Category: Class 11

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Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 4 Reproductive Morphology

Kickstart your preparation by using this Tamilnadu State Board Solutions for 11th Bio Botany Chapter 4 Reproductive Morphology Questions and Answers get the max score in the exams. You can cover all the topics of Chapter 4 Reproductive Morphology Questions and Answers easily after studying the Tamilnadu State Board 11th Bio Botany Textbook solutions pdf.

Samacheer Kalvi 11th Bio Botany Reproductive Morphology Text Book Back Questions and Answers

Choose the correct answer
Question 1.
Vexillary aestivation is characteristic of the family …………… .
(a) Fabaceae
(b) Asteraceae
(c) Solanaceae
(d) Brassicaceae
Answer:
(a) Fabaceae

Question 2.
Gynoecium with united carples is termed as …………… .
(a) apocarpous
(b) multicarpellary
(c) syncarpous
(d) none of the above
Answer:
(c) syncarpous

Question 3.
Aggregate fruit develops from …………… .
(a) multicarpellary, apocarpous ovary
(b) multicarpellary, syncarpous ovary
(c) multicarpellary ovary
(d) whole inflorescence
Answer:
(a) multicarpellary, apocarpous ovary

Question 4.
In an inflorescence where flowers are borne laterally in an acropetal succession the position of the youngest floral bud shall be …………… .
(a) proximal
(b) distal
(c) intercalary
(d) anywhere
Answer:
(b) distal

Question 5.
A true fruit is the one where …………… .
(a) only ovary of the flower develops into fruit
(b) ovary and calyx of the flower develops into fruit
(c) ovary, calyx and thalamus of the flower develops into fruit
(d) all floral whorls of the flower develops into fruit
Answer:
(a) only ovary of the flower develops into fruit

Question 6.
Find out the floral formula for a bisexual flower with bract, regular, pentamerous, distinct calyx and corolla, superior ovary without bracteole?
Answer:

Question 7.
Give the technical terms for the following:
(a) A sterile stamen
(b) Stamens are united in one bunch
(c) Stamens are attached to the petals
Answer:
(a) A sterile stamen – Staminode
(b) Stamens are united in one bunch – Monadelphous
(c) Stamens are attached to the petals – Epipetalous (petalostemonous)

Question 8.
Explain the different types of placentation with example.
Answer:
The different types of placentation with example:

1. Marginal: It is with the placentae along the margin of a unicarpellate ovary. Example: Fabaceae.
2. Axile: The placentae arises from the column in a compound ovary with septa. Example: Hibiscus, tomato and lemon.
3. Superficial: Ovules arise from the surface of the septa. Example: Nymphaeceae.
4. Parietal: It is the placentae on the ovary walls or upon intruding partitions of a unilocular, compound ovary. Example: Mustard, argemone and cucumber.
5. Free – central: It is with the placentae along the column in a compound ovary without septa. Example: Caryophyllaceae, Dianthus and primrose.
6. Basal: It is the placenta at the base of the ovary. Example: Sunflower (Asteraceae) Marigold.

Question 9.
Differentiate between aggregate fruit with multiple fruit.
Answer:
1. Aggregate fruit:
Aggregate fruits develop from a single flower having an apocarpous pistil. Each of the free carpel is developed into a simple fruitlet. A collection of simple fruitlets makes an aggregate fruit. An individual ovary develops into a drupe, achene, follicle or berry. An aggregate of these fruits borne by a single flower is known as an etaerio. Example: Magnolia, Raspberry, Annona and Polyalthia.

2. Multiple or Composite fruit: A multiple or composite fruit develops from the whole inflorescence along with its peduncle on which they are borne.

• Sorosis: A fleshy multiple fruit which develops from a spike or spadix. The flowers fused together by their succulent perianth and at the same time the axis bearing them become fleshy or juicy and the whole inflorescence forms a compact mass. Example: Pineapple, Jack fruit and Mulberry.
• Syconus: A multiple fruit which develops from hypanthodium inflorescence. The receptacle develops further and converts into fleshy fruit which encloses a number of true fruit or achenes which develops from female flower of hypanthodium inflorescence. Example: Ficus.

Question 10.
Explain the different types of fleshy fruit with suitable example?
Answer:
The fleshy fruits are derived from single pistil, where the pericarp is fleshy, succulent and differentiated into epicarp, mesocarp and endocarp. It is subdivided into the following:

1. Berry: Fruit develops from bicarpellary or multicarpellary, syncarpous ovary. Here the epicarp is thin, the mesocarp and endocarp remain undifferentiated. They form a pulp in which the seeds are embedded. Example: Tomato, date palm, grapes and brinjal.
2. Drupe: Fruit develops from monocarpellary, superior ovary. It is usually one seeded. Pericarp is differentiated into outer skinny epicarp, fleshy and pulpy mesocarp and hard and stony endocarp around the seed. Example: Mango and coconut.
3. Pepo: Fruit develops from tricarpellary inferior ovary. Pericarp terns leathery or woody which encloses, fleshy mesocarp and smooth endocarp. Example: Cucumber, watermelon, bottle gourd and pumpkin.
4. Hesperidium: Fruit develops from multicarpellary, multilocular, syncarpous, superior ovary. The fruit wall is differentiated into leathery epicarp with oil glands, a middle fibrous mesocarp. The endocarp forms distinct chambers, containing juicy hairs. Example: Orange and lemon.
5. Pome: It develops from multicarpellary, syncarpous, inferior ovary. The receptacle also develops along with the ovary and becomes fleshy, enclosing the true fruit. In pome the epicarp is thin skin like and endocarp is cartilagenous. Example: Apple and pear.
6. Balausta: A fleshy indehiscent fruit developing from multicarpellary, multilocular inferior ovary whose pericarp is tough and leathery. Seeds are attached irregularly with testa being the edible portion. Example: Pomegranate.

Textbook Activity Solved

Prepare a diet chart to provide balanced diet to an adolescent (a school going child) which includes food items (fruits, vegetable and seeds) which are non – expensive and are commonly available.
Diet Chart for an Adolescent:

Samacheer Kalvi 11th Bio Botany Reproductive Morphology Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer:
Question 1.
Placentation in tomato and lemon is …………… .
(a) parietal
(b) marginal
(c) free – central
(d) axile
Answer:
(d) axile

Question 2.
The coconut water and the edible part of coconut are equivalent to …………… .
(a) endosperm
(b) endocarp
(c) mesocarp
(d) embryo
Answer:
(a) endosperm

Question 3.
Geocarpic fruits are seen in …………… .
(a) carrot
(b) groundnut
(c) radish
(d) turnip
Answer:
(b) groundnut

Question 4.
Keel is characteristic petal of the flowers of …………… .
(a) Gulmohar
(b) Cassia
(c) Calotropis
(d) Bean
Answer:
(d) Bean

Question 5.
When the calyx is coloured and showy, it is called …………… .
(a) petaloid
(b) sepaloid
(c) bract
(d) spathe
Answer:
(a) petaloid

Question 6.
Bracts are modified leaves which bear flowers in their axils. Identify the plant which has a large showy brightly coloured bract …………… .
(a) Jasmine
(b) Euphorbia
(c) Hibiscus
(d) Bougainvillea
Answer:
(d) Bougainvillea

Question 7.
A flower which can be divided into equal vertical halves, by more than one plane of division is …………… .
(a) zygomorphic
(b) cyclic
(c) actinomorphic
(d) heteromorphic
Answer:
(c) actinomorphic

Question 8.
In Theobroma cocoa, the inflorescence arise from …………… .
(a) terminal shoot
(b) axillary part
(c) trunk of plant
(d) leaf node
Answer:
(c) trunk of plant

Question 9.
The type of inflorescence seen in Caesalpinia is …………… .
(a) corymb
(b) compound corymb
(c) capitulum
(d) umbel
Answer:
(a) corymb

Question 10.
Head is the characteristic of …………… family.
(a) Fabaceae
(b) Malvaceae
(c) Asteraceae
(d) Solanaceae
Answer:
(c) Asteraceae

Question 11.
Thyrsus is a type of …………… inflorescence.
(a) raceme
(b) cyme
(c) mixed
(d) special
Answer:
(c) mixed

Question 12.
Number of whorls in a complete flower is …………… .
(a) one
(b) two
(c) three
(d) four
Answer:
(d) four

Question 13.
Monoclinous flower will have …………… .
(a) androecium
(b) gynoecium
(c) both androecium & gynoecium
(d) none
Answer:
(c) both androecium & gynoecium

Question 14.
If unisexual and bisexual flowers are seen in same plant then the plant is said to be …………… .
(a) polyphyllous
(b) polygamous
(c) hermaphroditic
(d) dioecious
Answer:
(b) polygamous

Question 15.
…………… is a raceme of cymes.
(a) Verticil
(b) Cyathium
(c) Umbel
(d) Thyrsus
Answer:
(d) Thyrsus

Question 16.
Unit of perianth is …………… .
(a) petal
(b) sepal
(c) tepal
(d) stamen
Answer:
(c) tepal

Question 17.
Number of floral parts per whorl is called …………… .
(a) curosity
(b) atrocity
(c) merosity
(d) porosity
Answer:
(c) merosity

Question 18.
What is the green cap – like part of brinjal fruit?
(a) Corolla
(b) Perianth
(c) Calyx
(d) Pistil
Answer:
(c) Calyx

Question 19.
Butterfly shaped corolla is seen in …………… type.
(a) rosaceous
(b) caryophyllaceous
(c) cruciform
(d) papilionaceous
Answer:
(d) papilionaceous

Question 20.
Inflorescence seen in Daucas carota is …………… .
(a) umbel
(b) corymb
(c) compound umbel
(d) compound corymb
Answer:
(c) compound umbel

Question 21.
Arrangement of sepals and petals in flower bud is called …………… .
(a) adhesion
(b) aestivation
(c) placentation
(d) cohesion
Answer:
(b) aestivation

Question 22.
Which is not a part of pistil?
(a) Style
(b) Stigma
(c) Connective tissue
(d) carpel
Answer:
(c) Connective tissue

Question 23.
The type of calyx in brinjal is …………… .
(a) caducous
(b) deciduous
(c) persistent
(d) fugacious
Answer:
(c) persistent

Question 24.
Sterile stamen is called …………… .
(a) pistillode
(b) sessile
(c) staminode
(d) apostamen
Answer:
(c) staminode

Question 25.
Other name for gynoecium is …………… .
(a) carpel
(b) pistil
(c) style
(d) overy
Answer:
(b) pistil

Question 26.
Cavity found inside the ovary is called …………… .
(a) lobule
(b) locule
(c) lacuna
(d) labium
Answer:
(b) locule

Question 27.
Which part of saffron flower is used as flavouring agent?
(a) Carpel
(b) Anther
(c) Style
(d) Stigma
Answer:
(d) Stigma

Question 28.
If the ovary is inferior, then the flower is …………… .
(a) hypogynous
(b) epigynous
(c) perigynous
(d) epihypogynous
Answer:
(b) epigynous

Question 29.
Fabaceae members show …………… placentation.
(a) basal
(b) parietal
(c) superficial
(d) marginal
Answer:
(d) marginal

Question 30.
The side of the flower facing mother axis is called as …………… side.
(a) anterior
(b) posterior
(c) lateral
(d) dorsi – ventral
Answer:
(b) posterior

Question 31.
Which of the following option represents calyx?
(a) C
(b) Ca
(c) K
(d) Ka
Answer:
(c) K

Question 32.
…………… are the products of pollination & fertilization.
(a) Seeds
(b) Ovules
(c) Fruits
(d) Vegetables
Answer:
(c) Fruits

Question 33.
Study of fruits is called as …………… .
(a) honology
(b) pomology
(c) horticulture
(d) apology
Answer:
(b) pomology

Question 34.
Fruit wall can also be called as …………… .
(a) endocarp
(b) epicarp
(c) pericarp
(d) mericorp
Answer:
(c) pericarp

Question 35.
Which part of the apple fruit does we eat?
(a) Perianth
(b) Involucre
(c) Thalamus
(d) Bracteole
Answer:
(c) Thalamus

Question 36.
The false septum seen in siliqua fruits is …………… .
(a) frenulum
(b) micropyle
(c) raphae
(d) replum
Answer:
(d) replum

Question 37.
The type of fruit in Ricinus in …………… .
(a) lomentum
(b) cremocarp
(c) regma
(d) nut
Answer:
(c) regma

Question 38.
Jack fruit is an example for …………… .
(a) syconus
(b) siliqua
(c) sorosis
(d) nut
Answer:
(c) sorosis

Question 39.
Which of the following is not a schizocarpic fruit?
(a) Cremocarp
(b) Regma
(c) Samara
(d) Carcerulus
Answer:
(c) Samara

Question 40.
After fertilization …………… modifies into seed.
(a) ovary
(b) ovule
(c) carpel
(d) stigma
Answer:
(b) ovule

Question 41.
In groundnuts, which part nourishes the embryo?
(a) Endosperm
(b) Albumin
(c) Cotyledons
(d) Carpel
Answer:
(c) Cotyledons

Question 42.
…………… are the means for perpetuation of species.
(a) Fruits
(b) Seeds
(c) Corolla
(d) Flowers
Answer:
(b) Seeds

Question 43.
…………… is the second whorl of the flower.
(a) Calyx
(b) Corolla
(c) Gynoecium
(d) Perianth
Answer:
(b) Corolla

Question 44.
…………… is a ripened ovule.
(a) Carpel
(b) Pistil
(c) Seed
(d) Fruit
Answer:
(c) Seed

Question 45.
Imperfect flowers will have …………… essential whorl(s).
(a) only 1
(b) 2
(c) none
(d) 4
Answer:
(a) only 1

II. Very Short Answer Type Questions (2 Marks)

Question 1.
How will you define inflorescence?
Answer:
An inflorescence is a group of flowers arising from a branched or unbranched axis with a definite pattern.

Question 2.
Where does the inflorescence axis arise in cauliflorous type of inflorescence?
Answer:
In cauliflorous type, inflorescence developed directly from a woody trunk. Example: Theobroma cocoa.

Question 3.
Name any two mixed inflorescences.
Answer:
Two mixed inflorescences:

1. Thyrsus and
2. Verticillaster.

Question 4.
When a flower is said to be complete?
Answer:
A flower is said to be complete when it has all the four whorls (calyx, corolla, androecium & gynoecium).

Question 5.
What is a sessile flower?
Answer:
A flower without a pedicel or stalk is said to be sessile flower.

Question 6.
Define merosity.
Answer:
Number of floral parts per whorl is called merosity.

Question 7.
Write the units of (a) Perianth and (b) Calyx.
Answer:
The units of (a) Perianth and (b) Calyx:

1. (a) Perianth – tepals and
2. (b) Calyx – sepals

Question 8.
Name the three types of petals in papilionoceous corolla.
Answer:
The three types of petals in papilionoceous corolla:

1. Vexillum (standard)
2. wings (alae)
3. petals (carina)

Question 9.
What is a staminode? Give example.
Answer:
Sterile stamen is called staminode. e.g. Cassia

Question 10.
Define Pollinium.
Answer:
When the pollen grains are fused together as a single main, it is said to be pollinium.

Question 11.
List out the parts of a pistil.
Answer:
Ovary, style and stigma.

Question 12.
Define aestivation.
Answer:
Arrangement of sepals and petals in a floral bud.

Question 13.
What is mother axis?
Answer:
The branch that bears the flower is called mother axis.

Question 14.
What do you understand by the term “Pomology”?
Answer:
The branch of horticulture that deals with the study of fruits and their cultivation is called pomology.

Question 15.
How the seeds are classified based on endosperm?
Answer:
(a) Albuminous seed or Endospermous seed.
(b) Ex – Albuminous seed or non – Endospermous seed.

Question 16.
What is Spathe?
Answer:
In spadix, entire inflorescence is covered by a brightly coloured or hard bract called a spathe.

Question 17.
Differentiate Apocarpous and Syncarpous ovary.
Answer:
Apocarpous and Syncarpous ovary:

1. Apocarpous: A pistil contains two or more distinct carpels. Example: Annona
2. Syncarpous: A pistil contains two or more carpels which are connate. Example: Citrus and Tomato

Question 19.
Give examples for following fruit types: (a) Berry (b) Hesperidium
Answer:
(a) Berry: Tomato
(b) Hesperidium: Orange

III. Short Answer Type Questions (3 Marks)

Question 1.
Distinguish between Monoecious & Dioecious.
Answer:
Between Monoecious & Dioecious:

1. Monoecious: Both male and female flowers are present in the same plant, e.g., Coconut
2. Dioecious:  Male and female flowers are present on separate plants, e.g., Papaya

Question 2.
Explain Bilateral symmetry.
Answer:
In bilateral symmetry the flower can be divided into equal halves in only one plane. Zygomorphic flower can efficiently transfer pollen grains to visiting pollinators. Example: Pisum.

Question 3.
Differentiate Apopetalous from Sympetalous.
Answer:
Apopetalous from Sympetalous:

1. Apopetalous (or) Polypetalous: Petals are distinct, e.g., Hibiscus.
2. Sympetalous (or) Gamopetalous: Petals are fused, e.g., Datura.

Question 4.
Define Placentation & mention their types.
Answer:
The mode of distribution of placenta inside the ovary is called placentation. Types of placentation: Marginal, Axile, Superficial, Parietal, Free – central and Basal.

Question 5.
Write the floral formula for the Hibiscus rosasinensis.
Answer:

Question 6.
What are Parthenocarpic fruit?
Answer:
Development of fruits without fertilization are called Parthenocarpic fruit. They are seedless fruits. Example: Banana.

Question 7.
From which type of flowers does the aggregate fruit develops?
Answer:
Aggregate fruits develop from a single flower having an apocarpous pistil. Each of the free carpel is develops into a simple fruitlet. A collection of simple fruitlets makes an aggregate fruit.

Question 8.
Classify seeds based on their cotyledons.
Answer:
Based on the number of cotyledons present, two types of seeds are recognized.

1. Dicotyledonous seed: Seed with two cotyledons.
2. Monocotyledonous seed: Seed with one cotyledon.

Question 9.
List out any 3 significances of seed.
Answer:
3 significances of seed:

1. The seed encloses and protects the embryo for next generation.
2. Seeds of various plants are used as food, both for animals and human.
3. Seeds are the products of sexual reproduction so they provide genetic variations and recombination in a plant.

Question 10.
What is the importance of inflorescence.
Answer:
Function of inflorescence is to display the flowers for effective pollination and facilitate seed dispersal. The grouping of flowers in one place gives a better attraction to the visiting pollinators and maximize the energy of the plant.

Question 11.
Draw the line diagram for the following inflorescence.
Answer:
(a) Simple Dichasium:

(b) Compound Dichasium:

IV. Long Answer Type Questions (5 Marks)

Question 1.
Explain the various types of Schizocarpic fruit.
Answer:
This fruit type of intermediate between dehiscent and indehiscent fruit. The fruit instead of dehiscing rather splits into number of segments, each containing one or more seeds. They are of following types:

1. Cremocarp: Fruit develops from bicarpellary, syncarpous, inferior ovary and splitting into two one seeded segments known as mericarps. e.g., Coriander and Carrot.
2. Carcerulus: Fruit develops from bicarpellary, syncarpous, superior ovary and splitting into four one seeded segments known as nutlets, e.g., Leucas, Ocimum and Abutilon.
3. Lomentum: The fruit is derived from monocarpellary, unilocular ovary. A leguminous fruit, constricted between the seeds to form a number of one seeded compartments that separate at maturity, e.g., Desmodium, Arachis and Mimosa.
4. Regma: They develop from tricarpellary, syncarpous, superior, trilocular ovary and splits into one – seeded cocci which remain attached to carpophore, e.g., Ricinus and Geranium.

Question 2.
Explain the different types of flowers based on the position of ovary.
Answer:
Based on the position of ovary, a flower can be classified as:

1. Hypogynous: The term is used for sepals, petals and stamens attached at the base of a superior ovary, e.g. Malvaceae.
2. Epihypogynous: The term is used for sepals, petals and stamens attached at the middle of the ovary (half – inferior), e.g. Fabaceae and Rosaceae.
3. Epigynous: The term is used for sepals, petals and stamens attached at the tip of an inferior ovary, e.g. Cucumber, apple and Asteraceae.
4. Perigynous: The term is used for a hypanthium attached at the base of a superior ovary.
5. Epiperigynous: The term is used for hypanthium attached at the apex of an inferior ovary.

Question 3.
Classify the anthers based on their mode of attachment.
Answer:
The anthers based on their mode of attachment:

1. Basifixed: (Innate) Base of anther is attached to the tip of filament, e.g., Brassica, Datura
2. Dorsifixed: Apex of filament is attached to the dorsal side of the anther, e.g. Citrus, Hibiscus
3. Versatile: Filament is attached to the anther at midpoint, e.g., Grasses
4. Adnate: Filament is continued from the base to the apex of anther, e.g. Verbena, Ranunculus, Nelumbo.

Question 4.
Define aestivation. Explain its types with example.
Answer:

Question 5.
Distinguish between racemose and cymose inflorescence.
Answer:
Racemose inflorescence

1. Main axis of unlimited growth
2. Flowers arranged in an acropetal succession
3. Opening of flowers is centripetal
4. Usually the oldest flower at the base of the inflorescence axis.

Cymose inflorescence:

1. Main axis of limited growth
2. Flowers arranged in a basipetal succession
3. Opening of flowers is centrifugal
4. Usually the oldest flower at the top of the inflorescence axis.

Question 6.
Write in detail about head inflorescence.
Answer:
Head: A head is a characteristic inflorescence of Asteraceae and is also found in some members of Rubiaceae. Example: Neolamarkia cadamba and Mitragyna parvifolia; and in some members of Fabaceae – Mimosoideae, example: Acacia nilotica, Albizia lebbeck, Mimosa pudica (sensitive plant). Torus contains two types of florets:

1. Disc floret or tubular floret.
2. Ray floret or ligulate floret.

Heads are classified into two types:

1. Homogamous head: This type of inflorescence exhibits single kind of florets. Inflorescence has disc florets alone. Example: Vernonia, Ageratum or Ray florets alone. example: Launaea, Sonchus.
2. Heterogamous head: The inflorescence possesses both types of florets. Example: Helianthus, Tridax.

Disc florets at the centre of the head are tubular and bisexual whereas the ray florets found at the margin of the head which are ligulate pistilate (unisexual).

Question 7.
List out the significance of fruits.
Answer:
The significance of fruits:

1. Edible part of the fruit is a source of food, energy for animals.
2. They are source of many chemicals like sugar, pectin, organic acids, vitamins and minerals.
3. The fruit protects the seeds from unfavourable climatic conditions and animals.
4. Both fleshy and dry fruits help in the dispersal of seeds to distant places.
5. In certain cases, fruit may provide nutrition to the developing seedling.
6. Fruits provide source of medicine to humans.

Question 8.
Draw a flow chart depicting the classification of fruits.
Answer:

Question 9.
Explain in detail about any two special inflorescence.
Answer:
The inflorescences do not show any of the development pattern types are classified under special type of inflorescence.
1. Cyathium: Cyathium inflorescence consists of small unisexual flowers enclosed by a common involucre which mimics a single flower. Male flowers are organized in a scorpioid manner. Female flower is solitary and centrally located on a long pedicel. Male flower is represented only by stamens and female flower is represented only by pistil. Cyathium may be actinomorphic (Example: Euphorbia) or zygomorphic (Example: Pedilanthus) Nectar is present in involucre.

2. Hypanthodium: Receptacle is a hollow, globose structure consisting unisexual flowers present on the inner wall of the receptacle. Receptacle is closed except a small opening called ostiole which is covered by a series of bracts. Male flowers are present nearer to the ostiole, female and neutral flowers are found in a mixed manner from middle below. Example: Ficus sp. (Banyan and Pipal).

V. Higher Order Thinking Skills (HOTs)

Question 1.
Brinjal fruit has persistent calyx. Have you ever noticed the same in any other fruits? Name them.
Answer:
Tomato, Lady’s finger, guava and chilli also have persistent calyx.

Question 2.
Whether parthenocarpic fruits develop endosperm? Why?
Answer:
No, parthenocarpic fruits are developed without fertilization, but endosperm will form only after fertilization. So parthenocarpic fruits do not have endosperm.

Question 3.
Ovary develops into fruit after fertilization. While eating an Apple which part do you eat? Explain.
Answer:
Apple belongs to false fruit. In false fruits, apart from the ovary, non – carpellary parts also develop into fruit. In apple, the thalamus develops into fleshy edible part.

Question 4.
Cremocarp and Carcerulus both are schizocarpic fruits yet they differ. How?
Answer:
Cremocarp:

1. Fruit develops from syncarpous inferior ovary.
2. Ripened fruit split into two, one – seeded segments called mericarps.

Carcerulus:

1. Fruit develops from syncarpous superior ovary.
2. Ripened fruit split into four, one – seeded segments called nutlets.

Question 5.
Mango and coconut are ‘drupe’ type of fruits. In Mango, the edible part is fleshy mesocarp. What does the milk of tender coconut represent?
Answer:
Endosperm is the liquid (milk) potable part of tender coconut, which is rich in nutrients and is formed as a result of triple fusion.

Question 6.
Pick out correct ratio of the male flower to female flower in cyathium inflorescence and explain it?
(a) one : one
(b) one : many
(c) many : many and
(d) many : one.
Answer:
(a) One : one – Cyathium has single female flower represented by pistil and male flower represented by stamen.

Question 7.
Pollen differs from pollinium. How?
Answer:
Pollen are microspores which produces male gametes, whereas pollinium refers to the single mass of fused pollen grains.

Question 8.
Sunflower is not a flower – Justify your answer.
Answer:
Sunflower is actually an inflorescence not a single flower. The inflorescence of sunflower is capitulum composed of disc florets and ray florets.

Question 9.
Flower is a modified shoot for reproduction – Give possible evidence.
Answer:
Modified shoot for reproduction:

• Floral leaves (sepals and petals) are modified leaves.
• Floral and vegetative buds both emerge either in terminal or axillary position.
• Foliage leaves and floral leaves have identical arrangement on stem.

Question 10.
Is tomato a fruit or vegetable? Explain.
Answer:
Yes, tomato is a fruit. Because it develops from the ripened ovary and bears seeds, whereas vegetable refers to all other plant parts like root, stem and leaves.

Question 11.
What is caruncle? Where it is seen? How it helps the plant?
Answer:
Caruncle is the fleshy outgrowth at the base of seed. Usually caruncle helps in the seed dispersal, particularly by ants (Myrmecophily).

Question 12.
Both the prefixes (Uni – and Mono -) have the same meaning i.e. one in number. Does it mean that unisexual and monoecious are the same?
Answer:
That unisexual and monoecious:

1. Unisexual refers to the sex of flower (i.e. whether it has anther or carpel).
2. Monoecious refers to the plant bearing both the sexes in their flowers.

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Samacheer Kalvi 11th Bio Zoology Organ and Organ Systems in Animals Text Book Back Questions and Answers

Question 1.
The clitellum is a distinct part in the body of earthworm Lampito mauritii, it is found in …………….
(a) Segments 13 – 14
(b) Segments 14 – 17
(c) Segments 12 – 13
(d) Segments 14 – 16
Answer:
(b) Segments 14 -17

Question 2.
Sexually, earthworms are ……………..
(a) Sexes are separate
(b) Hermaphroditic but not self-fertilizing
(c) Hermaphroditic and self-fertilizing
(d) Parthenogenic
Answer:
(b) Hermaphroditic but not self-fertilizing

Question 3.
To sustain themselves, earthworms must guide their way through the soil using their powerful muscles. They gather nutrients by ingesting organic matter and soil, absorbing what they need into their bodies. Say whether the statement is true or false: The two ends of the earthworm can equally ingest soil.
(a) True
(b) False
Answer:
(b) False

Question 4.
The head region of Cockroach pairs of and shaped eyes occur.
(a) One pair, sessile compound and kidney shaped
(b) Two pairs, stalked compound and round shaped
(c) Many pairs, sessile simple and kidney shaped
(d) Many pairs, stalked compound and kidney shaped
Answer:
(a) One pair, sessile compound and kidney shaped

Question 5.
The location and numbers of malpighian tubules in Periplaneta ……………
(a) At the junction of midgut and hindgut, about 150.
(b) At the junction of foregut and midgut, about 150.
(c) Surrounding gizzard, eight.
(d) At the junction of colon and rectum, eight.
Answer:
(a) At the junction of midgut and hindgut, about 150.

Question 6.
The type of vision in Cockroach is ………………
(a) Three dimensional
(b) Two dimensional
(c) Mosaic
(d) Cockroach do not have vision
Answer:
(c) Mosaic

Question 7.
How many abdominal segments are present in male and female cockroaches?
(a) 10, 10
(b) 9, 10
(c) 8, 10
(d) 9, 9
Answer:
(a) 10,10

Question 8.
Which of the following have an open circulatory system?
(a) Frog
(b) Earthworm
(c) Pigeon
(d) Cockroach
Answer:
(d) Cockroach

Question 9.
Buccopharyngeal respiration in frog …………….
(a) is increased when nostrils are closed
(b) stops when there is pulmonary respiration
(c) is increased when it is catching fly
(d) stops when mouth is opened.
Answer:
(b) Stops when there is pulmonary respiration

Question 10.
Kidney of frog is ………………
(a) Archinephros
(b) Pronephros
(c) Mesonephros
(d) Metanephros
Answer:
(c) Mesanephros

Question 11.
Presence of gills in the tadpole of frog indicates that ………….
(a) fishes were amphibious in the past
(b) fishes evolved from frog-like ancestors
(c) frogs will have gills in future
(d) frogs evolved from gilled ancestor
Answer:
(d) frogs evolved from gilled ancestor

Question 12.
Choose the wrong statement among the following …………….
(a) In earthworm, a pair of male genital pore is present.
(b) Setae help in locomotion of earthworms.
(c) Muscular layer in the body wall of earthworm is made up of circular muscles and longitudinal muscles.
(d) Typhlosole is part of the intestine of earthworm.
Answer:
(a) In earthworm, a pair of male genital pore is present.

Question 13.
Which of the following are the sense organs of Cockroach?
(a) Antennae, compound eyes, maxillary palps, anal cerci
(b) Antennae, compound eye, maxillary palps and tegmina
(c) Antennae, ommatidia, maxillary palps, stemumv and anal style
(d) Antennae, eyes, maxillary palps, tarsus of walking legs and coxa
Answer:
(c) Antennae, ommatidia, maxillary palps, sternumv and anal style

Question n 14.
What characteristics are used to identify the earthworms?
Answer:
In gardens, earthworms can be traced by their fecal deposits known as worm castings on the soil surface. The earthworms can be identified using the following characteristics:

• Long and cylindrical narrow body.
• Bilateral symmetry
• It is light brown in colour with purple tinge at the anterior end.
• The division of body into many segments or metameres.
• The dorsal surface of the body is marked by a dark mid dorsal line.
• In mature worms, segments 14-17 may be found swollen with a glandular thickening of the skin called the clitellum.

Question 15.
What are earthworm casts?
Answer:
The undigested particles of food along with earth passed out through the anus of the earthworm are called worm casting.

Question 16.
How do earthworms breathe?
Answer:
In earthworms, respiration takes place through the body wall by the moist skin diffusion, oxygen diffuses through the skin into the blood while carbon dioxide from the blood diffuses out.

Question 17.
Why do you call cockroach a pest?
Answer:
Cockroaches destroy food and contaminates it with their offensive odour. They are carriers of a number of bacterial diseases. The cockroach allergen can cause asthma to sensitive people.

Question 18.
Comment on the functions of alary muscles?
Answer:
Alary muscles are the triangular muscles that are responsible for blood circulation in the cockroach. Each segment has one pair and a pumped anteriorly to sinuses again.

Question 19.
Name the visual units of the compound eyes of cockroach.
Answer:
Ommatidia

Question 20.
How does the male frog attracts the female for mating?
Answer:
Male frog has a pair of vocal sacs and a nuptial pad on the ventral side of the first digit of each forelimb. Vocal sacs assist in amplifying the croaking sound of frog. It makes a characteristic sound and attracts the female.

Question 21.
Write the types of respiration seen in frog.
Answer:
Frog respires on land and in the water by two different methods. In water, skin acts as aquatic respiratory organ (cutaneous respiration). Dissolved oxygen in the water gets exchanged through the skin by diffusion. On land, the buccal cavity, skin and lungs act as the respiratory organs. In buccal respiration on land, the mouth remains permanently closed while the nostrils remain open.

The floor of the buccal cavity is alternately raised and lowered, so air is drawn into and expelled out of the buccal cavity repeatedly through the open nostrils. Respiration by lungs is called pulmonary respiration. The lungs are a pair of elongated, pink coloured sac-like structures present in the upper part of the trunk region (thorax). Air enters through the nostrils into the buccal cavity and then to the lungs. During aestivation and hibernation gaseous exchange takes place through skin.

Question 22.
Differentiate between peristomium and prostoinium in earthworm.
Answer:
Peristomium:
The first segment of the body of earthworm is called peristomium.

Prostomium:
A small flap overhanging the mouth is called prostomium or upper lip.

Question 23.
Give the location of clitellum and spermathecal openings in Lampito mauritii.
Answer:
In mature earthworms, 14 – 17th segments are swollen with a glandular thickening of the skin called the clitellum. Permathecal openings are three pairs of small ventrolateral apertures lying intersegmentally between the grooves of the segment 6 / 7,  7 / 8 and 8 / 9.

Question 24.
Differentiate between tergum and a sternum
Answer:
Tergum:
Tergum is the covering each segment of cockroach on the dorsal side.

Sternum:
The sternum is the covering of each segment of cockroach on the ventral side.

Question 25.
Head of cockroach is called hypognathous. Why?
Answer:
The mouth parts of cockroach are directed downwards. The head is small, triangular lies at right angle to the longitudinal body axis. Hence it is called hypognathous.

Question 26.
What are the components of blood in frog?
Answer:
The blood consists of plasma [60%] and blood cells [40%], red blood cells, white blood cells, and platelets. RBCs are loaded with red pigment, nucleated and oval in shape. Leucocytes are nucleated, and circular in shape.

Question 27.
Draw a neat labeled diagram of the digestive system of frog.
Answer:
   

Question 28.
Explain the reproductive system of frog.
Answer:
The male frog has a pair of testes which are attached to the kidney and the dorsal body wall by folds of peritonium called mesorchium. Vasa efferentia arise from each testis. They enter the kidneys on both side and open into the bladder canal. Finally, it communicates with the urinogenital duct that comes out of kidneys and opens into the cloaca (Fig. 1).

Female reproductive system (Fig. 2) consists of paired ovaries, attached to the kidneys, and dorsal body wall by folds of peritoneum called mesovarium. There is a pair of coiled oviducts lying on the sides of the kidney. Each oviduct opens into the body-cavity at the anterior end by a funnel like opening called ostia. Unlike the male frog, the female frog has separate genital ducts distinct from ureters. Posteriorly the oviducts dilated to form ovisacs before they open into cloaca. Ovisacs store the eggs temporarily before they are sent out through the cloaca.

In-Text Questions Solved

Question 1.
How do the earthworms sense activity in their habitat without eyes, ears or a nose?
Answer:
The earthworm’s receptors are stimulated by a group of slender columnar cells connected with nerves. The Photoreceptors (sense of light) are found on the dorsal surface of the body. Gustatory (sense of taste) and olfactory receptors (sense of smell) are found in the buccal cavity. Tactile receptors (sense of touch), chemoreceptors (detect chemical changes) and thermoreceptors (changes in temperature) are present in the prostomium and the body wall.

Question 2.
Respiratory system of cockroach is formed of spiracles and tracheal interconnections. Why is it said to be more efficient than that of earthworm? Why inspiration of cockroach is said to be a passive process while it is an active process in man?
Answer:
The respiratory system of cockroach is well developed compared with other terrestrial insects. Branched tubes known as trachea open through 10 pairs of small holes called spiracles or stigmata

Question 3.
Arthropod eyes are called compound eyes because they are made up of repeating units, the ommatidia, each of which functions as a separate visual receptor. What is the difference between compound eyes and simple eyes? Why is mosaic vision with less resolution seen in cockroaches?
Answer:
f an eye is made of many eyes, it is a compound eye. It has many visual receptors unlike a simple eye which has only one visual receptor. The photoreceptors of the cockroach consists of a pair of compound eyes at the dorsal surface of the head. Each eye is formed of about 2000 simple eyes called the ommatidia (singular: ommatidium), through which the cockroach can receive several images of an object. This kind of vision is known as mosaic vision with more sensitivity but less resolution.

Question 4.
Why three chambered heart of frog is not as efficient has the four chambered heart of birds and mammals?
Answer:
A 4-chambered heart can pump blood more powerfully and efficiently. This helps in better oxygenation of the blood, better circulation and better purification of the blood.

Entrance Examination Questions Solved

Question 1.
The body cells in cockroach discharge their nitrogenous waste in the haemolymph mainly in the form of ……….. (NEET 2015)
(a) Calcium carbonate
(b) Ammonia
(c) Potassium urate
(d) Urea
Answer:
(c) Potassium urate

Question 2.
Frog’s heart when taken out of the body continues to beat for sometime. Select the best option from the following statements. (NEET 2017)
(a) Frog is a poikilotherm.
(b) Frog does not have any coronary circulation.
(c) Heart is “myogenic” in nature.
(d) Heart is autoexcitable Options
Answer:
(d) Heart is autoexcitable Options

Samacheer Kalvi 11th Bio Zoology Organ and Organ Systems in Animals Additional Questions & Answers

I. Multiple Choice Questions
Choose the correct answer

Question 1.
Which of the following is found in upper layers of the soil?
(a) Perionyx exavatus
(b) Octochaetona thurstoni
(c) Lampito mauritii
(d) Eudrilus eugeniae
Answer:
(c) Lampito mauritii

Question 2.
The body setae of earthworms are concerned with
(a) sensory function
(b) protection
(c) excretion
(d) locomotion
Answer:
(d) locomotion

Question 3.
The female genital aperture lies on the ventral side in the segment.
(a) 18^th
(b) 10^th
(c) 14^th
(d) 8^th
Answer:
(c) 14^th

Question 4.
The pair of male genital apertures are situated later ventrally in the segment.
(a) 18^th
(b) 10^th
(c) 14^th
(d) 9^th
Answer:
(a) 18^th

Question 5.
In earthworm, what is present in the 6th segment?
(a) oesophagus
(b) intestine
(c) gizzard
(d) anus
Answer:
(c) gizzard

Question 6.
Which of the following is not found in the prostomium?
(a) Tactile receptors
(b) chemoreceptors
(c) Thermoreceptors
(d) Olfactory receptors
Answer:
(d) Olfactory receptors

Question 7.
In protandrous condition, what is correct?
(a) ova mature earlier than sperms
(b) sperms mature earlier than ova
(c) both sperms and ova mature at the same time
(d) both sperms and ova do not mature at all
Answer:
(b) sperms mature earlier than ova –

Question 8.
The mouth parts of cockroach are of type.
(a) eating and chewing
(b) chewing and sicking
(c) sucking and chewing
(d) biting and chewing
Answer:
(d) biting and chewing

Question 9.
Which of the following is right?
(a) Anal styles are present only in male cockroach.
(b) Anal styles are present in both male and female cockroaches
(c) Anal cerei are present in male cockroach only
(d) Anal cerei are present in female cockroach only
Answer:
(a) Anal styles are present only in male cockroach.

Question 10.
Which is responsible for circulation of blood in cockroach?
(a) spiracular muscles
(b) alary muscles
(c) haemocytes
(d) ostia
Answer:
(b) alary muscles

Question 11.
Cockroach excretes as the nitrogenous waste from the body.
(a) urea
(b) ammonia
(c) uric acid
(d) creatinine
Answer:
(c) uric acid

Question 12.
In cockroach, maiphigian tubules are concerned with
(a) digestion
(b) respiration
(c) excretion
(d) secretion of fluids
Answer:
(c) excretion

Question 13.
The secretion of collaterial glands in cockroach is to produce
(a) ova
(b) sperms
(c) pigments
(d) oothoea
Answer:
(d) oothoea

Question 14.
Which of the following is not a feature of frog?
(a) presence of webbed deet
(b) absence of teeth
(c) smooth and moist skin
(d) slender body
Answer:
(b) absence of teeth

Question 15.
When the frog is ¡n water respiration occurs.
(a) pulmonary
(b) gill
(c) buccal
(d) cutaneous
Answer:
(d) cutaneous

Question 16.
In frog, aortic arch divides into three aortic arches namely …………….
(a) carotid, systemic and sinus venosus
(b) carotid, systemic and pulmo cutaneous
(c) carotid, dorsal aorta and pulmo cutaneous
(d) carotic, truneous arteriosus and pulmo cutaneous
Answer:
(b) carotid, systemic and pulmo cutaneous

Question 17.
Frogs excrete and hence they are called ……………
(a) urea, urecotelic
(b) uric acid. urecotelic
(c) ammonia, ammonotalic
(d) urea, urotelic
Answer:
(c) ammonia, ammonotalic

Question 18.
Which of the following change happens during metamorphosis of tadpole?
(a) gills disappear and lungs develop
(b) lungs disappear and gills develop
(c) gills remains as such
(d) lungs disappear
Answer:
(a) gills disappear and lungs develop

II. Fill In the blanks

Question 1.
………… are the earthworms found in deeper layers of the soil.
Answer:
Endogeics

Question 2.
The segments of the earthworm are ………..
Answer:
metameres

Question 3.
The first segment of earthworm is called …………..
Answer:
prostomium

Question 4.
The last segment of the earthworm is called ……………
Answer:
pygidium

Question 5.
The swollen segments 14 to 17 in mature earthworm is called the …………..
Answer:
clitellum

Question 6.
The body setac are absent in the , last segments and allatum ……………
Answer:
first

Question 7.
The are present from the O segment onwards in the earthworm …………..
Answer:
dorsal

Question 8.
Nephrogenous eliminate wastes …………….
Answer:
metabolic/nitrogenous

Question 9.
found in the coelomic fluid of earthworm plays an important role in regeneration, immunity and wound healing.
Answer:
Coelomocytes

Question 10.
The dorsal wall of the intestine of earthworm is tblded into the cavity as the …………..
Answer:
typhiosole

Question 11.
In earthworm, the vessel has no valves and non-contractile …………….
Answer:
Ventral

Question 12.
In earthworms excretion is effected by …………….
Answer:
nephridia

Question 13
nephridia are present in the 5th – 9th segments.
Answer:
Pharyngeal of tufted

Question 14.
Special cells on the coelomic wall of the intestine called are excretory in function in earthworms.
Answer:
chioragogen cells

Question 15.
In earthworms, the male genital aperture contains two pairs of for copulation.
Answer:
penial setae

Question 16.
The secretion of the prostate gland serves to cement the spermatozoa into bundles known as
Answer:
spermatophores

Question 17.
are present in the 7th, 8th and 9th segment of earthworm.
Answer:
Spermatheca

Question 18.
The process of producing compost using earthworm is called ……………
Answer:
vermi composting

Question 19.
The first pair of wings of cockroach is called ………..
Answer:
alytra of fegmina

Question 20.
In both male and female cockroach, genital apertures are surrounded by scierites called …………….
Answer:
gonapophysis

Question 21.
helps in grinding of food particles in cockroach.
Answer:
Gizzard

Question 22.
In between the foregut and midgut of cockroach are present.
Answer:
Aepatic caecae

Question 23.
In between the midgut and the hindgut of cockroach are present which help in …………….
Answer:
maiphigian tubules excretion

Question 24.
The respiratory organs of cockroach are ……………..
Answer:
trachea

Question 25.
The trachea of cockroach open through 10 pairs of small holes called ……………
Answer:
spiracles/stigmata

Question 26.
The opening and closing of spiracles are regulated by ……………..
Answer:
sphincter spiracular muscles

Question 27.
The blood of cockroach is called as ……………
Answer:
Haemolymph.

Question 28.
The blood from the sinuses enters the heart of cockroach through the …………..
Answer:
ostia

Question 29.
in cockroach, there is a accessory at the base of each antenna which also pumps blood.
Answer:
pulsatile vesicle

Question 30.
The receptors are located in the antenna, maxillary paips and cerci.
Answer:
thigmo

Question 31.
are found on the first four tarsal segments on the legs of cockroach.
Answer:
Thermoreceptors

Question 32.
The receptor chordotonal is found on the which respond lo air or earth borne vibrations.
Answer:
Anal cerci

Question 33.
The eye of cockroach is made up of simple’eyes called the ……………..
Answer:
ommatidia

Question 34.
The collaterial glands present behind the ovaries of female reproductive system cockroach produce ………….
Answer:
ootheca

Question 35.
The young cockroach are called …………….
Answer:
nymphs

Question 36.
The temperature of the frog varies with the varying environmental temperature and hence it is called …………..
Answer:
poikulothermic

III. Answer the following Questions

Question 1.
Classify earthworms based on their ecological strategies.
Answer:

• Earthworms are classified as epigeics, anecics and endogeics based on their ecological strategies.
• Epigeics are the surface dwellers e.g., Perionyx excavaus and Eudrilus eugeniae.
• Anecics are found in the upper layers of the soil e.g., Lampiro mauritii, Lumbricus terrestris.
• Endogeics are found in deeper layers of the soil e.g., Octochaetona thursoni.

Question 2.
Explain the morphology of earthworm, Lampito mauritii.
Answer:

• Lampilo mauritii has a long and cylindrical narrow body which is laterally symmetrical.
• it is light brown in colour with purplish length at the anterior end.
• The body is divided into 165 – 190 segments.
• The dorsal surface of the body is marked by a dark mid dorsal line along the longitudinal axis of the body.
• The ventral surface has genital opening.
• The mouth is found in the centre of the first segment called peristomium. Overhanging
• the mouth is a small flag called upper lip or prostomium.
• The last segment called pygidiurn has the anus.
• The segments 14-17 are swollen with a glandular thickening of the skin called the clitellum.
• The body of earthworm is divided into pre clitellar region (1 – 13), clitellar region (14 – 17) and post clitellar region (after 17).
• Chitinous body selae are present in all segments except the first, last and clitellar segments. They are locomoting in function.
• The dorsal pores are present from the 10th segment onwards. The coelomic fluid
• communicates to the exterior through these pores and keeps the body surface moist and free from harmful microorganism.
• Between the segment 6/7, 7/8 and 8/9, supernatural openings are present.
• The female genital aperture lies on the ventral side in the 14th segment and a pair of male genital apertures are situated determinately in the 1 8th segment.
• Nephridiopores are numerous and found throughout the body except a free anterior segments.

Question 3.
Explain the internal structure of the earthworm.
Answer:
The body wall of the earthworm is very moist, thin, soft, skinny, elastic and consists of the cuticle, epidermis, muscles and coelomic epithelium. The epidermis consists of supporting cells, gland cells, basal cells and sensory cells.

Question 4.
Write a note on coelom of earthworm.
Answer;
A spacious body cavity called the coelom is seen between the alimentary canal and the body wall. The coelom contains the coelomic fluid and serves as a hydrostatic skeleton, in which the coelomocytes are known to play a major role in regeneration. immunity and wound healing. The coelomic fluid of the earthworm is milky and alkaline, which consists of granulocytes or cicocytes. amoebocytes, mucocytes and leucocytes.

Question 5.
Explain the digestive system of earthworm.
Answer:
The digestive system of the earthworm consists of the alimentary canal and the digestive glands. The alimentary canal runs as a straight tube throughout the length of the body from the mouth to anus. The mouth opens into the buccal cavity which occupies the 1 and 2 segments.

The buccal cavity leads into a thick muscular pharynx,which occupies the 3^th and 4^th segments and is surrounded by the pharyngeal glands. A small narrow tube, oesophagus lies in the 5^th segment and continues into a muscular gizzard in the 6th segment. The gizzard helps in the grinding of soit particles and decaying leaves.

Intestine starts from the 7^th segment and continues till the last segment. The dorsal wall of the intestine is folded into the cavity as the typhlosole. This fold contains blood vessels and increases the absorptive area of the intestine. The inner epithehum consists of columnar cells and glandular cells. The alimentary canal opens to the exterior through the anus. The ingested organic rich soil passes through the digestive tract where digestive enzymes breakdown complex food into smaller absorbable units.

The simpler molecules are absorbed through the intestinal membrane and are utilized. The undigested particles along with earth are passed out through the anus, as worm castings or vermicasts. The pharyngeal or salivary gland cells and the glandular cells of the intestine are supposed to be the digestive glands which secrete digestive enzymes for digestion of food.

Question 6.
What is typhiosole?
Answer:

• The dorsal wall of the intestine of earthworm is folded into the cavity as the typhiosole.
• This fold contains blood vessels and increases the absorptive area of the intestine.

Question 7.
Explain the nervous system of earthworm.
Answer:
The bibbed mass of nervous tissue called supra-pharyngeal ganglia lies on the dorsal wall of the pharynx in the 3 segment. It is referred as the brain. In the 4th segment, the sub-pharyngeal ganglia is found. The brain and the sub-pharyngeal are connected by a pair of circum-pharyngeal connectives. The double ventral nerve cord runs backward from the sub-pharyngeal ganglion. The brain and other nerves in the ring integrates sensory inputs and command muscular responses of the body.

Question 8.
Explain the receptors in earthworm.
Answer:
The receptors are simulated by a group of slender columnar cells connected with nerves. The photoreceptor are found on the dorsal surface of the body. Gustatory and olfactory receptors are found in the buccal cavity. Tactile receptors, chernoreceptors and thermoreceptors are present in the prostomium and the body wall.

Question 9.
Explain the excretory system of the earthworm
Answer:
Excretion is the process of elimination of metabolic waste products from the body. In earthworm, excretion is effected by segmentally arranged, minute coiled, paired tubules called nephridia.
There are three types of nephridia:
1. pharyngeal or tufted nephridia – present as paired tufts in the 5^th segments.

2. Microncphridia or Integurnentary nephridia attached to the lining of the body wall from the 14th segment to the last which open on the body surface.

3. Megancphridia or septal nephridia – present as pair on both sides of intersegmental septa of the 1 9th segment to the last and open into intestine (Figure). The mcganephridium has an internal funnel like opening called the nephrostome. which is fully ciliated. The nephrostorne is in the preceding segment and the rest of the tube is in the succeeding segment. This tube consists of three distinct divisions, the ciliated, the glandular and the muscular region.

The waste material collected through the ciliated funnel is pushed into the muscular part of nephridium by the ciliated region. The glandular part extracts the waste from the blood and finally the wastes exit out through the nephridiopore. Bcsides nephridia. special cells on the coelomic wall of the intestine, called chloragogen cells arc present. They extract the nitrogenous waste from the blood of the intestinal wall. into the body cavity to he sent out through the ncphridia.

Question 10.
Explain the male reproductive system of the earthworm.
Answer:
In the male reproductive system, two pairs of testes are present in the 10^th and 11^th segments. The testes give rise to the germ cells or spermatogonia, which develops into spermatozoa in the two pairs of seminal vesicles. Two pairs of seminal ftrnnels called ciliary rosettes are situated in the same segments as the testes. The ciliated funnels of the same side are connected to a long tube called vas dcfcrcns.

The vasa deferentia run upto the 18^th segment where they open to the exterior through the male genital aperture. The male genital aperture contains Iwo pairs of penial setae for copulation. A pair of prostate glands lies in the 18 – 9th segments. The secretion of the prostate gland serves to cement the spermatozoa into bundles known as spermatophores.

Question 11.
Explain the female reproductive system of earthworm.
Answer:
The female reproductive system consists of a pair of ovaries lying in the 3’ segment. Each ovary has finger like projections which contain ova in linear series. Ovarian funnels are present beneath the ovaries which continue into the oviducts.

They join together and open on the ventral side as a single median female genital pore in the 14^th segment. Spermathecae or seminal receptacles are three pairs lying in segments 7^th, 8^th and 9^th opening to the exterior on the ventral side between 6th & 7^th 7^th & 8^th and 8^th & 9^th segments. They receive spermatozoa from the partner and store during copulation.

Question 12.
In earthworm, self fertilization does not take place though it has both male and female reproductive system. Why?
Answer:
The male and female sex organs mature at different times Sperms develops earlier than the production of ova (protandrous). Hence, self fertilization does not take place in earthworm.

Question 13.
Write a short note on the life cycle of earthworni.
Answer:
Lampito mauritii begins its life cycle, from the fertilized eggs. The eggs are held in a protective cocoon. These cocoons have an incubation period of about 14 – 18 days after which they hatch to release juveniles (Figure). The juveniles undergo changes into non clitellate forms in phase I after about 15 days.

which then develops a clitellum, called the clitellate at the end of the growth phase II taking 15 – 17 days to complete. During the reproductive stage, earthworms copulate, and later shed their cocoons in the soil after about 10 days. The life cycle of Lampito mauritii takes about 60 days to complete.

Question 14.
What are earthworm knowii as friends of farmers?
Answer:
They make burrows in the soil and make it porous which helps in respiration and penetration of developing plant roots. They help in recycling of dead and decayed plant material by feeding on them. Hence they arc called as friends of farmers.

Question 15.
Write a short note on vermitech and

1. Vermiculture
2.  Vermicomposting
3. Vermiwash
4.  Wormery

Answer:
1. Vermiculture :
Artificial rearing or cultivation of earthworms involves new technology for the betterment of human beings. This process is known as vermiculture.

2. Vermicomposting :
The process of producing compost using earthworms is called vermicomposting.

3. Vermiwash :
Vermiwash is a liquid manure or plant tonic obtained from earthworm It is used as a foliar spray and helps to induce plant growth. It is a collection of excretory products and mucus secretion of earthworms along with micronutrients from the soil organic molecules.

4. Wormery or wormbin :
Earthworm can be used for recycling of waste food, leaf, litter and biomass to prepare a good fertilizer in a container known as wormery or wormbin. It makes superior compost than conventional composting methods. Vermiculture, vericomposting, vermiwash and wormery are inter-related and independent process, collectively referred as vermitech.

Question 16.
Explain the morphology of cockroach.
Answer:
The body of cockroach is compressed dorso – ventrally, bilaterally symmetrical and segmented. The body is divisible into head, thorax and abdomen. The entire body is covered by a hard, brown, chitinous exoskeleton. In each segment.

exoskeleton has hardened plates called scierites which are joined by a delicate and elastic articular membrane or arthrodial membrane. The sclerites on the dorsal side are called tergites, and those on the ventral side are called stemites and those on the lateral sides are called pleurites.

Head :
Head is small, triangular lies at right angle to the longitudinal axis. The mouth parts are directed downwards so it is hypognathous. The head has a pair of large, sessile and reniform compound eyes, a pair of antennae and appendages around the mouth part.

Antennae have sensory organs. The cockroach has a biting and chewing type (mandibulate or orthopterus type) of mouth parts. It includes a labrum, a pair of mandibles, a pair of maxillae, a labium and a hypopharynx or lingua.

Thorax :
The thorax consists of three segments, prothorax, mesothorax and metathorax. The head is connected with thorax by a short neck or cervicum. Each thoracic segment has three pairs of walking legs. Each leg consists of five segments, coxa, trochanter, femur, tibia and tarsus.

The tarsus has five movable joints or podomeres or tarsomeres. A pair of forewings arises from the mesothorax called elytra or tegmina. It protects the hind wings when at rest. The second pair membranous wings arises from the metathorax and are used in flight.

Abdomen :
The abdomen has 10^th segments. Each segment has a dorsal tergurn, ventral sternum and between them a narrow membranous pleuron. In female the 7^th, segment is boat shaped and together with the gth and 9th sterna forms a brood pouch or genital pouch. Its anterior parts contain female gonopore, spermathecal pores, collaterial glands and the posterior parts constitutes the oothecal chamber in which the cocoons are formed.

in males, the genital pouch lies at the hind end of the abdomen bound dorsally by 9^th and 10^th terga and ventrally by the 9^th sternum. It contains dorsal anus and ventral male genital pore. In both sexes, genital apertures are surrounded by scierites called goriapophysis.

Male has a pair of short and slender anal styles in the 9th stenium which are absent in feniale. In both sexes, the 10th segment has a pair of jointed filanientous structures called anal cerci which is a sense organ that is receptive to vibrations in the air and land. The 7th sternum of male has a pair of large and oval apical lobes orgynovalvular plates which form a keel like structure which distinguishes the male from female.

Question 17.
Explain the digestive system of cockroach.
Answer:
The digestive system of cockroach consists of the alimentary canal and digestive glands. The alimentary canal is present in the body cavity and is divided into three regions: foregut, midgut and hindgut (Figure). The foregut includes pre-oral cavity, mouth, pharynx and oesophagus. This in turn opens into a sac like structure called the crop which is used for storing food.

The crop is followed by the gizzard or proventriculus which has an outer layer of thick circular muscles and thick inner cuticle forming six highly chitinous plates called ‘tecth’. gizzard helps in the grinding of the food particles. The midgut isa short and narrow tube behind the gizzard and is glandular in nature. At the junctional region of the gizzard are eight fingers like tubular blind processes called the hapatic caecae or enteric caecae.

The hindgut is marked by the presence of 100 – 150 yellow coloured thin filamentous maiphigian tubules which arc helpful in removal of the excretory products from the haemolymph. The hindgut is broader than the midgut and is differentiated into ileum. colon, and rectum.

The rectum opens out through the anus. Digestive glands of cockroach consist of the salivary glands. the glandular cells and hcpatic caccac. A pair of salivary glands is found on either side of the crop in the thorax. Thc.glandular cells of the midgut and hepatic or gastric caeeae produce digestive juices.

Question 18.
Explain the circulatory system of cockroach.
Answer:
Periplaneta has an open type of circulatory system (Figure). Blood vessels are poorly developed and opens into the haemocoel in which the blood or haemolymph flows freely. Visceral organs located in the haemocoel are bathed in blood. The haemolynìph is colourless and consists of plasma and haemocytes which are ‘phagocytic’ in nature. Heart is an elongated tube with muscular wall lying mid dorsally beneath the thorax.

The heart consists of 13 chambers with ostia on either side. The blood from the sinuses enters the heart through the ostia and is pumped anteriorly to sinuses again. The triangular muscles that are responsible for blood circulation in the cockroach are called alary muscles (13 pairs). One pair of these muscles is found in each segment on either side of the heart. In cockroach, there is an accessory pulsatile vesicle at the base of each antenna which also pumps blood.

Question 19.
Explain the nervous system of cockroach.
Answer:
The nervous system of cockroach consists of a nerve ring and a ganglionated double ventral nerve cord. suh-oesophageal ganglion. circum – oesophageaÌ connectives and double ventral nerve cord (see figure). The nerve ring is present around the ocsophagus in the head capsule and is formed by the supra-oesophagial ganglion called the ‘brain’. The brain is mainly a sensory and an endocrine centre and lies above the oesophagus. Sub – oesophageal ganglion is the motor centre that controls the movements of the mouth parts, legs and wings.

it lies below the oesophagus and formed by the fusion of the paired gangalia of mandibular, maxillary and labial segments of the head. A pair of circum – oesophageal connectives is present around the oesophagus. connecting the supra oesophageal ganglia with the sub-ocsophageal ganglion.

The double ventral nerve cord is solid. ganglionated and arises from the sub-oesophageal ganglion and extends up to the 7th abdominal segment. Three thoracic ganglia are one in each thoracic segment and six abdominal ganglia in the abdomen.

Question 20.
Write about the sense organs of cockroach.
Answer:
In cockroach, the sense organs are antennae, compound eyes, labrum, maxillary alps, labial paips and anal cerci. The receptor for touch (thigmo receptors) is located in the antenna, maxillary palps and cerci. The receptor for smell (olfactory receptors) is found on the antennae. The receptor for taste (gustatory receptors) is found on the paips of maxilla and labium. Thermoreceptors are found on the first four tarsal segments on the legs.

The receptor chordotonal is found on the anal cerci which respond to air or earth borne vibrations. The photoreceptors of the cockroach Consists of a pair of compound eyes at the dorsal surface of the head. Each eye is formed of about 2000 simple eyes called the ommatidia (singular: ommatidium), through which the cockroach can receive several images of an object. This kind of vision is known as mosaic vision with more sensitivity
but less resolution.

Question 21.
Differentiate male and female cockroaches.
Answer:

Question 22.
Explain excretion in cockroach.
Answer:
The Malpighian tubules are the main excretory organs of cockroach which help in eliminating the nitrogenous wastes from the body in the form of uric acid. Cockroach excretes uric acid, so it is uricotelic. In addition, fat body, nephrocytes. cuticle, and urecose glands are also excretory in function.

The malpighian tubules are thin, long, tilamentous, yellow coloured structures attached at the junction of midgut and hindgut. These are about 100-150 in number and are present in 6-9 bundles. Each tubule is lined by glandular and ciliated cells and the waste is excreted out through the hindgut.

The glandular cells of the malpighian tubules absorb water. salts, and nitrogenous wastes from the haernolymph and transfer them into the lumen of the tubules. The cells of the tubules reabsorb water and certain inorganic salts. By the contraction of the tubules nitrogenous waste is pushed into the ileum, where more water is reabsorbed. It moves into the rectum and almost solid uric acid is excreted along with the fecal matter.

Question 23.
Explain the male reproductive system of cockroach.
Answer:
Cockroach is dioecious or unisexual. They have well developed reproductive organs. The male reproductive system consists of a pair of testes, vasa deferentia, an ejaculatory duct, utricular gland, phallic gland and the external genitalia. A pair of three lobed testes lies on the lateral side of the 4th and 6th abdominal segments. From each testis anses a thin vas deferens.

which opens into the ejaculatory duct through the seminal vesicles. The ejaculatory duct is an elongated duct which opens out by the male gonopore lying ventral to the anus.

A utricular or mushroom shaped gland is a large accessory reproductive gland, which opens into the anterior part of the ejaculatory duct. The seminal vesicles are present on the ventral surface of the ejaculatory duct, These sacs store the sperms in the form of bundles called spermatophores.

The duct of phallic or conglobate gland also opens near the gonopore, whose function is uncertain. Surrounding the male genital opening are few chitinous and asymmetrical structures called phallomeres or gonapophyses which help in copulation.

Question 24.
ExplaIn the female reproductive system in cockroach.
Answer:
The female reproductive system of cockroach consists of a pair of ovaries, vagina, genital pouch, collaterial glands. spermathecae and the external genitalia. A pair of ovaries lies laterally in the 2^nd and 6^th abdominal segment. Each ovary is formed of a group of eight ovarian tubules or ovarioles, containing a chain of developing ova. The lateral oviducts of each ovary unite into a broad median common oviduct known as vagina, which opens into the genital chamber.

The vertical opening of the vagina is the female genital pore. A pair of spermathccae is present in the 6^th segment, which opens by a median aperture in the dorsal wall of the genital pouch. During copulation, the ova descend to the genital chamber, where they are fertilized by the sperms. A pair of white and branched collaterial glands present behind the ovaries forms a hard egg case called Ootheca around the eggs. Genital pouch is formed by the 7, 8^th and 9^th abdominal sterna.

The genital pouch has two chambers, a genital chamber into which the vagina opens and an oothecal chamber where oothecae are formed. Three pairs of plate like chitinous structures called gonapophyses are present around the female genital aperture. These gonapophyses guide the ova into the ootheca as ovipositors.

Question 25.
Write a short note on ootheca.
Answer:
A pair of white and branched collaterial glands present behind the ovaries forms a hard egg case called oothcca around the eggs. Ootheca is a dark reddish to blackish brown capsule about 12 mm long which contains nearly 16 fertilized eggs and dropped or glued to a suitable surface, usually in crack or crevice of high relative humidity near a food source. On an average, each female cockroach produces nearly 15 – 40 oothecae in its life span of about one to two years.

Question 26.
Explain the morphological features of frog.
Answer:
The body of a frog is streamlined to help in swimming. It is dorso-ventrally flattened and is divisible into head and trunk. Body is covered by a smooth, slimy skin loosely attached to the body wall. The skin is dark green on the dorsal side and pale ventrally.

The head is almost triangular in shape and has an apex which forms the snout. The mouth is at the anterior end and can open widely. External nostrils are present on the dorsal surface of the snout, one on each side of the median line. Eyes are large and project above the general surface of the body.

They lie behind the external nostrils and are protected by a thin movable lower eyelid, thick immovable upper eyelid and a third transparent eyelid called nictitating membrane. This membrane protects the eye when the frog is tinder water.

A pair of tympanic membranes forms the ear drum behind the eyes on either side. Frogs have no external ears, neck and tail are absent. Trunk bears a pair of fore limbs and a pair of hind limbs. At the posterior end of the dorsal side, between the hind limbs is the cloacal aperature.

This is the common opening for the digestive, excretory and reproductive systems. Fore limbs are short, stumpy, and helps to bear the weight of the body. They are also helpful for the landing of the frog aller leaping. Each forelimb consists of an upper arm.

fore arm and a hand. Hand bears four digits. Hind limbs are large, long and consist of thigh. shank and foot. Foot bears five long webbed toes and one small spot called the sixth toe. These are adaptations for leaping and swimming. When the animal is at rest, the hind limbs are kept folded in the form of letter ‘Z’ Sexual dimorphism is exhibited clearly during the breeding season.

The male frog has a pair of vocal sacs and a copulatory or nuptial pad on the ventral side of the first digit of each forelimb. ‘cal sacs assist in amplifying the croaking sound of frog. Vocal sacs and nuptial pads are absent in the female frogs.

Question 27.
Differentiate between a frog and toad.
Answer:

Question 28.
Explain the digestive system of frog.
Answer:
The alimentary canal consists of the buccal cavity, pharynx, oesophagus, duodenum, ileum and the rectum which leads to the cloaca and opens outside by the cloacal aperture. The wide mouth opens into the buccal cavity. On the floor of the buccal cavity lies a large muscular sticky tongue.

The tongue is attached in front and free behind. The free edge is forked. When the frog sights an insect it flicks out its tongue and the insect gets glued to the sticky tongue. The tongue is immediately withdrawn and the mouth closes. A row of small and pointed maxillary teeth is found on the inner region of the upper jaw.

In addition vomerine teeth are also present as two groups, one on each side of the internal nostrils. The lower jaw is devoid of teeth. The mouth opens into the buccal cavity that leads to the oesophagus through the pharynx. Oesophagus is a short tube that opens into the stomach and continues as the intestine, rectum and finally opens outside by the cloaca. Liver secretes bile which is stored in the gall bladder. Pancreas, a digestive gland produces pancreatic juice containing digestive enzymes.

Food is captured by the bilobed tongue. Digestion of food takes place by the action of hydrochloric acid and gastric juices secreted from the walls of the stomach. Partially digested food called chyme is passed from the stomach to the first part of the intestine, the duodenum. The duodenum receives bile from the gall bladder and pancreatic juices from the pancreas through a common bile duct.

Bile emulsifies fat and pancreatic juices digest carbohydrates, proteins and lipids. Final digestion takes place in the intestine. Digested food is absorbed by the numerous finger-like folds in the inner wall of intestine called villi and microvilli. The undigested solid waste moves into the rectum and passes out through the cloaca.

Question 29.
Explain the circulatory system of frog.
Answer:
Blood vascular system consists of a heart with three chambers, blood vessels and blood. Heart is covered by a double- walled membrane called pericardium. There are two thin walled anterior chambers called auricles (Atria) and a single thick walled posterior chamber called ventricle. Sinus venosus is a large, thin walled, triangular chamber, which is present on the dorsal side of the heart. Truncus arteriosus is a thick walled and cylindrical structure which is obliquely placed on the ventral surface of the heart.

It arises from the ventricle and divides into right and left aortic trunk, which is further divided into three aortic arches namely carotid, systemic and pulmo-cutaneous. The Carotid trunk supplies blood to the anterior region of the body. The systemic trunk of each side is joined posteriorly to form the dorsal aorta. They supply blood to the posterior part of the body. Pulmo-cutaneous trunk supplies blood to the lungs and skin.

Sinus venosus receives the deoxygenated blood from the body parts by two anterior precaval veins and one post caval vein. It delivers the blood to the right auricle; at the same time left auricle receives oxygenated blood through the pulmonary vein. Renal portal and hepatic portal systems are seen in frog.

Question 30.
Explain the nervous system of frog.
Answer:
The Nervous system is divided into the Central Nervous System [CNS], the Peripheral Nervous System [PNS], and the Autonomous Nervous System [ANS], Peripheral Nervous System consists of 10 pairs of cranial nerves and 10 pairs of spinal nerves. Autonomic Nervous System is divided into sympathetic and parasympathetic nervous system. They control involuntary functions of visceral organs.

Question 31.
Explain the structure of brain of frog ?
Answer:
Brain is situated in the cranial cavity and covered by two meninges called piamater and duramater. The brain is divided into forebrain, midbrain and hindbrain. Fore brain (Prosencephalon) is the anterior most and largest part consisting of a pair of olfactory lobes and cerebral hemisphere (as Telen-cephalon) and a diencephalon. Anterior part of the olfactory lobes is narrow and free but is fused posteriorly. The olfactory lobes contain a small cavity called olfactory ventricle.

The mid brain (Mesencephalon) includes two large, oval optic lobes and has cavities called optic ventricles. The hind brain (Rhombencephalon) consists of the cerebellum and medulla oblongata. Cerebellum is a narrow, thin transverse band followed by medulla oblongata. The medulla oblongata passes out through the foramen magnum and continues as spinal cord, which is enclosed in the vertebral column.

Question 32.
Explain the excretory system of Frog.
Answer:
Elimination of nitrogenous waste and salt and water balance are performed by a well developed excretory system. It consists of a pair of kidneys, ureters, urinary bladder and cloaca. Kidneys are dark red, long, flat organs situated on either sides of the vertebral column in the body cavity.

Kidneys are Mesonephric. Several nephrons are found in each kidney. They separate nitrogenous waste from the blood and excrete urea, so frogs are called ureotelic organisms. A pair of ureters emerges from the kidneys and opens into the cloaca. A thin walled unpaired urinary bladder is present ventral to the rectum and opens into the cloaca.

Question 33.
Write the short note on the development in frog.
Answer:
Within few days of fertilization, the eggs hatch into tadpoles. A newly hatched tadpole lives off the yolk stored in its body. It gradually grows larger and develops three pairs of gills. The tadpole grows and metamorphosis into an air-breathing carnivorous adult frog (Figure). Legs grow from the body, and the tail and gills disappear. The mouth broadens, developing teeth and jaws, and the lungs become functional.

Question 34.
Write a note on the economic importance of frog.
Answer:
Economic importance of Frog:
Frog is an important animal in the food chain; it helps to maintain our ecosystem. So ‘frogs should be protected’.
Frog are beneficial to man, since they feed on insects and helps in reducing insect pest population. Frogs are used in traditional medicine for controlling blood pressure and for its anti aging properties. In USA. Japan, China and North East of India, frogs are consumed as delicious food as they have high nutritive value.

Question 35.
Frog respires through gills, lungs, skin and buccal cavity’. Justify.
Answer:
Frog is amphibious. It respires through lungs (pulmonary respiration) when it is on land. Buccal respiration is there in frog when it is on land. When it is in water, it respires through skin (cutaneous respiration). The larva of frog, tadpole respires through gills. During development, gills disappear and lungs develop and the tadpole metamorphoses into an adult frog. Hence, frog respires through gills, lungs, buccal cavity and skin in its life cycle.

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Tamilnadu Samacheer Kalvi 11th Bio Zoology Solutions Chapter 3 Tissue Level of Organisation

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Samacheer Kalvi 11th Bio Zoology Tissue Level of Organisation Text Book Back Questions and Answers

Choose the correct answer:
Question 1.
The main function of the cuboidal epithelium is –
(a) Protection
(b) Secretion
(c) Absorption
(d) Both (b) and (c)
Answer:
(d) Both (b) and (c)

Question 2.
The ciliated epithelium lines the –
(a) Skin
(b) Digestive tract
(c) Gall bladder
(d) Trachea
Answer:
(d) Trachea

Question 3.
What type of fibres are found in connective tissue matrix?
(a) Collagen
(b) Areolar
(c) Cartilage
(d) Tubular
Answer:
(a) Collagen

Question 4.
Prevention of substances from leaking across the tissue is provided by –
(a) Tight junction
(b) Adhering junction
(c) Gap junction
(d) Elastic junction
Answer:
(a) Tight junction

Question 5.
Non-shivering thermogenesis in neonates produces heat through –
(a) White fat
(b) Brown fat
(c) Yellow fat
(d) Colourless fat
Answer:
(b) Brown fat

Question 6.
Some epithelia are pseudostratified. What does this mean?
Answer:
Pseudostratified epithelial cells are columnar, but unequal in size. Although the epithelium is single layered yet it appears to be multilayered due to the fact that nuclei lie at different levels in different cells.

Question 7.
Differentiate white adipose tissue from brown adipose tissue.
Answer:
Adipose tissue:

• Adipose tissue is the group of fat cells.
• It stores fats.
• It releases energy while fasting.

Brown Adipose Tissue:

• Adipose tissue which contains abundant mitochondria are called brown adipose tissue.
• It is used to warm the blood stream to warm the body.
• It produces heat by non-shivering thermogenesis.

Question 8.
Why blood is considered as a typical connective tissue?
Answer:
Blood is considered as a typical connective tissue because it is the fluid connective tissue containing plasma, RBCs, WBCs and platelets. It functions as the transport medium for the cardiovascular system carrying nutrients, nitrogenous wastes and respiratory gases throughout the body.

Question 9.
Differentiate between elastic fibres and elastic connective tissue.
Answer:
Elastic fibres:

• It contains elastin and other proteins and glycoproteins.
• It attaches muscles and bones and one bone to another bone.
• It withstands tensile stress when pulling force is applied in one direction or in many directions.

Elastic connective tissue:

• It contains high proportion of elastic fibres.
• It is found in the walls of large arteries, ligaments associated with vertebral column and within the walls of the bronchial tubes.
• It allows recoil of tissues after stretching.

Question 10.
Name any four important functions of epithelial tissue and provide at least one example of a tissue that exemplifies each function.
Answer:
1. Secretion and absorption :
Cuboidal epithelium in kidney tubules, ducts. Columnar epithelium found in the digestive tract.

2. Filtration :
Squamous epithelium found in the glomerulus of kidney.

3. Ciliated epithelium :
Found in the bronchi, uterine tubes propels the materials due to ciliary actions.

Question 11.
Write the classification of connective tissue and their functions.
Answer:
There are four main classes of connective tissues.

• Connective tissue proper
• Cartilage
• Bones
• Blood

The major functions of connective tissues are binding and support, protection, insulation and transportation of substances.

Question 12.
What is an epithelium? Enumerate the characteristic features of different epithelia.
Answer:
Epithelial tissue is sheet of cells that covers the body surface or lines the body cavity.

• Simple epithelium is single layered.
• Squamous epithelium is made of flattened cells with irregular boundaries.
• Columnar epithelium is made of column like cells with round to oval nuclei at the base.
• Ciliated epithelium has ciilia at the free end.
• Compound epithelium is made of multi-layered cells.

In-Text Questions Solved

Question 1.
Stratified epithelia are “built” for protection or to resist abrasion. What are the simple epithelia better at?
Answer:
The simple epithelia are better at absorption, secretion of mucus, enzymes and other substances.

Question 2.
What type of connective tissue is damaged when one get cut on his index finger accidently?
Answer:
The Areolar connective tissue is damaged when finger gets cut.

Question 3.
The stored lipids are in the form of adipose tissue. Are they coloured? Why?
Answer:
The white adipose tissue is called white fat. The adipose that has abundant mitochondria is called Brown fat.

Question 4.
You are looking at a slide of a tissue through the compound microscope and you see striped branching cells that connect with one another. What type of muscle are you viewing?
Answer:
I am viewing the skeletal muscle.

Question 5.
A player has sustained a severe injury during football practice and was told that he has a torn knee cartilage. Can he expect a quick uneventful recovery? Explain your response.
Answer:
The knee cartilage is an important connective tissue. Since the knee moves during locomotion, a quick, uneventful recovery cannot take place. Complete rest to the knee joint is necessary.

Question 6.
An overweight high school student, is overheard telling her friend that she is going to research how she can transform some of her white fat to brown fat. What is her rationale here (assuming it is possible)?
Answer:
The white fat stores nutrients while the brown fat warms the body. The student feels that she may bring down her weight by converting brown fat to white fat.

Textbook Activities Solved

Question 1.
Students are asked to identify the unlabelled slides of tissues and to classify them. Similar exercise can also be accomplished by projecting unlabelled histological images on a screen. They can identify the slides of different tissues through microscope.
Answer:
Do it yourself.

Question 2.
The preparation of smear of stratified squamous epithelia from the inner lining of cheek allows the students to make their own slides using biological stain. They will have the experience of examining their cheek cells.
Answer:
Do it yourself.

Entrance Examination Questions Solved

Question 1.
Transitional epithelium occurs in ………… (MHTCET 2008)
(a) Blood vessels
(b) Trachea
(c) Kidney
(d) Ureter/urinary bladder
Answer:
(d) Ureter/urinary bladder

Question 2.
The study of tissues is known as …………….. (MPPMT 2010)
(a) Physiology
(b) Ecology
(c) Histology
(d) Anatomy
Answer:
(c) Histology

Question 3.
Find out the wrong match :
(a) Eosinophils Allergic response
(b) Basophils Secrete histamine and serotonin
(c) Monocytes Secrete heparin
(d) Lymphocytes Immune response
Answer:
(c) Monocytes Secrete heparin

Question 4.
The outer covering of cartilage is called (WB 2010)
(a) Peritoneum
(b) Periosteum
(c) Endosteum
(d) Perichondrium
Answer:
(d) Perichondrium

Question 5.
Skin is …………… (CPMT2010)
(a) Cubiodal epithelium
(b) Stratified epithelium
(c) Columnar epithelium
(d) Pseudostratified epithelium
Answer:
(b) Stratified epithelium

Question 6.
Match the animals listed in column – I to blood listed in column – II. (KCET 2010)

Column -I	Column – II
(i) Plasma and cells are colourless	(P) Man
(ii) Plasma colourless and nucleated RBC	(Q) Earth worm
(ii) Plasma colourless and enucleated RBC	(R) Cockroach
(iv) Plasma red and nucleated colourless RBC	(S) Frog
(v) Plasma and RBS have haemoglobin	–

(a) (P – iii), (Q – iv), (R – i), (S – ii)
(b) (P – iv), (Q – v), (R – iii), (S – ii)
(c) (P – i), (Q – iv), (R – ii), (S – iii)
(d) (P – v), (Q – iii), (R – i), (S – iv)
Answer:
(a) (P – iii), (Q – iv), (R – i), (S – ii)

Question 7.
Matrix of bone and cartilage can be distinguished by the presence of –
(a) Lacunae
(b) Chromatophares
(c) Haversian canals
(d) Adipose cells
Answer:
(c) Haversian canals

Question 8.
Which type of tissue forms glands? (MPPMT- 2010)
(a) Epithelial
(b) Muscular
(c) Nervous
(d) Connective
Answer:
(a) Epithelial

Question 9.
Which of the following blood cells help in blood coagulation?
(a) RBCs
(b) Lymphocytes
(c) Thrombocytes
(d) Basophils
Answer:
(c) Thrombocytes

Question 10.
Fibroblasts macrophages and mast cells are present in –
(a) Cartilage tissue
(b) Areolar tissue
(c) Adipose tissue
(d) Glandular epithelium
Answer:
(b) Areolar tissue

Question 11.
Which type of epithelium is involved in a function to move particles or mucus in specific direction? (HPPMT 2010)
(a) Squamous epithelium
(b) Cuboidal epithelium
(c) Columnar epithelium
(d) Ciliated epithelium
Answer:
(d) Ciliated epithelium

Question 12.
Which of these is not found in connective tissue? (MPPMT2010)
(a) Collagen fibres
(b) Basement membrane
(c) Hyaluronic acid
(d) Fluid
Answer:
(b) Basement membrane

Question 13.
Multi-lobed nucleus and granular cytoplasm are characteristics of which of the WBCs?
(a) Neutrophils
(b) Monocytes
(c) Lymphocytes
(d) Eosinophils
Answer:
(a) Neutrophils

Question 14.
Which one of the following plasma proteins is involved in the coagulation of blood? (2011)
(a) globulin
(b) Fibrinogen
(c) albumin
(d) Serum amylase
Answer:
(b) Fibrinogen

Question 15.
Which of the following is not a connective tissue? (CPMT – 2010)
(a) Blood
(b) bone
(c) Lymph
(d) Nerve
Answer:
(d) Nerve

Question 16.
The ciliated columnar epithelial cells in humans are known to occur in –
(a) Bile duct and oesophagus
(b) Fallopian tubes and urethra
(c) Eustachian tube and stomach lining
(d) Bronchioles and fallopian tubes
Answer:
(d) Bronchioles and fallopian tubes

Samacheer Kalvi 11th Bio Zoology Tissue Level of Organisation Additional Questions & Answers

 Multiple Choice Questions
Choose the correct answer
Question 1.
Groups of cells that are similar in structure and perform a common function are called –
(a) tissues
(b) organs
(c) cells
(d) organ systems
Answer:
(a) tissues

Question 2.
Which of the following have flattened cells?
(a) cuboidal epithelium
(b) columnar epithelium
(c) squamous epithelium
(d) ciliated epithelium
Answer:
(c) squamous epithelium

Question 3.
Microvilli and Goblet cells are the modifications of –
(a) cuboidal epithelium
(b) columnar epithelium
(c) squamous epithelium
(d) ciliated epithelium
Answer:
(b) Columnar epithelium

Question 4.
Which of the following is not exocrine gland?
(a) Sweat glands
(b) Sebaceous glands
(c) Mammary glands
(d) Thyroid gland
Answer:
(d) Thyroid gland

Question 5.
Pancreas is the example of glands –
(a) Merocrine
(b) Holocrine
(c) Apocrine
(d) Epithelial
Answer:
(a) Merocrine

Question 6.
Which is the site of production of blood cells?
(a) Cartilage
(b) Blood
(c) PLasma
(d) Bone marrow
Answer:
(d) Bone marrow

Question 7.
Biceps and Triceps are the examples of –
(a) Smooth muscle
(b) Cardiac muscle
(c) Striped muscle
(d) Involuntary muscle
Answer:
(c) Striped muscle

Question 8.
The walls of internal organs are made up of –
(a) Smooth muscle
(b) involuntary muscle
(c) Skeletal muscle
(d) Cardiac muscle
Answer:
(a) Smooth muscle

Question 9.
Bone cells are called as –
(a) Neurons
(b) Epithelial cells
(c) Osteoblasts
(d) Chondrocytes
Answer:
(c) Osteoblasts

Question 10.
Cartilage is the –
(a) Loose connective tissue
(b) Dense connective tissue
(c) Areolar connective tissue
(d) Specialized connective tissue
Answer:
(d) Specialized connective tissue

Question 11.
Salivary gland is –
(a) Unicellular, glandular cells
(b) Multicellular, glandular cells
(c) Unicellular, sensory cells
(d) Multicellular, sensory cells
Answer:
(c) Unicellular, sensory cells

Question 12.
lines gall bladder.
(a) ciliated epithelium
(b) columnar epithelium
(c) non – ciliated epithelium
(d) pseudo – stratified epithel lurn
Answer:
(c) non – ciliated epithelium

Question 13.
Dry epidermis of the skin is formed as –
(a) keratinized stratified squamous epithelium
(b) non – keratinized stratified squamous epithelium
(c) stratified cuboidal epithelium
(d) stratified columnar epithelium
Answer:
(a) keratinized stratified squamous epithelium

Question 14.
The walls of the Bronchial tubes have –
(a) Dense irregular connective tissues
(b) Reticular connective tissue
(c) elastic connective tissue
(d) Adipose tissue
Answer:
(c) elastic connective tissue

Question 15.
Bones have –
(a) Osteocytes
(b) Fibroblasts
(c) Adipocytes
(d) Myofibrils
Answer:
(c) Adipocytes

II. Answer the following Questions

Question 1.
Define tissues.
Answer:
Group of cells that are similar in structure and perform a common or related functions are called tissues.

Question 2.
What is the study of tissues called?
Answer:
Histology.

Question 3.
Differentiate Simple epithelium and compound epithelium.
Answer:
Simple epithelium:

• It consists of a simple layer.
• It helps in protection, absorption, filtration, excretion, secretion and sensory reception.

Compound epithelium:

• It is multilayered.
• It provides protection against chemical and mechanical stresses.

Question 4.
Explain the types of simple epithelium.
Answer:
Simple epithelium is a simple layered sheet of cells that covers the body surface or lines the body cavity.
Types:
1. Squamous epithelium:
It is made of flattened cells with irregular boundaries. It is found in glomeruli, air sacs of lungs, lining of heart, blood vessels.

2. Cuboidal epithelium:
It is made of cube like cells. It is found in kidney tubules, ducts and glands. It is important for secretion and absorption.

3. Columnar epithelium :
It is made of column like cells. It lines the digestive tract. It is important for secretion and absorption.

4. Ciliated epithelium :
It has cilia at the free end. It is found in bronchi, uterine tubes. It is helpful in propelling materials.

5. Glandular epithelium :
Cuboidal or columnar epithelium specialized for secretion is called glandular epithelium. E.g., goblet cells and salivary gland.

Question 5.
Distinguish between exocrine glands and endocrine glands.
Answer:
Exocrine glands:

• These glands release their products through ducts.
• These secrete mucous, saliva, ear wax, oil, milk, digestive enzymes etc. e.g., salivary glands

Endocrine glands:

• These are ductless gland and their secretions are released directly into the blood.
• These secrete hormones, e.g., Pituitary gland

Question 6.
Classify multicellular exocrine glands based on their structure.
Answer:

Question 7.
Classify exocrine glands based on mode of secretion.
Answer:

Question 8.
Explain compound epithelium.
Answer:

• Compound epithelium is made up of multilayered cells.
• These protect organs against chemical and mechanical stresses.
• These cover the dry surface of the skin, moist surface of the buccal cavity, pharynx, inner lining of ducts of salivary glands and pancreatic ducts.

Question 9.
Classify compound epithelium.
Answer:

Question 10.
Write a short note on specialized junctions of epithelia.
Answer:
All cells of epithelia are held together with little intercellular material forming specialized junctions. These provide structural and functional links between the cells. Three types of cell junctions, tight, adhering and gap junctions are found in animal tissues.

Tight junctions help to stop substances from leaking across the tissue. Adhering junctions cement the neighbouring cells together. Gap junctions facilitate the transfer of ions, small and big molecules between the adjoining cells by connecting the cytoplasm of these cells.

Question 11.
Write a short note on connective tissue.
Answer:
Connective tissue develops from the mesoderm. Proper, cartilage, bones and blood are the four main classes of connective tissues. Binding, support, protection, insulation and transportation of substances are the major functions of connective tissue.

Question 12.
What are the types of proper connective tissues?
Answer:
Loose connective tissue and dense connective tissues.

Question 13.
Write a short note on loose connective tissues.
Answer:
In this tissue, the cells and fibres are loosely arranged in semifluid ground substances, e.g., fibroblasts, macrophages, fat cells and mast cells. Areolar connective tissue present beneath the skin acts as a support framework for epithelium. It acts as a reservoir of water and salts for the surrounding body tissues. Hence, these are called tissue fluid.

Adipose tissue is similar to areolar tissue in structure and function. It is located beneath the skin, surrounding the kidneys, eyeball, heart etc. Adipocytes store fat. It is called white fat. The adipose tissue which contains a lot of mitochondria is called brown fat or brown adipose tissue. Reticular connective tissue is filled with fibroblasts called reticular cells. These cells store fats and the excess nutrients.

Question 14.
Distinguish between tendons and ligaments.
Answer:
Tendons:
Tendons attach skeletal muscles to bones

Ligaments:
Ligaments attach one bone to another.

Question 15.
Explain specialised connective tissues.
Answer:
Cartilage :
The intercellular material of cartilage is solid and pliable and resists compression. Cells of cartilage (chondrocytes) are enclosed is small cavities within the matrix secreted by them. Cartilage is present in the tip of nose, outer ear joints, ear pinna, between adjacent bones of the vertebral column, limbs and hands on adults.

Bones :
Bones have a hard and non-pliable ground substance rich in calcium salts and collagen fibres. Bones support and protect softer tissues and organs. Osteoblasts are present in the spaces called lacunae.

Blood :
It is the fluid connective tissue. It contains RBCs, WBCs and platelets. It functions as a transport medium for nutrients, wastes and respiratory gases.

Question 16.
Explain the types of muscle.
Answer:
Each muscle is made of long, cylindrical fibres. They are composed of fine fibrils called myofibrils. Muscle fibres contract and relax. Skeletal muscle is attached to skeletal bones. It is striped or striated. It is the voluntary muscle. The smooth muscle fibres are fusiform and do not have striations. It is an involuntary muscle. Cardiac muscle tissue is present in the heart. It is striated and branched and involuntary.

Question 17.
Write a note on neural tissue.
Answer:

• Neurons are units of the neural system. The neuroglial cells protect and support the neurons.
• Neurons transmit sensations as electric impulses.

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Samacheer Kalvi 11th Chemistry Environmental Chemistry Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Environmental Chemistry Multiple Choice Questions

Question 1.
Th gaseous envelope around the earth is known as atmosphere. The region lying between an altitude of 11-50 km is –
(a) Troposphere
(b) Mesophere
(c) Thermosphere
(d) Stratosphere
Answer:
(d) Stratosphere

Question 2.
Which of the following is natural and human disturbance in ecology?
(a) Forest fire
(b) Floods
(c) Acid rain
(d) Greenhouse effect
Answer:
(a) Forest fire

Question 3.
Bhopal Gas Tragedy is a case of –
(a) thermal pollution
(b) air pollution
(c) nuclear pollution
(d) land pollution
Answer:
(b) air pollution

Question 4.
Haemoglobin of the blood forms carboxyhaemoglobin with –
(a) Carbon dioxide
(b) Carbon tetrachioride
(c) Carbon monoxide
(d) Carbonic acid
Answer:
(c) Carbon monoxide

Question 5.
Which sequence for greenhouse gases is based on GWP?
(a) CFC > N₂O > CO₂ > CH₄
(b) CFC > CO₂ > N₂O > CH₄
(c) CFC > N₂O > CH₄ > CO₂
(d) CFC > CH₄ > N₂O > CO₂
Answer:
(c) CFC > N₂O > CH₄ > CO₂

Question 6.
Photo chemical smog formed in congested metropolitan cities mainly consists of –
(a) Ozone, SO₂ and hydrocarbons
(b) Ozone, PAN and NO₂
(c) PAN, smoke and SO₂
(d) Hydrocarbons, SO₂ and CO₂
Answer:
(b) Ozone, PAN and NO₂

Question 7.
The pH of normal rain water is –
(a) 6.5
(b) 7.5
(c) 5.6
(d) 4.6
Answer:
(c) 5.6

Question 8.
Ozone depletion will cause –
(a) forest fires
(b) eutrophication
(c) bio magnification
(d) global warming
Answer:
(d) global warming

Question 9.
Identify the wrong statement in the following.
(a) The clean water would have a BOD value of more than 5 ppm
(b) Greenhouse effect is also called as Global warming
(c) Minute solid particles in air is known as particulate pollutants
(d) Biosphere is the protective blanket of gases surrounding the earth
Answer:
(a) The clean water would have a BOD value of more than 5 ppm

Question 10.
Living in the atmosphere of CO is dangerous because it –
(a) Combines with O₂ present inside to form CO₂
(b) Reduces organic matter of tissues
(c) Combines with haemoglobin and makes it incapable to absorb oxygen
(d) Dries up the blood
Answer:
(c) Combines with haemoglobin and makes it incapable to absorb oxygen

Question 11.
Release of oxides of nitrogen and hydrocarbons into the atmosphere by motor vehicles is prevented by using –
(a) grit hamber
(b) scrubbers
(c) trickling filters
(d) catalytic convertors
Answer:
(c) trickling filters

Question 12.
Biochemical oxygen Demand value less than 5 ppm indicates a water sample to be
(a) highly polluted
(b) poor in dissolved oxygen
(c) rich in dissolved oxygen
(d) low COD
Answer:
(c) rich in dissolved oxygen

Question 13.
Match the list I and list II and select the correct answer using the code given below in the list:

Answer:

Question 14.
Match the list I and list II and select the correct answer using the code given below in the list

Answer:

Question 15.
Assertion (A) : if BOD level of water in a reservoir is more than 5 ppm it is highly polluted.
Reason(R) : High biological oxygen demand means high activity of bacteria in water
(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)
Answer:
(d) (iv)

Question 16.
Assertion (A) : Excessive use of chlorinated pesticide causes soil and water pollution.
Reason (R) : Such pesticides arc non-biodegradable.
(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)
Answer:
(a) (i)

Question 17.
Assertion (A) : Oxygen plays a key role in the troposphere.
Reason (R) : Troposphere is not responsible for all biological activities
(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)
Answer:
(d) (iv)

Samacheer Kalvi 11th Chemistry Environmental Chemistry Short Answer Questions

Question 18.
Dissolved oxygen in water is responsible for aquatic life. What processes are responsible for the reduction in dissolved oxygen in water’?
Answer:

• Organic matter such as leaves, grass, trash can pollute water. Microorganisms present in water can decompose these organic matter and consume dissolved oxygen in water.
• Eutrophication is a process by which water bodies receive excess nutrients that stimulates excessive plant growth. This enhanced plant growth in water bodies is called algal bloom.
• The growth of algae in extreme abundance covers the water surface and reduces the oxygen concentration in water. Thus, bloom-infeded water inhibits the growth of other. living organisms in the water body.
• This process in which the nutrient rich water support a dense plant population, kills animal life by depriving it of oxygen and results in loss of biodiversity is known as eutrophication.

Question 19.
What would happen, if the greenhouse gases were totally missing in the earth’s atmosphere?
Answer:

• The primary greenhouse gases in Earth’s atmosphere are water vapour, carbon dioxide, methane, nitrous oxide and ozone.
• Naturally occurring greenhouse gases allow solar radiations to reach the earth’s surface, while trapping radiations from the earth on its way back out to space.
• There would he no life on Earth without the warmth provided by this natural greenhouse gases.
• In the absence of greenhouse gases. the average temperature of the earth will decrease drastically. As a result. life on Earth would be impossible.

Question 20.
Define smog.
Answer:

• Smog is a combination of smoke and fog which form droplets that remains suspended in the air.
• Smog is a chemical mixture of gases that forms a brownish yellow haze. It mainly consists of ground level ozone, oxides of nitrogen, volatile organic compounds, SO₂, acidic aerosols and some other gases.

Question 21.
Which is considered to be earth’s protective umbrella? Why?
Answer:

• At high altitudes in the atmosphere consists of a layer of ozone (O₂) which acts as an umbrella for harmful UV radiations. Ozone is considered to be earth’s protective umbrella.
• It protects us from harmful effects of UV-radiations of the sun such as skin cancer.
• Ozone layer prevent the UV radiations to reach the earth surface. So it acts as an umbrella for the Earth.

Question 22.
What are bio-degradable and non-biodegradable pollutants?
Answer:

• The pollutants which can be easily decomposed by the natural biological processes are called biodegradable pollutants. For example plant wastes, animal wastes.
• The pollutants which cannot be decomposed by the natural biological processes are called non-biodegradable pollutants. For example, metal wastes such as Hg and Pb, D.D.T. plastics, nuclear vastcs.

Question 23.
From where does ozone come in the photochemical smog?
Answer:

• Photochcrnical smog is formed by the combination of smoke, dust and fog with air pollutants in the presence of sunlight.
• Chemically it is oxidising in nature because of high concentration of oxidising agents such as NO₂ and O₂. So it is also called oxidising smog.
• Photochemical smog is formed by following reactions:
• N₂ + O₂ → 2NO
  2NO + O₂ → 2NO₂
  
  (O) + O₂ → O₃
  O₃ + NO → NO₂ + O₂
  
• NO and O₃ arc strong oxidizing agents and they can react with unburnt hydrocarbons in polluted air to form formaldehyde, acrolcin and PAN.

Question 24.
A person was using water supplied by corporation. Due to shortage of water he started using underground water. He felt laxative effect. What could be the cause?
Answer:
Drinking water containing moderate level of sulphatcs is harmless. But excessive concentration (>500 ppm) of suiphates in drinking water causes laxative effect.

Question 25.
What is green chemistry?
Answer:

• Green chemistry is a chemical philosophy encouraging the design of products and processes that reduces or eliminates the use and generation of hazardous substances.
• Efforts to control environmental pollution resulted in development of science for the synthesis of chemicals favorable to environment.
• Green chemistry means science of environmentally favorable chemical synthesis.

Question 26.
Explain how does greenhouse effect cause global warming.
Answer:

• The earth’s atmosphere allows most of the visible light from the sun to pass through and reach the earth’s surface. As earth’s surface is heated by sunlight, it radiates a part of this energy back towards the space as longer IR wavelengths.
• Some of the heat is trapped by CH₂, CO₂. CFCs and water vapour present in the atmosphere. They absorb IR radiations and block a large portion of earth’s emitted radiations.
• The radiations thus absorbed is partly remitted to the earth’s surface. Therefore the earth’s surface gets heated up by a phenomenon called greenhouse effect.
• Thus greenhouse effect is defined as the heating up of the earth surface due to trapping of infrared radiations reflected by earth’s surface by CO₂ layer in the atmosphere. The heating up of the earth through the greenhouse effect is called global warming.

Question 27.
Mention the standards prescribed by BIS for quality of drinking water.
Answer:
Standard characteristics prescribed for deciding the quality of drinking water by BIS are as follows:

Question 28.
How does classical smog differ from photochemical smog?
Answer:
Classical smog:

• Classical smog is caused by coal-smoke and fog.
• It occurs in cold humid climate.
• The chemical composition is the mixture of SO₂, SO₃ gases and humidity.
• Chemically it is reducing in nature because of high concentration of SO, and so it is also called reducing smog.
• It is primarily responsible for acid rain.
• It also causes bronchial irritation.

Photochemical smog:

• Photochemical smog is cause by photochemical oxidants.
• It occurs in warm, dry and sunny climate.
• The chemical composition is the mixture of NO₂ and O₃ gases.
• Chemically it is oxidising in nature because of high concentration of oxidising agents such as NO₂ and O₃
• and so it is also called oxidising smog.
• It causes irritation to eyes, skin and lungs and increase the chances of asthma.
• It causes corrosion of metals, stones

Question 29.
What are particulate pollutants? Explain any three.
Answer:
1. Particulate pollutants are small solid particles, and liquid droplets suspended in air.
Examples:
dust, pollen, smoke, soot and liquid aerosols.

2. Types of Particulates:
Particulates in the atmosphere may be of two types:

• viable particulate and
• non-viable particulate.

3.The viable particulates are small size living organisms such as bacteria, fhngi, moulds and algae which are dispersed in air.

4. The non-viable particulates are small solid particles and liquid droplets suspended in air. There are four types of non-viable particulates in the atmosphere. They are
(a) Smoke
(b) Dust
(c) Mist
(d) Fumes

5. Smoke:
Smoke particulate consists of solid particles formed by combustion of organic matter. For example, cigarette smoke, oil smoke, smokes from burning of fossil fuels, garbage and dry leaves.

6. Dust:
It is composed of fine solid particles produced during crushing and grinding of solid materials. For example, sand from sand blasting, saw dust from wood works and fly ash from power generating units.

7. Mist:
They are formed by particles of sprayed liquids and condensation of vapours in air. For example, sulphuric acid mist, herbicides and insecticides sprays can form mists.

8. Fumes:
They are obtained by condensation of vapours released during sublimation, distillation, boiling and calcination and by several other chemical reactions.
For example:
organic solvents, metals and metallic oxides.

Question 30.
Even though the use of pesticides increases the crop production, they adversely affect the living organisms. Explain the function and the adverse effects of the pesticides.
Answer:
1. Pesticides are the chemicals that are used to kill or stop the growth of unwanted organims. But these pesticides can affect the health of human beings. Pesticides are classified as
(a) insecticides,
(b) Fungicides and
(c) Herbicides.

(a) Insecticides:
Insecticides like DDT, BHC, Aidrin can stay in soil for a long period of time and are absorbed by soil. They contaminate root crops like carrot, radish.

(b) Fungicides:
Organomercury compounds dissociate in soil to produce mercury which is highly toxic.

(c) Herbicides:
They are used to control unwanted plants and are also known as weed killers. Eg, Sodium chlorate, sodium nitrate. They are toxic to mammals.

Question 31.
Ethane bums completely in air to give CO₂. while in a limited supply of air gives CO. The same gases are found in automobile exhaust. Both CO and CO₂ are atmospheric pollutants

1. What is the danger associated with these gases?
2. How do the pollutants affect the human body?

Answer:
Danger associated with CO and CO₂ & health hazards to human body
(a) Carbon monoxide binds with haemoglohin and form carboxyhaemoglobin which impairs normal oxygen transport by blood and hence the oxygen carrying capacity of blood is reduced. This oxygen deficiency results in headache, tension, dizziness, loss of consciousness, blurring of eyesight and cardiac arrest.

(b) Increase in CO₂ level in the atmosphere is responsible for global warming. It causes headache and nausea.

Question 32.
On the basis of chemical reactions involved, explain how do CFC’s cause depletion of ozone layer in stratosphere?
Answer:
(i) The chioro-Iluoro derivatives of methane and ethane are named Freons (CFC’s). They slowly pass from troposphere to stratosphere. They stay for a very longer period of about 50-100 years. In the presence of UV radiations, CFC’s break up into chienne free radicals.

Cl• + O₃ → { ClO }^{ \bullet } + O₂
ClO• + O → { Cl }^{ \bullet } + O₂

(iii) Chlorine radical is regenrated in the course of the reaction. Due to this continuous attack of Cl free radicals, thinning of ozone layer takes place which leads to the formation of ozone hole.

(iv) Li is estimated that for every reactive chlorine atom generated in the stratosphere 1,00,000 molecules of ozone are depleted.

Question 33.
How is acid rain formed? Explain its effect.
Answer:
1. Rain water has a pli of 5.6 due to the dissolution of CO., into it. Oxides of sulphur and nitrogen in the atmosphere may be absorbed by droplets of water that make up the clouds and get chemically converted into sulphuric acid and nitric acid. Due to this the pH of rain water drops below the level of 5.6. Hence it is called acid rain.

2. Acid rain is a by-product of sulphur and Nitrogen oxides in the atmosphere. Burning of fossil fuels in power stations, furnaces and petrol, diesel in motor engines produce SO₂ and NO₂ gases. They are converted into H₂SO₄ and HNO₃ by the reaction with oxygen and water.

3. 2SO₂ + O₃ + 2H₂O → 2H₂SO₄
4NO₂ + O₂+ 2H₂O → 4HNO₃

Harmful effects of acid rain:
1. Acid rain causes damage to buildings made us of marbles. This attack on marble is termed as stone leprosy.
CaCO₃ + H₂SO₂ CaSO₄ + H₂O + CO₂↑

2. Acid rain affects plant and animal life in aquatic ecosystem.

3. It is harmful For agriculture, as it dissolves in the earth and removes the nutrients needed for the growth of plants.

4. It corrodes water pipes resulting in the leaching of heavy metals such as iron, lead and copper into drinking water which have toxic effects.

5. it causes respiratory ailment in humans and animals.

Question 34.
Differentiate the following:

1. BOD and COD
2. Viable and non-viable particulate pollutants

Answer:
Biochemical oxygen demand (BOD):

• The total amount of oxygen (in milligrams) consumed by microorganisms in decomposing the waste in one litre ut water at 20°C for a period of 5 days is called biochemical oxygen demand (BOD).
• its value is expressed in ppm.
• DOD is used as a measure of degree ofwater pollution.
• BOD is only a measurement of consumed oxygen by microorganims to decompose the organic matter.
  Clean water would have BOD value less than 5 ppm.

Chemical ox gen demand (COD):

• Chemical oxygen demand is defined as the amount of oxygen required by the organic matter in a sample of water for its oxidation by a strong oxidising agent like K₇Cr₂O₇ in acidic medium for a period of 2 hours.
• Its value is expressed in mg / litre.
• COD is a measure of amount of organic compounds in a water sample.
• COD refers to the requirement of dissolved oxygen for both the oxidation of organic and inorganic constituents.
• Clean water would have COD value greater than 250 mg/litre.

2. Viable and non-viable particulate pollutants
Viable pollutants:

• The viable particulates are small size living organisms such as bacteria, fungi. moulds, algae which are dispersed in air.
• They are all organic particulates.
• They contain living organisms.
•  Eg. ftingi, bacteria, algae, moulds. Viable particles are the particles with at least one microorganism affecting the sterility of the product.

Non-viable pollutants:

• The non-viable particulates are small solid particles and liquid droplets suspended in air.
• They are all inorganic particulates.
• They contain non-living organisms.
• Eg. Smoke, dust, mist, fumes.
• Non-viable particles are the particles without microorganisms but act as transporting agent for viable particles.

Question 35.
Explain how oxygen deficiency is caused by carbon monoxide in our blood? Give its effect.
Answer:

1. Carbon monoxide binds with haemoglobin and form carboxy the haemoglobin which impairs normal oxygen transport by blood and hence the oxygen carrying capacity of blood is reduced.
2. This oxygen deficiency results in headache, dizziness, tension, loss of consciousness, blurring of eyesight and cardiac arrest.

Question 36.
What are the various methods you suggest lo protect our environment from pollution?
Answer:
Methods to control environmental pollution:

• Waste management Environmental pollution can be controlled by proper disposal of wastes.
• Recycling A large amount of disposed waste materials can he reused by recycling the waste, thus it reduces the landfill. –
• By substitution of less toxic solvents for highly toxic ones are used in industrial processes.
• By growing more trees.
• By using fuels with lower sulphur content.
• By control measures in vehicle emissions which are adequate.

Samacheer Kalvi 11th Chemistry Environmental Chemistry Additional Questions Solved

I. Choose the correct answer

Question 1.
Which one of the following is bio-degradable pollutant?
(a) DDT
(b) Plastics
(c) Mercury
(d) Wood
Answer:
(d) Wood

Question 2.
Which one of the following gases is not present in troposphere?
(a) N₂O₂
(b) CO₂
(c) N₂
(d) water vapours
Answer:
(c) N₂

Question 3.
Which of the following pair of’ oxides is responsible for acid rain?
(a) SO₃ + NO₂
(b) CO₂ + CO
(c) N₂O + CH₄
(d) O₂ + H₂
Answer:
(a) SO₃ + NO₂

Question 4.
Which one of the following is produced as a result of incomplete combustion of coal’?
(a) CO₂
(b) CO
(c) SO₂
(d) SO₃
Answer:
(b) CO

Question 5.
Which of the following is not a greenhouse gas’?
(a) CO
(b) O₃
(c) CH₄
(d) Water vapours
Answer:
(a) CO

Question 6.
Photochemical smog occurs in warm, dry and sunny climate. One of the following is not among-st the components of photochemical smog. Identify it.
(a) NO₂
(b) O₃
(c) SO₂
(d) Unsaturated hydrocarbons
Answer:
(c) SO₂

II. Match the following
1.

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

III. Fill in the blanks

Question 1.
An example for biodegradable pollutant is ………
Answer:
plant waste.

Question 2.
An example for non-biodegradable pollutant ………
Answer:
metal waste.

Question 3
……… is the lowest layer of atmosphere.
Answer:
Troposphere.

Question 4.
Ozone is present in ……… layer of the atmosphere.
Answer:
Stratosphere.

Question 5
……… is called blue planet.
Answer:
Earth.

Question 6.
About 80% of the mass of the atmosphere is present in ………
Answer:
Troposphere.

Question 7.
……… will reduce the oxygen carrying capacity of blood.
Answer:
CO

Question 8
……… gas is used in the process of photosynthesis.
Answer:
CO₂

Question 9.
……… gas potentially damage plant leaves and retard photosynthesis.
Answer:
NO₂

Question 10.
……… is formed by incomplete combustion of coal.
Answer:
CO₂

Question 11.
……… binds with haemoglobin and reduce the oxygen carrying capacity of blood.
Answer:
CO

Question 12.
……… is carcinogenic and causes irritation in eyes and mucous membrane.
Answer:
PAH

Question 13.
The earths surface get heated up by a phenomenon called ………
Answer:
Greenhouse effect

 

Question 14.
The main constituent of layer responsible for global warming is ………
Answer:
CO₂

Question 15.
Earth’s average surface temperature would be only about ………
Answer:
-18°C

Question 16.
The pH of rain water is normally ………
Answer:
5.6

Question 17.
……… pair of compounds is present in acid-rain water.
Answer:
HNO₃ + H₂SO₄

Question 18.
……… is an example of viable particulate.
Answer:
Bacteria.

Question 19.
……… is an example of non-viable particulate matter.
Answer:
Cement dust.

Question 20.
Particulate pollutants will result in the health hazard named as ………
Answer:
pneimoconiosis.

Question 21.
Coal miners may suffer from ………
Answer:
Black lung disease.

Question 22.
Textile workers may suifer from ………
Answer:
White lung disease.

Question 23.
………….. aftect childrcns brain. interfers with maturation of RBC’s and even causes cancer.
Answer:
Lead.

Question 24.
……… can be used to reduce particulate pollutant.
Answer:
electrostatic precipitator

Question 25.
……… is the combination of smoke and fog.
Answer:
Smog.

Question 26.
Classical smog called London smog contains ………
Answer:
Coal-smoke and fog.

Question 27.
Classical smog is otherwise called ………
Answer:
reducing smog.

Question 28.
Photochemical smog is otherwise called ………
Answer:
Los angeles smog.

Question 29.
The three main components of photochemical sinog are ………
Answer:
Hydrogen sulphide. dust and PAN.

Question 30.
……… plantation can metabolise nitrogen oxide and control photochemical smog.
Answer:
Pinus tree.

Question 31.
……… acts as an umbrella for the earth and prevent harmful UV radiations.
Answer:
ozone.

Question 32.
……… pair of compounds is found to be highly responsible for depletion of ozone layer.
Answer:
Nitric oxide + CFC

Question 33.
Freons are ………
Answer:
Chiorofluoroalkanes.

Question 34.
Ozone layer is depicted by the reactive ………
Answer:
Chlorine atom.

Question 35.
Examples for water borne diseases are ………
Answer:
Dysentery and cholera.

Question 36.
The standard pH of drinking water is ………
Answer:
6.5 to 8.5

Question 37.
The essential elements for soil are ………
Answer:
N,P,K

Question 38
……… is a better alternative for carnicogentic benzene.
Answer:
Xylene.

Question 39.
The alternate solvent used instead of tetrachioroethYlefle in dry cleaning is ………
Answer:
Liquefied CO₂

Question 40.
……… is used for bleaching clothes in laundry.
Answer:
H₂O₂

Question 41.
……… is used to bleach paper.
Answer:
H₂O₂

Question 42.
……… is the most safest pesticide.
Answer:
Neem based pesticide.

Question 43.
……… acid is most abundant ¡n acid rain.
Answer:
H₂SO₄

Question 44.
……… causes less pollution.
Answer:
CO₂

Question 45.
Besides CO₂ the other greenhouse gas is ………
Answer:
CH₄

Question 46.
BOD is a measure of ………
Answer:
Organic pollutant in water.

Question 47.
The pollutant released in Bhopal gas tragedy was ………
Answer:
Methyl isocyanate.

Question 48.
The greatest affinity for haemoglobin in shown by ………
Answer:
CO

Question 49.
Eutrophication causes reduction in ………
Answer:
dissolved oxygen.

IV Choose the odd one out

Question 1.
(a) Plant waste
(b) DDT
(c) Plastic
(d) Nuclear waster
Answer:
(a) Plant waste. It is biodegradable pollutant whereas others are non-biodegradable pollutants.

Question 2.
(a) Plant waste
(b) Animal wastes
(c) Paper
(d) Nuclear waste
Answer:
(d) Nuclear waste. It is non-biodegradable pollutant whereas others are biodegradable wastes.

Question 3.
(a) N₂O₂
(b) CO₂
(c) H₂O (Vap)
(d) O₃
Answer:
(d) O₃ It is present in stratosphere whereas others are present in troposphere.

Question 4.
(a) O₂⁺
(b) O⁺
(c) N₂
(d) NO⁺
Answer:
(c) N₂. ills present in mesosphere whereas others are present in thermosphere.

Question 5.
(a) N₂
(b) O₂
(c) O₃
(d) N₂O₂
Answer:
(d) N₂O₂ It is present in troposphere whereas others are present in stratosphere.

Question 6.
(a) CH₄
(b) CO
(c) CO₂
(d) CFC
Answer:
(b) CO. It is a poisonous gas whereas others are responsible for green house effect.

V. Choose the correct pair

Question 1.
(a) CCF : Green house effect
(b) CO : Carcinogenic
(c) PAH : Acid rain
(d) NO : Lung injury
Answer:
(a) CCF : Green house effect

Question 2.
(a) NO₂ : Green house effect
(b) CFC : asthma and lung injury
(c) PAH : carcinogenic
(d) CO₂ : green house effect
Answer:
(a) NO₂ : Green house effect

Question 3.
(a) Classical smog : NO₂ and O₂
(b) London smog : SO₂, SO₂ and humidity
(c) Photochemical smog : CO₂ and CO
(d) Los Angel smog : NO₂ and NO₃
Answer:
(b) London smog : SO₂ , SO₂ and humidity

VI. Choose the incorrect pair

Question 1.
(a) Photochemical smog : NO₂ and O₂
(b) Classical smog : SO₂, SO₃ and humidity
(c) Smog : Smoke and fog
(d) Non viable particulate : Algae, Fungi
Answer:
(d) Non viable particulate : Algae, Fungi

Question 2.
(a) Viable particulate : bacteria, fungi
(b) Non viable particulate : smoke, dust
(c) Acid rain : HCl + HNO₂
(d) Photochemical smog : NO₂ + O₃
Answer:
(c) Acid rain : HCl + HNO₂

Question 3.
(a) Lead : Damage to kidney, liver
(b) Sulphate : Laxative effect
(c) Nitrage : Blue baby syndrome
(d) TDS : Damage to bone and teeth
Answer:
(d) TDS : Damage to bone and teeth

Question 4.
(a) Insecticides : DDT. BCHs
(b) Herbicides : Organo mercury compounds
(c) Herbicides : Sodium chlorate, sodium arsenite
(d) Industrial waste : Mercury. copper
Answer:
(b) Herbicides : Organo mercuiy compounds

VII. Assertion & Reason
Assertion (A) : Depletion of ozone layer causes skin cancer.
Reason (R) : Depletion olozone layer will allow more UV rays to reach the earth surface and cause skin cancer.
(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 (R).
(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) : UV radiations damages the fish productivity.
Reason (R) : UV radiations affect the growth of phytoplankions as a result food chain in ocean is disturbed.
(a) Both (A) and (R) are correct hut (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) : The pH ol acid rain is less than 5.6.
Reason (R) : CO, present in the atmosphere dissolves in rain water and forms carbonic acid,
(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 correct explanation of (A).
(c) (A) is correct bitt (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 Environmental Chemistry 2 Mark Questions and Answers

Question 1.
What is meant by environmental pollution?
Answer:
Any undesirable change in our environment that have harmful effects on plants. animals and human beings is called environmental pollution.

Question 2.
Write a note on constituents of earth’s atmosphere.
Answer:
Earth’s atmosphere is a layer of gases retained by earth’s gravity. it contains roughly 78% Nitrogen, 21% Oxygen. 0.93% Argon, 0.04% Carbon dioxide. traces of other gases and little amount of water vapour. This mixture is commonly known as air.

Question 3.
What is called troposphere? How many layers are present in it? Give their names.
Answer:

1. The lowest layer of the atmosphere is called the troposphere and il extends from 0 10 km from the earth surface.
2. About 80% of the mass of the atmosphere is in this layer. 3 layers are present in it.
3. They are –
   • Hydrosphere
   • Lithosphere
   • Biosphere

Question 4.
Write about hydrosphere (or) Why Earth is called as Blue planer?
Answer:

• Hydrosphere include all types of water sources like oceans, seas, rivers, lakes. streams, underground water, polar ice-caps, clouds etc.
• It covers about 75% of the earth’s surface. Hence earth is called as Blue planet.

Question 5.
What is lithosphere?
Answer:
Lithosphere includes soil, rocks and mountains which are solid components of Earth.

Question 6.
What is biosphere?
Answer:
Biosphere includes the lithosphere. hydrosphere and atmosphere integrating the living organisms present in the lithosphere, hydrosphere and atmosphere.

Question 7.
What is air pollution?
Answer:
Air pollution is defined as any undesirable change in air which adversely affects living organisms. Air pollution is limited to troposphere and stratosphere.

Question 8.
What are the types of air pollutants? Give examples.
Answer:
1. Air pollutants exists in two major forms namely:
(a) Gaseous air pollutants
(b) Particulates

2. Gaseous air pollutants:
Oxides of sulphur, oxides of nitrogen, oxides of carbon and hydrocarbons are the gaseous air pollutants.

3. Particulate pollutants:
Particulate pollutants are small solid particles and liquid droplets suspended in air.
Exarnple:
dust, pollen, smoke, soot, aerosols.

Question 9.
Define greenhouse effect.
Answer:
Greenhouse effect is defined as the heating up of the earth’s surface due to trapping of infrared radiations reflected by earth’s surface by the CO₂ layer in the atmosphere.

Question 10.
What is Global warming?
Answer:
The heating up of the earth through the greenhouse effect is called Global warming.

Question 11.
When rain water is named as acid rain?
Answer:
Rain water normally has a pH of 5.6 due to dissolution of atmospheric CO₂ particles into it. Oxides of sulphur and nitrogen in the atmosphere are absorbed by droplets of water that make up clouds and get chemically converted into sulphuric acid and nitric acid. As a result the pH of rainwater drops below 5.6 and hence it is called acid rain.

Question 12.
What is stone leprosy? How is it formed?
Answer:

• The attack on the marble of buildings by acid rain is called stone leprosy.
• Acid rain causes extensive damage to buildings made up of marble CaCO + H₂SO₄ CaSO₄ + H₂O + CO₂↑

Question 13.
What are fumes? Give one example.
Answer:
Fumes are one of the non-viable particulate pollutants air. They are obtained by condensation of vapours released during sublimation, distillation, boiling and calcination. For example, organic solvents, metals and metallic oxides form fume particles.

Question 14.
What are the techniques to reduce particulate pollutants?
Answer:
The particulates from air can be removed by using electrostatic precipitators, gravity settling chambers, wet scrubbers or by cyclone collectors. These techniques are base on washing away or settling of the particulate matter.

Question 15.
How will you control photochemical smog?
Answer:
The formation of photochemical smog can be suppressed by preventing the release of nitrogen oxide and hydrocarbons into the atmosphere from the motor vehicles by using catalytic convertors in engines. Plantation of certain trees like Pinus, Pyrus Querus vitus and Juniparus can metabolise nitrogen oxide.

Question 16.
What is meant by water pollution’?
Answer:
Water pollution is defined as the addition of foreign substances or factors like heat which degrades the quality of water so that it becomes an health hazard or unfit for use.

Question 17.
What are the sources of water pollution? Give examples.
Answer:

• The water pollutants originate form both natural as well as human activities. The source of water pollution are classified as point and non-point sources.
• Easily identified source of water pollution is called a point source. For example, Municipal and industrial discharge pipes.
• Non-point source cannot be identified easily. For example, agricultural runoffs, mining wastes, acid rain, storm water drainage and construction sediments.

Question 18.
What is BOD’?
Answer:
The total amount of oxygen (in milligrams) consumed by microorganisms in decomposing the waste in one litre of water at 20°C for a period of 5 days is cal Led biochemical oxygen demand (BOD) and its value is expressed in ppm.

Question 19.
What is COD?
Answer:
Chemical oxygen demand (COD) is defined as the amount of oxygen required by the organic matter in a sample of water for its oxidation by a strong oxidising agent like K₂Cr₂O₇ in acidic medium for a period of 2 hours.

Question 20.
What are total dissolved solids (TDS)?
Answer:

• Most of the salts are soluble in water. It include cations like calcium, magnesium, sodium, potassium, iron and anions like carbonate, bicarbonate, chloride, sulphate, phosphate and nitrate.
• Use of drinking water having TDS (total dissolved solids) concentration higher than 500 ppm causes possibilities of irritation in stomach and intestine.

Question 21.
What are the constituents of soil?
Answer:
Soil is a thin layer of organic and inorganic material that covers the earth’s rocky surface. Soil constitutes the upper crust of the earth, which supports land, plants and animals.

Question 22.
Define – soil pollution.
Answer:
Soil pollution is defined as the build up of persistent toxic compounds, radioactive materials, chemical salts and disease causing agents in soil which have harmful effects on plant growth and animal health. Soil pollution affects the structure and fertility of soil, ground water quality and food chain in biological ecosystem.

Question 23.
Explain how industrial waste affects the soil.
Answer:

• Industrial activities have been the biggest contributor to soil pollution especially the mining and manufacturing activities.
• Industrial wastes include cyanides. chromates, acids, alkalis and metals like mercury, copper, zinc, cadmium and lead.
• These industrial wastes in the soil surface lies for a long time and makes it unsuitable for use.

Question 24.
What is green chemistry?
Answer:

• Efforts to control environment pollution have resulted in the development of chemicals favorable for environment and this branch of science is called green chemistry.
• It is a chemical philosophy encouraging the design of products and processes that reduces or eliminates the use and generation of hazardous substances.

Question 25.
Write a note about dry cleaning of clothes.
Answer:
Solvents like tetrachioroethylene is used in dry cleaning of clothes, pollute the ground water and are carcinogenic. In place of tetrachloroethylene liquefied CO₂ with suitable detergents an alternate solvent used. Liquefied CO₂ is not that much harmful for the groundwater. Nowadays H₂O₂ is used for bleaching clothes in laundry that give better results and utilises less water.

Question 26.
Carbon monoxide gas is more dangerous than carbon dioxide gas. Why?
Answer:
Carbon monoxide gas combines with haemoglobin to form a very stable compound known as carboxy haemoglobin. When its concentration in blood reaches 3-4%, the oxygen carrying capacity of the blood is greatly reduced. This results into headache, nervousness and sometimes death of the person. On the other hand CO₂ does not combine with haemoglobin and hence it is less harm ful than CO.

Question 27.
Which gases are responsible for greenhouse effect? List some of them.
Answer:
CO, is mainly responsible for greenhouse effect. Other greenhouse gases are methane, nitrous oxide, water vapours, CFCs and ozone.

Question 28.
A large number of fishes are suddenly found floating dead on a lake. There is no evidence of toxic dumping but you find an abundance of phytoplankton. Suggest a reason for the fish killing.
Answer:
Excessive phytoplankton (organic pollutants such as leaves, grass trash etc.) growth which is present in water is biodegradable. Bacteria decomposes this organic matter in water. During this process when large number of bacteria decomposes the organic matter, they consume the dissolved oxygen in water. When the level of dissolved oxygen falls below 6 ppm then the fishes cannot survive and they die.

Question 29.
How carbon monoxide acts as a poison for human beings’?
Answer:
Carbon monoxide is a poisonous gas because it combines with haemoglobin of RBC to form carboxyhaernoglobin as:
CO + Haemoglobin ⇌ Carboxyhaemoglobin
It inhibits the transport of oxygen to different pans of the body. Thus the body becomes oxygen-starved.

Question 30.
Give three examples in which the principles green chemistry has been applied.
Answer:

• In dry-cleaning, use of liquefied CO₂ in place of non-environment friendly tetrachioroethene (Cl₂C = CCl₂).
• Use of H₂O₂ in bleaching in place of chlorine.
• In the manufacture of chemicals like ethanal by using environment-friendly chemicals and conditions. ,

Samacheer Kalvi 11th Chemistry Environmental Chemistry 3 Marks Questions and Answers

Question 1.
Write about Bhopal gas tragedy.
Answer:

• On 3rd Decembers. 1984 at Bhopal city in India, by the early morning, an explosion at Union carbide pesticide plant released a cloud of toxic gas (Methyl isocyanate) CH₃NCO into the air.
• Since the gas is twice as heavy as air, it did not drift away but formed a blanket over the surrounding area.
• It attacked people’s lungs and affect their breathing staying there or in the nearby areas. Thousands of peopic died and lives of many were ruined. The lungs, brain, eyes, muscles as well as gastrointestinal, neurological and immune system of those people who survived were severely affected.

Question 2.
Explain how the oxides of sulphur pollute the atmospheric air. Give its haririful eflects.
Answer:

• Sulphur dioxide and sulphur trioxide are produced by burning sulphur containing fossil fuels and by roasting of suiphide ores.
• SO₂ is a poisonous gas for both animals and plants. SO₂ causes eye irritation, coughing and respiratory diseases like asthma, bronchitis.
• SO₂ is oxidised to more harmful SO₃ gas in the presence of particulate matter present in the polluted air:
  
  SO₂ combines with atmospheric water vapour to form H₂SO₄ which comes down along with rain in the form of acid rain:
• Acid rain causes stone leprosy. affect aquatic ecosystem, corrode water pipes and causes respiratory ailment in humans and animals.

Question 3.
How oxides of nitrogen are harmful?
Answer:
(i) Oxides of nitrogen are produced during high temperature combustion processes, by oxidation of nitrogen in air and it is formed the combustion of fuels such as coal, diesel and petrol.

(iii) The oxides of nitrogen are converted into nitric acid which comes down in the form of acid rain. They also form reddish brown haze in heay traffic.
(iv) Nitrogen dioxide potentially damage plant leaves and retard photosynthesis.
(v) NO₂ is a respiratory irritant and it can cause asthma and lung injury.

Question 4.
Explain how hydrocarbons pollute the atmospheric air.
Answer:

• The compounds composed of carbon and Hydrogen only are called hydrocarbons. They are mainly produced naturally and also by incomplete combustion of automobile fuels.
• They are potential cancer causing (carcinogenic) agents.
• For example polynuclear aromatic hydrocarbons (PAH) are carcinogenic, they cause irritation in eyes and mucous membranes.

Question 5.
Write flotes about great London smog.
Answer:

• The great smog of London (1952) was an instance of severe air pollution that affected the London from 5th December 9^th of December, 1952 and then dispersed quickly when the whether changed.
• It causes major disruption by reducing the visibility and even penetrating indoor areas.
• Government medical reports estimated that 4000 people had died as a direct result of smog and 100.000 were made ill by the smog effect on their respiratory tract.

Question 6.
What are the effects of classical smog?
Answer:

• Smog is primarily responsible for acid rain.
• Smog results in poor visibility and it affects air and road transport.
• It also causes bronchial irritation.

Question 7.
Explain how oxides of nitrogen are introduced directly into the stratosphere?
Answer:

• Nitrogen oxides are introduced directly into the stratosphere by the supersonic jet aircraft engines in the form of exhaust gases.
• These oxides arc released by combustion of fossil fuels and nitrogen fertilisers. Inert nitrous oxide in the stratosphere is photochemically converted into more reactive nitric oxide.
• Oxides of nitrogen catalyses the decomposition of ozone and are themselves regenerated. Ozone gets depleted as follows:
  
  Thus NO is regenerated in the chain reaction.

Question 8.
How chemical wastes pollute water?
Answer:

• A whole variety of chemicals from industries, such as metals and solvents are poisonous to fish and other aquatic life.
• Toxic pesticides can accumulate in fish and shell fish and poison the people who eat them.
• Detergents and oil float spoils the water bodies.
• Acids from mine drainage and salts from various sources can also contaminate water sources.

Question 9.
Explain the presence of fluoride in water and its hazardous effects?
Answer:

• Fluoride ion deficiency in drinking water causes tooth decay.
• The fluoride ions make the enamel on teeth much harder by converting hydroxyapatite
  [3(Ca₃(PO₄)₂. Ca(OFl)₂]. the enamel on the surface of the teeth into much harder fluorapatite [3(Ca₃(PO₃)₂. CaF₂]
• Fluoride ion concentration above 2 ppm causes brown mottling of teeth. Excess fluoride causes damage to bones and teeth.

Question 10.
Explain the harmful effects of

1. lead
2. Nitrate in drinking water.

Answer:

1. Drinking water containing lead contamination above 50 ppm can cause damage to liver, kidneys and reproductive system.
2. Use of drinking water having concentration of nitrates higher than 45 ppm may cause (blue baby syndrome) methemoglobinemia disease in chiidrens.

Question 11.
Explain how styrene is produced by traditional and greener routes?
Answer:
1. Traditional route:
This method involve two steps-Carcinogenic benzene reacts with ethylene to form ethyl benzene. After that ethyl benzene undergoes dehydrogenation using Fe₂O₃ / Al₂O₃ to give styrene.

2. Greener route:
To avoid carcinogenic benzene. greener routes to start with cheaper and
environment friendly xylene.

Question 12.
How acetaldehyde is commercially prepared by green chemistry?
Answer:
Acetaldehyde is commercially prepared by one step oxidation of ethene in the presence of ionic catalyst in aoiieniis medium with 9% yield.

Question 13.
What do you mean by ozone hole? What are its consequences?
Answer:
Depletion of ozone layer creates some sort of holes in the blanket of ozone which surrounds us in the atmosphere and this is known as ozone hole.

• With the depletion of the ozone layer, UV radiations filters into the troposphere which leads to ageing of skin. cataract, sunburn, skin cancer etc.
• By killing many of the phytoplanktons, it can damage the fish productivity.
• Evaporation rate increases through the surface and stomata of leaves which can decrease the moisture content of the soil.

Question 14.
What is photo chemical smog? What arc its effects? How can it be controlled?
Answer:
It is a kind of smog formed in warm, dry and sunny climate. It is formed when sunlight is absorbed by SO₂ oxides of nitrogen and hydrocarbons. It act as an oxidising agent.

Effects of photo chemical smog:

• It produces irritation in eyes and also in respiratory system.
• It can damage many materials such as metals, stones, building materials etc.
• NO₂ present in photochemical smog gives it brown colour which reduces the visibility.
• it is harmful to fabrics, crops and ornamental plants.

Control of photochemical smog:

• By using catalytic converters in automobilës.
• By spraying certain compounds into the atmosphere which generate free radicals that can easily combine with the free radicals that initiates the reaction forming toxic compounds of photochernical smog.
• Certain plants such as Pinus, Juniparus, Pyrus could be also helpful in this matter.

Question 15.
(a) Define eutrophication and pneumoconiosis.
(b) Write differences between photochemical smog and classical smog.
Answer:
(a) Eutrophication:
When the growth of algae increases in the surface of water, dissolved oxygen in water is greatly reduced. This phenomenon is known as eutrophication. Due to this growth of fishes gets inhibited.

(b) Pueumoconiosis:
It is a disease which irritates lungs. It causes scarring or fibrosis of the lungs.

Pholochemical smog:

• It is formed as a result of photochernical decomposition of nitrogen dioxide and chemical reactions involving hydrocarbons.
• It takes place during dry warm season in presence of sunlight.
• It is oxidising in nature.

Classical smog:

• It’s formed due to condensation of SO₂ vapours on particles of carbon in cold climate.
• it is generally formed during winter when there is severe cold.
• It is reducing in nature.

Samacheer Kalvi 11th Chemistry Environmental Chemistry 5 Marks Questions and Answers

Question 1.
Explain about the different layers of Earth’s atmosphere.
Answer:
Earth’s atmosphere can he divided into different layers with characteristic altitude and temperature.

Question 2.
What arc the health effects of particulate pollutants?
Answer:
1. Dust, mist, fumes etc. are air borne particles which are dangerous for human health. Particulate pollutants bigger than 5 microns are likely to settle in the nasal passage whereas particles of about 10 microns enter the lungs easily and causes scaring or fibrosis of lung lining.

They irritate the lungs and causes cancer and asthma. This disease is called pneumoconiosis. Coal miners may suffer from black lung disease. Textile workers may suffer from white lung disease.

2. Lead particulates affect children’s brain. interferes with maturation of RBC’s and even causes cancer.

3. Particulates in the atmosphere reduces the visibility by scattering and absorption of sun light. It is dangerous for aircraft and motor vehicles.

4. Particulates provide nuclei for cloud formation and increases fog and rain.

5. Particulates deposit on plant leaves and hinder the intake of CO₂ from the air and affect photosynthesis.

Question 3.
What arc the effects of photochernical smog?
Answer:

• The three main components of photochemical smog are nitrogen oxide, ozone and oxidized hydrocarbons like HCHO, CH₂ = CH – CRO and PAN.
• Photo chemical smog causes irritation to eyes. skin and lungs. It also increases the chance of asthma.
• high concentration of ozone and NO can cause nose and throat irritation, chest pain, difficulty in breathing.
• PAN is toxic to plants, attack younger leaves and cause bronzing and glazing of their surfaces.
• It causes corrosion of metals, stones, building materials and painted surfaces.

Question 4.
Explain about the environmental impacts of ozone depletion.
Answer:
1. The formation and destruction of ozone is a regular natural process, which never disturbs the equilibrium level of ozone in the stratosphere. Any change in the equilibrium level of ozone in the atmosphere will adversely affect the Life in biosphere in the following ways.
2. Depletion of ozone layer will allow more UV rays to reach the earth surface and would cause skin cancer and also decreases the immunity level in human beings.
3. UV radiations atlecis plant proteins which lead to harniful mutation in plant cells.
4. UV radiations affect the growth of phytoplankton and as a result ocean food chain is disturbed and it even damages the fish productivity.

Question 5.
Explain the list of major water pollutants and their sources.
Answer:

• Microorganisms – Domestic sewage, domestic waste water, dung heap
• Organic wastes – Domestic sewage, animal excreta, food processing factory wastc. detergents and decayed animals and plants
• Plant nutrients – Chemical fertilisers
• Heavy metals – Heavy metal producing factories
• Sediments – Soil erosion by agriculture and strip-mining
• Pesticides – Chemicals used for killing insects, fungi and weeds
• Radioactive substances – Mining of uranium containing minerals
• Heat – Water used for cooling in industries

Question 6.
Describe about the causes of water pollution.
Answer:
Causes of water pollution –
1. Microbiological pollutants:
(a) Disease causing microorganisms like bacteria, viruses and protozoa are most senous water pollutants. They come from domestic sewage and animal excreta.
(b) Fish and shellfish can become contaminated from them and people who eat them will also become ill.
(c) Dysentery and cholera are water borne diseases.
(d) Human excrcta contain bacteria such as Escherichia coll and Streptococcus farcical- -is which causes gastrointestinal diseases.

2. Organic wastes:
Organic matter such as leaves, grass, trash can also pollute water. Water pollution is cause by excessive phytoplankton growth within water.

3. Chemical wastes:
A whole variety of chemicals from industries such as metals and solvents are poisonous to fish and other aquatic life. Detergents and oils float spoils the water bodies. Acids from mine drainage and salts form various sources can also contaminate water sources.

Question 7.
What are the harmful effects of chemical water pollutants’?
Answer:

1. Cadmium and mercury can cause kidney damage.
2. Lead poisoning can lead to severe damage of kidneys. liver and brain. It also affects the central nervous system.
3. Polychiorinated biphenyl causes skin diseases and are carcinogenic in nature.

Question 8.
Explain about green chemistry in day-to-day life.
Answer:
1. Dry cleaning of clothes:
Solvents like tetrachloroethylene used in dry cleaning of clothes, pollute the ground water and are carcinogenic. In place of tetrachloro ethylene, liquefied CO₂ with suitable detergent is an alternate solvent used. Liquefied CO₂ is not harmful to the ground water. Nowadays H₂O₂ is used for bleaching clothes in laundry, gives better result and utilises less water.

2. Bleaching of paper:
Conventional method of bleaching was done with chlorine. Nowadays H₂O₂ can be used for bleaching paper in the presence of catalyst.

3. Synthesis of chemicals:
Acetaldehyde is commercially prepared by one step oxidation of ethene in the presence of ionic catalyst in aqueous medium with 90% yield.

4. Instead of petrol, methanol is used as a fuel in automobiles.
5. Neem based pesticides have been synthesis-ed, which are more safer than the chlorinated hydrocarbons.

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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  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  X + N₂, X is …………

Answer:

Question 8.
Which of the following compounds will give racemic mixture on nucleophilic substitution by OH^– ion?

(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 –

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:

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 S_N¹ condition is …………
(a) allyl chloride
(b) ethyl chloride
(c) isopropyl chloride
(d) benzyl chloride
Answer:
(a) benzyl chloride

Question 18.
The carbocation formed in S_N¹ 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?

Answer:

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 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

(d) CH₂ = CH – Cl
Answer:
(a) CH₃-CH = CH-Cl – Allylic halide
(b) C₆H₅CH₂ – Benzylic halide

(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 S_N² reaction is called Finkelstein reaction.

Question 29.
Which alkyl halide from the following pair is –
1. chiral
2. undergoes faster S_N² reaction?

Answer:
1.

Here the C*is chiral carbon atom and it is surrounded by four different groups. Br
2.  i.e., 1-chiorohutane undergoes faster S_N² reaction. CH₃-CH₂-CH₂Cl is a primary alkyl halide and so it undergoes faster S_N² 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.

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^–)
• 

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.

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.

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)

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.

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:

2. Reaction of alcohol with phosphorous halide:

3. Reaction of alcohol with thionyl chloride:

Question 38.
Compare S_N¹ and S_N² reaction mechanisms.
Answer:
S_N¹ 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

Step – II

S_N² 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.

Question 39.
Compare S_N¹ and S_N² reaction mechanisms.

Answer:

Question 40.
Discuss the aromatic nucicophilic substitution reactions of chiorobenzene.
Answer:
Aromatic nucleophilic substitution reactions:
Dow’s process:

Question 41.
Account for the following –
1. t-butyl chloride reacts with aqueous KOH by S_N¹ mechanism while n-butyl chloride reacts with S_N² 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 S_N¹ mechanism rather than S_N² mechanism because S_N¹ 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 S_N² 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.

(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.

Methyl magnesium iodide reacts with acetyl chloride to form acetone.

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”.

2. Carbon tetrachioride from carbon disuiphide:
Carbon disuiphide reacts with chlorine gas in the presence ofanhydrous AlCl₃ as catalyst to give carbon tetrachloridc.

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.

2. When bromoethane is treated with AgNO₂ nitroethane will be formed as the product.

Question 46.
Explain the mechanism of S_N¹ reaction by highlighting the stereochemistry behind it.
Answer:
1. S_N¹ stands for unimolecular nucleophilic substitution reaction of first order reaction.
2. The rate of the following S_N¹ 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 S_N¹ reaction follows first order kinetics and it occurs in two steps:
5. S_N¹ reaction mechanism takes place in tertiary butyl bromide with aqueous KOH a follows:

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.

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.

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.

2. Dow’s process:
Chiorobenzene is boiled with Sodium hydroxide to get Phenol. This reaction is called Dow’s process.

3. Darzens process:
Ethanol reacts with SOCl, in the presence of pyridine to form chloroethane. This reaction is called Darzens process.

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:

2. Acetone from Gngnard reagent:

3. Ethyl acetate from Grignard reagent:

4. Isopropyl alcohol from Grignard reagent:

5. Methyl cyanide form Grignard reagent:

Question 49.
Complete the following reactions:

Answer:

Question 50.
Explain the preparation of the following compounds:

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

Answer:
1. DUT

2. Chloroform:

3. Biphenyl:
Fittig reaction:

4. Chioropicrin.

5. Freon – 12.

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:

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.

3. ChIooethane reacts with alcoholic KOH to give ethene C₂H₄ as (C) by elimination reaction.


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:

1. The simplest alkene (A) is CH₂ = CH₂, ethene.

2. Ethene reacts with HCl to give Chioroethane CH₂-CH₂Cl as (B) by addition reaction.

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.

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:

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).

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)

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:

1. The compounds (A) and (B) with the molecular formula C₂H₄Cl₂ are and 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  as (D).

[Evaluate Yourself ]

Question 1.
Write the IUPAC name of the following:

Answer:

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

2. 1. 4-Dichioro but-2-ene

3. 2-Çhloro-3-methyl pentane

Question 3.
Write all possible chain isomers with molecular formula C₅H₁₁Cl
Answer:
C₅H₁₁Cl:
8 isomers arc possible. These are –

Question 4.
neo-pentyl bromide undergoes nucleophilic substitution reactions very slowly. Justify.
Answer:
1.  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.

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?

Answer:

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) S_N¹ mechanism
(b) E₁ mechanism
(c) S_N² 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

(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.

the product X is –

Question 28.
Which one of the following will undergo S_N¹ reaction faster?

Answer:

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.

Answer:

Question 2.

Answer:

Question 3.

Answer:

Question 4.

Answer:

Question 5.

Answer:
“17” />

Question 6.

Answer:

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:

Question 9.

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 is ………
Answer:
2-Bromo-3-chloro-2, 4-dimethylpentane.

Question 12.
The IUPAC name of 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:
S_N²

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:
S_N²

Question 29.
Tertiary butyl bromide reacts with aqueous KOH follows ………
Answer:
S_N² 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 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:

Question 64.
The most easily hydrolysed molecule under S_N² reaction is………..
Answer:
Ten. butyl chloride

Question 65.
. The product ‘X’ is ………..
Answer:
Ethanol

Question 66.
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 : S_N¹ reaction
(b) (CH₃)₃C-Cl + alcoholic KOH : E₁ reaction
(c) (CH₃)₃C-Cl + aqueous KOH : S_N² reaction
(cl) CH₃-Cl + aqueous KOH : S_N¹ reaction
Answer:
(b) (CH₃)₃C-Cl + alcoholic KOH : E₁ reaction

VI. Choose the incorrect pair.

Question 1.

Answer:

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

Answer:
(c) CH₃-CH₂-CH₂Cl <CH₃ CH₂Cl <CH₃Cl : increasing order of boiling point

Question 3.
(a) (CH₃)₃ C-Br + aqueous KOH : S_N¹ reaction
(b) (CH₃)₃ C-Br + alcoholic KOH: E₁ reaction
(c) CH₃Br – aqueous KOH : E₂ reaction
(d) CH₃Br + aqueous KOH : S_N² 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

2. 1 -Brorno-2, 2-dimethyipropane

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

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

 Ethylene dichioride (or) 1, 2-dichioroethane.

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

2. 2-bromo-3-chloro-2, 4-dimethy 1 pentane

Question 6.
What happens when HI reacts with ten. butyl alcohol?
Answer:

Question 7.
Explain the action of
(i) PCl₅
(ii) PCl₃ with ethanol.
Answer:

Question 8.
How does HBr react with propene?
Answer:
Propene react with HBr and follows Markovnikov’s rule.

Question 9.
How methane reacts with Cl₂ in the presence of light?
Answer:

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:

Question 11.
Explain Swans reaction.
Answer:
Chioro (or) bromoalkanes on heating with AgF give fluoroalkanes. This reaction is called Swans reaction.

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:

Question 18.
Complete the following reactions:

Answer:

Question 19.
Explain the action of sodium hydrogen suiphide with bromoethane?
Answer:

Question 20.
Explain Williamson’s synthesis.
Answer:
Williamson’s synthesis:
Haloalkanes when boiled with sodium alkoxide gives the corresponding ether.

Question 21.
Explain the action of alcoholic potash with bromoethane.
Answer:
Elimination reaction takes place when alcoholic potash reacts with bromoethane:

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:

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:

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.

Question 26.
What is tetraethyl Lead? How is it prepared?
Answer:

Question 27.
Convert bromoethane to ethane.
Answer:

Question 28.
How is iodoethane converted to ethane?
Answer:

Question 29.
Starting from CH₃MgI, how will you prepare acetaldehyde?
Answer:

Question 30.
How will you get acetone from methyl manesium iodide?
Answer:

Question 31.
Explain the action of Ethyl chioroformate with Methyl magnesium iodide.
Answer:
Ethyl chioroformate reacts with Grignard reagent to form esters as follows:

Question 32.
Write the IUPAC names of:

Answer:
1 – bromo-4-fluoro-2-iodobenzene
1 – bromo-2-fluoro-4-iodobenzene

Question 33.
How is benzene directly converted to chiorobenzene?
Answer:

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.

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.

Question 36.
How will you prepare iodobenzene?
Answer:

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.

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.

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.

Question 41.
Explain about the preparation of phenyl magnesium chloride.
Answer:

Question 42.
How will you prepare ethylidcne dichioride from acetylene?
Answer:

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:

Question 44.
Explain the action of zinc and HCl on chloroform.
Answer:

Question 45.
How does nickel react with chloroform?
Answer:

Question 46.
Convert methane to methylene chloride.
Answer:

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).

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.

Question 49.
How will you convert carbon tetrachioride to chloroform?
Answer:

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

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:

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 S_N² 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 S_N² reaction with OH^– ion.
2. On steric grounds 1° alkyl halides are more reactive than tert alkyl halide in S_N² reactions. Therefore, CH₃Cl will react at a faster rate than (CH₃)₃CCl in a S_N² 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:

2. Wurtz-Fittig reaction:
It involves the reaction of an aryl halide and alkyl halide to form the corresponding hydrocarbon.

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

(b) Secondary haloalkane

(c) Tertiary haloalkane

Question 2.
Write the possible isomers for the formula C₄H₉Cl with structures and names.
Answer:
C₄H₉Cl (4 isomers):

Question 3.

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:

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:

Question 5.
Explain how bromoethane reacts with –
1. alcoholic KCN
2. alcoholic AgCN
Answer:
1. Bromoethanc reacts with alcoholic KCN to form ethyl cyanide.

2. Bromoethane reacts with alcoholic AgCN to form alkyl isocyanide.

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

• 2-bromobutane is optically active and it undergoes S_N¹ 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:

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.
  

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.

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:
   

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:

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.
1,2-dichloroethane

Question 16.
Give two examples for –
1.  gem dihalide
2. vicinal dihalide.
Answer:
1. Gem dihalides:

2. Vicinal dihalides:

Question 17.
Explain two methods of preparation of ethylene diehioride.
Answer:
1. Addition of Cl₂ to Ethylene:

2. Action of PCl₅ on Ethylene glycol:

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.

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:

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.

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-

Step 2:
Chlorination-

Step 3:
Hydrolysis-

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):

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

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:

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.

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:

Question 31.
Give reasons for the following:

1. Ethyl iodide undergoes S_N² 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 S_N² 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?

Answer:

Samacheer Kalvi 11th Chemistry Haloalkanes and Haloarenes 5 Marks Question And Answers

Question 1.
Explain the classification of organic compounds with example.
Answer:

Question 2.
Explain S_N² mechanism with suitable examples.
Answer:

• S_N² reaction means bimolecular nucleophilic substitution reaction of second order.
• The rate of S_N² 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.
  

Question 3.
S_N² reaction of an optically active haloalkane is accompanied by inversion of configuration at the asymmetric centre. Prove it.
Answer:
1. S_N² 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.

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:
  

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:

5. Step 2.
Elimination of a proton from the fl-carbon to produce an alkene:

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.

2. 2-Propanol :
Acetaldehyde react with CH₃MgI to give an addition product which on hydrolysis yields 2-propanol.

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.

2. Methyl cyanide:
Grignard reagent reacts 4rith cyanogen chloride to form alkyl cyanide.

3.  Methane:
Grignard reagent reacts with waler to give methane as product.

Question 8.
Describe electrophilic substitution reaction of chiorohenzene with equations.
Answer:

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

3. 1-Bromopropane on reaction with aqueous KOH undergoes hydrolysis to give  Propan- 1-ol. CH₃-CH₂-CH₂OH as ©

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.

3. Ethyl chloride on treatment with alcoholic KOH, undergoes dehydrohalogenation to give C₂H₄, ethylene Ⓒ as the product.

4. Ethylene on reaction with Cl₂ yield ethylene dichioride as Ⓓ as the product.

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.

3. Chiorobenzene Ⓑ reacts with sodium hydroxide to give (C₆H₅OH) phenol, Ⓒ as the product.

4. Chiorobenzene reacts with ammonia to give Aniline, C₆H₅NH₂ Ⓓ as the product.
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.

3. Ethylidene dichioride on reaction with aqueous KOH will give acetaldehyde, © as the product.

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: Tert butyl bromide and

2. Ⓐ on reaction with alcoholic KOH gives iso butylene © as the product.

3. 2-bromobutane Ⓑ on reaction with alcoholic KOH follows Saytzeff s rule to give a mixture of olefins in different amounts.

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.

3. Chiorobenzene reacts with ethyl chloride to form ethyl benzene, © as the product

4. Chiorobenzene on reaction Na metal gives Biphenyl compound © as the product.

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.

3. Ethanol Ⓑ reacts with PCl₅ to give C₂H₅Cl, Ethyl chloride © as the product.

4. CH₃-CH₂Cl © on reaction with alcoholic KOH undergoes dehydrohalogenation to give ethylene CH₂=CH₃ Ⓓ as the product.

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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



Answer:

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
the major product obtained is ………

Answer:

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) C_nH_n
(b) C_nH_2n
(c) C_nH_2n-2
(d) C_nH_2n+2
Answer:
(b) C_nH_2n

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]

(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 ……….

(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 ………

(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.


Answer:

Question 17.
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?

Answer:
(d) both (a) and (b)

Question 20.
Propyne on passing through red hot iron tube gives ………

Answer:

Question 21.


Answer:

Question 22.
Which one of the following is non-aromatic?

Answer:

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.

(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)

(iii) (CH₃)₃ C-C≡C-CH (CH₃)₂
(iv) Ethyl – isopropyl – acetytene
(v) CH≡C – C = C – C≡CH
Answer:

Question 32.
Identify the compound A. B, C and D in the following series of reactions.

Answer:

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.

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:

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.

2. 3 – Bromo -4 methylpentane on dehydrogenation give two isomers X and Y as follows:

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.

Step-3:
Rearomatisation of arenium ion.

Overall Reaction:

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:

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.

2. Preparation 4-chiorotoluene from benzene:
Benzene undergoes Fnedel crafi’s alkylation followed by chlorination and it leads to the formation of 4-chiorotoluene.

3. Prepar2tion of Bromobenzene from benzene:
Bezene undergo bromination to give bromobenzene.

4. Preparation of m-dinitrobenzene from benzene:
Benzene undergo twice the time nitration to give m-dinitrobenzene.

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.

Question 40.
What happens when isobutylene is treated with acidified potassium permanganate?
Answer:
Isobutylene is treated with acidified KMnO₄ to give acetone.

Question 41.
how will you convert ethyl chloride in to –
1. ethane
2. n – butane
Answer:
1. Conversion of ethyl chloride into ethane:

2. Conversion of ethyl chloride into n-butane:
Wurtz reaction:

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.

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:

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:

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:

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

2. 5 – (2 – Ethylbutyl) – 3, 3- dimethyldecane

3. 5 – (1,2 – Dimethylpropyl) – 2- methylnonane

Question 47.
How will you prepare propane from a sodium salt of fatty acid?
Answer:

Sodium salt of butyric acid on heating with sodalime gives propane.

Question 48.

Answer:

Question 49.
How will you distinguish 1 – butyne and 2 – butyne?
Answer:

Answer:

Question 50.
How will you distinguish 1 – butyene and 2 – butyne?
Answer:

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 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:

Question 2.
Give the IUPAC name for the following alkane.


Answer:

Question 3.
Draw the structural formula for 4,5 -diethyl -3,4,5-trimethyloiane
Answer:

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:

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.

Question 8.
How toluene can be prepared by this method?
(i) From n-heptane,
(ii) From 2-methyihexane
Answer:

(ii)

Question 9.
Write the IUPAC names for the following alkenes.

Answer:

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:

2. 5 Bromo -4 – Chloro – 1 – heptene:

3. 2.5 – dimethyl – 4 – Octene:

4. 4 – methyl 2 pentene:

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

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


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

Therefore – 1 – Chloropropene has cis-trans isomers.

(b) 2 – Chloropropene:

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:

(b) CH₃ – CH = CH – CH – CH₃

Question 14.
How propene is prepared form 1, 2-dichloropropane?
Answer:

Question 15.
How ozone reacts with 2-methyl propene?
Answer:

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.


Geometrical isomers of 2 – Butene (A):

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.

Question 18.
Prepare propyne from its corresponding alkene.
Answer:
Preparation of propyne from propene:

Question 19.
Write the products A & B for the following reaction.

Answer:

Question 20.

Answer:

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

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:

Question 23.
Write all possible isomers Iòr a monosubstituted aromatic benzenoid compound having the molecular formula C₉H₁₂
Answer:
Possible isomers for C₉H₁₂:

Question 24.
how benzene can be prepared by Grignard Reagent?
Answer:
Phenyl magnesium bromde reagent reacts with water molecule to give benzene:

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:

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 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 C₂H_2n+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.
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?

Answer:

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.
A → B + H₂O. Identify A and B.

Answer:

Question 19.
Benzene undergoes Birch reduction to form .

Answer:

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.


Answer:

Question 2.

Answer:

Question 3.


Answer:

Question 4.


Answer:

Question 5.


Answer:

Question 6.


Answer:

Question 7.

Answer:

Question 8.


Answer:

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:

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

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.

Answer:

Question 4.

Answer:

VI. Choose the incorrect pair.

Question 1.

Answer:

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

Question 6.
Give the IUPAC name of the following compounds.

Answer:

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:

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.

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.

Question 10.
How will you prepare propane from Chloropropane?
Answer:
Nascent hydrogen reacts with chloropropane to give propane.

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:

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.

Question 13.
How will you prepare methane from grignard reagent’?
Answer:
Methyl magnesium chloride reacts with water to give methane.

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:

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.

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.

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.

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

Question 21.
cis-isomers are less stable than rans-isomers?
Answer:
Consider:

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.

Question 23.
How will you convert 1-brornopropane into popene?
Answer:

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.

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.

Answer:

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.

Question 30.
Complete the following reaction and identify A, B and C.

Answer:

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.

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.

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.

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).

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:

Answer:

• 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) Cyclo-octatetraene.

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

(c) 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.

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:

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.

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:

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:

3. Fill in ail the hydrogen atoms so that each carbon forms four bonds,

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

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.

Step 3:
Add hydrogen atoms to the carbon skeleton so that each carbon atoms have four bonds.

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.

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.

3. If the similar groups lie on the opposite side then the geometrical isomers are called as trans-isomers.

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.

(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.

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

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.

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.

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

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.

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:

Question 16.
Complete the following reactions:

Answer:

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:

(b) propyne:

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.

2. Cyclic polymerisation:
Acetylene undergoes cyclic polymerisation on passing through red hot iron tube. Three molecules of acetylene polymerises to form benzene.

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:

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.

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.

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.

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.

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.

It is found in cigarette smoke, in tobacco and and charcoal boiled food.

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:

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:

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:

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:

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.

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.

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.

(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.

7. Representation of benzene:
Hence, there are three ways is which benzene can be represented.

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).

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.

(b) The methyl free radicals thus obtained attacks the second molecule of chlorine to give chloromethane (CH₃Cl) and a chlorine free radical as follows –

(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.

Question 4.
Write the mechanism for following reaction.

Answer:

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.

Step 2.
The radicals abstracts a hydrogen atom from HBr thus generating bromine free radical.

Step 3.
The Bromine free radical adds lo the double bond in the way so as to form the more stable alkyl free radical,

Step 4.
Addition of HBr to secondary free radical.

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.

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

Step 2:
Attack of the electrophile on the benzene ring to form arenium ion:

Step 3:
Rearomatisation of arenium ion:

Question 7.
Describe the mechanism of suiphonation of benzene.
Answer:
Step 1:
Generation of SO3 electrophile:
2H₂SO₄ → H₃O^⊕ + SO₃ + HSO₄^Θ

Step 2:
Attack of the electrophile on benzene ring to form arenium ion:

Step 3:
Rearomatisation of Areniurn ion:

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

Step 2 :
Attack of the electrophile on benzene ring to form areniurn ion:

Step 3:
Rearornatisation of arenium ion:

Question 9.
Write the mechanism of Freidel craft’s acylation.
Answer:
Step 1:
Generation of CH₃CO electrophile:

Step 2 :
Attack of the electrophile on benzene ring to form arenium ion.

Step 3 :
Rearomatisation of arenium ion:

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).

3. Benzene (A) reacts with Cl₂ in presence of UV-light to give benzene hexachioride (C).

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:

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).

3. Benzene (A) is treated with sodium and liquid ammonia to give 1.4 – cyclohexadiene (C)

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.

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).

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).

2. Propyne (A) reacts with mercuric sulphate and dil. H₂SO₄ at 333 K to give acetone (B).

3. Propyne (A) on passing through red hot iron tube at 873 K will give mesitylene (C).

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