Question
$S_{N}1$ reaction on optically active substrates mainly gives$:$
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Answer
Since the nucleophile attacks the carbocation only after the leaving group has departed, in $S_N1$ there is no need for back$-$side attack.
The carbocation and its substituents are all in the same plane $($Figure $1),$ meaning that the nucleophile can attack from either side.
As a result, both enantiomers are formed in a $S_N1$ reaction, leading to a racemic mixture of both enantiomers.
The carbocation and its substituents are all in the same plane $($Figure $1),$ meaning that the nucleophile can attack from either side.
As a result, both enantiomers are formed in a $S_N1$ reaction, leading to a racemic mixture of both enantiomers.
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