Predict the major product formed when propene reacts with HI

The correct answer is Option (2) → 2-Iodopropane

HI adds to an unsymmetrical alkene by electrophilic addition, following Markovnikov’s rule. The \( \ce{H^{+}} \) adds first, to the double-bond carbon that already has more hydrogen atoms. This forms the more stable carbocation, which the \( \ce{I^{-}} \) ion then attacks.

With propene, \( \ce{H^{+}} \) can form a secondary or a primary carbocation. The secondary one is more stable and forms preferentially:

    \( \displaystyle \ce{CH3-CH=CH2 + H^{+} -> CH3-CH^{+}-CH3}\quad(2^{\circ}\text{, more stable}) \)

    \( \displaystyle \ce{CH3-CH^{+}-CH3 + I^{-} -> CH3-CHI-CH3} \)

    \( \displaystyle \text{Net: }\ce{CH3-CH=CH2 + HI -> CH3-CHI-CH3} \)

So the major product is 2-iodopropane.

NCERT (2026-27 edition):
From Alkenes