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JEE Advanced 2025Chemistry Organic Chemistry

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mcq
2025
Official previous-year question

Verified 30 May 2026.

Question

The major product of the reaction: $\text{CH}_3\text{CH}=\text{CH}_2 + \text{HBr}$ (in presence of peroxide) is:

Options

  1. A

    $\text{CH}_3\text{CHBr}\text{CH}_3$ (Markovnikov)

  2. B

    $\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}$ (Anti-Markovnikov)

  3. C

    $\text{CH}_2\text{Br}\text{CH}=\text{CH}_2$

  4. D

    $\text{CH}_3\text{CBr}=\text{CH}_2$

Solution

In the presence of peroxide, HBr addition follows the anti-Markovnikov rule (Kharasch effect).

The reaction proceeds via a free radical mechanism:

$$\text{ROOR} \xrightarrow{\Delta} 2\text{RO}^{\bullet}$$

$$\text{RO}^{\bullet} + \text{HBr} \rightarrow \text{ROH} + \text{Br}^{\bullet}$$

$$\text{Br}^{\bullet} + \text{CH}_3\text{CH}=\text{CH}_2 \rightarrow \text{CH}_3\text{CH}^{\bullet}\text{CH}_2\text{Br}$$

The $\text{Br}^{\bullet}$ adds to the terminal carbon, forming the more stable secondary radical, giving 1-bromopropane.

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