Appeared in: 2026 (3 sets)
The reaction of amines with mineral acids to form ammonium salts shows that these are basic in nature. Aliphatic amines are stronger bases than ammonia whereas aromatic amines are weaker bases than ammonia. Aliphatic and aromatic primary and secondary amines react with acid chlorides, anhydrides and esters by nucleophilic substitution reaction. The main problem encountered during electrophilic substitution reactions of aromatic amines is that of their high reactivity. Substitution tends to occur at ortho- and para-positions. Hinsberg reagent is used for the identification and distinction between primary, secondary and tertiary amines. Aryldiazonium salts, usually obtained from arylamines, undergo replacement of the diazonium group with a variety of nucleophiles to provide advantageous methods for producing aryl halides, cyanides, phenols and arenes.
Answer the following questions :
(a) (i) Why $ ext{CH}_3 - ext{NH}_2$ is a stronger base than $( ext{CH}_3)_3 ext{N}$ in aqueous solution ?
(ii) Write structural formulae of the compound A and B :
$$ ext{CH}_3 ext{CONH}_2 \xrightarrow{ ext{NaOBr}} ext{A} \xrightarrow[ ext{Base}]{ ext{C}_6 ext{H}_5 ext{COCl}} ext{B}$$
(b) A compound 'X' with molecular formula $ ext{C}_3 ext{H}_9 ext{N}$ reacts with Hinsberg reagent to give a product insoluble in alkali. Identify 'X'.
OR
(b) How can you convert aniline to benzonitrile ?
(c) Why is $- ext{NH}_2$ group of aniline acetylated before carrying out nitration ?
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CBSE Class 12 Chemistry: Amines Marking Scheme Solution
(a) (i) Why $\text{CH}_3\text{NH}_2$ is a stronger base than $(\text{CH}_3)_3\text{N}$ in aqueous solution? [1 Mark]
- Step 1 (Governing Concept): The basicity of amines in aqueous solution depends on the combination of inductive effect, solvation effect (H-bonding), and steric hindrance.
- Step 2 (Working): While $(\text{CH}_3)_3\text{N}$ has a higher +I effect, it experiences maximum steric hindrance, which prevents the approach of the proton. Furthermore, $\text{CH}_3\text{NH}_2$ is more effectively stabilized by solvation (H-bonding with water molecules) compared to $(\text{CH}_3)_3\text{N}$, which has fewer H-atoms available for bonding.
- Step 3 (Conclusion): Thus, the combined effect of solvation and steric hindrance makes $\text{CH}_3\text{NH}_2$ a stronger base than $(\text{CH}_3)_3\text{N}$ in aqueous medium.
(a) (ii) Write structural formulae of the compound A and B: [1 Mark]
- Step 1 (Reaction Identification): The first step is the Hofmann Bromamide Degradation reaction.
- Step 2 (Working): $$\text{CH}_3\text{CONH}_2 \xrightarrow{\text{NaOBr}} \text{CH}_3\text{NH}_2 (\text{A})$$ $$\text{CH}_3\text{NH}_2 + \text{C}_6\text{H}_5\text{COCl} \xrightarrow{\text{Base}} \text{CH}_3\text{NHCOC}_6\text{H}_5 (\text{B}) + \text{HCl}$$
- Step 3 (Final Answer): A: $\text{CH}_3\text{NH}_2$ (Methanamine); B: $\text{CH}_3\text{NHCOC}_6\text{H}_5$ (N-methylbenzamide).
(b) Identify 'X' (Molecular formula $\text{C}_3\text{H}_9\text{N}$): [1 Mark]
- Step 1 (Concept): Hinsberg reagent (Benzenesulphonyl chloride) reacts with primary amines to form a product soluble in alkali, secondary amines to form a product insoluble in alkali, and tertiary amines do not react.
- Step 2 (Working): The formula $\text{C}_3\text{H}_9\text{N}$ corresponds to an amine. Since the product is insoluble in alkali, 'X' must be a secondary amine. The only secondary amine with 3 carbons is diethylamine or ethylmethylamine. Given the formula $\text{C}_3\text{H}_9\text{N}$, it is Ethylmethylamine.
- Step 3 (Conclusion): $\text{X} = \text{CH}_3\text{NHC}_2\text{H}_5$ (N-methylethanamine).
(c) Why is $-\text{NH}_2$ group of aniline acetylated before nitration? [1 Mark]
- Step 1 (Governing Concept): Aniline is highly reactive, and direct nitration leads to oxidation and the formation of tarry products, along with meta-substitution due to protonation in acidic medium.
- Step 2 (Working): Acetylation with acetic anhydride reduces the electron-donating power of