Which of the following will form the major product in accordance with Markovnikov’s rule?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Propene
Markovnikov’s rule: when an unsymmetrical reagent (\( \ce{H2O} \) in acid, HBr, HCl) adds to an unsymmetrical alkene, the negative part of the reagent goes to the double-bond carbon that carries fewer hydrogen atoms. The reason is the mechanism: \( \ce{H^{+}} \) adds first to give the more stable carbocation (3° > 2° > 1°), which then picks up the nucleophile.
The rule decides between two possible products only when the two carbons of the double bond are different. Propene is the only unsymmetrical alkene given; protonation gives the secondary carbocation, so propan-2-ol is the major product:
\( \displaystyle \ce{CH3-CH=CH2 + H^+ -> CH3-C^+H-CH3} \)
\( \displaystyle \ce{CH3-C^+H-CH3 + H2O -> CH3-CH(OH)-CH3 + H^+} \)
So propene forms the major product in accordance with Markovnikov’s rule.
Why the other options are incorrect
Option (1) Ethene — Incorrect. It is symmetrical (\( \ce{CH2=CH2} \)); both carbons are alike, so only one product (ethanol) forms.
Option (2) But-2-ene — Incorrect. It is symmetrical (\( \ce{CH3CH=CHCH3} \)); addition at either carbon gives the same product, butan-2-ol.
Option (3) Cyclohexene — Incorrect. It is symmetrical about the double bond; only cyclohexanol forms.
NCERT (2026-27 edition):
From Alkenes
