An SN1 reaction at the asymmetric carbon of an enantiomerically pure chiral alkyl halide gives a product:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → with racemisation
An SN1 reaction goes through a planar carbocation. The nucleophile attacks it equally from either face, so the two enantiomers of the product form in a 50:50 ratio: an optically inactive racemic mixture. This is racemisation.
Why the other options are incorrect
Option (1) with retention of configuration — Incorrect. Retention means that only the product with the same configuration is formed. Attack on a planar carbocation is not limited to the side the halide left from.
Option (2) with inversion of configuration — Incorrect. Complete inversion is the mark of the SN2 mechanism, where the nucleophile attacks from the side opposite to the halogen in one step, with no carbocation.
Option (4) with conversion into an achiral alkene — Incorrect. An alkene comes from β-elimination (dehydrohalogenation, for example with alcoholic KOH), a different reaction. A substitution reaction replaces the halogen by the nucleophile.
NCERT (2026-27 edition):
6.7.1 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).”