The major product in the following reaction is C_6H_5CH=CH_2 + HI→ ?

The correct answer is Option (2) → \(C_6H_5CHICH_3\)

A hydrogen halide adds to an unsymmetrical alkene by Markovnikov’s rule: the proton goes to the double-bond carbon that already has more hydrogens, so that the more stable carbocation is formed, and the halide ion then joins that carbocation.

In styrene, \( \ce{C6H5CH=CH2} \), protonation of the terminal \( \ce{CH2} \) gives a secondary benzylic carbocation, which is stabilised by resonance with the benzene ring. Iodide ion then attacks this carbon:

    \( \displaystyle \ce{C6H5CH=CH2 + H^+ -> C6H5CH^+CH3} \)

    \( \displaystyle \ce{C6H5CH^+CH3 + I^- -> C6H5CHICH3} \)

The major product is 1-iodo-1-phenylethane, \( \ce{C6H5CHICH3} \).

Why the other options are incorrect

• Option (1) \(C_6H_5CH_2CH_2I\) — Incorrect. This is the anti-Markovnikov product. It would need attack through the much less stable primary carbocation; the peroxide effect, which can reverse the orientation, works only for HBr, not HI.

• Option (3) \(IC_6H_4CH=CH_2\) — Incorrect. Here iodine has replaced a ring hydrogen. HI does not substitute the ring; it adds across the C=C bond of the side chain.

• Option (4) \(C_6H_5CH_2CHI_2\) — Incorrect. A diiodide would need two molecules of HI, as with an alkyne; one double bond takes up only one HI.

NCERT (2026-27 edition):
From Alkenes – “(i) Addition of hydrogen halides: An alkene is converted to corresponding alkyl halide by reaction with hydrogen chloride, hydrogen bromide or hydrogen iodide. … Propene yields two products, however only one predominates as per Markovnikov’s rule.”