JEE Main 2026 — Chemistry Organic Chemistry

2026mcqmedium

Given below are two statements:

Statement I: $3$ -phenylpropene reacts with HBr and gives secondary alkyl bromide having a chiral carbon atom as the major product.

Statement II: Aryl chlorides and aryl cyanides can be prepared by Sandmeyer reaction as well as Gattermann reaction.

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

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Options

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

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Solution

Statement I:

When $3$ -phenylpropene reacts with $HBr$ , protonation of the alkene occurs to form a secondary carbocation.

$C_6H_5-CH_2-CH=CH_2 + H^+ \rightarrow C_6H_5-CH_2-CH^+-CH_3$

This secondary carbocation undergoes a $1,2$ -hydride shift to form a more stable, resonance-stabilized benzylic carbocation.

$C_6H_5-CH_2-CH^+-CH_3 \rightarrow C_6H_5-CH^+-CH_2-CH_3$

The bromide ion then attacks this benzylic carbocation to form the major product, $1$ -bromo- $1$ -phenylpropane.

$C_6H_5-CH^+-CH_2-CH_3 + Br^- \rightarrow C_6H_5-CH(Br)-CH_2-CH_3$

The carbon atom bonded to the bromine is attached to four different groups: a phenyl group ( $-C_6H_5$ ), an ethyl group ( $-CH_2CH_3$ ), a hydrogen atom ( $-H$ ), and a bromine atom ( $-Br$ ). Thus, it is a chiral carbon. The product is a secondary alkyl bromide. Therefore, Statement I is true.

Statement II:

The Sandmeyer reaction involves treating a diazonium salt with $Cu(I)$ salts ( $CuCl$ , $CuBr$ , $CuCN$ ) to prepare aryl chlorides, aryl bromides, and aryl cyanides, respectively.

The Gattermann reaction involves treating a diazonium salt with copper powder ( $Cu$ ) and halogen acids ( $HCl$ , $HBr$ ) to prepare aryl chlorides and aryl bromides. Aryl cyanides are not prepared using the Gattermann reaction. Therefore, Statement II is false.

Conclusion: Statement I is true but Statement II is false.

Answer: Statement I is true but Statement II is false

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Solution

Statement I:

When $3$ -phenylpropene reacts with $HBr$ , protonation of the alkene occurs to form a secondary carbocation.

$C_6H_5-CH_2-CH=CH_2 + H^+ \rightarrow C_6H_5-CH_2-CH^+-CH_3$

This secondary carbocation undergoes a $1,2$ -hydride shift to form a more stable, resonance-stabilized benzylic carbocation.

$C_6H_5-CH_2-CH^+-CH_3 \rightarrow C_6H_5-CH^+-CH_2-CH_3$

The bromide ion then attacks this benzylic carbocation to form the major product, $1$ -bromo- $1$ -phenylpropane.

$C_6H_5-CH^+-CH_2-CH_3 + Br^- \rightarrow C_6H_5-CH(Br)-CH_2-CH_3$

Statement II:

The Sandmeyer reaction involves treating a diazonium salt with $Cu(I)$ salts ( $CuCl$ , $CuBr$ , $CuCN$ ) to prepare aryl chlorides, aryl bromides, and aryl cyanides, respectively.

Conclusion: Statement I is true but Statement II is false.

Answer: Statement I is true but Statement II is false

JEE Main 2026 — Chemistry Organic Chemistry

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Options

Did you get this right?

Solution

Your note

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