In an SN1 (substitution nucleophilic unimolecular) reaction of an opt…

The correct answer is Option (4) → Both (1) and (2)

In an S_N1 reaction the halide ion leaves first, and the carbocation formed in the slow step is sp² hybridised and planar. The nucleophile can attack it from either side of the plane:

• Attack from the side the halide left gives the product with the same configuration (retention).

• Attack from the opposite side gives the product with the opposite configuration (inversion).

Both products form, so the optically active alkyl halide gives a racemic mixture: racemisation is due to both inversion and retention.

Why the other options are incorrect

Option (1) Inversion of configuration — Incorrect. Inversion alone gives only one enantiomer, as in the S_N2 reaction; it cannot give a racemic mixture.

Option (2) Retention of configuration — Incorrect. Retention alone also gives only one enantiomer, so the product would stay optically active.

Option (3) Rearrangement of the carbocation — Incorrect. Racemisation comes from attack on both faces of the planar carbocation, not from any change in its carbon skeleton.

NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes – “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).”