Chemistry · Hydrocarbons

JEE Main 2026 — 28 January, Morning Shift — Question 54

$\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}_{2} \xrightarrow[\mathrm{HBr}]{(\mathrm{PhCOO})_{2}}$ Product

Consider the above reaction

A. The reaction proceeds through a more stable radical intermediate.

B. The role of peroxide is to generate $\dot{H}$ (Hydrogen radical).

C. During this reaction, benzene is formed as a biproduct.

D. 1-Bromo-2-phenylethane is fanned as the minor product.

E. The same reaction in absence of peroxide proceeds via carbocation intermediate.

Identify the correct statements. Choose the correct answer from the options given below:

Option A:
A & E Only
Option B:
A, B & D Only
Option C:
C, D & E Only
Option D:
A, C & E Only
Correct

Answer: D

Step-by-step solution

$\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}_{2} \xrightarrow[\mathrm{HBr}]{(\mathrm{PhCOO})_{2}} \mathrm{Ph}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Br}$ Anti Markovnikov addition Reaction follow radical addition in presence of peroxide In absence of peroxide follow carbocation mechanism Benzene also formed

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Morning Shift
Subject: Chemistry
Chapter: Hydrocarbons
Topic: Properties & Uses of Alkenes and Dienes