In the following table, match the reactants given in List-I with the correct product in List-II as per the reaction of hydration of alkene under acidic condition.
| List-I (Reactants) | List-II (Products) |
(A)  | (I)  |
(B)  | (II)  |
(C)  | (III)  |
(D)  | (IV)  |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
In acid catalysed hydration, an alkene adds a molecule of water across its double bond and gives an alcohol. A proton first adds to one carbon of the double bond, which leaves a carbocation on the other carbon; water then attacks the carbocation, and loss of a proton gives the alcohol. The proton goes to the carbon that gives the more stable (more substituted) carbocation, so the –OH group ends up on the more substituted carbon (Markovnikov’s rule).
\( \displaystyle \ce{R2C=CR2 + H2O} \xrightarrow{\ce{H+}} \ce{R2C(OH)-CHR2} \)
(A) 3-methylhex-2-ene — This matches structure (II), 3-methylhexan-3-ol. The proton adds to C-2, which leaves a tertiary carbocation on C-3 (the carbon carrying the methyl group); the –OH group goes there.
\( \displaystyle \ce{CH3CH2CH2C(CH3)=CHCH3 + H2O} \xrightarrow{\ce{H+}} \ce{CH3CH2CH2C(OH)(CH3)CH2CH3} \)
(B) 1-methylcyclohexene — This matches structure (I), 1-methylcyclohexan-1-ol. The proton adds to the ring CH carbon, and the tertiary carbocation forms on the ring carbon that carries the methyl group.
\( \displaystyle \text{1-methylcyclohexene} + \ce{H2O} \xrightarrow{\ce{H+}} \text{1-methylcyclohexan-1-ol} \)
(C) but-2-ene — This matches structure (IV), butan-2-ol. The alkene is symmetrical, so either carbon of the double bond gets the –OH group and the product is the same.
\( \displaystyle \ce{CH3CH=CHCH3 + H2O} \xrightarrow{\ce{H+}} \ce{CH3CH(OH)CH2CH3} \)
(D) 2-cyclohexylidenebutane — This matches structure (III), 2-cyclohexylbutan-2-ol. Neither carbon of this double bond carries a hydrogen, so either carbocation would be tertiary. When the proton adds to the ring carbon, the carbocation and then the –OH group are on the side-chain carbon that carries the methyl and ethyl groups, which is structure (III); it is also the only product left in List-II.
\( \displaystyle \text{2-cyclohexylidenebutane} + \ce{H2O} \xrightarrow{\ce{H+}} \text{2-cyclohexylbutan-2-ol} \)
Shortcut: But-2-ene (C) can only give butan-2-ol (IV), and the pair (C)–(IV) appears only in Option (3).
NCERT (2026-27 edition):
Preparation of Alcohols – …
