Which of the given alkylhalide would undergo racemisation when hydrol…

The correct answer is Option (4) → sec-butyl Chloride

Racemisation is the conversion of one enantiomer into a 50:50 mixture of both enantiomers (a racemic mixture), which is optically inactive. It can happen only to an optically active (chiral) alkyl halide, that is, one in which the carbon carrying the halogen is attached to four different groups. When such a halide is hydrolysed by the S_N1 route, the planar carbocation formed in the slow step is attacked by \( \ce{OH^{-}} \) from either face, so both configurations of the alcohol are formed.

In sec-butyl chloride (2-chlorobutane) the C-2 carbon carries H, Cl, \( \ce{CH3} \) and \( \ce{C2H5} \): four different groups, so it is chiral. Its hydrolysis with aqueous KOH gives butan-2-ol through the planar sec-butyl carbocation, and the product is the racemic mixture:

    \( \displaystyle \ce{CH3CH2CH(Cl)CH3 + KOH(aq) -> CH3CH2CH(OH)CH3 + KCl} \)

So the halide that undergoes racemisation is sec-butyl chloride.

Why the other options are incorrect

• Option (1) Ethyl Chloride — Incorrect. \( \ce{CH3CH2Cl} \) has two H atoms on the carbon bearing Cl, so it is achiral and has no optical activity to lose.

• Option (2) n-propyl Chloride — Incorrect. \( \ce{CH3CH2CH2Cl} \) is also achiral (two H atoms on C-1); as a primary halide it is hydrolysed mainly by the S_N2 route.

• Option (3) isopropyl Chloride — Incorrect. In \( \ce{(CH3)2CHCl} \) the carbon bearing Cl carries two identical \( \ce{CH3} \) groups, so it is achiral; its hydrolysis gives propan-2-ol, which is not optically active either.

NCERT (2026-27 edition):
Stereochemical Aspects of Nucleophilic Substitution Reactions – “… In case of optically active alkyl halides, S_N1 reactions are accompanied by racemisation. Can you think of the reason why it happens? Actually the carbocation formed in the slow step being sp² hybridised is planar (achiral). The attack of the nucleophile may be accomplished from either side of the plane of carbocation resulting in a mixture of products, one having the same configuration (the –OH attaching on the same position as halide ion) and the other having opposite configuration (the –OH attaching on the side opposite to halide ion).”