JEE Chemistry 2026 Question with Solution

Given: Addition of HBr to styrene $\text{Ph–CH=CH}_2$ is carried out in presence of peroxide.

Find: Which statements among A to E are correct.

Concept: In presence of peroxide, addition of HBr follows the free radical anti-Markovnikov mechanism. In absence of peroxide, the reaction proceeds by the ionic Markovnikov mechanism.

For styrene, addition of $\mathrm{Br^\bullet}$ to the terminal carbon gives a benzylic radical, which is resonance-stabilized.

Peroxide first generates alkoxy radicals, and these ultimately produce $\mathrm{Br^\bullet}$ from HBr. It does not generate $\mathrm{H^\bullet}$ radical as the active chain carrier.

No step in this mechanism forms benzene as a byproduct.

Anti-Markovnikov addition of HBr to styrene gives $1$-bromo-$2$-phenylethane as the major product, not a minor product.

In absence of peroxide, addition of HBr occurs through the ionic pathway involving a carbocation intermediate.

Therefore, the correct statements are A and E only. The correct option is D.

Given: The alkene is styrene $\text{Ph–CH=CH}_2$ and reagent is HBr in presence of peroxide.

Find: Evaluate statements A to E.

1. Statement A: Radical addition in styrene leads to a benzylic radical intermediate. A benzylic radical is stabilized by resonance with the phenyl ring.
2. Statement B: Peroxide decomposes to radicals such as alkoxy radicals, which help generate $\mathrm{Br^\bullet}$ from HBr. Hence saying peroxide generates $\mathrm{H^\bullet}$ radical is incorrect.
3. Statement C: The mechanism is radical addition across the double bond. There is no pathway here for formation of benzene as a byproduct.
4. Statement D: Under peroxide effect, anti-Markovnikov product forms predominantly. Therefore $1$-bromo-$2$-phenylethane is the major product, so calling it minor is incorrect.
5. Statement E: Without peroxide, the reaction switches to the ordinary ionic addition of HBr, which proceeds through a carbocation intermediate.

Hence only A and E are correct, so the answer is D.