Given below are two statements, one is labelled as Statement I and the other is labelled as Statement II. Statement I : Bromination of phenol in solvent with low polarity such as CHCl 3 or CS 2 requires Lewis acid catalyst. Statement II : The lewis acid catalyst polarises the bromine to generate Br + . In the light of the above statements, choose the correct answer from the options given below :

Statement I is true but Statement II is false.

Both Statement I and Statement II are true

Both Statement I and Statement II are false.

Chemistry · Alcohols, Ethers and Phenols

JEE Main 2024 — 5 April, Shift 1 — Question 74

Given below are two statements, one is labelled as Statement I and the other is labelled as Statement II.

Statement I : Bromination of phenol in solvent with low polarity such as $\mathrm{CHCl}_{3}$ or $\mathrm{CS}_{2}$ requires Lewis acid catalyst.

Statement II : The lewis acid catalyst polarises the bromine to generate $\mathrm{Br}^{+}$ .

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

Option A:
Statement I is true but Statement II is false.
Option B:
Both Statement I and Statement II are true
Option C:
Both Statement I and Statement II are false.
Option D:
Statement I is false but Statement II is true.
Correct

Answer: D

Step-by-step solution

Statement I is False: Phenol is a strongly activating aromatic compound due to the $-\mathrm{OH}$ group. Therefore, bromination of phenol occurs readily even in low-polarity solvents such as $\mathrm{CHCl_3}$ or $\mathrm{CS_2}$ without the need for a Lewis acid catalyst.

Statement II is True: A Lewis acid catalyst polarises $\mathrm{Br_2}$ to generate $\mathrm{Br^+}$ , which acts as the electrophile in electrophilic substitution reactions.

Thus, Statement I is false but Statement II is true (Option D).

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 5 April, Shift 1
Subject: Chemistry
Chapter: Alcohols, Ethers and Phenols
Topic: Phenols