Which of the following compounds will give a pair of enantiomers upon…

The correct answer is Option (1) → 2-bromobutane

A substitution can give a pair of enantiomers only if the product has a chiral carbon, a carbon carrying four different groups. When the reaction goes through a carbocation (\( S_N1 \)), the carbocation is planar, and the hydroxide ion can attack it equally from either face. If the carbon is chiral in the product, a 50:50 mixture of the two enantiomers (racemic mixture) results.

In 2-bromobutane, C-2 carries H, \( \ce{CH3} \), \( \ce{C2H5} \) and Br; after substitution it carries H, \( \ce{CH3} \), \( \ce{C2H5} \) and OH, still four different groups. Attack from the two faces gives (+)- and (−)-butan-2-ol:

    \( \displaystyle \ce{CH3CH(Br)C2H5 + OH^- -> CH3CH(OH)C2H5 + Br^-} \)

So 2-bromobutane gives the pair of enantiomers, (±)-butan-2-ol.

Why the other options are incorrect

• Option (2) 2-bromo-2-methylbutane — Incorrect. The product, 2-methylbutan-2-ol, \( \ce{(CH3)2C(OH)C2H5} \), has two identical methyl groups on C-2, so it is not chiral.

• Option (3) 1-bromo-3-methylbutane — Incorrect. The product, 3-methylbutan-1-ol, \( \ce{(CH3)2CHCH2CH2OH} \), has no carbon with four different groups.

• Option (4) 1-bromobutane — Incorrect. The product, butan-1-ol, \( \ce{CH3CH2CH2CH2OH} \), is achiral; the substitution happens at a \( \ce{CH2} \) carbon.

NCERT (2026-27 edition):
Inversion, Retention and Racemisation – “… In case of optically active alkyl halides, S_N1 reactions are accompanied by racemisation. … 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).”