Compounds Containing Nitrogen — JEE Main practice

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Sample questions with solutions

Q1 · 2026

A hydrocarbon 'P' $(C_4H_8)$ on reaction with HCl gives an optically active compound 'Q' $(C_4H_9Cl)$ which on reaction with one mole of ammonia gives compound 'R' $(C_4H_{11}N)$ . 'R' on diazotization followed by hydrolysis gives 'S'. Identify P, Q, R and S.

Answer: B

Q2 · 2026

An organic compound (P) on treatment with aqueous ammonia under hot condition forms compound (Q) which on heating with $Br_2$ and KOH forms compound (R) having molecular formula $C_6H_7N$ . Names of P, Q and R respectively are.

A.
Toluic acid, methylbenzamide, 2-methylaniline
B.
Benzoic acid, 4-methylbenzamide, 4-methylaniline
C.
Phenylethanoic acid, phenylethanamide, benzamine
D.
Benzoic acid, benzamide, aniline

Answer: D

The final product R has molecular formula $C_6H_7N$ , giving a degree of unsaturation of 4, which corresponds to exactly one benzene ring with no other unsaturation — matching aniline ( $C_6H_5NH_2$ ). [IMAGE: Structure of aniline confirming the C6H7N molecular formula with one benzene ring]
Working backward: since R (aniline) is formed by Hoffmann bromamide degradation ( $Br_2$ /KOH) of an amide Q, Q must be benzamide ( $C_6H_5CONH_2$ ), since this degrades to aniline by losing the carbonyl carbon.
Since Q (benzamide) is formed by treating P with hot aqueous ammonia (a reaction that converts carboxylic acids to their ammonium salts, then to amides upon heating), P must be benzoic acid ( $C_6H_5COOH$ ). Hence, P, Q, and R are benzoic acid, benzamide, and aniline respectively, so the answer is option D.

Q3 · 2026

Consider the below sequence of reactions. The number of bromine atom(s) in the final product (P) will be :

A.
6
B.
3
C.
5
D.
1

Answer: C

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