NCERT Solutions for Class 12 Chemistry Chapter 13 Amines

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NCERT Solutions for Class 12 Chemistry Chapter 13 Amines

Q: What is the weightage of NCERT Class 12 Chemistry chapter 13 in JEE Mains?

Around 5-7 marks worth questions are asked from this chapter in JEE mains. The questions for Amines can be practiced from the NCERT book, NCERT exemplar and JEE Main previous year papers.

Q: Where can I find complete solutions of NCERT class 12 Chemistry?

For complete solutions : https://school.careers360.com/ncert/ncert-solutions-class-12-chemistry . Students can download the chapter wise solutions PDF.

Q: What is the weightage of NCERT Class 12 Chemistry chapter 13 in CBSE Board Exams?

This chapter holds weightage of 5-6 Marks in Board exams. Follow NCERT syllabus to get a good score in the CBSE board exam.

Q: What is the weightage of NCERT class 12 Chemistry chapter 14 in NEET?

Weightage of NCERT class 12 Chemistry chapter 13 in NEET is around 3 percent

Edited By Vishal kumar | Updated on Mar 06, 2023 11:54 AM IST | #CBSE Class 12th

NCERT Solutions for Class 12 Chemistry Chapter 13 Amines - You will find amines ncert solutions class 12 one of the important organic molecules which are present in proteins, vitamins, and hormones, etc. In Amines Class 12 chapter, you will deal with organic compounds of amines and learn about amines and diazonium salts. You may also have some doubts which can be resolved by going through NCERT solutions for Class 12 Chemistry chapter 13 Amines. Amines are also considered as derivatives of ammonia.

Also Read

In Amines Class 12 NCERT chapter, there are 14 questions in the exercise. The NCERT solutions for class 12 chemistry chapter 13 Amines are prepared in a very detailed manner which has all the answers to questions. Chemistry chapter 13 Amines holds four marks in the class 12 CBSE board exam. These amines class 12 ncert solutions help students in their preparation of Cclass 12 CBSE Board exam as well as various competitive exams like JEE Mains, NEET, etc. Scroll down to get NCERT solutions for Amines Class 12. By referring to the NCERT solutions for class 12 , students can understand all the important concepts and practice questions well enough before their examination.

NCERT Solutions for Class 12 Chemistry Chapter 13 Amines- Exercise Questions

Solutions to In-Text Questions Ex 13.1 to 13.9

Question 13.1 Classify the following amines as primary, secondary or tertiary:

(i)

Answer :

1596565129561

N atom is directly connected with only one C atom, so it is a primary aromatic amine

Question 13.1 Classify the following amines as primary, secondary or tertiary:

(ii)

Answer :

In this compound N atom directly connected with 3 carbon atoms. So, it is a tertiary aromatic amine

Question 13.1 Classify the following amines as primary, secondary or tertiary:

(iii) $(C_{2}H_{5})_{2}CHNH_{2}$

Answer :

$(C_{2}H_{5})_{2}CHNH_{2}$

1596565186466

Here N atom directly connected with only one C atom. So it is a primary aliphatic amine.

Question 13.1 Classify the following amines as primary, secondary or tertiary:

(iv) $(C_{2}H_{5})_{2}NH$

Answer :

$(C_{2}H_{5})_{2}NH$

1596565203549

In structure, we can clearly see that N is directly connected with 2 carbon atom. so, it is a secondary amine.

Question 13.2 (i) Write structures of different isomeric amines corresponding to the molecular formula,

$C_{4}H_{11}N$

Answer :

different isomeric amines corresponding to the molecular formula, $C_{4}H_{11}N$

(i ) $CH_{3}CH_{2}CH_{2}CH_{2}NH_{2}$ (v) $CH_{3}CH_{2}CH_{2}CH_{2}NHCH_{3}$

(ii) 1596565271605 (vi) 1596565310726

(iii) 1596565289920 (vii) 1596565320554

(iv) 1596565299578 (viii) 1596565330072

Question 13.2 (ii) Write IUPAC names of all the isomers.

Answer :

IUPAC name of all the isomers-

$CH_{3}CH_{2}CH_{2}CH_{2}NH_{2}$	Butanamine
	Butan-2-amine
	2-Methylpropanamine
	N-Methylpropan-2-amine
	2-Methylpropan-2-amine
	N,N-Dimethylethanamine
$CH_{3}CH_{2}CH_{2}CH_{2}NHCH_{3}$	N-Methylpropanamine
	N-Ethylethanamine

Question 13.2(iii) What type of isomerism is exhibited by different pairs of amines?

Answer :

$CH_{3}CH_{2}CH_{2}CH_{2}NH_{2}$	Butanamine (Chain isomerism + position isomerism)
	Butan-2-amine (chain isomerism + position isomerism)
	2-Methylpropanamine (chain isomerism)
	N-Methylpropan-2-amine (position isomerism + metamerism)
	2-Methylpropan-2-amine (chain isomerism)
	N,N-Dimethylethanamine
$CH_{3}CH_{2}CH_{2}CH_{2}NHCH_{3}$	N-Methylpropanamine (position isomerism)
	N-Ethylethanamine (no isomerism)

Question 13.3(i) How will you convert

Benzene into aniline

Answer :

1596565516861

Nitration of benzene gives nitrobenzene. And now reduce the nitro group by catalytic hydrogenation.

Question 13.3(ii) How will you convert

Benzene into N, N-dimethylaniline

Answer :

1596565527611

Nitration of benzene gives nitrobenzene and after catalytic hydrogenation of nitrobenzene, it gives aniline. Aniline on reacting with two moles of chloromethane to form N, N-dimethylaniline.

Question 13.3(iii) How will you convert

$Cl-(CH_{2})_{4}-Cl$ into $hexan-1$ , $6-diamine?$

Answer :

1596565536567

ON reacting the given reactant with ethanolic sodium cyanide, the CN molecules replace both chlorine atom. And after that catalytic hydrogenation, we get our desired product. (reduce the CN to $-CH_{2}NH_{2}$ in both sides)

Question 13.4(i) Arrange the following in increasing order of their basic strength:

$C_{2}H_{5}NH_{2},$ $C_{6}H_{5}NH_{2},$ $NH_{3},$ $C_{6}H_{5}CH_{2}NH_{2}$ and $(C_{2}H_{5})_{2}NH$

Answer :

Considering the inductive effect, solvation effect and steric hindrance of the alkyl group which decides the basic strength of alkylamines. order of basic strength in ethyl substituted amine is-

$NH_{3}<C_{2}H_{5}NH_{2}<(C_{2}H_{5})_{2}NH$

and order in benzene substituted ring-

$C_{6}H_{5}NH{2}<C_{6}H_{5}CH_{2}NH_{2}$

Due to the -R effect of benzene $C_{6}H_{5}NH{2}$ has less electron density than ammonia. So, the final order is -

$C_{6}H_{5}NH{2}<NH_{3}<C_{6}H_{5}CH_{2}NH_{2}<C_{2}H_{5}NH_{2}<(C_{2}H_{5})_{2}NH$

Question 13.4(ii) Arrange the following in increasing order of their basic strength:

$C_{2}H_{5}NH_{2},$ $(C_{2}H_{5})_{2}NH,$ $(C_{2}H_{5})_{3}N,$ $C_{6}H_{5}NH_{2}$

Answer :

$C_{6}H_{5}NH_{2}<C_{2}H_{5}NH_{2}<(C_{2}H_{5})_{3}N<(C_{2}H_{5})_{2}NH$

On considering the inductive effect, solvation effect and steric hindrance of the alkyl group the increasing order of basicity in ethyl as a substituted group is shown above.

Question 13.4(iii) Arrange the following in increasing order of their basic strength:

(iii) $CH_{3}NH_{2},$ $(CH_{3})_{2}NH,$ $(CH_{3})_{3}N,$ $C_{6}H_{5}NH_{2},$ $C_{6}H_{5}CH_{2}NH_{2}.$

Answer :

On considering the inductive effect, solvation effect and steric hindrance of the alkyl group which decides the basic strength of alkyl amines in the aqueous state. The order of basic strength in case of methyl substituted amines are -

$C_{6}H_{5}NH_{2}<C_{6}H_{5}CH_{2}NH_{2}<(CH_{3})_{3}N<CH_{3}NH_{2}<(CH_{3})_{2}NH$

Question 13.5(i) Complete the following acid-base reactions and name the products:

$CH_{3}CH_{2}CH_{2}NH_{2}+HCl\rightarrow$

Answer :

The above-given reaction is an acid-base reaction. so salt is formed.

$CH_{3}CH_{2}CH_{2}NH_{2}+HCl\rightarrow$ $CH_{3}(CH_{2})_{2}NH_{3}^{+}Cl^{-}$

1596565551481

Question 13.5(ii) Complete the following acid-base reactions and name the products:

$(C_{2}H_{5})_{3}N+HCl\rightarrow$

Answer :

$(C_{2}H_{5})_{3}N+HCl\rightarrow$ $(C_{2}H_{5})_{3}N^{+}HCl^{-}$

It is an acid-base reaction, so the N will accept H from $HCl$ and form a salt.

1596565560887

Question 13.6 Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.

Answer :

The methyl iodide reacts with aniline to give N, N-dimethylaniline.

1596565570028 With the excess of methyl iodide in the presence of sodium carbonate solution ( $Na_{2}CO_{3}$ ), N, N-dimethylaniline produces N, N, N-trimethyl anilinium carbonate.
1596565579864

Question 13.7 Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.

Answer :

When aniline reacts with benzoyl chloride HCl is produced as a by-product and N-phenyl benzamide is produced as a major product.
lone pair of N atom attacks the acidic carbon of benzoyl chloride.

1596565588671

Question 13.8 Write structures of different isomers corresponding to the molecular formula, $C_{3}H_{9}N.$ Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid.

Answer :

structures of different isomers corresponding to the molecular formula $C_{3}H_{9}N$ and their IUPAC name-

(i)		Propan-1-amine
(ii)		Propan-2-amine
(iii)		N – Methylethanamine
(iv)		N, N-Dimethylmethanamine

The structure (i) and (ii) will liberate nitrogen gas ( $N_{2}$ ) on treating with nitrous acid.

Question 13.9(i) Convert

3-Methylaniline into 3-nitrotoluene

Answer :

1596565637949

On diazotisation, reaction $NH_{2}$ will convert to $N_{2}Cl$ . And then reacts with fluoroboric acid followed by $NaNO_{2}/Cu/\Delta$ convert $N_{2}Cl$ into the nitro group ( $-NO_{2}$ ) and evolution of dinitrogen gas.

Question 13.9(ii) Convert

Aniline into 1,3,5 - tribromobenzene.

Answer :

1596565649772 Aniline on bromination in the presence of water gives 2,4,6 tribromoaniline, which on further reacting with nitrous acid converts $NH_{2}$ into $N_{2}Cl$ . Now, with the help of $H_{3}PO_{2}/water$ it removes the $N_{2}Cl$ group from the benzene ring and we get our final product 1,3,5 tribromobenzene.

NCERT Solutions for Class 12 Chemistry Chapter 13 Amines- Exercise Questions

Question 13.1 Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

(i) $(CH_{3})_{2}CHNH_{2}$

Answer :

The IUPAC name of the compound is 1-methylethanamine.

since $N$ (nitrogen ) is connected to only one carbon. Hence it is a primary amine.

1596565661754

structure of the compound $(CH_{3})_{2}CHNH_{2}$

Question 13.1(ii) Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

$CH_{3}(CH_{2})_{2}NH_{2}$

Answer :

IUPAC name is Propan-1-amine

It is a primary amine (nitrogen connects with only one carbon atom)

Here is the structure of the compound $CH_{3}(CH_{2})_{2}NH_{2}$ -

1596565672365

Question 13.1 (iii) Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

$CH_{3}NHCH(CH_{3})_{2}$

Answer :

$CH_{3}NHCH(CH_{3})_{2}$

N-Methyl-2-methylethanamine

since N atom is connected with two C atom, So it is a two-degree amine

Question 13.1(iv) Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

$(CH_{3})_{3}CNH_{2}$

Answer :

$(CH_{3})_{3}CNH_{2}$

2-Methylpropane-2-amine

It is a one-degree amine

1596565682994

Question 13.1(v) Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

$C_{6}H_{5}NHCH_{3}$

Answer :

$C_{6}H_{5}NHCH_{3}$

N-Methylbenzamine or N-methylaniline

Here N atom is connected with two C atom. So, it is a secondary amine.

1596565693795

Question 13.1(vi) Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

$(CH_{3}CH_{2})_{2}NCH_{3}$

Answer:

$(CH_{3}CH_{2})_{2}NCH_{3}$

N-Ethyl-N-methylethanamine

1596565707700

N is connected with three carbon atom so that it is a tertiary amine

Question 13.1(vii) Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

$m-BrC_{6}H_{4}NH_{2}$

Answer :

$m-BrC_{6}H_{4}NH_{2}$
3-bromoaniline or 3-bromobenzenamine

It is a primary amine

1596565717107

Question 13.2 Give one chemical test to distinguish between the following pairs of compounds.

(i) Methylamine and dimethylamine

Answer :

Carbylamine test

This test is used to distinguish between aliphatic and aromatic primary amines on heating with chloroform $(CHCl_{3})$ and potassium hydroxide ( $KOH$ ) gives isocyanides or carbylamines which has foul smelling. two degree and 3-degree amines do not show this reaction.

Here methylamine is primary and dimethylamine is a secondary amine.

1596565730642

first one is 1-degree and second one is 2-degree

Question 13.2 Give one chemical test to distinguish between the following pairs of compounds.

(ii) Secondary and tertiary amines

Answer :

Secondary and tertiary amines can be distinguished by reacting them with Hinsberg's reagent which is also called benzenesulphonyl chloride. $(C_{6}H_{5}SO_{2}Cl)$ In the case of primary amine the product is soluble in alkali but not in secondary amine case. And tertiary amine does not react with this reagent.

In the case of a secondary amine

1596565740393

Tertiary amine + benzenesulphonyl chloride $\rightarrow$ no reaction

Question 13.2 Give one chemical test to distinguish between the following pairs of compounds.

(iii) Ethylamine and aniline

Answer :

Ethylamine and aniline can be differentiated by the azo-dye test. On reacting with $(NaNO_{2}+dil.HCl)$ and followed by 2-naphthol gives a colored product. But when it is ethylamine it gives brisk effervescence due to the evolution of $N_{2}$ gas under the same conditions.

Question 13.2 Give one chemical test to distinguish between the following pairs of compounds.

(iv) Aniline and benzylamine

Answer :

Benzylamine reacts with nitrous acid to form diazonium salt, which is unstable and also gives alcohol with the evolution of nitrogen gas. On the other hand, aniline reacts with nitrous acid to form a stable diazonium salt.

Question 13.2 Give one chemical test to distinguish between the following pairs of compounds.

(v) Aniline and N-methylaniline.

Answer :

Aniline and N-methylaniline both can be distinguished by carbylamine test. aniline is the primary aromatic compound so it gives s positive test but N-methyl aniline is secondary and does not give this test.

structure of both compound-

1596565751583 primary aromatic amine secondary aromatic amine
13.2 Give one chemical test to distinguish between the following pairs of compounds. (v) Aniline and N-methylaniline.
Edit Q

Question 13.3 Account for the following:

(i) pKb of aniline is more than that of methylamine..

Answer :

Here is the structure of aniline and methylamine

1596565760265

Due to the resonance in aniline the electrons of nitrogen atoms delocalised over benzene ring. So, because of that, the electron density at N atom decreases and less available for donation.

1596565770185 On the other hand, In the case of methylamine, the methyl ( $CH_{3}$ ) group is electron donating group (+I effect) which increase the electron density at N atom. Hence $pK_{b}$ value of aniline is higher than that of methylamine.

Question 13.3 Account for the following:

(ii) Ethylamine is soluble in water whereas aniline is not.

Answer:

1596565779599 $R = C_{2}H_{5}$

The extent oh intermolecular hydrogen bonding in ethylamine is very high. Hence it is soluble in water. But aniline does not undergo hydrogen bonding with water to a large extent due to the presence of a bulky hydrophobic group $(C_{6}H_{6})$ . Hence it is insoluble in water.

Question 13.3 Account for the following:

(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.

Answer :

1596565789474

Due to +I effect of methyl ( $CH_{3}$ ) group methylamine is more basic than water. So in water $CH_{3}NH_{2}$ produces hydroxide ion by accepting $H^{+}$ ions from water.

$CH_{3}NH_{2}+H_{2}O\rightarrow CH_{3}N^{+}H_{3}+OH^{-}$

Ferric chloride ( $FeCl_{3}$ ) dissociates in water into- $FeCl_{3}\rightarrow Fe^{3+}+3Cl^{-}$ The hydroxide ion ( $OH^{-}$ ) react with $Fe^{3+}$ and form a precipitate of $Fe(OH)_{3}$ (reddish brown ppt) $Fe^{3+}+3OH^{-}\rightarrow Fe(OH)_{3}$

Question 13.3 Account for the following:

(iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions,aniline on nitration gives a substantial amount of m- nitroaniline.

Answer :

Nitration is carried out in strongly acidic medium, aniline is protonated to form the anilinium ion which is meta directing. Because of this reason, aniline on nitration gives a substantial amount of meta - nitroaniline 1596565798623 1596565806109

Question 13.3 Account for the following:

(v) Aniline does not undergo Friedel-Crafts reaction.

Answer :

In fridal craft reaction, we use $AlCl_{3}$ which is acidic in nature and aniline is also a show basic nature. Thus there is an acid-base reaction between them and form a salt.

1596565815066

Due to the positive charge on the N-atom. the electrophilic substitution in benzene ring is deactivated.

Question 13.3 Account for the following:

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

Answer :

In diazonium salt, the structure goes under resonance due to which the dispersal of positive charge is more and we know that higher is the resonance higher is the stability. Therefore diazonium salt of aromatic amines is more stable than those of aliphatic amines.

1596565822279

Question 13.3 Account for the following:

(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.

Answer :

Gabriel synthesis is used for primary amines. secondary and tertiary amines are not formed by this method. So, therefore to obtain pure and only 1-degree amine Gabriel phthalimide reaction is preferred.

Question 13.4 Arrange the following:

(i) In decreasing order of the pKb values:

$C_{2}H_{5}NH_{2},$ $C_{6}H_{5}NHCH_{3},$ $(C_{2}H_{5})_{2}NH$ and $C_{6}H_{5}NH_{2}$

Answer :

$C_{2}H_{5}NH_{2},$ and $(C_{2}H_{5})_{2}NH$ are aliphatic amines so they are more basic than rest two compound. In ethylamine, there is only one electron donating group (+I effect) but in diethylamine, there is two electron donating group so the electron density is much higher than ethylamine.

In between $C_{6}H_{5}NHCH_{3},$ and $C_{6}H_{5}NH_{2}$ they are the weak base because of delocalisation of lone pair electron in the benzene ring. Aniline is less basic than N-methylaniline( $C_{6}H_{5}NHCH_{3},$ ) due to the absence of an electron donating group. Thus the decreasing order of pKb values is-

$C_{6}H_{5}NH_{2}>C_{6}H_{5}NHCH_{3}>C_{2}H_{5}NH_{2}>(C_{2}H_{5})_{2}NH$

Question 13.4 Arrange the following:

(ii) In increasing order of basic strength:

$C_{6}H_{5}NH_{2},$ $C_{6}H_{5}N(CH_{3})_{2},$ $(C_{2}H_{5})_{2}NH$ and $CH_{3}NH$

Answer :

The increasing order of basic strength-

$C_{6}H_{5}NH_{2}<C_{6}H_{5}N(C_{2}H_{5})_{2}<CH_{3}NH_{2}<(C_{2}H_{5})_{2}NH$

Aliphatic amines are more basic than aromatic amines due to the presence of high electron density of electron at N atom so that it can donate an electron. On the other hand in aromatic amines the lone pair of electron delocalised in the benzene ring so the availability of electron is less.

$(C_{2}H_{5})_{2}NH$ is more basic because of +I effect of the two ethyl group (so electron density is highest in this compound) while methylamine has one ethyl group. In aromatic amines, diethylaniline is more basic due to the presence of more number of electron donating group.

Question 13.4 Arrange the following:

(iii) In increasing order of basic strength:

(a) Aniline, p-nitroaniline and p-toluidine

Answer :

Structure of the following compounds-

1596565835752 Aniline, p-nitroaniline p-toluidine

In p-toluidine, the methyl group increases the electron density at the N atom (+I effect of methyl), so it is more basic than aniline. In p-nitroaniline, the nitro group is an electron withdrawing group, so it decreases the electron density from N atom, so it is less basic than aniline.

Question 13.4 Arrange the following:

(iii) In increasing order of basic strength:

(b) $C_{6}H_{5}NH_{2},$ $C_{6}H_{5}NHCH_{3},$ $C_{6}H_{5}CH_{2}NH_{2}.$

Answer :

The increasing order of basic strength- $C_{6}H_{5}NH_{2}<C_{6}H_{5}NHCH_{3}<C_{6}H_{5}CH_{2}NH_{2}$

$C_{6}H_{5}NHCH_{3},$ it is more basic than aniline due to the presence of one electron donating group which increases the electron density at N atom. And also in this compound N atom is directly attached with the benzene ring, so due to -R effect $C_{6}H_{5}NHCH_{3},$ is less basic than $C_{6}H_{5}CH_{2}NH_{2}.$

Question 13.4 Arrange the following:

(iv) In decreasing order of basic strength in gas phase:

$C_{2}H_{5}NH_{2},$ $(C_{2}H_{5})_{2}NH,$ $(C_{2}H_{5})_{3}N,$ and $NH_{3}$

Answer :

Decreasing order of basic strength in the gas phase-

$(C_{2}H_{5})_{3}N>(C_{2}H_{5})_{2}NH>C_{2}H_{5}NH_{2}>NH_{3}$

In the gas phase, there is no solvation effect. So, the basicity directly depends on the number of electron donation group. More is the electron donating group, higher is the +I effect and if the size of the donating group is large +I effect should also high.

Question 13.4 Arrange the following:

(v) In increasing order of boiling point:

$C_{2}H_{5}OH,$ $(CH_{3})_{2}NH,$ $C_{2}H_{5}NH_{2}$

Answer :

The boiling point of the compounds depends on the extent of hydrogen bonding of the compound. More is the extent; higher is the boiling point. $C_{2}H_{5}NH_{2}$ , it contains two H atoms, so it has a higher boiling point than $(CH_{3})_{2}NH,$ . On the other hand, Oxygen atom is more electronegative than N atom, so the strength of hydrogen bonding is higher in $C_{2}H_{5}OH,$ than $C_{2}H_{5}OH,$

THus the increasing order of boiling point is-

$(C_{2}H_{5})_{2}NH<C_{2}H_{5}NH_{2}<C_{2}H_{5}OH$

Question 13.4 Arrange the following:

(vi) In increasing order of solubility in water:

$C_{6}H_{5}NH_{2},$ $(C_{2}H_{5})_{2}NH,$ $C_{2}H_{5}NH_{2}.$

Answer :

More extensive is the hydrogen bonding, more is the solubility in water. $C_{2}H_{5}NH_{2}.$ contains two H atom while $(C_{2}H_{5})_{2}NH,$ contains only one. So, the $C_{2}H_{5}NH_{2}.$ undergoes extensive hydrogen bonding. Hence its solubility is more than $(C_{2}H_{5})_{2}NH,$ in water. On increasing the molecular mass of amines decreases its solubility in water because of an increase in bulky hydrophobic groups.

Thus the increasing order of solubility in water is-

$C_{6}H_{5}NH_{2<}(C_{2}H_{5})NH<C_{2}H_{5}NH_{2}$

Question 13.5 How will you convert:

(i) Ethanoic acid into methanamine

Answer :

Ethanoic acid on reacting with $SOCl_{2}$ replaces the OH by $Cl$ and then reacts it with an excess of ammonia molecule to produce methanamide ( $CH_{3}CONH_{2}$ ) which by Hoffmann bromamide degradation reaction gives us desired product (methenamine)

$\\CH_{3}COOH+SOCl_{2}\rightarrow CH_{3}COCl\overset{NH_{3}(excess)}{\rightarrow}CH_{3}CONH_{2}$ $\overset{Br_{2}+KOH}{\rightarrow}CH_{3}NH_{2}$

Question 13.5 How will you convert:

(ii) Hexanenitrile into 1-aminopentane

Answer :

First, acid hydrolysis of hexanenitrile which convert CN group into COOH group and then react with $SOCl_{2}$ which replace COOH by $COCl$ and then the excess of ammonia which replace Cl by $NH_{2}$ and then Hoffmann bromamide degradation reaction which gives us desired product.

1596565852749

Question 13.5 How will you convert:

(iii) Methanol to ethanoic acid

Answer :

methanol on reacting with phosphorous pentachloride, OH is reduced by Cl to form chloromethane, which on reacting with ethanolic sodium cyanide gives $CH_{3}CN$ followed by acidic hydrolysis it produces ethanoic acid.( $CH_{3}COOH$ )

1596565860357

Question 13.5 How will you convert:

(iv) Ethanamine into methanamine

Answer :

1596565867668

React ethanamine with nitrous acid to form an azo compound, which further reacts with water to form ethanol, which on oxidising gives ethanoic acid. After treating with an excess of ammonia the ethanoic acid becomes ethanamide, which on further reacting with Bomine and a strong base (Hoffmann bromamide degradation reaction) to form methenamine.

Question 13.5 How will you convert:

(v) Ethanoic acid into propanoic acid

Answer :

1596565875265

Acetic acid on reduction with lithium aluminium hydride followed by acid hydrolysis gives ethanol, which on reacting with $PCl_{5}$ gives ethyl chloride. Ethyl chloride on further reacting with ethanolic sodium cyanide to form propanenitrile, which on acidic hydrolysis gives the desired output.

Question 13.5 How will you convert:

(vi) Methanamine into ethanamine

Answer :

1596565882589

Methenamine on diazotisation gives methane diazonium salt, which on further hydrolysis form methanol. Methanol on reacting with $PCl_{5}$ followed by ethanolic sodium cyanide ( $NaCN$ ) produces Ethane nitrile, which on reduction with $H_{2}/Pd$ gives ethenamine.

Question 13.5 How will you convert:

(vii) Nitromethane into dimethylamine

Answer :

Conversion of Nitromethane into dimethylamine is shown below:

$(Nitromethane)CH_{3}NO_{2} \xrightarrow[H]{Sn/HCL} (Aminomethane)CH_{3}NH_{2} \overset{CHCl_{3}/KOH}{\rightarrow}\\ (Methyl\ isocyanide)CH_{3}NC \xrightarrow [H]{Na/C_{2}H_{5}OH}(Dimethylamine)CH_{3}NHCH_{3}$

Question 13.5 How will you convert:

(viii) Propanoic acid into ethanoic acid?

Answer :

1596565896117

Propanoic acid on reacting with an excess of ammonia gives propanamide which on further react with bromine and potassium hydroxide (Hoffmann bromamide degradation reaction) gives ethenamine. On diazotisation, ethenamine converted into azo salt which on treating with water forms ethanol. Ethanol on oxidation provides ethanoic acid.

Question 13.6 Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.

Answer :

Primary, secondary and tertiary amines are distinguished by the Hinsberg's reagent test. In this test, the amines are allowing to react with benzene sulphonyl chloride(Hinsberg's reagent). All three types of amines give different products.

1.Primary amine reacts with benzene sulphonyl chloride $(C_{6}H_{5}SO_{2}Cl)$ gives N-ethylbenzene sulphonamide which is soluble in alkali.

1596566276020

2. when secondary amines react it gives N, N-diethyl benzene sulphonamide which is insoluble in alkali.

1596566321487

3. Tertiary amine does not react with Hinsberg's reagent.

Question 13.7 Write short notes on the following:

(i) Carbylamine reaction

Answer :

Carbylamine reaction-

Aliphatic and aromatic primary amines on reacting with chloroform and ethanolic KOH form isocyanides or carbylamine which is foul smelling substance. Secondary and tertiary amine does not give this reaction. This reaction is known as carbylamine reaction or isocyanide test. It is used to distinguish primary amine.

$R-NH_{2}+CHCl_{3}+KOH\xrightarrow[heat]{^\Delta }R-NC+3KCl+3H_{2}O$

Question 13.7 Write short notes on the following:

(ii) Diazotisation

Answer :

DIAZOTISATION-

Aromatic primary amines react with nitrous acid ( $NaNO_{2}+2HCl$ )at low temperature to form diazonium salts. This conversion of aromatic primary amines into diazonium salt is known as diazotisation.

$C_{6}H_{5}NH_{2}+NaNO_{2}+2HCl\overset{273-278K}{\rightarrow}C_{6}H_{5}N^{+}_{2}Cl^{-}+NaCl+2H_{2}O$

1596565915931 1596565910426

Question 13.7 Write short notes on the following:

(iii) Hofmann’s bromamide reaction

Answer :

Hofmann’s bromamide reaction-
When an amide is treated with bromine in aqueous solution or ethanolic solution of sodium hydroxide. It produces primary amines with one less carbon atom than the parent compound. This reaction is known as the Hoffmann bromide degradation reaction.
1596566544782 R = alkyl group like $CH_{3},C_{2}H_{5}....etc$

Question 13.7 Write short notes on the following:

(iv) Coupling reaction

Answer :

Coupling reaction-
The reaction of joining of diazonium salt to the aromatic ring (like phenol) through $-N=N-$ bond is known ass coupling reaction. Benzene diazonium chloride reacts with phenol in which the phenol molecule at its para position is coupled with the diazonium salt( $C_{6}H_{5}N^{+}_{2}Cl^{-}$ ) to give p-hydroxyazobenzene(orange in colour)

Reactions-

1596566555567 diazonium salt reacts with aniline to give p-aminoazobenzene (yellow colour)

Question 13.7 Write short notes on the following:

(v) Ammonolysis

Answer :

Ammonolysis-
When an alkyl or benzyl halide is going to react with the solution of ammonia(an ethanolic) undergoes nucleophilic substitution reaction in which the halogen atom is replaced by an amino ( $-NH_{2}$ ) group. This process of cleavage of the carbon-halogen(C-X) bond by ammonia molecule is known as ammonolysis.

1596566372072 When this substituted ammonium salt is treated with a strong base it gives primary amine
$RN^{+}H_{3}X^{-}+NaOH\rightarrow R-NH_{2}+H_{2}O+NaX$

Above primary amine is behaves as a nucleophile and can further react with an alkyl halide to form secondary and tertiary amines and finally quaternary ammonium salt.

Question 13.7 Write short notes on the following:

(vi) Acetylation

Answer :

Acetylation-
The introduction of an acetyl group ( $CH_{3}CO$ )in a molecule is known as acetylation.
1596565937421

Example-

1596566385638 Aliphatic and aromatic primary and secondary amines undergo acetylation reaction when treated with acid chlorides, anhydrides or esters by nucleophilic substitution. Here Hydrogen atom of $NH_{2}$ or $NH$ is replaced by $CH_{3}CO$ the group.
1596566392759

Question 13.7 Write short notes on the following:

(vii) Gabriel phthalimide synthesis.

Answer :

Gabriel phthalimide synthesis.-
It is used for mainly aliphatic primary amine preparation. Phthalimide reacts with ethanolic potassium hydroxide and produces potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis gives corresponding primary amine. Aromatic amines cannot be produced by this method due to the inability of aryl halides to do nucleophilic substitution reaction with anion produced by phthalimide.

1596565944101

Question 13.8 Accomplish the following conversions:

(i) Nitrobenzene to benzoic acid

Answer :

First, reduce nitrobenzene into aniline with the help of $H_{2}/Pd$ . Now convert aniline to diazonium salt followed by sandmeyers reaction ( $CuCN/KCN$ ) to benzonitrile and after acid hydrolysis benzonitrile becomes benzoic acid.

1596565952459

Question 13.8 Accomplish the following conversions:

(ii) Benzene to m-bromophenol

Answer :

First, We do nitration of benzene so that it can direct the bromine to meta position and then bromination of nitrobenzene after that reduce it with the help of $Sn/HCl$ followed by diazotisation reaction and then hydrolysis in warm water.
1596565960886

Question 13.8 Accomplish the following conversions:

(iii) Benzoic acid to aniline

Answer :

On reacting benzoic acid with $SOCl_{2}$ we get benzene sulphonyl chloride and when it reacts with ammonia the chlorine is replaced by $NH_{2}$ and after that do Hoffmann bromide degradation reaction to remove -CO group.
screenshot-388

Question 13.8 Accomplish the following conversions:

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

Answer :

First, we do bromination of aniline which gives 2,4,6-tribromoaniline after that do diazotisation reaction which converts $NH_{2}$ group into $N_{2}Cl$ and then react it with fluoroboric acid( $HBF_{4}+\Delta$ ) gives us desired product.

1596565973725

Question 13.8 Accomplish the following conversions:

(v) Benzyl chloride to 2-phenylethanamine

Answer :

On reacting benzyl chloride with the ethanolic sodium cyanide ( $NaCN$ ), it replaces the Chlorine with CN molecule. And then reduction with the help of $H_{2}/Ni$

1596565980143

Question 13.8 Accomplish the following conversions:

(vi) Chlorobenzene to $p-chloroaniline$

Answer :

On nitration of chlorobenzene, it gives p-nitrochlorobenzene which is on further reacting with $H_{2}/Pd$ , it reduces $NO_{2}$ to $NH_{2}$

screenshot-394

Question 13.8 Accomplish the following conversions:

(vii) Aniline to $p-bromoaniline$

Answer :

Aniline on reacting with the acetic anhydride in the presence of pyridine it gives N-phenylethanamide which on further reacting with $Br_{2}/CH_{3}COOH$ the $NHCOCH_{3}$ group direct the bromine to the para- position and after hydrolysis, we can recover the original $NH_{2}$ group at the same position

1596565991838

Question 13.8 Accomplish the following conversions:

(viii) Benzamide to toluene

Answer :

Use the Hoffmann bromamide degradation reaction so that it gives aniline and then do diazotisation reaction so that $NH_{2}$ group is replaced by $N_{2}Cl$ and then reduction with the help of ( $H_{2}O/H_{3}PO_{2}$ ) so that it reduced into the benzene ring. And then alkylation of benzene gives toluene as a final product.

1596566000675

Question 13.8 Accomplish the following conversions:

(ix) Aniline to benzyl alcohol

Answer :

Aniline on treating with nitrous acid followed by sandmeyers reaction gives benzonitrile ( $C_{6}H_{5}C\equiv N$ ), which on acidic hydrolysis form benzoic acid ( $C_{6}H_{5}COOH$ ) and then reduction process with $LiAlH_{4}$ (lithium aluminium hydride) to get desired product.

1596564302098

Question 13.9 Give the structures of A, B and C in the following reactions:

(i) $CH_{3}CH_{2}I\xrightarrow[]{NaCN}A\xrightarrow[Partial hydrolysis]{OH^{-}}B\xrightarrow[]{NaOH+Br_{2}}C$

Answer :

In the first reaction, the iodine is replaced by CN so the structure of (A) is $CH_{3}CH_{2}CN$ ( propionitrile) . On partial hydrolysis, the CN converted in to $CONH_{2}$ , So the structure of B is $CH_{3}CH_{2}CONH_{2}$ (propanamide) and the third reactant is Hoffmann bromide degradation reaction, So just remove the -CO group and it becomes a primary aliphatic amine. Thus the structure of C is $CH_{3}CH_{2}NH_{2}$ (ethanamine)

Question 13.9 Give the structures of A, B and C in the following reactions:

(ii) $C_{6}H_{5}N_{2}Cl\xrightarrow[]{CuCN}A\xrightarrow[]{H_{2}O/H^{+}}B\xrightarrow[\Delta ]{NH_{3}}C$

Answer :

The first part is the Sandmeyers reaction so CN replaces the $N_{2}Cl$ of the reactant and the product (A) is cynobenzene $(C_{6}H_{5}CN)$ . On acidic hydrolysis of cynobenzene give product (B) benzoic acid $(C_{6}H_{5}COOH)$ and benzoic acid react with ammonia (heat) it produces product (C)benzamide $(C_{6}H_{5}CONH_{2})$
The structure of A, C and B are-
screenshot-389

Question 13.9 Give the structures of A, B and C in the following reactions:

(iii) $CH_{3}CH_{2}Br\xrightarrow[]{KCN}A\xrightarrow[]{LiAlH_{4}}B\xrightarrow[0^{\circ}C]{HNO_{2}}C$

Answer :

The structure of A, B, C are respectively $CH_{3}CH_{2}CN,CH_{3}CH_{2}CH_{2}NH_{2}$ and $CH_{3}CH_{2}CH_{2}OH$

In the first reaction, the bromine is replaced by cyanide molecule ( $CN$ ) and after a reduction by $LiAlH_{4}$ the CN, molecules become $CH_{2}NH_{2}$ .

Question 13.9 Give the structures of A, B and C in the following reactions:

(iv) $C_{6}H_{5}NO_{2}\xrightarrow[]{Fe/HCl}A\xrightarrow[273K]{NaNO_{2}+HCl}B\xrightarrow[H_{2}O/H^{+}]{\Delta }C$

Answer :

A- $C_{6}H_{5}NH_{2}$

B- $C_{6}H_{5}N^{+}_{2}Cl^{-}$

C- $C_{6}H_{5}OH$

The First nitrobenzene reduced to aniline and then it reacts with $(NaNO_{2}/HCl)$ to form benzene diazonium salt and is hydrolysed to form phenol( $C_{6}H_{5}OH$ )

Question 13.9 Give the structures of A, B and C in the following reactions:

(v) $CH_{3}COOH\xrightarrow[\Delta ]{NH_{3}}A\xrightarrow[]{NaOBr}B\xrightarrow[]{NaNO_{2}/HCl}C$

Answer :

A- ethanamide ( $CH_{3}CONH_{2}$ )

B- methenamine ( $CH_{3}NH_{2}$ )

C- methanol ( $CH_{3}OH$ )

Acetic acid reacts with ammonia, gives ethanamide and the second reaction is Hoffmann bromamide degradation reaction so it removes -CO group and form methenamine and after that the last product is methanol.

Question 13.9 Give the structures of A, B and C in the following reactions:

(vi) $C_{6}H_{5}NO_{2}\xrightarrow[]{Fe/HCl}A\xrightarrow[273K]{HNO_{2}}B\xrightarrow[]{C_{6}H_{5}OH}C$

Answer :

Initially, product A is the reduction of nitrobenzene means it is an aniline. when A react with $HNO_{2}$ it gives benzene diazonium chloride(B). And when B reacts with another aromatic ring (phenol) it's a coupling reaction so it gives azodye (p-Hydroxyazobenzene)

A= Aniline B = benzene diazonium chloride C= p-Hydroxyazobenzene

1596566051588 1596566042855 1596566034767

Question 13.10 An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with $Br_{2}$ and $KOH$ forms a compound ‘C’ of molecular formula $C_{6}H_{7}N.$ . Write the structures and IUPAC names of compounds A, B and C.

Answer :

The molecular formula of C = $C_{6}H_{7}N$

$A\xrightarrow[aq NH_{3}]{\Delta }B$

$B+Br_{2}+KOH\rightarrow C(C_{6}H_{7}N)$ it is a Hoffmann bromamide degradation reaction so, the product should be $C_{6}H_{5}NH_{2}$ (aniline)

So B should be benzamide ( $C_{6}H_{5}CONH_{2}$ ) and A should be benzoic acid ( $C_{6}H_{5}COOH$ )

Therefore the above reaction proceeds as-

screenshot-391

Question 13.11 Complete the following reactions:

(i) $C_{6}H_{5}NH_{2}+CHCl_{3}+alc.KOH\rightarrow$

Answer :

The given above reaction is carbylamine reaction in which primary aromatic amines react with chloroform and alc. potassium hydroxide to give isocyanide or carbylamine as a product which is foul-smelling substance.

$C_{6}H_{5}NH_{2}+CHCl_{3}+alc.KOH\rightarrow$ $C_{6}H_{5}NC+KCl +H_{2}O$

Question 13.11 Complete the following reactions:

(ii) $C_{6}H_{5}N_{2}Cl+H_{3}PO_{2}+H_{2}O\rightarrow$

Answer :

When benzene diazonium chloride reacts with $H_{3}PO_{2}$ it reduced into benzene and $H_{3}PO_{2}$ oxidised into $H_{3}PO_{3}(+3)$ , HCl produced as a by-product in this reaction.

$C_{6}H_{5}N_{2}Cl+H_{3}PO_{2}+H_{2}O\rightarrow$ $C_{6}H_{6}+H_{3}PO_{3}+HCl$

Question 13.11 Complete the following reactions:

(iii) $C_{6}H_{5}NH_{2}+H_{2}SO_{4}(conc.)\rightarrow$

Answer :

Aniline is a base and $H_{2}SO_{4}$ is acid. So, basically, it is acid-base reaction aniline accepts $H^{+}$ ions from sulphuric acid and forms a product anilinium hydrogen sulphate.

$C_{6}H_{5}NH_{2}+H_{2}SO_{4}(conc.)\rightarrow$ $C_{6}H_{5}NH_{3}^{+}HSO_{4}^{-}$

Question 13.11 Complete the following reactions:

(iv) $C_{6 } H_{5}N_{2}Cl+C_{2}H_{5}OH\rightarrow$

Answer :

Ethanol reduces the benzene diazonium chloride to benzene and oxidises into ethanal ( $CH_{3}CHO$ ), hydrochloric acid produces as a by-product and evolution of nitrogen gas is also occurring.

$C_{6 } H_{5}N_{2}Cl+C_{2}H_{5}OH\rightarrow$ $C_{6}H_{6}+CH_{3}CHO+N_{2}+HCl$

screenshot-392

Question 13.11 Complete the following reactions:

(v) $C_{6}H_{5}NH_{2}+Br_{2}(aq)\rightarrow$

Answer :

Aniline on bromination in the presence of water at room temperature gives a white ppt of 2, 4, 6- tribromoaniline

1596564428916

Question 13.11 Complete the following reactions:

(vi) $C_{6}H_{5}NH_{2}+(CH_{3}CO)_{2}O\rightarrow$

Answer :

Aniline reacts with acetic anhydride by a nucleophilic substitution reaction. Here the replacement of H atom from $NH_{2}$ group is by an acyl group. The product formed is N-phenylethanamide.p. The reaction is carried out in the presence of a base stronger than the amine, like pyridine

$C_{6}H_{5}NH_{2}+(CH_{3}CO)_{2}O\rightarrow$ $C_{6}H_{5}NHCOCH_{3}+CH_{3}COOH$

1596564455558

Question 13.11 Complete the following reactions:

(vii) $C_{6}H_{5}N_{2}Cl\xrightarrow[(ii)NaNO_{2}/Cu,\Delta )]{(i) HBF_{4})}$

Answer :

In this reaction, the chloride ion is replaced by $BF_{4}^{-}$ ion. When diazonium fluoroborate is heated with aqueous sodium nitrite in the presence of copper, the diazonium group is replaced by –NO2 group.

$C_{6}H_{5}N_{2}Cl\xrightarrow[(ii)NaNO_{2}/Cu,\Delta )]{(i) HBF_{4})}$ $C_{6}H_{5}NO_{2}+N_{2}+NaBF_{4}$

1596566636152

Question 13.12 Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

Answer :
Gabriel phthalimide synthesis is used for mainly aliphatic primary amine preparation. Phthalimide reacts with ethanolic potassium hydroxide and produces potassium salt of phthalimide

which on heating with alkyl halide followed by alkaline hydrolysis gives corresponding primary amine.

1596566079767

Aromatic amines cannot be produced by this method due to the inability of aryl halides to do nucleophilic substitution reaction with anion produced by phthalimide.

Hence aromatic amines cannot be produced by this method.

1596566087216 $No \ reaction$ {if R = Aromatic compound}

Question 13.13 Write the reactions of

(i) aromatic primary amines with nitrous acid.

Answer :

Aromatic primary amines react with nitrous acid( $NaNO_{2}+HCl$ ) at 273-278 K to form stable diazonium salts. aniline reacts with nitrous acid to form benzene diazonium chloride.
1596566096520 1596566104222

Question 13.13 Write the reactions of

(ii) aliphatic primary amines with nitrous acid.

Answer :

When aliphatic primary amines react with nitrous acid to form unstable diazonium salts which on further produces alcohol and hydrochloric acid ( $HCl$ ) with the evolution of dinitrogen gas ( $N_{2}$ ).
1596566113004

Question 13.14 Give a plausible explanation for each of the following:

(i) Why are amines less acidic than alcohols of comparable molecular masses?

Answer :

Protonation of amines gives- $RNH_{2}\rightarrow RNH^{-}+H^{+}$ (amide ion)
protonation of alcohol gives - $R-OH\rightarrow RO^{-}+H^{+}$ (alkoxide ion)

the negative charge is easily accommodated by O atom because it is more electronegative than N atom. So, the amide ion is less stable than the alkoxide ion. Hence amines are less acidic than alcohols.

Question 13.14 Give a plausible explanation for each of the following:

(ii) Why do primary amines have a higher boiling point than tertiary amines?

Answer :

The extent of inter-molecular hydrogen bonding in primary amines is higher than the secondary amines. Secondary amine has only one hydrogen, and tertiary amines have no hydrogen atom for inter-molecular hydrogen bonding whereas primary amine has two hydrogen atom available for inter-molecular hydrogen bonding. And we know that higher the extent of hydrogen bonding, higher is the boiling point.

1596566122228 (In $1^{0}$ amine), 1596566129657 (In $2^{0}$ amines), 1596566136983 (In $3^{0}$ amine)

Question 13.14 Give plausible explanation for each of the following:

(iii) Why are aliphatic amines stronger bases than aromatic amines?

Answer :

Aliphatic amines stronger bases than aromatic amines due to resonance in aromatic base the lone pair of electron of N atom delocalised over the benzene ring, which makes it less available for electron donation. On the other hand, in aliphatic amines( $R-NH_{2}$ ), the electron donor group is attached which increase the electron density at N atom. Thus the aliphatic amines are more basic than aromatic amines.

1596566484941

Features of Amines Class 12 NCERT Chapter

In NCERT solutions for class 12 chemistry chapter 13 Amines, there are ten sub-topics from which questions are generally asked that covers important concepts of chemistry such as the structure, nomenclature, and classification of amines and also covers its chemical and physical properties and method of preparation of diazonium salts. After studying NCERT solutions for Class 12 chemistry chapter 13 Amines you will be able to describe amines as derivatives of ammonia; classify amines as primary, secondary and tertiary; explain IUPAC nomenclature of amines; describe methods of preparation of amines and properties of amines.

Topics and Sub-topics of Amines Class 12 NCERT Chapter

13.1 Structure of Amines

13.2 Classification

13.3 Nomenclature

13.4 Preparation of Amines

13.5 Physical Properties

13.6 Chemical Reactions

13.7 Method of Preparation of Diazonium Salts

13.8 Physical Properties

13.9 Chemical Reactions

13.10 Importance of Diazonium Salts in the Synthesis of Aromatic Compounds

NCERT Solutions for Class 12 Chemistry

Chapter 1	NCERT solutions for class 12 chapter 1 The Solid State
Chapter 2	NCERT solutions for class 12 chemistry chapter 2 Solutions
Chapter 3	NCERT Solutions for class 12 chemistry chapter 3 Electrochemistry
Chapter 4	NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics
Chapter 5	NCERT Solutions for class 12 chemistry chapter 5 Surface chemistry
Chapter 6	NCERT solutions for class 12 chemistry General Principles and Processes of isolation of elements
Chapter 7	NCERT solutions for class 12 chemistry chapter 7 The P-block elements
Chapter 8	NCERT Solutions for class 12 chemistry chapter 8 The d and f block elements
Chapter 9	NCERT solutions for class 12 chemistry chapter 9 Coordination compounds
Chapter 10	NCERT Solutions for class 12 chemistry chapter 10 Haloalkanes and Haloarenes
Chapter 11	NCERT solutions for class 12 chemistry Alcohols, Phenols, and Ethers
Chapter 12	NCERT solutions for class 12 chemistry chapter 12 Aldehydes, Ketones and Carboxylic Acids
Chapter 13	NCERT solutions for class 12 chemistry chapter 13 Amines
Chapter 14	NCERT solutions for class 12 chemistry chapter 14 Biomolecules
Chapter 15	NCERT solutions for class 12 chemistry chapter 15 Polymers
Chapter 16	NCERT solutions for class 12 chemistry chapter 16 Chemistry in Everyday life

NCERT Solutions for Class 12 Subject wise

NCERT solutions for class 12 biology

NCERT solutions for class 12 maths

NCERT solutions for class 12 chemistry

NCERT solutions for class 12 physics

Benefits of NCERT Solutions for class 12 chemistry chapter 13 Amines

The answers presented in a step-by-step manner in the NCERT solutions for class 12 chemistry chapter 13 Amines will help in understanding chapter easily.
It will be easy to revise because the detailed solutions will help you to remember the concepts and fetch you good marks.
Homework problems won't bother you anymore, all you need to do is check the detailed NCERT solutions for class 12 chemistry chapter 13 Amines and you are ready to sail.

If you have a doubt or question that is not available here or in any of the chapters, contact us. You will get all the answers that will help you score well in your exams.

Also check NCERT Exemplar Class 12 Solutions

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Question (FAQs)

1. What is the weightage of NCERT Class 12 Chemistry chapter 13 in JEE Mains?

Around 5-7 marks worth questions are asked from this chapter in JEE mains. The questions for Amines can be practiced from the NCERT book, NCERT exemplar and JEE Main previous year papers.

2. Where can I find complete solutions of NCERT class 12 Chemistry?

For complete solutions : https://school.careers360.com/ncert/ncert-solutions-class-12-chemistry. Students can download the chapter wise solutions PDF.

3. What is the weightage of NCERT Class 12 Chemistry chapter 13 in CBSE Board Exams?

This chapter holds weightage of 5-6 Marks in Board exams. Follow NCERT syllabus to get a good score in the CBSE board exam.

4. What is the weightage of NCERT class 12 Chemistry chapter 14 in NEET?

Weightage of NCERT class 12 Chemistry chapter 13 in NEET is around 3 percent

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i couldn't clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

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Upasana Barman 24 August,2022

i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

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Mayankjha.student2007 23 August,2022

if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

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Nuthalapati Vyshnavi 21 August,2022

I will have my boards in Feb 2023 (Class 12) and I still haven't completed 1 chapter in any subject (Physics, chemistry and maths) , My 11th concepts are not that great but its okish, I wanted to know that If I do start studying from 15th of August Can I get above 85?

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

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khushi.thakurbbk 07 September,2022

I want to take a admission in class 12 privately.. so can I???

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

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Parvati Solanki 02 June,2022

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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms^-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×10⁷J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms⁻² :

Option 1)

2.45×10⁻³ kg

Option 2)

6.45×10⁻³ kg

Option 3)

9.89×10⁻³ kg

Option 4)

12.89×10⁻³ kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

A particle is projected at 60⁰ to the horizontal with a kinetic energy $K$ . The kinetic energy at the highest point

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

In the reaction,

$2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, $Mg_{3}(PO_{4})_{2}$ will contain 0.25 mole of oxygen atoms?

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol^-1) is

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t²) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m² , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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