Ncert Chemistry Class 12 intext questions solutions chapter 9 Amines

Amines intext questions solutions: Class 12 Chemistry Chapter 9 intext questions solutions

Textbook	Ncert
Class	Class 12
Subject	Chemistry
Chapter	Chapter 9
Chapter Name	Amines Class 12 intext questions solutions
Category	Intext questions
Medium	English

Are you looking for Ncert Chemistry Class 12 intext questions solutions chapter 9 Amines? Now you can download Amines intext questions solutions from here.

Question 9.1: Classify the following amines as primary, secondary or tertiary:

(iii) (C₂H₅)₂CHNH₂

(iv) (C₂H₅)₂NH

Solution 9.1:

• (i) Primary
• (ii) Tertiary
• (iii) Primary
• (iv) Secondary

Question 9.2: Write the structures of different isomeric amines corresponding to the molecular formula, C₄H₁₁N.
(i) Write the IUPAC names of all the isomers
(ii) What type of isomerism is exhibited by different pairs of amines?

Solution 9.2: (i), (ii) The structures and their IUPAC names of different isomeric amines corresponding to the molecular formula, C₄H₁₁N are given below:

(a) Butanamine (1°)

CH₃-CH₂-CH₂-CH₂-NH₂

(b) Butan-2-amine (1°)

(c) 2-Methylpropanamine (1°)

(d) 2-Methylpropan-2-amine (1°)

(e) N-Methylpropanamine (2°)

CH₃-CH₂-CH₂-NH-CH₃

(f) N-Ethylethanamine (2°)

CH₃-CH₂-NH-CH₂-CH₃

(g) N-Methylpropan-2-amine (2°)

(h) N, N-Dimethylethanamine (3°)

(iii) The pairs (a) and (b) and (e) and (g) exhibit position isomerism.

The pairs (a) and (c); (a) and (d); (b) and (c); (b) and (d) exhibit chain isomerism.

The pairs (e) and (f) and (f) and (g) exhibit metamerism.

All primary amines exhibit functional isomerism with secondary and tertiary amines and vice-versa.

Question 9.3: How will you convert:
(i) Benzene into aniline
(ii) Benzene into N, N-dimethylaniline
(iii) Cl-(CH₂)₄-Cl into Hexane -1,6-diamine?

Solution 9.3:

(i) Benzene into aniline

(ii) Benzene into N, N-dimethylaniline

(iii) Cl−(CH₂)₄−Cl into hexan-1, 6-diamine

Question 9.4: Arrange the following in increasing order of their basic strength :
(i) C₂H₅NH₂, C₆H₅NH₂, NH₃, C₆H₅CH₂NH₂ and (C₂H₅)₂NH
(ii) C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N, C₆H₅NH₂
(iii) CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂, C₆H₅CH₂NH₂.

Solution 9.4: Part (i)

Compounds:

• C₂H₅NH₂ (Ethylamine)
• C₆H₅NH₂ (Aniline)
• NH₃ (Ammonia)
• C₆H₅CH₂NH₂ (Benzylamine)
• (C₂H₅)₂NH (Diethylamine)

Basic Strength Concept: Basicity in amines depends on the availability of the lone pair of electrons on the nitrogen atom for protonation. Electron-donating groups (like alkyl groups) increase basicity, while electron-withdrawing groups (like phenyl) decrease it.

Order of basic strength:

• Diethylamine (C₂H₅)₂NH: The presence of two electron-donating ethyl groups increases the electron density on nitrogen, making it the most basic.
• Ethylamine (C₂H₅NH₂): Only one ethyl group is present, so its basicity is lower than diethylamine but higher than the others.
• Benzylamine (C₆H₅CH₂NH₂): The phenyl group in the -CH₂ position has a less direct electron-withdrawing effect, so its basicity is lower than ethylamine but higher than aniline.
• Ammonia (NH₃): It has no alkyl or phenyl groups, so its basicity is lower than amines with electron-donating groups.
• Aniline (C₆H₅NH₂): The phenyl group directly attached to nitrogen withdraws electron density from the nitrogen atom, reducing basicity.

Increasing order of basic strength:
C₆H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ < C₂H₅NH₂ < (C₂H₅)₂NH

Part (ii)

Compounds:

• C₂H₅NH₂ (Ethylamine)
• (C₂H₅)₂NH (Diethylamine)
• (C₂H₅)₃N (Triethylamine)
• C₆H₅NH₂ (Aniline)

Basic Strength Concept:

• Triethylamine (C₂H₅)₃N has three ethyl groups donating electron density to nitrogen, making it the most basic.
• Diethylamine (C₂H₅)₂NH is next because it has two ethyl groups.
• Ethylamine (C₂H₅NH₂) has one ethyl group, making it less basic than diethylamine.
• Aniline (C₆H₅NH₂), as discussed earlier, has a phenyl group that withdraws electron density from nitrogen, making it the least basic.

Increasing order of basic strength:
C₆H₅NH₂ < C₂H₅NH₂ < (C₂H₅)₂NH < (C₂H₅)₃N

Part (iii)

Compounds:

• CH₃NH₂ (Methylamine)
• (CH₃)₂NH (Dimethylamine)
• (CH₃)₃N (Trimethylamine)
• C₆H₅NH₂ (Aniline)
• C₆H₅CH₂NH₂ (Benzylamine)

Basic Strength Concept:

• Dimethylamine (CH₃)₂NH has two electron-donating methyl groups, making it more basic than both methylamine and trimethylamine.
• Trimethylamine (CH₃)₃N is less basic than dimethylamine due to steric hindrance, which hinders protonation.
• Methylamine (CH₃NH₂) has one methyl group, making it less basic than dimethylamine but more basic than the aromatic amines.
• Benzylamine (C₆H₅CH₂NH₂) is more basic than aniline due to the less direct influence of the phenyl group.
• Aniline (C₆H₅NH₂) is the least basic because the phenyl group withdraws electron density from nitrogen.

Increasing order of basic strength:
C₆H₅NH₂ < C₆H₅CH₂NH₂ < CH₃NH₂ < (CH₃)₃N < (CH₃)₂NH

In summary, here are the increasing order of basic strength for each part:

• (i): C₆H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ < C₂H₅NH₂ < (C₂H₅)₂NH
• (ii): C₆H₅NH₂ < C₂H₅NH₂ < (C₂H₅)₂NH < (C₂H₅)₃N
• (iii): C₆H₅NH₂ < C₆H₅CH₂NH₂ < CH₃NH₂ < (CH₃)₃N < (CH₃)₂NH

Question 9.5: Complete the following acid-base reactions and name the products:

• (i) CH₃CH₂CH₂NH₂ + HCl →
• (ii) C₂H₅)₃N + HCl →

Solution 9.5:

Question 9.6: Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.

Solution 9.6: Aniline reacts with methyl iodide to produce N, N-dimethylaniline.

With excess methyl iodide, in the presence of Na₂CO₃ solution, N, N-dimethylaniline produces N, N, N−trimethylanilinium carbonate.

Question 9.7: Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.

Solution 9.7:

Question 9.8: Write structures of different isomers corresponding to the molecular formula, C₃H₉N. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid.

Solution 9.8:

The structures of different isomers corresponding to the molecular formula, C₃H₉N are given below:

(a) Propan-1-amine (1°)

CH₃-CH₂-CH₂-NH₂

(b) Propan-2-amine (1°)

(c) N-Methylethanamine (2°)

CH₃-NH-C₂H₅

(d) N, N-Dimethylmethanamine (3°)

1° amines, (a) propan-1-amine, and (b) Propan-2-amine will liberate nitrogen gas on treatment with nitrous acid.

Question 9.9: Convert:
(i) 3-Methylaniline into 3-nitrotoluene.
(ii) Aniline into 1,3,5 – tribromobenzene.

Solution 9.9:

(i) 3-Methylaniline into 3-nitrotoluene

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