Complete the following analogy:
\( S_N2 \): Inversion of configuration; \( S_N1 \): ______
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Racemisation
The two nucleophilic substitution mechanisms give different stereochemical results with an optically active alkyl halide:
\(S_N2\): one step; the nucleophile attacks from the side opposite to the leaving group, so the product has the inverted configuration (inversion of configuration).
\(S_N1\): two steps; the halide first leaves and forms a planar carbocation (sp2, achiral). The nucleophile can attack this flat ion from either face equally, giving a 50:50 mixture of the two enantiomers, an optically inactive racemic mixture.
\( \displaystyle \ce{R3C-X -> R3C+ + X-} \quad (\text{slow}) \)
\( \displaystyle \ce{R3C+ + OH- -> R3C-OH} \quad (\text{fast, from both faces}) \)
So the analogy is completed by racemisation.
Why the other options are incorrect
Option (2) Retention — Incorrect. Retention means the product keeps the same configuration as the reactant; it is seen when no bond to the stereocentre is broken, not when a planar carbocation is formed.
Option (3) Optical rotation — Incorrect. This is the rotation of plane-polarised light by a chiral substance, a property, not a stereochemical result of a mechanism; the racemic product in fact shows zero rotation.
Option (4) Chiral — Incorrect. This word describes a molecule that is not superimposable on its mirror image; it is not an outcome of a reaction.
NCERT (2026-27 edition):
Inversion, Retention and Racemisation – “… 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.”