. The white matter has myelinated nerve cells with myelin sheath made of fat.

Question 2.
Human brain is formed of a large number of parts like cerebrum, thalamus, hypothalamus, pons, cerebellum and medulla oblongata. Each part performs some specialized function and all the parts are essential for the survival of a person. Discuss the following statements:
(a) Thalami are called relay centres of the brain.
(b) Damage to medulla may cause the death of organism.
Answer:
(a) Thalamus is composed of grey mater which serves as a relay centre for impulses between the spinal cord, brain stem and cerebrum. Thus acting as a major coordinating centre for sensory and motor signaling. Within the thalamus, information is sorted and edited and plays a key role in learning and memory.

(b) Medulla contains vital centres that control cardio vascular reflexes, respiration and gastric secretions. Therefore, damage to medulla may cause the death of an organism.

Question 3.
Your friend is returning home after his visit to USA. All at home are waiting for his arrival. How would you feel? State the division of ANS that predominates and mention few changes that take place in your body?
Answer:
I would feel excited. Sympathetic neural system of the ANS predominates and the various changes taking place inside the body includes; excess secretion of “adrenaline” to the blood stream from the medulla of the adrenal gland. This in turn, dilates pupil, inhibits salivation, accelerates heartbeat, inhibites digestion, etc.

Question 4.
Name the parts of the organ of equilibrium involved in the following functions?
(a) Linear movement of the body.
(b) Changes in the body position.
(c) Rotational movement of the head.
Answer:
(a) The utricle and saccule are two membranous sacs, found nearest the cochlea and contain equilibrium receptor regions called maculae that are involved in detecting the linear movement of the head.

(b) Two fluids, perilymph and endolymph, respond to the mechanical forces, during changes occurring in body position and acceleration.

(c) The anterior, posterior and lateral canals detects the rotational movement of the head.

Samacheer Kalvi 11th Bio Zoology Neural Control and Coordination Additional Questions & Answers

I. Choose The Correct Answer

Question 1.
Which of the following acts as a phagocytic cells to engulf the foreign particles at the time of any injury to the brain?
(a) Neuron
(b) Neurilemma
(c) Neuroglia
(d) Axolemma
Answer:
(b) Neurilemma

Question 2.
The granular endoplasmic reticulum of the cell body and dendrites are …………………..
(a) Schwann cells
(b) Myelin sheath
(c) Nissl’s granules
(d) Cytoplasm
Answer:
(c) Nissl’s granules

Question 3.
Which of the following has bipolar neurons?
(a) Intemeurons
(b) Cranial nerves
(c) Spinal nerves
(d) Inner ear
Answer:
(d) Inner ear

Question 4.
Which of the following is present more in the extra cellular fluid found outside the axolemma?
(a) Sodium chloride
(b) Potassium
(c) Magnesium phosphate
(d) Organic molecules
Answer:
(a) Sodium chloride

Question 5.
The interior of the cell of the resting neuron is negative due to ……………….
(a) Greater efflux of Na+ outside the cell than,K+ influx into the cell.
(b) Only greater efflux of Na+ outside the cell.
(c) Greater efflux of K+ outside the cell than Na+ influx into the cell.
(d) Only greater efflux of K+ outside the cell.
Answer:
(c) Greater efflux of K+ outside the cell than Na+ influx into the cell.

Question 6.
When a nerve fibre is stimulated, the axolemma is permeable to Na+ ions in which of the following process?
(a) Opening sodium voltage-gate
(b) Opening potassium voltage-gate
(c) Opening sodium voltage-gate and closing potassium voltage-gate
(d) Opening neurolemma
Answer:
(c) Opening sodium voltage-gate and closing potassium voltage-gate

Question 7.
When the membrane potential shoots.rapidly upto +45 mV, it is called the ………………….
(a) Threshold potential
(b) Spike potential
(c) Repolarization
(d) Hyperpolarization
Answer:
(b) Spike potential

Question 8.
When the membrane potential reaches the spike potential, what happens?
(a) The potential again goes to the spike potential
(b) The potential reaches to the threshold potential
(c) The potential falls back towards the resting potential
(d) The potential remains the same
Answer:
(c) The potential falls back towards the resting potential

Question 9.
The subarachnoid space is present in between ………………..
(a) Piamater and arachnoid mater
(b) Arachnoid mater and duramater
(c) Brain and Piamater
(d) Spinal cord and duramater
Answer:
(a) Piamater and arachnoid mater

Question 10.
The hormone melatonin which regulates sleep and wake cycle is secreted by ………………..
(a) Choroid plexus
(b) Pituitary gland
(c) Infundibulum
(d) Pineal body
Answer:
(d) Pineal body

Question 11.
Which of the following plays a key role in learning and memory?
(a) Hypothalamus
(b) Pons varolii
(c) Thalamus
(d) Medulla oblongata
Answer:
(c) Thalamus

Question 12.
Which of the following controls and coordinates the muscular movements and body equilibrium?
(a) Cerebrum
(b) Cerebellum
(c) Pons
(d) Medulla oblongata
Answer:
(b) Cerebellum

Question 13.
The reflex action is effected by ……………..
(a) Brain
(b) Medulla oblongata
(c) Effector organs
(d) Spinal cord
Answer:
(d) Spinal cord

Question 14.
The number of lumbar spinal nerves is ………………..
(a) 8
(b) 12
(c) 5
(d) 1
Answer:
(c) 5

Question 15.
Which of the following is produced at the terminal ends of the post ganglionic fibres at the effector organs of parasympathetic neural system?
(a) Noradrenaline
(b) Acetylcholine
(c) Adrenalin
(d) Melatonin
Answer:
(b) Acetylcholine

Question 16.
The eye lens is made up of long ……………….
(a) Ciliated epithelial cells
(b) Squamous epithelial cells
(c) Germinal epithelial cells
(d) Columnar epithelial cells
Answer:
(d) Columnar epithelial cells

Question 17.
Which of the following absorbs light to prevent internal reflection in the eye?
(a) Sclera
(b) Retina
(c) Chorid
(d) Cornea
Answer:
(c) Chorid

Question 18.
What of the following does not happen in the bright light?
(a) Size of the pupil increases
(b) Size of the pupil decreases
(c) Lens light enters the eye
(d) The circular muscle of the iris contract
Answer:
(a) Size of the pupil increases

Question 19.
The defect hypermetropia can be overcome by using ……………
(a) Concave lens
(b) Convex lens
(c) Cylindrical glass
(d) Surgical procedures
Answer:
(b) Convex lens

Question 20.
Which are the sensory cells in the ear?
(a) Ossicles
(b) Endolymph
(c) Cochlea
(d) Organs of corti
Answer:
(d) Organs of corti

Question 21.
The olfactory impulses are transmitted to the ……………….. lobe of the brain.
(a) Parietal
(b) Temporal
(c) Occipital
(d) Frontal
Answer:
(d) Frontal

Question 22.
Which of the following is a wrong statement?
(a) Gustatory hairs project from the tip of the gustatory cells.
(b) Gustatory cells are sensory portion of the taste.
(c) Basal epithelial cells are stem cells which divide and differentiate into new gustatory cells.
(d) Basal epithelial cells are sensitive portions of the taste.
Answer:
(d) Basal epithelial cells are sensitive portions of the taste.

Question 23.
Which of the following are present in the finger tips and soles of the feet?
(a) Pacinian corpuscles
(b) Meissner’s corpuscles
(c) Ruffini endings
(d) Krause end bulbs
Answer:
(b) Meissner’s corpuscles

Question 24.
The blind spot is called so because
(a) It has only cones
(b) It has only rods
(c) It has neither rods nor cones
(d) It is present beyond lens
Answer:
(c) It has neither rods nor cones

Question 25.
The protein part of the photo pigment is ………………..
(a) Retinal
(b) Opsin
(c) Macula lutea
(d) Fovea centralis
Answer:
(A) Opsin

II. Fill in the Blanks

Question 1.
The structural and functional unit of the nervous system are ………………..
Answer:
Neurons

Question 2.
……………….. neurons that take sensory impulses from the sense organs to the central nervous system.
Answer:
Afferent

Question 3.
The efferent neurons carry ……………….. impulses from the central nervous system to the effector organs.
Answer:
Motor

Question 4.
The plasma membrane covering the neuron is called ………………..
Answer:
Neurilemma

Question 5.
Myelin sheath acts as an ………………..
Answer:
Insulator

Question 6.
The synaptic vesicles of the synaptic knob are filled with ………………..
Answer:
Neurotransmitters

Question 7.
……………….. neurons are found in the retina of the eye, inner ear and the olfactory area of the brain.
Answer:
Bipolar

Question 8.
The neurons which have only one process are called ………………..
Answer:
Unipolar

Question 9.
During the resting potential, the interior of the cell is negative due to greater efflux of ……………….. ions outside the cell.
Answer:
potassium

Question 10.
The normal value of resting membrane potential is ………………..
Answer:
-70 mV

Question 11.
Due to the rate of flow of Na+ ions into the axoplasm, more than the rate of flow of K+ ions to the outside fluid makes the neurilemma ……………….. charged inside.
Answer:
Positively

Question 12.
The threshold potential is ………………..
Answer:
-55 mV

Question 13.
The spike potential is ………………..
Answer:
+45 mV

Question 14.
In the neurilemma, the synaptic vesicles release neurotransmitters into the synaptic cleft by ………………..
Answer:
Exocytosis

Question 15.
……………….. is the layer which is closely adhered to the brain.
Answer:
Piamater

Question 16.
The folds on the surface of the cerebrum are called
Answer:
Gyri

Question 17.
The grooves between the gyri are called ………………..
Answer:
Sulci

Question 18.
The cerebral hemispheres are connected by a tract of nerve fibres called ………………..
Answer:
Corpus callosum

Question 19.
The medulla acts as a nerve tract between cortex and the ………………..
Answer:
Diencephalon

Question 20.
……………….. forms the roof of the diencephalon.
Answer:
Epithalamus

Question 21.
The pineal body secretes the hormone ……………….. which regulates sleep and wake cycle.
Answer:
Melatonin

Question 22.
……………….. is composed of grey matter which serves as a relay centre for impulses between the spinal cord, brain system and cerebrum.
Answer:
Thalamus

Question 23.
……………….. plays a key role in learning and memory.
Answer:
Thalamus

Question 24.
The downward extension of the hypothalamus, ……………….. connects the hypothalamus with the pituitary gland.
Answer:
Infundibulum

Question 25.
……………….. acts as the satiety centre.
Answer:
Hpothalamus

Question 26.
The ……………….. system is called emotional brain.
Answer:
Limbic

Question 27.
……………….. is the second largest part of the brain.
Answer:
Cerebellum

Question 28.
……………….. forms the posterior most part of the brain.
Answer:
Medulla oblongata

Question 29.
……………….. connects the spinal cord with various parts of the brain.
Answer:
Medulla oblongata

Question 30.
……………….. contains vital centres that control cardiovascular reflexes, respiration and gastric secretions.
Answer:
Medulla oblongata

Question 31.
The ……………….. connects the lateral ventricles with the III ventricle.
Answer:
Foramen of Monro/Interventricular foramen

Question 32.
The choroid plexus found in the roof of the ventricles forms ………………..
Answer:
Cerebro spinal fluid

Question 33.
……………….. provide information about position and movements of the body.
Answer:
Proprioceptors

Question 34.
The receptors of taste and smell are called ………………..
Answer:
Chemoreceptors

Question 35.
The sebaceous glands at the base of eyelashes are called ……………….. glands.
Answer:
Ciliary

Question 36.
……………….. is the outermost layer of the eyeball.
Answer:
Sclera

Question 37.
……………….. is the highly vascularized pigmented layer that nourishes all the eye layers.
Answer:
Choroid

Question 38.
……………….. is the coloured protein of the eye lying between the cornea and lens.
Answer:
Iris

Question 39.
The aperture at the centre of the iris is the ………………..
Answer:
Pupil

Question 40.
The ……………….. muscle alters the convexity of the eye lens.
Answer:
Ciliary

Question 41.
The ……………….. optic nerve arises from the
Answer:
Blind spot

Question 42.
The protein part of the photo pigment is ………………..
Answer:
Opsin

Question 43.
Myopia can be corrected by using ……………….. lens.
Answer:
Concave

Question 44.
……………….. is the defect of the eye due to a shortened eyeball or thin lens.
Answer:
Hypermetropia

Question 45.
……………….. is due to the rough curvature of cornea or lens.
Answer:
Astigmatism

Question 46.
The opaqueness of the lens is called ………………..
Answer:
Cataract

Question 47.
……………….. connects the middle ear cavity with the pharynx.
Answer:
Eustachian tube

Question 48.
The scala vestibuli and scala media are separated by a membrane called ………………..
Answer:
Reisner’s membrane

Question 49.
Protruding from the apical part of each hair cell is hair like structures known as ………………..
Answer:
Stereocilia

Question 50.
The hair cells are embedded in a gelatinous otolithic membrane that contains small calcareous particles called ………………..
Answer:
Otoliths

Question 51.
The swollen area of each semicircular canal is called ………………..
Answer:
Ampulla

Question 52.
The tongue has many small projections called ………………..
Answer:
Papillae

Question 53.
……………….. is the largest sense organ.
Answer:
Skin

Question 54.
……………….. are numerous in hairless skin areas such as finger tips and soles of the feet.
Answer:
Meissner’s corpuscles

Question 55.
……………….. detect different textures, temperature, hardness and pain.
Answer:
Pacinian corpuscles

Question 56.
……………….. which lie in the dermis respond to continuous pressure. .
Answer:
Ruffini endings

Question 57.
……………….. are thermoreceptors that sense temperature.
Answer:
Krause end bulbs

Question 58.
Melanocytes are the cells responsible for producing the skin pigment called ………………..
Answer:
Melanin

Question 59.
……………….. is a condition in which the melanin pigment is lost from areas of the skin, causing white patches.
Answer:
Vitiligo (Leucoderma)

Question 60.
The sense of taste is recognized by the ………………..
Answer:
Gustatory receptor

III. Short Answer Questions

Question 1.
What are neurons? What are their functions?
Answer:
The structural and functional unit of the nervous system is neurons. They detect, receive, process and transmit different kinds of stimuli. They are sensory, motor and autonomic in functions.

Question 2.
What are neuroglia?
Answer:
The non-nervous special supporting cells of the nervous tissue are called neuroglia.

Question 3.
Differentiate between afferent neurons and efferent neurons?
Answer:

Afferent Neurons

Efferent Neurons

1. These take sensory impulses to the central nervous system from the sense organs.1. These carry motor impulses from the central nervous system to the effectors.

Question 4.
What are the functions of neuroglia?
Answer:

  1. They provide nourishment to the surrounding neurons.
  2. They involve in the memory process.
  3. They repair the injured tissue due to their dividing and regenerating capacity.
  4. They engulf the foreign particles at the time of any injury to the brain.

 

Question 5.
Distinguish between Axon and Dendrites?
Answer:

Axon

Dendrites

1.An axon is a long fibre that arises from a cone shaped area of the cell body called the Axon hillock and ends at the branched distal end.1. Dendrites are the repeatedly branched short fibres coming out of the cell body.
2.It does not have Nissl’s granules and Golgi apparatus.2. It has Nissl’s granules and Golgi apparatus.
3.It is myelinated.3. It is non-myelinated.

Question 6.
What is neurilemma?
Answer:
The plasma membrane covering the neuron is the neurilemma.

Question 7.
What is axolemma?
Answer:
The plasma membrane covering the axon is the axolemma.

Question 8.
What are Nodes of Ranvier?
Answer:
The Schwann cells covering the axon are not continuous. There are gaps in the myelin sheath between adjacent Schwann cells. These gaps are called Nodes of Ranvier.

Question 9.
What is Synapse?
Answer:
The junction between two neurons is called a Synapse through which a nerve impulse is transmitted.

Question 10.
What is Synaptic Cleft?
Answer:
A small gap between the pre and post synaptic membranes is called Synaptic Cleft. It forms a structural gap and a functional bridge between neurons.

Question 11.
What are meninges?
Answer:
The brain is covered by outer Duramater, the median Arachnoid mater and the inner Piamater. These membranes are called meninges.

Question 12.
What is subdural space?
Answer:
The space between the duramater and arachnoid mater is called subdural space.

Question 13.
What is subarachnoid space?
Answer:
The space between the piamater and arachnoid mater is called subarachnoid space.

Question 14.
What is corpus callosum?
Answer:
The cerebral hemispheres are connected by a tract of nerve fibres called corpus callosum.

Question 15.
What are mammillary bodies?
Answer:
A pair of small rounded body in the hypothalamus is called mammillary bodies. These are involved in olfactory reflexes and emotional response to odour.

Question 16.
What are corpora quadrigemina?
Answer:
The dorsal portion of the mid brain consists of four rounded bodies called corpora quadrigemina. It acts as a reflex centre for vision and hearing.

Question 17.
What is septum pellucidum?
Answer:
A thin membrane which separates the lateral ventricles I and II is called the septum pellucidum.

Question 18.
What is foramen of Monro?
Answer:
The lateral ventricle communicates with the III ventricle in the diencephalon through an opening called interventricular foramen or foramen of Monro.

Question 19.
What is cerebral aqueduct or aqueduct of Sylvius?
Answer:
The ventricle III is continuous with the ventricle IV in the hind brain through a canal called aqueduct of Sylvius or cerebral aqueduct.

Question 20.
What is Choroid plexus?
Answer:
Choroid plexus is a network of blood capillaries found in the roof of the ventricles. This forms cerebro spinal fluid from the blood.

Question 21.
What is cauda equina?
Answer:
The thick bundle of elongated nerve roots within the lower vertebral canal is called the cauda equina.

Question 22.
What are Cranial nerves?
Answer:
The 12 pairs of nerves which arise from the brain are called cranial nerves.

Question 23.
What are spinal nerves?
Answer:
The 31 pairs of nerves which emerge out from the spinal cord through spaces called the intervertebral foramina found between the adjacent vertebrae are the spinal nerves.

Question 24.
What is a mixed nerve?
Answer:
A nerve which contains both sensory (afferent) and motor (efferent) fibres is called a mixed nerve.

Question 25.
What are Exteroceptors?
Answer:
Exteroceptors are located at or near the surface of the body. These are sensitive to external stimuli and receive sensory impulses for hearing, vision, touch, taste and smell.

Question 26.
What are Interoceptors?
Answer:
Interoceptors are located in the visceral organs and blood vessels. These are sensitive to internal stimuli.

Question 27.
What are Lacrymal glands?
Answer:
Tears secreting glands located in the upper lateral region of each orbit are called Lacrymal glands.

Question 28.
What is Lysozyme?
Answer:
The enzyme present in tears which destroys bacteria is lysozyme.

Question 29.
What is canal of schlemm?
Answer:
At the junction of the sclera and the cornea, there is a channel called ‘canal of schlemm’. It continuously drains out the excess of aqueous humour.

Question 30.
What is accommodation?
Answer:
The ability of the eyes to focus objects at varying distances is called accommodation.

Question 31.
What is macula lutea?
Answer:
The yellow flat spot at the centre of the posterior region of the retina is called macula lutea. It is responsible for sharp detailed vision.

Question 32.
What is fovea centralis?
Answer:
A small depression present in the centre of the yellow spot is called fovea centralis.

Question 33.
Write the difference between Rod cells and Cone cells.
Answer:

Rod cells

Cone cells

1. Rods are responsible for vision in dim light.1. The cones are responsible for colour vision and works best in bright light.
2. The pigment present in the rods is rhodopsin, formed of a protein scotopsin and retinal (an aldehyde of vitamin A).2. The pigment present in the cones is photopsin, formed of opsin protein and retinal.
3. There are about 120 million rod cells.3. There may be 6-7 million cone cells.
4. Rods are predominant in the extra fovea region.4. Cones are concentrated in the fovea region.

Question 34.
What are ceruminous glands?
Answer:
The wax producing sebaceous glands in the external auditory meatus are ceruminous glands.

Question 35.
What is Eustachian tube?
Answer:
A tube called Eustachian tube connects the middle ear cavity with the pharynx. It helps in equalizing the pressure of air on either sides of the ear drum.

Question 36.
What is crista ampullaris?
Answer:
The lower end of each semicircular canal is swollen and it is called ampulla. Each ampulla has a sensory area known as crista ampullaris.

Question 37.
What are Meissner’s corpuscles?
Answer:
Meissner’s corpuscles are small light pressure receptors found just beneath the epidermis in the dermal papillae.

Question 38.
What are Pacinian corpuscles?
Answer:
Pacinian corpuscles are the large egg shaped receptors found scattered deep in the dermis and monitoring vibration due to pressure.

Question 39.
What are Krause end bulbs?
Answer:
These are thermoreceptors of the skin that sense temperature.

Question 40.
What are Tactile merkel disc?
Answer:
Tactile merkel disc is light touch receptor lying in the deeper layer of epidermis.

IV. Long Answer Questions

Question 1.
Explain the structure of neuron?
Answer:
A neuron is a microscopic structure composed of three major parts namely cell body (soma), dendrites and axon. The cell body is the spherical part of the neuron that contains all the cellular organelles as a typical cell (except centriole). The plasma membrane covering the neuron is called neurilemma and the axon is axolemma.

The repeatedly branched short fibres coming out of the cell body are called dendrites, which transmit impulses towards the cell body. The cell body and the dendrites contain cytoplasm and granulated endoplasmic reticulum called Nissl’s granules.

An axon is a long fibre that arises from a cone shaped area of the cell body called the Axon hillock and ends at the branched distal end. Axon hillock is the place where the nerve impulse is generated in the motor neurons.

The axon of one neuron branches and forms connections with many other neurons. An axon contains the same organelles found in the dendrites and cell body but lacks Nissl’s granules and Golgi apparatus.

The axon, particularly of peripheral nerves is surrounded by Schwann cells (a type of glial cell) to form myelin sheath, which act as an insulator. Myelin sheath is associated only with the axon; dendrites are always non-myelinated.

Schwann cells are not continuous along the axon; so there are gaps in the myelin sheath between adjacent Schwann cells. These gaps are called Nodes of Ranvier.

Large myelinated nerve fibres conduct impulses rapidly, whereas non-myelinated fibres conduct impulses quite slowly. Each branch at the distal end of the axon terminates into a bulb like structure called synaptic knob which possesses synaptic vesicles filled with neurotransmitters. The axon transmits nerve impulses away from the cell body to inter neural space or to a neuro-muscular junction.

Question 2.
Classify neurons on the basis of number of axon and dendrites?
Answer:
The neurons are divided into three types based on number of axon and dendrites they possess:

  • Multipolar neurons have many processes with one axon and two or more dendrites. They j are mostly intemeurons.
  • Bipolar neurons have two processes with one axon and one dendrite. These are found in the retina of the eye, inner ear and the olfactory area of the brain.
  • Unipolar neurons have a single short process and one axon. Unipolar neurons are located in the ganglia of cranial and spinal nerves.

Samacheer Kalvi 11th Bio Zoology Solutions Chapter 10 Neural Control and Coordination img 4

Question 3.
Tabulate the ionic channels in the axolemma?
Answer:
Samacheer Kalvi 11th Bio Zoology Solutions Chapter 10 Neural Control and Coordination img 5

Question 4.
Explain the transmission of nerve impulses?
Answer:
The transmission of impulse involves two main phases; Resting membrane potential and Action membrane potential. Resting membrane Potential: The electrical potential difference across the plasma membrane of a resting neuron is called the resting potential during which the interior of the cell is negative due to greater efflux of K+ outside the cell than Na+
influx into the cell.

When the axon is not conducting any impulses i.e. in resting condition, the axon membrane is more permeable to K+ and less permeable to Na+ ions, whereas it remains impermeable to negatively charged protein ions. The axoplasm contains high concentration of K+ and negatively charged proteins and low concentration of Na+ ions.

In contrast, fluid outside the axon (ECF) contains low concentration of K+ and high concentration of Na+, and this forms a concentration gradient. This ionic gradient across the resting membrane is maintained by ATP driven Sodium-Potassium pump, which exchanges 3Na+ outwards for 2K+ into the cells.

In this state, the cell membrane is said to be polarized. In neuron, the resting membrane potential ranges from -40 mV to -90 mV, and its normal value is -70 mV. The minus sign indicates that the inside of the cell is negative with respect to the outside.

Action membrane potential:
An action potential occurs when a neuron sends information down an axon, away from the cell body. It includes following phases, depolarization, repolarisation and hypopolarization.

Depolarization – Reversal of polarity:
When a nerve fibre is stimulated, sodium voltage-gate opens and makes the axolemma permeable to Na+ ions; meanwhile the potassium voltage-gate closes. As a result, the rate of flow of Na+ ions into the axoplasm exceeds the rate of flow of K+ ions to the outside fluid [ECF]. Therefore, the axolemma becomes positively charged inside and negatively charged outside.

This reversal of electrical charge is called Depolarization. During depolarization, when enough Na+ ions enter the cell, the action potential reaches a certain level, called threshold potential [-55 mV], The particular stimulus which is able to bring the membrane potential to threshold is called threshold stimulus.

The action potential occurs in response to a threshold stimulus but does not occur at subthreshold stimuli. This is called all or none principle. Due to the rapid influx of Na+ ions, the membrane potential shoots rapidly up to + 45 mV which is called the Spike potential.

Repolarisation [Falling Phase]: When the membrane reaches the spike potential, the sodium voltage-gate closes and potassium voltage-gate opens. It checks influx of Na+ ions and initiates the efflux of K+ ions which lowers the number of positive ions within the cell.’Thus, .the potential falls back towards the resting potential. The reversal of membrane potential inside the axolemma to negative occurs due to the efflux of K+ ions. This is called Repolarisation.

Hyperpolarization:
If repolarization becomes more negative than the resting potential -70 mV to about -90 mV, it is called Hyperpolarization. During this, K+ ion gates are more permeable to K+ even after reaching the threshold level as it closes slowly; hence called Lazy gates. The membrane potential returns to its original resting state when K+ ion channels close completely. During hyperpolarization the Na+ voltage gate remains closed.

Conduction Speed of a nerve impulse: The conduction speed of a nerve impulse depends on the diameter of axon. The greater the axon’s diameter, the faster is the conduction.. The myelinated axon conducts the impulse faster than the non-myelinated axon.

The voltage-gated Na+ and K+ channels are concentrated at the nodes of Ranvier. As a result, the impulse jumps node to node, rather than travelling the entire length of the nerve fibre. This mechanism of conduction is called Saltatory Conduction. Nerve impulses travel at the speed of 1-300 m/s.

Question 5.
Explain the Synaptic transmission?
Answer:
The junction between two neurons is called a Synapse through which a nerve impulse is transmitted. The first neuron involved in the synapse forms the pre-synaptic neuron and the second neuron is the post-synaptic neuron.

A small gap between the pre- and post-synaptic membranes is called Synaptic Cleft that forms a structural gap and a functional bridge between neurons. The axon terminals contain synaptic vesicles filled with neurotransmitters.

When an impulse [action potential] arrives at the axon terminals, it depolarizes the pre-synaptic membrane, opening the voltage-gated calcium channels. Influx of calcium ions stimulates the synaptic vesicles towards the pre-synaptic membrane and fuses with it.

In the neurilemma, the vesicles release their neurotransmitters into the synaptic cleft by exocytosis. The released neurotransmitters bind to their specific receptors on the post-synaptic membrane, responding to chemical signals.

The entry of the ions can generate a new potential in the post-synaptic neuron, which may be either excitatory or inhibitory. Excitatory post-synaptic potential causes depolarization whereas inhibitory post-synaptic potential causes hyperpolarization of post- synaptic membrane.

Question 6.
Write a short note on meninges?
Answer:
Brain is covered by three cranial meninges. The outer thick layer is Duramater which lines the inner surface of the cranial cavity; the median thin layer is Arachnoid mater which is separated from the duramater by a narrow subdural space. The innermost layer is Piamater which is closely adhered to the brain but separated from the arachnoid mater by the subarachnoid space.

Question 7.
Explain the structure of fore brain?
Answer:
Fore brain comprises the following regions: Cerebrum and Diencephalon. Cerebrum is the ‘seat of intelligence’ and forms the major part of the brain. The cerebrum consists of an outer cortex, inner medulla and basal nuclei.

The superficial region of the cerebrum is.called cerebral cortex, which looks grey due to the presence of unmyelinated nerve cells. Cerebral cortex – consists of neuronal cell body, dendrites, associated glial and blood vessels.

The surface of the cerebrum’shows many convolutions (folds) and grooves. The folds are called gyri, the shallow grooves between the gyri are called sulci and deep grooves are called fissures. These sulci and gyri increase the surface area of the cerebral cortex. Several sulci divide the cerebrum into eight lobes; a pair of frontals, parietals, temporals and occipital lobes.

A median longitudinal fissure divides the cerebrum longitudinally into two cerebral hemispheres. A transverse fissure separates the cerebral hemispheres from the cerebellum.

The hemispheres are connected by a tract of nerve fibres called corpus callosum. Cerebral cortex has three functional areas namely sensory areas occur in the parietal, temporal and occipital lobes of the cortex. They receive and interpret the sensory impulses.

Motor area of the cortex which controls voluntary muscular movements lies in the posterior part of the frontal lobes. The areas other than sensory and motor areas are called Association areas that deal with integrative functions such as memory, communications, learning and reasoning. Inner to the cortex is medulla which is white in colour and acts as a nerve tract between the cortex and the diencephalon.

Diencephalon consists largely of following three paired structures. Epithalamus forms the roof of the diencephalon and it is a non-nervous tissue. The anterior part of epithalamus is vascular and folded to form the choroid plexus. Just behind the choroid plexus, the epithalamus forms a short stalk which ends in a rounded body called pineal body which secretes the hormone, melatonin which regulates sleep and wake cycle.

Thalamus is composed of grey mater which serves as a relay centre for impulses between the spinal cord, brain stem and cerebrum. Within the thalamus, information is sorted and edited and plays a key role in learning and memory. It is a major coordinating centre for sensory and motor signalling.

Hypothalamus forms the floor of the diencephalon. The downward extension of the hypothalamus, the infundibulum connects the hypothalamus with the pituitary gland. The hypothalamus contains a pair of small rounded body called mammillary bodies that are involved in olfactory reflexes and emotional responses to odour.

Hypothalamus maintains homeostasis and has many centres which control the body temperature, urge for eating and drinking. It also contains a group of neurosecretory cells which secrete the hypothalamic hormones. Hypothalamus also acts as the satiety centre.

Limbic system: The inner part of the cerebral hemisphere constitutes the limbic system. The main components of limbic system are olfactory bulbs, cingulate gyrus, mammillary body, amygdala, hippocampus and hypothalamus.

The limbic system is called ‘emotional brain’ because it plays a primary role in the regulation of pleasure, pain, anger, fear, sexual feeling and affection. The hippocampus and amygdala also play a role in memory. Brain stem is the part of the brain between the spinal cord and the diencephalon. It consists of mid brain, pons varolii and medulla oblongata.

Question 8.
Explain the mid brain?
Answer:
The mid brain is located between the diencephalon and the pons. The lower portion of the midbrain consists of a pair of longitudinal bands of nervous tissue called cerebral peduncles which relay impulses back and forth between cerebrum, cerebellum; pons and medulla. The dorsal portion of the midbrain consists of four rounded bodies called corpora quadrigemina which acts as a reflex centre for vision and hearing.

Question 9.
Explain the structure of Hind brain?
Answer:
Rhombencephalon forms the hind brain. It comprises of cerebellum, pons varolii and medulla oblongata. Cerebellum is the second largest part of the brain. It consists of two cerebellar hemispheres and central worm shaped part, the vermis. The cerebellum controls and coordinates muscular movements and body equilibrium. Any damage to cerebellum often results in uncoordinated voluntary muScle movements.

Pons varoli lies infront of the cerebellum between the midbrain and the medulla oblongata. The nerve fibres in the pons varolii form a bridge between the two cerebellar hemispheres and connect the medulla oblongata with the other region of the brain. The respiratory nuclei found in the pons cooperate with the medulla to control respiration.

Medulla oblongata forms the posterior most part of the brain. It connects the spinal cord with various parts of the brain. It receives and integrates signals from spinal cord and sends it to the cerebellum and thalamus. Medulla contains vital centres that control cardio vascular reflexes, respiration and gastric secretions.

Question 10.
Explain the Ventricles of the brain?
Answer:
The brain has four hollow, fluid filled spaces. The C- shaped space found inside each cerebral hemisphere forms the lateral ventricles I and II which are separated from each other by a thin membrane called the septum pellucidum. Each lateral .ventricle communicates with the . narrow III ventricle in the diencephalon through an opening called interventricular foramen (foramen of Monro).

The ventricle III is continuous with the ventricle IV in the hind brain through a canal called aqueduct of Sylvius (cerebral aqueduct). Choroid plexus is a network of blood capillaries found in the roof of the ventricles and forms cerebro spinal fluid (CSF) from the blood. CSF provides buoyancy to the CNS structures; CSF acts as a shock absorber for the brain and spinal cord; it nourishes the brain cells by transporting constant supply of food and oxygen; it carries harmful metabolic wastes from the brain to the blood; and maintains a constant pressure inside the cranial vessels.

Question 11.
Explain the structure of Spinal cord?
Answer:
The spinal cord is a long, slender, cylindrical nervous tissue. It is protected by the vertebral column and surrounded by the three membranes as in the brain. The spinal cord that extends from the brain stem into-the vertebral canal of the vertebral column up to the level of 1st or 2nd lumbar vertebra. So the nerve roots of the remaining nerves are greatly elongated to exit the vertebral column at their appropriate space. The thick bundle of elongated nerve roots within the lower vertebral canal is called the cauda equina (horse’s tail) because of its appearance.

In the cross section of spinal cord, there are two indentations: the posterior median sulcus and the anterior median fissure. Although there might be slight variations, the cross section of spinal cord is generally the same throughout its length. In contrast to the brain, the grey matter in the spinal cord forms an inner butterfly shaped region surrounded by the outer white matter.

The grey matter consists of neuronal cell bodies and their dendrites, intemeurons and glial cells. White matter consists of bundles of nerve fibres. In the center of the grey matter there is a central canal which is filled with CSF. Each half of the grey matter is divided into a dorsal horn, a ventral horn and a lateral hom.

The dorsal hom contains cell bodies of intemeurons on which afferent neurons terminate. The ventral hom contains cell bodies of the efferent motor neurons supplying the skeletal muscle.

Autonomic nerve fibres, supplying cardiac and smooth muscles and exocrine glands, originate from the cell bodies found in the lateral horn. In the white matter, the bundles of nerve fibres form two types of tracts namely ascending tracts which carry sensory impulses to the brain and descending tracts which carry motor impulses from the brain to the spinal nerves at various levels of the spinal cord.

The spinal cord shows two enlargements, one in the cervical region and another one in the lumbosacral region. The cervical enlargement serves the upper limb and lumbar enlargement serves the lower limbs.

Question 12.
Write a short note on Reflex action and Reflex arc?
Answer:
Reflex action is a fast, involuntary unplanned sequence of actions that occurs in response to a particular stimulus, e.g., closing the eyelids when dust falls in the eyes, withdrawing hand on touching a hot pan. It is brought about by the spinal cord. The nervous elements involved in carrying out the reflex actions constitute a reflex arc.

Reflex arc-has:-

  1. Sensory receptors: It is a sensory structure that responds to a specific stimulus.
  2. Sensory neuron: This neuron takes sensory impulse to the grey matter of the spinal cord through the dorsal root of the spinal cord.
  3. Interneurons: one or two intemeurons serve to transmit impulses from the sensory neuron to the motor neuron.
  4. Motor neuron: It transmits impulse from the central nervous system to the effector organ.
  5. Effector organs: It may be a muscle or gland which responds to the impulse received.

Samacheer Kalvi 11th Bio Zoology Solutions Chapter 10 Neural Control and Coordination img 6

Question 13.
Explain the type of reflexes?
Answer:
There are two types of reflexes. They are:

1. Unconditional reflex is an inborn reflex for an unconditioned stimulus. It does not need any past experience, knowledge or training to occur; Ex: blinking of an eye when a dust particle is about to fall into it, sneezing and coughing due to foreign particle entering the nose or larynx.

2. Conditioned reflex is a respone to a stimulus that has been acquired by learning. This does not naturally exists in animals. Only an experience makes it a part of the behaviour. Example: excitement of salivary gland on seeing and smelling a food. Thb conditioned reflex was first demonstrated by the Russian physiologist Pavlov in his classical conditioning experiment in a dog. The cerebral cortex controls the conditioned reflex.

Question 14.
Tabulate the Cranial nerves and its function?
Answer:

Cranial nerves

Nature of nerve

Function

1.Olfactory nerveSensorySense of smell
2.Optic nervesSensorySense of sight
3.Oculomotor nervesMotorMovement of the eye
4.Trochlear nerveMotorRotation of the eye ball
5.Trigeminal nerveSensory and motor (mixed)Functioning of facial parts
6.Abducens nerveMotorRotation of the eye ball
7.Facial nerveMixedFunctioning of facial parts
8.Auditory/

Vestibulocochlear nerve

SensoryMaintains the equilibrium of the body /Auditory function
9.Glossopharyngeal nerveMixedTaste and touch
10.VagusMixedRegulation of the visceral organs
11.Spinal accessoryMotorMuscular movement of Pharynx, larynx, neck and shoulder
12.HypoglossalMotorSpeech and swallowing

Question 15.
Explain the peripheral nervous system?
Answer:
Peripheral Neural System (PNS) consists of all nervous tissue outside the CNS. Components of PNS include nerves, ganglia, enteric plexuses and sensory receptors. A nerve is a chord like structure that encloses several neurons inside. Ganglia (singular-ganglion) are small masses of nervous tissue, consisting primarily of neuron cell bodies and are located outside the brain and spinal cord.

Enteric plexuses are extensive networks of neurons located in the walls of organs of the gastrointestinal tract. The neurons of these plexuses help in regulating the digestive system. The specialized structure that helps to respond to changes in the environment i.e. stimuli are called sensory receptor which triggers nerve impulses along the afferent fibres to CNS. PNS comprises of cranial nerves arising from the brain and spinal nerves arising from the spinal cord.

(a) Cranial nerves:
There are 12 pairs of cranial nerves, of which the first two pairs arise from the fore brain and the remaining 10 pairs from the mid brain. Other than the Vagus nerve, which extends into the abdomen, all cranial nerves serve the head and face.

(b) Spinal nerves:
31 pairs of spinal nerves emerge out from the spinal cord through spaces called the intervertebral foramina found between the adjacent vertebrae. The spinal nerves are named according to the region of vertebral column from which they originate

  1. Cervical nerves (8 pairs)
  2. Thoracic nerves (12 pairs)
  3. Lumbar nerves (5 pairs)
  4. Sacral nerves (5 pairs)
  5. Coccygeal nerves (1 pair)

Each spinal nerve is a mixed nerve containing both afferent (sensory) and efferent (motor) fibres. It originates as two roots: 1) a posterior dorsal root with a ganglion outside the spinal cord and 2) an anterior ventral root with no external ganglion.

Somatic neural system (SNS) or voluntary neural system is the part of the peripheral neural system associated with the voluntary control of body movements via skeletal muscles. The sensory and motor nerves that innervate striated muscles form the somatic neural system. Major functions of the somatic neural system include voluntary movement of the muscles and organs, and reflex movements.

Question 16.
Explain the Autonomic Neural System?
Answer:
The autonomic neural system is auto functioning and self governed. It is a part of peripheral neural system that innervates smooth muscles, glands and cardiac muscle. This system controls and coordinates the involuntary activities of various organs. ANS controlling centre is in the hypothalamus.

Autonomic neural system comprises the following components:

  1. Preganglionic neuron whose cell body is in the brain or spinal cord; its myelinated axon exits the CNS as part of cranial or spinal nerve and ends in an autonomic ganglion.
  2. Autonomic ganglion consists of axon of preganglionic neuron and cell bodies of postganglionic neuron.
  3. Postganglionic neuron conveys nerve impulses from autonomic ganglia to visceral effector organs.
  4. The autonomic neural system consists of Sympathetic neural system and Parasympathetic neural system.

Question 17.
Tabulate the differences between sympathetic and parasympathetic neural system?
Answer:

Sympathetic Neural system (SNS)

Parasympathetic Neural system (PNS)

1.SNS originates in the thoracic and lumbar region of the spinal cord.PNS originates in the cranial region of the brain and the sacral region of the spinal cord.
2.Sympathetic ganglia are linked up to form a chain.Its ganglia remain isolated.
3.Preganglionic fibres are short and the postganglionic fibres are long.Preganglionic fibres are long and the postganglionic fibres are short.
4.Noradrenaline is produced at the terminal ends of the postganglionic fibres at the effector organs. Hence the system is adrenergic.Acetylcholine is produced at the terminal ends of the postganglionic fibres at the effector organs. Hence the system is cholinergic.
5.Active during stressful conditions preparing the body to face them.Active during relaxing times restoring normal activity after a stress.
6.The overall effect is excitatory and stimulating.The overall effect is inhibitory.
7.It is considered as the flight or fight system.It is considered as ‘The Rest and Digest System’ or ‘The Feed and Breed System’.

 

Question 18.
Explain the structure of Photoreceptor?
Answer:
Eye is the organ of vision; located in the orbit of the skull and held in its position with the help of six extrinsic muscles. They are superior, inferior, lateral, median rectus muscles, superior oblique and inferior oblique muscles.

These muscles aid in movement of the eyes and they receive their nerve innervation from III, IV and VI cranial nerves. Eyelids, eye lashes and eye brows are the accessory structures useful in protecting the eyes.

The eye lids protect the eyes from excessive light and foreign objects and spread lubricating secretions over the eyeballs. Eyelashes and the eyebrows help to protect the eyeballs from foreign objects, perspiration and also from the direct rays of sunlight.

Sebaceous glands at the base of the eyelashes are called ciliary glands which secrete a lubricating fluid into the hair follicles. Lacrymal glands, located in the upper lateral region of each orbit, secrete tears. Tears are secreted at the rate of 1 ml/day and it contains salts, mucus and lysozyme enzyme to destroy bacteria.

The conjunctiva is a thin, protective mucous membrane found lining the outer surface of the eyeball. The eye has two compartments, the anterior and posterior compartments. The anterior compartment has two chambers, first one lies between the cornea and iris and the second one lies between the iris and lens.

These two chambers are filled with watery fluid called aqueous humor. The posterior compartment lies between the lens and retina and it is filled with a jelly like fluid called vitreous humor that helps to retain the spherical nature of the eye. Eye lens is transparent and biconvex, made up of long columnar epithelial cells called lens fibres. These cells are accumulated with the proteins called crystalline.

The eye ball is spherical in nature. The anterior one- sixth of the eyeball is exposed; the remaining region is fitted well into the orbit. The wall of the eye ball consists of three layers: fibrous Sclera, vascular Choroid and sensory Retina.

The outer coat is composed of dense non-vascular connective tissue. It has two regions: the anterior cornea and the posterior sclera. Cornea is a non-vascular transparent coat formed of stratified squamous epithelium which helps the cornea to renew continuously as it is very vulnerable to damage from dust.

Sclera forms the white of the eye and protects the eyeball. Posteriorly the sclera is innervated by the optic nerve. At the junction of the sclera and the cornea, is a channel called ‘canal of schlemm’ which continuously drains out the excess of aqueous humor.

Samacheer Kalvi 11th Bio Zoology Solutions Chapter 10 Neural Control and Coordination img 7

Choroid is highly vascularized pigmented layer that nourishes all the eye layers and its pigments absorb light to prevent internal reflection. Anteriorly the choroid thickens to form the ciliary body and iris. Iris is the coloured portion of the eye lying between the cornea and lens. The aperture at the centre of the iris is the pupil through which the light enters the inner chamber. Iris is made of two types of muscles the dilator papillae (the radial muscle) and the sphincter papillae (the circular muscle).

In the bright light, the circular muscle in the iris contract; so that the size of pupil decreases and less light enters the eye. In dim light, the radial muscle in the iris contract; so that the pupil size increases and more light enters the eye. Smooth muscle present in the ciliary body is called the ciliary muscle which alters the convexity of the lens for near and far vision.

The ability of the eyes to focus objects at varying distances is called accommodation which is achieved by suspensory ligament, ciliary muscle and ciliary body. The suspensory ligament extends from the ciliary body and helps to hold the lens in its upright position. The ciliary body is provided with blood capillaries that secrete a watery fluid called aqueous humor that fills the anterior chamber.

Retina forms the inner most layer of the eye and it contains two regions:-

  • A sheet of pigmented epithelium (non visual part) and neural visual regions.
  • The neural retina layer contains three types of cells:-
  • photoreceptor cells – cones and rods, bipolar cells and ganglion cells.
  • The yellow flat spot at the centre of the posterior region of the retina is called macula lutea which is responsible for sharp detailed Vision.

A small depression present in the centre of the yellow spot is called fovea centralis which contains only cones. The optic nerves and the retinal blood vessels enter the eye slightly below the posterior pole, which is devoid of photo receptors; hence this region is called blind spot.

Question 19.
Explain the mechanism of vision?
Answer:
When light enters the eyes, it gets refracted by the cornea, aqueous humor and lens and it is focused on the retina and excites the rod and cone cells. The photo pigment consists of Opsin, the protein part and Retinal, a derivative of vitamin A. Light induces dissociation of retinal from opsin and causes the structural changes in opsin.

This generates an action potential in the photoreceptor cells and is transmitted by the optic nerves to the visual cortex of the brain, via bipolar cells, ganglia and optic nerves, for the perception of vision.

Question 20.
Explain the Refractive errors of eye or defects of human eye?
Answer:
Myopia (near sightedness): The affected person can see the nearby objects but not the distant objects. This condition may result due to an elongated eyeball or thickened lens; so that the image of distant object is formed in front of the yellow spot. This error can be corrected using concave lens that diverge the entering light rays and focuses it on the retina.

Hypermetropia (long sightedness): the affected person can see only the distant objects clearly but not the objects nearby. This condition results due to a shortened eyeball and thin lens; so the image of closest object is converged behind the retina. This defect can be overcome by using convex lens that converge the entering light rays on the retina.

Presbyopia:
Due to aging, the lens loses elasticity and the power of accommodation. Convex lenses are used to correct this defect. Astigmatism is due to the rough (irregular) curvature of cornea or lens. Cylindrical glasses are used to correct this error.

Cataract:
Due to the changes in nature of protein, the lens becomes opaque. It can be corrected by surgical procedures.

Question 21.
Explain the structure of human ear or Phonoreceptor?
Answer:
The ear is the site of reception of two senses namely hearing and equilibrium. Anatomically, the ear is divided into three regions: the external ear, the middle ear and internal ear.

The external ear consists of pinna, external auditory meatus and ear drum. The pinna is flap of elastic cartilage covered by skin. It collects the sound waves. The external auditory meatus is a curved tube that extends up to the tympanic membrane [the ear drum].

The tympapic membrane is composed of connective tissues covered with skin outside and with mucus membrane inside. There are very fine hairs and wax producing sebaceous glands called ceruminous glands in the external auditory meatus. The combination of hair and the ear wax [cerunien] helps in preventing dust and foreign particles from entering the ear.

The middle ear is a small air-filled cavity in the temporal bone. It is separated from the external ear by the eardrum and from the internal ear by a thin bony partition; the bony partition contains two small membrane covered openings called the oval window and the round window.

The middle ear contains three ossicles: malleus [hammer bone], incus [anyil bone] and stapes [stirrup bone] which are attached to one another. The malleus is attached to the tympanic membrane and its head articulates with the incus which is the intermediate bone lying between the malleus and stapes.

The stapes is attached to the oval window in the inner ear. The ear ossicles transmit sound waves to the inner ear. A tube called Eustachian tube connects the middle ear cavity with the pharynx. This tube helps in equalizing the pressure of air on either sides of the ear drum.

Inner ear is the fluid filled cavity consisting of two parts, the bony labyrinth and the membranous labyrinths. The bony labyrinth consists of three areas: cochlea, vestibule and semicircular canals.

The cochlea is a coiled portion consisting of 3 chambers namely: scala vestibuli and scala tympani- these two are filled with perilymph; and the scala media is filled with endolymph. At the base of the cochlea, the scala vestibule ends at the ‘oval window’ whereas the scala tympani ends at the ‘round window’ of the middle ear.

The chambers scala vestibuli and scala media are separated by a membrane called Reisner’s membrane whereas the scala media and scala tympani are separated by a membrane called Basilar membrane. Organ of corti.

The organ of corti is a sensory ridge located on the top of the Basilar membrane and it contains numerous hair cells that are arranged in four rows along the length of the basilar membrane. Protruding from the apical part of each hair cell is hair like structures known as stereocilia. During the conduction of sound wave, stereocilia makes a contact with the stiff gel membrane called tectorial membrane, a roof like structure overhanging the organ of corti throughout its length.

Question 22.
Explain the mechanism of hearing?
Answer:
Sound waves entering the external auditory meatus fall on the tympanic membrane. This causes the ear drum to vibrate, and these vibrations are transmitted to the oval window through the three auditory ossicles. Since the tympanic membrane is 17-20 times larger than the oval window, the pressure exerted on the oval window is about 20 times more than that on the tympanic membrane.

This increased pressure generates pressure waves in the fluid of perilymph. This pressure causes the round window to alternately bulge outward and inward meanwhile the basilar membrane along with the organ of Corti move up and down. These movements of the hair alternately open and close the mechanically gated ion channels in the base of hair cells and the action potential is propagated to the brain as sound sensation through cochlear nerve.

Question 23.
Write a short note on defects of ear?
Answer:
Deafness may be temporary or permanent. It can be further classified into conductive deafness and sensory-neural deafness. Possible causes for conductive deafness may be due to

  1. Blockage of ear canal with earwax
  2. Rupture of eardrum
  3. Middle ear infection with fluid accumulation
  4. Restriction of ossicular movement.

In sensory-neural deafness, the defect may be in the organ of Corti or the auditory nerve or in the ascending auditory pathways or auditory cortex.

Question 24.
Explain the organ of equilibrium or proprioception?
Answer:
Balance is part of a sense called proprioception, which is the ability to sense the position, orientation and movement of the body. The organ of balance is known as the vestibular system which is located in the inner ear next to the cochlea.

The vestibular system is composed of a series of fluid filled sacs and tubules.These sacs and tubules contain endolymph and are kept in the surrounding perilymph. These two fluids, perilymph and endolymph, respond to the mechanical forces, during changes occurring in body position and acceleration.

The utricle and saccule are two membraiious sacs, found nearest the cochlea and contain equilibrium receptor regions called maculae that are involved in detecting the linear movement of the head.

The maculae contain the hair cells that act as mechanoreceptors. These hair cells are embeded in a gelatinous otolithic membrane that contains small calcareous particles called otoliths. This membrane adds weight to the top of the hair cells and increase the inertia.

The canals that lie posterior and lateral to the vestibule are semicircular canals; they are anterior, posterior and lateral canals oriented at right angles to each other.

At one end of each semicircular canal, at its lower end has a swollen area called ampulla. Each ampulla has a sensory area known as crista ampullaris which is formed of sensory hair cells and supporting cells. The function of these canals is to detect rotational movement of the head.

Question 25.
Explain the sensory function of skin?
Answer:
Skin is the sensory organ of touch and is also the largest sense organ. This sensation conies from millions of microscopic sensory receptors located all over the skin and associated with the general sensations of contact, pressure, heat, cold and pain. Some parts of the body, such as the finger tips have a large number of these receptors, making them more sensitive. Some of the sensory receptors present in the skin are:

  • Tactile merkel disc is light touch receptor lying in the deeper layer of epidermis.
  • Hair follicle receptors are light touch receptors lying around the hair follicles.
  • Meissner’s corpuscles are small light pressure receptors found just beneath the epidermis in the dermal papillae. They are numerous in hairless skin areas such as finger tips and soles of the feet.
  • Pacinian corpuscles are the large egg shaped receptors found scattered deep in the dermis and monitoring vibration due.to pressure. It allows to detect different textures, temperature, hardness and pain.
  • Ruffini endings which lie in the dermis responds to continuous pressure.
  • Krause end bulbs are thermoreceptors that sense temperature.

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Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

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Samacheer Kalvi 11th Bio Botany Plant Growth and Development Text Book Back Questions and Answers

Question 1.
Select the wrong statement from the following:
(a) Formative phase of the cells retain the capability of cell division.
(b) In elongation phase development of central vacuole takes place.
(c) In maturation phase thickening and differentiation takes place.
(d) In maturation phase, the cells grow further.
Answer:
(d) In maturation phase, the cells grow further.

Question 2.
If the diameter of the pulley is 6 inches, length of pointer is 10 inches and distance travelled by pointer is 5 inches. Calculate the actual growth in length of plant:
(a) 3 inches
(b) 6 inches
(c) 12 inches
(d) 30 inches
Answer:
(a) 3 inches

Question 3.
In uni sexual plants, sex can be changed by the application of:
(a) ethanol
(b) cytokinins
(c) ABA
(d) auxin
Answer:
(c) ABA

Question 4.
Select the correctly matched one:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development
(a) A – (iii), B – (iv), C – (v), D – (vi), E – (i), F – (ii)
(b) A – (v), C – (ii), D – (iv), E – (vi), F – (iii)
(c) A – (iii), B – (v), C – (vi),D – (i), E – (ii), F – (iv)
(d) A – (ii), B – (iii), C – (v), D – (vi), E – (iv), F – (i)
Answer:
(b) A – (v), C – (ii), D – (iv), E – (vi), F – (iii)

Question 5.
Seed dormancy allows the plants to:
(a) overcome un favorable climatic conditions
(b) develop healthy seeds
(c) reduce viability
(d) prevent deterioration of seeds
Answer:
(a) overcome unfavorable climatic conditions

Question 6.
What are the parameters used to measure growth of plants?
Answer:

  • Increase in length or girth (roots and stems)
  • Increase in fresh or dry weight
  • Increase in area or volume (fruits and leaves)
  • Increase in number of cells produced.

Question 7.
What is plasticity?
Answer:
Plants follow different pathways in response to environment or phases of life to form different kinds of structures. This ability is called plasticity,
eg : Heterophylly in cotton and coriander. In such plants, the leaves of the juvenile plant are different in shape from those in mature plants.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development
On the other hand, the difference in shapes of leaves produced in air and those produced in water in buttercup also represent he heterophyllous development due to the environment. This phenomenon of heterophylly is an example of plasticity.

Question 8.
Write the physiological effects of Cytokinins.
Answer:

  1. Cytokinin promotes cell division in the presence of auxin (IAA).
  2. Induces cell enlargement associated with IAA and gibberellins
  3. Cytokinin can break the dormancy of certain light-sensitive seeds like tobacco and induces seed germination.
  4. Cytokinin promotes the growth of lateral bud in the presence of apical bud.
  5. Application of cytokinin delays the process of aging by nutrient mobilization. It is known as Richmond Lang effect.
  6. Cytokinin:
    • increases rate protein synthesis
    • induces the formation of inter-fascicular cambium
    • overcomes apical dominance
    • induces formation of new leaves, chloroplast and lateral shoots.
  7. Plants accumulate solutes very actively with the help of cytokinins.

Question 9.
Describe the mechanism of photoperiodic induction of flowering.
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development
Photoperiodic stimulus is perceived by the leaves. Floral hormone is synthesised in leaves and translocated to the apical tip to promote flowering. This can be explained by a simple experiment on Cocklebur (Xanthium pensylvanicum), a short day plant. Usually Xanthium will flower under short day conditions. If the plant is defoliated and kept under short day conditions it will not flower.

Flowering will occur even when all the leaves are removed except one leaf. If a cocklebur plant is defoliated and kept under long day conditions, it will not flower. If one of its leaves is exposed to short day condition and rest are in long day condition, flowering will occur.

Question 10.
Give a brief account on Pr grammed Cell Death (PCD).
Answer:
Senescence is controlled by plants own genetic program and death of the plant or plant part consequent to senescence is called Programmed Cell Death. In short senescence of an individual cell is called PCD. The proteolytic enzymes involving PCD in plants are phytases and in animals are caspases. The nutrients and other substrates from senescing cells and tissues are remobilized and reallocated to other parts of the plant that survives.

The protoplasts of developing xylem vessels and tracheids die and disappear at maturity to make them functionally efficient to conduct water for transport. In aquatic plants, aerenchyma is normally formed in different parts of the plant such as roots and stems which encloses large air spaces that are created through PCD. In the development of unisexual flowers, male and female flowers are present in earlier stages, but only one of these two completes its development while other aborts through PCD.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Samacheer Kalvi 11th Bio Botany Plant Growth and Development Additional Questions & Answers

I. Choose T he correct answer (1 Mark)
Question 1.
Open form of the growth occurs in:
(a) leaves and flowers
(b) stem and root
(c) leaves and stem
(d) stem and flowers
Answer:
(b) stem and root

Question 2.
Bamboo is classified under:
(a) monocarpi c annual plants
(b) polycarpic perennials
(c) monocarpic perennials
(d) polycarpic annual plants
Answer:
(c) monocarpic perennials

Question 3.
Primary growth of the plant is due to the activity of:
(a) phloem parenchyma
(b) phloem meristem
(c) vascular cambium
(d) apical meristem
Answer:
(d) apical meristem

Question 4.
One single maize root apical meristem can give rise to more than:
(a) 17,500 hew cells per hour
(b) 18,500 new cells per hour
(c) 19,000 new cells per hour
(d) 500 new cells per hour
Answer:
(a) 17,500 hew cells per hour

Question 5.
Thickening and differentiation of cells take place during:
(a) elongation phase
(b) formative phase
(c) maturation phase
(d) flowering phase
Answer:
(c) maturation phase

Question 6.
When die total growth of a plant is plotted against time, the shape of the curve obtained is:
(a) hyperbolic curve
(b) ‘S’ shaped sigmoid curve
(c) linear curve
(d) none of the above
Answer:
(b) ‘S’ shaped sigmoid curve

Question 7.
The total growth of the plant consists of four phases in the following order.
(a) Log phase, lag phase, decelerating phase and maturation phase
(b) Log phase, lag phase, maturation phase and decelerating phase
(c) Lag phase, log phase, maturation phase and decelerating phase
(d) Lag phase, log phase, decelerating phase and maturation phase
Answer:
(d) Lag phase, log phase, decelerating phase and maturation phase

Question 8.
Internal factors, that influences the growth of the plant is:
(a) nutrition
(b) light
(c) C / N ratio
(d) oxygen
Answer:
(c) C / N ratio

Question 9.
Absence of light may lead to yellowish in color in plants and this is called:
(a) venation
(b) etiolation
(c) estivation
(d) vernation
Answer:
(b) etiolation

Question 10.
Differentiated cells, after multiplication again lose the ability to divide and mature to perform specific functions. This is called:
(a) plasticity
(b) differentiation
(c) dedifferentiation
(d) redifferentiation
Answer:
(d) redifferentiation

Question 11.
Indicate a plant growth regulator from the following:
(a) cytocin
(b) cytokinins
(c) acetic acid
(d) methylene
Answer:
(b) cytokinins

Question 12.
Some of the polyamines are known to behave like:
(a) growth inhibitors
(b) plant hormones
(c) flowering inhibitors
(d) fruit ripening agent
Answer:
(b) plant hormones

Question 13.
The activity of synergistic effect involves the activity of:
(a) auxin and gibberellins
(b) auxin and ethylene
(c) ABA and gibberellins
(d) none of the above
Answer:
(a) auxin and gibberellins

Question 14.
Phytohormones are usually produced to in tips of:
(a) root alone
(b) stem alone
(c) leaves alone
(d) root, stem and leaves
Answer:
(d) root, stem and leaves

Question 15.
The term auxin was first coined by:
(a) Charles Darwin
(b) Kogl
(c) F.W. Went
(d) Smith
Answer:
(c) F.W. Went

Question 16.
Indole Acetic Acid (IAA) is a:
(a) growth inhibitor
(b) hetero auxin
(c) root inhibitor
(d) synthetic auxin
Answer:
(b) hetero auxin

Question 17.
Indicate a synthetic auxin.
(a) Indole Acetic Acid
(b) Phenyl Acetic Acid
(c) Indole Butyric Acid
(d) Napthalene Acetic Acid
Answer:
(d) Napthalene Acetic Acid

Question 18.
Auxin has a similar chemical structure of:
(a) Indole acetic acid
(b) Napthalene acetic acid
(c) Phenyl acetic acid
(d) 2,4 – Dichloro phenoxy
Answer:
(a) Indole acetic acid

Question 19.
Auxin stimulates:
(a) transpiration
(b) respiration
(c) flowering
(d) none of the above
Answer:
(b) respiration

Question 20.
The term gibberllin was named by:
(a) Brain
(b) Yabuta
(c) Sumiki
(d) kurosawa
Answer:
(b) Yabuta

Question 21.
Who established the structure of gibberellic acid?
(a) Brain etal
(b) Kurosawa
(c) Cross et al
(d) Yabuta and Sumiki
Answer:
(c) Cross et al

Question 22.
Formation of seedless fruits without fertilization is induced by:
(a) auxin
(b) cytokinin
(c) ethylene
(d) gibberellin
Answer:
(d) gibberellin

Question 23.
Cytokinins inducing cell division was first demonstrated by:
(a) Haberlandt
(b) Charles Darwin
(c) Clarke
(d) Hubert
Answer:
(a) Haberlandt

Question 24.
Zeatin is first isolated from unripe grains of:
(a) paddy
(b) wheat
(c) maize
(d) com
Answer:
(c) maize

Question 25.
Indicate correct statements.
(i) Genes are intracellular factors for growth.
(ii) Temperature has no role in the growth of plant.
(iii) Oxygen has a vital role in the growth of plants.
(iv) CIN ratio of soil does not affect the growth of plant.
(a) (i) and (iv)
(b) (ii) and (iv)
(c) (i) and (iii)
(d) (ii) and (iii)
Answer:
(c) (i) and (iii)

Question 26.
Aspartic acid is classified under:
(a) freeauxin
(b) precursor of auxin
(c) chemical structure of auxin
(d) bound auxin
Answer:
(d) bound auxin

Question 27.
The stress phytohormones (Abscisic acid) was first isolated by:
(a) Linn et al
(b) Addicott et al
(c) Edward et al
(d) Stone and Black
Answer:
(b) Addicott et al

Question 28.
The chemical structure of abscisic acid resembles the structure of:
(a) indole Acetic Acid
(b) malanic acid
(c) carotenoid
(d) xanthophyll
Answer:
(c) carotenoid

Question 29.
Pick out the correct statement from the following:
(i) Abscisic acid is found abundantly inside the chloroplast of green cells.
(ii) ABA is a powerful growth promotor.
(iii) ABA is formed from pentose phosphate pathway.
(iv) ABA has anti-auxih and anti-gibberellin property.
(a) (i) and (iv)
(b) (i) and (ii)
(c) (ii) and (iii)
(d) (ii) and (iv)
Answer:
(a) (i) and (iv)

Question 30.
Abscisic acid induces male flower formation on female plants of:
(a) potato
(b) Cannabis sativa
(c) Vinca rosea
(d) Delomix regia
Answer:
(b) Cannabis sativa

Question 31.
Pea and barley are classified under:
(a) short day plants
(b) short long day plants
(c) long day plants
(d) long short day plants
Answer:
(c) long day plants

Question 32.
The term ‘photoperiodism’ was coined by:
(a) Miller and Amald
(b) Gamer and Allard
(c) Michael and Edward
(d) Darwin and Lamark
Answer:
(b) Gamer and Allard

Question 33.
Usually Xanthiumpensylvanicum will flower under:
(a) long day condition
(b) short long day condition
(c) photoneutral condition
(d) short day condition
Answer:
(d) short day condition

Question 34.
Phytochrome is a:
(a) reddish xanthophyll pigment
(b) bluish biliprotein pigment
(c) rodopsin pigment
(d) none of the above
Answer:
(b) bluish biliprotein pigment

Question 35.
Who found out the phytochrome in plants?
(a) Butler et al
(b) Michell et al
(c) Boumick et al
(d) Gamers and Allard
Answer:
(a) Butler et al

Question 36.
The term “vernalization” was first used by:
(a) Gamer
(b) Michell
(c) Lysenko
(d) Kawasacki
Answer:
(c) Lysenko

Question 37.
Pick out the wrong statement from the following:
(a) Vernalization increases the cold resistance of plants
(b) It increase the resistance of plants to fungal disease
(c) Vemalizatiqn increase the vegetative period of the plant
(d) It accelerates the plant breeding
Answer:
(c) Vemalizatiqn increase the vegetative period of the plant

Question 38.
In Oxalis, the seed viability ranges from:
(a) 10 to 15 years
(b) a few days
(c) more than 100 years
(d) upto 100 years
Answer:
(b) a few days

Question 39.
In apple and plum, the method of breaking seed dormancy involves the process of:
(a) impaction
(b) Scarification
(c) exposing to red light
(d) Stratification
Answer:
(d) Stratification

Question 40.
The proteolytic enzymes involved in – programmed cell death in plants are:
(a) phytochrome
(b) caspases
(c) phytaspases
(d) protolysis
Answer:
(c) phytaspases

II. Answer the following (2 Marks)

Question 1.
Define closed form of growth in plants.
Answer:
Leaves, flowers and fruits are limited in growth or of determinate or closed form growth.

Question 2.
What is meant by grand period of growth in plants?
Answer:
The total period from initial to the final stage of growth is called the grand period of growth. The total growth is plotted against time and ‘S’ shaped sigmoid curve (Grand period curve) is obtained.

Question 3.
Name the phases of growth in ‘S’ shaped growth curve.
Answer:

  • Lag phase
  • Log phase
  • Decelerating phase
  • Maturation phase

Question 4.
Define arithmetic growth rate in plant organ.
Answer:
If the length of a plant organ is plotted against time, it shows a linear curve and this growth is called arithmetic growth.

Question 5.
Distinguish between absolute growth rate and relative growth, rate.
Answer:
Absolute growth rate:
Increase in total growth of two organs measured and compared per unit time is called absolute growth rate.

Relative growth rate:
The growth of the given system per unit time expressed per unit initial parameter is called relative growth rate.

Question 6.
Define the term etiolation
Answer:
Light has its own contribution in the growth of the plant. Light is important for growth and photosynthesis. Light stimulates healthy growth. Absence of light may lead to yellowish in colour. This is called etiolation.

Question 7.
What is meant by redifferentiation of plant cells?
Answer:
Differentiated cells, after multiplication again lose the ability to divide and mature to perform specific functions. This is called redifferentiation, eg: Secondary xylem and Secondary phloem.

Question 9.
Mention any two synthetic auxins.
Answer:

  • 2,4 – Dichloro Phenoxy Acetic Acid (2,4 – D)
  • 2, 4, 5 – Trichloro Phenoxy Acetic Acid (2,4,5 – T)

Question 10.
Explain the synergistic effect of phytochromes.
Answer:
The effect of one or more substance in such a way that both promote each others activity, eg: Activity of auxin and gibberellins or cytokinins.

Question 11.
Name the natural auxins present in plants.
Answer:

  • Indole Acetic Acid (IAA)
  • Indole Propionic Acid (IPA)
  • Indole Butyric Acid (IBA)
  • Phenyl Acetic Acid (PAA)

Question 12.
Mention any two physiological effect of auxins in plant.
Answer:

  • They promote cell elongation in stem and coleoptile.
  • At higher concentrations auxins inhibit the elongation of roots but induce more lateral roots. Promotes growth of root only at extremely low concentrations.

Question 13.
Match the following.

(i) Indole acetic acid(a) bolting
(ii) Napthalene acetic acid(b) anti-auxin
(iii) Gibberellins(c) synthetic auxin
(iv) Abscisic acid(d) Natural auxin

Answer:
(i) – (d) Natural auxin
(ii) – (c) synthetic auxin
(iii) – (a) bolting
(iv) – (b) anti-auxin

Question 14.
Where do you find cytokinin hormone in plants?
Answer:
The distribution of cytokinin in plants is not as wide as those of auxin and gibberellins but found mostly in roots. Cytokinins appear to be translocated through xylem.

Question 15.
What is Richmond Lang effect?
Answer:
Application of cytokinin delays the process of aging by nutrient mobilization. It is known as Richmond Lang effect.

Question 16.
What is meant by non-climacteric fruits?
Answer:
All fruits cannot be ripened by exposure to ethylene. Such fruits are called nonclimacteric fruits and are insensitive to ethylene, eg: Grapes, Watermelon, Orange.

Question 17.
Why do you call Abscisic acid (ABA) as stress hormone?
Answer:
It inhibits the shoot growth and promotes growth of root system. This character protect the plants from water stress. Hence, ABA is called as stress hormone.

Question 18.
What is meant by short day plants?
Answer:
The plants that require a short critical day length for flowering are called short day plants or long night plants, eg: Tobacco, Cocklebur, Soybean, Rice and Chrysanthemum.

Question 19.
Write down the importance of photoperiodism in plants.
Answer:

  • The knowledge of photoperiodism plays an important role in hybridisation experiments.
  • Photoperiodism is an excellent example of physiological pre-conditioning that is using an external factor to induce physiological changes in the plant.

Question 20.
Define the term vernalization.
Answer:
Besides photoperiod certain plants require a low temperature exposure in their earlier stages for flowering. Many species of biennials and perennials are induced to flower by low temperature exposure (0°C to 5°C). This process is called Vernalization.

Question 21.
What is meant by Epigeal germination?
Answer:
During epigeal germination cotyledons are pushed out of the soil. This happens due to the elongation of the hypocotyl. eg: Castor and Bean.

Question 22.
Explain the term seed dormancy.
Answer:
The condition of a seed when it fails to germinate even in suitable environmental condition is called seed dormancy.

Question 23.
Define the term phytogerontology.
Answer:
The branch of botany which deals with ageing, abscission and senescence is called Phytogerontology.

Question 24.
Explain the term programmed cell death.
Answer:
Senescence is controlled by plants own genetic programme and death of the plant or plant part consequent to senescence is called Programmed Cell Death. In short senescence of an individual cell is called PCD.

Question 25.
Define the term “Abscission”.
Answer:
Abscission is a physiological process of shedding of organs like leaves, flowers, fruits and seeds from the parent plant body.

III. Answer the following (3 Marks)

Question 1.
Explain the different phases of growth in plants.
Answer:
There are three phases of growth.
1. Formative phase:
Growth in this phase occurs in. meristematic cells of shoot and root tips. These cells are small in size, have dense protoplasm, large nucleus and small vacuoles. Cells divide continuously by mitotic cell division. Some cells retain capability of cell division while other cells enter the next phase of growth.

2. Elongation Phase:
Newly formed daughter cells are pushed out of the meristematic zone and increases the volume. It requires auxin and food supply, deposition of new cell wall materials (intussusception), addition of protoplasm and development of central vacuole take place.

3. Maturation Phase:
During this stage cells attain mature form and size. Thickening and differentiation takes place. After differentiation, the cells do not grow further.

Question 2.
Draw the ‘S’ shaped growth curve and mark the different phases of growth
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 3.
Mention the internal factors, that affect the growth of plant.
Answer:

  • Genes are intracellular factors for growth.
  • Phytohormones are intracellular factors for growth, eg: auxin, gibberellin, cytokinin.
  • C/N ratio.

Question 4.
What are the characteristic features of phytohormones?
Answer:

  • Usually produced in tips of roots, stems and leaves.
  • Transfer of hormones “from one place to another takes part through conductive systems.
  • They are required in trace quantities.
  • AH hormones are organic in nature.
  • There are no specialized cells or organs for their secretion.
  • They are capable of influencing physiological activities leading to promotion, inhibition and modification of growth.

Question 5.
List out the agricultural applications of auxins.
Answer:

  • It is used to eradicate weeds, eg: 2,4 – D and 2,4,5 – T.
  • Synthetic auxins are used in the formation of seedless fruits (Parthenocarpic fruit).
  • It is used to break the dormancy in seeds.
  • Induce flowering in Pineapple by NAA & 2,4 – D.
  • Increase the number of female flowers and fruits in cucurbits.

Question 6.
Mention any three physiological effects of gibberellins.
Answer:

  • It produces extraordinary elongation of stem caused by cell division and cell elongation.
  • Rosette plants (genetic dwarfism) plants exhibit excessive inter modal growth when they are treated with gibberellins. This sudden elongation of stem followed by flowering is called bolting.
  • Gibberellin breaks dormancy in potato tubers.

Question 7.
What are the uses of ethylene in agriculture?
Answer:

  • Ethylene normally reduces flowering in plants except in Pine apple and Mango.
  • It increases the number of female flowers and decreases the number of male flowers.
  • Ethylene spray in cucumber crop produces female flowers and increases the yield.

Question 8.
What is meant by climacteric fruits?
Answer:
In most of the plants, there is sharp rise in respiration rate near the end of the development of fruit, called climacteric  rise. Such fruits are called climacteric fruits. The ripening on demand can be induced in these fruits by exposing them to normal air containing about 1 ppm of ethylene. A liquid called ethephon is being used in fruit ripening as it continuously releases ethylene, eg: Tomato, Apples, Banana, Mango.

Question 9.
Give the classification of plants based on photoperiodism.
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 10.
Explain the term photoperiodic induction.
Answer:
An appropriate photoperiod in 24 hours cycle constitutes one inductive cycle. Plants , may require one or more inductive cycles for flowering. The phenomenon of conversion of leaf primordia into flower primordia under the influence of suitable inductive cycles is called photoperiodic induction.
eg: Xanthium (SDP) – 1 inductive cycle and Plantago (LDP) – 25 inductive cycles.

Question 11.
What are the practical applications of vernalization in plants?
Answer:

  • Vernalization shortens’ the vegetative period and induces the plant to flower earlier.
  •  It increases the cold resistance of the plants.
  • It increases the resistance of plants to fungal disease.
  • Plant breeding can be accelerated.

Question 12.
Write down the internal factors, that affect seed germination.
Answer:
1. Maturity of embryo:
The seeds of some plants, when shed will contain immature embryo. Such seeds germinate only after maturation of embryo.

2. Viability:
Usually seeds remain viable or living only for a particular period. Viability of seeds range from a few days (eg: Oxalis) to more than hundred years. Maximum viability (1000 years) has been recorded in lotus seeds. Seeds germinate only within the period of viability.

3. Dormancy:
Seeds of many plants are dormant at the time of shedding.

Question 13.
Mention the factors causing dormancy of seeds.
Answer:

  • Hard, tough seed coat causes barrier effect as impermeability of water, gas and restriction of the expansion of embryo prevents seed germination.
  • Many species of seeds produce imperfectly developed embryos called rudimentary embryos which promotes dormancy.
  • Lack of specific light requirement leads to seed dormancy.
  • A range of temperatures either higher or lower cause dormancy.
  • The presence of inhibitors like phenolic compounds which inhibits seed germination cause dormancy.

Question 14.
What are the significances of abscission?
Answer:

  • Abscission separates dead parts of the plant, like old leaves and ripe fruits.
  • It helps in dispersal of fruits and continuing the life cycle of the plant.
  • Abscission of leaves in deciduous plants helps in water conservation during summer.
  • In lower plants, shedding of vegetative parts like gemmae or plantlets help in vegetative reproduction.

IV. Answer the following (5 Marks)

Question 1.
Describe the geometric growth rate in plants with suitable diagram.
Answer:
Geometric growth rate:
This growth occurs in many higher plants and plant organs and is measured in size or weight. In plant growth, geometric cell division results if all cells of an organism or tissue are active mitotically. eg: Round three in the given figure, produces 8 cells as 23 – 8 and after round 20 there are 220 = 1,048,576 cells. The large plant or animal parts are produced this way. In fact, it is common in animals but rare in plants except when they are young and small. Exponential growth curve can be expressed as,
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

W1 = W0ert
W1 = Final size (weight, height and number)
W0 = Initial size at the beginning of the period
r = Growth rate
t = Tittle of growth
e = Base of the natural logarithms

Here V is the relative growth rate and also a measure of the ability of the plant to produce new plant material, referred to as efficiency index. Hence, the final size of W1 depends on the initial size W0.

Question 2.
Describe the experiment to measure the increase in length of the stem tip using an arc auxanometer.
Answer:
The increase in the length of the stem tip can easily be measured by an arc auxanometer which consists of a small pulley to the axis of which is attached a long pointer sliding over a graduated arc. A thread one end of which is tied to the stem tip and another end to a weight passes over the pulley tightly.

As soon as the stem tip increases in length, the pulley moves and the pointer slide over the graduated arc (Figure). The reading is taken. The actual increase in the length of the stem is then calculated by knowing the length of the pointer and the radius of the pulley. If the radius of the pulley is 4 inches and the length of pointer 20 inches the actual growth is measured as follows:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 3.
Write an essay on the phytochrome, Gibberelf ns in plants.
Answer:
1. Discovery:
The effect of gibberellins had been known in Japan since early 1800 where certain rice plants were found to suffer from ‘Bakanae’ or foolish seedling disease. This disease was found . by Kurosawa (1926) to be caused by a fungus Gibberella fujikuroi. The active substance was separated from fungus and named as gibberellin by Yabuta (1935). These are more than 100 gibberellins reported from both fungi and higher plants. They are noted as GA1, GA2, GA3 and so on. GA3 is the first discovered gibberellin. In 1938, Yabuta and Sumiki isolated gibberellin in crystalline form. In 1955, Brain et al., gave the name gibberellic acid. In 1961, Cross et al., established its structure.

2. Occurrence:
The major site of gibberellin production in plants is parts like embryo, roots and young leaves near the tip. Immature seeds are rich in gibberellins.

3. Precursors:
The gibberellins are chemically related to terpenoids (natural rubber, carotenoids and steroids) formed by 5 – C precursor, an Isoprenoid unit called Iso Pentenyl Pyrophosphate (IPP) through a number of intermediates. The primary precursor is acetate.

4. Chemical structure:
All gibberellins have gibbane ring structure.

5. Transport in plants:
The transport of gibberellins in plants is non-polar. Gibberellins are translocated through phloem and also occur in xylem due to lateral movement between vascular bundles.

6. Bioassay (Dwarf Pea. assay):
Seeds of dwarf pea are allowed to germinate till the formation of the coleoptile. GA solution is applied to some seedlings. Others are kept under control. Epicotyle length is measured and as such, GA stimulating epicotyle growth can be seen.

7. Physiological Effects:
It produces extraordinary elongation of stem caused by cell division and cell elongation.

  • Rosette plants (genetic dwarfism) plants exhibit excessive intermodal growth when they are treated with gibberellins. This sudden elongation of stem followed by flowering is called bolting.
    Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development
  • Gibberellin breaks dormancy in potato tubers.
  • Many biennials usually flower during second year of their growth. For flowering to take place, these plants should be exposed to cold season. Such plants could be made to flower without exposure to cold season in the first year itself, when they are treated with gibberellins.

8. Agricultural role:

  • Formation of seedless fruits without fertilization is induced by gibberellins eg: Seedless tomato, apple and cucumber.
  • It promotes the formation of male flowers in cucurbitaceae. It helps in crop improvement.
    Uniform bolting and increased uniform seed production.
  • Improves number and size of fruits in grapes. It increase yield.
  • Promotes elongation of inter-node in sugarcane without decreasing sugar content.
  • Promotion of flowering in long day plants even under short day conditions. .
  • It stimulates the seed germination

Question 4.
What are their physiological effects of Abscisic acid in – plants and its role in agriculture?
Answer:
Physiological effects:

  • It helps in reducing transpiration rate by closing stomata. It inhibits K+ uptake by guard cells and promotes the leakage of malic acid. It results in closure of stomata;
  • It spoils chlorophylls, proteins and nucleic acids of leaves making them yellow.
  •  Inhibition of cell division and cell elongation.
  •  ABA is a powerful growth inhibitor. It causes 50% inhibition of growth in Oat coleoptile.
  • It induces bud and seed dormancy.
  • If promotes, the abscission of leaves, flowers and fruits by forming abscission layers.
  • ABA plays an important role in plants dtiring water stress -and during drought conditions. It results in loss of turgor and closure of stomata,
  • It has anti-auxin and anti-gibberellin property.
  • Abscisic acid promotes senescence in leaves’ by causing loss of chlorophyll pigment decreasing the rate of photosynthesis and changing the rate of proteins and nucleic acid synthesis

Agricultural Role:

  • In Cannabis sativa, induces male flower formation on female plants.
  • Induction of flowers in short day plants.
  • It promotes sprouting in storage organs like Potato.
  • ABA plays an important role in plants during water stress drought conditions.
  • It inhibits the shoot growth and promotes growth of root system. This character protect the plants from water stress. Hence, ABA is called as stress hormone.

Question 5.
Write an essay on the role of ethylene on plant physiology and agriculture.
Answer:
Almost all plant tissues produce ethylene gas in minute quantities.
1. Discovery:
In 1924, Denny found that ethylene stimulates the ripening of lemons. In 1934, R. Gane found that ripe bananas contain abundant ethylene. In 1935, Cocken et al., identified ethylene as a natural plant hormone.

2. Occurrence:
Maximum synthesis occurs during climacteric ripening of fruits and tissues undergoing senescence. It is formed in almost all plant parts like roots, leaves, flowers, fruits and seeds.

3. Transport in plants:
Ethylene can easily diffuse inside the plant through intercellular spaces.

4. Precursor:
It is a derivative of amino acid methionine, linolenic acid and fumaric acid.

5. Bioassay (Gas Chromatography):
Ethylene can be measured by gas chromatography. This technique helps in the detection of exact amount of ethylene from different plant tissues like lemon and orange.

6. Physiological Effects:

  • Ethylene stimulates respiration and ripening in fruits.
  • It stimulates radial growth in sterft and roof and inhibits linear growth.
  • It breaks the dormancy of buds, seeds and storage organs.
  • It stimulates formation of abscission zone in leaves, flowers and fruits. This makes the leaves to shed prematurely.
  • Inhibition of stem elongation (shortening the intemode).
  • In low concentration, ethylene helps in root initiation.
  • Growth of lateral roots and root hairs. This increases the absorption surface of the plant roots.
  • The growth of fruits is stimulated by ethylene in some plants. It is more marked in climacteric fruits.
  • Ethylene causes epinasty.

7. Agricultural role:

  • Ethylene normally reduces flowering in plants except in Pine apple and Mango.
  • It increases the number of female flowers and decreases the number of male flowers.
  • Ethylene spray in cucumber crop produces female flowers and increases the yield.

Question 6.
Explain the two hypothesis of explaining the mechanism of vernalization.
Answer:
Two main theories to explain the mechanism of vernalization are:

1. Hypothesis of phasic development:
According to Lysenko, development of an annual seed plant consists of two phases. First phase is thermostage, which is vegetative phase requiring low temperature and suitable moisture. Next phase is photo stage which requires high temperature for synthesis of florigen (flowering hormone).

2. Hypothesis of hormonal involvement:
According to Purvis (1961), formation of a substance A from its precursor is converted into B after chilling. The substance B is unstable. At suitable temperature B is converted into stable compound D called Vemalin. Vernalin is converted to F (Florigen). Florigen induces flower formation. At high temperature B is converted to C and devemalization occurs.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 7.
Write an essay on the types of senescence, its physiology and the factors affecting senescence.
Answer:
1. Types of Senescence:
Leopold (1961) has recognised four types of senescence.

(a) Overall senescence:
This kind of senescence occurs in annual plants when entire plant gets affected and dies, eg: Wheat and Soybean. It also occurs in few perennials also, eg: Agave and Bamboo.

(b) Top senescence:
It occurs in aerial parts of plants. It is common in perennials, underground and root system remains viable, eg: Banana and Gladiolus.

(c) Deciduous senescence:
It is common in deciduous plants and occurs only in leaves of plants, bulk of the stem and root system remains alive, eg: Elm and Maple.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

(d) Progressive senescence:
This kind of senescence is gradual. First it occurs in old leaves followed by new leaves f then stem and finally root system. It is common in annuals.

2. Physiology of Senescence:

  • Cells undergo changes in structure.
  • Vacuole of the cell acts as lysosome and secretes hydrolytic enzymes.
  • The starch content is decreased in the cells.
  • Photosynthesis is reduced due to loss of chlorophyll accompanied by synthesis and accumulation of anthocyanin pigments, therefore the leaf becomes red.
  • There is a marked decrease in protein content in the senescing organ.
  • RNA content of the leaf particularly rRNA level is decreased in the cells due to increased activity of the enzyme RNAase.
  • DNA molecules in senescencing leaves degenerate by the increased activity of enzyme DNAase.

3. Factors affecting Senescence:

  •  ABA and ethylene accelerate senescence while auxin and cytokinin retard senescence.
  • Nitrogen deficiency increases senescence whereas nitrogen supply retards, senescence. High temperature senescence but low retards senescence.
  • Senescence is rapid in dark than in light.
  • Water stress leads to accumulation of ABA leading to senescence.

Question 8.
Describe the methods of breaking dormancy of seeds in plants.
Answer:
The dormancy of seeds can be broken by different methods. These are:
1. Scarification:
Mechanical and chemical treatments like cutting or chipping of hard tough seed coat and use of organic solvents to remove waxy or fatty compounds are called as Scarification.

2. imp-action:
in some seeds water and oxygen are unable to penetrate micropyle due to blockage by cork cells. These seeds are shaken vigorously to remove the plug which is called Imp-action.

3. Stratification:
Seeds of rosaceous plants (Apple, Plum, Peach and Cherry) will not germinate until they have been exposed to well aerated, moist condition under low temperature (0°C to 10°C) for weeks to months. Such treatment is called Stratification.

4. Alternating temperatures: Germination of some seeds is strongly promoted by alternating daily temperatures. An alternation of low and high temperature improves the germination of seeds.

5. Light:
The dormancy of photoblastic seeds can be broken by exposing them to red light.

Solution To Activity

Textbook Page No : 164

Question 1.
Demonstration of phases of growth.
Answer:
To demonstrate and study the phases of growth, germinate a few seeds of bean on a circular filter paper soaked with water in a petridish. After two days of growth, select a few seedlings with straight radical of 2 to 3 cm length. Dry the surface of radical with a blotting paper and mark the radical from tip to base with at least 2 mm gap using water proof ink. Replace the seedlings in filter paper and observe further growth.

Observation:
The marked area in the radical will grow and increase in length and hence the marked area of 2mm is found to be grow beyond 2mm size due to the growth in the radical.

Textbook Page No: 169

Question 2.
Measurement of growth by direct method.
Answer:
Step 1: Take ordinary scale.
Step 2: Measure ground stem up to the growing point of the plant.
Step 3: Use Indian ink and mark at regular intervals to measure the length of root, stem, and girth of the trunk.

Observation:
At regular intervals measure the increase in length and girth of the trunk and it can be observed that the length of the root, stem and girth of the trunk increased with the increase in the period of growth.

 

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Textbook Evaluation Solved
Question 1.
Muscles are derived from ………………..
(a) Ectoderm
(b) Mesoderm
(c) Endoderm
(d) Neuro ectoderm
Answer:
(b) Mesoderm

Question 2.
Muscles are formed by …………………
(a) Myocytes
(b) Leucocytes
(c) Osteocytes
(d) Lymphocytes
Answer:
(a) Myocytes

Question 3.
The muscles attached to the bones are called …………………
(a) Skeletal muscle
(b) Cardiac muscle
(c) Involuntary muscle
(d) Smooth muscles
Answer:
(a) Skeletal muscle

Question 4.
Skeletal muscles are attached to the bones by ………………….
(a) Tendon
(b) Ligament
(c) Pectin
(d) Fibrin
Answer:
(a) Tendon

Question 5.
The bundle of muscle fibres is called ………………….
(a) Myofibrils
(b) Fascicle
(c) Sarcomere
(d) Sarcoplasm
Answer:
(b) Fascicle

Question 6.
The pigment present in the muscle fibre to store oxygen is ……………………
(a) Myoglobin
(b) Troponin
(c) Myosin
(d) Actin
Answer:
(a) Myoglobin

Question 7.
The functional unit of a muscle fibre is …………………..
(a) Sarcomere
(b) Sarcoplasm
(c) Myosin
(d) Actin
Answer:
(a) Sarcomere

Question 8.
The protein present in the thick filament is …………………
(a) Myosin
(b) Actin
(c) Pectin
(d) Leucin
Answer:
(a) Myosin

Question 9.
The protein present in the thin filament is ……………….
(a) Myosin
(b) Actin
(c) Pectin
(d) Leucin
Answer:
(b) Actin

Question 10.
The region between two successive Z-discs is called a …………………
(a) Sarcomere
(b) Microtubule
(c) Myoglobin
(d) Actin
Answer:
(a) Sarcomere

Question 11.
Each skeletal muscle is covered by …………………
(a) Epimysium
(b) Perimysium
(c) Endomysium
(d) Hypomysium
Answer:
(a) Epimysium

Question 12.
Knee joint is an example of ………………..
(a) Saddle joint
(b) Hinge joint
(c) Pivot joint
(d) Gliding joint
Answer:
(b) Hinge joint

Question 13.
Name of the joint present between the atlas and axis is ……………….
(a) Synovial joint
(b) Pivot joint
(c) Saddle joint
(d) Hinge joint
Answer:
(b) Pivot joint

Question 14.
ATPase enzyme needed for muscle contraction is located in ………………
(a) Actinin
(b) Troponin
(c) Myosin
(d) Actin
Answer:
(c) Myosin

Question 15.
Synovial fluid is found in ……………..
(a) Ventricles of the brain
(b) Spinal cord
(c) Immovable joint
(d) Freely movable joints
Answer:
(d) Freely movable joints

Question 16.
Inflammation of joints due to accumulation of uric acid crystals is called as ……………..
(a) Gout
(b) Myasthenia gravis
(c) Dsteoporosis
(d) Osteomalacia
Answer:
(a) Gout

Question 17.
Acetabulum is located in ……………..
(a) Collar bone
(b) Hip bone
(c) Shoulder bone
(d) Thigh bone
Answer:
(b) Hip bone

Question 18.
Appendicular skeleton is ………………..
(a) Girdles and their limbs
(b) Vertebrae
(c) Skull and vertebral column
(d) Ribs and sternum
Answer:
(a) Girdles and their limbs

Question 19.
The type of movement exhibited by the macrophages are …………………
(a) Flagellar
(b) Ciliary
(c) Muscular
(d) Amoeboid
Answer:
(d) Amoeboid

Question 20.
The pointed portion of the elbow is ………………….
(a) Acromion process
(b) Glenoid cavity
(c) Olecranon process
(d) Symphysis
Answer:
(c) Olecranon process

Question 21.
Name the different types of movement?
Answer:

  1. Amoeboid movement
  2. Ciliary movement
  3. Flagellar movement
  4. Muscular movement

Question 22.
Name the filaments present in the sarcomere?
Answer:
Thick and thin filaments are the two types of filaments present inside the sarcomere.

Question 23.
Name the contractile proteins present in the skeletal muscle?
Answer:
Actin and myosin are contractile proteins present in the skeletal muscle.

Question 24.
When describing a skeletal muscle, what does “striated” mean?
Answer:
Each skeletal muscle fibre has a repeated series of dark and light bands. The dark A-bands and light I-bands give a striated appearance to the muscle.

Question 25.
How does an isotonic contraction take place?
Answer:
In isotonic contraction the length of the muscle changes but the tension remains constant. The force produced is unchanged, e.g., lifting dumbbells and weight lifting.

Question 26.
How does an isometric contraction take place?
Answer:
In isometric contraction the length of the muscle does not change but the tension of the muscle changes. The force produced is changed, e.g., pushing against a wall, holding a heavy bag.

Question 27.
Name the bones of the skull?
Answer:
The skull is composed of two sets of bones – cranial and facial bones. It consists of 22 bones of which 8 are cranial bones and 14 are facial bones.

Question 28.
Which is the only jointless bone in human body?
Answer:
Hyoid bone.

Question 29.
List the three main parts of the axial skeleton?
Answer:
The skull, the vertebral column and the ribcage are the three main parts of the axial skeleton.

Question 30.
How is tetany caused?
Answer:
Tetany is caused when rapid muscle spasms occur in the muscles due to deficiency of parathyroid hormone resulting in reduced calcium levels in the body.

Question 31.
How does rigor mortis happen?
Answer:
After the death of an individual, the membrane of muscle cells become more permeable to calcium ions. This happens due to partial contraction of skeletal muscles. The contracted muscles are unable to relax. This condition is known as rigor mortis.

Question 32.
What are the different types of rib bones that form the rib cage?
Answer:
Thoracic vertebrae ribs and sternum together constitute the ribcage.

Question 33.
What are the bones that make the pelvic girdle?
Answer:
Ilium, ischium and pubis make the pelvic girdle.

Question 34.
List the disorders of the muscular system?
Answer:

  1. Myasthenia gravis
  2. Tetany
  3. Muscle fatigue
  4. Atrophy
  5. Muscle pull
  6. Muscular dystrophy

Question 35.
Explain the sliding- filament theory of muscle contraction?
Answer:
Andrew F.Huxley and Rolf Niedergerke proposed the sliding filament theory to explain muscle contraction. According to this theory, overlapping actin and myosin filaments of fixed length slide past one another in an energy requiring process, resulting in muscle contraction.

Question 36.
What are the benefits of regular exercise?
Answer:

  1. The benefits of regular exercise are:
  2. The muscles used in exercise grow larger and stronger.
  3. The resting heart rate goes down.
  4. More enzymes are synthesized in the muscle fibre.
  5. Ligaments and tendons become stronger.
  6. Joints become more flexible.
  7. Protection from heart attack.
  8. Influences hormonal activity.
  9. Improves cognitive functions.
  10. Prevents obesity.
  11. Promotes confidence, esteem.
  12. Aesthetically better with good physique.
  13. Over all well-being with good quality of life.
  14. Prevents depression, stress and anxiety.

In-Text Questions Solved

Question 1.
Which myofilament has the binding sites for calcium? Name the specific molecule that binds with calcium?
Answer:
Actin filament has the binding sites for calcium. Troponin binds to calcium on thin filaments.

Question 2.
All muscles produce movement, but only skeletal muscle is responsible for locomotion. What is meant by this statement?
Answer:
All the muscles, skeletal, smooth and cardiac, produce movement. Smooth muscles control the activities of internal organs like intestine, stomach, lungs, bladder etc. and their actions are involuntary. Cardiac muscles help in the functioning of heart. But only skeletal muscles are responsible for locomotion.

Locomotion is effected by both bones and muscles attached to the bones and takes place by the coordination between muscular contraction and relaxation and skeletal system. Moreover, the functioning of those skeletal muscles is voluntary.

Question 3.
The pelvic girdle is a heavy, strong girdle. How does its structure reflect its function?
Answer:
The pelvic girdle is a heavy structure specialized for weight bearing. This feature enables it to connect the trunk and the legs, support and balance the trunk, and contain and supports the intestines, the urinary bladder, and the internal sex organs.

Question 4.
An exhausted student was attending a lecture. After 30 minutes or so, he lost interest and he let go with a tremendous yawn. To his great distress he couldn’t close his mouth -his lower jaw was locked open. What do you think would have caused it?
Answer:
When he opened his mouth very wide, the mandibular condyle might have slid forward to the point that the joint might have got dislocated.

Samacheer Kalvi 11th Bio Zoology Locomotion and Movement Additional Questions & Answers

I. Choose The Correct Answer

Question 1.
Which of the following is not related to skeletal muscle?
(a) It is attached to the bone
(b) It is striated
(c) It is an involuntary muscle
(d) It brings about movement of the organ
Answer:
(c) It is an involuntary muscle

Question 2.
The skeletal system is derived from
(a) Ectoderm
(b) Endoderm
(c) Mesoderm
(d) Mesoglea
Answer:
(c) Mesoderm

Question 3.
The cytoplasm of the muscle fibre is ……………….
(a) Sarcolemma
(b) Sarcoplasm
(c) Ectoplasm
(d) Endoplasm
Answer:
(b) Sarcoplasm

Question 4.
The thick filament of muscle fibre is made up of ……………….
(a) Actin
(b) Myosin
(c) Tropomyosin
(d) Troponin
Answer:
(b) Myosin

Question 5.
The cranial bones are ………………..
(a) 22
(b) 14
(c) 8
(d) 3
Answer:
(c) 8

Question 6.
Where is the hyoid bone present?
(a) Cranium
(b) Appendicular skeleton
(c) Pectoral girdle
(d) Base of the buccal cavity
Answer:
(d) Base of the buccal cavity

Question 7.
The number of vertebrates is ………………..
(a) 8
(b) 12
(c) 5
(d) 33
Answer:
(d) 33

Question 8.
How many thoracic vertebrates are there?
(a) 7
(b) 12
(c) 5
(d) 4
Answer:
(b) 12

Question 9.
Rib cage protects ……………….
(a) Brain
(b) Kidney
(c) Lungs, heart, liver
(d) Heart
Answer:
(c) Lungs, heart, liver

Question 10.
The fore arm bones are the ………………
(a) Tibia and fibula
(b) Radius and ulna
(c) Carpals and metacarpals
(d) Tarsal and metatarsals
Answer:
(b) Radius and ulna

Question 11.
The joint between carpal and metacarpals is ………………..
(a) Pivot joint
(b) Ball and socket joint
(c) Saddle joint
(d) Hinge joint
Answer:
(c) Saddle joint

Question 12.
Which of the following allows movement in only one direction?
(a) Pivot joint
(b) Ball and socket joint
(c) Saddle joint
(d) Hinge joint
Answer:
(d) Hinge joint

Question 13.
Which of the following disorders is related to endocrine gland?
(a) Myasthenia gravis
(b) Tetany
(c) Atrophy
(d) Muscular dystrophy
Answer:
(b) Tetany

Question 14.
Which of the following arthritis is related to protein metabolism?
(a) Osteoarthritis
(b) Rheumatoid arthritis
(c) Gouty arthritis
(d) Osteoporosis
Answer:
(c) Gouty arthritis

Question 15.
Which of the following is deficiency disorder?
(a) Osteoarthritis
(b) Rheumatoid arthritis
(c) Gouty arthritis
(d) Osteoporosis
Answer:
(d) Osteoporosis

II. Fill in the blanks

Question 1.
The sperm cells show …………………… movement.
Answer:
Flagellar

Question 2.
Skeletal muscle is attached to the bone by a bundle of collagen fibres known as ……………………
Answer:
Tendon

Question 3.
The cytoplasm of the muscle fibre is called the ……………………
Answer:
Sarcoplasm

Question 4.
…………………… is a red coloured respiratory pigment of the muscle fibre.
Answer:
Myoglobin

Question 5.
…………………… are the granules of stored glycogen.
Answer:
Glycosomes

Question 6.
…………………… is the functional unit of the skeletal muscle.
Answer:
Sarcomere

Question 7.
The thick filaments are composed of the protein ……………………
Answer:
Myosin

Question 8.
The monomer of the myosin molecule is ……………………
Answer:
Meromyosin

Question 9.
The study of muscle is called ……………………
Answer:
Myology

Question 10.
The junction between the motor neuron and the sarcolemma of the muscle fibre is called the ……………………
Answer:
Neuromuscular junction or motor end plate

Question 11.
When nerve impulse reaches a neuromuscular junction, …………………… is released.
Answer:
Acetylcholine

Question 12.
In …………………… contraction of the length of the muscle changes but the tension remains constant.
Answer:
Isotonic

Question 13.
In …………………… contraction of the length of the muscle does not change but the tension of the muscle changes.
Answer:
Isometric

Question 14.
The oxidative fibres are called as …………………… fibres.
Answer:
Red muscle

Question 15.
Glycolytic fibres or white muscle fibres lack ……………………
Answer:
Myoglobin

Question 16.
The skeletal system is derived from the ……………………
Answer:
Mesoderm

Question 17.
The RBCs and WBCs are produced in the ……………………
Answer:
Bone marrow

Question 18.
The large hole in the temporal bone is the ……………………
Answer:
External auditory meatus

Question 19.
The lower jaw bone is called ……………………
Answer:
Mandible

Question 20.
…………………… is the only bone without any joint.
Answer:
Hyoid bone

Question 21.
The upper jaw is formed of the ……………………
Answer:
Maxilla

Question 22.
…………………… is the large opening found at the posterior base of the skull.
Answer:
Foramen magnum

Question 23.
Through foramen magnum the medulla oblongata continues as the ……………………
Answer:
Spinal cord

Question 24.
The first vertebra is called as the ……………………
Answer:
Atlas

Question 25.
The second vertebra is called as the ……………………
Answer:
Axis

Question 26.
The cranium protects the ……………………
Answer:
Brain

Question 27.
…………………… is the flat bone on the mid ventral line of the thorax.
Answer:
Sternum

Question 28.
The first seven pairs of ribs are called ……………………
Answer:
True ribs/vertebro-sternal

Question 29.
The 11th and 12th pairs of ribs are called as …………………… ribs.
Answer:
Floating /vertebral

Question 30.
The 8th, 9th and 10th pairs of ribs are called …………………… ribs.
Answer:
Flase /vertebro-chondral

Question 31.
Rib cage plays a role in ……………………
Answer:
Breathing

Question 32.
…………………… is a depression in the pectoral girdle on which the head of the humerus form the shoulder joint.
Answer:
Glenoid cavity

Question 33.
…………………… is situated at the upper end of the ulna which forms the pointed portion of the elbow.
Answer:
Olecranon process

Question 34.
…………………… is the largest, longest and strongest bone in the body.
Answer:
Femur

Question 35.
The head of femur articulates with the …………………… of the pelvis to form the hip joint.
Answer:
Acetabulum

Question 36.
The bone forming cells are called ……………………
Answer:
Osteoblasts

Question 37.
The bone destroying cells are called the ……………………
Answer:
Osteoclasts

Question 38.
…………………… cells give rise to the osteoblasts.
Answer:
Osteogenic

Question 39.
The internal bone surfaces are covered with a delicate connective tissue membrane called the ……………………
Answer:
Endosteum

Question 40.
Between the epiphysis and diaphysis …………………… is present.
Answer:
Epiphyseal plate/growth plate

Question 41.
The …………………… are points of contact between the bones.
Answer:
Joints

Question 42.
Sutures of the flat skull bones are …………………… joints.
Answer:
Fibrous

Question 43.
…………………… are the freely movable joints.
Answer:
Synovial joints/Diarthroses joints

Question 44.
Between atlas and axis …………………… joint is present.
Answer:
Pivot

Question 45.
Between the carpals …………………… joint is present.
Answer:
Plane/gliding

Question 46.
Between the carpal and metacarpal …………………… joint is present.
Answer:
Addle

Question 47.
Between humerus and pectoral girdle …………………… joint is seen.
Answer:
Ball and socket

Question 48.
…………………… is an autoimmune disorder affecting the action of acetylcholine.
Answer:
Myasthenia gravis

Question 49.
The removal of parathyroid gland results in ……………………
Answer:
Tetany

Question 50.
A traumatic pulling of the fibres produces a tear known as ……………………
Answer:
Sprain

Question 51.
…………………… is an inflammatory or degenerative disease that damages the joints.
Answer:
Arthritis

Question 52.
The wearing away of the bone ends of the knees and other movable joints is called ……………………
Answer:
Osteoarthritis

Question 53.
The inflammation of the synovial membranes is known as ……………………
Answer:
Rheumatoid arthritis

Question 54.
Inflammation of joints due to accumulation of uric acid crystals is known as ……………………
Answer:
Gout

Question 55.
The deficiency of vitamin D and hormonal imbalance cause ……………………
Answer:
Osteoporosis

Question 56.
The deficiency of vitamin D causes …………………… in children.
Answer:
Rickets

Question 57.
The deficiency of vitamin D causes …………………… in adults.
Answer:
Osteomalacia

Question 58.
…………………… increase the breathing and heart rate.
Answer:
Endurance or aerobic activities

Question 59.
…………………… protects us from heart attack.
Answer:
Exercise

III. Short Answer Questions

Question 1.
What is amoeboid movement?
Answer:
The movement of cells by streaming movements of the cytoplasm forming pseudo-podia is known as amoeboid movement, e.g., macrophages.

Question 2.
What is ciliary movement?
Answer:
The movement caused by the cilia is ciliary movement, e.g., ciliated epithelial cells of respiratory passage and genital tracts.

Question 3.
What is flagellar movement?
Answer:
The movement due to the lashing of flagella is known as flagellar movement, e.g., sperm cells.

Question 4.
What is muscular movement?
Answer:
The movement of hands, legs, jaws, tongue are caused by the contraction and relaxation of the muscle which is known as muscular movement.

Question 5.
What is fascicle?
Answer:
Each muscle is made up of bundles of muscle fibres called fascicle.

Question 6.
What are myofibrils?
Answer:
Each muscle fibre contains hundreds to thousands of rod-like structures called myofibrils that run parallel to its length.

Question 7.
What is epimysium?
Answer:
The connective tissue covering the whole muscle is the epimysium,

Question 8.
What is perimysium?
Answer:
The connective tissue covering around each fascicle is the perimysium.

Question 9.
What is endomysium?
Answer:
The connective tissue surrounding the muscle fibre is called the endomysium.

Question 10.
Distinguish between voluntary muscle and involuntary muscle.
Answer:

Voluntary muscle

Involuntary muscle

1. The muscle whose functions are under the control of our will is voluntary muscle.1. The muscle whose functions are not under the control of our will is the involuntary muscle.
2. It is striated and and hence called striated or striped muscle.2. It has no stripes and hence it is called smooth muscles or unstriped or non-striated muscles.

Question 11.
What is tendon?
Answer:
Skeletal muscle is attached to the bone by a bundle of collagen fibres known as tendon.

Question 12.
What is myoglobin?
Answer:
Myoglobin is a red-coloured respiratory pigment of the muscle fibre.

Question 13.
What is sarcoplasm?
Answer:
The cytoplasm of the muscle fibre is called the sarcoplasm.

Question 14.
What are Glycosomes?
Answer:
Glycosomes are the granules of stored glycogen that provide glucose during the period of muscle fibre activity.

Question 15.
What is sarcomere?
Answer:
The functional unit of the skeletal muscle is known as sarcomere.

Question 16.
Distinguish between thick filaments and thin filaments?
Answer:

Thick filaments

Thin filaments

 1. The thick filaments are composed of the protein myosin.1. The thin filaments are composed of protein actin.

Question 17.
What is meromyosin?
Answer:
The monomer of myosin molecule is called meromyosin.

Question 18.
Name the proteins which regulates the contraction of muscles?
Answer:
Actin, myosin, tropomyosin and troponin.

Question 19.
What is myology?
Answer:
The study of muscle is called myology.

Question 20.
What are oxidative fibres?
Answer:
The muscle fibres that contain numerous mitochondria and have a high capacity for oxidative phosphorylation are classified as oxidative fibres. They are also called red muscle fibres.

Question 21.
What are glycolytic fibres?
Answer:
The muscle fibres which contain a few mitochondria but possess a high concentration of glycolytic enzymes and large stores of glycogen are called glycolytic fibres. These lack myoglobin and hence these are pale. These muscle fibres are termed as white muscle fibres.

Question 22.
What is hydrostatic skeleton?
Answer:
The skeleton found in soft-bodied invertebrates is called hydrostatic skeleton. It is a fluid filled cavity encircled by muscles, e.g., earthworm.

Question 23.
Distinguish between exoskeleton and endoskeleton.
Answer:

Exoskeleton

Endoskeleton

1. The rigid hard case present outside the body of animals is exoskeleton.1. The rigid support structure found inside the body of vertebrates is endoskeleton.
2. e.g., Cockroach2. e.g., Human being

Question 24.
What is a metaphysis?
Answer:
The bone region where the diaphysis and epiphyses meet is called the metaphysis.

Question 25.
What is periosteum?
Answer:
The external surface of the bone is covered by a double layered membrane called periosteum. What is endosteum?

Question 26.
What is endosteum?
Answer:
Internal bone surfaces are covered with a delicate connective tissue membrane called the endosteum.

Question 27.
What is sternum?
Answer:
Sternum is a flat bone on the mid ventral line of the thorax. It provides space for the attachment of the thoracic ribs and abdominal muscles.

IV. Give Long Answers

Question 1.
Explain the structure of a skeletal muscle fibre?
Answer:

  1. Each muscle fibre is thin and elongated.
  2. It has multiple oval nuclei beneath sarcolemma.
  3. The cytoplasm of the muscle fibre is called as sarcoplasm.
  4. It contains glycosomes, the stored glycogen granules, myoglobin, respiratory pigment and sarcoplasmic reticulum.
  5. Actin and myosin are muscle proteins present in the muscle fibre.
  6. Each myofibril has a repeated series of dark and light bands called A-bands and I-bands.
  7. Each dark band has a lighter region in its middle called the H-zone.
  8. Each H-zone is bisected vertically by a dark line called the M-line.
  9. Each light I-band has a darker mid line area called the Z-disc.
  10. Sarcomere is the functional unit of the skeletal muscle. It is a region of a myofibril between two successive Z-discs.
  11. Sarcomere has thick and thin filaments. The thick filaments extend the entire length of the A-band, the thin filaments extend across the I-band and partly into the A-band.
  12. The invagination of the sarcolemma forms transverse tubules (T-tubules) and they penetrate into the junction between the A and I-bands.

 

Question 2.
Write the schematic presentation of muscle contraction?
Answer:
Samacheer Kalvi 11th Bio Zoology Solutions Chapter 9 Locomotion and Movement img 1

Question 3.
Explain the structure of contractile proteins?
Answer:
The thick filaments of the muscle has the protein mysoin. Each myosin molecule is made up of a monomer called meromyosin. It has a globular head, a short arm and a tail. The short arm constitutes the heavy meromyosin (HMM). The tail portion forms the light meromyosin (LMM). The head bears an actin-binding site and an ATP- binding site. It also contains ATPase enzyme that split ATP to generate energy for the contraction of muscle.

The thin filaments are composed of two intertwined actin molecules. It has polypeptide subunits called globular actin or G-actin and filamentous form or F-actin.

Each thin filament is made of two F-actins helically wound to each other. Each F-actin is a polymer of monomeric G-actins. Tropomyosin and troponin are other proteins which help in regulating the contraction of muscles along with actin and myosin.

Question 4.
Explain the types of skeletal muscle fibres?
Answer:
Skeletal muscle fibres are classified into three types. They are:

  1. Slow – oxidative fibres
  2. Fast – oxidative fibres
  3. Fast – glycolytic fibres

1. Slow – oxidative fibres have low rates of myosin ATP hydrolysis but have the ability to make large amounts of ATP. These fibres are used for prolonged, regular activity such as long distance swimming. Long – distance runners have a high proportion of these fibres in their leg muscles.

2. Fast – oxidative fibres have high myosin ATPase activity and can make large amounts of ATP. They are particularly suited for rapid actions.

3. Fast – glycolytic fibres have myosin ATPase activity but cannot make as much ATP as oxidative fibres, because their source of ATP is glycolysis. These fibres are best suited for rapid, intense actions, such as short sprint at maximum speed.

Question 5.
Write the functions of skeletal system?
Answer:
Functions of skeletal system

  1. Support -It forms a rigid framework and supports the weight of the body against gravity.
  2. Shape – It provides and maintains the shape of the body.
  3. Protection – It protects the delicate internal organs of the body.
  4. Acts as reservoir – It stores minerals such as calcium and phosphate. Fat (triglyceride) is stored in yellow bone marrow and represents a source of stored energy for the body.
  5. Locomotion – It acts as lever along with the muscles attached to it.
  6. Strength – It can withstand heavy weight and absorbs mechanical shock.
  7. Asa haemopoietic tissue – Red and white blood cells are produced in the bone marrow of the ribs, spongy bones of vertebrae and extremities of long bones.

 

Question 6.
Explain the bones that form the skull?
Answer:
The skull is composed of two sets of bones – cranial and facial bones. It consists of 22 bones of which 8 are cranial bones and 14 are facial, bones. The cranial bones form the hard protective outer covering of the brain and called the brain box. The capacity of the cranium is 1500 cm3.

These bones are joined by sutures which are immovable. They are a paired parietal, paired temporal and individual bones such as the frontal, sphenoid, occipital and ethmoid. The large hole in the temporal bone is the external auditory meatus. In the facial bones maxilla, zygomatic, palatine, lacrimal, nasal are paired bones whereas mandible or lower jaw and vomer are unpaired bones. They form the front part of the skull.

A single U-shaped hyoid bone is present at the base of the buccal cavity. It is the only one bone without any joint. Each middle ear contains three tiny bones- malleus, incus and stapes collectively are called ear ossicles. The upper jaw is formed of the maxilla and the lower jaw is formed of the mandible.

The upper jaw is fused with the cranium and is immovable. The lower jaw is connected to the cranium by muscles and is movable. The most prominent openings in the skull are the orbits and the nasal cavity. Foramen magnum is a large opening found at the posterior base of the skull. Through this opening the medulla oblongata of the brain descends down as the spinal cord.

Question 7.
Write a short note on the vertebral column?
Answer:
Vertebral column is also called the back bone. It consists of 33 serially arranged vertebrae which are interconnected by cartilage known as intervertebral disc. The vertebral column extends from the base of the skull to the pelvis and forms the main frame work of the trunk. The vertebral column has five major regions.

They are, the cervical, thoracic, lumbar, sacrum (5 sacral vertebrae found in the infant which are fused to form one bone in the adult) and coccyx (4 coccygeal vertebrae found in the infant which are fused to form one bone in the adult).

Each vertebra has a central hollow portion, the neural canal, through which the spinal cord passes. The first vertebra is called as the atlas and the second vertebra is called as the axis. Atlas is articulated with the occipital condyles. The vertebral column protects the spinal cord, supports the head and serves as the point of attachment for the ribs and musculature of the back.

Question 8.
Write a short note on Rib cage?
Answer:
There are 12 pairs of ribs. Each rib is a thin flat bone connected dorsally to the vertebral column and ventrally to the sternum. It has two articulation surfaces on its dorsal end, hence called bicephalic.

The first seven pairs of ribs are called ‘true ribs or vertebro-stemal ribs. Dorsally they are attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilages.

The 8th, 9th and 10th pairs of ribs do not articulate directly with the sternum but joined with the cartilaginous (hyaline cartilage) part of the seventh rib. These are called ‘false ribs’ or vertebro-chondral ribs.

The last 11th and 12th pairs of ribs are not connected ventrally. Therefore, they are called as ‘floating ribs’ or vertebral ribs. Thoracic vertebrae, ribs and sternum together form the ribcage. Rib cage protects the lungs, heart, liver and also plays a role in breathing.

Question 9.
Write a note on Pectoral girdle?
Answer:
The upper limbs are attached to the pectoral girdles. These are very light and allow the upper limbs a degree of mobility not seen anywhere else in the body. The girdle is formed of two halves. Each half of the pectoral girdle consists of a clavicle or collar bone and a scapula.

The scapula is a large, thin, triangular bone situated in the dorsal surface of the ribcage between the second and seventh ribs. It has a slightly elevated ridge called the spine which projects as a flat, expanded process called the acromion. The clavicle articulates with this process.

Below the acromion is a depression called the glenoid cavity which articulates with the head of the humerus to form the shoulder joint. Each clavicle is a long slender bone with two curvatures which lies horizontally and connects axial skeleton with appendicular skeleton.

Question 10.
Write a note on the bones of the upper limb?
Answer:
The upper limb consists of 30 separate bones and is specialized for1 mobility. The skeleton of the arm, the region between the shoulder and elbow is the humerus. The head of humerus articulates with the glenoid cavity of the scapula and forms the shoulder joint. The distal end of humerus articulates with the two forearm bones the radius and ulna. The forearm is the region between the elbow and the wrist.

Olecranon process is situated at the upper end of the ulna which forms the pointed portion of the elbow. The hand consists of carpals, metacarpals and phalanges. Carpals, the wrist bones, 8 in number are arranged in two rows of four each.

The anterior surface of the wrist has tunnel-like appearance, due to the arrangement of carpals with the ligaments. This tunnel is termed as carpal tunnel. Metacarpals, the pajm bones are 5 in number and phalanges the digits bones are 14 in number.

Question 11.
Explain the structure of Pelvic Girdle?
Answer:
The pelvic girdle is a heavy structure specialized for weight bearing. It is composed of two hip bones called coxal bones that secure the lower limbs to the axial skeleton. Together, with the sacrum and coccyx, the hip bones form the basin-like bony pelvis.

Each coxal bone consists of three fused bones, ilium, ischium and pubis. At the point of fusion of ilium, ischium, and pubis a deep hemispherical socket called the acetabulum is present on the lateral surface of the pelvis.

It receives the head of the femur or thigh bone at the hip joint and helps in the articulation of the femur. Ventrally the two halves of the pelvic girdle meet and form the pubic symphysis containing fibrous cartilage..

The ilium is the superior flaring portion of the hip bone. Each ilium forms a secure joint with the sacrum posteriorly. The ischium is a curved bar of bone.

The V-shaped pubic bones articulate anteriorly at the pubic symphysis. The pelvis of male is deep and narrow with larger heavier bones and the female is shallow, wide and flexible in nature, and this helps during pregnancy which is influenced by female hormones.

Question 12.
Write a note on the bones of lower limb?
Answer:
The lower limb consists of 30 bones which carries the entire weight of the erect body and is subjected to exceptional forces when we jump or run. The bones of the lower limbs are thicker and stronger than the upper limbs.

The three segments of each.lower limb are the thigh, the leg or the shank and the foot. The femur is the single bone of the thigh. It is the largest, longest and strongest bone in the body.

The head of femur articulates with the acetabulum of the pelvis to form the hip joint. Two parallel bones, the tibia and fibula, form the skeleton of the shank.

A thick, triangular patella forms the knee cap, which protects the knee joint anteriorly and improves the leverage of thigh muscles acting across the knee. The foot includes the bones of ankle, the tarsus, the metatarsus and the phalanges or toe bones.

The foot supports our body weight and acts as a lever to propel the body forward, while walking and running. The tarsus is made up of seven bones called tarsals. The metatarsus consists of five bones called metatarsals. The arrangement of the metatarsals is parallel to each other. There are 14 phalanges in the toes which are smaller than those of the fingers.

Question 13.
Explain the structure of a typical long bone?
Answer:
A typical long bone has a diaphysis, epiphyses and membranes. A tubular diaphysis or shaft, forms the long axis of the bone. It is constructed of a thick collar of compact bone that surrounds a central medullary cavity or marrow cavity. The epiphyses are the bone ends.

Compact bone forms the exterior of epiphyses and their interior contains spongy bone with red marrow. The region where the diaphysis and epiphyses meet is called the metaphysis. The external surface of the entire bone except the joint surface is covered by a double-layered membrane called the periosteum.

The outer fibrous layer is dense irregular connective tissue. The inner osteogenic layer consists of osteoblasts (bone- forming cells) which secrete bone matrix elements and osteoclasts (bone-destroying cells). In addition, there are primitive stem cells, osteogenic cells, that give rise to the osteoblasts.

The periosteum is richly supplied with nerve fibres, lymphatic vessels and blood vessels. Internal bone surfaces are covered with a delicate connective tissue membrane called the endosteum. The endosteum covers the trabeculae of spongy bone and lines the canals that pass through the compact bone. It also contains both osteoblasts and osteoclasts. Between the epiphysis and diaphysis epiphyseal plate or growth plate is present.

Question 14.
What are joints?
Answer:
The joints are points of contact between bones.

Question 15.
Explain the types of joints?
Answer:
(I) Fibrous joints or Synarthroses: They are immovable fixed joints in which no movement between the bones is possible. Sutures of the flat skull bones are fibrous joints.

(II) Cartilaginous joints or Amphiarthroses: They are slightly movable joints in which the joint surfaces are separated by a cartilage and slight movement is only possible, e.g., Joints of adjacent vertebrae of the vertebral column.

(III) Synovial joints or Diarthroses joints: They are freely movable joints, the articulating bones are separated by a cavity which is filled with synovial fluid e.g., Pivot joint – between atlas and axis plane/gliding joint – between the carpals

  • Saddle j oint – between the carpal and metacarpal
  • Ball and socket joint – between humerus and pectoral girdle
  • Hinge joint – knee joint
  • Condyloid or angular or ellipsoid-joint between radius and carpal.

Question 16.
Write a short note on myasthenia gravis?
Answer:
Myasthenia gravis: An autoimmune disorder affecting the action of acetylcholine at neuromuscular junction leadihg to fatigue, weakening and paralysis of skeletal muscles. Acetylcholine receptors on the sarcolemma are blocked by antibodies leading to weakness of muscles. When the disease progresses, it can make chewing, swallowing, talking and even breathing difficult.

Question 17.
Explain muscle fatigue?
Answer:
Muscle fatigue is the inability of a muscle to contract after repeated muscle contractions. This is due to lack of ATP and accumulation of lactic acid by anaerobic breakdown of glucose.

Question 18.
Explain muscle atrophy?
Answer:
A decline or cessation of muscular activity results in the condition called atrophy which results in the reduction in the size of the muscle and makes the muscle to become weak, which occurs with lack of usage as in chronic bedridden patients.

Question 19.
Write a short note on muscle pull?
Answer:
Muscle pull is actually a muscle tear. A traumatic pulling of the fibres produces a tear known as sprain. This can occur due to sudden stretching of muscle beyond the point of elasticity. Back pain is a common problem caused by muscle pull due to improper posture with static sitting for long hours.

Question 20.
Write a short note on muscular dystrophy?
Answer:
The group of diseases collectively called the muscular dystrophy are associated with the progressive degeneration of skeletal muscle fibres, weakening the muscles and leading to death from lung or heart failure. The most common form of muscular dystrophy is called ‘ Duchene Muscular Dystrophy (DMD).

Question 21.
Explain the disorders of skeletal system?
Answer:
Arthritis and osteoporosis are the major disorders of skeletal system.
1. Arthritis: Arthritis is an inflammatory or degenerative disease that damages the joints. There are several types of arthritis.

(I) Osteoarthritis: The bone ends of the knees and other freely movable joints wear away as a person ages. The joints of knees, hip, fingers and vertebral column are affected.

(II) Rheumatoid arthritis: The synovial membranes become inflamed and there is an accumulation of fluid in the joints. The joints swell and become extremely painful. It can begin at any age but symptoms usually emerge before the age of fifty.

(III) Gouty arthritis or gout: Inflammation of joints due to accumulation of uric acid crystals or inability to excrete it. It gets deposited in synovial joints.

2. Osteoporosis: It occurs due to deficiency of vitamin D and hormonal imbalance. The bone becomes soft and fragile. It causes rickets in children and osteomalacia in adult females. It can be minimized with adequate calcium intake, vitamin D intake and regular physical. activities.

Question 22.
Explain the basic categories of exercise and physical activity?
Answer:
Exercise and physical activity fall into four basic categories. Endurance, strength, balance and flexibility. Endurance or aerobic activities increase the breathing and heart rate. They keep the circulatory system healthy and improve overall fitness.

Strength exercises make the muscles stronger. They help to stay independent and cany out everyday activities such as climbing stairs and carrying bags.

Balance exercises help to prevent falls which is a common problem in older adults. Many strengthening exercises also improves balance.

Flexibility exercises help to stretch body muscles for more freedom of joint movements.

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Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

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Samacheer Kalvi 11th Chemistry Basic Concepts of Organic Reactions Textual Evaluation Solved

I. Choose The Best Answer:

Question 1.
For the Following reactions –
(A) CH3 CH2 CH2 Br + KOH → CH3 – CH + KBr + H2O
(B) (CH3 )3 CBr + KOH → (CH3 )3 COH + KBr
(C) Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Which of the following statement is correct’?
(a) (A) is elimination, (B) and (C) are substitution
(b) (A) is substitution, (B) and (C) are elimination
(c) (A) and (B) are elimination and (C) is addition reaction
(d) (A) is elimination, B is substitution and (C) is addition reaction.
Answer:
(d) (A) is elimination, B is substitution and (C) is addition reaction.

Question 2.
What is the hybridisation state of benzyl carbonium ion?
(a) sp2
(b) spd2
(c) sp3
(d) sp2d
Answer:
(a) sp2

Question 3.
Decreasing order of nucleophilicity is –
(a) OH> RNH2 > OCH3> RNH2
(b) NH2> OH> OCH3> RNH2
(c) NH,> CH3O >OH> RNH2
(d) CH3O> NH2> OH> RNH2
Answer:
(b) NH2> OH> OCH3> RNH2+

Question 4.
Which of the tòllowing species is not eLectrophilic in nature?
(a) Cl+
(b) BH3
(c) H3O+
(d) +NO2
Answer:
(c) H3O+

Question 5.
Homolytic fission of covalent bond leads to the formation of –
(a) electrophile
(b) nucicophile
(c) Carbocation
(d) free radical
Answer:
(d) free radical

Question 6.
Hyper Conjugation is also known as –
(a) no bond resonance
(b) Baker – nathan effect
(c) both (a) and (b)
(d) none of these
Answer:
(c) both (a) and (b)

Question 7.
Which of the group has highest +I effect?
(a) CH3
(b) CH3CH2
(c) (CH3)2 -CH-
(d) (CH3)3-C-
Answer:
(d) (CH3)3-C-

Question 8.
Which of the following species does not exert a resonance effect?
(a) C6H5OH
(b) C6H5Cl
(c) C6H5NH2
(d) C6H5NH3
Answer:
(d) C6H5NH3

Question 9.
-I effect is shown by –
(a) -Cl
(b) -Br
(c) both (a) and (b)
(d) -CH3
Answer:
(c) both (a) and (b)

Question 10.
Which of the following carbocation will be most stable?
(a) \({ PH }_{ 3 }\overset { + }{ C- } \)
(b) CH3–\(\overset { + }{ C{ H }_{ 2 } } \)
(c) (CH3)2–\(\overset { + }{ C{ H } } \)
(d) CH2 = CH-CH3
Answer:
(a) \({ PH }_{ 3 }\overset { + }{ C- } \)

Question 11.
Assertion : Tertiary Carbocations are generally formed more easily than primary Carbocations.
Reason : Hyper conjugation as well as inductive effect due to additional alkyl group stabilize tertiary carbonium ions.
(a) both assertion and reason arc true and reason is the correct explanation of assertion.
(b) both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false
(d) Both assertion and reason are false
Answer:
(a) both assertion and reason are true and reason is the correct explanation of assertion.

Question 12.
Heterolytic fission of C-C bond results in the formation of –
(a) free radical
(b) Carbanion
(c) Carbocation
(d) Carbanion and Carbocation
Answer:
(e) Carbocation

Question 13.
Which of the following represent a set of nucleophiles?
(a) BF3, H2O, NH2-
(b) AlCl3, BF3, NH3
(c) CN, RCH2, ROH
(d) H+, RNH+2, CCl2
Answer:
(c) CN, RCH2, ROH

Question 14.
Which of the following species does not acts as a nucleophile?
(a) ROH
(b) ROR
(c) PCl3
(d) BF3
Answer:
(d) BF3

Question 15.
The geometrical shape of carbocation is
(a) Linear
(b) tetrahedral
(c) Planar
(d) Pyramidal
Answer:
(c) Planar

II. Write brief answer to the following questions.

Question 16.
Write short notes on:
(a) Resonance
(b) Hyper conjugation
Answer:
(a) Resonance:
1. Certain organic compounds can be represented by more than one structure and they differ only in the position of bonding and lone pair of electrons. Such structures are called resonance structure and this phenomenon is called as resonance. This phenomenon is also called as mesomerism or mesomeric effect.

2. For example, the structure of aromatic compounds such as benzene and conjugated system like 1,3 butadiene cannot be represent by a single structure and their observed properties can be explained on the base of a resonance hybrid.

3. Resonance structure of benzene.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
(I) and (II) are called as resonance hybridš of benzcnc.

4. For 1,3 butadiene:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

(I), (II) and (III) are called as resonance hybrids of 1,3 butadiene.

(b) Hyper conjugation:
1. The denationalization of electrons of σ bond is called as hyper conjugation. it is a special stabilizing effect that results due to the interaction of electrons of a σ bond with the adjacent empty non-bonding P-orbitais resulting in an extended molecular orbital.

2. Hyper conjugation is a permanent effect.

3. For example, in propane. the σ- electrons of C-H bond of methyl group can be delocalised into the it- orbital of doubly bonded carbon as represented below.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 17.
What are electrophiles and nucleophiles? Give suitable examples for each.
Answer:
Electrophiles:
Electrophiles are reagents that are attracted towards negative charge or electron rich center. They are either positively charged ions or electron deficient neutral molecules.
Example:
CO2, AlCl3, BF3, FeCl3, NO+, NO+2,, etc.

Nucleophiles:
Nucleophiles are reagents that has high affinity for electropositive centers. They possess an atom has a unshared pair of electrons. They are usually negatively charged ions or electron rich neutral molecules.
Example.
NH3, R-NH2, R-SH, H2O, R-OH, CN. OH etc.

Question 18.
Show the heterolysis of covalent bond by using curved arrow notation and complete the following equations. Identify the nucleophile In each case.
1. CH3-Br+KOH →
2. CH3-OCH3+HI →
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
In this case, \(\overset { \ominus }{ Br } \) and \(\overset { \ominus }{ OH } \) are nucleophiles.

Another mechanism:
Step – I
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Step-II
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

In this case, OCH3 and -I are nucleophiles
Another mechanism:
Step – I
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Step – II

Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 19.
Explain inductive effect with suitable example.
Inductive effect:
Answer:
1. it is defined as the change ¡n the polarization of a covalent bond due to the presence of adjacent bonded atoms or groups in the molecule. It is denoted as I-effect.

2. Atoms or groups which lose electron towards a carbon atom are said to have a +I effect.
Example:
CH3-,(CH3)2 CH-,(CH3)2 C- etc.

3. Atoms or groups which draw electrons away from a carbon atom are said to have a -I effect.
Example:
-NO2. -I, -Br, -OH, C6H5 etc.

4. For example, consider ethane and ethyl chloride. The C-C bond in ethane is non polar while the C-C bond in ethyl chloride is polar. We know that chlorine is more electronegative than carbon and hence it attracts the shared pair of electrons between C-Cl in ethyl chloride towards it self.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
This develops a slight negative charge on chlorine and a slight positive charge on carbon to which chlorine is attached. To compensate it, the C1 draws the shared pair of electron between itself and C2. This polarization effect is called inductive effect.

Question 20.
Explain electromeric effect.
Electromeric effect:
Answer:
1. The electromeric effect refers to the polarity produced in a multiple bonded the compound when it is attacked by a reagent when a double or a triple bond is exposed to an attack by an electrophile the two π electrons which from the π bond are completely transferred to one atom or the other.

2. When a nucleophile approaches the carbonyl compound, the π-electrons between C and O is instantaneously shifted to the more electronegative oxygen. This make the carbon electron deficient and thus facilitating the formation of a new bond between the incoming nucleophile and the carbonyl carbon atom.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

3. When an electrophile such as H approaches an alkene molecule their electrons are instantaneously shifted to the electrophiie and a new bond is formed between carbon and hydrogen. This makes the other carbon electron deficient and hence it acquires a positive charge.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

4. This effect denotes as E-effect.

Question 21.
Give examples for the following types of organic reactions
1. – elimination
2. Electrophilic substitution.
Answer:
1. β-eIimination:
Elimination reactions involve the cleavage of a a bond and tórmation of air bond. A nucleophilic pair of electrons heads into a new it bond as a leaving group departs. This process is called 3-elimination because the bond Í to the nucleophilic pair of electrons breaks.
Example:
(a) n-propyl bromide on reaction with alcoholic KOH give propene. in this reaction hydrogen and Br are eliminated.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

(b) Acid-catalysed dehydration of alcohols
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

2. Electrophilic substitution:
Substitution reactions when are brought about by electrophiles are called electrophilic substitution reaction.
Example:
(a) Nitration of benzene
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

(b) Bromination of benzene:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Samacheer Kalvi 11th Chemistry Basic Concepts of Organic ReactionsAdditional Questions Solved

I. Choose The Correct Answer.

Question 1.
Statement-I: All the organic molecules contain covalent bonds.
Statement-II: Organic molecules are formed by the mutual sharing of electrons between atoms.
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.
(b) Statement-I and II are correct but Statement-II is not correct explanation of statement-I.
(c) Statement-I is correct but statement-II is wrong.
(d) Statement-I is wrong but statement-II is correct.
Answer:
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.

Question 2.
Statement-I : Homolytic cleavage is symmetrical one.
Statement-II : A single covalent bond breaks and each of the bonded atoms retains one electron.
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.
(b) Statement-I and II are correct but statement-II is not correct explanation of statement-I.
(c) Statement-I is correct but statement-II is wrong.
(d) Statement-I is wrong but statement-II is correct.
Answer:
(a) Statement-I and II are correct and statement-Il is correct explanation of statement-I.

Question 3.
Which one of the following is an example for free radical initiators?
(a) Benzoyl peroxide
(b) Benzyl alcohol
(c) Benzyl acetate
(d) Benzaldehyde.
Answer:
(a) Benzoyl peroxide.

Question 4.
Which one of the following ¡s correct order of stability of alkyl free radicals?
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 5.
Statement-I: Ileterolytic cleavage is unsymmetrical one.
Statement-II: A covalent bond breaks and one of the bonded atom retains the bond pair of electrons.
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.
(b) Statement-I and II are correct but statement-II is not correct explanation of statement-I.
(c) Statement-I is correct but statement-II is wrong.
(d) Statement-I is wrong but statement-II is correct
Answer:
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.

Question 6.
Which of the following is correct order of the stability of carbocations?
(a) +CH3 > +CH2 CH3> +CH(CH3)2 > +C(CH3)3
(b) +CH2CH3 > +CH3 >+ CH(CH3)2 > +C(CH3)3
(c) +C(CH3)3 > +CH(CH3)2 >+CH2CH3CH3
(d) +CH(CH3)2 > +CH3 > +CH2CH3 > +C(CH3)3
Answer:
(c) +C(CH3)3 > +CH(CH3)2 >+CH2CH3CH3

Question 7.
Which one of the following is correct order of the stability of carbanions?
(a) C(CH3)3> CH(CH3)2> CH2-CH3> CH3
(b) CH3> CH2-CH3> CH(CH3)2> C(CH3)3
(c) CH(CH3), > CH3> CH2-CH3 > C(CH3)3
(d) CH2-CH3 > CH(CH3)2> CH3> C(CH3)3
Answer:
(b) CH3> CH2-CH3> CH(CH3)2> C(CH3)3

Question 8.
Which one of the following is not electrophile’?
(a) NH3
(b) AlCl3
(c) FeCl3
(d) R-X
Answer:
(a) NH3

Question 9.
Which one of the following is not nucleophile
(a) H2O
(b) NH3
(c) R-OH
(d) FeCl3
Answer:
(d) FeCl3

Question 10.
Which one of the following is positively charged electrophiles?
(a) CO2
(b) AlCl3
(c) BF3
(d) RX
Answer:
(d) RX

Question 11.
Which one of the following is nucleophile?
(a) BF3
(b) AlCl3
(c) CO2
(d) R-SH
Answer:
(d) R-SH

Question 12.
Which one of the following species has tendency to show-I effect?
(a) -CH3
(b) -CH2-CH3
(c) -CH(CH3)2
(d) -C6H5
Answer:
(d) -C6H5

Question 13.
Which one of the following species has tendency to show +I effect?
(a) -NH2
(b) -Cl
(c) -C6H5
(d) -CH3
Answer:
(d) -CH3

Question 14.
Which one of the following has strongest acidic character?
(a) HCOOH
(b) CH3COOH
(c) CH2ClCOOH
(d) CCl3COOH
Answer:
(d) CCl3COOH

Question 15.
Which one of the following has least acidic character?
(a) HCOOH
(b) CH3COOH
(c) CH2ClCOOH
(d) CCl3COOH
Answer:
(b) CH3COOH

Question 16.
Pick out the correct order of acid strength.
(a) CH3-CH2-COOH > CH3COOH > CH2CICOOH
(b) CH3COOH > CH3-CH2-COOH >CH2CICOOH
(c) CH2ClCOOH > CH3COOH > CH3-CH2-COOH
(d) CH2ClCOOH > CH3-CH2-COOH > CH3COOH
Answer:
(c) CH2ClCOOH > CH3COOH > CH3—CH2-COOH

Question 17.
Statement-I: Fluoro acetic acid is stronger acid than acetic acid
Statement-II: Fluorine has high electronegativity and it is facilitate to dissociate the OH bond easily.
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.
(b) Statement-I and II are correct but statement-II is not correct explanation of statement-I.
(c) Statement-I is correct but statement-II is wrong.
(d) Statement-I is wrong but statement-Il is correct
Answer:
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.

Question 18.
Which one of the following is an example for negative mesomeric effect?
(a) -SH
(b) -SR
(c) -NH2
(d) -NO2
Answer:
(d) -NO2

Question 19.
Which one of the following electrophile used for nitration of benzene?
(a) \({ Br }^{ \oplus }\)
(b) \({ { NO }_{ 2 }^{ \ominus } }\)
(c) -NH2
(d) \({ NO }^{ \ominus }\)
Answer:
(b) \({ { NO }_{ 2 }^{ \ominus } }\)

Question 20.
Identify the one which does not come under the organic addition reaction
(a) Hydration
(b) Dehydration
(c) Halogenation
(d) Hydro halogenation
Answer:
(b) Dehydration

Question 21.
Primary alcohols undergo which type of reaction to form alkenes?
(a) Elimination
(b) Oxidation
(c) Reduction
(d) Hydrolysis
Answer:
(a) Elimination

Question 22.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions Identify the type of reaction
(a) Addition reaction
(b) Elimination reaction
(c) Reduction reaction
(d) Oxidation reaction
Answer:
(d) Oxidation reaction

II. Match the following.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Answer:
Samacheer Kalvi 11th Chemistry Chapter 11 Fundamentals of Organic Chemistry

III. Fill in the blanks.
Question 1.
The slowest step in the mechanism determines ………..
Answer:
Rate of reaction

Question 2.
Homolytic cleavage occurs under the conditions of ………….
Answer:
High temperature.

Question 3.
During the cleavage of Azobisisobutyronitrile produces species ……….
Answer:
Free radical

Question 4.
The cleavage of C-Br bond in tert-butyl bromide leads to formation of ………..
Answer:
Carbocation

Question 5.
The cleavage of C-H bond in aldehydes Leads to formation of ………….
Answer:
Carbanion

Question 6.
Electron displacement occurring in saturated compounds along a carbon chain is termed as …………
Answer:
Inductive effect

Question 7.
The addition of H to alkene is an example of effect ……….
Answer:
+E

Question 8.
Example for positive mesomeric effect is ………..
Answer:
-OH

Question 9.
Acidity of phenol was explained by …………
Answer:
R-effect

Question 10.
Hydrolysis of alkyl halide is an example for ……….
Answer:
Nucleophilic substitution

Question 11.
4-hydroxy phenol reacts with acidified potassium dichromate to gives
Answer:
p-Benzoquinone

Question 12.
Enzyme present in apple is
Answer:
Poly phenol oxidase

Question 13.
Benzene reacts with H2 in the presence of Pt to give
Answer:
Cyclohexane

Question 14.
Alcohol on refluxing with K2Cr2O7 gives……..
Answer:
Carboxylic acid

Question 15.
Carbonyl compounds especially ketones undergo reduction to form ………..
Answer:
Secondary alcohols

Question 16.
Ethane undergo thrombolytic cleavage to form ………..
Answer:
Methyl free radical

IV. Choose the odd one out.
Question 1.
(a) NH3
(b) H2O
(c) CN
(d) RSH
Answer:
(c) CN It is a negatively charged nucleophile whereas others are neutral nucleophiles.

Question 2.
(a) CO2
(b) RX
(c) MCl3
(d) FeCl3
Answer:
(b) RX. It is a positive charged electrophiles whereas others are neutral clectrophiles.

Question 3.
(a) -F
(b) – Cl
(c) – COOH
(d) CH3O
Answer:
(d) CH3O. It is a group carrying negative charge one electron donating (or) +I group whereas others are -I group.

Question 4.
(a) – C(CH3)3
(b) – COOK
(c) – CH(CH3)2
(d) – COO
Answer:
(b) – COOH. It has -I effect whereas other have +I effect.

Question 5.
(a) -OH
(b) -NH2
(c) -SH
(d) -COOH
Answer:
(d) – COOH. It has negative mesomeric effect whereas others have positive mesomeric effect.

V. Choose the correct pair.
Question 1.
(a) Homolytic clevage : carbocation
(b) Homolytic clevage : carbanion
(c) Homolytic clevage : free radicals
(d) Heterolytic clevage : free radicals
Answer:
(c) Homolytic clevage : free radicals

Question 2.
(a) + C(CH3)3 > +CH(CH3)2> +CH2-CH3 > +CH3 : Relative stability of carbocations.
(b) – C(CH3)3 > CH(CH3)2 > CH2-CH3 > CH3 : Relative stability of carbanion.
(c) OH,RO, RCOO : Neutral nucleophile
(d) AlCl3, BF3, FeCl3 : Positively charged nucleophile
Answer:
(a) + C(CH3)3 > +CH(CH3)2> +CH2-CH3 > +CH3 : Relative stability of carbocations.

Question 3.
(a) CH3 – CH2 – CH2Br + Alcoholic KOH : Substitution reaction
(b) CH3 – CH2 – CH2Br + Alcoholic KOH : Elimination reaction
(c) CH3CHO + Acidic dichromate : Reduction reaction
(d) Benzene + Pt + H2 : Oxidation reaction
Answer:
(b) CH3 – CH2 – CH2Br + Alcoholic KOH : Elimination reaction

VI. Choose the incorrect pair.
Question 1.
(a) – OH,- SH, -NH2 : Positive mesometic effect
(b) – NO2 >CO, – COOH : Negative mesomeric effect
(c) – F, – Cl, – NO2 : Electron withdrawing group
(d) – C(CH3)3, – CH(CH3)3 – CH2 – CH3 : Electron withdrawing group
Answer:
(c) – F, – Cl, – NO2 : Electron withdrawing group

Question 2.
(a) NH3 andAmines : Neutral nucleophile
(b) OH and RCOO : Negative nucleophile
(c) RX and H3O3 : Positive electrophile
(d) MCl3, BF3 : Negative electrophile
Answer:
(d) AlCl3, BF3 : Negative electrophile

Question 3.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

VII. Assertion and Reason.
Question 1.
Assertion (A) : Neutral molecule SnCl4 can act as an electrophile.
Reason (R) : it has vaccant ‘d’ orbitais which can accommadate the electrons from others.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 2.
Assertion (A) : The C-C bond in ethane is non-polar while the C-C bond in ethyl chloride is polar.
Reason (R) : Chlorine is more electronegative than carbon and hence it attracts the shared pair of electron between C-C in ethyl chloride and it develops a negative charge on Cl and positive charge on Carbon.
(a) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 3.
Assertion (A): Phenol is more acidic than aliphatic alcohols.
Reason (R): The phenoxide ion is more stabilized than phenol by resonance effect and hence resonance favours ionization of phenol to form H and shows acidity.
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).

Question 4.
Assertion (A): The cut apple turns brown.
Reason (R): Cut apple exposes its cells to atmospheric oxygen and the oxidizes the phenolic compound present in it. Due to this enzymatic browning, the cut apple turns brown.
(a) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Samacheer Kalvi 11th Chemistry Basic Concepts of Organic Reactions 2 Marks Question and Answers

II. Answer briefly.

Question 1.
What are organic reactions?
Answer:
Substrate is an organic molecule reacts with reagent, which may be an organic, inorganic or any agent like heat, photon etc, that brings about the chemical change to form a product, this is known as organic reactions.

Question 2.
What is mechanism of the reaction?
Answer:
In the organic reactions, that series of simple steps which collectively represent the chemical change, from substrate to product, this is called as the mechanism of the reaction. The slowest step in the mechanism determines the overall rate of the reaction.

Question 3.
Mention the types of fission of a covalent bond?
Answer:
There are three types of fission of a covalent bond. They are,

  • Homolytic cleavage
  • Heterolytic cleavage

Question 4.
Explain the homolytic fission of a covalent bond?
Answer:

  • Homolytic cleavage is the process in which a covalent bond breaks symmetrically in such way that each of the bonded atoms retains one electron.
  • This type of cleavage occurs under high temperature or in the presence of UV light.
  • in a compound containing non-polar covalent bond formed between atoms of similar electronegativity, in such molecules the cleavage of bonds results into free radicals.
  • For example, ethane undergo homolytic fission to produce, two methyl free radicals.
    Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 5.
What are free radical initiators?
Answer:
The types of reagents that promote homolytic cleavage in substrate are called as free radical initiators. They are short lived and are highly reactive.

Question 6.
Mention any two examples for free radical initiators?
Answer:
Two examples for free radical initiators

  • Azobisisobutyronitnie (AIBN)
  • Benzoyl peroxide

Question 7.
Explain the heterolytic fission of a covalent bond?
Answer:

  • Heterolytic cleavage is the process in which a covalent bond breaks unsymmetrically such that one of the bonded atom retains the bond pair of electron.
  • It results in the formation of a cation and an anion of the two bonded atoms the most electronegative.
  • For example in tert-butyibromide, the C-Br bond is polar as bromine is more electronegative than carbon. Hence the C-Br undergoes heterolytic cleavage to form a terl-butyl carbocation and bromide anion.
    Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 8.
What are carbocations?
Answer:
1. Let us consider the heterolytic fission of the bond C-X present in organic molecule. If the atom X has greater electronegativity than the carbon atom, the former takes away the bonding electron pair and becomes negatively charged while the carbon will lose its electron and thus acquire a positive charge.
2. Such a cationic species carrying a positive charge on carbon are known as carbocation or carhonitim ion.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 9.
What are carbanions?
Answer:
Let us consider the heterolytic fission of the bond C-X present in an organic molecule. if the carbon atom has greater electronegativity than the atom X, the former takes away the bonding electron pair and acquires a negative charge. The resulting carbon anion is known as carbanion.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 10.
Differentiate the carbocation and carbanion
Answer:
Carbocation:

  • In a carbocation carbon bearing positive charge.
  • Carbon bearing positive charge has sp2 hybridization.
  • It has a planar structure.
  • Example:
    CH3+ (CH3)2\({ C }^{ \ominus }\)H, (CH3)3C+ etc.

Carbanion:

  • In a carbanion carbon bearing negative charge.
  • Carbon bearing negative charge has sp3 hybridization.
  • It has a pyramidal structure.
  • Example:
    CH3–\({ C }^{ \ominus }\)H2, CH3-CHO, (CH3)2\({ C }^{ \ominus }\)H etc.

Question 11.
Identify which of the following are electrophiles and nucleophiles?

  1. NH3
  2. AlCl3
  3. R-SH
  4. R-X
  5. R-O-R
  6. BF3

Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 12.
How will you distinguish between electrophiles and nucleophiles?
Answer:
Electrophiles:

  • They are electron deficient.
  • They are cations.
  • They are lewis acids.
  • Accept an electron pair.
  • Attack on electron rich sites.

Nucleophites:

  • They are electron rich.
  • They are anions.
  • They are lewis bases.
  • Donate an electron pair.
  • Attack on electron deficient sites.

Question 13.
What are all the sources for human body that produces free radicals?
Answer:
Sources for human body produces free radicals,

  • Human body is exposed to X-rays.
  • Cigarette smoke.
  • Industrial chemicals.
  • Air pollutants.

Question 14.
In what way free radical affect the human body?
Answer:

  • Free radicals can disrupt cell membranes.
  • Increase the risk of many forms of cancer.
  • Damage the interior lining of blood vessels.
  • leads to a high risk of heart disease and stroke.

Question 15.
How to reduce the effect of free radicals?
Answer:

  • Use vitamins and minerals to counter the effects of free radicals.
  • Fruits contains antioxidants which decrease the effects of free radicals.

Question 16.
Identify which of the following shows +I and -I effect?

  1. -NO2
  2. -SO2H
  3. -I
  4. -OH
  5. CH2O-
  6. CH2

Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 17.
Why chloro acetic acid is stronger acid than acetic acid?
Chioro acetic acid:
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Chioro acetic acid has Cl-group and it has high electronegativity and shows -I effect. Therefore Cl-atom to faciLitate the dissociation of O-H bond very fastly, whereas in the case of acetic acid, has CH3 group and it shows +I effect, theretòre dissociation of O-H bond will be more difficult. Thus chloro acetic acid is stronger acid than acetic acid.

Question 18.
Explain the positive and negative electromeric effects?
Answer:
1. When an electrophile such as W approaches an alkene molecule, the π electrons are instantaneously shifted to the electrophile and a new bond is formed between carbon and hydrogen. This makes the other carbon electron deficient and hence it acquires a positive charge.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
2. When the it electron is transferred towards the attacking reagent, it is called positive electromeric (+E) effect.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
3. Whèn the t electron is transferred away from the attacking reagent it is called negative electromenc (-E) effect.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
For example : The attack of C\({ N }^{ \ominus }\) on a carbonyl carbon.
Samacheer Kalvi 11th Chemistry Chapter 11 Fundamentals of Organic Chemistry

Question 19.
Write a short notes on positive mesomeric effect?
Answer:

  • Positive mesomeric effect occurs, when the electrons move away from substituent attached
    to the conjugated system.
  • It occurs, if the electron releasing substituents are attached to the conjugated system.
  • The attached group has a tendency to release electrons through resonance, these electron
    releasing groups are usually denoted as +R or +M groups
  • Examples: -OH, -SH, -OR, -SR, -NH2 etc.

Question 20.
Write a short notes on negative mesomeric effect?
Answer:

  • Negative mesomeric effect occurs, when the electrons move towards the substituent attached to the conjugated system.
  • It occurs if the electron withdrawing substituents are attached to the conjugated system.
    The attached group has a tendency to withdraw electrons through resonance, these electron withdrawing groups are usually denoted as -R or -M groups.
  • Examples:
    Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 21.
What are addition reactions? Give an example.
Answer:
All organic compounds having double or triple bond adopt addition reactions in which two substances unite to form a single compound. During the addition reaction the hybridization of the substrate changes as only one bond breaks and two new bonds are formed.
Example:

Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 22.
What are elimination reaction? Give an example.
Answer:
In these reactions, two atoms or groups are removed from a molecule without being substituted by other atoms or groups. A new C-C double bond is formed between the carbon atoms to which the eliminated atoms or groups are previously attached. It is always accompanied with change in hybridization.
Example:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 23.
What are organic oxidation reactions? Give an example.
Answer:
Most of the oxidation reaction of organic compounds involves gain of oxygen or loss of hydrogen.
Example:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 24.
What are organic reduction reactions? Give an example.
Answer:
Most of the reduction reaction of organic compounds involves gain of hydrogen or loss of oxygen.
Example:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 25.
Why cut apple turns a brown colour?
Answer:

  • Apples contains an enzyme called polyphenol oxidase (PPO) also known as tyrosinase.
  • Cutting an apple exposes its cells to the atmospheric oxygen and oxidizes the phenolic compounds present in apples. This is called the “enzymatic browning” that turns a cut apple brown.
  • In addition to apples, enzymatic browning is also evident in bananas, pears, avocados and even potatoes.

Question 26.
What are functional group inter conversions?
Answer:
Organic synthesis involves functional group inter conversions. A particular functional group can be converted into other functional group by reacting it with suitable reagents. For example, the carboxylic acid group (-COOH) present in organic acids can be transformed into -COCl by treating the acid with SOCl2 reagent..

Question 27.
How will you convert alcohol into aldehyde?
Answer:
When primary alcohol reacts with acidified potassium dichromate to gives aldehyde.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 28.
What happen when nitrile undergoes acid hydrolysis?
Answer:
When alkyl nitrile undergoes acid hydrolysis to give amide, which on further hydrolysis to give carboxylic acid.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 29.
Complete the following reactions and identify the products?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 30.
Predict the product for the following reaction.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 31.
How will you convert benzene into cyclohexane?
Answer:
Benzene undergo catalytic hydrogenation in presence of platinum to give cyclohexane.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 32.
Complete the reaction and name the product
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 33.
Identify the product and mention the type of organic reaction.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
This is one of the elimination reaction.

Question 34.
Complete the following reaction and identify the product.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 35.
What are substitution reaction?
Answer:
In this reaction an atom or a group of atoms attached to a carbon atom is replaced by a new atom or a group of atoms.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Here, -Br is replaced by -OH group.

Question 36.
How will substitution reactions are classified?
Answer:
Substitution reactions can be classified as,

  • Nucleophilic substitution reaction.
  • Electrophilic substitution reaction.
  • Free radical substitution reaction.

Question 37.
Draw the resonance structures for the following compounds?
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 38.
Identify the reagents shown in box in the following equations as nucleophiles or electrophiles.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
In (a) \(\overset { \ominus }{ OH } \) is nucleophile
In (b) \(\overset { \ominus }{ CN } \) is nucleophile
In (c) CH3\(\overset { \ominus }{ CO } \) is electrophile

Question 39.
Classify the following reactions in one of the reaction type of organic reaction.
Answer:
(a) CH3CH2Br + \({ SH }^{ \ominus }\) → CH3CH2SH + \({ Br }^{ \ominus }\)
(b) (CH3)2C = CH2 + HCI (CH3)2CCl – CH3
(c) CH3CH2Br + \({ OH }^{ \ominus }\) → CH2 = CH2 + H2O + \({ Br }^{ \ominus }\)
(a) Nucleophilic substitution reaction.
(b) Electrophilic addition reaction.
(c) Elimination reaction.

Question 40.
Which electron displacement effect explain the following correct orders of acidity of the carboxylic acids?
Answer:
CH3CH2COOH > (CH3)2 CHCOOH > (CH3)3C COOH
As the number of alkyl group increases, +I effect increases and it strengthen the O-H bond. i.e., O-H bond dissociation is very difficult.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Therefore +I effect explain the correct orders of acidity of the carboxylic acids.

Question 41.
Which of the two, NO2CH2CH2\({O}^{ \ominus }\) or CH3CH2\({O}^{ \ominus }\) is expected to be more stable and why?
Answer:
v
(I) is more stable than (II) because NO2 group has -I effect and hence it tends to disperse the -ve charge on the 0-atom. In contrast., CH3CH, has +I effect. It therefore, tends to intensify the -ve charge and hence destabilizes it.

Samacheer Kalvi 11th Chemistry Basic Concepts of Organic Reactions 5 Marks Questions and Answers

Question 1.
Explain electron movement in organic reactions.
Answer:
All organic reactions can be understood by following the electron movements.

  • Lone pair becomes a bonding pair.
  • Bonding pair becomes a lone pair.
  • A bond breaks and becomes another bond.

The electron movement depends on the nature of the substrate, reagent and the prevailing conditions.
Type – 1.
A lone Pair to a bonding pair
v
Type – 2.
A bonding pair to a lone Pair
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Type – 3.
A bonding pair to an another bonding pair
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 2.
How does inductive effect influence the reactivity and acidity of carboxylic acids?
Answer:
(a) Reactivity:

  • When a highly electronegative atom such as halogen is attached to a carbon then it makes the C—X bond polar.
  • In such cases the -I effect of halogen facilitates the attack of an incoming nucleophile at the polarized carbon and hence increases the reactivity.
    Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
  • If a -I group is attacher neared to a carbonyl carbon, it decreases the availability of electron density on the carbonyl carbon and hence increases the rate of the nucleophilic addition reaction.

(b) Acidity of carboxylic acid:

  • When a halogen atom is attached to the carbon which is neared to the carboxylic acid group, its -I effect withdraws the bonded electrons towards itself and makes the ionization of H+ easy.
  • The acidity of various chioro acetic acid is in the following order. Cl3C-COOH > Cl2CHCOOH > ClCH2COOH The strength of the acid increases with increase in the effect of the group attached to the carboxyl group.
  • Similarly the following order of acidity in the carboxylic acids is due to the +1 effect of alkyl group. (CH3)3CCOOH < (CH3)2CHCOOH <CH3COOH

Question 3.
Explain the acidic nature of phenol.
Answer:

  • Resonance is useful in explaining certain properties such as acidic of phenol.
  • The phenoxide ion is more stabilized than phenol by resonance effect.
  • The resonance favours ionisation of phenol to form H+ and shows acidity.
    Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
    phenoxide ion resonance structures.
    Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
  • The above structures shows that there is a charge separation in the resonance structure of phenol which needs energy whereas there is no such hybrid structures in the case of phenoxide ion. This increases stability accounts for the acidic character of phenol.

Question 4.
How does hyper conjugation effect explain the stability of alkenes?
Answer:

  • The relative stability of various classes of carhonium ions may be explained by the number of no bond resonance structures that can be written for them.
  • Such structures are arrived by shifting the bonding electrons from an adjacent C-H bond to the electron deficient carbon.
  • in this way, the positive charge originally on carbon is dispersed to the hydrogen. This manner of electron release by assuming no bond character in the adjacent C-H bond is called hyper conjugation or Baker- Nathan effect.
  • The greater the hyper conjugation, the greater will be the stability of the compound. The increasing order of stability can be shown as:
    Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
  • Alkyl group increases in the C=C double bond carbon, hyper conjugation increases and stability of that organic compound also increases.

Question 5.
Explain the types of addition reactions?
Answer:
Addition reactions are classified into three types. They are,

  1. Electrophi lic addition reaction
  2. Nucleoph lije addition reaction
  3. Free radical addition reaction

1. Electrophilic addition reaction:
An electrophilic addition reaction can be described as an addition reaction in which a reactant with multiple bonds as in a double or triple bond undergoes has it air bond broken and two new a bond are formed.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
2. Nucleophilic addition reaction:
A nucleophilic addition reaction is an addition reaction where a chemical compound with an electron deficient or electrophilic double or triple bond, a it bond, reacts with a nucleophilic which is an electron rich reactant with the disappearance of the double bond and creation of two new single or a bonds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

3. Free radical addition reaction:
it is an addition reaction in organic chemistry involving free radicals. The addition may occur between a radical and a non radical or between two radicals.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 6.
Explain the types of substitution reaction?
Answer:
Substitution reactions are classified into three types. They are,

  1. Nucleophilic substitution reaction.
  2. Electrophilic substitution reaction.
  3. Free radical substitution reaction.

1. Nucleophilic substitution reaction:
A nucleophilic substitution reaction in organic chemistry is a type of reaction where a nucleophilic gets attached to the positive charged atoms or molecules of the other substance. A good example of a nucleophilic substitution reaction is the hydrolysis of alkyl bromide, under the basic conditions where in the nucleophile is nothing but the base H. whereas the leaving group is the Br°. The reaction for the following is as give below.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

2. Electrophilic substitution reaction:
The electrophilic substitution reaction involves the electrophiles. The electrophilic reactions occur mostly with the aromatic compounds. This types of substitution reaction arc basically defined as those chemical reactions where the electrophile replaces the functional group in a compound but not the hydrogen atom, sometimes hydrogen atom can be also replaced by electrophiles.
For example:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

3. Free radical substitution reaction:
Free radical substitution reaction involving free radicals arc a reactive intermediate.
A – X + Y → A – Y + X
CH4 + Cl → CH + HCl
CH3 + Cl → CH3Cl (methyl chloride)

Question 7.
For the following bond cleavages use curved-arrows to show the electron flow and classify each as homolytic or heterolytic fission. Identify reactive intermediate produced as free radical,, carbocation and carbanion?
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Question 8.
An organic compound (A) has a molecular formula C2H6O it is one of the primary alcohol. A reacts with acidified potassium dichromate to give B. B on further undergoes to oxidation reaction to give C. C on reacts with SOCl3 to give D which ¡s chlorinated product. Identify A,B,C and D, explain with equation.
Answer:
1. C2H6O is CH3-CH2-OH which is a primary alcohol (A)
2. CH3-CH2-OH (A) reacts with H+/ K2Cr2O7 to give acetaldehyde (B)
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
3. CH3CHO (B) which on further oxidation to give acetic acid (C).
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
4. Acetic acid reacts with SOCl2 to give acetyl chloride (D).
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 9.
An organic compound (A) of a molecular formula C2H4 which is a simple alkene. A reacts with dil H2S04 to give B. A again reacts with Cl2 to give C. Identify AB and C and write the equations.
Answer:
1. C2H2 is CH2=CH2 is a simple alkene. A is ethylene.
2. Ethylene (A) reacts with dii H2SO4 to give ethanol (B)
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
3. Ethylene (A) reacts with Cl, to give 1.2 dichloro ethane (C)
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 10.
Complete the reactions and identify the products.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Students looking for Chapter 11 Fundamentals of Organic Chemistry Concepts can find them all in one place from our Tamilnadu State Board Fundamentals of Organic Chemistry. Simply click on the links available to prepare the corresponding topics of Chemistry easily. Samacheer Kalvi 11th Chemistry Chapter wise Questions and Answers are given to you after sample research and as per the latest edition textbooks. Clarify all your queries and solve different questions to be familiar with the kind of questions appearing in the exam. Thus, you can increase your speed and accuracy in the final exam.

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry Multiple Choice Questions

Question 1.
Select the molecule which has only one ir bond.
(a) CH3-CH=CH-CH3
(b) CH3-CH=CH-CHO
(c) CH3-CH=CH-COOH
(d) All of these
Answer:
(a) CH3-CH=CH-CH3

Question 2.
In the hydrocarbon Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistrythe state of hybridisation of carbon 1,2,3,4 and 7 are in the following sequence.
(a) sp, sp, sp3, sp2, sp3
(b) sp2, sp, sp3, sp2, sp3
(c) sp, sp, sp2, sp, sp3
(d) none of these
Answer:
(a) sp, sp, sp3, sp2, sp3

Question 3.
The general formula for alkadiene is ……….
(a) CnH2n
(b) CnH2n-1
(c) CnH2n-2
(d) CnHn-2
Answer:
(c) CnH2n-2

Question 4.
Structure of the compound whose IUPAC name is 5, 6 – dimethylhept-2-ene is ……….
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 5.
The IUPAC name of the compound is …………
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
(a) 2,3 – Diemethyiheptane
(b) 3 – Methyl – 4 – ethyloctane
(c) 5 – ethyl – 6 – methyloctanc
(d) 4 – Ethyl -3 – methyloctane.
Answer:
(d) 4 – Ethyl – 3 – methyloctane.

Question 6.
Which one of the following names does not fit a real name?
(a) 3 – Methyl – 3 – hexanone
(b) 4- Methyl – 3 – hexanone
(c) 3 – Methyl – 3 – hexanol
(d) 2 – Methyl cyclo hexanone.
Answer:
(a) 3 – Methyl – 3 – hexanone

Question 7.
The TUPAC name of the compound CH3— CII = CH -C CH is ……………
(a) Pent – 4 – yn – 2 – ene
(b) Pent -3-en – 1- yne
(c) pent – 2 – en – 4 – yne
(d) Pent – 1 – yn – 3 – ene
Answer:
(b) Pent -3-en – 1- yne

Question 8.
IUPAC name of Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry is ……….
(a) 3, 4, 4 – Trimethylheptane
(b) 2 – Ethyl – 3, 3 – dimethyl heptane
(c) 3, 4, 4 – Trimethyloctane
(d) 2 Butyl – 2 -methyl – 3 – ethyl-butane.
Answer:
(c) 3, 4, 4 – Trimethyloctane

Question 9.
The IUPAC name of Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry is ………
(a) 2 – Hydroxypropionic acid
(b) 2, 4. 4 – Trimethylpent -3-ene
(c) Propan – 2 – ol 1 – oie acid
(d) 2, 2, 4 – Trimethylpent -2-ene
Answer:
(b) 2 – Hydroxy Propanoic acid

Question 10.
The IUPAC name of the compound Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry is ………
(a) 3 – Ethyl – 2- hexene
(b) 3 – Propyl – 3. hexene
(c) 4 – Ethyl – 4 – hexene
(d) 3 – Propyl -2-hexenc
Answer:
(a) 3 – Ethyl – 2- hexene

Question 11.
The IUPAC name of the compound Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry is ……..
(a) 2 – Hydroxypropionic acid
(b) 2 – Hydroxy Propanoic acid
(c) Propan 2 – ol – 1 – oic acid
(d) 1 – Carboxyethanol.
Answer:
(b) 2 – Hydroxy Propanoic acid

Question 12.
The IUPAC name of Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry is ………
(a) 2 – Bromo – 3 – methylbutanoic acid
(b) 2 – methyl – 3 – bromobutanoic acid
(c) 3 – Bromo – 2 – methylbutanoic acid
(d) 3 – Bromo – 2. 3 – dimethyl propanoic acid.
Answer:
(c) 3 – Bromo – 2 – methylbutanoic acid

Question 13.
The structure of isobutyl group in an organic compound is ………
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 14.
The number of stereoisomers of 1, 2-dihydroxycyclopentane is ……..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 15.
Which of the following is optically active?
(a) 3 – Chioropentane
(b) 2 – Chioropropane
(c) Meso – tat-tat-ic acid
(d) Glucose
Answer:
(d) Glucose

Question 16.
The isomer of ethanol is ……….
(a) acetaldehyde
(b) dimethylether
(c) acetone
(d) methyl carbinol
Answer:
(b) dimethylether

Question 17.
How many cyclic and acyclic isomers are possible for the molecular formula C3H6O?
(a) 4
(b) 5
(c) 9
(d) 10
Answer:
(c) 9

Question 18.
Which one of the following shows functional isomerism?
(a) ethylene
(b) Propane
(c) ethanol
(d) CH2Cl2
Answer:
(c) ethanol

Question 19.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry are ………
(a) resonating structure
(b) taulomers
(c) optical isomers
(d) conformers
Answer:
(b) tautomers

Question 20.
Nitrogen detection in an organic compound is earned out by Lassaigne’s test. The blue colour formed is due to the formation of ………….
(a) Fe3[Fe(CN)6]2
(b) Fe4[Fe(CN)6]3
(c) Fe4[Fe(CN)6]2
(d) Fe3[Fe(CN)6]3
Answer:
(b) Fe4[Fe(CN)6]3

Question 21.
Lassaigne’s test for the detection of nitrogen fails in ………..
(a) H2N -CO – NH. NH2. HCl
(b) NH2 – NH2. HCl
(c) C6H5 – NH – NH2. HCl
(d) C6H5CONH2
Answer:
(c) C6H5 – NH – NH2. HCl

Question 22.
Connect pair of compounds which give blue colouration/precipitate and white precipitate respectively, when their Lassaigne’s test is separately done.
(a) NH2NH2HCl and CICH2 – CHO
(b) NH2CS NH2 and CH3 – CH2Cl
(c) NH2CH2 COOH and NH2CONH2
(d) C6H5NH2 and ClCH2 – CHO
Answer:
(d) C6H5NH2 and ClCH2 – CHO

Question 23.
Sodium nitropruside reacts with suiphide ion to give a purple colour due to the formation of ………..
(a) [Fe(CN)5N0]3-
(b) [Fe(NO)5CN]+
(c) [Fe(CN)5NOS]4-
(d) [Fe(CN)5NOS]3-
Answer:
(c) [Fe(CN)5NOS]4-

Question 24.
An organic compound weighing 0.15 g gave on carius estimation, 0.12 g of silver bromide. The percentage of bromine in the compound will be close to ……….
(a) 46%
(b) 34%
(c) 3.4%
(d) 4.6%
Answer:
(b) 34%

Question 25.
A sample of 0.5g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 mL of 0.5M H2SO4. The remaining acid after neutralisation by ammonia consumed 80 mL of 0.5 M NaOH, The percentage of nitrogen in the organic compound is ……….
(a) 14%
(b) 28%
(c) 42%
(d) 56%
Answer:
(b)28%

Question 26.
In an organm compound, phosphorus is estimated as ……….
(a) Mg2P2O7
(b) Mg3(PO4)2
(c) H3PO4
(d) P2O5
Answer:
(a) Mg2P2O7

Question 27.
Ortho and para-nitro phenol can be separated by ………….
(a) azeotropic distillation
(b) destructive distillation
(c) steam distillation
(d) cannot be separated
Answer:
(c) steam distillation

Question 28.
The purity of an organic compound is determined by …………
(a) Chromatography
(b) Crystallisation
(c) melting or boiling point
(d) both (a) and (c)
Answer:
(d) both (a) and (c)

Question 29.
A liquid which decomposes at its boiling point can be purified by …………
(a) distillation at atmospheric pressure
(b) distillation under reduced pressure
(c) fractional distillation
(d) steam distillation
Answer:
(b) distillation under reduced pressure

Question 30.
Assertion: Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry is 3-carbethoxy -2- butenoicacid.
Reason: The principal functional group gets lowest number followed by double bond (or) triple bond.
(a) both the assertion and reason are true and the reason is the correct explanation of assertion.
(b) both assertion and reason are true and the reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d both the assertion and reason are false.
Answer:
(a) both the assertion and reason are true and the reason is the correct explanation of assertion.

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry Short Answer Questions

Question 31.
Give the general characteristics of organic compounds.
Answer:

  • All organic compounds are covalent compounds of carbon and are insoluble in water and soluble in organic solvents.
  • They are inflammable (except CCl4).
  • They possess low boiling and melting points due to their covalent nature.
  • They are characterised by functional groups.
  • They exhibit isomerism.

Question 32.
Describe the classification of organic compounds based on their structure.
classification of organic compounds based on the structure
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 33.
Write a note on homologous series.
Answer:

  • A series of organic compounds each containing a characteristic functional group and the successive members differ from each other in molecular formula by a CH2 group is called homologous series.
  • e.g., Alkanes, Methane (CH4) ethane (C2H6), Propane (C3H8) etc.
  • Compounds of the homologous series are represented by a general formula. e.g., Alkanes: C2H2nAlkene: CnH2n
  • They can be prepared by general methods.
  • They show regular gradation in physical properties but have almost similar chemical properties.

Question 34.
What is meant by a functional group? Identify the functional group in the following compounds.
(a) acetaldehyde
(b) oxalic acid
(c) dimethyl ether
(d) methylamine
Answer:
1. A functional group ¡s an atom or a specific combination of bonded atoms that react in a characteristic way, irrespective of the organic molecule in which it is present.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 35.
Give the general formula for the following classes of organic compounds
(a) Aliphatic monohydric alcohol
(b) Aliphatic ketones
(c) Aliphatic amines.
Answer:
(a) Aliphatic monohydric alcohol – CnH2n+1 + OH
(b) Aliphatic ketones – CnH2nO
(c) Aliphatic amines – C2H2n+1NH2

Question 36.
Write the molecular formula of the first six members of homologous series of nitro – alkanes.
Nitroalkanes:
Answer:

  • CH2NO2 Nitromethane
  • CH2-CH2NO2 Nitroethane
  • CH3-CH2-CH2NO2 1- nitropropane
  • CH3-CH2-CH2-CH2-NO2 1- nitrobutane
  • CH3-CH2-CH2-CH2-CH2-NO2 1 – nitropentane
  • CH2-CH2-CH2-CH2-CH2-CH2-CH2-NO2 1- nitrohexane

Question 37.
Write the molecular and possible structural formula of the first four members of homologous series of carhoxylic acids.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 38.
Give the IUPAC names of the following compounds.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
(i) 2,3,5-tnmethylhexane
(ii) 2-bromo-3-methylbutane
(iii) methoxymethane
(iv) 2-hydroxybutanal
(v) buta-1,3-diene
(vi) 4-chioropent-2-yne
(vii) 1 -bromobut-2-ene
(viii) 5-oxohexanoic acid
(ix) 3-ethyl-4-ethenylheptane
(x) 2, 4, 4-trimethylpent-2-ene
(xi) 2- methyl-I -phenyipropan- I -amine
(xii) 2,2- dimethyl-4-oxopentanenitrile
(xiii) 2-ethoxypropane
(xiv) I -fluoro-4-methyl-2-nitrobenzene
(xv) 3-bromo-2-methylpentanal

Question 39.
Give the structure for the following compound.
Answer:
(i) 3 – ethyl – 2 methyl -1 – pentene
(ii) 1, 3, 5 – Tnmethyl cyclohex – 1 – ene
(iii) tertiary butyl iodide
(iv) 3 – Chlorobutanal
(y) 3 – Chlorobutanol .
(vi) 2 – Chloro – 2 – methyl propane
(vii) 2, 2-dimethyl- 1 – chioropropane
(viii) 3 – methylbut – 1 – ene
(ix) Butan – 2, 2 – diol
(x) Octane – 1 ,3 – diene
(xi) 1 ,5 – Dimethylcyclohexane
(xii) 2 – Chlorobut – 3 – ene
(xiii) 2 – methylbutan – 3 – ol
(xiv) acetaldehyde

Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 40.
Describe the reactions involved in the detection of nitrogen in an organic compound by Lassaigne method.
Answer:
Detection of Nitrogen:
The following reactions are involved in the detection of nitrogen with formation of prussian blue precipitate conforming the presence of nitrogen in an organic compound.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 41.
Give the principle involved in the estimation of halogen in an organic compound by Carius method.
Estimation of halogens:
Answer:
carius method:
1. A known mass of the organic compound is heated with fuming HNO3 and AgNO3.
2. C, H and S gets oxidised to CO2, H2O and SO2 and halogen combines with AgNO3 to form a precipitate of silver halide
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
3. The precipitate AgX is filtered, washed, dried and weighed.
4. From the mass of AgX and the mass of organic compound taken, the percentage of halogens are calculated.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 42.
Give a brief description of the principles of:
1. Fractional distillation
2. Column Chromatography
Answer:
1. Fractional distillation:
This method is used to purify and separate liquids present in the mixture having their boiling point close to each other. The process of separation of the components in liquid mixture at their respective boiling points in the form of vapours and the subsequent condensation of those vapours is called fractional distillation. This method is applied in distillation of petroleum. coal tar and crude oil.

2. Column chromatography:
(a) The principle behind chromatography is selective distribution of mixture of organic substances between two phases a stationary phase and a moving phase. The stationary phase can be a solid or liquid, while the moving phase is a liquid or a gas. when the stationary phase is a solid, the moving phase is a liquid or gas.

(b) If the stationary phase is solid, the basis is adsorption, and when it is a liquid, the basis is partition.

(c) Chromatography is defined as a technique for the separation of a mixture brought about by differential movement of the individual component through porous medium under the influence of moving solvent.

(d) In column chromatography, the above principle is carried out in a long glass column.

Question 43.
Explain paper chromatography.
Answer:
Paper chromatography:
1. It is an example of partition chromatography. A strip of paper acts as an adsorbent. This method involves continues differential partioning of components of a mixture between stationary and mobile phase. In paper chromatography, a special quality paper known as chromatographic paper is used. This paper act as a stationary phase.

2. A strip of chromatographic paper spotted at the base with the solution of the mixture is suspended in a suitable solvent which acts as the mobile phase. The solvent rises up and flows over the spot. The paper selectivity retains different components according to their different partition in the two phases where a chromatograrn is developed.

3. The spots of the separated coloured components are visible at different heights from the position of initial spots on the chromatogram. The spots of the separated colourless compounds may be observed either under ultraviolet light or by the use of an appropriate spray reagent.

Question 44.
Explain various types of constitutional isomerism (structural isomerism) in organic compounds.
Answer:
Constitutional isomers:
These isomers having the same molecular formula but differ in their bonding sequence. It is classified into 6 types:
1. Chain (or) nuclear (or) skeletal isomerism:
The phenomenon in which the isomers have similar molecular formula but differ in the nature of carbon skeleton (i.e., straight (or)
branched)
e.g., C5H12:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

2. Position isomerism:
If different compounds belonging to same homologous series with the same molecular formula and carbon skeleton but differ in the position of substituent or functional group or an unsaturated linkage are said to exhibit position isomerism.
e.g., C5H10:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

3. Functional isomerism:
Different compounds having same molecular formula but different functional groups are said to exhibit functional isomerism.
e.g., C3H6O:
(i) CH3-CH2-CH2-CH=CH2 propanal (Aldehyde group)
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

4. Metamerism:
This isomerism anses due to the unequal distribution of carbon atoms on either side of the functional group or different alkyl groups attached to either side of the same functional group and having same molecular formula.
e.g., C4H10O:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

5. Tautomerism:
It is an isomerism in which a single compound exists in two readily inter convertible structures that differ markedly in the relative position of atleast one atomic nucleus generally hydrogen.
e.g., C2H4O:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

6. Ring chain isomerism:
It is an isomerism in which compounds having same molecular formula but differ in terms of bonding of carbon atom to form open chain and cyclic structures.
e.g., C2H6:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 45.
Describe optical isomerism with suitable example.
Answer:
Optical isomerism:
1. Compounds having same physical and chemical property but differ only in the rotation of plane of polarised light are known as optical isomers and the phenomenon is known as optical isomerism.

2. Glucose have the ability to rotate the plane of plane polarised light and it is said to be an optically active compound and this property of any compound is called optical activity.

3. The optical isomer which rotates the plane of plane polarised light to the right or in clockwise direction is said to be dextrorotatory and is denoted by the sign (+).

4. The optical isomer which rotates the plane of plane polarised light to the left or in anti- clockwise direction is said to be laveo rotatory and is denoted by the sign (-).

5. Dextrorotatory compounds are represented as ‘d’ (or) by (+) sign and leave rotatory compounds are represented as l (or) by (-) sign.

6. The optical isomers which are non-superimposible mirror images of each other are called enantiomers.

Question 46.
Briefly explain geometrical isomerism in alkenes by considering 2- butene as an example.
Answer:
2-butene: Geometrical isomerism : CH3  -CH = CH – CH3

1. Geometrical isomers are the stereoisomers which have different arrangement of groups or atoms around a rigid framework of double bonds. This type of isomerism occurs due to restricted rotation of double bonds or about single bonds in cyclic compounds.

2. In 2-butene, the carbon-carbon double bond is sp2 hybridised. The carbon-carbon double bond consists of a a bond and a it bond. The presence of it bond lock the molecule in one position. Hence, rotation around C = C bond is not possible.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
4. These two compounds are termed as geometrical isomers and are termed as cis and trans form.

5. The cis-isomer is the one in which two similar groups arc on the same side otthe double bond. The trans-isomer is that in which two similar groups are on the opposite side of the double bond. Hence, this type of isomerism is called cis-Irans isomerism.

Question 47.
0.30 g of a substance gives 0.88 g of carbon dioxide and 0.54 g of water calculate the percentage of carbon and hydrogen in it.
Answer:
Weight of organic compound = 0.30 g
Weight of carbon-dioxide = 0.88 g
Weight of water = 0.54 g
Percentage of hydrogen:
18 g of water contains 2 g of hydrogen
0.54 g of water contain = \(\frac {2}{18}\) × 0.54 g of hydrogen
% of hydrogen = \(\frac {2}{18}\) × \(\frac {0.54}{0.30}\) × 100 = \(\frac {2}{18}\) × \(\frac {54}{0.3}\)
% of H = 0.111 × 180 = 19.888 ≈ 20%

Percentage of carbon:
44 g of CO2 contains 12 g of carbon
0.88 g of CO2 contains = \(\frac {12}{44}\) × 0.88 g of carbon
% of carbon = \(\frac {12}{44}\) × \(\frac {0.88}{0.30}\) × 100 = \(\frac {12}{44}\) × \(\frac {88}{0.3}\) = \(\frac {24}{0.3}\)
% of carbon = 80 % .

Question 48.
The ammonia evolved form 0.20 g of an organic compound by kjeldahl method neutralised 15m1 of N/20 Sulphuric acid solution. Calculate the percentage of Nitrogen.
Answer:
Weight of organic compound = 0.20 g
Volume of sulphuric acid taken = 15 ml
Strength of sulphuric acid taken = \(\frac {N}{20}\) = 0.05 N
Percentage of nitrogen = \(\frac{14 \times \mathrm{NV}}{1000 \times 10} \times 100\)
= \(\frac{14 \times 0.05 \times 15}{1000 \times 0.20} \times 100\)
= \(\frac {1050}{200}\) = 5.25
% of nitrogen = 5.25%

Question 49.
0.32 g of an organic compound. after heating with fuming nitric acid and barium nitrate crystals is a scaled tube gave 0.466 g of barium sulphate. Determine the percentage of sulphur in the compound.
Answer:
Weight of organic compound = 0.32 g
Weight of BaSO4 formed = 0.466 g
233 g of BaSO4 contains = 32 g of sulphate
0.466 g of l3aSO4 contain = \(\frac {32}{233}\) x \(\frac {0.466}{2.32}\) x 100
= \(\frac {32}{233}\) x \(\frac {46.6}{0.32}\) = 19.999 g of sulphur
% of sulphur = 20 %

Question 50.
024 g of an organic compound gave 0.287 g of silver chloride in the carius method. Calculate the percentage of chlorine in the compound.
Answer:
Weight of organic compound = 0.24 g
Weight of silver chloride = 0.287 g
143.5 g of AgCl contains = 35.5 g of Cl
0.287 g of AgCl contains = \(\frac {35.5}{143.5}\) x 0.287 g of Cl
% of chlorine = \(\frac {35.5}{143.5}\) x \(\frac {0.287}{0.24}\) x 100 = 29.58 %

Question 51.
In the estimation of nitrogen present in an organic compound by Dumas method 0.35 g yielded 20.7 mL of nitrogen at 15°C and 760 mm Hg pressure. Calculate the percentage of nitrogen ¡n the compound.
Answer:
Weight of organic compound = 0.35 g
Volume of moist nitrogen (V1) = 20.7 ml = 20.7 x 10-3 L
Temperature = T1 = 15°C = 273 + 15°C = 288K
Pressure of moist nitrogen P1 = 760 mmHg
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
V0 = 19.62 x 10-3L
Percentage of nitrogen = \(\frac {28}{22.4}\) x \(\frac{\mathrm{V}_{0}}{\mathrm{W}}\) x 100
= \(\frac {28}{22.4}\) x \(\frac{19.62 \times 10^{-3}}{0.35}\) x 100
= \(\frac {28}{22.4}\) x = \(\frac {19.62}{0.35}\) x 10-1
= 56.05 x 10-3 x 100 = 7.007%
Percentage of nitrogen = 7.007%

In Text Questions – Evaluate Yourself

Question 1.
Give two examples for each of the following type of organic compounds.
1. non-benzonoid aromatic
2. aromatic heterocyclic
3.  alicycic
4. aliphatic open chain.
Answer:
1. Non benzenoid aromatic compounds
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

2. Aromatic heterocyclic compounds
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

3. Alicyclic compounds
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

4. Aliphatic open chain compounds

  • CH3-CH2-CH2-CH2-CH3 n-pentane
  • CH3-CH2-CH2OH 1-propanol

Question 2.
Write structural formula for the following compounds
1. Cyclohexa-1, 4-diene
2. Ethynykyclohexane
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 3.
Write structural formula for the following compounds
1. m – dinitrobenzene
2. p-dichlorobenzene
3. 1, 3, S- Trimethytbeuzene
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 4.
Write all the possible isomers of molecular formula C4H10O and identify the isomerisms found in them.
Answer:
C4H10O isomers:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 5.
0.2346 g of an organic compound containing C, H & O, o comhution giweb 0.2754 g of H2O and 0.4488 g CO2. Calculate the % composition of C, H & O in the organic compound.
Answer:
Weight of organic substance (w) = 0.2346 g
Weight of water (x) = 0.2754 g
Weight of CO2 (y) = 0.4488 g
Percentage of carbon = \(\frac {12}{44}\)  x  \(\frac {y}{w}\) x 100
= \(\frac {12}{44}\) – \(\frac {0.4488}{0.2346}\) x 100 = 52.17%
Percentage of hydrogen = \(\frac {2}{18}\) x \(\frac {y}{w}\) x 100
= \(\frac {2}{18}\) x \(\frac {0.2754}{0.2346}\) x 100 = 13.04%
Percentage of oxygen = [100- (52.17 +13.04)] = 100 – 65.21 = 34.79%

Question 6.
0.16 g of an organic compound was heated in a carlus tube and H2SO4 acid formed was precipitated with BaCl4. The mass of BaSO4 was 0.35 g. Find the percentage of sulphur.
Answer:
Weight of organic substance (w) = 0.16 g
Weight of Barium sulphate (x) = 0.35 g
Percentage of Sulphur = \(\frac {32}{233}\) x \(\frac {x}{w}\) x 100
= \(\frac {32}{233}\) x \(\frac {0.35}{0.16}\) x 100 = 30.04%

Question 7.
0.185 g of an organic compound when treated with Conc. HNO3 and silver nitrate gave 0.320 g of silver bromide. Calculate the % of bromine in the compound.
Answer:
Weight of organic substance (w) 0.185 g ;
Weight of silver bromide (x) = 0.320 g
Percentage of bromine = \(\frac {80}{188}\) x \(\frac {x}{w}\) x 100 = \(\frac {80}{188}\) x \(\frac {0.32}{0.185}\) x 100 = 73.6%

Question 8.
0.40 g of an iodo-substituted organic compound gave 0.235 g of Agi by carius method. Calculate the percentage of iodine in the compound. (Ag = 108, I = 127).
Answer:
Weight of organic substance (w) = 0.40 g
Weight of silver iodide (x) = 0.235 g
127 x 127 0.235
Percentage of iodine = \(\frac {127}{235}\) x \(\frac {x}{w}\) x 100
= \(\frac {x}{w}\) x \(\frac {0.235}{0.40}\) x 100 = 31.75%

Question 9.
0.33 g of an organic compound containing phosphorous gave 0.397 g of Mg2P2O7 by the analysis. Calculate the percentage of P in the compound.
Answer:
Weight of organic compound = 0.33g ;
Weight of Mg,P,07 = 0.397g
222 g of Mg2P2O7 contains 62 g of phosphorous.
0.397 g of Mg2P2O7 will contain \(\frac {62}{222}\) x 0.397 g of P.
0.33 g of organic compound contains \(\frac {62}{222}\) x 0.397 g of P
100 g of organic compound will contain \(\frac {62}{222}\) x \(\frac {0.397}{0.33}\) x 100
= \(\frac {2,461.4}{73.26}\) = 33.59 %
Percentage of phosphorous = 33.59 %

Question 10.
0.3 g of an organic compound on Kjeldahl’s analysis gave enough ammonia to just neutralise 30 mL of 0.1N H2SO4. Calculate the percentage of nitrogen in the compound.
Answer:
Weight of organic compound (w) = 0.3 g
Strength of sulphuric acid used (N) = 0.1 N
Volume of sulphuric acid used (V) = 30 mL
30 ml of 0.1 N sulphuric acid 30 ml of 0.1 N ammonia
Percentage of nitrogen = \(\left(\frac{14 \times \mathrm{NV}}{1000 \times w}\right)\) x 100
= \(\left(\frac{14 \times 0.1 \times 30}{1000 \times 0.3}\right)\) x 100 = 14%

Example Problems

Question 1.
Classify the following compounds based on the structure
1. CH≡C-CH2-C≡CH
2. CH3-CH2-CH2-CH2-CH3
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:

  1. Unsaturated open chain compound
  2. Saturated open chain compound
  3. Aromatic benzenoid compound
  4. Alicyclic compound

Question 2.
0.26 g of an organic compound gave 0.039 g of water and 0.245 g of carbon dioxide on combustion. Calculate the percentage of C & H.
Answer:
Weight of organic compound = 0.26 g
Weight of water = 0.039 g
Weight of CO2 = 0.245 g
Percentage of hydrogen:
18 g of water contains 2 g of hydrogen
0.039 g of water contains = \(\frac {2}{18}\) x \(\frac {0.039}{0.26}\) of H
% of hydrogen = \(\frac {0.039}{0.26}\) x \(\frac {2}{18}\) x 100 = 1.66%
Percentage of carbon:
44 g of CO2 contains 12 g of C
0.245 g of CO2 contains = \(\frac {12}{44}\) x \(\frac {0.245}{0.26}\) g of C
% of Carbon = \(\frac {12}{44}\) x \(\frac {0.245}{0.26}\) x 100 = 25.69%

Question 3.
In an estimation of sulphur by Carius method, 0.2 175 g of the substance gave 0.5825 g of BaSO4, calculate the percentage composition of S ¡n the compound.
Answer:
Weight of organic compound = 0.2 175 g
Weight of BaSO4 = 0.5825 g
233 g of BaSO4 contains = 32 g of S
0.5825 g of BaSO4 contains = \(\frac {32}{233}\) x \(\frac {0.5825}{0.2175}\) g of S
Percentage of S = \(\frac {32}{233}\) x \(\frac {0.5825}{0.2175}\) x 100 = 36.78%

Question 4.
0.284 g of an organic substance gave 0.287 g AgCl in Carius method for the estimation of halogen. Find the percentage of Cl in the compound.
Answer:
Weight of the organic substance = 0.284 g
Weight of AgCl = 0.287 g
143.5 g of AgCl contains 35.5 g of chlorine
0.287 g of AgCl Contains = \(\frac {35.5}{143.5}\) x \(\frac {0.287}{0.284}\)
% of chlorine = \(\frac {35.5}{143.5}\) x \(\frac {0.287}{0.284}\) x 100 = 24.98%

Question 5.
0.24 g of organic compound containing phosphorous gave 0.66 g of Mg2P2O7 by the usual analysis. Calculate the percentage of phosphorous ¡n the compound
Answer:
Weight of an organic compound = 0.24 g
Weight of Mg2P2O7 = 0.66 g
222 g of Mg2P2O7 contains = 62 g of P
o. 66 g contains = \(\frac {62}{222}\) x 0.66 g of P
Percentage of P = \(\frac {62}{222}\) x \(\frac {0.66}{0.24}\) x 100 = 76.80%

Question 6.
0.1688 g when analysed by the Dumas method yield 31.7 mL of moist nitrogen measured at 14°C and 758mm mercury pressure. Determine the % of N in the substance (Aqueous tension at 14°C =12 mm of Hg).
Answer:
Weight of Organic compound = 0.168 g
Volume of moist nitrogen (V1) = 31.7 ml. = 31.7 x 10-3 L
Temperature (T1) = 14°C = 14 + 273 = 287 K
Pressure of Moist nitrogen (P) = 758 mm Hg
Aqueous tension at 14°C = 12 mm of Hg
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{0} V_{0}}{T_{0}}\)
V0 = \(\frac{746 \times 31.7 \times 10^{-3}}{287} \times \frac{273}{760}\)
V0 = 29.58 x 10-3 L
Percentage of nitrogen:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
= 21.90 %

Question 7.
0.6 g of an organic compound was Kjeldhalised and NH3 evolved was absorbed into 50 mL of semi-normal solution of H2SO4. The residual acid solution ws diluted with distilled water and the volume made up to 150 mL. 20 mL of this diluted solution required 35 mL of \(\frac {N}{2}\) NaOH solution for complete neutralisation. Calculate the % of N in the compound.
Answer:
Weight of Organic compound = 0.6 g
Volume of sulphuric acid taken = 50 mL
Strength of sulphuric acid taken = 0.5 N
20 ml of diluted solution of unreacted sulphuric acid was neutralised by 35 mL of 0.05 N Sodium hydroxide
Strength of the diluted sulphuric acid = \(\frac {35 x 0.05}{20}\) = 0.0875 N
Volume of the sulphuric acid remaining after reaction with ammonia = V1 mL
Strength of H2SO4 = 0.5 N
Volume of the diluted H2SO4 = 150 mL
Strength of the diluted sulphuric acid = 0.0875 N
V1 = \(\frac {150 x 0.087}{0.5}\) = 26.25 mL
Volume of H2SO2consumed by ammonia = 50 – 26.25 = 23.75 mL
23.75 mL of 0.5 N H2SO4 = 23.75 mL of 0.5N NH3
The amount of Nitrogen present in the 0.6 g of organic compound
= \(\frac{14 \mathrm{g}}{1000 \mathrm{mL} \times 1 \mathrm{N}}\) x 23.75 x 0.5 N = 0.166 g
Percentage of Nitrogen \(\frac {0.166}{0.6}\) x 100 = 27.66%

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry Additional Questions Solved

I. Choose the correct answer.
Question 1.
Statement 1. The tendency of an atom to form a chain of bonds with the atoms of the same element is called catenation.
Statement 2. The high strength of C-C bond is responsible for its catenation property.
(a) Statement 1 & 2 are correct and statement 2 is the correct explanation of statement 1.
(b) Statement 1 & 2 arc correct but statement 215 not the correct explanation of statement 1.
(c) Statement 1 is correct but statement 2 is wrong.
(d) Statement 1 is wrong but statement 2 is correct.
Answer:
(a) Statement 1 & 2 are correct and statement 2 is the correct explanation of statement 1.

Question 2.
Which of the following is not an organic compound?
(a) DNA
(b) Lipid
(c) Glycogen
(d) Bronze
Answer:
(d) Bronze
Solution:
It is an alloy and a mixture of metals and all other are organic compounds.

Question 3.
Which of the following is an example of an organic reaction?
(a) Rusting of iron
(b) Combustion of magnesium
(c) Biochemical reactions
(d) All the above
Answer:
(c) Biochemical reactions

Question 4.
Which of the following is an example of heterocylic aromatic compound?
(a) THF
(b) Cyclopropane
(c) Pyridine
(d) Azulene
Answer:
(c) Pyridine

Question 5.
Which of the following is an example of non-benzenoid aromatic compound?
(a) Tolucnc
(b) Phenol
(c) Benzyl alcohol
(cl) azulene
Answer:
(d) azulene

Question 6.
Which of the following pair is an example of aromatic compounds?
(a) Benzene, Toluene
(b) Cyclopropane, Cyclobuane
(c) Pyridine, Pyrrole
(d) Propane, Butane
Answer:
(a) Benzene, Toluene

Question 7.
Which of the following is an example of carbocyclic alicyclic compound?
(a) Pyrrole
(b) Thiophene
(c) Cyclopropane
(d) Phenol
Answer:
(c) Cyclopropane

Question 8.
Which one of the following is the functional group of ketone?
(a)-CHO
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
(c) -O-
(d)-OH
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 9.
Which one of the following indicates isothiocyanate functional group?
(a) -NC
(b) -NCS
(c) -SCN
(d) -NCO
Answer:
(b) -NCS

Question 10.
Which of the following represent thiol?
(a) -SH
(b) -OH
(c) -SR
(d) -SCN
Answer:
(a) -SH

Question 11.
Which structure ¡s named as 3-chlorocyclobut-1-ene?
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry-

Question 12.
Which one of the following is called 2-cyclobutyipropanal?
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 13.
Which one of the following is called cyclopentylbenzene’?
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 14.
Which one of the following is commonly called mesitylene?
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 15.
Which one of the following is called benzylchloride?
(a) C6H5CH2Cl
(b) C6H5CHCl2
(c) C6H5CCl3
(d) C6H5Cl
Answer:
(a) C6H5CH2Cl

Question 16.
Which of the following pair are called functional isomers?
(a) methyl propyl ether and diethyl ether
(b) 2-pentanone & 3-pentanone
(c) propanoic acid and methyl acetate
(d) I -butanol and 2-butanol
Answer:
(c) propanoic acid and methyl acetate

Question 17.
Which of the following does not show optical isomerism’?
(a) Glucose
(b) Tartane acid
(c) Lactic acid
(d) Methane
Answer:
(d) Methane

Question 18.
Which metal is used to prepare Lassaigne’s extract?
(a) Copper
(b) Sodium
(c) Aluminium
(d) Iron
Answer:
(b) Sodium

Question 19.
Which colour is formed in the Lassaigne’s test for nitrogen?
(a) Purple
(b) Black
(c) Prussian blue
(d) Violet
Answer:
(c) Prussian blue

Question 20.
Which one of the following is called feme ferrocyanide?
(a) Na4[Fe(CN)6]
(b) Na4[Fe(CN)6]3
(c) Fe4[Fe(CN)6]
(d) Fe4[Fe(CN)6]3
Answer:
(d) Fe4[Fe(CN)6]3

Question 21.
What is the colour formed in Lassaigne’s test of an organic compound containing N and S?
(a) Prussian blue colour
(b) Blood red colour
(c) Purple colour
(d) Canary yellow colour
Answer:
(b) Blood red colour

Question 22.
Which one of the following is the formula of sodium nitroprusside?
(a) Na4[Fe(CN)5N05]
(b) Na4[Fe(CN)5SON]
(c) Na4[Fe(CN)6]
(d) Fe4[Fe(CN)6]3
Answer;
(a) Na4[Fe(CN)5N05]

Question 23.
Identify the colour formed when Lassigne’s extract of sulphur containing organic compound is mixed with sodium nitroprusside solution?
(a) Prussian blue
(b) Black
(c) Green
(d) Purple
Answer:
(d) Purple

Question 24.
Which one of the following solutions are added to Lassaigne’s extract to identify halogens?
(a) Acetic acid + Lead acetate
(b) dil HNO3 + AgNO3
(c) Fe(OH)2 + FeCl3
(d) Na2CO3 + KNO3
Answer:
(b) dil HNO3 + AgNO3

Question 25.
Which one of the following is not identified by Lassaigne’s test?
(a) nitrogen
(b) sulphur
(c) halogens
(d) phosphorous
Answer:
(d) phosphorous

Question 26.
Which one of the following test is used to detect phosphorous in an organic compound?
(a) Silver nitrate test
(b) Copper oxide test
(c) Ammonium molybdate test
(d) Lassaigne’s test
Answer:
(c) Ammonium molybdate test

Question 27.
Identify the colour formed in the test kr phosphorous using ammonium molybciate.
(a) Crimson red colour
(b) Deep violet colour
(c) Prussian blue colour
(d) Canary yellow colour
Answer:
(d) Canary yellow colour

Question 28.
Which of the following will absorb CO2?
(a) Conc. H2SO4
(b) KOH
(c) HCl
(d) Copper
Answer:
(b) KOH

Question 29.
Which of the following is used as moisture absorbent?
(a) Potash
(b) Soda
(c) Conc. H2SO4
(d) Na3CO4
Answer:
(c) Conc. H2SO4

Question 30.
Which method is used to estimate sulphur?
(a) Lassaigne’s test
(b) Oxide test
(c) Carius method
(d) Kjedahl’s method
Answer:
(c) Cari us method

Question 31.
Which method is used to estimate nitrogen?
(a) Dumas method and Kjeldahl’s method
(b) Carius method & Oxide method
(c) Lassaignes test & Copper oxide test
(d) Ammonium molybdate test & Silver nitrate test
Answer:
(a) Dumas method and Kjeldahl’s method

Question 32.
Which of the following is not purified by sublimation method?
(a) Camphor
(b) Benzoic acid
(c) Naphthalene
(d) Nitrobenzene
Answer:
(d) nitrobenzene

Question 33.
Which of the following is used to decolourise the organic compounds?
(a) Chlorine
(b) Bleaching powder
(c) Animal charcoal
(d) Iodine
Answer:
(c) Animal charcoal

Question 34.
Which method is used to extract essential oils from plants and flowers?
(a) Crystallization
(b) Sublimation
(c) Steam distillation
(d) Differential extraction
Answer:
(c) Steam distillation

Question 35.
Which of the following is used as adsorbent?
(a) silica gel and alumina
(b) glass wool and cotton
(c) glass plate and paper
(d) glucose and fructose
Answer:
(a) silica gel and alumina

Question 36.
Which of the following compounds gives prussian blue colour in Lassaigne’s test?
(a) CH4 and CH3OH
(b) CH3NH2 and CH3NO2
(c) CH3Cl and CHCl3
(d) CH3CHO and CH3COCH3
Answer:
(b) CH3NH2 and CH3NO2

Question 37.
Which of the following compounds gives curdy white precipitate in Lassaigne’s test?
(a) CH3Br
(b) C2H5I
(c) CH3Cl
(d) C6H5NO2
Answer:
(c) CH3Cl

Question 38.
Which one of the following is not used as air adsorbent in chromatography?
(a) Alumina
(b) Silica gel
(c) Magnesia
(d) Sucrose
Answer:
(d) Sucrose

Question 39.
The IUPAC name of Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry ……..
(a) 2-methyl butanal
(b) butan-2-aldehyde
(c) 2-ethyipropanal
(d) 3-methyl isobutraldehyde
Answer:
(c) 2-ethyipropanal

Question 40.
Which of the following compounds will exhibit cis-trans isomerism?
Answer:
(a) 2-Buiene
(b) 2-Butyne
(c) 1-Butene
(d) 2-Butanol
Answer:
(a) 2-Butene

Question 41.
Which of the following sodium fusion extract of organic compound gives brilliant violet colour with sodium nitroprusside solution?
(a) Urea
(b) Thiourea
(c) Benzoic acid
(d) Aniline
Answer:
(b) Thiourea

Question 42.
Which of the following reagent is used to distinguish between halogens (Cl, Br, I) in an organic compound?
(a) NaOH
(b) FeCl3
(c) H2SO4
(d) NH4OH
Answer:
(d) NH4OH

Question 43.
In which of the following, functional group isomerism is not possible?
(a) Alcohols
(b) Aldehydes
(c) Alkyl halides
(d) Cyanides
Answer:
(c) Alkyl halides

Question 44.
Which one of the following is used as a column in the separation of pigments of chlorophyll by chromatography technique?
(a) Petroleum ether
(b) CaCO3
(c) Activated charcoal
(d) Ethanoic acid
Answer:
(b) CaCO3

Question 45.
Which one of the following compound does not give Prussian blue colour in Lassaigne’s test?
(a) C6H5NH2
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
(c) C6H5CONH2
(d) C6H5COCl
Answer:
(d) C6H5COCl

Question 46.
Which one of the following shows geometrical isomerism?
(a) n-Butane
(b) 1-butene
(c) 2-butene
(d) butyne
Answer:
(c) 2-butene

Question 47.
Which one of the following shows functional group isomerism?
(a) Ethene
(b) Acetone
(c) Ethane
(a) Propane
Answer:
(b) Acetone

Question 48.
Which of the following pair gives curdy white precipitate and yellow precipitate respectivety in their Lassaignes test?
(a) C2H5I and C2H5Br
(b) C2H5NO2 and C2H5NH2
(c) C6H5Cl and CH3
(d) CH4 and CH3OH
Answer:
(c) C6H5Cl and CH3

II. Match the following.

Question 1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 3.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 4.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 5.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 6.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 7.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 8.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

III. Fill in the blanks.

Question 1.
The property catenation is possible in ……..
Answer:
Carbon.

Question 2.
Acetic acid was synthesised by ………
Answer:
Kolbe

Question 3.
Methane was synthesised in laboratory by ………
Answer:
Berthiot
CH3

Question 4.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry is an example of ………
Answer:
aromatic benzenoid compound

Question 5.
2-butene is an example of compound.
Answer:
unsaturated open chain

Question 6.
The IUPAC name of Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistryis …….
Answer:
2, 2, 5-trimethyl heptane

Question 7.
The IUPAC name of Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistryis ……..
Answer:
3, 4-diethyl, 4-methyiheptane

Question 8.
The name of Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistryis ……….
Answer:
1-cyclobutyl-2-cyclopropylethene

Question 9.
The name of Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
2-(cyclobut-2-en-1-yl)-propanal

Question 10.
The name ofSamacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistryis …………
Answer:
2-cyclopentyipropanal

Question 11.
Esopentane and neopentane are the examples for ………
Answer:
chain isomerism

Question 12.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry are called ……….
Answer:
functional isomers

Question 13.
Copper oxide test is used to detect ………..
Answer:
Carbon & Hydrogen

Question 14.
The formula of feme suiphocyanide is ………
Answer:
Fe(CNS)3

Question 15.
In Lassaigne’s test for halogens, if colour of the precipitate is curdy white, the halogen present is ……….
Answer:
chlorine

Question 16.
The formula of ammonium phospho molybate is ……….
Answer:
(NH4)3.PO4. MO3

Question 17.
Silver nitrate test is used to detect the presence of ……….
Answer:
Halogens

Question 18.
During the estimation of carbon and hydrogen, presence of nitrogen can be avoided by using ………
Answer:
a spiral of copper

Question 19.
In Carius method, the sulphur in an organic compound is oxidised to ……….
Answer:
H2SO4

Question 20.
The method used to estimate nitrogen in foods and fertilisers is ………
Answer:
Kjeidahl’s method

Question 21.
The mixture of diethyl ether and ethanol can he purified by ………
Answer:
simple distillation

Question 22.
The method used to purify petroleum. coal-tar and crude oil is ………
Answer:
fractional distillation

Question 23.
The method used in the manufacture of aniline and turpentine is ……….
Answer:
steam distilation

Question 24.
The mixture of exhanol and water are separated by ………..
Answer:
azeotropic distillation

Question 25.
The different coloured constituents of chlorophyll are separated by ………
Answer:
chromatography

Question 26.
The large number of organic compounds is due to of carbon ………..
Answer:
catenation

Question 27.
The IUPAC name of the compound shown below is:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Answer:
1, 1-dichloropropane

Question 28.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistryname of this compound is ………..
Answer:
2-chloro-3-ethyl- 1, 4-pentadiene

Question 29.
Carboxylic acids are isomenc with ……….
Answer:
esters

Question 30.
Alcohols are isomeric with ……….
Answer:
ethers

Question 31.
The correct IUPAC name for the following structure is ……….
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
5-hex-i -en-3-ol

Question 32.
The Prussian blue colour confirms the presence of nitrogen in an organic compound is due to the formation of ………
Answer:
Fe4[Fe(CN)6]I

Question 33.
The principle involved in paper chromatography is ……….
Answer:
partition

Question 34.
Steam distillation is used for the extraction of ……..
Answer:
essential oils

Question 35.
In chromatography, if the stationary phase is solid, the basis is ……….
Answer:
adsorption

Question 36.
In chromatography, if the stationary phase is liquid, the basis is ………
Answer:
partition

Question 37.
The isomer of acetaldehyde is ……….
Answer:
acetone

Question 38.
The general formula of alkyne is ……….
Answer:
CnH2n-2

Question 39.
The IUPAC name of (CH3)2CH-CH2-CH(CH3)2-CH(CH3)2 is ……..
Answer:
2, 3, 5-trimethyl hexane

IV. Choose the odd one out.

Question 1.
(a) THF
(b) Pyridine
(c) Phenol
(d) Thiophen
Answer:
(c) Phenol. ft is a homocyclic compound whereas others are heterocyclic compounds.

Question 2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 3.
(a) Azulene
(b) Propane
(c) Butane
(d) Ethene
Answer:
(a) Azulene. It is a non benzenoid aromatic homocyclic compound whereas others are aliphatic compounds.

Question 4.
(a) Dyes
(b) Polymers
(c) Cosmetics
(d) Common salt
Answer:
(d) Common salt. It is an inorganic compound whereas others are organic compound.

Question 5.
(a) Renzene
(b) Water
(c) Ether
(d) Chloroform
Answer:
(b) Water. It is a polar solvent whereas others are non-polar solvents.

V Choose the correct pair.

Question 1.
(a) Benzene : Aliphatic compound
(b) Propane : Aromatic compound
(c) Pyridine : Heterocyclic compound
(d) Cyclohexane : Polycyclic compound
(c) Pyridine : Heterocyclic compound

Question 2.
(a)-OH : Ketone
(b)-CHO : Carboxylic acid
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
(d)-NO2 : Amine
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 3.
(a) Organic compounds : inflammable
(b) Organic compounds : ionic compound
(c) Organic compounds : high boiling point and high melting point
(d) Organic compounds : soluble in water
Answer:
(a) Organic compounds : inflammable

Question 4.
(a) CnH2n+2 : C2H4
(b) CnH2n : C3H6
(c) CnH2n-2 : C2H6
(d) CnH2n+2 : C3H4
Answer:
(b) CnH2n : C3H6

Question 5.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 6.
(a) n-pentane and iso pentane : position isomerism
(b) neopentane and n-pentane : chain isomerism
(c) propanal and propanone : position isomerism
(d) propañoic acid and methyl acetate : chain isomerism
Answer:
(b) neopentane and n-pentane : chain isomerism

VI. Choose The incorrect pair.

Question 1.
(a) Dumas method : Estimation of nitrogen
(b) Kjeldahls method : Estimation of nitrogen
(c) Carius method : Estimation of halogens
(d) Dumas method : Estimation of sulphur
Answer:
(d) Dumas method : Estimation of sulphur

Question 2.
(a)-CHO : Aldehyde
(b)-COOH : Carboxylic acid
(c)-NH2 : Nitro group
(d)-O- : Ether
Answer:
(c)-NH2 : Nitro group

Question 3.
(a) Benzene and nitro benzene : Distillation
(b) Coal tar and crude oil : Fractional distillation
(c) Aniline and turpentine : Steam distillation
(d) Naphthalcne and benzoic acid : Crystallization
Answer:
(d) Naphthalene and benzoic acid: Crystallization

Question 4.
(a) BaSO4 : White colour precipitate
(b) Ag2S : Black colour precipitate
(c) Fe(CNS)3 : Prussian blue colour
(d) PbS : Black colour precipitate
Answer:
(c) Fe(CNS)3 : Prussian blue colour

Question 5.
(a) propanal and propropane : Functional isomerism
(b) Nitrite fòrrn and nitro form : Tautomerism
(c) Pent- 1-ene and pcnt-2-ene : Chain isomerism
(d) Propanoic acid and methyl acetate: Functional isomerism
Answer:
(c) Pent-1-ene and pent-2-ene : Chain isomerism

VII. Assertion & Reason.

Question 1.
Assertion (A) : Carbon cannot form ionic bond.
Reason (R) : It is not possible for the carbon to form either C4t or C ions, as it requires large amount of energy.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct hut R is not the correct explanation of A.
(c) A is.correct but R is wrong.
(d) A is wrong but Ris correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 2.
Assertion (A) : Simple distillation can help in separating a mixture of propan-1-ol (boiling point 97°C) and propanone (boiling point 56°C).
Reason (R) : Liquids with a dîftèrence of more than 30°C in their boiling points can be separated by simple distillation.
(a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
(c) Assertion is correct but Reason is wrong.
(d) Assertion is wrong but Reason is correct.
Answer:
(a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Question 3.
Assertion (A) : Pent-1 -ene and pent-2-ene are position isomers.
Reason (R) : Position isomers ditTer in the position of functional group or substituent.
(a) Both assertion and reason are correct and reason is the correct explanation of assertion.
(b) Both assertion and reason are correct but reason is not the correct explanation of assertion.
(c) Assertion is correct but reason is wrong.
(d) Assertion is wrong but reason is correct.
Answer:
(a) Both assertion and reason are correct and reason is the correct explanation of assertion.

VIII. Choose the correct statement.

Question 1.
(a) All organic compounds are ionic compounds.
(b) All organic compounds have high boiling point and high melting point.
(c) Many of the organic compounds are inflammable.
(d) Organic compounds are mostly soluble in water.
Answer:
(c) Many of the organic compounds are inflammable.

Question 2.
(a) Propane is heterocyclic compound.
(b) Azulene is a non benzenoid and aromatic homocyclic compound.
(c) Pyridine is a homocyc lic compound.
(d) Cyclopropane is an aromatic compound.
Answer:
(b) Azulene is a non henzenoid and aromatic homocyclic compound.

Question 3.
(a) CH≡CH-CH2-C≡CH is a saturated open chain compound.
(b) CH3-CH2-CH2-CH2-CH3 is an aromatic benzenoid compound.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistryis an aromatic benzenoid compound.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistryis an aromatic benzenoid compound.

Question 4.
(a) Organic compounds are covalent and generally insoluble in water.
(b) Organic compounds are ionic but generally soluble in water.
(c) Organic compounds non-inflammable
(d) Organic compounds do not show catenation.
Answer:
(a) Organic compounds are covalent and generally insoluble in water.

Question 5.
(a) Fe4[Fe(CN)6] is prussian blue precipitate.
(b) Ag2S is a white precipitate.
(c) PbS is a blood red colour precipitate.
(d) BaSO4 is a black colour precipitate.
Answer:
(a) Fe4[Fe(CN)6] is prussian blue precipitate.

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry 2 Mark Question and answers

Write brief answer to the following questions:

Question 1.
What is catenation?
Answer:
The tendency of an atom to form a chain of bonds with the atoms of the same element is called catenation. The high strength of C-C bond is responsible for its catenation property.

Question 2.
Almost all compounds of carbon form covalent bonds. Give reason.
Answer:
Carbon (Z = 6) have electronic configuration of is2 2s2 2p2. It is not possible for the carbon to form either C4+ or C4- ions to attain the nearest noble gas configuration as it requires large amount of energy. This implies that carbon cannot form ionic bond. So almost in all compounds of carbon, it form four covalent bonds.

Question 3.
What is meant by functional group? Give two example.
Answer:
A functional group is an atom or a specific combination of bonded atoms that react in a characteristic way. irrespective of organic molecule in which it is present. The reaction of an organic compound takes place at the functional group.
e.g.. Alcohol -OH group
Ether -O- group

Question 4.
Classify the following compounds based on the structure.
(i) CH2=CH-CH=CH2
(ii) CH3-CH2-CH2-CH2-CH2-CH2-CH3
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(i) Unsaturated open chain compound
(ii) Saturated open chain compound
(iii) Aromatic benzenoid compound
(iv) Alicylic compound

Question 5.
Give one example for each of the following type of organic compounds.

  1. Non-benzeoid
  2. Aromatic heterocclic
  3. Alicyclic
  4. Aliphatic open chain

Answer:
1. Non-benzenoid aromatic compound
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

2. Alicyclic compound
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

3. Aromatic heterocyclic Compound
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

4. Atiphatic open chain compound
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 6.
Give two examples for each of the following type of organic compounds.

  1. Aromatic bomocyclic compound
  2. Aromatic heterocyclic compound

Answer:
1. Aromatic homocyclic compound
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

2. Aromatic heterocyclic compound
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 7.
Write the functional group of the following compounds

  1. Aldehyde
  2. Ester
  3. Ether
  4. alcohol

Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 8.
Write the functional group of –

  1. cyanide
  2. socyanide
  3. cyanate
  4. isocyanate

Answer:

  1. cyanide -CN
  2. isocyanide -NC
  3. cyanate -CNO
  4. isocyanate -NCO

Question 9.
Write the functional group of –

  1. thiocyanate
  2. isothiocyanate
  3. thiols
  4. thioether

Answer:

  1. Thio cyanale -SCN
  2. Isothiocyanate -NCS
  3. thiols -SH
  4. thicethers -S-

Question 10.
Write the IUPAC names of the following compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Answer:

  1. 3-methylpentane
  2. 2, 2, 5-trimethylheptane

Question 11.
Write the IUPAC names of the following compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:

  1. 3-ethyl-2-methyl pentane
  2. 2-methylbutanal

Question 12.
What are the IUPAC names of the following compounds?
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:

  1. 2-ethyl-but-4-ene-oic acid
  2. 2, 2-dimethyl-hexanoic acid

Question 13.
Write the IUPAC names of the following compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:

  1. 2-ethyl-3-propyl pentane-dioic acid
  2. 3-methy-1 hexane

Question 14.
Predict the IUPAC names of the following compounds
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:

  1. cyclo butane
  2. cyclo pentane
  3. cyclo hutene
  4. cyclo octane

Question 15.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:

  1. Ethyl cyclobutanc
  2. Methyl cyclohexane
  3. Cyclo hexanol

Question 16.
Write the ¡UPAC names of the following compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:

  1. 2-cyclobutyl propanol
  2. 3-cyclohexyl pentan-2-one

Question 17.
Write the structural formula for the following compounds.
(i) cyclohexa- 1, 3-diene
(ii) methyl cyclopentane
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 18.
Draw the structures of –
(i) o-xylene
(ii) m-xylene
(iii) p-xylene
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 19.
Draw the structure of –
(i) Mesitylene
(ii) 1,2, 3-trimethyl benzene
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 20.
Write the structure of –
(i) p-dinitrobenzene
(ii) o-dichlorobenzene
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 21.
Draw the structure of –
(i) 2-cyclopentyl propanal
(ii) 2-(cyclo-but-cnyl) propanal
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 22.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:

  1. N, N-dimethylbenzene amine
  2. N-ethyl-N-methylpropan-I-amine

Question 23.
Draw the structure of 4-hydroxy-3(2-hydroxy ethyl) hexanoic acid.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 24.
Explain about dash line structure with a suitable example.
Answer:
The line bond structure is obtained by representing the two electron covalent bond by a dash or line (-) in a lewis structure. A single line or dash represents a single covalent bond.
e.g., n- propanal:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 25.
What ¡s meant by condensed structure? Explain with an example.
Answer:
The bond line structure can be further abbreviated by omitting all these dashes representing covalent bonds and by indicating the number of identical groups attached to an atom by a subscript. The resulting expression of the compound is called a condensed structural formula.
e.g., 1, 3-butadiene.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 26.
What are bond line structures? Give one example.
Answer:
Condensed structural formula is simplified in which only lines are used. In this type of representation of organic compounds, carbon and hydrogen atoms are not shown and the lines representing carbon-carbon bond are shown in zig zag fashion. The only atoms specially written are oxygen, chlorine, nitrogen etc. Example, Ten. butyl chloride
condensed structure
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
bond line structure
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 27.
Define isomerism. Give example.
Answer:
Isomerism represents the existence of two or more compounds with the same molecular formula but different structure and properties. Compounds exhibiting this isomerism are called isomers.
e.g., C2H6O:

  1. CH3-CH2OH Ethanol
  2. CH3-O-CH3 Methoxyrnethane

Question 28.
Write the possible isomers for the formula C5H10 with their name and type of isomerism present in it.
Answer:
C5H10:

  1. CH3-CH2-CH2-CH = CH2 (Pent-1-ene)
  2. CH3-CH2-CH=CH-CH3 (Pent-2-ene)
    The type of isomerism present above is position isomerism.

Question 29.
Write the possible isomers for the formula C5H10O with their name indicating position isomerism.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Question 30.
Draw the functional isomers for the formula C3H6O2 with their names.
Answer:
C3H6O2:

  1. CH3-CH2-COOH (Propanoic acid)
  2. CH3-COOCH2 (Methyl acetate)

Question 31.
What is metamerism? Give an example.
Answer:
Metamerism is one kind of structural isomerism which arises due to the unequal distribution of carbon atoms on either side of the functional group or diffèrent alkyl groups attached to the either side of the same functional group and having the same molecular formula.
e.g., C4H10O:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 32.
What is tautomerism?
Answer:
It is a special type of functional isomerism in which single compound exists in two readily interconvertible structures that differ markedly in the relative position of atleast one atomic hydrogen. The two different structures are known as tautomers.

Question 33.
What ¡s meant by dyad system? Explain with example.
Answer:
In this system, hydrogen atom oscillates between two directly linked polyvalent atoms.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 34.
What is triad system? Give example.
Answer:
1. In this system hydrogen atom oscillates between three polyvalent atoms. It involves 1. 3-migration of hydrogen atom from one polyvalent atom to other with in the molecule.
2. The most important type of triad system is keto-enol tautomerism and the two groups of tautomers are keto form and enol form.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 35.
What is enolisation? What is labile form?
Answer:
Enolisation is a process in which keto form is converted into eno! form. Both tautomeric forms are equally stable. The less stable form is known as labile form.

Question 36.
Give the structures of Nitro-aci tautomerism.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 37.
Explain ring chain isomerism with the formula C4H8.
Answer:
In ring chain isomerism, compounds have the same molecular formula but differ in terms of bonding of carbon atoms to form open chain and cyclic structures.
C4H8:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 38.
Define stereo-isomerism.
Answer:
The isomers which have same bond connectivity but different arrangement of groups or atoms in space are known as stereoisomers. This phenomenon is known as stereoisomerism.

Question 39.
Define geometrical isomerism with an example.
Answer:
Geometrical isomers are the stereoisomers which have different arrangement of groups or atoms around a rigid framework of double bonds. This type of isomerism occurs due to restricted rotation of double bonds or about single bonds in cyclic compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 40.
Trans isomer is more stable than cis isomer. Justify this statement.
Answer:
Trans isomer is more stable than cis isomer. This is because in the cis isomer, the bulky groups are on the same side of the double bond. The steric repulsion of the groups makes the cis isomers less stable than the trans isomers in which bulky groups are on the opposite side.

Question 41.
Draw the cis, trans isomeric structures of 1, 3-butadiene.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 42.
What are the condition for optical isomerism (or) enantiomerism.
Answer:
1. A carbon atom whose tetravalency in satisfied by four different substituents (atoms (or) groups) is called asymmetric carbon (or) chiral carbon. The optical isomer should have one or more chiral carbon to show optical activity.

2. The molecule possessing chiral carbon atom and is non-superimposable its own mirror image is said to be chiral ntolecule and the property is called chirality or dissymmetry.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 43.
How will you prepare Lassaigne’s extract?
Lassagine’s extract preparation:
Answer:

  • A small piece of Na dried by pressing between the folds of filter paper is taken in a fusion tube and it is gently heated. When it melts to a shining globule, a pinch of organic compound is added.
  • The tube is heated till reaction ceases arid become red hot. Then it is plunged in 50 ml of distilled water taken in a china dish and the bottom of the tube is broken by striking it against the dish.
  • The contents of the dish is boiled for 10 minutes and then it is filtered. The filtrate is known as Lassaigne’s extract.

Question 44.
What Is the need for purification of organic compounds?
Answer:
In order to study the structure, physical properties, chemical properties and biological properties of organic compounds, they must be in the pure state. So organic compounds must be purified.

Question 45.
Define sublimation. Give two examples.
Answer:
The process of conversion of solid to vapour without melting or heating and on cooling the vapours getting back solids, such phenomenon is known as sublimation.
e.g., Naphthalene, Camphor.

Question 46.
Explain the process of chromatography in chlorophyll.
Answer:
The separation of different coloured constituents of chlorophyll is done by chromatography by M.S. Tswelt. He achieved it by passing a petroleum ether solution of chlorophyll present in leaves through a column of CaCO3 firmly packed into a narrow glass tube. Different components of the pigments got separated and lands to form zones of different colours.

Question 47.
Draw the first six members of the carboxylic acid homologous series.
Answer:

  1. HCOOH
  2. CH3COOH
  3. CH3-CH2-COOH
  4. CH3-CH2-CH2-COOH
  5. CH3-CH2-CH2-CH2COOH
  6. CH3-CH2-CH2-CH2-CH2-COOH

Question 48.
Give the condensed formula and bond line formula of 2, 2,4- trimethylpentane.
Answer:
2, 2, 4 –  irimethylpentane
(CH3)3CCH2CH(CH3)2-Condensed formula
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 49.
Differentiate between the principle of estimation of nitrogen in an organic compound by

  1. Dumas method
  2. Kjeldahl’s method.

Answer:
1. Dumas method:
The organic compound is heated strongly with excess of CuO (Cupic Oxide) in an atmosphere of CO2 where free nitrogen, CO2 and H2O are obtained.

2. Kjeldahl’s method:
A known mass of the organic compound is heated strongly with conc. H2SO4, a little amount of potassium sulphate and a little amount of mercury (as catalyst). As a result of reaction, the nitrogen present in the organic compound is converted to ammonium sulphate.

Question 50.
Explain the principle of paper chromatography.
Answer:
This is the simplest form of chromatography. Here a strip of paper acts as an adsorbent. It is based on the principle which is partly adsorption. The paper is made of cellulose fibres with molecules of water adsorbed on them. This acts as stationary phase. The mobile phase is the mixture of the components to be identified whose solution is prepared in a suitable solvent.

Question 51.
Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.
Answer:
Organic compound is fused with sodium metal so as to convert organic compounds into NaCN, Na2S, NaX and Na3PO4. Since these are ionic compounds and become more reactive and thus can be easily tested by suitable reagents.

Question 52.
Name a suitable technique of separation of the components from a mixture of calcium, sulphate and camphor.
Answer:
Sublimation. Because camphor can sublime whereas CaSO4 does not.

Question 53.
Explain, why an organic liquid vapourises at a temperature below its boiling point on steam distillation?
Answer:
It is because in steam distillation the sum of vapour pressure of organic compound and steam should be equal to atmospheric pressure.

Question 54.
Will CCl4give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.
Answer:
No. CCl4 is a completely non-polar covalent compound whereas AgNO3 is ionic in nature. Therefore they are not expected to react and thus a white ppt. of silver chloride will not be formed.

Question 55.
Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?
Answer:
CO2 is acidic in nature and therefore it reacts with the strong base KOH to form K2CO3:
2KOH + CO2 – K2CO2 + H2O

Question 56.
Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?
Answer:
Sulphur sodium extract is acidified with acetic acid because lead acetate is soluble and does not interfere with the test.
Pb(OCOCH3)2 + H2SO4 → PbSO4-+ 2CH3COOH

Question 57.
Why is an organic compound fused with sodium for testing nitrogen, halogens and sulphur?
Answer:
On Ilising with sodium metal the clements present in an organic compound are converted into sodium salts which are water soluble which can be filtered and detected by the respective tests.

Question 58.
Under what conditions can the process of team distillation is used?
Answer:
Steam distillation is used to purify the Liquids which are steam volatile and not miscible with water.

Question 59.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
1, 2-dichloropropane

Question 60.
Write bond-line formulas for: Isopropyl alcohol, 2, 3-dimethvlbutanal, Heptan-4-one.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry 3-Mark Questions

Question 1.
Write the functional group of the following comopounds:
(i) carboxylic acid
(ii) Acid anhydride
(iii) Acyichioride
(iv) Amide
(v) imines
(vi) Nitroso compound
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 2.
What are the general molecular formula and functional group of the following compounds?
Answer:
(i) Ilydrazines
(ii) Hydrazo compound
(iii) Imide
(iv) Phenols
(v) Amine
(vi) Nitroalkane
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 3.
Write the tUPAC names of the following compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(i) Pentan-2a1
(ii) Pentan-(2-ene-2-propyl)- i -oic acid
(iii) 4-methyl-i -cyanohexane

Question 4.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(I) Hex-4-ene-2-oI
(ii) 3-ethyl-5-methyl heptane

Question 5.
Draw the structure of
(i) 1-ethyl-2-methyl cyclopentane
(ii) 1-ethyl-2, 3-dimethyl cyclohexane
(iii) 5-ethyl-2-methylcyclohcx- I -ene
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 6.
Draw the structures of:
(i) 2-cyclobutyl propane
(ii) 2-cyctopropyl butane
(iii) chiorocyclo but-2-eue
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 7.
Give the IUPAC name of the following compounds:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(i) 2-(cyclo but-2-en-1-yl)-propanal
(ii) 4-(cyclopent-3-en- 1-yl )-3-methylbutanoic acid
(iii) 3-(3-nitro cyclopenryl)-prop-2-enoic acid

Question 8.
Write the IUPAC names of the following compounds
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(I) 2-(2-hydroxypropyl) cyclohexan- 1-01
(ii) CyclopentyÍ benzene
(iii) Cyclohexane carboxyl Ic acid

Question 9.
Draw the structure
(i) 1-(cyclo bytyl)-2 (cylopropyl) ethane
(ii) 2-carbamyl cyclobutane-1-carboxylic acid
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 10.
Draw the structures of:
(i) Bromohenzene
(ii) 1, 2-dichlorobenzene
(iii) 1-chloro-3-methvlbenzene
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 11.
Draw the structures of –
(i) Benzvl chloride
(ii) Benzal dichloride
(iii) Benzotrichloride
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 12.
Write the IUPAC names of the following compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(i) 3-methylpentane
(ii) 2, 2. 5-trimethylheptane
(iii) 2-methylbutanal

Question 13.
Write the IUPAC names of the following compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(i) 3-ethyl-2-methylpentane
(ii) 2-ethyl but-3-enoic acid
(iii) 2-forrnyl-2-methylheptanoic acid

Question 14.
Draw the structures of
(i) 3-methylpentanal
(ii) 5-hydroxy 2,2-dimethyl heptanoic acid
(iii) 2-ethyl-4-propy Ipentane-d ioic acid
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 15.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(i) 3-methylhexane
(ii) 2-methylbutanal
(iii) 2-ethylbut-3-enoic acid

Question 16.
Give the IUPAC name Of –
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(i) 4-methyl/hexanenitrite
(ii) 2-methyl but-3-en-amide
(iii) I-Iex-4-en-2-ol

Question 17.
Draw the structures of –
(i) 3-ethyl-5-methylheptane
(ii) 3-ethyl-2-methylhexane
(iii) 2, 4-dimethylpent-2-ene
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 18.
Draw the structures of:
(i) 3-methylhcpta 1, 3, 5-triene
(ii) pent-1-yne
(iii) 2-methylpropan-2-oI
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 19.
Give the IUPAC name of the following compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(i) 4-mcthylpcntan-1-ol
(ii) 2, 2-dimethylpropan-1-ol
(iii) Propanoic acid

Question 20.
Draw the structural formula of:
(i) 4-methylpent-3-en-2rone
(ii) pent-I-yne-3-one
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 21.
Write the IUPAC names of the following compounds.
CH3-CH2-CH2-NH-CH3

Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(i) N-methylpropan- 1 -amine
(ii) N-rnethylpropan-2-amine
(iii) N, N-dimethylpropan- 1-amine

Question 22.
Draw the structurai formula of the following compounds.
(i) N-cthyl-N-methylpropan-1-amine
(ii) N, N-dimethyl benzenamine
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 23.
Draw the complete structural formula, condensed structure and bond line structure of
(i) n-propanol
(ii) 1, 3-butadiene.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 24.
Draw the dash line structure, condensed structure and bond line structure of 1, 3-dimethyl cyclopentane.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 25.
What is wedge formula? Explain with suitable example.
Answer:
1. The simplest convention is solid and dashed wedge formula in which 3-D image of a molecule can be perceived from two dimensional picture.

2. In this representation, a tetrahedral molecule with four atoms or groups a, b, e and d bonded to it can be represented by wedge formula as follows.

3. A solid wedge Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
or a heavy line is used to indicate a bond projecting above the plane of the paper and dashed wedge Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
or a dashed line is used to depict the bond below the plane. The bonds lying in the plane of the paper are shown by normal lines.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 26.
Draw the fisher projection formula for tartaric acid.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 27.
Explain the advantage of sawhorse projection formula over the fisher projection formula with an example.
Answer:
1. The fisher projection fonnula inadequately portrays the spatial relationship between ligands attached to the atoms. The sawhorse projection attempts to clarify the relative location of the groups.

2. In sawhorse projection formula, the bond between two carbon atoms ¡s drawn diagonally and slightly elongated. The lower left hand carbon is considered lying towards the front and the upper right hand carbon towards the back.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 28.
Explain about the Newmann projection formula with an example.
Answer:
1. In this method, the molecules are viewed from the front along the carbon-carbon bond axis.

2. The two carbon atom forming the G bond is represented by two circles. One behind the other so that only the front carbon is seen. The front carbon atom is shown by a point where as the carbon lying farther from the eye is represented by the origin of the circle.

3. Therefore the C-H bonds of the front carbon are depicted from the circle while the C H bonds of the back carbon are drawn from the circumferance of the circle with an angle of 120 to each other.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 29.
Write the possible isomers for the formula C5H12 with their names and structures.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
(i), (ii) and (iii) are chain isomers.

Question 30.
What are the possible isomers for the formula C4H9Cl? Give their structures and IUPAC names.
Answer:
C4H9Cl:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 31.
Write the metamers for the formula C5H10O with their IUPAC names.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 32.
Explain about the geometrical isomerism possible in oximes.
Answer:
1. Restricted rotation around C = N (oximes) gives rise to geometrical isomerism in oximes. Here syn and anti are used instead of cis and trans respectively.

2. In the syn isomer the H atom of a doubly bonded carbon and OH group of doubly bonded nitrogen lie on the same side of the double bond, while in the anti isomer, they lie on the opposite side of the double bond.

3. for e.g.,
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 33.
What are enantiomers?
Answer:
1. An optically active substance may exist in two or more isomeric forms which have same physical and chemical properties but diflèr in terms of direction of rotation of plane polarised light, such optical isomers which rotate the plane polarised light with equal angle but in opposite directions are known as enantiomers and (he phenomenon is known as enantiomerism.

2. Isomers which are non-super impossible mirror images of each other are called enantiomers.

Question 34.
How would you detect sulphur?
Answer:
1. To a portion of the Lassaigne’s extract, freshly prepared sodium nitroprusside solution is added. If deep violet or purple colour is formed, the presence of sulphur is confirmed.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

2. To another portion of Lassaignes extract, acetic acid and lead acetate solution are added. If black precipitate is formed, sulphur presence is confirmed.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 35.
Explain about the oxidation test for sulphur.
Answer:
1. Oxidation test:
The organic substances are fused with a mixture of KNO3 and Na2CO3. The sulphur if present is oxidised to sulphate.
Na2CO3 + S + 3(O) Na2SO4 + CO2

2. The fused mass is extracted with water, acidified with HCl and the BaCl2 solution is added to it. A while precipitate indicates the presence of sulphur.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 36.
How would you detect the halogens in an organic compound?
Answer:

  • To a portion of the Lassaigne’s filtrate. dii. HNO3 is added, warmed gently and AgNO3 solution is added.
  • Appearance of curdy white precipitate soluble in ammonia solution indicates the presence of chlorine.
  • Appearance of pale yellow precipitate sparingly soluble in ammonia solution indicates the presence of bromine.
  • Appearance of yellow precipitate insoluble in ammonia solution indicates the presence of iodine.
    Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 37.
Why nitric acid is added in the Lassaigne’s test for halogens?
Answer:
1. if N (or) S is present in the organic compound along with the halogen, we might obtain Na2S and NaCN in the solution which interfere with the detection of the halogen in the AgNO3 test.

2. Therefore we boil the Lassiagne’s extract with HNO3 which decomposes NaCN and Na2 S as follows:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 38.
Explain about the test for phosphorous in an organic compound.
Answer:

  • A solid organic compound is strongly heated with a mixture of Na2CO3 and KNO3. Phosphorous present in the compound is oxidised to sodium phosphate.
  • The residue is extracted with water and boiled with cone. HNO3. A solution of ammonium molyixiate is added to this solution.
  • A canary yellow precipitate shows the presence of phosphorous.

Question 39.
Explain about principle and reactions involved in carius method of estimation of sulphur. Carius method:
Answer:
1. Principle:
A known mass of the organic substance is heated strongly with fuming HNO3. C and H get oxidized to CO2. and H2O while sulphur is oxidised to sulphuric acid as follows.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

2. The resulting solution is treated with excess of BaCl2 solution, H2SO4 present in the solution is converted into BaSO4. From the mass of BaSO4, the percentage of sulphur can be calculated.

Question 40.
Explain about the procedure and calculation behind the carius method of estimation of sulphur.
Answer:
Carius method:
(I) Procedure:
A known mass of the organic compound is taken in a clean carius tube and few mL of fuming HNO3 is added and then the tube is sealed. It is then placed in an iron tube and bcatcd for 5 hours. The tube is allowed to cool and a hole is made to allow gases to escape. The carius tube is broken and the content collected in a beaker. Excess of BaCl, is added to the beaker. BaSO4 furmed is converted to BaSO4 (white ppt.) The precipitate is filtered. washed, dried and weight. From the mass of BaSO4, percentage of S is calculated.

(ii) Calculation:
Mass of organic compound = Wg
Mass of BaSO4 formed = r g
233 g of BaSO4 contains 32 g of sulphur
∴ x g of BaSO4 contain \(\frac {32}{233}\) x x g of sulphur
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 41.
How are naphthalene and camphor purified?
Answer:
1. Naphthalcne, camphor and benzoic acid when heated, pass directly from solid to vapour without melting. On cooling the vapours will give back solid. This phenomenon is known as sublimation. This technique is used to purify naphthalene, camphor from non volatile impurities.

2. Substances to he purified is taken in a beaker. It is covered with a watch glass. The beaker is heated for a while and the resuJting vapours condense on the bottom of the watch glass. Then the watch glass is removed and the crystals are collected.

Question 42.
Explain how nitrobenzene and benzenc arc separated and purified. (or) How will you separate the mixture of diethyl ether and ethanol?
Answer:
Distillation:

  • The process of distillation involves the impure liquid when boiled gives out vapour and the vapour so formed is collected and condensed to give back the pure liquid in the receiver.
  • This method ¡s to purify liquids from non-volatile impurities and used for separating the constituents of a liquid mixture which differ in their boiling points.
  • In this simple distillation process, liquids with large difference in boiling point (about 40K) and do not decompose under ordinary pressure can be purified.
  • e.g., the mixture of C6H5NO2 nitrohenzene (h.p. 484 K) and C6H6 benzene (b.p. 354 K) can be urified and separated. Similarly the mixture of diethyL ether (b.p. 308K) and ethyl alcohol (h.p. 351 K) can he purified and separated.

Question 43.
Explain about differential extraction.
Answer:

  • The process of removing a substance from its aqueous solution by shaking with a suitable organic solvent is termed extraction.
  • When an organic substance present as solution in water can be recovered from the solution by means of a separating funnel.
  • The aqueous solution is taken in a separating funnel with little quantity of ether or chloroform (CHCl3). The organic solvent immiscible with water will form a separate layer and the contents are shaken gently.
  • The solute being more soluble in the organic solvent is transferred to it.
  • The solvent layer is then separated by opening the tap of separating funnel and the substance is recovered.

Question 44.
Explain about the principle involved in chromatography. Give its types.
Answer:

  1. The principle behind chromatography is selective distribution of the mixture of organic substances between two phases-a stationary phase and a moving phase. The stationary phase can be a solid or liquid while the moving phase is a liquid or a gas.
  2. if the stationary phase is solid, the basis is adsorption and when it is a liquid, the basis is partition.
  3. Chromatography is defined as technique for the separation of a mixture brought about by differential movement of the individual compound through porous medium under the influence of moving solvent.
  4. The various methods of chromatography are:
    • Column chromatography (CC)
    • Thin layer chromatography (TLC)
    • Paper chromatography (PC)
    • Gas liquid chromatography (GLC)
    • Ion exchange chromatography

Question 45.
Describe about adsorption chromatography.
Answer:

  • The principle involved is different compounds are adsorbed on an adsorbent to different degree.
  • Silica gel and alumina are the commonly used adsorbent. The components of the mixture move by varying distances over the stationary phase.

Question 46.
What are hybridisation states of each carbon atom ¡n the following compounds?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 47.
Give the IUPAC names of the following compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(a) Propylbenzene
(b) Methylpentanenitrile
(c) 2, 5-Dimethyllieptanc
(d) 3-Bromo-3-chloroheptane
(e) 3-Chioropropanal
(f) 2, 2-Dichloroethanol

Question 48.
Write the formulas for the first five members of each homologous series beginning with the following compound. CH3COCH3
Answer:
CH3COCH3
CH3COCH2CH2CH3
CH3COCH2CH2CH2CH3
CH3COCH2CH2CH2CH3
CH3CO(CH2)4CH3

Question 49.
Write the formulas for the first five members of each homologous series beginning with the following compound: H-CH=CH2
Answer:
H-CH=CH2
CH3CH=CH2
CH3CH2CH=CH2
CH3CH2CH2CH=CH2
CH3CH2CH2CH2CH=CH2

Question 50.
Identify the functional groups ¡n the following compounds.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 51.
What is the difference between distillation, distillation under reduced pressure and steam distillation?
Answer:
Distillation:
it is used in case of volatiLe liquid mixed with a non-volatile impurities.

Distillation under reduced pressure:
This method is used to purify such liquids which have very high boiling points and which decompose at or below their boiling points.

Steam distillation:
it is used to purify steam volatile liquids associated with water immiscible impurities.

Question 52.
(a) What is Lassaigne’s extract? Will NaCN give a positive Lassaigne’s test for nitrogen?
(b) Which colour will appear in the Lassaigne’s test if the compound contains both nitrogen and sulphur.
(c) Why is Lassaigne’s extract prepared in distilled water? Can we detect oxygen iii a compound by Lassaigne’s test?
Answer:
(a) When organic compounds is fused with sodium metal and then extracted by water it is called Lassaigne’s extract. Yes.
(b) Blood red colour.
(c) Lassaigne’s extract is prepared in distilled water since ta water contains Cl’ ions. No, oxygen cannot be detected by Lassaigne’s test.

Question 53.
0.3780 g of an organic compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine in the compound.
Answer:
Mass of the compound = 0.3780
Mass of silver chloride 0.5740 g
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 54.
In an estimation of sulphur by Carius method, 0.468 of an organic sulphur compound gave 0.668 g of barium sulphate. Find the percentage of sulphur in the compound.
Answer:
Percentage of sulphur
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry 5 marks Questions and Answers

Question 1.
Explain about the steps involved in naming an organic compound as per IUPAC nomenclature.
Answer:
The following steps should be followed for naming an organic compound as per IUPAC nomenclature.

  • Choose the longest carbon chain (Root word). Consider all other groups attached to this chain as substituents.
  • Numbering of the longest carbon chain.
  • Naming the substituents (prefixes) or (suffixes).
  • Arrange the substituents in the alphabetical order.
  • Write the name of the compound as below.

Prefix + Root word Primary suffix + r Secondary suffix

Question 2.
How will you detect the presence of carbon and hydrogen in an organic compound?
Answer:
Copper oxide test:
1. The organic substance is mixed with three times its weight of dry copper oxide by grinding. The mixture is placed in a hard glass test rube fitted with a bent delivery tube. The other end of which is dipping into lime water in an another test tube. The mixture is heated strongly.

2. 2CuO + C → CO2 + 2Cu
CuO + 2H → H2O + Cu

3. Thus if carbon is present, it is oxidized to CO2 which turns lime water milky. If hydrogen is also present, it will be oxidized to water and condenses in small droplets on the cooler wall of the test tube and inside the bulb. Water is collected on white anhydrous CuSO4 which turns blue.

4. This confirms the presence of C and H in the compound.

Question 3.
Explain about lassaigne’s test for detection of nitrogen ¡n an organic compound.
Answer:
I step:
Preparation of sodium fusion extract:
A small piece of Na dried by pressing between the folds of filter paper is taken in a fusion tube and it is heated. When it melts to a shining globule, a pinch of organic compound is added to it. The tube is then heated till the reaction ceases and becomes red hot.

Then the test tube is plunged in about 50 ml of distilled water taken in a china dish and break the bottom of the tube by striking against the dish. The contents of the dish is boiled for about 10 minutes and then filtered. This filtrate is known as lassaigne’s extract (or) sodium fusion extract.

II step :
Test for Nitrogen:
If Nitrogen is present, it gets converted to sodium cyanide which reacts with freshly prepared ferrous sulphate and ferric ion followed by cone. HCl and gives a Prussian blue colour (or) green coloured precipitate, it confirms the presence of nitrogen. HCl is added to dissolve the greenish precipitate of ferrous hydroxide produced by the action of NaOH on FeSO4 which would otherwise mark the Prussian blue precipitate.

Reactions involved:
Na + C + N → NaCN
FeSO4 + 2NaOH → Fe(OH)2 + Na2SO4
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
if both N & S are present. a blood red colour is obtained due to the following reactions
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 4.
Explain about the estimation of carbon and hydrogen.
1. Principle:
A known weigIt of organic substance is brunt in excess of oxygen and the carbon and hydrogen present in it are oxidised to CO2 and H2O respectively.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
The weight of carbon dioxide and water thus formed are determined and the amount of carbon and hydrogen in the organic substance are calculated.

2. Description of the apparatus:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
(a) The oxygen supply (b) combustion tube (c) Absorption tube

Oxygen supply:
To remove the moisture from oxygen. it is allowed to bubble through sulphuric acid and then passed through a U-tube containing sodalime to remove CO2. The oxygen gas free from moisture and CO2 enters the combustion tube.

Combustion tube:
A hard glass tube open at both ends used for the combustion. It contains (i) an oxidized copper gauze to prevent the backward diffusion of the products of combustion (ii) a porcelain boat containing a known weight of the organic substance (iii) coarse copper oxide on either side and (iv) an oxidized copper gauze placed towards the end of the combustion tube. The combustion tube is heated by a gas burner.

Absorption apparatus:
The combustion products containing moisture and CO2 are then passed through the absorption apparatus which consists of(i) a weighed U-tube packed with pumice soaked in conc. H2SO4 to absorb water (ii) a set of bulbs containing a Strong solution of KOH to absorb CO2 and finally (iii) a guard tube filled with anhydrous CaCl2 to prevent the entry of moisture from atmosphere.

3. Procedure:
The combustion tube is heated strongly to dry its content. It is then cooled and connected to absorption apparatus. The other end of the combustion tube is open for a while and the boat containing weighed organic substance is introduced. The tube is again heated strongly till all the substance in the boat is burnt away. This takes about 2 hours. Finally a strong current of oxygen is passed. Then the U-tube and potash bulbs are then
detached and increase in weight of each of them is determined.

4. Calculation:
Weight of organic substance = W g
Increase in weight of H2O = x g
Increase in weight of CO2 = y g
18g of H2O contains 2 g of hydrogen
∴ x g of H2O contain \(\frac {1}{2}\) × x g of hydrogen.
∴ Percentage of hydrogen = (\(\frac {2}{18}\) x \(\frac {x}{w}\) x 100
44g of CO2 contains 12 g of carbon
∴ y g of CO2 contain \(\frac {12}{44}\) × y g of carbon
∴ Percentage of carbon = (\(\frac {12}{44}\) × \(\frac {12}{44}\) × 100)%

Question 5.
Explain about the estimation of halogens by carius method.
Answer:
Carius method:
A known mass of the substance is taken along with fuming HNO3 and AgNO3 taken in a clean carius tube. The open end of the carius tube is sealed and placed in a iron tube for 5 hours in a range at 530 to 540 K. Then the tube is allowed to cool and a small hole is inade in the tube to allow the gases to excape. The tube is broken and the precipitate is filtered, washed, dried and weighed. From the mass of AgX produced percentage of halogen in the organic compound is calculated.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Calculation:
Weight of the organic compound = Wg
Weight of AgCl = a g
143.5 g AgCl contains 35.5g of Cl
a g of AgCl contain \(\frac {35}{143.5}\) x 9 g of Cl
Wg of organic compound contains \(\frac {35}{143.5}\) × 9 g of Cl
% of chlorine = (\(\frac {35}{143.5}\) × \(\frac {a}{w}\) × 100)%
Weight of silver bromide = b g
188 g of AgBr contains 80 g of Br
b g of AgBr contain = \(\frac {80}{188}\) × b g of Br
% of Bromine = (\(\frac {80}{188}\) × \(\frac {b}{w}\) × 100)
Weight of silver iodide = c g
235 g of AgI contains 127 g of I
c g of AgI contain = \(\frac {80}{188}\) × c g of I
% of Iodine = (\(\frac {80}{188}\) × \(\frac {c}{w}\) × 100)

Question 6.
How will you estimate phosphorous in an organic compound?
Answer:
Carius method:
Procedure:
A known mass of organic compound (wg) containing phosphorus is heated with fuming HNO3 in a sealed tube where C is converted into CO2 and H to H2O. Phosphorous present in the organic compound is oxidised to phosphoric acid which is precipitated as ammonium phospho molybdate by heating with conc. HNO3 and by adding ammonium molybdate.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
The precipitate of ammonium phrospho molybdate is filtered, washed, dried and weighed.

Calculation:
Mass of organic compound = Wg
Mass of ammonium phospho molybdate = x g
Molar mass of ammonium phospho molybdate 1877 g
1877 g of ammonium phospho molybdate contains 31 g of phosphorous
x g of ammonium phospho molybdate contain = \(\frac {31}{1877}\) × x g of phosphorous
% of phosphorous = (\(\frac {31}{1877}\) × \(\frac {x}{w}\) × 100)% of phosphorous
In an alternate method, phosphoric acid is precipitated as magnesium-ammoniumphosphate by adding magnesia mixture. The ppt. is washed dried and ignited to get magnesium pyrophosphate which is washed, dried and weighed.

Weight of magnesium pyrophosphate = y g
Molar mass of magnesium pyrophosphate = 222 g
222 g of magnesium pyrophosphate contains 62 g of P
y g of magnesium pyrophosphate contain = \(\frac {62}{222}\) × y g of P
% of phosphorous = (\(\frac {31}{1877}\) × \(\frac {y}{w}\) × 100)%

Question 7.
Explain Dumas method of estimation of nitrogen.
Answer:
Dumas method:
Principle:
This method is based on the fact that nitrogeneous compound when heated with cupric oxide in an atmosphere of CO2 yields free nitrogen.
Traces of nitrogen are reduced to elemental nitrogen by passing over heated copper spiral.
Description of the apparatus:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
CO2 Generator:
CO2 needed in this process’is prepared by heating magnetite or sodium bicarbonate contained in a hard glass tube (or) by the action of dil. HCl on marble in a kipps apparatus. The gas is passed through the combustion tube after dried by bubbling through cone. H2SO4.

Combustion tube:
The combustion tube is heated in a furnace is charged with (a) A roll of oxidised copper gauze to prevent the back diffusion of products of combustion and to heat the organic substance mixed with CuO by radiation (b) a weighed amount of organic substance mixed with excess of CuO (c) a layer of CitO packed in about 2/3 length of the tube and kept in position by loose asbestos plug on either side and (d) a reduced copper piral which reduces any oxides of nitrogen formed during combustion of nitrogen.

Schiff’s nitromctc:
The nitrogen gas obtained by the decomposition of the substance in the combustion tube is mixed with considerable excess of CO2. It is estimated by passing nitro meter when CO2 is absorbed by KOH and the nitrogen gas gets collected in the upper part of the graduated tube.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Calculation:
Weight of the substance taken Wg
Volume of nitrogen = V1L
Room temperature = T1K
Atmospheric pressure = p mm Hg
Aqueous tension al room temperature P’ nun of Hg
Pressure of dry nitrogen = P—P’ = P’1 mm Fig
P0, V0 and T0 be the pressure, volume and temperature respectively of dry nitrogen at S.T.P.
Then, \(\frac{P_{0} V_{0}}{T_{0}}=\frac{P_{1} V_{1}}{T_{1}}\)
V0 = \(\frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{0}}{P_{0}}\)
V0 = \(\frac{P_{1} V_{1}}{T_{1}} \times \frac{273 \mathrm{K}}{760 \mathrm{mm} \mathrm{Hg}}\)
22.4 L of N2 at STP weigh 28 g of N2
V0L of N2 at STP weigh \(\frac {28}{22.4}\) x V0
W g of organic compound contain \(\frac {28}{22.4}\) x V0 g of N2
100 g of organic contain \(\frac {28}{22.4}\) x \(\frac{V_{o}}{w}\) x 100 = % of Nitrogen

Question 8.
Explain Kjeldahl’s method.
Answer:
Principle:
This method is based on the fact that an organic compound containing nitrogen is heated with cone. H2SO4. the nitrogen is conerted Lo ammonium sulphate. The resultant liquid is heated with excess of alkali and then liberated ammonia gas is absorbed in excess of standard acid. The amount of ammonia (nitrogen) is determined by finding the amount of acid neutralised by back titration with same standard alkali.

Procedure:
A weighed quantity of the substance 0.3 to 0.5g is placed in a special long necked Kjeldahl flask made of pyrex glass. About 25 ml of cone. H2SO4 together with a little K2SO4 and CuSO4 [catalyst] are added to il. the flask is loosely stoppered by a glass bulb and heated gently in an inclined position.

The heating is continued till the brown colour of the liquid disappears leaving the content clear as before. At this point all the nitrogen is converted to ammonium sulphate. The kjeldahl flask is cooled aiid its contents are diluted with distilled water and carefully transferred into a I litre round bottom flask. An excess NaOH is poured down the side of the flask and it is filled with a kjeldhals trap and a water condenser.

The lower end of the condenser dips in a measured volume of excess of \(\frac {N}{20}\) H2SO4 solution. The liquid in the round bottom flask is heated and liberated ammonia is distilled to sulphuric acid. When no more ammonia passes over (test the distillate with red litmus) the receiver is removed. The excess of acid is then determined by titration with alkali, using phenolphthalein as the indicator.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Calculation:
Weight of the substance = Wg
Volume of H2SO4 required for the complete neutralisation of evolved NH3 = V ml
Strength H2SO4 used to neutralise NH3 = N.
Let the volume and strength of NH3 formed are V1 and N1 respectively.
V1N1 = VN
The amount of nitrogen present in W g of organic compound = \(\frac{14 \times \mathrm{NV}}{1 \times 1000 \times w}}\)
Percentage of nitrogen = \(\frac{14 \times \mathrm{NV}}{1000 \times w} \times 100\) = \(\frac {1.4NV}{w}\) %

Question 9.
Explain the various steps involved in crystallization method.
Answer:
Most solid organic compounds are purified by crystallization method. This process is carried out by the following steps.
1. Selection of solvent:
rganic substances being covalent do not dissolve in water, hence selection of suitable solvent becomes necessary. Hence the powdered organic substance is taken in a test tube and the solvent is added little by little with constant stirring and heating, till the amount added is sufficient to dissolve the organic compound.

If the solid dissolves upon heating and throws out maximum crystals on cooling, then the solvent is suitable. This process is repeated with benzene, ether, acetone and alcohol various solvent. till the most suitable one is sorted out.

2. Preparation of solution:
The organic compound is dissolved in minimum quantity of suitable solvent small amount of animal charcoal can he added to decolonize any colored substance. The solution may be prepared by heating over a wire gauze or water bath.

3. Filtration of hot solution:
The hot solution so obtained is filtered through a fluted filter paper placed in a funnel.

4. Crystallization:
The hot filtrate is then allowed to cool. Most of the impurities are removed on the filter paper. the pure solid substance separate as crystal. If the rate of crystallization slow, it is induced either by scratching the walls of the beaker with a glass rod or by adding a few crystals of pure compounds to the solution.

5. Isolation and drying of crystals:
The crystals are separated from the mother liquor by filtration is done under reduced pressure using a Buchner funnel. Finally the crystals are washed with small amount of pure cold solvent and dried.

Question 10.
Explain about steam distillation (or) How is essential oils are recovered from plants and flowers.
Steam distillation:
This method is applicable for solids and liquids. If the compound to be steam distilled and it should not decompose at any steam temperature should have a fairly high vapour pressure at 273 K, it should be insoluble in water and the impurities present should be non-volatile.

The impure liquid along with little water is taken in a round bottomed flask which is connected to a boiler on one side and water condenser on the other side, the flask is kept in a slanting position so that no droplets of the mixture will enter into the condenser on the brisk boiling and bubbling of steam.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
The mixture in the flask is heated and then a current of steam passed into it. The vapours of the compound mix up with the steam and escape into the condenser. The condensate obtained is a mixture of water and organic compound which can be separated. This method is used to recover essential oils from plants and flowers also in the manufacture of aniline and turpentine oil.

Question 11.
Explain about azeotropic distillation.
Answer:
The mixture of liquids that cannot be separated by fractional distillation can be purified by azeotropic distillation. The mixture are called azeotropes. These azeotropes are constant boiling mixture which distill as a single component at a fixed temperature for example ethanol and water in the ratio of 95.87: 4.13. In this method, the presence of a third component C6H6, CCl4, ether, glycol glycerol which act as dehydrating agent depress the partial pressure of one component of azeotropic mixture and raises the boiling point of that component and thus the other component will distil over. Substance like C6H6, CCl4 have low b.pt. and reduce the partial vapour pressure of alcohol more than that of water while substance like glycerol and glycol have high boiling point and reduce the partial vapour pressure of water more than that of alcohol.

Question 12.
Explain about thin layer chromatography.
Answer:
A sheet of glass is coated with a thin layer of adsorbent (cellulose, silica gel (or) Alumina). This sheet of glass is called chromplate or thin layer chromatography plate. After drying the plate, a drop of the mixture is placed just above one edge and the plate is then placed in a closed jar containing eluant (solvent) .

The eluant is drawn up the adsorbent layer by capillary action. The components of the mixture move up along with the eluent to different distances depending upon their degree of adsorption of each component of the mixture. It is expressed in terms of its retention factor (Rf) value.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

The spots of coloured compounds are visible on TLC plate due to their original colour. The colourless compounds are viewed under Uy light or in another method using Iodine crystals or by using appropriate reagent.

Question 13.
An organic compound contains 69% carbon and 4.9% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this compound is subjected to complete combustion.
Answer:
Step I.
Calculation of mass of CO2 produced
Mass of compound = 0.20 g
Percentage of carbon = 69 g
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Mass of CO2 formed = \(\frac{69 \times 44 \times(0.20 \mathrm{g})}{12 \times 100}\) = 0.506g

Step II.
Calculation of mass of H2O produced
Mass of compound = 0.20 g
Percentage of hydrogen = 4.8%
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Mass of H2O formed = \(\frac{4.8 \times 18 \times(0.20 \mathrm{g})}{2 \times 100}\) = 0.0864 g

Question 14.
0.50 g of an organic compound was Kjeldahlished. The ammonia eolved was passed in 50cm3 of IN H2SO2. The residual acid required 60 cm3 of N/2 NaOH solution. Calculate the percentage of nitrogen in the compound.
Answer:
Step 1.
Calculation of volume of unused acid
Volume of NaOli solution required = 60 cm3
Normality of NaOH solution = \(\frac {1}{2}\) N
Normality of H2SO4 solution = \(\frac {1}{N}\)
Volume of unused acid can be calculated by applying normality equation
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
1 x V = \(\frac {1}{2}\) x 60 = 30cm3

Step II.
Calculation of volume of acid used
Volume of acid added = 50 cm3
Volume of unused acid = 30 cm3
Volume of acid used = (50 – 30) = 20 cm3

Step III.
Calculation of percentage of nitrogen
mass of compound = 0.05 g
Volume of acid used = 20 cm3
Normality of acid used = 1 N
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
percentage of nitrogen = \(\frac{1.4 \times 20 \times 1}{0.50}=56 \%\)

Question 15.
In a Dumas nitrogen estimation method, 0.30 g of an organic compound gave 50 cm3 of N2 collected at 300 K and 715 mm Hg pressure. Calculate the percentage composition of nitrogen ¡n the compound. (‘apour pressure of water at 300 K is 15 mm Hg)
Answer:
P1 = 715 – 15 = 700mm Hg, P1 = 760 mm Hg
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)
T1 = 300 K, T2 = 273K, V1 = 50 cm3, V2 = ?
V2 = \(\frac{700 \times 50 \times 273}{300 \times 760}\) = 41.9 cm3
% of N = \(\frac {28}{22400}\) x 41.9 x \(\frac {100}{W}\) = 17.46%

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Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

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Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

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Samacheer Kalvi 11th Physics Waves Textual Evaluation Solved

Samacheer Kalvi 11th Physics Waves Multiple Choice Questions
Question 1.
A student tunes his guitar by striking a 120 Hertz with a tuning fork, and simultaneously plays the 4th string on his guitar. By keen observation, he hears the amplitude of the combined sound oscillating thrice per second. Which of the following frequencies is the most likely the frequency of the 4th string on his guitar?
(a) 130
(b) 117
(c) 110
(d) 120
Answer:
(b) 117

Question 2.
A transverse wave moves from a medium A to a medium B. In medium A, the velocity of the transverse wave is 500 ms-1 and the wavelength is 5 m. The frequency and the wavelength of the wave in medium B when its velocity is 600 ms-1, respectively are
(a) 120 Hz and 5 m
(b) 100 Hz and 5 m
(c) 120 Hz and 6 m
(d) 100 Hz and 6 m
Answer:
(d) 100 Hz and 6 m

Question 3.
For a particular tube, among six harmonic frequencies below 1000 Hz, only four harmonic frequencies are given : 300 Hz, 600 Hz, 750 Hz and 900 Hz. What are the two other frequencies missing from this list?
(a) 100 Hz, 150 Hz
(b) 150 Hz, 450 Hz
(c) 450 Hz, 700 Hz
(d) 700 Hz, 800 Hz
Answer:
(b) 150 Hz, 450 Hz
Hint:
If the tube is open at both ends so the harmonic frequencies are based on 150 Hz.
1st = 150 Hz ; 2nd = 300 Hz ; 3rd = 450 Hz ; 4th = 600 Hz ; 5th = 750 Hz ; 6th = 900 Hz
The above frequencies the missing frequency in the list 150 Hz, 450 Hz

Question 4.
Which of the following options is correct?
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1
Options for (1), (2) and (3), respectively are
(a) (B), (C) and (A)
(b) (C), (A) and (B)
(c) (A), (B) and (C)
(d) (B), (A) and (C)
Answer:
(a) (B), (C) and (A)

Question 5.
Compare the velocities of the wave forms given below, and choose the correct option.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 2
where, vA, vB, vC and vD are velocities given in (A), (B), (C) and (D), respectively.
(a) VA > VB > VD > VC
(b) VA < VB < VD < VC
(c) VA = VB = VD = VC
(d) VA > VB = VD > VC
Answer:
(c) VA = VB = VD = VC

Question 6.
A sound wave whose frequency is 5000 Hz travels in air and then hits the water surface. The ratio of its wavelengths in water and air is …….
(a) 4.30
(b) 0.23
(c) 5.30
(d) 1.23
Answer:
(a) 4.30
Hint.
Frequency of sound, f = 5000 Hz
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 5

Question 7.
A person standing between two parallel hills fires a gun and hears the first echo after t1 sec and the second echo after t2 sec. The distance between the two hills is …..
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 6
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 7
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 8

Question 8.
An air column in a pipe which is closed at one end, will be in resonance with the vibrating body of frequency 83 Hz. Then the length of the air column is ………
(a) 1.5 m
(b) 0.5 mSamacheer Kalvi 11th Physics Solutions Chapter 11 Waves
(c) 1.0 m
(d) 2.0 m
Answer:
(c) 1.0 m
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 9
Velocity of sound in air y = 343 ms-1
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 10

Question 9.
The displacement y of a wave travelling in the x direction is given by Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 11 where x and y are measured in metres and t in second. The speed of the wave is ………
(a) 150 ms-1
(b) 300 ms-1
(c) 450 ms-1
(d) 600 ms-1
Answer:
(a) 150 ms-1
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 111

Question 10.
Consider two uniform wires vibrating simultaneously in their fundamental notes. The tensions, densities, lengths and diameter of the two wires are in the ratio 8 : 1, 1 : 2, x : y and 4 : 1 respectively. If the note of the higher pitch has a frequency of 360 Hz and the number of beats produced per second is 10, then the value of x : y is ……….
(a) 36 : 35
(b) 35 : 36
(c) 1 : 1
(d) 1 : 2
Answer:
(a) 36 : 35

Question 11.
Which of the following represents a wave?
(a) (x – vt)3
(b) x(x + vt)
(c) \(\frac{1}{(x+v t)}\)
(d) sin (x + vt)
Answer:
(d) sin (x + vt)

Question 12.
A man sitting on a swing which is moving to an angle of 60° from the vertical is blowing a whistle which has a frequency of 2.0 k Hz. The whistle is 2.0 m from the fixed support point of the swing. A sound detector which detects the whistle sound is kept in front of the swing. The maximum frequency the sound detector detected is …….
(a) 2.027 kHz
(b) 1.947 kHz
(c) 9.74 kHz
(d) 1.011 kHz
Answer:
(a) 2.027 kHz

Question 13.
Let. y = \(\frac{1}{1+x^{2}}\) at t = 0s be the amplitude of the wave propagating in the positive x-direction. At t = 2s, the amplitude of the wave propagating becomes Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 13. Assume that the shape of the wave does not change during propagation. The velocity of the wave is …..
(a) 0.5 ms-1
(b) 1.0 ms-1
(c) 1.5 ms-1
(d) 2.0 ms-1
Answer:
(b) 1.0 ms-1
Hint.
The general expression y in terms of x
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 131
The shape of wave does not change, also wave move in 2 sec, 2m in positive ‘x’ direction. So, wave moves 2m in 2 sec.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 14

Question 14.
A uniform rope having mass m hangs vertically from a rigid support. A transverse wave pulse is produced at the lower end. Which of the following plots shows the correct variation of speed v with height h from the lower end?
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 15
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 16

Question 15.
An organ pipe A closed at one end is allowed to vibrate in its first harmonic and another pipe B open at both ends is allowed to vibrate in its third harmonic. Both A and B are in resonance with a given tuning fork. The ratio of the length of A and B is …….
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 17
Answer:
(c) \(\frac{1}{6}\)
Hint:
(c) \(\frac{1}{6}\)

Samacheer Kalvi 11th Physics Waves Short Answer Questions

Question 1.
What is meant by waves?
Answer:
The disturbance which carries energy and momentum from one point in space to another point in space without the transfer of the medium is known as a wave.

Question 2.
Write down the types of waves.
Answer:
Waves can be classified into two types:
(a) Transverse waves
(b) Longitudinal waves

Question 3.
What are transverse waves? Give one example.
Answer:
In transverse wave motion, the constituents of the medium oscillate or vibrate about their mean positions in a direction perpendicular to the direction of propagation (direction of energy transfer) of waves.
Example: light (electromagnetic waves)

Question 4.
What are longitudinal waves? Give one example.
Answer:
In longitudinal wave motion, the constituent of the medium oscillate or vibrate about their mean positions in a direction parallel to the direction of propagation (direction of energy transfer) of waves.
Example: Sound waves travelling in air.

Question 5.
Define wavelength.
Answer:
For transverse waves, the distance between two neighbouring crests or troughs is known as the wavelength. For longitudinal waves, the distance between two neighbouring compressions or rarefactions is known as the wavelength. The SI unit of wavelength is meter.

Question 6.
Write down the relation between frequency, wavelength and velocity of a wave.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 19

Question 7.
What is meant by interference of waves?
Answer:
Interference is a phenomenon in which two waves superimpose to form a resultant wave of greater, lower or the same amplitude.

Question 8.
Explain the beat phenomenon.
Answer:
When two or more waves superimpose each other with slightly different frequencies, then a sound of periodically varying amplitude at a point is observed. This phenomenon is known as beats. The number of amplitude maxima per second is called beat frequency. If we have two. sources, then their difference in frequency gives the beat frequency. Number of beats per second n = | f1 – f2| per second

Question 9.
Define intensity of sound and loudness of sound.
Answer:

  1. The loudness of sound is defined as “the degree of sensation of sound produced in the ear or the perception of sound by the listener”.
  2. The intensity of sound is defined as “the sound power transmitted per unit area taken normal to the propagation of the sound wave”.

Question 10.
Explain Doppler Effect.
Answer:
When the source and the observer are in relative motion with respect to each other and to the medium in which sound propagates, the frequency of the sound wave observed is different from the frequency of the source. This phenomenon is called Doppler Effect.

Question 11.
Explain red shift and blue shift in Doppler Effect.
Answer:
If the spectral lines of the star are found to shift towards red end of the spectrum (called as red shift) then the star is receding away from the Earth. Similarly, if the spectral lines of the star are found to shift towards the blue end of the spectrum (called as blue shift) then the star is approaching Earth.

Question 12.
What is meant by end correction in resonance air column apparatus?
Answer:
The antinodes are not exactly formed at the open end, we have to include a correction, called end correction e, by assuming that the antinode is formed at some small distance above the open end. Including this end correction, the first resonance is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 20
Again taking end correction into account, we have
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 21

Question 13.
Sketch the function Y = x + a. Explain your sketch
Answer:
When a = 0, y = x
when a = 1; x = 1: y = 1 + 1 = 2
when a = 2; x = 2; y = 2 + 2 = 4
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 22
Explanation: This implies, when increasing the value of a, the line shifts towards right side at a = 0, and line shifts towards left side at a = 1, 2, ….For a = vt, y = x – vt satisfies the differential equation. Though this function satisfies the differential equation, it is not finite for all values of x and t. Hence it does not represent a waves.

Question 14.
Write down the factors affecting velocity of sound in gases.
Answer:
(a) Effect of pressure
(b) Effect of temperature
(c) Effect of density
(e) Effect of wind

Question 15.
What is meant by an echo? Explain.
Answer:
Echo: An echo is a repetition of sound produced by the reflection of sound waves from a wall, mountain or other obstructing surfaces.
Explanation: The speed of sound in air at 20°C is 344 m s-1. If we shout at a wall which is at 344 m away, then the sound will take 1 second to reach the wall. After reflection, the sound will take one more second to reach us. Therefore, we hear the echo after two seconds. Scientists have estimated that we can hear two sounds properly if the time gap or time interval between each sound is \(\left(\frac{1}{10}\right)^{\text {th }}\) of a second (persistence of hearing) i.e., 0.1 s. Then,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 221
2d= 344 × 0.1 = 34.4m ;d= 17.2m
The minimum distance from a sound reflecting wall to hear an echo at 20°C is 17.2 meter.

Samacheer Kalvi 11th Physics Waves Long Answer Questions

Question 1.
Discuss how ripples are formed in still water.
Answer:
Suppose we drop a stone in a trough of still water, we can see a disturbance produced at the place where the stone strikes the water surface. We find that this disturbance spreads out (diverges out) in the form of concentric circles of ever increasing radii (ripples) and strike the boundary of the trough. This is because some of the kinetic energy of the stone is transmitted to the water molecules on the surface. Actually the particles of the water (medium) themselves do not move outward with the disturbance. This can be observed by keeping a paper strip on the water surface. The strip moves up and down when the disturbance (wave) passes on the water surface. This shows that the water molecules only undergo vibratory motion about their – mean positions.

Question 2.
Briefly explain the difference between travelling waves and standing waves.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 23

Question 3.
Show that the velocity of a travelling wave produced in a string is v =\(\sqrt{\frac{\mathrm{T}}{\mu}}\)
Answer:
Velocity of transverse waves in a stretched string: Let us compute the velocity of transverse travelling waves on a string. When a jerk is given at one end (left end) of the rope, the wave pulses move towards right end with a velocity v. This means that the pulses move with a velocity v with respect to an observer who is at rest frame. Suppose an observer also moves with same velocity v in the direction of motion of the wave pulse, then that observer will notice that the wave pulse is stationary and the rope is moving with pulse with the same velocity v. Consider an elemental segment in the string. Let A and B be two points on the string at an instant of time. Let dl and dm be the length and mass of the elemental string, respectively. By definition, linear mass density, μ is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 24
The elemental string AB has a curvature which looks like an arc of a circle with centre at O, radius R and the arc subtending an angle θ at the origin O. The angle θ can be written in terms of arc length and radius as θ = \(\frac{d l}{R}\). The centripetal acceleration supplied by the tension in the string is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 25
Then, centripetal force can be obtained when mass of the string (dm) is included in equation (3)
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 26
The centripetal force experienced by elemental string can be calculated by substituting equation (2) in equation (4) we get
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 27
The tension T acts along the tangent of the elemental segment of the string at A and B. Since the arc length is very small, variation in the tension force can be ignored. We can resolve T into horizontal component T cos \(\left(\frac{\theta}{2}\right)\) and vertical component T sin \(\left(\frac{\theta}{2}\right)\) The horizontal component at A and B are equal in magnitude but opposite in direction; therefore, they cancel each other. Since the elemental arc length AB is taken to be very small, the vertical components at A and B appears to acts Vertical towards the centre of the arc and hence, they add up. The net radial force Fr is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 28
Since the amplitude of the wave is very small when it is compared with the length of the spring, the sine of small angle is approximated as sin \(\left(\frac{\theta}{2}\right) \approx \frac{\theta}{2}\). Hence equation (6) can be written as
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 29
Applying Newton’s second law to the elemental string in the radial direction, under equilibrium, the radial component of the force is equal to the centripetal force. Hence equating equation (5)
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 30

Question 4.
Describe Newton’s formula for velocity of sound waves in air and also discuss the Laplace’s correction.
Answer:
Newton’s formula for speed of sound waves in air: Sir Isaac Newton assumed that when sound propagates in air, the formation of compression and rarefaction takes place in a very slow manner so that the process is isothermal in nature. That is, the heat produced during compression (pressure increases, volume decreases), and heat lost during rarefaction (pressure decreases, volume increases) occur over a period of time such that the temperature of the medium remains constant. Therefore, by treating the air molecules to form an ideal gas, the changes in pressure and volume obey Boyle’s law, Mathematically
PV = constant …(1)
Differentiating equation (1), we get
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 31
where, BT is an isothermal bulk modulus of air. Substituting equation (2) in equation the speed of sound in air is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 32
Since P is the pressure of air whose value at NTP (Normal Temperature and Pressure) is 76 cm of mercury, we have
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 33
Here ρ is density of air, then the speed of sound in air at Normal Temperature and Pressure (NTP) is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 34
But the speed of sound in air at 0°C is experimentally observed as 332 m s-1 which is close upto 16% more than theoretical value (Percentage error is Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 35). This error is not small.
Laplace’s correction: In 1816, Laplace satisfactorily corrected this discrepancy by assuming that when the sound propagates through a medium, the particles oscillate very rapidly such that the compression and rarefaction occur very fast. Hence the exchange of heat produced due to compression and cooling effect due to rarefaction do not take place, because, air (medium) is a bad conductor of heat. Since, temperature is no longer considered as a constant here, sound propagation is an adiabatic process. By adiabatic considerations, the gas obeys Poisson’s law (not Boyle’s law as Newton assumed), which is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 351
Since air contains mainly, nitrogen, oxygen, hydrogen etc, (diatomic gas), we take γ = 1.47.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 36

Question 5.
Write short notes on reflection of sound waves from plane and curved surfaces. Reflection of sound through the plane surface
Answer:
When the sound waves hit the plane wall, they bounce off in a manner similar to that of light. Suppose a loudspeaker is kept at an angle with respect to a wall (plane surface), then the waves coming from the source (assumed to be a point source) can be treated as spherical wave fronts (say, compressions moving like a spherical wave front). Therefore, the reflected wave front on the plane surface is also spherical, such that its centre of curvature (which lies on the other side of plane surface) can be treated as the image of the sound source (virtual or imaginary loud speaker) which can be assumed to be at a position behind the plane surface.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 37
Reflection of sound through the curved surface: The behaviour of sound is different when – it is reflected from different surfaces-convex or concave or plane. The sound reflected from a convex surface is spread out and so it is easily attenuated and weakened. Whereas, if it is reflected from the concave surface it will converge at a point and this can be easily amplified. The parabolic reflector (curved reflector) which is used to focus the sound precisely to a point is used in designing the parabolic mics which are known as high directional microphones.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 38
We know that any surface (smooth or rough) can absorb sound. For example, the sound produced in a big hall or auditorium or theatre is absorbed by the walls, ceilings, floor, seats etc. To avoid such losses, a curved sound board (concave board) is kept in front of the speaker, so that the board reflects the sound waves of the speaker towards the audience. This method will minimize the spreading of sound waves in all possible direction in that hall and also enhances the uniform distribution of sound throughout the hall. That is why a person sitting at any position in that hall can hear the sound without any disturbance.

Question 6.
Briefly explain the concept of superposition principle.
Answer:
Superposition Principle: When a jerk is given to a stretched string which is tied at one end, a wave pulse is produced and the pulse travels along the string. Suppose two persons holding the stretched string on either side give a jerk simultaneously, then these two wave pulses move towards each other, meet at some point and move away from each other with their original identity. Their behaviour is very different only at the crossing/meeting points; this behaviour depends on whether the two pulses have the same or different shape.
When the pulses have the same shape, at the crossing, the total displacement is the algebraic sum of their individual displacements and hence its net amplitude is higher than the amplitudes of the individual pulses. Whereas, if the two pulses have same amplitude but shapes are 180° out of phase at the crossing point, the net amplitude vanishes at that point and the pulses will recover their identities after crossing.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 39
Only waves can possess such a peculiar property and it is called superposition of waves. This means that the principle of superposition explains the net behaviour of the waves when they overlap. Generalizing to any number of waves i.e., if two are more waves in a medium move simultaneously, when they overlap, their total displacement is the vector sum of the individual displacements. We know that the waves satisfy the wave equation which is a linear second order homogeneous partial differential equation in both space coordinates and time. Hence, their linear combination (often called as linear superposition of waves) will also satisfy the same differential equation. To understand mathematically, let us consider two functions which characterize the displacement of the waves, for example,
y1 = A1 sin (kx – ωt) and y2= A2 cos (kx – ωt)
Since, both y1 and y2 satisfy the wave equation (solutions of wave equation) then their algebraic sum
y = y1 + y2
also satisfies the wave equation. This means, the displacements are additive. Suppose we multiply y1 and y2 with some constant then their amplitude is scaled by that constant Further, if C1 and C2 are used to multiply the displacements y1 and y2, respectively, then, their net displacement y is
C = C1y1 + C2y2
This can be generalized to any number of waves. In the case of n such waves in more than one dimension the displacements are written using vector notation. Here, the net displacement \(\vec{y}\) is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 42
The principle of superposition can explain the following :
(a) Space (or spatial) Interference (also known as Interference)
(b) Time (or Temporal) Interference (also known as Beats)
(c) Concept of stationary waves
Waves that obey principle of superposition are called linear waves (amplitude is much smaller than their wavelengths). In general, if the amplitude of the wave is not small then they are called non-linear waves. These violate the linear superposition principle, e.g., laser. In this chapter, we will focus our attention only on linear waves.

Question 7.
Explain how the interference of waves is formed.
Answer:
Consider two harmonic waves having identical frequencies, constant phase difference cp and same wave form (can be treated as coherent source), hut having amplitudes A1 and A2, then
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 43
Suppose they move simultaneously in a particular direction, then interference occurs (i.e., overlap of these two waves). Mathematically
y = y1 + y2 …. (3)
Therefore, substituting equation (1) and equation (3) in equation (3), we get
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 46
By squaring and adding equation (5) and equation (6), we get
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 461
Since, intensity is square of the amplitude (I = A2), we have
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 462
This means the resultant intensity at any point depends on the phase difference at that point.

(a) For constructive interference:
When crests of one wave overlap with crests of another wave, their amplitudes will add up and we get constructive interference. The resultant wave has a larger amplitude than the individual waves as shown in figure (a).
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 4611
The constructive interference at a point occurs if there is maximum intensity at that point, which means that
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 47
This is the phase difference in which two waves overlap to give constructive interference. Therefore, for this resultant wave,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 48
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 49

(b) For destructive interference: When the trough of one wave overlaps with the crest of another wave, their amplitudes “cancel” each other and we get destructive interference as shown in figure (b). The resultant amplitude is nearly zero. The destructive interference occurs if there is minimum intensity at that point, which means cos φ = – 1 ⇒ φ = π, 3π, 5π,… = (2n – 1) K, where n = 0, 1, 2, …. i.e. This is the phase difference in which two waves overlap to give destructive interference. Therefore,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 50
Hence, the resultant amplitude A = |A1 – A2|

Question 8.
Describe the formation of beats.
Answer:
Formation of beats: When two or more waves superimpose each other with slightly different frequencies, then a sound of periodically varying amplitude at a point is observed. This phenomenon is known as beats. The number of amplitude maxima per second is called beat frequency. If we have two sources, then their difference in frequency gives the beat frequency. Number of beats per second
n = |f1 – f2|

Question 9.
What are stationary waves?. Explain the formation of stationary waves and also write down the characteristics of stationary waves.
Answer:
Explanation of stationary waves: When the wave hits the rigid boundary it bounces back to the original medium and can interfere with the original waves. A pattern is formed, which are known as standing waves or stationary waves.
Explanation: Consider two harmonic progressive waves (formed by strings) that have the same amplitude and same velocity but move in opposite directions. Then the displacement of the first wave (incident wave) is
y1 = A sin (kx – ωt) (waves move toward right) …(1)
and the displacement of the second wave (reflected wave) is
y2 = A sin (kx + ωt) (waves move toward left) …(2)
both will interfere with each other by the principle of superposition, the net displacement is
y = y1 + y2 …… (3)
Substituting equation (1) and equation (2) in equation (3), we get
y = A sin (kx – ωt) + A sin (kx + ωt) …(4)
Using trigonometric identity, we rewrite equation (4) as
y(x, t) = 2A cos (ωt) sin (kx) …(5)
This represents a stationary wave or standing wave, which means that this wave does not move either forward or backward, whereas progressive or travelling waves will move forward or backward. Further, the displacement of the particle in equation (5) can be written in more compact form,
y(x, t) = A’ cos (ωt)
where, A’ = 2A sin (foe), implying that the particular element of the string executes simple harmonic motion with amplitude equals to A’. The maximum of this amplitude occurs at positions for which
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 52
where m takes half integer or half integral values. The position of maximum amplitude is known as antinode. Expressing wave number in terms of wavelength, we can represent the anti-nodal positions as
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 53
The distance between two successive antinodes can be computed by
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 54
Similarly, the minimum of the amplitude A’ also occurs at some points in the space, and these points can be determined by setting
sin (kx) = 0 ⇒ kx = 0, π, 2π, 3π, … = nπ
where n takes integer or integral values. Note that the elements at these points do not vibrate (not move), and the points are called nodes. The nth nodal positions is given by,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 55

Characteristics of stationary waves:

  1. Stationary waves are characterised by the confinement of a wave disturbance between two rigid boundaries. This means, the wave does not move forward or backward in a medium (does not advance), it remains steady at its place. Therefore, they are called “stationary waves or standing waves”.
  2. Certain points in the region in which the wave exists have maximum amplitude, called as anti-nodes and at certain points the amplitude is minimum or zero, called as nodes.
  3. The distance between two consecutive nodes (or) anti-nodes is \(\frac{\lambda}{2}\)
  4. The distance between a node and its neighbouring anti-node is \(\frac{\lambda}{4}\)
  5. The transfer of energy along the standing wave is zero.

Question 10.
Discuss the law of transverse vibrations in stretched strings.
Answer:
Laws of transverse vibrations in stretched strings: There are three laws of transverse vibrations of stretched strings which are given as follows:
(i) The law of length: For a given wire with tension T (which is fixed) and mass per unit length µ (fixed) the frequency varies inversely with the vibrating length. Therefore,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 58
⇒ l × f = C, where C is a constant

(ii) The law of tension: For a given vibrating length l (fixed) and mass per unit length p , (fixed) the frequency varies directly with the square root of the tension T,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 59

(iii) The law of mass: For a given vibrating length l (fixed) and tension T (fixed) the frequency varies inversely with the square root of the mass per unit length µ,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 60

Question 11.
Explain the concepts of fundamental frequency, harmonics and overtones in detail. Fundamental frequency and overtones in detail.
Answer:
Fundamental frequency and overtones: Let us now keep the rigid boundaries at x = 0 and x = L and produce a standing waves by wiggling the string (as in plucking strings in a guitar). Standing waves with a specific wavelength are produced. Since, the amplitude must vanish at the boundaries, therefore, the displacement at the boundary must satisfy the following conditions
x(x = 0, t) = 0 and y(x = L, t) = 0
Since, the nodes formed at a distance \(\frac{\lambda_{n}}{2}\) apart, we have \(n\left(\frac{\lambda_{n}}{2}\right)=\mathrm{L}\), where n is an integer, L is the length between the two boundaries and λn is the specific wavelength that satisfy the specified boundary conditions. Hence,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 63
Therefore, not all wavelengths are allowed. The (allowed) wavelengths should fit with the specified boundary conditions, i.e., for n = 1, the first mode of vibration has specific wavelength λ1 = 2L. Similarly for n = 2, the second mode of vibration has specific wavelength
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 64
For n = 3, the third mode of vibration has specific wavelength
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 65
The frequency of each mode of vibration (called natural frequency) can be calculated.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 66
The lowest natural frequency is called the fundamental frequency.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 67
The second natural frequency is called the first over tone.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 68
The third natural frequency is called the second over tone.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 69
Therefore, the nth natural frequency can be computed as integral (or integer ) multiple of fundamental frequency, i.e.,
fn = nf1 where n is an integer …(5)
If natural frequencies are written as integral multiple of fundamental frequencies, then the frequencies are called harmonics. Thus, the first harmonic is f1 = f1 (the fundamental frequency is called first harmonic), the second harmonic is f2 = 2f1, the third harmonic is f3 = 3f1 etc.

Question 12.
What is a sonometer? Give its construction and working. Explain how to determine the frequency of tuning fork using sonometer.
Answer:
Stationary waves in sonometer: Sono means sound related, and sonometer implies sound-related measurements.
It is a device for demonstrating the relationship between the frequency of the sound produced in the transverse standing wave in a string, and the tension, length and mass per unit length of the string. Therefore, using this device, we can determine the following quantities:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 70
(a) the frequency of the tuning fork or frequency of alternating current
(b) the tension in the string
(c) the unknown hanging mass
Construction: The sonometer is made up of a hollow box which is one meter long with a uniform metallic thin string attached to it. One end of the string is connected to a hook and the other end is connected to a weight hanger through a pulley as shown in figure. Since only one string is used, it is also known as monochord. The weights are added to the free end of the wire to increase the tension of the wire. Two adjustable wooden knives are put over the board, and their positions are adjusted to change the vibrating length of the stretched wire.
Working: A transverse stationary or standing wave is produced and hence, at the knife edges P and Q, nodes are formed. In between the knife edges, anti-nodes are formed. If the length of the vibrating element is then
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 71
Let f be the frequency of the vibrating element, T the tension of in the string and p the mass per unit length of the string. Then using equation, we get
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 72
Let ρ be the density of the material of the string and d be the diameter of the string. Then the mass per unit length μ
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 73

Question 13.
Write short notes on intensity and loudness.
Answer:
Intensity and loudness: Consider a source and two observers (listeners). The source emits sound waves which carry energy. The sound energy emitted by the source is same regardless of whoever measures it, i.e., it is independent of any observers standing in that region. But the sound received by the two observers may be different; this is due to some factors like sensitivity of ears, etc. To quantify such thing, we define two different quantities known as intensity and loudness of sound.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 74
Intensity of sound: When a sound wave is emitted by a source, the energy is carried to all possible surrounding points. The average sound energy emitted or transmitted per unit time or per second is called sound power.
Therefore, the intensity of sound is defined as “the sound power transmitted per unit area taken normal to the propagation of the sound wave ”.
For a particular source (fixed source), the sound intensity is inversely proportional to the square of the distance from the source.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 75
This is known as inverse square law of sound intensity.
the power output does not depend on the observer and depends on the baby. Therefore, Loudness of sound: Two sounds with same intensities need not have the same loudness. For example, the sound heard during the explosion of balloons in a silent closed room is very loud when compared to the same explosion happening in a noisy market. Though the intensity of the sound is the same, the loudness is not. If the intensity of sound is increased then loudness also increases. But additionally, not only does intensity matter, the internal and subjective experience of “how loud a sound is” i.e., the sensitivity of the listener also matters here. This is often called loudness. That is, loudness depends on both intensity of sound wave and sensitivity of the ear (It is purely observer dependent quantity which varies from person to person) whereas the intensity of sound does not depend on the observer. The loudness of sound is defined as “the degree of sensation of sound produced in the ear or the perception of sound by the listener”.

Question 14.
Explain how overtones are produced in a:
(a) Closed organ pipe
(b) Open organ pipe
Answer:
(a) Closed organ pipes: Clarinet is an example of a closed organ pipe. It is a pipe with one end closed and the other end open. If one end of a pipe is closed, the wave reflected at this closed end is 180° out of phase with the incoming wave. Thus there is no displacement of the particles at the closed end. Therefore, nodes are formed at the closed end and anti-nodes are formed at open end.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 76
Let us consider the simplest mode of vibration of the air column called the fundamental mode. Anti-node is formed at the
open end and node at closed end. From the figure, let L be the length of the tube and the wavelength of the wave produced. For the fundamental mode of vibration, we have,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 77
which is called the fundamental note
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 771
The frequencies higher than fundamental frequency can be produced by blowing air strongly at open end. Such frequencies are called overtones.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 78
The figure (b) shows the second mode of vibration having two nodes and two antinodes,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 79
is called first over tone, since here, the frequency is three times the fundamental frequency it is called third harmonic.
The figure (c) shows third mode of vibration having three nodes and three anti-nodes.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 80
is called second over tone, and since n = 5 here, this is called fifth harmonic. Hence, the closed organ pipe has only odd harmonics and frequency of the nth harmonic is fn = (2n + 1)f1. Therefore, the frequencies of harmonics are in the ratio
f1 : f2 : f3 : f4 …… = 1 : 3 : 5 : 7 : …… ……… (3)

(b) Open organ pipes: Flute is an example of open organ pipe. It is a pipe with both the ends open. At both open ends, anti-nodes are formed. Let us consider the simplest mode of vibration of the air column called fundamental mode. Since anti-nodes are formed at the open end, a node is formed at the mid-point of the pipe.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 81
From figure (d), if L be the length of the tube, the wavelength of the wave produced is given by
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 82
The frequency of the note emitted is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 83
which is called the fundamental note. The frequencies higher than fundamental frequency can be produced by blowing air strongly at one of the open ends. Such frequencies are called overtones.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 84
The Figure (e) shows the second mode of vibration in open pipes. It has two nodes and three anti-nodes, and therefore,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 85
is called first over tone. Since n = 2 here, it is called second harmonic.
The Figure (f) above shows the third mode of vibration having three nodes and four anti-nodes.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 86
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 87
is called second over tone. Since n = 3 here, it is called the third harmonic.
Hence, the open organ pipe has all the – harmonics and frequency of nth harmonic is fn = nf1. Therefore, the frequencies of harmonics are in the ratio
f1 : f2 : f3 : f4 … = 1 : 2 : 3 : 4 : … …(6)

Question 15.
How will you determine the velocity of sound using resonance air column apparatus?
Answer:
Resonance air column apparatus:
The resonance air column apparatus and first, second and third resonance The resonance air column apparatus is one of the simplest techniques to measure the speed of sound in air at room temperature. It consists of a cylindrical glass tube of one meter length whose one end A is open and another end B is connected to the water reservoir R through a rubber tube as shown in figure. This cylindrical glass tube is mounted on a vertical stand with a scale attached to it. The tube is partially filled with water and the water level can be adjusted by raising or lowering the water in the reservoir R. The surface of the water will act as a closed erid and other as the open end. Therefore, it behaves like a closed organ pipe, forming nodes at the surface of water and antinodes at the closed end. When a vibrating tuning fork is brought near the open end of the tube, longitudinal waves are formed inside the air column. These waves move downward as shown in Figure, and reach the surfaces of water and get reflected and produce standing waves. The length of the air column is varied by changing the water level until a loud sound is produced in the air column. At this particular length the frequency of Waves in the air column resonates with the frequency of the tuning fork (natural frequency of the tuning fork). At resonance, the frequency of sound waves produced is equal to the frequency of the tuning fork. This will occur only when the length of air column is proportional to \(\left(\frac{1}{4}\right)^{t h}\) of the wavelength of the sound waves produced. Let the first resonance occur at length L1, then
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 88
But since the antinodes are not exactly formed at the open end, we have to include a correction, called end correction e, by assuming that the antinode is formed at some small distance above the open end. Including this end correction, the first resonance is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 881
Now the length of the air column is increased to get the second resonance. Let L2 be the length at which the second resonance occurs. Again taking end correction into account, we have
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 90
In order to avoid end correction, let us take the difference of equation (3) and equation (2)
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 91
The speed of the sound in air at room temperature can be computed by using the formula
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 92
Further, to compute the end correction, we use equation (2) and equation (3), we get
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 93

Question 16.
What is meant by Doppler effect? Discuss the following cases
(1) Source in motion and Observer at rest
(a) Source moves towards observer
(b) Source moves away from the observer
(2) Observer in motion and Source at rest.
(a) Observer moves towards Source
(b) Observer resides away from the Source
(3) Both are in motion
(a) Source and Observer approach each other
(b) Source and Observer resides from each other
(c) Source chases Observer
(d) Observer chases Source
Answer:
Doppler Effect: When the source and the observer are in relative motion with respect to each other and to the medium in which sound propagates, the frequency of the sound wave observed is different from the frequency of the source. This phenomenon is called Doppler Effect.
1. Source in motion and the observer at rest
(a) Source moves towards the observer: Suppose a source S moves to the right (as shown in figure) with a velocity vs and let the frequency of the sound waves produced by the source be fs. We assume the velocity of sound in a medium is v.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 94
The compression (sound wave front) produced by the source S at three successive instants of time are shown in the figure. When S is at position x1 the compression is at C1. When S is at position x2, the compression is at C2 and similarly for x3 and C3. Assume that if reaches the observer’s position A then at that instant C2 reaches the point B and C3 reaches the point C as shown in the figure. It is obvious to see that the distance between compressions C2 and C3 is shorter than distance between C1 and C2. This means the wavelength decreases when the source S moves towards the observer O (since sound travels longitudinally and wavelength is the distance between two consecutive compressions). But frequency is inversely related to wavelength and therefore, frequency increases.
Let λ be the wavelength of the source S as measured by the observer when S is at position x1 and λ’ be wavelength of the source observed by the observer when S moves to position x2. Then the change in wavelength is ∆λ = λ – λ’ = vs t, where t is the time taken by the source to travel between x1 and x2. Therefore,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 95
On substituting equation (2) in equation (1), we get Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 96
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 961

(b) Source moves away from the observer: Since the velocity here of the source is opposite in direction when compared to case (a), therefore, changing the sign of the velocity of the source in the above case i.e, by substituting (vs ➝ – vs) in equation (1), we get
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 962
Using binomial expansion again, we get,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 963

2. Observer in motion and source at rest:
(a) Observer moves towards Source:
Let us assume that the observer O moves towards the source S with velocity vo. The source S is at rest and the velocity of sound waves (with respect to the medium) produced by the source is v. From the figure, we observe that both vo and v are in opposite direction. Then, their relative velocity is vr = v + vo. The wavelength of the sound wave is \(\lambda=\frac{v}{f}\), which means the frequency observed by the observer O is f’ = \(\). Then
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 964
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 9641

(b) Observer recedes away from the Source: If the observer O is moving away (receding away) from the source S, then velocity v0 and v moves in the same direction. Therefore, their relative velocity is vr = v – vr. Hence, the frequency observed by the observer O is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 965

3. Both are in motion:
(a) Source and observer approach each other:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 9651
Let vs and vo be the respective velocities of source and observer approaching each other as shown in figure. In order to calculate the apparent frequency observed by the observer, as a simple calculation, let us have a dummy (behaving as observer or source) in between the source and observer. Since the dummy is at rest, the dummy (observer) observes the apparent frequency due to approaching source as given in equation (3) as
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 966
At that instant of time, the true observer approaches the dummy from the other side. Since the source (true source) comes in a direction opposite to true observer, the dummy (source) is treated as stationary source for the true observer at that instant. Hence, apparent frequency when the true observer approaches the stationary source (dummy source), from equation (7) is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 967
Since this is true for any arbitrary time, therefore, comparing equation (9) and equation (10), we get
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 958

(b) Source and observer resides from each other
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 969
Here, we can derive the result as in the previous case. Instead of a detailed calculation, by inspection from figure, we notice that the velocity of the source and the observer each point in opposite directions with respect to the case in (a) and hence, we substitute (vs ➝ – vs) and (v0 ➝ – vo) in equation (11), and therefore, the apparent frequency observed by the observer when the source and observer recede from each other is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 970

(c) Source chases the observer
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 971
Only the observer’s velocity is oppositely directed when compared to case (a).
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 972

(d) Observer chases the source
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 973
Only the source velocity is oppositely directed when compared to case (a). Therefore, substituting vs ➝ – vs in equation (12), we get
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 9731

Samacheer Kalvi 11th Physics Waves Numerical Problems

Question 1.
The speed of a wave in a certain medium is 900 m/s. If 3000 waves passes over a certain point of the medium in 2 minutes, then compute its wavelength?
Answer:
Speed of the wave in medium v = 900 ms-1
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 974

Question 2.
Consider a mixture of 2 mol of helium and 4 mol of oxygen. Compute the speed of sound in this gas mixture at 300 K.
Answer:
The mixture of helium and oxygen.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 975
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 976

Question 3.
A ship in a sea sends SONAR waves straight down into the seawater from the bottom of the ship. The signal reflects from the deep bottom bed rock and returns to the ship after 3.5 s. After the ship moves to 100 km it sends another signal which returns back after 2s. Calculate the depth of the sea in each case and also compute the difference in height between two cases.
Answer:
Speed of SONAR waves in water c = 1500 ms-1
Time taken to reflect from the bottom of the sea, 2t = 3.5 sec
∴ t = 1.75 sec
Distance covered in forward and reflected backward (d1) = c × t
d2 = 1500 × 1.75 = 2625 m
After ship moves in a distance = 150 km
Time taken to reflect by the waves 2t = 2s
t = 1s
Distance covered by the waves (d2) = c × t = 1500 × 1 = 1500 m
The different between the height of two cases = 2625 – 1500
hdifference = 1124 m

Question 4.
A sound wave is transmitted into a tube as shown in figure. The sound wave splits into two waves at the point A which recombine at point B. Let R be the radius of the semicircle which is varied until the first minimum. Calculate the radius of the semi-circle if the wavelength of the sound is 50.0 m
Answer:
The sound travelling in the curved path distance = πR
L1 = πR
The sound travelling in the straight path distance = 2R
L2 = 2R
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 977
The path distance of straight and curved path AP = L1 – L2
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 978

Question 5.
N tuning forks are arranged in order of increasing frequency and any two successive tuning forks give n beats per second when sounded together. If the last fork gives double the frequency of the first (called as octave), Show that the frequency of the first tuning fork is f = (N – 1)n.
Answer:
Total number of fork = N
The frequency of the 1st fork = f
The frequency of the last fork = 2f
∴ an = a + (n – 1)d
2f = f + (N – 1)n
2f – f = (N – 1)n
∴ f = (N – 1)n

Question 6.
Let the source propagate a sound wave whose intensity at a point (initially) be I. Suppose we consider a case when the amplitude of the sound wave is doubled and the frequency is reduced to one-fourth. Calculate now the new intensity of sound at the same point?
Answer:
Intensity of sound wave (old) = I1
Amplitude of sound wave (A2) = 2A1
Frequency of the sound wave I2 = ?
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 979
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 9781

Question 7.
Consider two organ pipes of same length in which one organ pipe is closed and another organ pipe is open. If the fundamental frequency of closed pipe is 250 Hz. Calculate the fundamental frequency of the open pipe.
Answer:
Fundamental frequency of closed organ pipe
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 9791
Fundamental frequency:f open organ pipe \(f_{o}=\frac{v}{2 l}=?\)
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 980

Question 8.
A police in a siren car moving with a velocity 20 ms chases a thief who is moving in a car with a velocity v0 ms-1. The police car sounds at frequency 300 Hz, and both of them move towards a stationary siren of frequency 400 Hz. Calculate the speed in which thief is moving.
(Assume the thief does not observe any beat)
Answer:
Velocity of sound v = 330 ms-1
Velocity of car (vs ) = 20 ms-1
Frequency of car (f1) = 300 Hz
Frequency of stationary siren (f2) = 400 Hz
The speed of the thief (vo ) = ?
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 981
330 – vo = 413.325 + 1.2525vo
2.2525vo = – 83. 325
vo = – 36.99
∴ speed of the thief in moving = 36.99 ms-1

Question 9.
Consider the following function:
(a) y = x2 + 2 α tx
(b) y = (x + vt)2 which among the above function can be characterized as a wave?
Answer:
(a) y = x2 + 2 α tx. This expression is not a wave equation.
(b) y = (x + vt)2 . This expression is satisfies the wave equation.

Samacheer Kalvi 11th Physics Waves Conceptual Questions

Question 1.
Why is it that transverse waves cannot be produced in a gas? Can the transverse waves can be produced in solids and liquids?
Answer:
Transverse waves travel in the form of crests and through. They involve changes in the shape of the medium. As gas has no elasticity of shape, hence transverse waves cannot be produced in it. So, they can be transmitted through media which sustain shearing stress such as solids, strings and liquid surface.

Question 2.
Why is the roar of our national animal different from the sound of a mosquito?
Answer:
Both sounds travel at the speed of sound. The speed of sound varies according to the density and temperature of the air, but not according to the loudness of the sound,at least not for the levels of loudness we talking about here.
The roaring of the lion will be audible a lot further away, but that’s simply because it’s louder.

Question 3.
A sound source and listener are both stationary and a strong wind is blowing. Is there a Doppler effect?
Answer:
Yes, It does not matter whether the sound source or the transmission media are in motion, vibrations will be compressed in the direction of convergence and dilated in the direction of divergence.

Question 4.
In an empty room why is it that a tone sounds louder than in the room having things like furniture etc.
Answer:
Because in a furniture room will absorb the sound waves, hence there went be any echo. But in an empty room reflect the sound. Therefore there will be echo hence we hear sound louder.

Question 5.
How do animals sense impending danger of hurricane?
Answer:
Some animals are believed to be sensitive to be low frequency sound waves emitted by hurricanes, they can also detect the slight drops in air and water pressure that signal a storm’s approach.

Question 6.
Is it possible to realize whether a vessel kept under the tap is about to fill with water?
Answer:
The frequency of the note produced by an air column is inversely proportional to its length. As the level of water is the vessel rises, the length of the air column above it decreases. It produces sound of decreasing frequency, i.e., the sound becames shorter. From the shrillness of sound, it is possible to realize whether the vessel is filled which water.
vmin = 11.71 ms-1

Samacheer Kalvi 11th Physics Waves Textual Evaluation Solved Additional Questions Solved

I. Choose the correct answer from the following:

Question 1.
Mechanical Waves
(a) are longitudinal only
(b) are transverse only
(c) can be both longitudinal and transverse.
(d) are neither longitudinal for transverse waves.
Answer:
(c) can be both longitudinal and transverse.

Question 2.
Sound whose frequency is 50 Hz?
(a) has a relatively short wavelength.
(b) has a relatively long wavelength
(c) is very loud
(d) is very intense
Answer:
(a) has a relatively short wavelength.

Question 3.
Sound travels fastest in …….
(a) Steel
(b) air
(c) water
(d) vaccum
Answer:
(a) steel

Question 4.
A boat at anchor is rocked by waves of velocity 25m/s, having crests 100 m apart. The boat bounches up once in every
(a) 4.0s
(b) 2500s
(c) 0.25s
(d) 75s
Answer:
(a) 4.0s
Hint:
λ = distance between crests = 100 m frequency v = \(\frac{25}{100}=\frac{1}{4} \mathrm{s}^{-1}\)
Therefore, the crests reach the boat once every 4 seconds.

Question 5.
Choose the correct statement:
(a) sound waves are transverse waves
(b) sound travels fastest through vaccum.
(c) sound travels faster in solids than in gases.
(d) sound travels faster in gases than in liquids.
Answer:
(c) sound travels faster in solids than in gases.

Question 6.
Transverse waves can propagate ……
(a) both in a gas and in a metal
(b) in a gas but not in a metal
(c) not in a gas but in a metal
(d) neither in a gas nor in a metal
Answer:
(a) not in a gas but in a metal.

Question 7.
The speed of the wave represented by y = A sin(ωt – kx) is ……..
(a) k/ω
(b) ω/k
(c) ωk
(d) 1/ωk
Answer:
(b) ω/k

Question 8.
The equation of a wave travelling in a string can be written as y = 3 cos {π(100t – x)} where y and x are in cm and t is in seconds. Then the value of wavelength is …….
(a) 100 cm
(b) 2 cm
(c) 50 cm
(d) 4 cm
Answer:
(b) 2 cm
Hint:
On comparing given equation with y = A cos (kx – ωt), we get
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 985

Question 9.
A wave of frequency 500 Hzhas a velocity 300 m/s. The distance between two nearest points which are 60° out of phase, is ……
(a) 0.2 m
(b) 0.1 m
(c) 0.4 m
(d) 0.5 m
Answer:
(a) 0.1 cm
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 986

Question 10.
The equation of a wave travelling on a string is Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 9792, where x, y are in cm and t in seconds. The velocity of the waves is ……..
(a) 64 cm/s in – x direction
(b) 32 cm/s in – x direction
(c) 32 cm/s in +x direction
(d) 64 cm/s in + x direction
Answer:
(d) 64 cm/s in + x direction to S
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 9793

Question 11.
The equation of a wave is Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 9794 where y, x are in cm and t in seconds. The amplitude wavelength, velocity and frequency of the wave are, respectively, …….
(a) 4 cm, 32 cm, 16 cm/s, 0.5 Hz
(b) 8 cm, 16 cm, 32 cm/s, 1.0 Hz
(c) 4 cm, 32 cm, 32 cm/s, 0.5 Hz
(d) 8 cm, 16 cm, 16 cm/s, 1.0 Hz
Answer:
(a) 4 cm, 32 cm, 16 cm/s, 0.5 Hz

Question 12.
The diagram shows the profile of a wave, which of the following pairs of points are in phase?
(a) A, B
(b) B, C
(c) B, D
(d) B, E
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 120
Answer:
(d) B, E

Question 13.
Ultrasonic waves are those waves which ………
(a) human beings cannot hear
(b) human beings can hear
(c) have high velocity
(d) have large amplitude
Answer:
(a) human beings cannot hear

Question 14.
A transverse wave of amplitude 0.5m, wavelength 1 m and frequency 2Hz is propogating in a string in the negative x direction. The equation of this wave is ….
(a) y = 0.5 sin (2πx – 4πt)
(b) y = 0.5 sin (2πx + 4πt)
(c) y = 0.5 sin (πx – 2πt)
(d) y = 0.5 cos (kx – 2πt)
Answer:
(b) y = 0.5 sin (2πx + 4πt)
Hint:
y = A sin(kx + ωt)
Here A = 0.5 m
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 121

Question 15.
With the rise of temperature, the speed of sound in a gas ……..
(a) increases
(b) decreases
(c) remain the same
(d) may increase or decrease depending on the corresponding change in pressure.
Answer:
(a) increases

Question 16.
Speed of sound in a gas in proportional to …….
(a) square root of isothermal elasticity
(b) square root of adiabatic elasticity
(c) isothermal elasticity
(d) adiabatic elasticity
Answer:
(b) square root of adiabatic elasticity

Question 17.
The velocity of sound in are is not affected by change in the …….
(a) atmospheric pressure
(b) moisture content of air
(c) temperature of air
(d) composition of air
Answer:
(a) atmospheric pressure

Question 18.
A longitudinal wave is described by the equation y = y0 sin 2π (ft – x/λ). The maximum particle velocity is equal to four times the wave velocity if ……..
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 122
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 123

Question 19.
If v0 and v denote the sound velocity and the rms velocity of the molecules in a gas, then ……..
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 124
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 125
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1256

Question 20.
With the propagation of a longitudinal wave through a material medium, the quantities transferred in the direction of propagation are ……..
(a) energy, momentum and mass
(b) energy and momentum
(c) energy and mass
(d) energy
Answer:
(b) energy and momentum

Question 21.
If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity will ………
(a) increase by a factor of 2
(b) decrease by a factor of 2
(c) decrease by a factor of 4
(d) remain unchanged
Answer:
(c) decrease by a factor of 4
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1241

Question 22.
When a source of sound is in motion towards a stationary observer, the effect observed is
(a) increase in the velocity of sound only
(b) decrease in the velocity of sound only
(c) increase in frequency of sound only
(d) increase in both the velocity and the frequency of sound
Answer:
(c) increase in frequency of sound only

Question 23.
The apparent wavelength of the light from a star, moving away from the earth, is 0.01% more than its real wave length. The speed of the star with respect to the earth is ……
(a) 10 km/s
(b) 15 km/s
(c) 30 km/s
(d) 60 km/s
Answer:
(c) 30 km/s
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1253

Question 24.
The frequency of a radar is 780 MHz. When it is reflected from an approaching aeroplane the opponent frequency is more than the actual frequency by 2.6 kHz. The speed of the aeroplane is ……
(a) 0.25 km/s
(b) 0.5 km/s
(c) 1.0 km/s
(d) 2.0 km/s
Answer:
(b) 0.5 km/s
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 126

Question 25.
The temperature at which the speed of sound in air becomes double its value at 27°C is …….
(a) 54°C
(b) 327°C
(c) 927°C
(d) -123°C
Answer:
(c) 927°C
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 127

Question 26.
The equation of a transverse wave is given by y = 10 sin{π(0.01x – 2t)} where y and x are in cm and t is in seconds. Its frequency is …….
(a) 10 s-1
(b) 2 s-1
(c) 1 s-1
(d) 0.01 s-1
Answer:
(c) 1 s-1
Hint:
Comparing with the standard equation y = A sin(kx – ωt),
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 128

Question 27.
When sound waves travel from air to water, which of the following remains constant?
(a) velocity
(b) frequency
(c) wavelength
(d) all of these
Answer:
(b) frequency

Question 28.
The speed of sound in oxygen is 332 m/s at STP. The speed of sound in hydrogen at STP will be ……..
(a) 53/2 m/s
(b) 2546 m/s
(c) 1328 m/s
(d) 664 m/s
Answer:
(c) 1328 m/s
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 129

Question 29.
If va, vh, and vm are the speeds of sound in air, hydrogen and a metal at the same temperature, then ……..
(a) vh > va > vm
(b) vm > vh > va
(c) vh > vm > va
(d) va > vh > vm
Answer:
(b) vm > vh > va

Question 30.
Ultrasonic waves can be detected by ……
(a) telephone
(b) Hebb’s method
(c) Kundt’s tube
(d) Quincke’s tube
Answer:
(c) Kundt’s tube

Question 31.
The velocity of sound in a gas depends on ….
(a) Wavelength only
(b) density and elasticity of gas
(c) intensity only
(d) amplitude and frequency
Answer:
(b) density and elasticity of gas

Question 32.
When sound waves travel from air to water, which of these remains constant?
(a) velocity
(b) wavelength
(c) frequency
(d) all the above
Answer:
(c) frequency

Question 33.
When a wave goes from one medium to another, there is a change in
(a) velocity
(b) amplitude
(c) wavelength
(d) all the above
Answer:
(d) all the above

Question 34.
The equation of a sound wave is y = 0.0015 sin (62.8x + 316t). Find the wave length of the above ……
(a) 0.2 units
(b) 0.3 units
(c) 0.1 units
(d) 0.15 units
Answer:
(c) 0.1 units
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 130

Question 35.
Red shift is an illustration of ……….
(a) low temperature emission
(b) high frequency absorption
(c) Doppler effect
(d) Same unknown Phenomenon.
Answer:
(c) Doppler effect

Question 36.
The ratio of the velocity of sound in a monatomic gas to that in a triatomic gas having same molar mass, under similar conditions of temperature and pressure, is ………
(a) 1.12
(6) 1.25
(c) 1.50
(d) 1.6
Answer:
(a) 1.12
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1312

Question 37.
Doppler shift in frequency does not depend upon …….
(a) the actual frequency of the wave
(b) the velocity of the source from the listener.
(c) the velocity of the source.
(d) the velocity of the observer.
Answer:
(b) the velocity of the source from the listener.

Question 38.
If the density of oxygen is 16 times that of hydrogen, what will be the ratio of the velocities of sound in them?
(a) 1 : 4
(b) 4 : 1
(c) 2 : 1
(d) 1 : 16
Answer:
(a) 1 : 4
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 132

Question 39.
Pitch of sound depends on ……
(a) frequency
(b) wavelength
(c) amplitude
(d) speed
Answer:
(a) frequency

Question 40.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 133
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 134
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 135

Question 41.
Which of the following equations represents a wave?
(a) y = A(ωt – kx)
(b) y = A sin ωt
(c) y = A cos kx
(d) y = A sin (at – bx + c)
Answer:
(d) y = A sin (at -bx + c)

Question 42.
A wave travels in a medium according to the equation of displacement given by y(x, t) = 0.03 sin{π(2t – 0.01 x)} where y and x are in metres and t in seconds. The wave length of the wave is …..
(a) 200 m
(b) 100 m
(c) 20 m
(d) 10 m
Answer:
(a) 200 m
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 140

Question 43.
The equation of a wave moving on string is y = 8 sin{π(0.002 x – 4t)} where x, y are in centimeter and t in seconds. The velocity of the wave is ……
(a) 100 cm/s
(b) 0.2π cm/s
(c) 4π cm/s
(d) 200 cm/s
Answer:
(d) 200 cm/s
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 141

Question 44.
If the velocity of sound in air is 340 ms-1, a person singing a note of frequency 250 cps is producing sound waves with a wavelength of ……..
(a) 0.7
(b) 1.36 cm
(c) 1.36 m
(d) 85 km
Answer:
(c) 1.36
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 142

Question 45.
Asa transverse wave strikes against a fixed end …….
(a) its phase changes by 180°, but velocity does not change.
(b) its phase does not change, but velocity changes
(c) its velocity changes and phase too changes by 180°
(d) nothing can be predicted about changes in its velocity and phase.
Answer:
(a) its phase changes by 180°, but velocity does not change

Question 46.
A source of sounds is travelling with a velocity of 40 km/hr towards an observer and emits sound of frequency 2000 Hz. If the velocity of sound is 1220 km/hr, then what is the apparent frequency heard by the observer?
(a) 2068 Hz
(b) 2180 Hz
(c) 2000 Hz
(d) 1980 Hz
Answer:
(a) 2068 Hz
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 143

Question 47.
A vehicle with a horn of frequency n is moving with a velocity of 30m/s in a direction perpendicular to the straight line joining the observer and the vehicle. The observer perceives the sound to have a frequency n + n1. Then (if the sound velocity in air is 300 m/s)
(a) n1 = 10 n
(b) n1 = 0
(c) n1 = – 0.1 n
(d) n1 = 0.1 n
Answer:
(b) n1 = 0
Hint:
No Doppler effect is observed if the source moves perpendicular to the line joining the source and the observer. Therefore, the correct choice is (b).

Question 48.
The Doppler effect is applicable for ……..
(a) light waves
(b) sound waves
(c) space waves
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 49.
The speed of a wave in a medium is 760 m/s. If 3600 waves are passing through a point in the medium in 2 minutes, then its wavelength is ……
(a) 13.8 m
(b) 25.3 m
(c) 41.5 m
(d) 57.2 m
Answer:
(b) 25.3 m
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 150

Question 50.
If a sound wave travels from air to water, the quantity that remain unchanged is …….
(a) velocity
(b) wavelength
(c) frequency
(d) amplitude
Answer:
(c) frequency

Question 51.
Asa spherical wave propagates, …….
(a) the wave intensity remains constant
(b) the wave intensity decrease as the inverse of the distance from the source
(c) the wave intensity decreases as the inverse square of the distance from the source.
(d) The wave intensity decreases as the inverse cube of the distance from the source.
Answer:
(c) The wave intensity decreases as the inverse square of the distance from the source.

Question 52.
A source of sound and a listener are approaching each other with a speed of 40ms-1.The apparent frequency of a note produced by the source is 400 Hz. Then its true frequency is (velocity of sound in air = 360 ms-1)
(a) 320 Hz
(b) 400 Hz
(c) 360 Hz
(d) 420 Hz
Answer:
(a) 320 Hz
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 160

Question 53.
Sound waves of wavelength greater than that of audible sound are called ……
(a) infrasonic waves
(b) ultrasonic waves
(c) sonic waves
(d) seismic waves
Answer:
(a) infrasonic waves

Question 54.
The frequency of a sound wave is/and its velocity is v. If the frequency is increased to 4 f the velocity of the wave will be:
(a) v
(b) 2v
(c) 4 v
(d) v/4
Answer:
(a) v
Hint:
The velocity is a characteristic of the medium and, therefore, it remains constant.

Question 55.
Which of the following statement is untrue? The velocity of sound in a gas …….
(a) is independent of pressure
(b) increases with increase in temperature
(c) is dependent on molecular weight
(d) is greater in dry air than in moist air
Answer:
(d) is greater in dry air than in moist air

Question 56.
When a stone is dropped on the surface of still water, the waves produced are …….
(a) transverse
(b) longitudinal
(c) Stationary
(d) partly longitudinal and partly transverse.
Answer:
(d) Partly longitudinal and partly transverse.

Question 57.
The equation of a wave is y = 0.1 sin (100πt – kx) where x, y are in metres and t in seconds. If – the velocity of the wave is 100 m/s, then the value of k is
(a) 1 m-1
(b) 2m-1
(c) πm-1
(d) 2πm-1
Answer:
(c) πm-1

Question 58.
A transverse wave propagating on a stretched string of linear density 3 × 10-4 kg m-1 is represented by the equation, y = 0.2 sin (1.5x + 60t)
Where x is in metres and t is in seconds. The tension in the string (in newtons) is:
(a) 0.24
(b) 0.48
(c) 1.20
(d) 1.80
Answer:
(a) 0.48
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 500

Question 59.
A transverse wave propagating along x-axis is represented by
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 170
where x is in metres and t is in seconds. The speed of the wave is ……….
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 171
Answer:
(c) 8m/s
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 172

Question 60.
Two waves represented by the following equation are travelling in the same medium: y1 = 5 sin 2π (75t – 0.25 x) and y2 = 10 sin 2π (150 – 0.25x) The intensity ratio of the two waves is ……..
(a) 1 : 2
(b) 1 : 4
(c) 1 : 8
(d) 1 : 16
Answer:
(b) 1 : 4
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 173

Question 61.
A point source emits sound equally in all direction is a non-absorbing medium. Two points P and Q are at distances of 2m and 3m, respectively, from the source. The ratio of the intensities of the waves at P and Q is …….
(a) 3 : 2
(b) 4 : 9
(c) 2 : 3
(d) 9 : 4
Answer:
(d) 9 : 4
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 174

Question 62.
The waves produced by a motor boat sailing in water are ……..
(a) transverse
(b) longitudinal
(c) longitudinal and transverse
(d) stationary
Answer:
(c) longitudinal and transverse

Question 63.
Doppler effect in sound is due to ………
(a) motion of source
(b) motion of observer
(c) relative motion of source and observer
(d) none of the above
Answer:
(c) relative motion of source and observer

Question 64.
The velocity of sound in air at NTP is 330m/s. What will be its value when temperature is doubled and pressure is halved?
(a) 165 m/s
(b) 330 m/s
(c) 330 /\(\sqrt{2}\)
(d) 300/\(\sqrt{2}\) m/s
Answer:
(c) 330 /\(\sqrt{2}\)
Hint:
There is no effect of change of pressure on the velocity of sound in air. Further, v ∝ \(\sqrt{\mathrm{T}}\)

Question 65.
Sound waves travel at 350 m/s through warm air and at 3500 m/s through brass. The wavelength of a 700 Hz acoustic wave as it enters brass from warm air
(a) increases by a factor 20
(b) increases by a factor 10
(c) decreases by a factor 20
(d) decreases by a factor 10
Answer:
(b) increase by a factor 10
Hint:
Since the frequency remains the same, we have
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1701

Question 66.
A train moving at a speed of 220 m/s towards a stationary object, emits a sound of frequency 1000 Hz. Some of the sound reaching the object gets reflected back to the train as echo. The frequency of the echo as detected by the driver of the train is
(a) 3000 Hz
(b) 3500 Hz
(c) 4000 Hz
(d) 5000 Hz
Answer:
(d) 5000 Hz
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1711

Question 67.
A source of sounds emitting waves of frequency 100 Hz and an observer O are located at same distance from each other The source is moving with a speed of 19.4 ms-1 at an angle of 60° with the source-observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air 330 ms-1) is ……
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1721
(a) 97 Hz
(b) 100 Hz
(c) 103 Hz
(d) 106 Hz
Answer:
(c) 103 Hz
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1723

Question 68.
Beats occur because of
(a) interference
(b) reflection
(c) refraction
(d) Doppler effect
Answer:
(a) interference

Question 69.
A vibrating stretched string resonates with a tuning fork of frequency 512 Hz when the length of the string is 0.5 m. The length of the string required to vibrate resonantly with a tuning fork of frequency 256 Hz would be ……
(a) 0.25 m
(b) 0.75 m
(c) 1.0 m
(d) 2.0 m
Answer:
(c) 1.0 m
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 600

Question 70.
A cylindrical tube, open at both ends has a fundamental frequency f in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now …….
(a) f/2
(b) f
(c) 3f/4
(d) 2f
Answer:
(a) f
Hint:
When the tube is dipped in water, it become a closed pipe of length L/2. Its fundamental frequency is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1734

Question 71.
With the increase in temperature, the frequency of the found from an organ pipe
(a) decrease
(b) increase
(c) remains unchanged
(d) changes erractically
Answer:
(b) increase
Hint:
Frequency ∝ v/L. Now v and L both increase with temperature but increase of v is much more than the increase of L which is negligible. Thus frequency increases with temperature.

Question 72.
Two waves of the same frequency and amplitude super impose to produce a resultant disturbance of the same amplitude. The phase difference between the waves is ……
(a) zero
(b) π/3
(c) π/4
(d) 2π/3
Answer:
(d) 2π/3
Hint:
Let the amplitude of each wave be A and phase difference between them be φ. Then,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1742

Question 73.
A sonometer wire is vibrating in the second overtone. In the wire there are ……
(a) two nodes and two antinodes
(b) one node and two antinodes
(c) four nodes and three antinodes
(d) three nodes and three antinodes
Answer:
(c) four nodes and three antinodes

Question 74.
If a resonance tube is sounded with a tuning fork of frequency 256 Hz, resonance occurs at 35 cm and 105 cm. The velocity of sound is about ……
(a) 358 m/s
(b) 512 m/s
(c) 524 m/s
(d) none of these
Answer:
(a) 358 m/s
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 175

Question 75.
A wave of frequency 100 Hz is sent along a string towards a fixed end when this wave travels back after reflection, a node is formed at a distance of 10 cm from the fixed end of the string. The speed of the incident wave is ……
(a) 40 m/s
(b) 20 m/s
(c) 10 m/s
(d) 5 m/s
Answer:
(b) 20 m/s
Hint:
The fixed end is also a node distance between two nodes = \(\frac{\lambda}{2}\) = 10 cm
or λ = 20 cm = 0.2 cm
Speed v = fλ = 100 × 0.2 = 20 m/s

Question 76.
A standing wave is represented by y = A sin (100t) cos (0.01x) where y and A are in millimetres, t in seconds and x in metres. The velocity of the wave is ………
(a) 104 m/s
(b) 1 m/s
(c) 10-4 m/s
(d) not derivable from the above information
Answer:
(a) 104 m/s
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 176

Question 77.
Two waves of the same frequency and intensity superimpose with each other in opposite phases. Then after superposition the ……
(a) intensity increases to four times
(b) intensity increase to two times
(c) frequency increases to four times
(d) none of the above
Answer:
(d) none of the above
Hint:
Since the waves are in opposite phases, the resultant intensity will be zero. The frequency remains the same. So, the correct choice is (d).

Question 78.
Two open organ pipes of lengths 50 cm and 50.5 cm produce 3 beats/s. Then the velocity of sound is …….
(a) 300 m/s
(b) 30 m/s
(c) 303 m/s
(d) 30.3 m/s
Answer:
(c) 303 m/s
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1865

Question 79.
If the ratio of the amplitudes of two waves is 4 : 3, then the ratio of maximum and minimum intensities is …….
(a) 16 : 9
(b) 49 : 16
(c) 7 : 1
(d) 49 :1
Answer:
(d) 49 : 1

Question 80.
An air column in a pipe, which is closed at one end, will be in resonance with a vibrating tuning fork of frequency 256 Hz, if the length of the column in centimeter is (velocity of sound in air = 340 m/s)
(a) 21.25
(b) 125
(c) 62.50
(d) 33.2
Answer:
(d) 33.2
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 187

Question 81.
Two sound waves with wavelengths 5.0 cm and 5.5 cm, respectively each propagate in a gas with velocity 330 m/s. The number of beats per second will be ……..
(a) 0
(b) 1
(c) 6
(d) 12
Answer:
(c) 6
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 188

Question 82.
Two vibrating tuning forks produce progressive waves given be y1 =4 sin 500 πt and y2 = 2 sin 506 πt where t is in seconds number of beats produced per minute is ……..
(a) 60
(b) 3
(c) 369
(d) 180
Answer:
(d) 180
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 189

Question 83.
The ratio of intensities of two waves is 16 : 9. If they produce interference, then the ratio of maximum and minium intensities will be ……..
(a) 4 : 3
(b) 49 : 1
(c) 64 : 27
(d) 81 : 49
Answer:
(b) 49 : 1
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 190

Question 84.
A closes organ pipe of length 20 cm is sounded with a tuning fork in resonance. What is the frequency of the tuning fork? (v = 332 m/s)
(a) 300 Hz
(b) 350 Hz
(c) 375 Hz
(d) 415 Hz
Answer:
(d) 415 Hz So,
Hint: In resonance, the frequency of the fork is equal to the frequency of the organ pipe,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 191

Question 85.
In a resonance tube, the first resonance is obtained at 40 cm length, using a tuning fork of frequency 450 Hz. Ignoring end correction, the velocity of sound in air is
(a) 620 m/s
(b) 720 m/s
(c) 820 m/s
(d) 1020 m/s
Answer:
(b) 720 m/s
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 192

Question 86.
If we study the vibration of a pipe open at both ends, then which of the following statement is not true?
(a) open end will be antinode
(b) odd harmonics of the fundamental frequency will be generated
(c) all harmonics of the fundamental
(d) pressure change will be maximum at both ends.
Answer:
(d) pressure change will be maximum at both ends.
Hint:
Pressure change at open ends is zero.

Question 87.
The fundamental frequency of a closes organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of the organ pipe open at both the ends is ……
(a) 80 cm
(b) 100 cm
(c) 120 cm
(d) 140 cm
Answer:
(c) 120 cm
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 200

Samacheer Kalvi 11th Physics Waves 2 Mark Questions

Question 1.
Define the term wave motion?
Answer:
Wave motion is a kind of disturbances which travels through a medium due to repeated vibrations of the particles of the medium about their mean positions, the disturbance being handed over from one particle to the next.

Question 2.
What is a progressive wave?
Answer:
A wave that travels from one point of the medium to another is called a progressive wave.

Question 3.
What is a plane progressive harmonic wave?
Answer:
If during the propagation of a wave through a medium the particles of the medium vibrate simple harmonically about their mean positions, than the wave is said to be plane progressive harmonic wave.

Question 4.
What do you mean by phase of a wave?
Answer:
The phase of a harmonic is a quantity that gives complete information of the wave at any time and at any position.

Question 5.
Define wave velocity or phase velocity?
Answer:
The distance covered by a wave in the direction of its propagation per unit time is called the wave velocity.

Question 6.
What are stationary waves?
Answer:
When two identical waves of same amplitude and frequency travelling in opposite directionals with the same speed along the same path superpose each other, the resultant wave does not travel in the either direction and is called stationary or standing waves.

Question 7.
What is meant by threshold of heating?
Answer:
The lowest intensity of sound that can be perceived by the human ear is called threshold of hearing. For a sound of frequency 10 kHz, the threshold of hearing is 10-12 Wm-2

Question 8.
What is meant by reverberation?
Answer:
The persistence of audible sound after the source has ceased to emit sound is called reverberation.

Question 9.
What is musical scale?
Answer:
A series of notes whose fundamental frequencies have definite ratios and which produce a pleasing effect on the ear when sounded in succession constitute a musical scale.

Question 10.
Define reverberation time?
Answer:
It is defined as the time which sound takes to fall in intensity to one millionth (10-6) part of its original intensity after it was stopped.

Samacheer Kalvi 11th Physics Waves Numerical Problems

Question 1.
The fundamental frequency in an open organ pipe is equal to the 3rd harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm. What is the length of the open organ pipe.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 210

Question 2.
Two cars moving in opposite directions approach each other with speed of 22 ms-1 and 16.5 ms-1 respectively. The driver of the first car blows a horn having a frequency 400 Hz. To find the frequency heard by the driver of the second car.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 211
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 212

Question 3.
The second overtone of an open organ pipe has the same frequency as the 1st overtone of a closed pipe L metre long. Then what will be the length of the open pipe.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 213
The Length of the open pipe is two times of the length of the closed pipe.

Question 4.
A steel wire 0.72 m long has a mass of 5.0 × 10-3 kg. If the wire is under a tension of 60 N. What is the speed of transverse waves on the wire?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 214

Question 5.
Estimate the speed of sound ¡n air at standard temperature and pressure by using
(i) Newton’s formula and
(ii) Laplace formula. The mass of 1 moIe of air = 29. 0 × 10-3 kg. For air, γ = 1.4
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 215
(ii) According to Newton’s formula speedof sound in air at S.T.P is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 216

Question 6.
An observer moves towards a stationary source of sound with a velocity one-fifth of the velocity of sound. What Is the percentage increase in the apparent frequency?
Answer:
Here observer moves towards the stationary source.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 217
The percentage increase in apparent frequency
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 218

Question 7.
Tube A has both ends open, while B has on end closed otherwise the two tubes are identical. What Is the ratio of fundamental frequency of the tubes A and B?
Answer:
The fundamental frequency for tube A with both ends open is fA = \(\frac{v}{2 \mathrm{L}}\)
The fundamental frequency for tube B with one end closed is fB = \(\frac{v}{4 \mathrm{L}}\)
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 219

Question 8.
A train moves towards a stationary observer with speed 34 mIs. The train sounds a whistle and its frequency registered by the observer is f1. If the train’s speed is reduced to 17 m/s, the frequency registered f2. If the speed of sound is 340 m/s, then find the ratio f1/f2
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 220

Question 9.
A police car with a siren of frequency 8 kHz ¡s moving with uniform velocity 36 km/h towards a tall building which reflect the sound waves. The speed of sound In air is 320m/s. What is the frequency of the siren heard by the car driver?
Answer:
(a) Frequency received by the building.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 2211
The wall (source) reflect this frequency, So frequency heard by the car driver is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 222

Question 10.
The displacement y of a wave travelling in the x-direction ¡s given by y = 10-4 sin (600t – 2x + π/3)
Where x is expressed in metres and t is seconds. What is the speed of the wave motion (in ms-1)?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 223

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