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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.

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