NCERT Solutions for Class 12 Chemistry Chapter 13 Amines – Free PDF

NCERT Solutions for Class 12 Chemistry Chapter 13 – Amines – FREE PDF Download

CoolGyan provides you Free PDF download of NCERT Solutions for Class 12 Chemistry Chapter 13 – Amines solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 13 – Amines Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

Revision Notes Class 12 chemistry
NCERT Solutions Class 12 chemistry

Subtopics of Class 12 Chemistry Chapter 13 – Amines

1. Structure of Amines
2. Classification
3. Nomenclature
4. Preparation of Amines
5. Physical Properties
6. Chemical Reactions
7. Methods of preparation of Diazonium Salts
8. Physical Properties
9. Chemical reactions
10. Importance of Diazonium salts in Synthesis of aromatic Compounds

Class 12 Chemistry NCERT Solutions (Chapter 13 Amines ) – Important Questions

INTEXT Questions

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

 

Amines class 12 NCERT Solutions: 

(a) Primary
(b) Tertiary
(c) Primary
(d) Secondary

Question 2.
(a) Write structures of different isomeric amines corresponding to the molecular formula, C₄H₁₁N.
(b) Write IUPAC names of all the isomers.
(c) What type of isomerism is exhibited by different pairs of amines?
 

Amines class 12 NCERT Solutions:


 

Question 3.
How will you convert: (i) Benzene into aniline, (ii) Benzene into N,N-dimethylaniline, (iii) Cl(CH₂)₄Cl into hexan-1,6-diamine?
 

Solution:


 

Question 4.
Arrange the following in increasing order of their basic strength. 

(a) C₂H₅NH₂, C₆H₅NH₂, NH₃, C₆H₅CH₂NH₂ and (C₂H₅)₂NH
(b) C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N, C₆H₅NH₂
(c) CH₃NH₂< (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂> C₆H₅CH₂NH₂.

 

Amines class 12 NCERT Solutions:
(a) The order of basicity is C₆H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ < C₂H₅NH₂ < (C₂H₅)₂NH
The basic nature of amine arises from their ability to donate the lone pair of electrons on N to electrophiles. The availability of this l.p. depends on two factors :
(i) Electron donating/withdrawing effect of the alkyl groups attached to the N atom.
(ii) Steric hindrance posed by alkyl groups around N.
The presence of alkyl groups on the N atom increases the electron density and makes the l.p. more available for electrophiles. This happens due to the +1 effect of the alkyl groups. But, if the alkyl groups are too bulky or too many in number, they tend to sterically hinder the incoming proton and the basic strength decreases. These two factors working in opposing directions, tend to balance out each other in 2° amines, making them most basic and the basic strength follows the order :
 

Question 5.
Complete the following acid-base reactions and name the products :

1. CH₃CH₂CH₂NH₂ + HCl →
2. (C₂H₅)₃N + HCl →

Solution:

Question 6.
Write the reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
Solution:

Question 7.
Write the chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Solution:

Question 8.
Write structures of different isomers corresponding to the molecular formula, C₃H₉N. Write IUPAC name of the isomers which will liberate nitrogen gas on treatment with nitrous acid.
Solution:

Question 9.
Convert

1. 3-Methylaniline into 3-nitrotoluene
2. Aniline into 1,3,5-tribromobenzene.

Solution:

Amines class 12 NCERT Solutions – NCERT Exercises

Question 1.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

1. (CH₃)₂CHNH₂
2. CH₃(CH₂)₂NH₂
3. CH₃NHCH(CH₃)₂
4. (CH₃)₃CNH₂
5. C₆H₅NHCH₃
6. (CH₃CH₂)₂NCH₃
7. m-BrC₆H₄NH₂

Solution:

1. Propan-2-amine (primary),
2. Propan-l-amine (primary),
3. N-Methylpropan-2-amine (secondary),
4. 2-Methylpropan-2-amine (primary),
5. N-Methylbenzenamine or N-Methylaniline (secondary),
6. N-Ethyl-N-methylethanamine (tertiary),
7. 3-Bromobenzenamine or 3-Bromoaniline (primary).

Question 2.
Give one chemical test to distinguish between the following pairs of compounds.

1. Methylamine and dimethylamine
2. Secondary and tertiary amine
3. Ethylamine and W-methylaniline
4. Aniline and benzylamine
5. Aniline and N-methylaniline

Solution:

Question 3.
Account for the following :
(i) pK_b of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o- and p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(vii) Gabriel phthalimide synthesis is preferred .for synthesising primary amines.
Solution:
(i) If the pK_b value of any base or compound is higher than that of another, it implies that the former is a weaker base than the latter. In aniline, the N-atom is attached to the benzene ring and therefore the lone pair on N is delocalised over the entire benzene ring. As a result, it cannot accept a proton or any other electrophile.

This is why it has a lower K_b value (lower basic strength) and high corresponding pK_b value.

In methylamine, CH₃NH₂, the electron density on nitrogen is greater than that in case of aniline. This is because -CH₃ group in methylamine, by virtue of its +1 effect, increases electron density on N, which is more available for protonation.

(ii) Any compound capable of forming hydrogen bonds with water, dissolves in it. Ethylamine is able to do the same and hence its solubility.

However, in aniline, the bulky hydrocarbon part – C₆H₅ prevents the formation of effective hydrogen bonding and therefore it is not soluble.

(iii) The formation of hydrated ferric oxide may be understood by taking into consideration the basic strength of CH₃NH₂. In presence of CH₃NH₂, water hydrolyses as

(iv) During nitration, the nitration mixture used (cone. HNO₃ and cone. H₂SO₄) protonates the NH₂ group to produce anilinium ion as

But when this reaction is carried out with aniline, no electrophile generation takes place. The reason being the presence of aniline as a base.

Aniline is a Lewis base, reacts with AlCl₃ and hence deactivates it. The Lewis acid is therefore no more available for electrophile generation and hence reaction does not take place.

(vi) Diazonium salts carry a N atom with a positive charge. This positive charge is well dispersed in aromatic diazonium salts through resonance as shown below :

Such a charge delocalisation is not possible in aliphatic amines and hence they are less stable.

(vii) Gabriel phthalimide reaction gives pure primary amines without any contamination of secondary and tertiary amines. Therefore, it is preferred for synthesising primary amines.

Question 4.
Arrange the following :
(i) In decreasing order of the pK_b values: C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂
(ii) In increasing order of basic strength: C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH and CH₃NH₂
(iii) In increasing order of basic strength:
(a) Aniline, p-nitroaniline and p-toluidine
(b) C₆H₅NH₂, C₆H₅NHCH₃, C₆H₅CH₂NH₂.
(iv) In decreasing order of basic strength in gas phase : C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N and NH₃
(v) In increasing order of boiling point: C₂H₅OH, (CH₃)₂NH, C₂H₅NH₂
(vi) In increasing order of solubility in water: C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂.
Solution:


The availability of l.p. on N of p-nitroaniline is drastically reduced by presence of electron withdrawing -NO₂ group on it.

In contrast, presence of electron releasing -CH₃ group increases the electron density on N atom and improves basicity in p-toluidine.

(b) C₆H₅NH₂ < C₆H₅NHCH₃ < C₆H₅CH₂NH₂ Involvement of l.p. of N in resonance causes aniline to have low basicity. In II, the -Me group through its +I effect improves the electron density on N and therefore its basic strength increases. In III, the -NH₂ is farther off from benzene ring and hence l.p. is localized on it and hence the basic strength is highest.

(iv) In gas phase, basicity follows the order : (C₂H₅)₃N > (C₂H₅)₂NH > C₂H₅NH₂ > NH₃ In gas phase, the stabilization by solvation is not present and hence basic strength follows the expected order based on +I effect of alkyl groups.

(v) (CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH

(vi) C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂

Amines can form hydrogen bonds with water and are therefore soluble in it. However, the solubility decreases if the mass of the hydrocarbon part increases.

Question 5.
How will you convert :

1. Ethanoic acid into methanamine
2. Hexanenitrile into 1-aminopentane
3. Methanol to ethanoic acid
4. Ethanamine into methanamine
5. Ethanoic acid into porpanoic acid
6. Methanamine into ethanamine
7. Nitromethane into dimethylamine
8. Propanoic acid into ethanoic acid?

Solution:

Question 6.
Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.
Solution:
1°,2° and 3° amines can be distinguished by Hinsberg’s reagent.

Question 7.
Write short notes on the following :

1. Carbylamine reaction
2. Diazotisation
3. Hofmann’s bromamide reaction
4. Coupling reaction
5. Ammonolysis
6. Gabriel phthalimide synthesis
7. Acetylation

Solution:
(i) Carbylamine 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 : The conversion of primary aromatic amines into diazonium salts is known as diazotisation.

The conversion is brought about by reacting the amine with HNO₂ which is prepared in situ.

(iii) Hofmann’s bromamide reaction : Primary amides when heated with Br₂ and (aqueous or ethanoic solution of) NaOH lose a carbon atom and are converted to the corresponding amines. It is an example of step-down reaction.

(iv) Coupling reaction : The reaction of diazonium salts with phenols and aromatic amines to form azo compounds having an extended conjugate 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 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 process of cleavage of the C — X bond in alkyl halides by ammonia molecule is called ammonolysis. 1° amine thus obtained behaves as a nucleophile and further reacts with alkyl halide to form 2°, 3° and finally quaternary ammonium salt.

(vi) Gabriel 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 :

1. Nitrobenzene to benzoic acid
2. Benzene to m-bromophenol
3. Benzoic acid to aniline
4. Aniline to 2,4,6-tribromofluorobenzene
5. Benzyl chloride to 2-phenylethanamine
6. Chlorobenzene to p-chloroaniline
7. Aniline to p-bromoaniline
8. Benzamide to toluene
9. Aniline to benzyl alcohol.

Solution:

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


Solution:

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.
Solution:

Question 11.
Complete the following reactions :

Solution:

Question 12.
Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
Solution:
(i) Gabriel phthalimide reaction involves the nucleophilic attack of the phthalimide on the alkyl halide.

(ii) Such a nucleophilic substitution reaction is not possible if the substrate is an aryl halide.
(iii) The reason for it can be explained on the basis of
(a) Partial double bond character of C — X bond in aryl halide. Consider the following structures :

From the resonance structures we see that the C — X bond has a double bond character in structures II, III and IV and this makes cleavage of C — X bond difficult.

(b) Also the steric hindrance by the bulky aryl group prevents the incoming nucleophile.

Question 13.
Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.
Solution:
Amines do not directly react with nitrous acid, rather they react with a mixture of dil. HCl and NaNO₂ and HNO₂ is produced in situ.
The reactions are :

Question 14.
Give plausible explanation for each of the following :
(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why do primary amines have higher boiling point than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?
Solution:
(i) Loss of a proton from an amine gives RNH^– ion while loss of a proton from alcohol gives RO- ion as shown below :

As O is more electronegative than N,RO^– can accommodate the negative charge more easily than the RNH^– can.

As, RO^– is more stable than RNH^– the former is formed more. As a result, amines are less acidic than alcohols.

(ii) (a) At boiling point, the molecules in a compound break free from their inter molecular forces and escape into the vapour phase.

Weaker the inter-molecular forces, lower will be the boiling point.

(b) In 1 ° amines, there is strong H-bonding that binds the amine molecules together. Whereas in 3° amine absence of H on N atom prevents hydrogen bonding completely.

This is why 1° amines have higher boiling point.

(iii) (a) The basic nature of amines is a result of the presence of l.p. of electron on the N atom. Also the electron density is increased on N due to the +I effect of alkyl group.
(b) In aryl amines the l.p. on N is involved in resonance with the benzene ring and hence less available for protonation.

(c) In aliphatic amines there is no such delocalisation and hence it is more basic.