What is the IUPAC name of the product of hydration of propene in the presence of dilute sulphuric acid?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Propan-2-ol
Alkenes react with water in the presence of an acid catalyst to form alcohols (acid catalysed hydration). For an unsymmetrical alkene the addition follows Markovnikov’s rule: the hydrogen goes to the double-bond carbon that already has more hydrogens, and the –OH to the other carbon.
\( \displaystyle \ce{CH3-CH=CH2 + H2O} \xrightarrow{\text{dil. }\ce{H2SO4}} \ce{CH3-CH(OH)-CH3} \)
The reason lies in the mechanism. The alkene is first protonated by \( \ce{H3O^{+}} \). Adding \( \ce{H^{+}} \) to the terminal \( \ce{CH2} \) gives a secondary carbocation, which is more stable than the primary carbocation the other way round would give:
\( \displaystyle \ce{CH3-CH=CH2 + H3O+ -> CH3-CH^+-CH3 + H2O} \)
Water then attacks this carbocation, and loss of a proton gives the alcohol:
\( \displaystyle \ce{CH3-CH^+-CH3 + H2O -> CH3-CH(OH2^+)-CH3 -> CH3-CH(OH)-CH3 + H+} \)
The –OH is on C-2 of a three-carbon chain, so the IUPAC name is propan-2-ol (isopropyl alcohol).
Why the other options are incorrect
Option (1) 2-Methylpropan-2-ol — Incorrect. This has four carbons; it is formed by hydration of 2-methylpropene, \( \ce{(CH3)2C=CH2} \).
Option (2) Propan-1-ol — Incorrect. This is the anti-Markovnikov product. It is obtained from propene by hydroboration–oxidation, not by acid catalysed hydration:
\( \displaystyle \ce{CH3CH=CH2} \xrightarrow[\text{(ii) }\ce{H2O2}/\ce{OH-}]{\text{(i) }\ce{B2H6}} \ce{CH3CH2CH2OH} \)
Option (4) 2-Methylbutan-2-ol — Incorrect. This has five carbons; it would come from hydration of 2-methylbut-2-ene. Hydration adds only H and OH and cannot change the carbon skeleton.
NCERT (2026-27 edition):
From Alkenes
