JEE Main 2026 · 5 Apr · Shift 2 · Chemistry Q67

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ChemistrySingle correct+4 / -1Question 67 of 75

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 $\ce{HNO3}$ / $\ce{H2SO4}$ 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 :

Both Statement I and Statement II are true

Both Statement I and Statement II are false

Statement I is true but Statement II is false

Statement I is false but Statement II is true

Answer and worked solution

Official answer: D

Statement I. Hofmann bromamide degradation of benzamide, $\ce{C6H5CONH2}$ , loses the carbonyl carbon. It gives aniline, not benzylamine. False.

Statement II. In the nitrating mixture, aniline is largely protonated to anilinium, which directs meta. The products are about 51% p, 47% m and only 2% o. So m-nitroaniline exceeds o-nitroaniline. True.

(D)

Topics:Amines

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JEE Main 2026 · 5 Apr · Shift 2, Chemistry question 67