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NCERT Solutions for Class 7 English Honeycomb Chapter 2 A Gift of Chappals are part of NCERT Solutions for Class 7 English. Here we have given NCERT Solutions for Class 7 English Honeycomb Chapter 2 A Gift of Chappals.

NCERT Solutions for Class 7 English Honeycomb Chapter 2 A Gift of Chappals

IMPORTANT PASSAGES FOR COMPREHENSION
Read the following extracts and answer the questions that follow by choosing the correct option :

[I]

Question 1.
“People are always telling us to be kind to animals, but when we are, they scream. Ooh don’t bring that dirty creature here !” said Ravi. “Do you know how hard it is just to get a little milk from the kitchen ?       (Page 19)
Multiple Choice Questions
Question 1.
Identify ‘people’ and ‘we’
(a) colony members and the family
(b) family members and the group
(c) elders and children
(d) environmentalists and children
Answer.
(c) elders and children

Question 2.
Meena shares with Mridu
(a) the biryani cooked by Rukku Mani
(b) the secret about the cat in the backyard
(c) the chocolate Ravi brought
(d) the advice given by the beggar that
Answer.
(b) the secret about the cat in the backyard

Question 3.
Ravi poured the milk for the kitten
(a) the kitchen
(b) the fridge
(c) the market
(d) the dairy
Answer.
(a) the kitchen

Question 2.
“She’ll never learn a thing. The train whizzing on and on, while Lalli’s all the time-derailing ! Going completely off track !”   (Page 21)
Question 1.
Who is the speaker of the above extract ?
Answer.
The speaker of the above extract is Ravi.

Question 2.
What is the music master trying to do ?
Answer.
The music master is trying to teach Lalli music.

Question 3.
Is he successful in his effort ?
Answer.
The music master is unsuccessful. Lalli is not able to leam.

[II]

Question 3.
“He has been coming here every day for the past week and it’s time he found another house to beg from !” Paati explained to Tapi.       (Page 24)
Multiple Choice Questions
Question 1.
Paati explained to Tapi that the beggar
(a) was very notorious
(b) should beg from some other house
(c) should find some other person
(d) never listened to her
Answer.
(b) should beg from some other house

Question 2.
The beggar raised his voice to
(a) beg for money
(b) beg for alms
(c) beg for food and rest
(d) beg for rest
Answer.
(c) beg for food and rest

Question 3.
Rukku Manni told Ravi to tell the beggar
(a) not to come again
(b) to take food
(c) to rest under the tree
(d) to find food elsewhere
Answer.
(a) not to come again

Question 4.
In two minutes he’ll be fiying his feet on that road.    (Page 25)
Question 1.
Who is the speaker of the above line ?
Answer.
The speaker of the above line is Ravi.

Question 2.
Who is ‘he’ in the above line ?
Answer.
‘He’ refers to the beggar.

Question 3.
Why should ‘he’ be frying his feet on the road ?
Answer.
The beggar has no shoes or chappals for his feet. As it is a hot day, Ravi thinks that the poor beggar will be frying his (bare) feet on the road.

Question 5.
“These should fit you, Sir. Please put these on. I gun so sorry. My son has been naughty.”   (Page 27)
Multiple Choice Questions
Question 1.
Rukku Manni offered chappals that were
(a) old and worn out
(b) new
(c) big in size
(d) small in size
Answer.
(b) new

Question 2.
On seeing the chappals, the music-master
(a) was too happy
(b) grew angiy
(c) picked them to keep in his bag
(d) sat down to wear them
Answer.
(a) was too happy

Question 3.
Ravi’s chappals were in question as
(a) he would not have given his chappals if they were a perfect fit
(b) he would have given his own chappals if they were a perfect fit
(c) his chappals fitted the beggar’s feet
(d) his chappals did not fit the beggar’s feet
Answer.
(b) he would have given his own chappals if they were a perfect fit

TEXTUAL QUESTIONS

Comprehension Check (Page 22)
Question 1.
What is the secret that Meena shares with Mridu in the backyard ?
Answer.
Meena shares with Mridu the secret about the cat in the backyard. The secret was that they had a kitten hidden behind the bitter berry bush.

Question 2.
How does Ravi get milk for the kitten ?
Answer.
Ravi said that he was hungry. So he got some milk from the kitchen. Ravi poured milk into the coconut shell. He then washed the tumbler and put it back. Thus Ravi got milk for the kitten.

Question 3.
Who does he say the kitten’s ancestors are ? Do you believe him ?
Answer.
He says that the kitten’s ancestors are the Pallavas. No, I don’t believe him.

Question 4.
Ravi has a lot to say about M.P. Poonai This shows that

1. he is merely trying to impress Mridu.
2. his knowledge of history is sound.
3. he has a rich imagination.
4. he is an intelligent child.

Which of these statements do you agree I disagree to ?
Answer.
We agree with the statements (1) he is merely trying to impress Mridu and (3) he has a rich imagination.

Question 5.
What was the noise that startled Mridu and frightened Mahendran ?
Answer.
A ‘kreeching’ sound startled Mridu and frightened Mahendran.

Comprehension Check (Page 28)
Question 1.
The music master is making lovely music. Read aloud the sentence in the text that expresses this idea.
Answer.
The music master’s notes seemed to float up and settle perfectly into the visible tracks of the melody.

Question 2.
Had the beggar come to Rukku Manni’s house for the first time ? Give reasons for your answer.
Answer.
No. The beggar had not come to Rukku Manni’s house for the first time. The beggar himself says that he had been coming there regularly for a week.

Question 3.
“A sharp V-shaped line hadformed between her eyebrows. ” What does it suggest to you about Rukku Manni’s mood ?
Answer.
A sharp V-shaped line between her eyebrows suggests that Rukku Manni was losing patience. She was getting angry.

Working with the Text
Question 1.
Complete the following sentences :
(i) Ravi compares Lalli’s playing the violin to ………………………………………
(ii) Trying to hide beneath the tray of chillies. Mahendran ……………………
(iii) The teacher played a few notes on his violin, and Lalli ……………………
(iv) The beggar said that the kind ladies of the household …………………….
(v) After the lesson was over, the music teacher asked Lalli if ………………..
Answers.

1.  a railway train which is all the time derailing, going completely off the track.
2.  tipped a few chillies over himself.
3.  stumbled behind him on her violin without much success.
4.  had helped him with food for a whole week.
5.  she had seen his chappals.

Question 2.
Describe the music teacher, as seen from the window.
Answer.
The music teacher sat in front of Lalli with most of his back to the window. He was bony. His head was bald. A fringe of oiled black hair fell around his ears. He had an old fashioned tuft. He had a thick neck. Round it he wore a gold chain. He had a diamond on his hand. He had a scrawny big toe.
He was playing on the violin with his hands. He beat time on the floor with his toe.

Question 3.
(i) What makes Mridu conclude that the beggar has no money to buy chappals ?   (Imp.)
(ii) What does she suggest to show her concern ?
Answers.

1.  The beggar showed the children his bare feet. There were large blisters on them. So Mridu concluded that the beggar had no money to buy chappals.
2.  To show her concern for the beggar she suggests that an old pair of chappals should be given to him.

Question 4.
“Have you children…” she began, and then, seeing they were curiously quiet, went on more slowly, “seen anyone lurking around the verandah ?”   (Imp.)
(i) What do you think Rukku Manni really wanted to ask ?
(ii) Why did she change her question ?
(iii) What did she think had happened ?
Answers.

1.  Rukku Manni really wanted to ask if the children had hidden the chappals.
2.  Seeing the children curiously quiet, she felt something more serious had happened.
   So she changed her question.
3.  She thought that the chappals had gone for good.

Question 5.
On getting Gopu Mama’s chappals, the music teacher tried not to look too happy. Why ?
Answer.
The music master did not like to show his greed. So, although his eyes lit up, he tried not to look too happy.

Question 6.
On getting a gift of chappals, the beggar vanished in a minute. Why was he in such a hurry to leave ?
Answer.
The beggar had realised that the children had given him chappals of their own. They had not sought the permission of the elders. He feared that the elders could take it back. So he vanished in a minute.

Question 7.
Walking towards the kitchen with Mridu and Meena, Rukku Manni began to laugh. What made her laugh ?  (Imp.)
Answer.
It was the mental picture of Gopu Mama which made Rukku Manni laugh. She knew he would feel very uncomfortable. On coming home, it was his habit to throw off his shoes and get into chappals as soon as possible.

Working with Language
Question 1.
Read the following sentences :
(a) If she knows we have a cat, Paati will leave the’ house. ‘
(b) She won’t be so upset if she knows about the poor beggar with sores on his feet.
(c) If the chappals do fit, will you really not mind ?

Notice that each sentence consists of two parts. The first part begins with ‘if. It is known as if-clause.
Rewrite each of the following pairs of sentences as a single sentence. Use ‘if at the beginning of the sentence.
(a) Walk fast. You’ll catch the bus.
If you walk fast, you’ll catch the bus.
(b) Don’t spit on the road. You’ll be fined.
If you spit on the road, you’ll be fined.

(i) Don’t tire yourself now. You won’t be able to work in the evening.
(ii) Study regularly. You’ll do well in the examination.
(iii) Work hard. You’ll pass the examination in the first division.
(iv) Be polite to people. They’ll also be polite to you.
(v) Don’t tease the dog. It’ll bite you.

Answers.

1.  If you tire yourself now, you won’t be able to work in the evening.
2.  If you study regularly, you’ll do well in the examination.
3.  If you work hard, you’ll pass the examination in the first division.
4.  If you are polite to people, they’ll also be polite to you.
5.  If you tease the dog, it’ll bite you.

Question 2.
Fill in the blanks in the following paragraph :
Today is Sunday. I’m wondering whether I should stay at home or go out. If I ……….. (go) out, I ……… (miss) the lovely Sunday lunch at home. If I …….. (stay) for lunch, I ……… (miss) the Sunday film showing at Archana Theatre. I think I’ll go out and see the film, only to avoid getting too fat.
Answer.
Today is Sunday. I’m wondering whether I should stay at home or go out. If I go out, I’ll miss the lovely Sunday lunch at home. If I stay for lunch, I’ll miss the Sunday film showing at Archana Theatre. I think I’ll go out and see the film, only to avoid getting too fat.

Question 3.
Complete each sentence below by appropriately using any one of the following :
if you want to/if you don’t want to/if you want him to
(i) Don’t go to the theatre ……………
(ii) He’ll post your letter ………………
(iii) Please use my pen ………………
(iv) He’ll lend you his umbrella ………..
(v) My neighbour, Ramesh, will take you to the doctor …………
(vi) Don’t eat it …………………
Answers.

1.  if you don’t want to.
2.  if you want him to.
3.  if you want to.
4.  if you want him to.
5.  if you want him to.
6.  if you don’t want to.

Speaking and Writing
Question 1.
Discuss in small groups
• If you want to give away something of your own to the needy, would it be better to ask your elders first ?
Answers.
A   :   If there is something which is our own, we needn’t seek permission to use it.
B   :   I don’t agree. We do not earn anything. What we have is given to us by our parents. So it is necessary to seek their permission to give it to someone else.
C  :   I am afraid, both of you are running to extremes. We have to take into consideration the value of the thing as well. For example, if we have two pencils, we may give one to a friend. However, we cannot do the same with a pair of shoes or a suit of clothes. We must seek the permission of our elders before offering it to someone. After all, it is they who will be burdened with extra expenditure.

• Is there someone of your age in the family who is very talkative ? Do you find her/him interesting and impressive or otherwise ? Share your ideas with oth¬ers in the group.
Answers.
A   :  I have a cousin. She is very talkative. Since we live in the same house, I have to tolerate her somehow. However, she causes much annoyance. I find it difficult to concentrate on my books.
B   :  I too have a talkative cousin. However, I love to hear her. Her knowledge is so vast and her voice so sweet. Whenever I am tired or bored with study, I go to her. Her conversation gives me a lot of joy. I feel refreshed and am ready to work again.
C  :   My sister is very talkative. However, I tell her not to talk much these days. These are examination days and I want to devote much time to study. Thankfully, she has accepted my request.

• Has Rukku Manni done exactly the same as children? In your opinion, then, is it right for one party to blame the other ?
Answers.
A   :   Yes. Rukku Manni did only what Ravi had done. So it was not at all right for her to scold Ravi.
B    :   I don’t agree. Rukku Manni was forced to do what she did. The music teacher’s chappals had gone. How to compensate him ? The only option left to her was to give Gopu Mama’s chappals to him. So we cannot equate the two acts. She was quite justified in blaming and scolding the children.
C    :  The chidren had certainly done something wrong. It is not correct and Rukku Manni’s act can’t be equated with theirs. However, I think, such children need a more sympathetic handling. They had taken pity on a beggar. They must be taught how far they can allow their sympathy to go. They should not have touched music teacher’s chappals.

Question 2.
Read the following :
• A group of children in your class are going to live in a hostel.
• They have been asked to choose a person in the group to share a room with.
• They are asking each other questions to decide who they would like to share a room with.
Ask one another questions about likes/dislikes/preferences/hobbies/personal characteristics. Use the following questions and sentence openings.
(i) What do you enjoy doing after school ?
I enjoy…
(ii) What do you like in general ?
I like…
(iii) Do you play any game ?
I don’t like…
(iv) Would you mind if I listened to music after dinner ?
I wouldn’t…
(v) Will it be all right if I… ?
It’s fine with me…
(vi) Is there anything you dislike, particularly ?
Well, I can’t share…
(vii) Do you like to attend parties ?
Oh, I…
(viii) Would you say you are… ?
I think…
Answer.
Please attempt yourself.

We hope the NCERT Solutions for Class 7 English Honeycomb Chapter 2 A Gift of Chappals help you. If you have any query regarding NCERT Solutions for Class 7 English Honeycomb Chapter 2 A Gift of Chappals, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 English An Alien Hand Chapter 2 Bringing up Kari are part of NCERT Solutions for Class 7 English. Here we have given NCERT Solutions for Class 7 English An Alien Hand Chapter 2 Bringing up Kari.

NCERT Solutions for Class 7 English An Alien Hand Chapter 2 Bringing up Kari

EXERCISES
(Page 14)

Answer the following questions.
Question 1.
The enclosure in which Kari lived had a thatched roof that lay on thick tree stumps. Examine the illustration of Kari’s pavilion on page 8 and say why it was built that way.
Answer.
The enclosure was so built as to make it suitable for Kari. Kari bumped against the poles as he moved about. Since these poles were thick tree stumps, they did not give way.

Question 2.
Did Kari enjoy his morning bath in the river ? Give a reason for your answer.
Answer.
Kari enjoyed his morning bath. It was clear from the fact that he lay in the water for a long time. On coming out, he would squeal with pleasure.

Question 3.
Finding good twigs for Kari took a long time. Why?
Answer.
Finding good twigs for Kari took a long time. First the author would sharpen his hatchet which would take half an hour. It was necessary because the elephant would not touch mutilated twigs. Then the author had to climb all kinds of trees. He did it to get the most delicate and tender twigs. All this naturally took a long time.

Question 4.
Why did Kari push his friend into the stream ? (Imp.)
Answer.
Kari pushed his friend into the stream to save the life of a boy. The author fell into the stream and he saw a boy lying on the bottom. He dived and pulled the boy to the surface. But the author was not a swimmer. The current of the water began to drag him down. Kari saw it. He came fast into the water. He caught the author by his trunk. Then Kari pulled both of them ashore.

Question 5.
Kari was like a baby. What are the main points of comparison ?
Answer.
Kari was like a baby. Like a baby, he was to be scolded when he was naughty. Again like a baby he learnt very quickly. He sometimes did mischief like a baby. But he quietly accepted punishment when he was wrong.

Question 6.
Kari helped himself to all the bananas in the house without anyone noticing it. How did he do it ?
Answer.
Bananas were kept on a large plate on a table in the dining room. The table was close to window. Kari put his trunk through the window on the fruit plate. He took all the bananas in one attempt. Nobody knew about it. But one day the author found him doing so.

Question 7.
Kari learnt the commands to sit and to walk. What were the instructions for each command ?
(Imp.)
Answer.
The command to sit was to say ‘Dhať and pull Kari by the ear. The command to walk was to say ‘Mali’ and pull his trunk forward.

Question 8.
What is “the master call” ? Why is it the most important signal for an elephant to learn ?
(V. Imp.)
Answer.
To teach the master call to an elephant is the most difficult thing. Yet it is the most important signal for an elephant to learn. The master call is a strange hissing howling sound. It seems as if a snake and a tiger were fighting.
A trained elephant can be given the master call in its ears in a forest. The elephant knows that his master has lost his way. He therefore starts uprooting trees. Thus, he creates a path through the forest to the master’s house.

We hope the NCERT Solutions for Class 7 English An Alien Hand Chapter 2 Bringing up Kari help you. If you have any query regarding NCERT Solutions for Class 7 English An Alien Hand Chapter 2 Bringing up Kari, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 English Honeycomb Chapter 3 Gopal and the Hilsa-Fish are part of NCERT Solutions for Class 7 English. Here we have given NCERT Solutions for Class 7 English Honeycomb Chapter 3 Gopal and the Hilsa-Fish.

NCERT Solutions for Class 7 English Honeycomb Chapter 3 Gopal and the Hilsa-Fish

TEXTUAL QUESTIONS

Working with the Text (Page 42)
Answer the following questions :
Question 1.
Why did the king want no more talk about the hilsa-fish ?  (Imp.)
Answer.
There was so much talk about the hilsa-fish that the king was fed up with it. So he wanted no more talk about hilsa-fish.

Question 2.
What did the king ask Gopal to do to prove that he was clever ?
Answer.
The king asked Gopal to buy a huge hilsa and bring that to the palace without anyone asking him a word about it.

Question 3.
What three things did Gopal do before he went to buy his hilsa-fish ?   (Imp.)
Answer.
Gopal half shaved his face, smeared ash and wore rags before he went to buy a hilsa-fish.

Question 4.
How did Gopal get inside the palace to see the king after he had bought the fish ?
Answer.
At first, the gatekeeper would not let Gopal in. But Gopal began to dance and sing loudly. The king heard the noise. He asked the man making the noise to be brought before him. Thus Gopal got inside the palace to see the king after he had bought the fish.

Question 5.
Explain why no one seemed to be interested in talking about the hilsa-fish which Gopal had bought. (Imp.)
Answer.
Gopal seemed more worth talking about than the fish in his hand or anything else. So no one seemed to be interested in talking about the hilsa-fish which Gopal had bought.

Question 6.
Write ‘True’ or False’ against each of the following sentences.
(i) The king lost his temper easily.
(ii) Gopal was a madman.
(iii) Gopal was a clever man.
(iv) Gopal was too poor to afford decent clothes.
(v) The king got angry when he was shown to be wrong.
Answers.

1.  True
2.  False
3.  True
4.  False
5.  False

Working with Language
Question 1.
Notice how in a comic book, there are no speech marks when characters talk. Instead what they say is put in a speech ‘bubble’. However, if we wish to repeat or ‘report’ what they say, we must put it into reported speech.
Change the following sentences in the story to reported speech. The first one has been done for you.
(i) How much did you pay for that hilsa ?
The woman asked the man how much he had paid for that hilsa.
(ii) Why is your face half-shaven ?
Gopal’s wife asked him ………………….
Answer.
…….. why his face was half-shaven.
(iii)I accept the challenge. Your Majesty.
Gopal told the king ………………
Answer.
…….. respectfully that he accepted the challenge.
(iv) I want to see the king.
Gopal told the guards ……………
Answer.
……. that he wanted to see the king.
(v) Bring the man to me at once.
The king ordered the guard ………………
Answer.
…… to bring the man to him at once.

Question 2.
Find out the meaning of the following words by looking them up in the dictionary. Then use them in sentences of your own.

Answers.

Picture Reading
Question 1.
Look at the pictures and read the text aloud.
Answer.
Do it yourself.

Question 2.
Now ask your partner questions about each picture.
(i) Where is the stag ?
(ii) What is he doing ?
(iii) Does he like his antlers (horns) ?
(iv) Does he like his legs ?
(v) Why is the stag running ?
(vi) Is he able to hide in the bushes ?
(vii) Where Eire the hunters now ?
(viii) Are they closing in on the stag ?
(ix) Is the stag free ?
(x) What does the stag say about his horns and his legs ?
Answers.

1.  The stag is by the side of a pond.
2.  He is drinking water.
3.  He likes his antlers. They are very beautiful.
4.  He does not like his legs. They are thin and ugly.
5.  The stag is running as he has seen hunters.
6.  No. He is not able to hide in the bushes.
7.  The hunters are too close for safety of the stag.
8.  Yes. they are closing in on the stag.
9.  Yes. The stag is free.
10. The stag says that he was proud of his horns which could cause his death.
    About his legs, he says, he was ashamed but the same legs saved him.

Question 3.
Now write the story in your own words. Give it a title.
Answer.

Title : The Proud Stag

There lived a stag in a certain forest. Once while drinking water, he saw his image in the pond. He liked his beautiful horns. Then he saw his legs. They were veiy thin. They looked ugly. He felt proud of his horns and sad about his feet. Just then he saw some hunters. He tried to hide in bushes. His whole body was hidden but his horns showed him. The hunters came too close for safety. The stag ran for life. His legs saved him. He understood his folly. He was proud of his horns but they could cause his death. He was ashamed of his legs but they saved his life.

Question 4.
Complete the following word ladder with the help of the clues given below:

Clues

1.  Mother will be very ……….. if you don’t go to school.
2.  As soon as he caught ………. of the teacher, Mohan started writing.
3.  How do you like my ……….. kitchen garden ? Big enough for you, is it ?
4.  My youngest sister is now …….. old.
5.  Standing on the ………. , he saw children playing on the road.
6.  Don’t make such a ………. . Nothing will happen.
7.  Don’t cross the ……… till the green light comes on.

Answer.

We hope the NCERT Solutions for Class 7 English Honeycomb Chapter 3 Gopal and the Hilsa-Fish help you. If you have any query regarding NCERT Solutions for Class 7 English Honeycomb Chapter 3 Gopal and the Hilsa-Fish, drop a comment below and we will get back to you at the earliest.

Class 12 Chemistry NCERT Solutions Chapter 14 provides a detailed insight into the concepts related to the chapter “Biomolecules”. Chemistry is an important subject and the students need to be thorough with its concepts if they are preparing for boards or NEET and JEE.

NCERT Solutions also contains solutions to the questions provided in the textbook. The students can use these solutions to answer in the exam and score well.

NCERT Solutions for Class 12 Chemistry Chapter 14 Biomolecules

Biomolecules is a very important chapter and requires detailed understanding of the concepts. This topic is often asked in the examination and the students need to be thorough with the concepts.

The students can learn the structure, properties and classification of various biomolecules such as carbohydrates, nucleic acids, etc. These concepts are also taught in further studies. Therefore, the students need to be thorough with the basics.

NCERT INTEXT QUESTIONS

Question 1.
Glucose and sucrose are soluble in water but cyclohexane and benzene (simple six-membered ring compounds) are insoluble in water. Explain.
Answer:
Both glucose (C₆H₁₂O₆) and sucrose (C₁₂H₂₂O₁₁) are organic compounds and are expected to be insoluble in water. But quite surprisingly, they readily dissolve in water. This is due to the presence of a number of OH groups (five in the case of glucose and eight in sucrose) which are of polar nature. These are involved in the intermolecular hydrogen bonding with the molecules of H₂O (water). As a result, both of them readily dissolve in water.
Benzene (C₆H₆) and cyclohexane (C₆H₁₂) are hydrocarbons which don’t have any polar group. They, therefore, don’t dissolve in water since there is hardly any scope of hydrogen bonding in their molecules with those of H₂O (water).

Question 2.
What are the expected products of hydrolysis of lactose?
Answer:
The hydrolysis of lactose (disaccharide) can be done either with dilute HC1 or with enzyme emulsin. D-glucose and D-galactose are the products of hydrolysis. Both of them are monosaccharides with the molecular formula C6Hi₂0₆.

Question 3.
How do you explain the absence of an aldehydic group in the pentaacetate of D-glucose?
Answer:
Glucose, as we know is an aldohexose and it is expected to give the characteristic reactions of the aldehydic group e.g., action with NH₂OH, HCN, Tollen’s reagent, Fehling reagent etc. However, the pentadactyl glucose formed by the acylation of glucose with acetic anhydride does not give these reactions.

This means that the aldehydic group is either absent or is not available in the penta acetyl glucose for chemical reactions. In fact, the aldehydic group is a part of the hemiacetal structure which the penta acetyl derivative has. It is, therefore, not free or available to take part in these reactions.

Question 4.
The melting points and solubility of amino acids in water are generally higher than those of corresponding haloacids. Explain.
Answer:
The amino acids exist as zwitter ions, H₃N⁺ — CHR-COO-. Due to this dipolar salt like character, they have strong dipole-dipole attractions. Therefore, their melting points are higher than corresponding haloacids which do not have salt-like character.
Due to their salt-like character, amino acids interact strongly with water. As a result, their solubility in water is higher than corresponding haloacids which do not have a salt-like character.

Question 5.
Where does the water in the egg go after boiling the egg?
Answer:
Upon boiling the egg, denaturation of globular protein present in it occurs. Water present probably gets either absorbed or adsorbed during denaturation and disappears.

Question 6.
Explain why vitamin C can not be stored in the body.
Answer:
Vitamin C is mainly ascorbic acid which is water-soluble. It is readily excreted through urine and cannot be stored in the body.

Question 7.
What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
Answer:
The products obtained are 2-deoxy-D-ribose, phosphoric acid, and thymine.

Question 8.
When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained? What does this fact suggest about the structure of RNA?
Answer:
As we know, a molecule of DNA has a double-strand structure, and the four complementary bases pair each other. Cytosine (C) always pairs up with guanine (G) while thymine (T) is paired up with adenine (A). Because of the presence of the double-strand structure, when a molecule of DNA is hydrolysed, in each pair the molar ratio of the bases remains the same. However, this is not seen when RNA is subjected to hydrolysis. This suggests that RNA has not a double-strand structure like DNA. It exists as a single strand.

NCERT Exercises

Question 1.
What are monosaccharides?
Answer:
Monosaccharides are carbohydrates which cannot be hydrolysed to smaller molecules. Their general formula is (CH₂O)_n where n = 3 → 7. These are of two types. Those which contain an aldehyde (-CHO) group are called aldose and those which contain a keto (C = 0) group are called ketose. They are further classified as triose, tetrose, pentose etc. according to the no. of carbon atoms present (3, 4, 5 respectively).

Question 2.
What are reducing sugars?
Answer:
Carbohydrates which reduce Fehling’s solution to red precipitate of Cu₂0 or Tollen’s reagent to metallic Ag are called reducing sugars. All monosaccharides (both aldoses and ketoses) and disaccharides except sucrose are reducing sugars. Thus, D – (+) – glucose, D-(-)-fructose, D – (+) – maltose and D – (+) – lactose are reducing sugars.

Question 3.
Write two major functions of carbohydrates in plants. (C.B.S.E. Delhi 2008)
Answer:

1. Structural material for plant cell walls: The polysaccharides cellulose acts as the chief structural material of the plant’s cell walls.
2. Reserve food material: The polysaccharide starch is the major reserve food material in the plants. It is stored in seeds and acts as the reserve food material for the tiny plant till its capable of making food on its own by photosynthesis.

Question 4.
Classify the following into monosaccharides and disaccharides:
Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
Answer:
Monosaccharides: Ribose, 2-deoxyribose, galactose, fructose.
Disaccharides: Maltose, lactose.

Question 5.
What do you understand by the term glycosidic linkage?
Answer:
Glycosidic linkage is used to link different monosaccharides in disaccharides and polysaccharides through an oxygen atoms. For example, the glycosidic linkage is present in sucrose, lactose, maltose, etc. These are all disaccharides.

Question 6.
What is glycogen? How is it different from starch?
Answer:

1. The carbohydrates are stored in the animal body as glycogen. It is present in the liver, muscles, and brain. Enzymes break the glycogen down to glucose when the body needs glucose.
2. Glycogen is more highly branched than amylopectin (starch) glycogen chain consists of 10-14 glucose units, whereas amylopectin (starch) glycogen chain consists of 20-25 glucose units.

Question 7.
What are the hydrolysis products of starch and lactose?
Answer:
Starch upon hydrolysis gives α – D(+) glucose which is the constituent of both amylose and amylopectin. Lactose upon hydrolysis gives galactose and glucose.
Upon hydrolysis, cellulose gives only D(+) glucose. This means that only D(+) glucose units are present in cellulose but unlike starch these are -D(+) glucose molecules and not a – D(+) glucose molecules. The X – ray analysis has shown that there are large linear chains of 3 – D( +) glucose molecules lying side by side in the form of bundles held together by hydrogen bonding in the neighbouring hydroxyl groups. 

Question 8.
What is the basic structural difference between starch and cellulose?
Answer:
Starch is not a single component. It consists of amylose and amylopectin. In contrast, cellulose is a single compound. Amylose is a linear polymer of α – D glucose while cellulose is a linear polymer of β -D glucose. In amylose, C1 – C4 α- glycosidic linkage is present, whereas in cellulose C1 – C4 β- glycosidic linkage is present. Amylopectin has a highly branched structure.

Question 9.
What happens when D-glucose is treated with
(i) HI
(ii) Bromine water
(iii) HNO₃?
Answer:
(i) Reaction with HI

(ii) Reaction with bromine water

(iii) Reaction with HNO₃

Question 10.
Enumerate the reactions of D-glucose which cannot be explained by its open-chain structure.
Answer:
The following reactions of D- glucose cannot be explained by its open-chain structure.
1. D – the glucose does not undergo certain characteristic reactions of aldehydes. For example, glucose does not form NaHSO₃ addition product, aldehyde-ammonia adduct, 2, 4, DNP derivative and does not respond to Schiff’s reagent test.

2. Glucose reacts with NH₂OH to form an oxime but glucose pentaacetate does not. This implies that the aldehyde group is absent in glucose pentaacetate.

3. D (+) – Glucose exists in two stereoisomeric forms ie. α – glucose and β- glucose, α – D (+) – glucose is obtained when a concentrated aqueous or alcoholic solution is crystallised at 303K. It has a melting point of 419K and has a specific rotation of +111° in a freshly prepared aqueous solution. However when glucose is crystallised from the water above 371 K β – D (+) glucose is obtained.

4. Both α – D glucose and β – D glucose undergoes mutarotation in an aqueous solution.

Question 11.
What are essential and non-essential amino acids?
Answer:
α – Amino acids which are needed for the health and growth of human beings but are not synthesised by the human body are called essential amino acids. For example, valine, leucine phenylalanine etc. On the other hand α – amino acids which are needed for the health and growth of human beings and are synthesised by the human body are called non-essential amino acids. For example glycine, alanine, aspartic acid etc.

Question 12.
Define the following as related to proteins
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation.
Answer:
(i) Peptide Linkage: Proteins are the polymers of a-amino acids which are connected to each other by peptide bond or peptide linkage. Chemically, peptide linkage is an amide formed between -COOH group and -NH2 group. The reaction between two molecules of similar or different amino acids, proceeds through the combination of the amino group of one molecule with the carboxyl group of the other This results in the elimination of a water molecule and formation of a peptide bond -CO-NH-. The product of the reaction is called a dipeptide because it is made up of two amino acids. For example, when carboxyl group of glycine combines with the amino group of alanine we get a dipeptide, glycylalanine.

(ii) Primary Structure: Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence and it is this sequence of amino acids that is said to be the primary structure of that protein. Any change in this primary structure i.e.; the sequence of amine acids creates a different protein.

(iii) Denaturation: Protein found in a biological system with a unique three-dimensional structure and biological activity is called a native protein. When a protein in its native form, is subjected to physical change like change in temperature or chemical change in pH, the hydrogen bonds are disturbed. Due to this; globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein. During denaturation 2° and 3° structures are destroyed but 1° structure remains intact. The coagulation of egg white on boiling is a common example of denaturation. Another example is the curdling of milk which is caused due to the formation of lactic acid by the bacteria present in milk.

Question 13.
What are the common types of secondary structures of proteins?
Answer:
Secondary structure of a protein refers to the shape in which a long polypeptide chain can exist. These are found to exist in two types :

• α-helix structure
• β-pleated sheet structure.

Secondary Structure of Proteins:
The long, flexible peptide chains of proteins are folded into the relatively rigid regular conformations called the
secondary structure. It refers to the conformation which the polypeptide chains assume as a result of hydrogen bonding
between the > C= O and > N-H groups of different peptide bonds.
The type of secondary structure a protein will acquire, in general, depends upon the size of the R-group. If the size of the
R-groups are quite large, the protein will acquire a ct-helix structure. If on the other hand, the size of the R-groups are relatively
smaller, the protein will acquire a β – flat sheet structure.

(a) α-Helix structure: If the size of the R-groups is quite large, the hydrogen bonding occurs between > C = O group
of one amino acid unit and the > N-H group of the fourth amino acid unit within the same chain. As such the polypeptide
chain coils up into a spiral structure called right-handed ct- helix structure. This type of structure is adopted by most of the
fibrous structural proteins such as those present in wool, hair, and muscles. These proteins are elastic i.e., they can be
stretched. During this process, the weak hydrogen bonds causing the α- helix are broken. This tends to increase the length of
the helix like a spring. On releasing the tension, the hydrogen bonds are reformed, giving back the original helical shape.

(b) β—Flat sheet or β—Pleated sheet structure: If R-groups are relatively small, the peptide chains lie side by side in a zig
zag manner with alternate R-groups on the same side situated at fixed distances apart. The two such neighbouring chains are held together by intermolecular hydrogen bonds. A number of such chains can be inter-bonded and this results in the formation of a flat sheet structure These chains may contract or bend a little in order to accommodate moderate-sized R-groups. As a result, the sheet bends into parallel folds to form a pleated sheet structure known as β – pleated sheet structure. These sheets are then stacked one above the other like the pages of a book to form a three-dimensional structure. The protein fibrion in silk fibre has a β – pleated sheet structure. The characteristic mechanical properties of silk can easily be explained on the basis of its β – sheet structure. For example, silk is non-elastic since stretching leads to pulling the peptide covalent bonds. On the other hand, it can be bent easily like a stack of pages because, during this process, the sheets slide over each other.

Question 14.
What type of bonding helps in stabilising the α-helix structure of proteins?
Answer:
The α-helix structure of proteins is stabilized by intramolecular H-bonding between C = O of one amino acid residue and the N – H of the fourth amino acid residue in the chain. This causes the polypeptide chain to coil up into a spiral structure called the right-handed α- helix structure.

Question 15.
Differentiate between globular proteins and fibrous proteins. (Jharkhand Board 2014; C.B.S.E. Delhi 2015)
Answer:

Question 16.
How do you explain the amphoteric behaviour of amino acids?
Answer:
Amino acids have basic (NH₂) and acidic (COOH) groups. These are, therefore, amphoteric in nature. However, they exhibit these characters in the amino acid molecule itself i.e., NH₂ group (basic) accepts a proton from COOH group (acidic) in the same molecule. Therefore, a-amino acid exists as a dipolar ion.

Question 17.
What are enzymes?
Answer:
We have learned in the study of carbohydrates that these are body fuels i.e., they provide the necessary energy to the body and
keeps it working. Actually, the human body is just like a furnace in which chemical reactions take place and are responsible for the digestion of food, absorption of appropriate molecules, and production of energy. The entire process involves a series of reactions that are catalyzed by biocatalysts known as enzymes. Thus, enzymes may be defined as:
biological or biocatalysts which catalyse the reactions in living beings.
All enzymes are basically globular proteins. Enzymes are very specific for a particular reaction as well as for a particular
substrate. These are generally named after the compounds or the class of substances with which they are linked or work.

Question 18.
What is the effect of denaturation on the structure of proteins?
Answer:
As a result of denaturation, globules get unfolded and helixes get uncoiled. Secondary and tertiary structures of the protein are destroyed, but the primary structures remain unaltered. It can be said that during denaturation, secondary and tertiary – structured proteins get converted into primary – structured proteins. Also, as the secondary and tertiary structures of a protein are destroyed, the enzyme loses its activity.

Question 19.
How are vitamins classified? Name the vitamin responsible for the coagulation of blood? (C.B.S.E. Outside Delhi 2015)
Answer:
On the basis of their solubility in water or fat, vitamins are classified into two groups:
1. Fat-soluble vitamins:
Vitamins that are soluble in fat and oils, but not in the water, belong to their group.
For example Vitamins A, D, E, and K.

2. Water-soluble vitamins:
Vitamins that are soluble in water belong to their group.
For example, B group vitamins (B₁, B₂, B₆, B₁₂, etc) and vitamin C.
However, biotin or vitamin H is neither soluble in water nor in fat.
Vitamin K is responsible for the coagulation of blood.

Question 20.
Why are vitamin A and vitamin C essential to us? Mention their sources.
Answer:
vitamin A: Soluble in water but insoluble in oils and fas. Destroyed by cooking or prolonged exposure to air. it increases the resistance of the body towards diseases. Maintains healthy skin and helps in the healing of cuts and abrasions. It is available in

vitamin C:  Soluble in oils and fats but insoluble in water, stable to heat. Promotes growth and improves vision. ¡t also increases resistance to disease. It is available in Citrus fruits (oranges, lemon, grapefruit, lime, etc.), amia, cabbage. guava etc.

Question 21.
What are nucleic acids? Mention their two important functions.
Answer:
Nucleic acids are biologically important polymers which are present in all living cells.

• Nucleic acids play a vital role in the transmission of heredity characteristics.
• Nucleic acids help in the biosynthesis of proteins.

Two important biological functions of nucleic acids are (i) Replication and (ii) protein synthesis.
These are briefly discussed.
Replication: Replication may be defined as the process by which a single DNA molecule produces two identical copies of itself.

Replication is an enzyme-catalyzed process.  The process of replication starts with the partial unwinding of the two strands of the DNA double helix through the breaking of the hydrogen bonds between pairs of bases. Each strand then acts as the template (or pattern) for the synthesis of two new strands of DNA in the celllix environment. The specificity of base-pairing ensures that each new strand is complementary to its old template strand. As a result, two identical copies of DNA from the original DNA are produced. Each of these two copies is then passed on to the two new cells resulting from cell division. In this way, hereditary characters are transmitted from one cell to another.

Question 22.
What is the difference between a nucleoside and a nucleotide?
Answer:
A nucleoside contains a pentose sugar and base (purine or pyrimidine) while in the nucleotide, a phosphoric acid component is also present.
Arrangement of constituents in Nucleic Acids. These are, intact, three building blocks in nucleic acid. A combination ol
base and sugar are known as a nucleoside. Similarly, base, sugar, and phosphates from nucleotides while nucleic acids are
polynucleotides which means that these are the polymers of nucleotides.

Question 23.
The two strands in DNA are not identical but are complementary. Explain.
Answer:
In the helical structure of DNA, the two strands are held together by hydrogen bonds between specific pairs of bases. Cytosine forms a hydrogen bond with guanine, while adenine forms a hydrogen bond with thymine. As a result, the two strands are complementary to each other.

Question 24.
Write the important structural and functional differences between DNA and RNA.
Answer:

Question 25.
What are the different types of RNA found in the cell?
Answer:

• Messenger RNA (m – RNA)
• Ribosomal RNA (r – RNA)
• Transfer RNA (t – RNA)

We hope the NCERT Solutions for Class 12 Chemistry Chapter 14 Biomolecules help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 14 Biomolecules, drop a comment below and we will get back to you at the earliest.

Class 12 NCERT Solutions for Chemistry Chapter 13 contains solved answers for the questions provided in the textbook. These answers are provided by the subject experts and are accurate to the best of our knowledge. The students can use these answers along with the diagrammatic representation in the examination and score well.

NCERT Solutions are beneficial for the students appearing for UP board, MP Board, CBSE, Gujarat board, etc., and also for competitive exams such as NEET and JEE.

NCERT Solutions for Class 12 Chemistry Chapter 13 Amines

This chapter explains the importance, structure, physical and chemical properties, and the methods of preparation of amines. This chapter also explains the diazonium salts and the methods of their preparation. This will help you to understand the importance of these salts in the synthesis of aromatic compounds.

The students can use NCERT Solutions for better preparations of the concepts mentioned in this chapter. Basic understanding is very important for advanced concepts.

NCERT INTEXT QUESTIONS

Question 1.
Classify the following amines as primary, secondary, and tertiary amines

Answer:
(i) primary
(ii) tertiary
(iii) primary
(iv) secondary.

Question 3.
(i)Write the structures of different isomeric amines corresponding to the molecular formula, C₄H₁₁N.
(ii) Write 1UPAC names of all the isomers.
(iii) What type of isomerism is exhibited by different pairs of amines?
Answer:

Chain isomers: (a) and (c); (a) and (d); (b) and (c); (b) and (d)
Metamers: (e) and (g); (f) and (g)
Functional isomers: All 10 amines are functional isomers of 2° and 3° amines and vice-versa.

Question 3.
How will you convert:
(i) Benzene into aniline
(ii) Benzene into N, N-dimethylaniline
(iii) C1(CH₂)₆Cl into hexane-1, 6-diamine
Answer:
(i) Benzene into aniline:

(ii) Benzene into N, N-dimethylaniline:

(iii) Cl(CH₂)₆Cl into hexane-1, 6-diamine:

Question 4.
Arrange the following in increasing order of their basic strength:
(i) C₂H₅NH₂,C₆H₅NH₂,NH₃,C₆H₅CH₂NH₂ and(C₂H₅)₂ NH
(ii) C₂H₅NH₂,(C₂H₅)₂NH,(C₂H₅)₃N,C₆H₅NH₂
(iii) CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂, C₆H₅CH₂NH₂
Answer:
(i) C₆H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ <  C₂H₅NH₂<(C₂H₅)₂NH
(ii) C₆H₅NH₂ < C₂H₅NH₂ < (C₂H₅)₃N <(C₂H₅)₂NH
(iii) C₆H₅NH₂ < C₆H₅CH₂NH₂ < (CH₃)₃ N < CH₃NH₂<(CH₃)₂NH

Question 5.
Complete the following acid-base reactions and name the products
(i) CH₃CH₂CH₂NH₂ + HC1 →
(ii) (C₂H₅)₃N + HC1  →
Answer:

Question 6.
Write the reactions of the final alkylation product of aniline with excess methyl iodide in the presence of sodium carbonate solution.
Answer:
Aniline is a primary amine. It will react with excess methyl iodide to form quaternary ammonium salt as the final product. The reaction is known as Hoffmann’s ammonolysis.

Question 7.
Write the chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Answer:
Aniline will undergo benzoylation to form a benzoyl derivative. The reaction will take place in the presence of aqueous alkali.

Question 8.
Write the structures of the different isomers corresponding to the molecular formula C₃H₉N. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid.
Answer:
Four isomeric aliphatic amines are represented by the molecular formula C₃H₉N. These are:

Only the primary amines will evolve N₂ gas on reacting with nitrous acid (HONO) and form corresponding primary alcohols.

Question 9.
How will you convert:
(i) 3-Methylaniline into 3-nitrotoluene
(ii) Aniline into 1,3,5-tribromobenzene ?
Answer:
(i) 3-Methylaniline into 3-nitrotoluene

(ii) Aniline into 1,3,5-tribromobenzene

NCERT EXERCISE

Question 1.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH₃)₂CHNH₂
(ii) CH₃(CH2)₂NH₂
(iii) CH₃NHCH(CH₃)₂
(iv) (CH₃)₃CNH₂
(v) C₆H₅NHCH₃
(vi) (CH₃CH₂)₂NCN₃
(vii) m-BrC₆H₄NH₂
Answer:

Question 2.
Give one chemical test to distinguish between the following pairs of compounds :
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-methylaniline. (C.B.S.E. Sample Paper 2015)
Answer:
(i) Methylamine on reaction with nitrous acid evolves N₂ gas with brisk effervescence while dimethylamine does not. Methylamine also gives carbylamine reaction upon warming with chloroform and alcoholic KOH while dimethylamine does not.
(ii) Secondary amines, both aliphatic and aromatic respond to Libermann’s nitroso reaction while tertiary amines do not.
(iii) Aniline responds to diazotisation and coupling reactions to form a dye while ethylamine does not.
(iv) Aniline gives diazotisation coupling reaction while benzylamine does not.
(v)  Aniline gives carbyl amine test with an extremely unpleasant smell while N-Methyl aniline does not.

Question 3.
Account for the following :
(i) pK_b of aniline is more than that of methylamine. (C.B.S.E. Delhi 2008, 2011)
(ii) Ethylamine is soluble in water whereas aniline is not. (C.B.S.E. Delhi 2008, 2011)
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide. (C.B.S.E. Delhi 2008)
(iv) Although the amino group is o- and p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline. (C.B.S.E. Sample Paper 2010)
(v) Aniline does not undergo Friedel Crafts reaction. (C.B.S.E. Delhi 2008, Sample Paper 2010, C.B.S.E. Outside Delhi 2015)
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Answer:
(i) pK_b of aniline is more than that of methylamine because aniline is less basic. In aniline, the electron pair on the nitrogen atom is involved in conjugation with the ring and is less available for protonation than in methylamine. Therefore, aniline has more pK_b.

(ii) Ethylamine is water-soluble due to hydrogen bonding. However, in aniline, the phenyl (C₆H₅) group is bulky in size and has -I effect. As a result, its hydrogen bonding with water is negligible and is therefore not soluble or miscible with water.

(iii) Methylamine forms a soluble hydroxide on reacting with water. The OH^– ions released by the hydroxide combine with  Fe³⁺ ions of ferric chloride to give ferric hydroxide or hydrated ferric oxide which is brown in colour.

(iv) Amino group (-NH₂) is an electron releasing or activating group when present on the benzene ring. It activates the ortho and para positions in the ring towards electrophilic substitution due to its +M or +R effect. The nitration of aniline carried by nitrating mixture (cone. HN0₃ + cone. H₂SO₄) is electrophilic in nature. The expected product of nitration is a mixture of ortho and para nitroaniline. However, in this case, a substantial amount of metanitroaniline is also formed. In fact, aniline being a base gets protonated in the acidic medium to form anilinium cation which is no longer activating. Rather, it is deactivating in nature and deactivates the ring. The substitution takes place at the meta position.

Thus, the nitration of aniline as such gives a significant amount of m-nitroaniline (47%). In addition to this, p-nitroaniline
is the major constituent (51%) while ortho isomer is in negligible amount (2%) mainly due to the reason that the ortho position
is sterically hindered because of the —NH2 group.

In order to check the activation of the ring by an amino group, the nitration of aniline is carried out indirectly by first
acetylating with acetic anhydride (or acetyl chloride) to form acetanilide. The compound formed is nitrated by the nitrating
mixture and the isomeric nitro derivatives are then hydrolysed in the acidic medium as discussed under halogenation.

(v) Aniline does not undergo Friedel Crafts reaction. Actually, aniline being a Lewis base forms a complex with AICI3 which is a Lewis acid. The amino group is not in a position to activate the benzene ring towards electrophilic substitution i.e., alkylation or acylation which leads to Friedel Crafts reaction. Therefore, the reaction is not possible. The same problem arises in phenols as well.

(vi) The diazonium salts of aromatic amines are more stable than those of aliphatic amines because these are resonance stabilised while no such resonance stabilisation is possible in the corresponding diazonium salts of aliphatic amines.

(vii) Gabriel phthalimide synthesis is generally preferred over other methods for the synthesis of primary aliphatic amines. Potassium phthalimide formed by reacting phthalimide with alcoholic KOH reacts with an alkyl halide such as C₂H₅-I to form N-alkyl derivative which undergoes hydrolysis to form the primary amine. However, no reaction is possible with aryl halide such as C₆H₅-I. Therefore, primary aromatic amines are not formed in the reaction.

Question 4.
Arrange the following:
(a) In decreasing order of the pK_b values:
C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂

(b) In decreasing order of basic strength:
C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH and CH₃NH₂ (C.B.S.E. Delhi 2011, Haryana Board 2013, C.B.S.E. Outside Delhi 2015)

(c) Increasing order of basic strength
Aniline, p-nitroaniline and p-toluidine

(d) Decreasing order of basic strength in gas phase
C₂H₅NH₂, (C₂H_s)₂NH, (C₂H5)₃N and NH₃

(e) Increasing order of boiling point
C₂H₅OH, (CH₃)₂NH, C₂H_sNH₂

(f) Increasing order of solubility in water
C6H₅NH₂, (C₂H_s)₂NH, C₂H5NH₂.
Answer:
From K_b  and PK_b values of some Amines:

(a) The decreasing order of pK_b values or increasing order of basic strength is:
C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂NH

(b) The decreasing order of basic strength is:
(C₂H₅)₂NH > CH₃NH₂ > C₆H₅N(CH₃)₂ > QH5NH2

(c) The increasing order of basic strength is:
p-nitroaniline < Aniline < p-Toluidine

(d) The decreasing order of basic strength in gaseous phase.
(C₂H₅)₃N > (C₂H₅)₂NH > C₂H₅NH₂ > NH₃

(e) Increasing order of boiling point is:
(C₂H₃)₂NH < C₂H₅NH₂ < C₂H₅OH

(f) Increasing order of solubility in water is :
C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂.

Question 5.
How will you convert
(i) Ethanoic acid to methanamfrie,
(ii)Hexanenitrile to pentan-l-amine,
(iii)Methanol to ethanoic acid,
(iv) thatfaminft to methanamine,
(v)Ethanoic acid to propanoic acid,
(vi)Methanamine to ethanamine,
(vii)Nitromethane into dimethylamine,
(viii)Propanoic acid into ethanoic acid ?
Answer:
(i) Ethanoic acid to methanamfrie

(ii) Hexanenitrile to pentan-l-amine

(iii) Methanol to ethanoic acid

(iv) thatfaminft to methanamine

(v) Ethanoic acid to propanoic acid

(vi) Methanamine to ethanamine

(vii) Nitromethane into dimethylamine

(viii) Propanoic acid into ethanoic acid

Question 6.
Describe a method for the identification of primary, secondary, and tertiary amines. Also, write chemical equations of the reactions involved.
Answer:
The three type of amines can be distinguished by Hinsberg test. In this test, the amine is shaken with benzene sulphonyl chloride (C₆H₅SO₂Cl) in the presence of excess of aqueous NaOH or KOH. A primary amine reacts to give a clear solution, which on acidification yields an insoluble compound.

Question 7.
Write short notes on the following:
(i) Carbylamine reaction
(ii) Diazotisation
(iii) Hoffmann’s bromamide reaction
(iv) Coupling reaction
(v) Ammonolysis
(vi)Acetylation
(vii)Gabriel phthalimide synthesis. (C.B.S.E. Delhi 2011, C.B.S.E. Outside Delhi 2015)
Answer:
(i) Carbyl amine reaction:
Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines which are foul-smelling substances. Secondary and tertiary amines do not show this reaction. This reaction is known as carbylamine reaction or isocyanide test and is used as a test for primary amines.

(ii) Diazotisation:
Primary aromatic amines such as aniline react with nitrous acid under ice-cold conditions (273 – 278 K) to form benzene diazonium salt. The reaction is known as diazotisation reaction.

In case, the temperature is allowed to rise above 278 K, benzene diazonium chloride is decomposed by water to form phenol.

Aliphatic primary amines also react with nitrous acid to form alkyl diazonium salts in a similar manner. But these are quite unstable and decompose to form a mixture of alcohols, alkenes, and alkyl halides along with the evolution of N₂ gas.

Aliphatic primary amines also react with nitrous acid to form alkyl diazonium salts in a similar manner. But these are quite unstable and decompose to form a mixture of alcohols, alkenes, and alkyl halides along with the evolution of N₂ gas.

(iii) Hoffmann’s bromamide reaction:
By Hoffmann degradation of Acid Amides. (Hoffmann Bromamide Reaction). When a primary acid amide is heated with an aqueous or ethanolic solution of sodium hydroxide and bromine, it gives a primary amine with one carbon
atom less.

The reaction is, therefore, regarded as a degradation reaction. For example.

(iv) Coupling reaction:
The reaction of diazonium salts with phenols and aromatic amines to form azo compounds having an extended conjugated system with both aromatic rings joined through the — N = N — bond, is called coupling reaction. In this reaction; the nitrogen atoms of the diazo group are retained in the product. The coupling with phenols takes place in a mildly alkaline medium while that with amines occurs under faintly acidic conditions. For example;

Coupling generally occurs at the p-position with respect to the hydroxyl or the amino group, if free, otherwise it takes place at the o-position.

(v) Ammonolysis:
The mechanism involves the nucleophilic attack of NH₃ molecule (through lone pair) on alkyl halide by an SN₂ mechanism. Amine salt is formed which reacts with ammonia to give primary amine and ammonium halide as follows:

The primary amine formed now acts as the nucleophile and reacts with another molecule of the alkyl halide to form secondary amine.

The reaction is repeated to form tertiary amine and quaternary ammonium salt as follows :

(vi) Acetylation:
Acylation of Amines Both aliphatic and aromatic amines form acyl derivatives (substituted acid amides) with reagents such as acid chlorides, esters, or acid anhydrides. The acylation is carried out in the presence of a base stronger than pyridine (e.g., NaOH) which can remove the acid formed in the reaction by neutralising it.
(a) Acylation of Aliphatic Amines: Both primary and secondary aliphatic amines from acyl derivatives as follows:

(b) Acylation of Aromatic Amines: Aromatic amines such as aniline can be acylated in the same manner with both acid
chloride and acid anhydride.

(vii) Gabriel’s phthalimide synthesis:
In this reaction phthalimide is converted into its potassium salt by treating it with alcoholic potassium hydroxide. Then potassium phthalimide is heated with an alkyl halide to yield an N-alkylpthalimide which is hydrolysed to phthalic acid and primary amine by alkaline hydrolysis

This synthesis is very useful for the preparation of pure aralkyl and aliphatic primary amines. However, aromatic primary amines cannot be prepared by this method.

Question 8.
Accomplish the following conversions:
(i) Nitrobenzene to benzoic acid
(ii) Benzene to m-bromophenol
(iii) Benzoic acid to aniline
(iv) Aniline to 2, 4, 6-tribromofluorobenzene
(v) Benzyl chloride to 2-phenylethanamine
(vi) Chlorobenzene to p-chloroaniline
(vii) Aniline to p-bromoaniline
(viii)Benzamide to toluene
(ix)Aniline to benzyl alcohol.
Answer:
(i) Nitrobenzene to benzoic acid: 

(ii) Benzene to m-bromophenol: 

(iii) Benzoic acid to aniline:

(iv) Aniline to 2, 4, 6-tribromofluorobenzene: 

(v) Benzyl chloride to 2-phenylethanamine: 

(vi) Chlorobenzene to p-chloroaniline:

(vii) Aniline to p-bromoaniline: 

(viii)Benzamide to toluene :

(ix)Aniline to benzyl alcohol:

Question 9.
Give the structures of A, B, and C in the following reactions:

Answer:

Question 10.
An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br₂ and KOH forms a compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B, and C.
Answer:
Since the compound ‘C’ with molecular formula C₆H₇N is formed from compound ‘B’ on treatment with Br₂ KOH, therefore, compound ‘B’ must be an amide and ‘C’ must be an amine.
The only amine having the molecular formula C₆H₇N, i. e., C₆H₅NH₂ is aniline.
Since ‘C’ is aniline, therefore, the die amide from which it is formed must be benzamide (C₆H₅CONH₂). Thus, compound ‘B’ is benzamide. Since compound ‘B’ is formed from compound ‘A’ with aqueous ammonia and heating, therefore, compound ‘A’ must be benzoic acid.

Question 11.
Complete the following reactions:

Answer:

Question 12.
Why cannot aromatic amines be prepared by Gabriel’s phthalimide reaction? (C.B.S.E. Sample Question Paper 2012, H.P. Board2017)
Answer:
In Gabriel phthalimide reaction, the potassium salt of phthalimide is formed. It readily reacts with an alkyl halide to form the corresponding alkyl derivative.

But it is not in a position to react with the aryl halide in case primary aromatic amine is to be prepared. Actually, the cleavage of C – X bond in haloarene or aryl halide is quite difficult due to partial double bond character. Therefore, aromatic amines cannot be prepared by this method.

Question 13.
How do aromatic and aliphatic primary amines react with nitrous acid?
Answer:
Reaction with nitrous acid. All three types of amines, aliphatic as well as aromatic, react with nitrous acid under different conditions to form a variety of products. Since nitrous acid is highly unstable, it is prepared in situ by the action of dilute hydrochloric acid on sodium nitrite.

(a) Primary aliphatic amines react with nitrous acid at low temperature (cold conditions) to form primary alcohol and nitrogen gas accompanied by brisk effervescence. Nitrous acid is unstable in nature and is prepared in situ by reacting sodium nitrite with dilute hydrochloric acid. For example,

The reaction ¡s used as a rest for primary aliphatic amines as no other amine evolves nitrogen with nitrous acid.

(b) Primary aromatic amines such as aniline react with nitrous acid under ice cold conditions (273—278 K) to form benzene diazonium salt. The reaction is known as diazotisation reaction.

In case, the temperature is allowed to rise above 278 K, benzene diazonium chloride is decomposed by water to form phenol.

Aliphatic primary amines also react with nitrous acid to form alkyl diazonium salts in a similar manner. But these are quite unstable and decompose to form a mixture of alcohols, alkenes, and alkyl halides along with the evolution of N2 gas.

Question 14.
Give plausible explanation for each of the following :
(i) Why are amines less acidic than alcohols of comparable molecular masses ?
(ii) Why are primary amines higher boiling than tertiary amines ?
(iii) Why are aliphatic amines stronger bases than aromatic amines ? (H.P. Board 2008)
Answer:
(i) The acidic character in the both in cases is due to the release of H⁺ ion. Now, the anion in case of amine

has a negative charge on the nitrogen atom while the anion formed in case of alcohol has negative charge on the oxygen atom. Since oxygen is more electronegative than nitrogen atom, the negative charge can be accomodated easily on oxygen than on nitrogen in these anions. In other words, RO” ion is more stable than RNH“ ion. Consequently, alcohol is a stronger acid than amine. Please remember that even alcohols are very weakly acidic so much so that they donot turn blue litmus red.

(ii) Primary amines are higher boiling than tertiary amines due to the presence of intermolecular hydrogen bonding in their molecules. Since tertiary amines (R₃N) have no hydrogen atom present, these are not involved in any such hydrogen bonding. For example, the boiling point of n-butylamine (CH₃CH₂CH₂CH₂NH₂) is 322 K while that of trimethylamine (CH₃)₃ N is 276 K.

(iii) In the aromatic amines, (lie secondary and tertiary amines are more basic than aniline.
Actually, the basic strength or the electron releasing tendency of an amine depends upon the following factors.

1. The ability of the nitrogen atom to donate a pair of electrons.
2.  The stability of cation by accepting the pair of electrons.

Any factor which tends lo increase the electron releasing tendency of amine or increase the stability of the cation, will tend to increase the basic strength of amine.

We hope the NCERT Solutions for Class 12 Chemistry Chapter 13 Amines help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 13 Amines, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Chemistry Chapter 16 is the best study material for all the students. It not only prepares them for the board exams but also strengthens their concepts for competitive exams such as NEET and JEE.

The solutions are provided by subject mater experts and are accurate. The diagrammatic representations make it even easier for the students to understand. The students appearing for UP board, MP board, Gujarat board, CBSE find the NCERT Solutions beneficial while preparing for the exams.

NCERT Solutions for Class 12 Chemistry Chapter 16 Chemistry in Every Day Life

This chapter deals with the principles of chemistry in everyday life. All the products such as soaps, detergents have an organic composition. All our daily activities are controlled by chemicals. This chapter gives some interesting facts about the products we use in our daily lives and how are they controlled by the chemicals.

The students are advised to go through the NCERT solutions for Class 12 Chemstry for better understanding of the concepts provided in the chapter.

NCERT INTEXT QUESTIONS

Question 1:
Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it is not advisable to take their doses without consultation with the doctor. Why?
Answer:
Sleeping pills contain drugs which may be tranquilizers or antidepressants. They affect the nervous system and induce sleep. However, if these doses are not properly controlled, they may create havoc. They even adversely affect the vital organs of the body. It is advisable to take these sleeping pills under the strict supervision of a doctor.

Question 2.
With refrence to which classification has the statement “ranitidine is an antacid”, been given?
Answer:
This statement refers to the classification of drugs according to pharmacological effect because any drug which will be used to neutralise the excess acid present in the stomach will be called an antacid.

Question 3.
Why do we require artificial sweetening agents?
Answer:
The commonly used sweetening agent i.e., sucrose is a carbohydrate with molecular formula C₁₂H₂₂O₁₁. Since it has high calorific value, it is not recommended to the patients, diabetics in particular which require low calorie diet. Most of the artificial sweeteners are better than sucrose but hardly provide any calories to the body. These are being used as substitutes of sugar.

Question 4.
Write chemical equations for preparing sodium soap from glyceryl oleate and glyceryl palmitate. Structural formulas of these compounds are given:
(i) (C₁₅H₃₁COO)₃C₃H₅ (Glyceryl palmitate)
(ii) (C₁₇H₃₃COO)₃ C₃H₅ (Glyceryl oleate)
Answer:

Question 5.
Label the hydrophilic and hydrophobic parts in the following molecule which is a detergent. Also identify the functional group(s) present.

Answer:

NCERT Exercise

Question 1.
Why do we need to classify the drugs in different ways?
Answer:
Drugs have been classified in different ways depending

• upon their pharmacological effect
• upon their action on a particular biochemical process
• on the basis of their chemical structure
• on the basis of molecular targets.

For example, the classification of the drugs based on pharmacological effect is useful for doctors. The classification of drugs based on molecular targets is the most useful classification for medicinal chemists. Thus, drugs are classified in different ways to serve different purposes.

Question 2.
Explain the term, target molecules or drug targets as used in medicinal chemistry.
Answer:
In medicinal chemistry, drug targets refer to the key molecules involve in certain metabolic pathways that result in specific diseases Carbohydrates, proteins, lipids, and nucleic acids are examples of drug targets.
Drugs are chemical agents designed to inhibit these target molecules by binding with the active sites of the key molecules.

Question 3.
Name the macromolecules that are chosen as drug targets.
Answer:
The different macromolecules or biomolecules which are used as drug targets are carbohydrates, proteins, enzymes, nucleic acids. Out of these, enzymes are the most significant because their deficiency leads to many disorders in the body.

Question 4.
Why should not medicines be taken without consulting doctors?
Answer:
A medicine can bind to more than one receptor site, thus a medicine may be toxic for some receptor site Further, in most cases, medicines cause harmful effects when taken in a higher dose than recommended As a result, medicines may be poisonous in such cases Hence, medicines should not be taken without consulting doctors.

Question 5.
Define the term chemotherapy. (C.B.S.E. Delhi 2008)
Answer:
The branch of chemistry which deals with the treatment of diseases using chemicals is called chemotherapy.

Question 6.
Which forces are involved in holding the drugs to the active sites of enzymes?
Answer:
These are different intermolecular forces like dipolar forces, hydrogen bonding, van der Waals’ forces etc. The receptor targets have specific roles to perform. They help in transferring message from messengers to the cell. The messengers are in fact chemical compounds which are received by the active sites of the receptor proteins that project out of the surface. In order to accommodate these, the receptors may undergo a change in shape. The receptors are held by the active sites also called binding sites. Once the message is transferred to the cells.

Question 7.
While antacids and antiallergic drugs interfere with the function of histamines, why do these not interfere with the function of each other?
Answer:
They do not interfere with the functioning of each other because they work on different receptors in the body. Secretion of histamine causes allergy and acidity while antacid removes only acidity.

Question 8.
Low level of noradrenaline is the cause of depression. What type of drugs are needed to cure this problem ? Name two drugs (C.B.S.E. Outside Delhi 2008 Supp.)
Answer:
Low level of noradrenaline which acts as a neurotransmitter reduces the signal sending ability to the nerves and the patient suffers from depression. Antidepressants are needed to give relief from depression. These are also called tranquilizers or neurologically active drugs. The two specific drugs are iproniazid and phenelzine.

Question 9.
What is meant by the term ‘broad-spectrum antibiotic? Explain.
Answer:
Broad-spectrum antibiotics are the drugs which are effective against a large number of harmful micro-organisms causing diseases.
Chloramphenicol It is a broad-spectrum antibiotic, isolated in 1947. It is rapidly absorbed from the gastrointestinal tract and hence can be given orally. It is very effective against typhoid, dysentery, acute fever, certain form of urinary infections, meningitis and pneumonia.

Chloramphenicol is quite easy to synthesise. Therefore, most of the chloramphenicol available in the market is synthetic.

Question 10.
How do antiseptics differ from disinfectants? Give one example of each.
Answer:

Antiseptics and disinfectants are effective against micro-organisms However antiseptics are applied to the living tissues such as wounds, cuts, ulcers, and diseased skin surfaces, while disinfectants are applied to inanimate objects such as floors, drainage system, instruments, etc Disinfectants are harmful to the living tissues.

Iodine is an example of a strong antiseptic Tincture of iodine (2-3 percent of the solution of iodine in the alcohol-water mixture) is applied to wounds 1 percent solution of phenol is used as a disinfectant.

Question 11.
Why are cimetidine and ranitidine better antacids than sodium bicarbonate or magnesium or aluminium hydroxides?
Answer:
Both sodium bicarbonate and hydroxides of magnesium or aluminium are very good antacids since they neutralise the acidity in the stomach. But their prolonged use can cause the secretion of excessive acid in the stomach. This may be quite harmful and may lead to the formation of ulcers. Both cimetidine and ranitidine are better salts without any side effects.

Question 12.
Name a substance which can be used as an antiseptic as well as a disinfectant.
Answer:
Phenol can be used as an antiseptic as well as a disinfectant 0.2 percent solution of phenol is used as an antiseptic, while 1 percent of its solution is used as a disinfectant.

Question 13.
What are the main constituents of Dettol?
Answer:
The main constituents of antiseptic Dettol are chloroxylenol and terpenol.

Question 14.
What is the tincture of iodine? What is its use?
Answer:
Tincture of iodine is a 2-3 percent solution of iodine in an alcohol-water mixture It is applied to wounds as an antiseptic.

Question 15.
What are food preservatives?
Answer:
Chemical substances which are used to protect food against bacteria, yeasts and moulds are called preservatives. For example, sodium benzoate and sodium metabisulphite.

Question 16.
Why is the use of aspartame restricted to cold foods and drinks?
Answer:
Aspartame becomes unstable at cooking temperature This is the reason why its use is limited to cold foods and drinks.

Question 17.
What are artificial sweetening agents? Give two examples.
Answer:
Carbohydrates in the form of sugar (sucrose) are the traditional sweeteners and are the essential constituents of our diet. In the present lifestyle, people lack physical activities and exercise and it becomes rather difficult to burn the extra calories that are produced by the carbohydrates. Chemists have provided certain chemicals known as artificial sweeteners which provide the desired sweet taste io the food articles but hardly affect the calorie intake by the body. The most popular among the artificial sweeteners is saccharin which is nearly 550 times more sweet than the cane sugar. It is a boon for diabetic patients who don’t want to take carbohydrates (sugar) which is likely to increase the calories. It is in fact, a life saviour for these patients and is in the form of sodium or calcium salt which is water-soluble. These days, a number of other sweeteners are also available, e.g.. Aspartame, Alitame, Sucrolose etc.

(a) Aspartame: It is a very successful and commonly used artificial sweetener. As stated above, it is nearly 100 times as
sweet as cane sugar. However, it can be used in soft drinks and cold foods only since it decomposes upon heating. Chemically
aspartame is the methyl ester of dipeptide formed by the action of aspartic acid with phenylalanine.
(b) Sucrolose: The artificial sweetener as the name suggests is a trichioroderivative of sucrose. It is better, than
aspartame in the sense that it can be used in hot food at the cooking temperature, since it does not decompose on heating.
Moreover, it does not provide calories.

Question 18.
Name the sweetening agent used in the preparation of sweets for a diabetic patient.
Answer:
Artificial sweetening agents such as saccharin, alitame, and aspartame can be used in preparing sweets for diabetic patients.

Question 19.
What problem arises by using alitame as an artificial sweetener?
Answer:
Alitame is no doubt, a very potent sweetener. Its sweetening capacity is more than 2000 times as compared to ordinary cane sugar or sucrose. But sometimes, it becomes quite difficult to control the sweetness level in the food which is actually desired.

Question 20.
Why are detergents called soapless soaps?
Answer:
Soaps work in soft water, they are not effective in hard water In contrast synthetic detergents work both in soft water and hard water Therefore, synthetic detergents are better than soaps.

Question 21.
Explain the following terms with suitable examples.
(a) Cationic detergents
(b) Anionic detergents
(c) Neutral detergents.
Answer:
1. Anionic Detergents: These detergents contain anionic hydrophilic groups. These are generally made from long-chain alcohols which are reacted with concentrated sulphuric acid to form alkyl hydrogen sulphates. These are then neutralised with alkali to give water-soluble salts.
A few examples are listed below :
Sodium Alkyl Sulphates: These are the sodium salts of sulphonic acid esters of long-chain aliphatic alcohols which normally contain 10 to 15 carbon atoms. The alcohols are formed from fats and oils as a result of hydrogenolysis.

Sodium Alkyl Benzene Suiphonates: A common detergent belonging to this class is Sod – p – dodecyl benzene sulphonate. It is obtained from benzene by reacting with dodecyl chloride in the presence of anhydrous AlCl₃ acting as a catalyst.
The different steps involved are as follows :

2. Cationic Detergents: In these detergents, the hydrophilic group is of cationic nature. These are generally acetates, chlorides or bromides of quaternary ammonium CH CH CH B salts. The cationic part enclosed in the bracket contains a long hydrocarbon chain. These  detergents have germicidal qualities and are quite expensive as well, The cationic CH₃ detergents are present in hair conditioners. Cetyl trimethyl ammonium bromide

3. Non-ionic or neutral Detergent: These detergents are simply long-chain organic compounds and are esters in nature. For example, stearic acid and polyethylene glycol react to form a non-ionic detergent.

Non-ionic detergents contain polar groups and form hydrogen bonds with water. Some dishwashing liquids contain non-ionic detergents.
The field of detergents is very vast because of their immense utility. Companies engaged in their manufacture are spending huge amounts of money to bring products of better quality.

Question 22.
What are biodegradable and non-biodegradable detergents? Give an example of each. (C.B.S.E. Delhi 2008, 2009)
Answer:
Bio-degradable detergents are degraded by bacteria. In them, the hydrocarbon chain is unbranded. They do not cause water pollution and are bitter. Example: Sodium lauryl sulphate.

Non-biodegradable detergents possess highly branched hydrocarbon chain so bacteria cannot degrade them easily. They cause water pollution. Example: Sodium 4-(l, 3, 5, 7-tetramethyl-actyl) benzene sulphonate.

Question 23.
Why do soaps not work in hard water?
Answer:
Hard water contains calcium and magnesium salts. Therefore, in hard water soaps get precipitated as calcium and magnesium soaps which being insoluble stick to the clothes as gummy mass.

Question 24.
Can you use soaps and synthetic detergents to check the hardness of water?
Answer:
Soaps can be used to check hardness of water as they will form insoluble precipitates of calcium and magnesium salts on reacting with hard water. Since detergents do not form any precipitate, they cannot check hardness of water.

Question 25.
Explain the cleansing action of soaps.
Answer:
In order to understand the cleansing action of soaps let us try to analyse how the clothes become dirty. They first become oily because of the perspiration coming out of the skin and also from the organic matter dispersed in the atmosphere. Dust particles stick to oil drops and the clothes become dirty. In order to wash these, they are dipped in water and soap is applied.

In solution, it dissociates to give carboxylate ions (RCOO^–) and the cations (Na⁺). The alkyl portion which contains a long chain of hydrocarbons is a tail pointing towards the oil drops while the COO portion is the head directed towards water. This is quite evident from the figure where the solid circles (.) represent the polar groups and the wavy lines represent the alkyl portions. This formation is known as micelle and helps in forming a stable emulsion of oil and water by acting as a bridge between the two. The oil droplets along with the particles of the dirt get detached from the fibres of the clothes and pass into the emulsion. In this manner, the clothes become free from dust or dirt. The cleansing action of the soap is depicted in the Fig. 5.15.

Question 26.
If the water contains dissolved calcium bicarbonate, out of soaps and synthetic detergents, which one will you use for cleaning clothes?
Answer:
Synthetic detergents are preferred for cleaning clothes When soaps are dissolved in water containing calcium ions, these ions form insoluble salts that are of no further use, however when synthetic detergents are dissolved in water containing calcium ions, these ions form soluble salts that act as cleaning agents.

Question 27.
Label the hydrophilic and hydrophobic parts in the following compounds.
(a) CH₃(CH₂)₁₀CH₂OSO₃Na⁺
(b) CH₃(CH₂)₁₅-N⁺(CH₃)₃Br^–
(c) CH₃(CH₂)₁₆-COO(CH₂CH₂O)_nCH₂CH₂OH
Answer:

We hope the NCERT Solutions for Class 12 Chemistry Chapter 16 Chemistry in Every Day Life help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 16 Chemistry in Every Day Life, drop a comment below and we will get back to you at the earliest.

NCERT Class 12 Chemistry solutions for Chapter 12 contains solved answers for the questions provided in the textbook. The solutions are provided stepwise in an easy language. The students find it easy to understand and can prepare well for the examination.

The students appearing for UP board, Maharashtra board, MP board, CBSE, Gujarat board, etc. can practice using NCERT Solutions. It also helps in preparing well for competitive exams such as JEE and NEET.

NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes

Haloalkanes and Haloarenes is an important chapter from examination perspective. This chapter defines haloalkanes and haloarenes according to the IUPAC system of nomenclature. The reactions involved in the preparation of haloalkanes and haloarenes are also explained here. These are very important for the examination.

The chapter also explains the correlation between haloalkanes and haloarenes. NCERT Solutions for Class 12 Chapter 10 acts as a guide for the students preparing this chapter. The students can refer to these for better practice.

NCERT IN-TEXT QUESTIONS

Question 1.
Write the structures of the following compounds : (C.B.S.E. Delhi 2010)
(i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcyclohexane
(iii) 4-tert. butyl-3-iodoheptane
(iv) 1,4-Dibromobut-2-ene
(v) 1-Bromo-4-sec butyl-2-methylbenzene. (C.B.S.E. Sample paper 2011)
Answer:

Question 2.
Why is sulphuric acid not used during the reaction of alcohols with KI?
Answer:
H₂SO₄ is an oxidising agent. It oxidises HI produced during the reaction to I₂ and thus prevents the reaction between an alcohol and HI to form alkyl iodide. To prevent this, a non¬oxidising acid like H₃PO₃ is used.

Question 3.
Write the structures of different dihalogen derivatives of propane.
Answer:
Propane (CH₃CH₂CH₃) has two primary and one secondary hydrogen atoms present. Four isomeric dihalogen derivatives are possible. Let the halogen X be Br.

Question 4.
Among the isomeric alkanes of molecular formula C₅H₁₂, identify the one which on photochemical chlorination yields
(i) A single monochloride
(ii) Three isomeric monochlorides
(iii) Four isomeric monochlorides.
Answer:
The molecular formula C₅H₁₂ represents three structural isomers which are chain isomers.

(i) The isomer is symmetrical with four primary (1°) carbon atoms and one quaternary (4°) carbon atom. Since all the hydrogen atoms are equivalent, it will yield only one monochloride upon photochlorination i.e., chlorination carried in the presence of ultra-violet light.

(ii) In the straight chain isomer pentane, there are three groups of equivalent hydrogen atoms. As a result, three isomeric monochlorides are possible.

(iii) The branched chain isomer has four types of equivalent hydrogen atoms present. It will give four isomeric monochlorides upon chlorination.

Question 5.
Draw the structures of the major monohaloproducts in each of the following reactions:

Answer:

The reaction is carried in the presence of dry acetone upon heating. It is called Finkelstein reaction. In this reaction, I^– ion being a stronger nucleophile displaces Br^– ion. NaBr formed is insoluble in dry acetone whereas Nal dissolves. This shifts the equilibrium in the forward direction.

Under the reaction conditions allylic halogenation will take place. Addition of bromine can be possible in case the reaction is carried at room temperature.

Question 6.
Arrange each set of compounds in order of increasing boiling points.
(i)Bromomethane, Bromoform, Chloromethane, Dibromomethane.
(ii)1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.
Answer:
(i) Chloromethane < Bromomethane <
Dibromomethane < Bromoform
The reason is:
(a)for same alkyl group, B.Pt increases with size of halogen atom.
(b)B.Pt increases as number of halogen atoms increase.
(ii)Isopropyl chloride < 1 – Chloropropane < 1 – Chlorobutane
Reason :
(a)For same halogen, B.Pt. increases as size of alkyl group increases.
(b)B.Pt. decreases as branching increases.

Question 7.
Which alkyl halide from the following pairs would you expect to react more rapidly by S_N² mechanism ? Explain your answer.

Answer:
If the leaving group is the same in different isomers of a particular molecular formula, the reactivity of the isomers towards S_N² mechanism decreases with the increase in steric hindrance. In the light of above, the reactivity order in different cases is :

(i) CH₃CH₂CH₂CH₂Br is a primary alkyl halide (1°). It is more reactive than the other isomer which is a secondary (2°) alkyl halide because less steric hindrance is caused by primary alkyl group as compared to secondary alkyl group.
(ii)

is a secondary alkyl halide (2°). It is more reactive than the other isomer which is a tertiary alkyl halide (3°). The explanation is the same.
(iii) Here both the isomers are primary alkyl halides (1°). However, the isomer with CH₃ group at C₂ atom exerts more steric hindrance to the attacking nucleophile at C₁ atom as compared to the other isomer in which a CH₃ group is attached to C₃ atom. It is, therefore, less reactive.

Question 8.
In the following pairs of halogen compounds, which compound undergoes reaction faster ? (C.B.S.E. Delhi 2008, Outside Delhi 2010, 2013)

Answer:
The reactivity of a particular halogen compound towards S_N¹ reaction depends upon the stability of the carbocation formed as a result of ionisation. This is a slow step and is called rate determining step. The order of relative stabilities of different carbocations is in the order : tertiary > secondary > primary. In the light of this, the order of reactivity in the two cases is explained.

1. The isomer (a) is a tertiary alkyl chloride while the other isomer (b) is a secondary alkyl chloride. The isomer (a) is more reactive towards S i reaction since the tertiary carbocation formed in this case is more stable than the secondary carbocation which is likely to be formed in the other case.
2.  The isomer (a) is a secondary alkyl chloride while the other isomer (b) is primary in nature. The secondary alkyl chloride (a) is expected to react faster since the secondary carbocation formed is more stable than the primary carbocation which is likely to be formed in the other case.

Question 9.
Identify A, B, C, D, E, R and R’ in the following :

Answer:

NCERT EXERCISE

Question 1.
Name the following compounds according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary) vinyl or aryl halides.
(i) (CH₃)₂CHCH(C1)CH₃ (C.B.S.E. Delhi 2013)
(ii) CH₃CH₂CH(CH₃)CH(C₂H₅)Cl
(iii) CH₃CH₂C(CH₃)₂CH₂I
(iv) CH3C(C1)(C₂H5)CH₂CH₃
(v) CH₃.C(C₂H₅)₂CH₂Br
(vi) CH₃CH=C(C1)CH₂CH(CH₃)₂
(vii) CH₂=CH-CH₂-Br
(viii) CH₃CH=CHC(Br)(CH₃)₂
(ix) m-C1CH₂C₆H₄CH₂C(CH₃)3
(x) o-BrC₆H₄CH(CH₃)CH₂CH₃
(xi) (CH₃)₃CCH₂CH(Br)C5H₅
(xii) p-ClC₆H₄CH₂CH(CH₃)₂
Answer:

Question 2.
Give the IUPAC names of the following compounds :

Answer:

Question 3.
Write the structures of the following compounds :
(i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcyclohexane.
(iii) 2-(2-Chlorophenyi)-1-iodooctane
(iv) 4-tert. butyl -3-iodooctane
(v) 1, 4-Dibromobut-2-ene
(vi) 1-Bromo-4-sec.butyl-2-methylbenzene.
(vii) p-Bromochlorobenzene
(viii) Perfluorobenzene
Answer:

Question 4.
Which one of the following has highest dipole moment?
(a) CH₂Cl₂
(b) CHCl₃
(c) CCl₄
Answer:

CCl₄ is a symmetrical molecule. Therefore, the dipole moments of all four C-Cl bonds cancel each other. Hence its resultant dipole moment is zero.

As shown in the above figure, in CHCl₃, the resultant dipole moments of two C-Cl bonds is opposed by the resultant dipole moments of one C-H and one C-Cl bond. Since the resultant of one C-H and one C-Cl bond is smaller than the resultant of the two C-Cl bonds dipole moments, the opposition is to a small extent. As a result CHCl₃ has a small net dipole moment.

On the other hand, in case of CH₂,Cl₂ the resultant of the dipole moments of two C-Cl bonds is strengthened by the resultant of the dipole moments of two C-H bonds. As a result, CH₂Cl₂ has a higher dipole moment. Hence CH₂Cl₂ has the highest dipole moments among the three compounds.
Hence, the given compounds can be arranged in the increasing order of their dipole moments as
CCl₄ < CHCl₃ < CH₂Cl₂

Question 5.
A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monobromo compound in bright sunlight. Identify the hydrocarbon.
Answer:
A hydrocarbon with the molecular formula, C₅H₁₀ belongs to the group with a general molecular form C_nH_2n. therefore, it may either be an alkene or a cycloalkane since hydrocarbon does not react with chlorine in the dark, it cannot be alkene. Further, the hydrocarbon gives a single monochloro compound, C₅H₉Cl by reacting with chlorine in might sunshine since the formed compound is monochloro one all the C-H bonds should be equivalent. Hence the compound should be a cycloalkane. Hence the compound is C₅H₁₀ (cyclopentane).

Question 6.
Write the isomers of the compound having the formula C₄H₉Br. (Haryana Board 2013)
Answer:
The compound has the following structural isomers.

2-Bromobutane has a chiral carbon and it is expected to exhibit optical isomerism.

Question 7.
Write equations for the preparation of 1-Iodobutane from :
(a) Butan-1- ol
(b) 1-Chlorobutane
(c) But-1-ene.
Answer:

Question 8.
What are ambident nucleophiles? Explain with an example.
Answer:
Nucleophiles which can attack through two different sites are called ambident nucleophiles. For example, cyanide ion exists as a hybrid of the following two structures. It can attack either

through carbon to form cyanides (or nitriles) or through nitrogen to form isocyanides (or carbyl amines). For more details, consult section 11.7.

Question 9.
Which compound in the following pairs will react faster in the Sn² reaction?
(1) CH₃Br or CH₃I
(1) (CH₃)₃CCl or CH₃Cl (C.B.S.E. 2008)
Answer:
1. In the S_N2 mechanism the reactivity of halides for the same alkyl group increase in order. This happens because as the size increases the halide ion becomes a better leaving group.
R-F << R-Cl < R – Br < R-I
Therefore, CH₃I will react faster than CH₃Br in S_N2 reaction with image 17.

2. The S_N2 mechanism involves the attack of the nucleophile at the atom bearing the leaving group. But, in the case (CH₃)₃ CCl, the attack of the nucleophile at the carbon atom is hindered by the presence of the bulky substituents on that carbon atom bearing-the leaving the group in CH₃Cl. Hence CH₃Cl reacts faster than (CH₃)₃ CCl in S_N2 reaction with

Question 10.
Predict all the alkenes that would be formed by dehydrohalogenation of following alkyl halides with sodium ethoxide in ethanol.
(i) 1-Bromo-l-metbylcyclohexane
(ii) 2-Chloro-2-methyl butane
(iii) 3-Bromo-2, 2, 3-trimethylpentane.
Answer:
(i) 1-Bromo-l-methylcyclohexane has two β-hydrogen atoms. This will give a mixture of two alkenes as a result of dehydrohalogenation. Since alkene (B) is more substituted according to SaytzefFs rule, it is more stable and will be the major product. The same rule applies to the other alkyl halides also.

(ii) The compound has two sets of β-hydrogen atoms. Therefore, two elimination products are formed. However, a more substituted alkene is formed in greater proportion as compared to a less substituted alkene.

The explanation is similar. More substituted alkene is formed in preference to less substituted alkene.

Question 11.
How will you bring about the following conversions?   (Haryana Board 2011)
(i) Ethanol to but-1-yne
(ii) Ethane to bromoethane
(iii) Propene to 1-nitropropane
(iv) Toluene to benzyl alcohol
(v) Propene to propyne
(vi) Ethanol to ethyl fluoride
(vii) Bromometbane to propanone
(viii) But-1-ene to but-2-ene
(ix) 1-Chiorobutane to n-octane
(x) Benzene to biphenyl
Answer:

Question 12.
Explain why:
(i) Dipole moment of chlorobenzene is lower than that of cyclohexyl chloride (C.B.S.E 2016)
(ii) Alkyl halides though polar, are immiscible with water.
(iii) Grignard reagents should be prepared under anhydrous conditions.
Answer:

The polarity of C- Cl bond in chlorobenzene is less than that of same bond in cyclohexyl chloride because of carbon atom involved in chlorobenzene is more electronegative (greater s-character) as compared to the carbon atom in case of cyclohexyl chloride (lesser s-character). Therefore, the dipole moment of chlorobenzene is less with respect to cyclohexyl chloride.

(ii) In water, H₂O molecules are linked to each other by intermolecular hydrogen bonding. Although alkyl halides also contain polar C – X bonds, they cannot break the hydrogen bonding in H₂0 molecules. This means that there is hardly any scope for the association between molecules of alkyl halides and water. They, therefore, exist as separate layers and are immiscible with each other. For more details, consult section 11.6.

(iii) Grignard reagents (R – Mg – X) should be prepared under anhydrous conditions because these are readily decomposed by water to form alkanes.

That is why ether used as solvent in the preparation of Grignard reagent is completely anhydrous in nature.

Question 13.
Give the uses of freon-12, D.D.T., carbon tetrachloride and iodoform?
Answer:

1. Freons are the trade names for the commercially used fluoro chloromethanes with the formula CF_xCl_y (x + y = 4). A few examples are:
   CF₄ (Freon-14), CF₃C1 (Freon-13), CF₂Cl₂ (Freon-12), CFCl₃ (Freon-11)
   Out of the various freons mentioned, Freon- 12 is the most common refrigerant. It is prepared by passing hydrogen fluoride
   through carbon tetrachionde in the presence of antimony trichioride catalyst.
   
   in addition to their use as refrigerants in place of highly toxic liquid sulphur dioxide (SO₂) and ammonia (NH₃), large amount of CFCs are also used in the manufacture of disposable foam products such as cups and plates, as aerosol propellants in spray cans and as solvents to clean freshly soldered electronic circuit boards.
2. D.D.T. is the abbreviated form of p, p’-dichlorodiphenyltrichloroethane and its actual IUPAC naine has been given above. It is
   prepared by heating chiorobenzene with chlorai (trichioroacetaldehyde) in the presence of conc. H₂S0₄
   
3. Carbon tetrachloride (CC14) is also a colourless oily liquid just like chloroform. It is completely immiscible with water but
   dissolves in organic solvents.
   Carbon tetrachloride is a very useful solvent for oils, fats, resins etc. Ills used as a cleansing agent both in industry and in home because it can easily dissolve grease and other organic matter. But it mainly finds application for the manufacture of refrigerants, propellants for aerosol cans and some pharmaceuticals.
4. lodoform is a yellow crystalline solid with a characteristic unpleasant smell. It is insoluble in water but dissolves in alcohol, ether and other organic solvents.
   lodoform can be prepared in the laboratory by treating ethyl alcohol or acetone with sodium hydroxide and iodine. The reaction is known as haloform or iodoform reaction.
5. Physiological effects: lodoform is used as an antiseptic, particularly for dressing wounds. Actually, on coming in contact with skin (organic mater) it decomposes and slowly loses iodine which accounts for the antiseptic properties of iodoform.

Question 14.
Write the structures of the major products in each of the following reactions :

Answer:

Question 15.
Explain the following reaction :    (C.B.S.E. Delhi 2009 Comptt.)

Answer:
KCN is a resonance hybrid of two contributing structures :

This shows that the cyanide ion is an ambident nucleophile and the nucleophile attack is possible either through carbon atom or nitrogen atom resulting in cyanides and isocyanides respectively. In this case, in the presence of polar solvent, KCN readily ionises to furnish ions. The nucleophile attack takes place predominantly through a carbon atom and not through nitrogen atom as C- C bond is more stable than C -N bond.

Question 16.
Arrange the compounds of each set in order of decreasing reactivity towards (S_N²) displacement:
(a) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
(b) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane
(c) 1-Bromobutane, l-Bromo-2, 2-dimethylpropane, l-Bromo-2-methylbutane, l-Bromo-3-methylbutane. (C.B.S.E. Outside Delhi 2011)
Answer:
The reactivity of a particular haloalkane towards S_N² reaction is inversely proportional to the steric hindrance around the carbon atom involved in C – X bond. More the steric hindrance, lesser will be the reactivity. In the light of this, the decreasing order of reactivity in all the three cases is as follows :

Question 17.
Out of  C₆H₅CH₂Cl and C₆H₅CH(C1)C₆H₅ which is more easily hydrolysed by aqueous KOH?
Answer:
The compound C₆H₅CH₂Cl is a primary aralkyl halide while C₆H₅CH(Cl)C₆H₅ is secondary in nature. The hydrolysis of both these compounds with aqueous KOH (polar) is likely to proceed by S_N¹ mechanism due to the following reasons.
(a) The carbocations formed in both the cases as a result of ionisation are resonance stabilised due to the presence of phenyl groups at the a-position(s).
(b) As water is a polar solvent, it is expected to favour ionisation of the two halogen-substituted compounds leading to S_N¹ mechanism.
The carbocations that are formed as a result of ionisation in the slow steps are shown :

The ease of hydrolysis depends upon the relative stability of the carbocation/s that are formed in two cases. The secondary carbocation is more stable since the positive charge on the carbocation is delocalised on two phenyl groups that are present at the a-positions. On the other hand, there is only one phenyl group in primary carbocation available for charge delocalisation.
Thus, we may conclude that C₆H₅CHClC₆H₅ is more easily hydrolysed by aqueous KOH as compared to C₆H₅CH₂Cl.

Question 18.
p-Dichlorobenzene has higher m.p. and lower solubility than those of o-and m-isomers. Discuss.
Answer:
p-dichlorobenzene has a higher melting point than its o-isomer due to the symmetry of the p-isomer that fits in the crystal lattice better than the o- or m- isomer. Therefore, it has stronger intermolecular forces of attraction than o- and m- isomers, and thus greater energy are required to break crystal lattice to melt or dissolve the p-isomer than the corresponding o- and m- isomers. In other words, the melting point of the p-isomer is higher and its solubility is lower than corresponding m- and o- isomers.

Question 19.
How the following conversions can be carried out?
(i) Propene to propan-1-ol
(ii) Ethanol to but-1-yne
(iii) 1-Bromopropane to 2-bromopropane
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenyl ethanoic acid
(vii) Ethanol to propane nitrite
(viii) Aniline to chlorobenzene
(ix) 2-Chlorobutane to 3, 4 – dimethyl hexane
(x) 2-Methylpropene to 2-chloro-2-methylpropane
(xi) Ethyl chloride to propanoic acid
(xii) But-l-ene to n-butyl iodide
(xiii) 2-Chloropropane to propan-l-ol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenyl isocyanide.
Answer:

Question 20.
The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.
Answer:
If aqueous solution, KOH is almost completely ionized to give OH^– ions which being a strong nucleophile brings about a substitution reaction on alkyl halides to form alcohols. Further in the aqueous solution, OH^– ions are highly solvated (hydrated). This solvation reduces the basic character of OH^– ions which, therefore, fails to abstract a hydrogen from the P-carbon of the alkyl chloride to form alkenes. In contrast, an alcoholic solution of KOH contains alkoxide (RO^–) ion which being a much stronger base than OH^– ions preferentially eliminates a molecule of HCl from an alkyl chloride to form alkenes.

Question 21.
Primary alkyl halide (a) C₄H₉Br was reacted with alcoholic KOH to give compound (b). Compound (b) was reacted with HBr to give (c) which was an isomer of (a). When (a) was reacted with sodium metal, it gave a compound (d) C₈H₁₈, that was different than the compound when n-butyl bromide was reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.
Answer:
The two primary alkyl bromides are possible from the molecular formula (a) C₄H₉Br. These are:

According to the available information, the isomer (I) does not represent the correct compound because this on reacting with sodium metal (Wurtz reaction) will give n-octane. (C₈H₁₈) which is not actually formed

Question 22.
What happens when
(i) n-butyl chloride is treated with alcoholic KOH,
(ii) bromobenzene is treated with Mg in the presence of dry ether,
(iii) chlorobenzene is subjected to hydrolysis,
(iv) ethyl chloride is treated with (aq.) KOH,
(v) methyl bromide is treated with sodium in the presence of dry ether,
(vi) methyl chloride is treated with KCN?
Answer:
(i) But-l-ene is formed as the product as a result of dehydrohalogenation.

(ii) Phenyl magnesium bromide (Grignard reagent) is formed as a result of the reaction.

(iii) Chlorobenzene will not get hydrolysed on boiling with NaOH. No product will be formed.
(iv) Ethyl alcohol is formed as the product

(v) Ethane is formed as a result of Wurtz reaction

(vi) Methyl cyanide is formed.

We hope the NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes drop a comment below and we will get back to you at the earliest.

NCERT Class 12 Chemistry Solutions for Chapter 11 provides an insight into the various concepts related to alcohols, phenols and ethers. This is an important chapter and hence requires an indepth knowledge of the topics. The subject experts have provided accurate explanations and step wise solutions for the questions provided in the textbook. This will help the students prepare well during the exams.

NCERT Solutions not only help the students appearing for UP board, MP board, CBSE, Maharashtra board, Gujarat board, etc. but also prepares them for competitive exams such as JEE and NEET. The students should refer to the NCERT Solutions to score well in the examinations.

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ehers

Alcohols, phenols, and ethers is an important chapter from the examination perspective. The chapter explains the structure, properties, and applications of alcohols, phenols, and ethers. It also explains the correlation and differences between the three.

NCERT Solutions for Class 12 Chemistry Chapter 11 gives a better idea of the related concepts. The students can refer to these for better practice.

NCERT INTEXT QUESTIONS

Question 1.
Classify the following into primary, secondary, and tertiary alcohols

Answer:
(i) Primary alcohol
(ii) Primary alcohol
(iii) Primary alcohol
(iv) Secondary alcohol
(v) Secondary alcohol
(vi) Tertiary alcohol

Question 2.
Identify allylic alcohol in the above examples.
Answer:
allylic alcohols:

Question 3.
Give the IUPAC names of the following compounds : (C.B.S.E 2008)

Answer:

Question 4.
Show how the following alcohols can be prepared by the action of suitable Grignard reagent on methanal ?

Answer:
(i) The structure of alcohol suggests that the Grignard reagent that reacts with methanal is isopropyl magnesium halide.

(ii) The structure of alcohol suggests that the Grignard reagent that reacts with methanal is cyclohexyl magnesium halide.

Question 5.
Write the structures of the products of the following reactions :

Answer:
(i) The acidic hydration of propene gives propan-2-ol (isopropyl alcohol)

(ii) NaBH₄ is a weak reducing agent. It brings about the reduction of the ketonic group present in cyclohexane ring to secondary alcoholic group. However, it does not affect ester group.

(iii) NaBH₄ reduces aldehydic group to a primary alcoholic group.

Question 6.
Give structures of the products you would expect when each of the following alcohols reacts with (a) HCl/ZnCl₂ (b) HBr (c) SOCl₂ :
(i) Butan-l-ol
(ii) 2-Methylbutan-2-ol
Answer:
(i) Reactions of Butan-1-ol

The reaction of butan-l-ol (primary alcohol) can take place only upon heating. At room temperature, the reaction does not occur.

(ii) Reactions of 2-Methylbutan-2-ol

The alcohol being tertiary in nature reacts immediately with Lucas Reagent at room temperature.

Question 7.
Predict the major product of add catalysed dehydration of :
(i) 1-Methylcyclohexanol
(ii) Butan-1-ol
Answer:
(i) Acid catalysed dehydration of 1-methylcyclohexanol can give two products. However, 1-methylcyclohexene will be  preferably formed according to Satyzeff s rule since it is more substituted.

(ii) Butan-1-o1 upon acid dehydration will give but-2-ene as the major product along with but-l-ene as the minor product. Actually, primary carbocation formed can either lose a H⁺ to form but-1-ene or may undergo rearrangement and shift to secondary carbocation, which is more stable. The latter then forms but-2-ene by losing a H⁺.

Question 8.
Ortho and para nitrophenols are more acidic than phenol. Draw the resonating structures of the corresponding phenoxide ions.
Answer:
We know that the nitrophenols are more acidic than phenol. Their acidic strength can be compared in terms of the relative stabilities of the corresponding phenoxide ions based on resonance. For example,
(i) Phenoxide ion :


(ii) p-Nitrophenoxide ion :


(iii) p-Nitrophenoxide ion :

In case of nitrosubstituted phenoxides, the contributing structures that are enclosed in boxes have negative charge on the carbon atom to which die electron withdrawing nitro group is attached. They therefore, contribute more towards the acidic character than the rest of the contributing structures. Consequently, both ortho and para nitrophenol are stronger acids than phenol.

Question 9.
Write the equations involved in the following reactions :
(i) Reimer Tiemann Reaction
(ii) Kolbe’s Reaction.
Answer:
(i) In this reaction phenol is heated with chloroform alongwith aqueous NaOH (or KOH) to about 340 K. This is followed by acidification with dilute HCl when 2 – hydroxyhenzaldehyde (salicylaldehyde) is formed as the major product.

A small amount of para isomer is also formed in the reaction. In case, chLoroform is replaced by carbon tetrachioride, then 2—hydroxybenzoic acid (salicylic acid) is formed as the main product.

(ii) In this reaction, CO₂ gas is passed through sodium phenate at 400 K under a pressure of 4 to 7 atmospheres. This is followed by acidification with dilute HCI when salicylic acid is formed. This method is commonly used for the commercial preparation of saucy lic acid.

Question 10.
Write the reactions of Williamson’s synthesis of 2-ethoxy-3-methoxypentane starting from ethanol and 3-methylpentan- 2-ol.
Answer:
In the Williamson’s synthesis, the reactants are alkyl halide and sodium salt of an alcohol. In order to avoid the formation of alkene during the reaction, the alkyl halide should be primary while sodium salt must be of branched chain alcohol. In the present case, alkyl halide must be derived from ethanol upon heating with halogen acid (e.g HBr).

Question 11.
Which of the following is an appropriate set of reactants for the preparation of l-methoxy-4-nitrobenzene and why ?

Answer:
The second set of reactants is more appropriate to give the products i.e., l-methoxy-4-nitrobenzene. In the first set, cleavage of C – Br bond is involved. It is rather difficult since the carbon atom is sp² hybridised and the bond has partial double bond character as well. The product is formed as a result of Williamson’s synthesis.

Question 12.
Predict the products of the following reactions :

Answer:
(i) The reaction involves the cleavage of C – 0 bond. The Br atom of HBr is to combine with the smaller alkyl group to give the following products.

For more details and explanation, consult section 12.24.
(ii) This reaction also proceeds in the same manner. The Br atom of HBr is expected to combine with ethyl group (smaller in size) and not with phenyl group (bigger in size).

(iii) Nitrating mixture brings about the nitration of benzene ring. The ethoxy group (OC₂H₅) is an activating group and increases the electron density at the ortho and para positions due to +M effect. As a result, a mixture of o-nitro and p-nitro derivatives is formed. Out of these, the p-isomer is in excess since there is less steric hindrance due to OC₂H₅ group at the para position than at the ortho position in the ring.

(iv) In this reaction, the ether is initially protonated by H⁺ ion of the acid HI. to accomodate I^– ion (nucleophile). The reaction follows S_n1 mechanism.

NCERT EXERCISE

Question 1.
Write IUPAC names of the following compounds :

Answer:
(i) 2, 2, 4 – Trimethylpentan-3-ol
(ii) 5 – Ethylheptan-2, 4-diol
(iii) Butane – 2, 3, diol
(iv) Propane – 1, 2, 3-triol
(v) 2 – Methylphenol
(vi) 4 – Methylphenol
(vii) 2, 5 – Dimethylphenol
(viii) 2, 6 – Dimethyiphenol
(ix) 1 -Methoxy – 2 – methy lpropane
(x) Ethoxybenzene
(xi) 1 – Phenoxyheptane
(xii) 2 – Ethoxybutane.

Question 2.
Write the structure of the compounds, whose IUPAC names are as follows
(i) 2-Methylbutan – 2 – ol
(ii) I-Phenylpropan – 2 – ol
(iii) 3-Phenylhexane – l, 3, 5 – triol
(iv) 2, 3 – Diethylphenol
(v) 1 – Ethoxypropane
(vi) 2- Ethoxy – 3- methylpentane
(vii) Cyclohexylmethanol
(viii) 3-Cyclohexylpentan – 3 – ol
(ix) Cyclopent – 3 – en –  – ol
(x) 3 – Chloromethvlpentan – l – ol
Answer:

Question 3.
(i) Draw the structures of all isomeric alcohols of molecular formula C₅HI₂0 and give their IUPAC names.
(ii) Classify the isomers of alcohols in question 11.3 (i)as primary, secondary, and tertiary alcohols.
Answer:
Eight isomers are possible. These are:

Question 4.
Explain why is propanol higher boiling than butane?
Answer:
Propanol (Propan-l-ol) and butane are of comparable molecular masses 60m and 58u respectively but the boiling point of propanol is higher (391 K) because of the presence of intermolecular hydrogen bonding in the molecules. However, it is not present in butane due to the absence of polar OH group. The only attractive forces are weak van der Waals forces. Therefore, the boiling point of propanol is more than that of butane (309 K).

Question 5:
Explain why are alcohols more soluble in water than the corresponding hydrocarbons?
Answer:
The solubility of alcohols in water may be attributed to two factors :
(i) Both of them are of polar nature.
(ii) Molecules of both of them are involved in the intermolecular hydrogen bonding.
However, hydrocarbons are non-polar and are also not involved in any hydrogen bonding with alcohols. Alcohols readily dissolve in a water while the hydrocarbons are almost insoluble.

Question 6.
What is meant by hydroboration oxidation reaction ? Illustrate with an example.
Answer:
By hydroboration- oxidation of alkenes. Indirect hydration of alkenes can also bedone by hvdroboration-oxidation which is completed in two steps. In the first step. alkene reacts ith diborane (B’1{6) as boron hydride (BH3) dissolved in tetrahydrofuran (THF) to form an alkyl horane. In fact. the boron atom along i th the hydrogen gets attached to the double
bonded carbon atom with more number of hydrogen atoms less sterically hindered side). One hydrogen is then transferred to the other carbon atom. In this manner, all the three hydrogen atoms of boron are transferred to alkene molecule to form
trialkyl borane as the product. In the next step. the alkyl borane is oxidised by alkaline 11202 to form an alcohol. The indirect hydration proceeds according to Antimarkownikov s rule. For example.

Question 7.
Give the structures and IUPAC names of the phenols of molecular formula C₇H₈O.
Answer:

Question 8.
In separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which is steam volatile? Give reason.
Answer:
o-nitrophenol is steam volatile while p-nitrophenol is not. This is on account of intramolecular hydrogen bonding in the molecules of o-nitrophenol. As a result, its boiling point is less than that of p-nitrophenol in which the molecules are linked by intermolecular hydrogen bonding.
It is interesting to note that in the substituted phenols, the nature and position of the substituent influences the boiling point of phenol.
For example .o-nitrophenol is steam volatile while p-nitrophenol is not. This is supported by the fact that the boiling point temperature of o-nitrophenol (100°C) is less than that of p-nitrophenol, (279°C). In o-nitrophenol, there is intramolecular hydrogen bonding in OH and NO₂ groups placed in a adjacent positions. However, these are linked by intermolecular hydrogen bonding in the p-isomers. It is quite obvious that extra energy is needed to cleave the hydrogen bonds in the p-isomer. Consequently, its boiling point is more.

o-nitrophenol with lower boiling point is steam volatile while p-nitrophenol is not. This helps in the separation of the two isomers present in the liquid mixture. On passing steam, o-nitrophenol volatilises, and its vapours rise alongwith steam and after condensation, collect in the receiver p-nitrophenol is left behind in the distillation flask. e-nkrophenol p-nnrophenol.

Question 9.
Give the equations of reactions for the preparation of phenol from cumene.
Answer:

Question 10.
Write chemical equations involved in the preparation of phenol from chlorobenzene.
Answer:
From chlorobenzene, phenol is prepared by Dow’s process.In this method, chlorobenzene is heated with 6 to 8% solution of NaOH to about 623 K under a pressure of 300 atmospheres to form sodium phenate which upon acidification with dilute HCI gives phenol as follows:

Question 11.
Write the mechanism of hydration of ethene to yield ethanol.
Answer:
Ethene does not react with water as such. Water being little polar, is not in a position to provide H⁺ ion for initial electrophilic attack on ethene. The reaction is carried in the presence of H₂S0₄. The acid provides proton (H⁺) for the initial electrophilic attack.

In the second step, H₂0 attacks the carbocation in preference to HS0₄ ion as a nucleophile

Question 12.
You are given benzene, cone. H₂S0₄ and NaOH. Write equations for the preparation of phenol using these reagents.
Answer:

Question 13.
Show how well you synthesise:
(i) 1-phenyl ethanol from a suitable alkene
(ii) Cyclohexylmethanol using an alkyl halide by S_N² reaction
(iii) Pentan-1-ol using a suitable alkyl halide.
Answer:

Question 14.
Give two reactions to show the acidic nature of phenol. Compare the acidity of phenol with that of ethanol.
Answer:
Acidic nature of phenols. Phenols are weakly acidic in nature. The liquid form of phenol containing about 5 percent water is known as carbolic acid. The dissociation constant (K_a) for phenol is 10^-10 at 298 K (room temperature). The corresponding pK_a* value is 10.0. The acidic character is exhibited by the following properties:
(i) Reaction with active metals. Phenols evolve hydrogen with active metals such as sodium and potassium.

(ii) Reaction with alkalies. Phenols neutralise caustic alkalies such as NaOH or KOH to form salt and water.

In addition to these, phenols turn blue litmus red which is the characteristic property or acids. However, phenols do not
react with either alkali metal carbonates or bicarbonates since these are quite weak acids.

Question 15.
Explain why is ortho-nitrophenol more acidic than ortho-methoxy phenol?
Answer:
Due to strong -R and – I-effect of the -NO₂ group, electron density of the O – H bond decreases and hence the loss of a proton becomes easy.

Question 16.
Explain how does -OH group attached to a carbon atom of benzene ring activates it towards electrophilic substitution ?
Answer:
The OH group exerts a +M (or + R) effect on the ring under the influence of attacking electrophile.

As a result, there is an increase in the electron density in the ring particularly at the ortho and para positions. The electrophilic substitution readily takes place at these positions when electrophile attacks.

Question 17.
Give equations for the following chemical reactions :
(i) Oxidation of propan-1-ol with alkaline KMnO₄
(ii) Reaction of bromine in CS₂ with phenol
(iii) Action of dilute HNO₃ on phenol
(iv) Treating phenol with chloroform in the presence of aqueous NaOH at 343 K.
Answer:

Question 18.
Write short notes on (i) Kolbe reaction (ii) Reimer-Tiemann reaction.
Answer:
(i) In this reaction phenol is heated with chloroform alongwith aqueous NaOH (or KOH) to about 340 K. This is followed by acidification with dilute HCl when 2 – hydroxyhenzaldehyde (salicylaldehyde) is formed as the major product.

A small amount of para isomer is also formed in the reaction. In case, chLoroform is replaced by carbon tetrachloride, then 2—hydroxybenzoic acid (salicylic acid) is formed as the main product.

(ii) In this reaction, CO₂ gas is passed through sodium phenate at 400 K under a pressure of 4 to 7 atmospheres. This is followed by acidification with dilute HCl when salicylic acid is formed. This method is commonly used for the commercial preparation of salicylic acid.

Question 19.
Write a mechanism for the acid dehydration of ethanol to ethene.
Answer:
Mechanism of dehydration. The mechanism is illustrated with ethanol which is a primary alcohol.

The relative ease of dehydration of different alcohols i.e. primary, secondary and tertiary can be further justified on the basis of the relative stabilities of the carbocations formed in the slow step. Since tertiary carbocation is maximum stable while primary is the least, the tertiary alcohols are maximum reactive while the primary are the least reactive in nature.
In other words, greater the stability of carbocation formed, more is the reactivity of the alcohol.

Question 20.
How are the following conversions carried out?
(i) Propene → Propan -2-ol
(ii) Benzyl chloride → Benzyl alcohol
(iii) Ethyl magnesium chloride → Propan-1-ol
(iv) Methyl magnesium bromide → 2-Methylpropan-2-ol
Answer:

Question 21.
Name the reagents used in the following reactions :
(i) Oxidation of primary alcohol to carboxylic acid
(ii) Oxidation of primary alcohol to an aldehyde
(iii) Bromination of phenol to 2, 4, 6-tribromophenol
(iv) Benzyl alcohol to benzoic acid
(v) Dehydration of propan-2-ol to propene
(vi) Butan-2-one to butan-2-ol.
Answer:
(i) Acidified K₂Cr₂O₇
(ii) Pyridine chlorochromate (PCC)
(iii) Bromine water (Br₂/H₂O)
(iv) Alkaline KMnO₄
(v) 60% H₂S0₄ at 373 K
(vi) LiAlH₄ or NaBH₄.

Question 22.
Give a reason for the higher boiling point of ethanol in comparison to methoxymethane.
Answer:
Ethers are isomeric with alcohols but their boiling points are comparatively low due to the lack of hydrogen bonding. For example, boiling points of isomeric n – butyl alcohol (nC₄H₉OH) and diethyl ether (C₂H₅ – O – C₂H₅) are, 390 K and 308 K respectively.

Question 23:
Give the IUPAC names of the following ethers :


Answer:
(i) 1-Methoxy-2-methylpropane
(ii) 1-Chloro-2-methoxy ethane
(iii) 4-Nitroanisole
(iv) 1-Methoxypropane
(v) 4-Ethoxy-1, 1-dimethyl cyclohexane
(vi) Ethoxybenzene

Question 24.
Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
(i) 1-Propoxypropane
(ii) Ethoxybenzene
(iii) 2-Methoxy-2-methylpropane
(iv) 1-Methoxyethane
Answer:

Question 25.
Illustrate with examples the limitations of Williamson’s synthesis for the preparation of certain types of ethers.
Answer:
Preparation from Alkyl Halides
From alkyl halides, ethers can be prepared by the following methods
By Williamson’s synthesis. It is the best method for the laboratory preparation of both simple and mixed ethers and involves the action of sodium alkoxide (formed by reaction between alcohol and sodium metal) on a suitable alkyl halide.

Limitations of the reaction. In the preparation of unsymmetrical ethers, a proper choice of the reactants is necessary.
Elimination leading to alkene can take place since alkoxide ion can also abstract one of the 3—hydrogen atom alongwith acting as a nucleophile. Thus, in order ro prepare ethyl tertiary butyl ether, we must use ethyl halide (primary) and sodium tertiary
butoxide.

In case, the alkyl halide is tertiary and sodium ethoxide is employed, then C2H5O ion will cause the elimination of alkyl halide to form an alkene as the main product.

Since secondary and tertiary alkyl halides prefer to undergo elimination rather substitution, symmetrical ethers containing secondary and tertiary alkyl groups are obtained only in poor yields by Willamson’s synthesis. For example,

This method is also not successful for preparing aryl alkyl ethers by reacting sod. alkoxide with aryl halides because the cleavage of C – X bond is not so easy due to partial double bond character. In such cases, we must react sodium phenoxide with
alkyl halide as follows:

However, diaryl ethers (both the groups are aryl or phenyl groups) cannot be prepared since aryl halides fail to participate in the nucleophilic substitution reactions.

Question 26.
How is 1-propoxypropane synthesised from propan-1-ol? Write the mechanism of this reaction.
Answer:
(a) Williamson’s synthesis

Question 27.
Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Answer:

Question 28.
Write the equation for the reaction of HI with :
(i) 1-Propoxypropane
(ii) Methoxybenzene
(iii) Benzyl ethyl ether.
Answer:

Question 29.
Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in benzene ring.
Answer:
In aryl alkyl ethers, the +R-effect of the alkoxy (OR) group increases the electron density in the benzene ring, thereby activating the benzene ring towards electrophilic substitution reaction.

Since the electron density increases more at the two ortho and one para position as compared to meta position therefore, electrophilic substitution reactions mainly occur at o-and -positions.

Question 30.
Write the mechanism of the reaction of Hl with methoxymethane.
Answer:

Question 31.
Write equations for the following reactions :
(i) Friedel Crafts reaction-alkylation in anisole.
(ii) Nitration of anisole
(iii) Bromination of anisole in ethanolic medium
(iv) Friedel Crafts acetylation of anisole.
Answer:
(i) The halogenation of the benzene ring occurs at the ortho and para positions. However, the para isomer is
formed in excess. For example, the bromination of anisole in ethanoic acid gives nearly 90 percent p-bromoanisole.

(ii)  The nitration of anisole carried with a nitrating mixture of conc. UNO3 and conc. H2SO4 upon heating gives a
a mixture of ortho and para nitro derivatives.

(iii) Sulphonation: Anisole upon sulphonation gives a mixture of isomeric sulphonic acid derivatives.

(iv) Friedel Craft’s reaction: Both alkylation and acylation are carried in the presence of anhydrous Aid3 catalyst which
behaves as a Lewis acid.

Question 32.
Show how would you synthesise the following alcohols from appropriate alkanes?

Answer:

Question 33.
When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place :

Give the mechanism of the reaction.
Answer:
The mechanism is explained as follows :

We hope the NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ehers help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ehers, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Chemistry Chapter 15 provides answers to the questions provided in the textbook. The answers are accompanied by diagrammatic representations for better understanding. Also, the solutions are explained in such a language that the students find easy to understand.

NCERT Solutions are beneficial for the students appearing in UP board, MP board, CBSE, Gujarat board, etc. Also the students appearing for competitive exams such as NEET and JEE will find the NCERT Solutions for Class 12 Chemistry beneficial.

NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers

Polymers are very large molecules having high molecular mass. These are also known as macromolecules. This chapter explains the mechanical properties and applications of polymers. The classification of polymers is also described in detail.

The chapter gives an overview of different polymerization reactions. The chapter explains that the polymers are the backbone of four major industries, plastic, elastomers, fibres and paints.

NCERT INTEXT QUESTIONS

Question 1.
What are polymers?
Answer:
The word polymer has a Greek origin where poly means many and mers or meros stands for unit or part. Thus, the polymer may be defined as a substance of a high molecular mass (10³ – 10⁷ u) formed by the combination of a large number of simple
molecules called monomers by chemical bonds. The process by which monomers are converted into polymers is called polymerization.

Question 2.
How are polymers classified on the basis of structure?
Answer:
On the basis of structure, polymers are classified as:
(i) Linear polymers in which the monomers are joined together to form long straight chains of polymer molecules. Forex: HDPE, PVC, nylons, etc.
(ii) Branched-chain polymers in which the monomers not only join in a linear fashion but also form branches of different lengths along the main chain. For ex: LDPE, glycogen, etc.
(iii) Cross-linked polymers in which the initially formed linear polymer chains join together to form a 3D network structure. For ex: bakelite, Urea-formaldehyde resin, etc.

Question 3.
Write the names of the monomers of the following polymers:

Answer:
(i) The polymer is Nylon 66. Its monomer units are hexamethylenediamine and adipic acid.
(ii) The polymer is Nylon 6. Its monomer units are of caproilactum.
(iii) The polymer is Teflon. Its monomer units are of tetrafluoroethylene.

Question 4.
Classify the following as addition and condensation polymers:
(i) Terylene
(ii) Bakelite
(iii) Polyvinyl chloride
(iv) Polythene
Answer:
(i) Condensation polymer
(ii) Condensation polymer
(iii) Addition polymer
(iv) Addition polymer

Question 5.
Explain the difference between Buna-N and Buna-S.
Answer:
Both Buna-N and Buna-S are synthetic rubber and are co-polymers in nature. They differ in their constituents.
Buna-N: Constituents are : buta-1, 3-diene and acrylonitrile.
Buna-S: Constituents are : buta-1, 3-diene, and styrene. They condense in the presence of Na.

Buna – S: It is a co—polymer of 1. 3 – butadiene and styrene and is prepared by the polymerisation of these components in the
ratio of 3 : 1 in the presence of sodium.

Buna-N (Nitrile rubber):  h is a co-polymer of buta-1. 3-diene and acrylonitrile. It is formed as follows:

Question 6.
Arrange the following polymers in increasing order of their intermolecular forces :
(i) Nylon-66, Buna-S, Polythene
(ii) Nylon-6, Neoprene, Polyvinyl chloride
Answer:
We have studied the classification of polymers based upon intermolecular forces. These follow the order :
Elastomers < Plastics < Fibres.

The polymers listed may be arranged in increasing order of intermolecular forces as follows :
(i) Buna-S (Elastomer) < Polythene (Plastic) < Nylon-66 (Fibre)
(ii) Neoprene (Elastomer) < Polyvinyl chloride (Plastic) < Nylon-6 (Fibre).

NCERT Exercise

Question 1.
Explain the terms polymers and monomers.
Answer:
Polymers are a high molecular mass of macromolecules composed of repeating structural units derived from monomers. Polymers have a high molecular mass (10³-10⁷u). In a polymer, various monomer units are joined by strong covalent bonds. Polymers can be natural as well as synthetic. Polythene, rubber, and nylon 6,6 are examples of polymers.

Monomers are simple, reactive molecules that combine with each other in large numbers through covalent bonds to give rise to polymers. For example ethene, propene, styrene, vinyl chloride.

Question 2.
What are natural and synthetic polymers? Give two examples of each.
Answer:
Natural polymers:
Natural polymers are high molecular mass macromolecules and are found in nature mainly in plants and animals. For example, protein, nucleic acids, starch, cellulose, etc.

Synthetic polymer:
Synthetic polymers are man-made high molecular mass macromolecules. For example, plastics (Polyethene, P.V.C.), synthetic fibers (Polyesters, Nylon-6,6), synthetic rubber (Neoprene, Buna-S), etc.

Question 3.
Distinguish between the terms homopolymer and copolymer and give one example of each. (C.B.S.E. Outside Delhi 2008, Pb. Board 2008)
Answer:
A polymer in which the monomers are the same is called a homopolymer. For example, polystyrene is made from styrene only.
A polymer in which the monomers are different is known as a copolymer. For example, in Nylon-66 the monomers are adipic acid and hexamethylenediamine.

Question 4.
How do you explain the functionality of a monomer?
Answer:
The functionality of a monomer is the number of binding sites that is/are present in that monomer.
For example, the functionality of monomers such as ethene and propene is one and that of 1, 3-butadiene, and adipic acid is two.

Question 5.
Define the term polymerisation. (C.B.S.E. Delhi 2008)
Answer:
A very widely used polymer is polyethylene which is formed by the polymerisation of ethylene molecules (monomers) by heating under pressure in the presence of oxygen.

It may be noted that all the monomer units in a polymer may or may not be the same, In case, they are the same as in case of polyethylene, then the polymer is called homopolymer. On the other hand, if they happen to be different, then the polymer is known as copolymer or mixed polymer. For example, the monomer units in terylene are ethylene glycol and terephthalic acid. Similarly, hexamethylenediamine and adipic acid are the monomer units of nylon-66. Both are co-polymers or mixed polymers. We shall discuss these at a later stage in the present unit.

Question 6.

Answer:
It is a homopolymer in nature. It has only one type of monomer units i.e., NH₂ – CHR – COOH. These area-amino acids.

Question 7:
In which classes, are the polymers classified on the basis of molecular forces?
Answer:
Polymers are classified into four classes on the basis of molecular forces. These are:
elastomers, fibres, thermoplastic polymers, and thermosetting polymers.

1. Elastomers: In these polymers, the intermolecular forces are the weakest. As a result, they can be readily stretched by applying small stress and regain their original shape when the stress is removed. The elasticity can be further increased by introducing some cross-links in the polymer chains. Natural rubber is the most popular example of elastomers. A few more examples are of: buna-S, buna-N and neoprene.

2. Fibres: Fibres represent a class of polymers which are thread-like and can be woven into fabrics in a number of ways. These are widely used for making clothes, nets, ropes, gauzes etc. Fibres possess high tensile strength because the chains possess strong intermolecular forces such as hydrogen bonding. These forces are also responsible for close packing of the chains. As a result, the fibres are crystalline in nature and have also sharp melting points. A few common polymers belonging to this class are nylon – 66, terylene and polyacrylonitrile etc.

3. Thermoplastics: These are linear polymers and have weak van der Waals forces acting in the various chains and are intermediate of the forces present in the elastomers and in the fibres. When heated, they melt and form a fluid which sets into a hard mass on cooling, Thus, they can be cast into different shapes by using suitable moulds. A few common examples are polyethylene and polystyrene polyvinyls etc. These can be used for making toys, buckets, telephone apparatus, television cabinets etc.

4. Thermosetting plastics: These are normally semifluid substances with low molecular masses. When heated, they become hard and infusible due to the cross-linking between the polymer chains. As a result, they also become three-dimensional in nature. They do not melt when heated. A few common thermosetting polymers are bakelite, melamine-formaldehyde, urea-formaldehyde and polyurethane etc.

Question 8:
How can you distinguish between addition and condensation polymerisation?
Answer:
In addition polymerization: The monomers are unsaturated in nature i.e., they have atleast one double or triple bond in their molecules. They represent the functionality of the monomers which combine with each other at these sites. The addition polymerisation is generally chain growth polymerisation in nature. For example, polythene, polystyrene, PVC, Teflon etc. are all formed as a result of additional polymerisation.

In condensation polymerization: The monomer units have specific functional groups present which represent their functionality. The monomers combine through these functional groups and the polymerisation is step growth polymerisation in nature. For example, terylene, nylon-6, nylon-66, bakelite etc. are all formed as a result of condensation polymerisation.

Question 9:
Explain the term co-polymerization and give two examples.
Answer:
Co-polymerisation is a process in which a mixture of more than one different monomeric units are allowed to polymerise. The copolymer thus formed contains multiple units of each monomer in the chain. The formation of buna-S copolymer in which the monomers are : buta-1, 3-diene, styrene and sodium is an example of co­polymerisation.
Another Example of co-polymer: Buna-S is a co-polymer of 1 : 3 – butadiene and styrene in the presence of sodium metal.

Thus, as a result of addition polymerisation, the double bonds change to single bonds. A large number of monomer units can be linked in this way.
The addition polymerisation is normally a chain reaction in which the chain is built up by the successive addition of the monomer units. It involves a number of intermediates such as free radicals, carbocations or carbanions. Let us illustrate the chain growth polymerisation based upon different mechanisms.

Question 10:
Write the free radical mechanism for the polymerisation of ethene.
Answer:
The polymerisation of ethene to polythene consists of heating or exposing to light a mixture of ethene with a small amount of benzoyl peroxide initiator The process starts with the addition of phenyl free radical formed by the peroxide to the ethene double bond thus regenerating a new and larger free radical. This step is called chain initiating step. As this radical reacts with another molecule of ethene, another bigger sized radical is formed. The repetition of this sequence with new and bigger radicals carries the reaction forward and the step is termed as chain propagating step. Ultimately; at some stage the product radical thus formed reacts with another radical to form the polymerised product. This step is called the chain terminating step. The sequence of steps may be depicted as follows :
Chain initiating steps:

Chain terminating step:
For termination of the long-chain, these radicals can combine in different ways to form polythene. One mode of termination of the chain is shown as under:

Question 11:
Define thermosetting and thermoplastic polymers with two examples of each.
Answer:
Thermoplastics are polymers which can be easily softened repeatedly on heating and hardened on cooling. Therefore, it can be used again and again. For example, polyethylene and polyvinyl chloride. Thermosetting polymers are those which undergo permanent change on heating. They become hard and infusible on heating and cannot be softened again. For example, Bakelite, and Melamine formaldehyde.

Question 12:
Write the monomers used for getting the following polymers :

1. Polyvinyl chloride
2. Teflon
3. Bakelite

Answer:

1. vinyl chloride
2. tetrafluoroethylene
3. phenol and formaldehyde.

Question 13:
Write the name and structure of one of the common initiators used in free radical addition polymerisation.
Answer:

Question 14:
How does the presence of double bonds in rubber molecules influence their structure and reactivity?
Answer:
Chemically natural rubber is polyisoprene in which monomer units are of isoprene i.e., 2-methyl-i, 3-butadiene. It is in fact, 1, 4 polymer in which monomers are linked through CH₂ groups located at 1.4 positions. The residual double bonds are located at C₂ and C₃ positions in the isoprene units. All the double bonds have cis configurations. Thus, natural rubber is cis polyisoprene.

The high elasticity of natural rubber is due to the absence of polar groups and the cis-configurations about double bonds do I not allow the polymer chains to come closer. Therefore, only weak van der Waals’ forces are present. Since the chains are not linear, they can be stretched just like springs and exhibit elastic properties.

Question 15:
Discuss the main purpose of vulcanisation of rubber.
Answer:
The main purpose of vulcanization of rubber is to improve the following draw-back of natural rubber:

• At high temperature (T >335K) natural rubber becomes soft.
• At low temperature (T< 283K) natural rubber becomes brittle.
• Natural rubber is soluble in non-polar solvents.
• It is non-resistant to attack by oxidizing agents.

Question 16:
What are the monomers repeating units of Nylon 6 and Nylon 66?
Answer:
The monomeric repeating unit of nylon 6 is [NH – (CH₂)₅ – CO], which is derived from caprolactam.
The monomeric repeating unit of nylon 6, 6 is [NH – (CH₂)₆ – NH – CO – (CH₂)₄ – CO], which is derived from hexamethylene diamine and adipic acid.

Question 17:
Write the names and structures of the monomers of the following polymers:
(i) Buna-S
(ii) Buna-N
(iii) Dacron
(iv) Neoprene (C.B.S.E. Delhi 2008)
Answer:

Question 18:
Identify the monomers in the following polymeric structures:

Answer:
(i) Decanedioic acid: HOOC(CH₂)₈COOH
Octamethylenediamine: H₂N(CH₂)₈NH₂

(ii)

Question 19:
How is dacron obtained from ethylene glycol and terephthalic acid?
Answer:

Question 20:
What is a biodegradable polymer? Give an example of an aliphatic biodegradable polyester.
Answer:
A polymer that can be decomposed by bacteria is called a biodegradable polymer. Poly – β- hydroxybutyrate – Co -β – hydroxy valerate (PHBV) is a biodegradable aliphatic polyester.

We hope the NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers, drop a comment below and we will get back to you at the earliest.

NCERT Class 12 Chemistry Solutions for Chapter 12 provides answers for the questions provided in the textbook. The answers are accurate to the best of our knowledge and are provided by subject experts. The students can refer to these or sure shot success.

The students appearing for various boards and competitive exams can find these solutions helpful for practice. Most of the questions asked in UP board, MP board, Gujarat board, CBSE, etc. are asked from these. To score well in the examinations, the students should go through these solutions atleast once after finishing the entire syllabus.

NCERT Solutions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids

This chapter explains the structure, physical and chemical properties and applications of aldehydes, ketones and carboxylic acid. It also explains a correlation between the three. It also explains the mechanism of different reactions of aldehydes, ketones and carboxylic acids. The factors affecting the acidity of carboxylic acids and their reactions help in understanding the advanced concepts related to this chapter.

NCERT Solutions for Class 12 Chapter 12 will help you revise the chapter during the examination. It also helps in clearing doubts if any.

NCERT INTEXT QUESTIONS

Question 1:
Write the structures of the following compoimds :
(i) α -Methoxypropionaldehyde
(ii) 3-Hydroxybutanal
(iii) 2-Hydroxycyclopentanecarbaldehyde
(iv) 4-Oxopentanal
(v) Di-sec butylketone
(vi) 4-Fluoroacetophenone
Answer:

Question 2.
Write the structures of products of following reactions:

Answer:

Question 3.
Arrange the following compounds in increasing order of their boiling points :
CH₃CHO,  CH₃CH₂OH,  CH₃OCH₃, CH₃CH₂CH₃
Answer:
The increasing order of boiling points of all these compounds of comparable molecular masses is :

Explanation: We know that the boiling points of liquids are directly related to the magnitude of the intermolecular forces of attraction.

1. Hydrocarbons (alkanes) are completely non-polar. The only attractive forces in their molecules are Van der Waals forces which are quite weak. That is why propane (CH₃CH₂CH₃) has the least boiling point. It is a gas at room temperature.
2. Ethers have bent structures and are also polar. However, there is no hydrogen bonding in their molecules. The only attractive forces are dipolar forces. Therefore, boiling point of dimethyl ether (CH₃OCH₃) is higher than that of propane. However, it is also a gas at room temperature.
3. Aldehydes contain polar carbonyl group and have strong dipolar interactions in their molecules. It is more than in ethers. Therefore, the boiling point of acetaldehyde (CH₃CHO) is more than that of dimethyl ether.
   
4. Out of all the families listed, alcohols have maximum intermolecular forces in the form of hydrogen bonding
   
   As a result, ethyl alcohol (C₂H₅OH) has the maximum boiling point.

Question 4.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions
(i) Ehtanal, propanaL, propanone, butanone
(ii) Benzaldehyde,p-Tolualdehyde, p-Nitrobenzaldehyde, acetophenone.
Ans: (i) Butanone < Propanone < Propanal < Ethanal .This is because as the no. of alkyl groups attached to carbonyl carbon increases, +I-effect increases. As a result, e^– density

Question 5.
Predict the products of the following reactions :

Answer:

Question 6.
Give the IUPAC names of the following compounds:

Answer:

Question 7.
Show how each of the following compounds can be converted into benzoic acid?
(i) Ethylbenzene (C.B.S.E. Outside Delhi 2017)   
(ii) Acetophenone  (C.B.S.E. Outside Delhi 2017)
(iii) Bromobenzene
(iv) Phenylethene (Styrene)
Answer:

Question 8.
Which acid of each pair would you expect to be stronger?
(i) CH₃CO₂H or FCH₂CO₂H
(ii) FCH₂CO₂H or ClCH₂CO₂H
(iii) FCH₂CH2CH₂CO₂H or CH₃CH(F)CH₂CO₂H

Answer:

NCERT EXERCISE

 Question 1.
What is meant by the following terms? Give an example in each case.
(a) Cyanohydrin
(b) Semicarbazone
(c) Acetal
(d) Oxime
(e) Cyanohydrin
(f) Ketal
(g) Aldol
(h) Schiff’s base
(i) 2, 4-D.N.P.
(ii) Imine
Answer:

Question 2.
Name the following compounds according to IUPAC system of nomenclature.
(i) CH₃CH(CH₃)CH₂CH₂CHO
(ii) CH₃CH=CHCHO
(iii) CH₃CH(CH₃)CH₂C(CH₃)₂COCH₃
(iv) OHCC₆H₄CHO(-p)
(v) CH₃CH₂COCH(C₂H₅)CH₂CH₂Cl
(vi) CH₃COCH₂COCH₃
(vii) (CH₃)₃CCH₂COOH
(viii) (CH₃)₂CHCH(CH₃)COCl
Answer:
(i) 4-Methylpentanal
(ii) But-2-enal
(iii) 3 3 5-Trimethylhexan-2-one
(iv) Benzene- 1, 4-dicarbaldehyde
(v) 6-Chloro-4-ethylhexan-3-one
(vi) Pentane-2, 4-dione
(vii) 3, 3-Dimethylbutanoic acid
(viii) 2, 3-Dimethylbutanoyl chloride

Question 3.
Draw the structures of the following compounds:
(i) 3-Methylbutanal
(ii) p-Nitropropiophenone
(iii) p-Methylbenzaldehyde
(iv) 4-Methylpent-3-en-2-one
(v) 4-Chloropentan-2-one
(vi) 3-Bromo-4-phenylpentanoic acid
(vii) pp’-Dihydroxybenzophenone
(viii) Hex-2-en-4-ynoic acid
Answer:

Question 4.
Write the IUPAC names of the following aldehydes and ketones. Also give the common names wherever possible.

Answer:

Answer:

Question 7.
Which of the following will undergo Aldol condensation, which Cannizzaro’s reaction, and which neither of these? Write the structures of the expected products in each case
(i) Methanal
(ii) 2-Methylpentanal
(iii) Benzaldehyde
(iv) Benzophenone
(v) Cyclohexanone
(vi) 1-Phenylpropanone
(vii) Phenylacetaldehyde
(viii) Butan-1-ol
(ix) 2, 2-Dimethyl butanal
Answer:
(i) Methanal (HCHO): It will give Cannizzaro’s reaction since the α-hydrogen atom is absent.

(ii) 2-Methylpentanal [CH₃CH₂CH₂CH (CH₃)CHO]: It will give Aldol condensation since the α-hydrogen atom is present.

(iii) Benzaldehyde (CgH₅CHO): It will give Cannizzaro’s reaction since a-hydrogen is not present.

(iv) Benzophenone (C₆H₅COC₆H₅): It will not give any of the two reactions. Being ketone, does not take part in Cannizzaro’s reaction. Without a-hydrogen, it fails to participate in Aldol condensation.

(vi) 1-Phenylpropanone (C₆H₅COCH₂CH₃): It will undergo Aldol condensation since the α-hydrogen atom is present.

(vii) Phenylacetaldehyde (C₆H₅CH₂CHO): It will give Aldol condensation since the α-hydrogen atom is present.

(viii) Butan-1-ol: It will not give any of the reactions.

Question 8.
How will you convert ethanal to the following compounds?
(i) Butane-1, 3-diol
(ii) But-2-enal
(iii) But-2-enoic acid.
Answer:
(i) Ethanal to butane -1, 3-diol

(ii) Ethanal to but-2-enal

(iii) Ethanal into but-2-enoic acid

Question 9.
Write the structural formulae and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde serves as a nucleophile and which as an electrophile.
Answer:
Both propanal and butanal have a-hydrogen atoms present. These can undergo self aldol condensation as well as cross aldol condensation to give four compounds as follows:
(i) Condensation involving propanal: It is a case of a self aldol condensation.

(ii) Condensation involving butanal: It is self aldol condensation.

(iii) Condensation involving butanal (electrophile) and propanal (nucleophile):  It is cross-aldol condensation.

(iv) Condensation involving propanal (electrophile) and butanal (nucleophile): It is cross-aldol condensation.

Question 10.
An organic compound with molecular formula CgHjoO forms 2, 4-DNP derivative, reduces Tollen’s reagent, and undergoes Cannizzaro’s reaction. On vigorous oxidation, it gives 1, 2 benzene dicarboxylic acid. Identify the compound.      (C.B.S.E. Outside Delhi 2012; Haryana Board 2013)
Answer:
Since the compound forms 2, 4-DNP derivative on reacting with 2, 4-DNP, it is a carbonyl compound. As the compound reduces Tollen’s reagent and undergoes Cannizzaro’s reaction, it is an aldehyde and not a ketone. The data further reveals that the compound on vigorous oxidation gives 1, 2-benzene dicarboxylic acid. This clearly shows that in the compound which is of aromatic nature, CHO group is present at position-1 and C₂H₅ side chain at position-2. The given compound is 2-ethyl benzaldehyde. The reactions involved are given below :

Question 11.
An organic compound [A] with molecular formula C₈H₁₆O₂ was hydrolysed with dilute sulphuric acid to give a carboxylic acid [B] and alcohol [C]. Oxidation of [C] with chromic acid produced [B]. The alcohol [C] on dehydration gave but-1-ene. Write equations for the reactions involved. (C.B.S.E. 2008 Supp., C.B.S.E. 2009)
Answer:
(i) The available data shows that the compound [A] upon hydrolysis gave a carboxylic acid [B] and alcohol [C]. It must be an ester.
(ii) Since the alcohol [C] upon oxidation with chromic acid gave back the carboxylic acid [B], both the acid and alcohol must have the same number of carbon atoms (four each).
(iii) The alcohol [C] upon dehydration gave an alkene.
The equations for the reactions are given:

Question 12.
Arrange the following compounds in increasing order of their property as indicated:
(i) Acetaldehyde, Acetone, Di-tert butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN)
(ii) CH₃CH₂CH(Br)COOH, CH₃CH(Br)CH₂COOH, (CH₃)₂CH COOH, CH₃CH₂CH₂COOH (acid strength).
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic add, 4-Methoxybenzok acid (acid strength).
Answer:
(i) The reactivity of aldehydes and ketones towards HCN addition decreases as the +1 – effect of the alkyl groups increases. Secondly, it decreases with increase in steric hindrance to the nucleophilic attack by CN^– at the carbonyl carbon. Thus the decreasing order of reactivity towards HCN is,

(ii) We know that the + I-effect decreases while -I-effect increases the acidic strength of carboxylic acids. Since + I-effect of isopropyl group is more than that of propyl group, therefore, (CH₃)₂CHCOOH is a weaker acid than CH₃CH₂CH₂COOH. Further since -I-effect decreases with distance, therefore CH₃CH₂CHBrCOOH is a stronger acid than CH₃CHBrCH₂COOH. Thus, the overall acid strength increases in the order:

(iii) Since electron-donating groups decrease the acidic strength, therefore, 4-methoxy benzoic acid is a weaker acid than benzoic acid. Further, since electron-withdrawing groups increase the acidic strength, therefore, both 4-nitrobenzoic acid and 3,4-dinitrobenzoic acid are stronger acids than benzoic acid. Further due to the presence of an additional -NO₂ group at /w-position with respect to -COOH group, 3,4-dinitrobenzoic acid is a stronger acid than 4-nitrobenzoic acid. Thus, the overall acidic strength increases in the order:4-methoxy benzoic acid < benzoic acid < 4-nitrobenzoic acid < 3,4-dinitrobenzoic acid.

Question 13.
Give chemical tests to distinguish between the following pairs of compounds :
(i) Propanal and propanone  (C.B.S.E. Delhi 2011, 2012)
(ii) Phenol and benzoic acid
(iii) Acetophenone and benzophenone
(iv) Benzoic acid and ethyl benzoate  (C.B.S.E. Outside Delhi 2009, 2011)
(v) Pentan-2-one and pentane-3-one
(vi) Benzaldehyde and acetophenone (C.B.S.E. Outside Delhi 2015)
(vii) Ethanal and propanal (C.B.S.E. Outside Delhi 2009, 2011, 2012)
Answer:
(i) Propanal and propanone:  Propanal will give a silver mirror upon heating with Tollen’s reagent but propanone will not respond.
(ii) Phenol and benzoic acid: Benzoic acid will give brisk effervescence with sodium hydrogen carbonate (NaHC0₃) but phenol will not respond.
(iii) Acetophenone and benzophenone: Acetophenone is a methyl ketone. It will give a yellow precipitate upon heating with I₂ and NaOH. Benzophenone will not respond.
(iv) Benzoic acid and ethyl benzoate: Benzoic acid will give brisk effervescence with sodium hydrogen carbonate (NaHC0₃) but ethyl benzoate (ester) will not respond.
(v) Pentan-2-one and pentan-3-one: Pentan-3-one is a methyl ketone and will give a yellow precipitate upon heating with I₂ and NaOH. Pentan-3-one will not respond.
(vi) Benzaldehyde and acetophenone: The distinction can also be made by iodoform test. Acetophenone will give yellow precipitate while benzaldehyde will not react.
(vii) Ethanal and propanal: Ethanal will respond to iodoform test and give yellow precipitate. Propanal will not react.

Question 14.
How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom.
(i) Methylbenzoate
(ii) m-Nitrobenzoic acid
(iii) p-Nitrobenzoic acid
(iv) Phenylacetic acid
(v) p-nitrobenzaldehyde
Answer:
(i) Benzene to methylbenzoate

(ii) Benzene to m-nitrobenzoic acid


(iii) Benzene to p-nitrobenzoic acid


(vi) Benzene to phenylacetic acid


(v) Benzene to p-nitrobenzaldehyde

Question 15.
How will you bring about the following conversions in not more than two steps?
(i) Propanone to propene   (C.B.S.E. Delhi 2009, Uttarakhand Board 2009)
(ii) Propanal to butanone
(iii) Ethanol to 3-hydroxybutanal (C.B.S.E. Outside Delhi 2012)
(iv) Benzaldehyde to benzophenone (C.B.S.E. Outside Delhi 2012)
(v) Benzaldehyde to 3-PhenyIpropan-1-ol
(vi) Benzaldehyde to a-Hydroxyphenylacetic acid
(vii) Benzoic acid to benzaldehyde (C.B.S.E. Delhi 2009, Outside Delhi 2017)
(viii) Benzene to m-nitroacetophenone
(ix) Benzoic acid to /n-nitrobenzyl alcohol.      (C.B.S.E. Delhi 2012)
Answer:
(i) Propanone to propene

(ii) Propanal to butanone

(iii) Ethanol to 3-hydroxybutanal

(iv) Benzaldehyde to benzophenone

(v) Benzaldehyde to 3-phenylpropan-1-ol

(vi) Benzaldehyde to a-Hydroxyphenylacetic acid

(vii) Benzoic acid to benzaldehyde

(viii) Benzene to m-nitroacetophenone

(ix) Benzoic acid to m-nitrobenzyl alcohol

Question 16:
Describe the following:
(i) Acylation
(ii) Cross-aldol condensation
(iii) Cannizzaro’s reaction
(iv) Decarboxylation.
Answer:

Question 17:
Complete each synthesis by giving missing starting material, reagent, or products.


Answer:

Question 18:
Give a plausible explanation for each of the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2, 4, 6-trimethylcyclohexanone does not.
(ii) There are two – NH₂ groups in semicarbazide. However, only one is involved in the formation of semicarbazone.
(iii) During the preparation of esters from carboxylic acid and alcohol in the presence of an acid catalyst, the water or the ester should be removed as fast as it is formed.
Answer:

In cyclohexanone, the attack of CN“ ion (nucleophile) can easily take place at the carbonyl carbon atom. However, in 2, 4, 6-trimethylcyclohexanone, the three CH₃ groups being electron releasing in nature (+ I effect) will considerably increase the electron density on the carbonyl carbon atom and the nucleophile attack does not seem to be feasible. Moreover, the two —CH₃ substituents at the ortho positions will also hinder the attack of nucleophile CN^– ion on the carbonyl group.

(ii) The structural formula of semi-carbazide is NH₂NHCONH₂. Although both the amino groups have lone electron pairs, one of these is in conjugation with the electron-withdrawing carbonyl group and acquires a positive charge. Therefore, it is not in a position to act as the nucleophile, and only one -NH₂ group is involved in the formation of semicarbazone.

(iii) The esterification carried in the presence of acid is of reversible nature and the reverse reaction is called ester hydrolysis.

In order that the reaction may proceed in the forward direction, ester or water formed in the reaction must be removed. Sulphuric acid added in esterification helps in removing molecules of H₂0 as it is a dehydrating agent.

Question 19:
An organic compound contains 69-77% carbon, 11-63% hydrogen and the rest is oxygen. The molecular mass of the compound is 86. It does not reduce Tollen’s reagent but forms an addition compound with sodium hydrogen sulphite and gives a positive iodoform test. On vigorous oxidation, it gives ethanoic acid and propanoic acid. Write the possible structure of the compound. (C.B.S.E. Delhi 2008, 2009, Uttarakhand Board 2015)
Answer:
Step I: Calculation of molecular formula of the compound
Percentage of oxygen = 100 – (% C + % H) = 100 – (69.77 + 11.63) = 18.6%

Step II. Predicting the structure of the compound

• Since the compound forms an addition compound with NaHS0₃, it must be a carbonyl compound.
• As the compound does not reduce Tollen’s reagent but gives a positive iodoform test, it must contain in it a methyl
  

Keeping in view these characteristics, the compound is CH₃CH₂CH₂COCH₃ (Pentan-2-one).
All the reactions in which pentan-2-one participates, are given for the benefit of the students.

Question 20.
Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?
Answer:

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