Which rule is followed when water adds to an unsymmetrical alkene in the presence of an acid catalyst?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Markovnikov's rule
In acid-catalysed hydration the H of water goes to the double-bond carbon that already carries more hydrogen atoms, and the OH to the other carbon, so propene gives propan-2-ol:
\( \displaystyle \ce{CH3CH=CH2 + H2O} \xrightarrow{\ce{H^+}} \ce{CH3CH(OH)CH3} \)
NCERT (2026-27 edition):
7.4.1 Preparation of Alcohols
