Propene reacts with HI to form major product as 2-bromopropane. Suggest the rule applied / effect used for predicting the major product .
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Markovnikov's rule
Markovnikov’s rule predicts the major product when an unsymmetrical reagent HX adds to an unsymmetrical alkene: the negative part (X) goes to the carbon atom of the double bond that carries fewer hydrogen atoms. The reason is the mechanism. \( \ce{H^{+}} \) adds first to give a carbocation, and the more stable carbocation is formed:
\( \displaystyle \ce{CH3-CH=CH2 + H^+ -> CH3-CH^+-CH3}\quad\text{(2}^\circ\text{, more stable)} \)
The halide ion then attacks the secondary carbocation, so the halogen ends up on the middle carbon:
\( \displaystyle \ce{CH3-CH=CH2 + HX -> CH3-CHX-CH3} \)
So the major product (a 2-halopropane) is predicted by Markovnikov’s rule.
Why the other options are incorrect
Option (1) Anti- Markovnikov's rule — Incorrect. This applies only to the addition of HBr in the presence of a peroxide (Kharasch effect), which goes through a free radical and gives 1-bromopropane. It does not work with HI or HCl.
Option (2) Saytzeff rule — Incorrect. This predicts the major alkene in elimination reactions (the more substituted alkene forms), not the product of an addition.
Option (3) Popoff's rule — Incorrect. This applies to the oxidation of unsymmetrical ketones, where the carbonyl group stays with the smaller alkyl group; it has nothing to do with addition to alkenes.
Note: It seems that there is a mistake in the original NTA paper, which prints that propene reacts with HI to form 2-bromopropane when it should read HBr (with HI the product is 2-iodopropane).
NCERT (2026-27 edition):
From Alkenes
