JEE Main 2026 Apr 5 Shift 2, Chemistry Q17: Preparation of Amines

JEE Main2026Apr 5, Shift 2Chemistry

Given below are two statements:

Statement-I: Heating benzamide with bromine in an ethanolic solution of sodium hydroxide will give benzylamine.

Statement-II: Nitration of aniline with $H N O_{3} / H_{2} S O_{4}$ at 288 K produces $m$ -nitroaniline in higher amount than $o$ -nitroaniline (pH adjusted).

In the light of the above statements, choose the correct answer from the options given below:

A
Both Statement-I and Statement-II are true.
B
Both Statement-I and Statement-II are false.
C
Statement-I is true but Statement-II is false.
D
Statement-I is false but Statement-II is true.

Solution

Statement-I: This is the Hofmann bromamide rearrangement, $P h C O N H_{2} B r_{2}, N a O H P h N H_{2}$ . The product is aniline (one less carbon), not benzylamine ( $P h C H_{2} N H_{2}$ ). False.

Statement-II: In strong acid ( $H_{2} S O_{4}$ ), aniline is mostly protonated to the anilinium ion, $P h N H_{3}^{+}$ , in which the $- N H_{3}^{+}$ group is a strong meta-director by induction. Hence at 288 K the $m$ -isomer (~47–51%) dominates over the $o$ -isomer (~2%), with a substantial $p$ -isomer also formed. True.

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JEE Main 2026 Apr 5 Shift 2, Chemistry Q17: Preparation of Amines

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