Which of the following statement is correct about the following reaction :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Product will be optically inactive
Tertiary alkyl halides undergo nucleophilic substitution by the SN1 mechanism. The C–I bond breaks first to give a carbocation, which is sp2 hybridised and planar (achiral). The nucleophile can attack it from either face with equal chance, so both enantiomers of the product form in equal amounts: a racemic mixture.
\( \ce{CH3-C(C6H5)(C2H5)-I} \xrightarrow{\text{slow}} \ce{CH3-C+(C6H5)(C2H5) + I-} \)
\( \ce{CH3-C+(C6H5)(C2H5) + CN-} \xrightarrow{\text{fast}} \ce{CH3-C(C6H5)(C2H5)-CN} \) (50% each enantiomer)
Each product molecule is chiral (\( \ce{CH3} \), \( \ce{C6H5} \), \( \ce{C2H5} \) and CN on one carbon), but the rotations of the two enantiomers cancel. So the product is optically inactive.
Why the other options are incorrect
Option (1) Reactant and product will rotate plane polarised light in the same direction — Incorrect. The racemic product does not rotate plane polarised light at all.
Option (3) Product will be dextrorotatory in nature — Incorrect. A dextrorotatory product would need one enantiomer in excess, which SN1 attack on a planar carbocation does not give.
Option (4) Product will be laevorotatory in nature — Incorrect. A laevorotatory product would also need one enantiomer in excess; attack from both faces of the planar carbocation gives the two enantiomers in equal amounts.
NCERT (2026-27 edition):
Reactions of Haloalkanes – “… In case of optically active alkyl halides, SN1 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). This may be illustrated by hydrolysis of optically active 2-bromobutane, which results in the formation of (±)-butan-2-ol.”