Behavior of a chiral alkyl halide in $S_N1$ reaction

In an $S_N1$ reaction, the leaving group first departs and forms a carbocation intermediate.

$R^* - X \rightarrow R^{*+} + X^-$

The carbocation is planar ($sp^2$-hybridised), so the nucleophile can attack from either side.

Therefore, both configurations are formed:

• attack from one side gives one enantiomer,
• attack from the other side gives the opposite enantiomer.

Hence, $S_N1$ generally leads to racemisation.

Behavior of a chiral alkyl halide in $S_N2$ reaction

In an $S_N2$ reaction, substitution occurs in one step by backside attack of the nucleophile.

Because the nucleophile attacks from the side opposite to the leaving group, the stereochemistry gets inverted.

This is called Walden inversion.

Hence, $S_N2$ leads to inversion of configuration.

Evaluate the options

• A: Retention in $S_N1$ and inversion in $S_N2$
  ❌ Wrong, because $S_N1$ gives racemisation, not retention.

• B: Racemisation in $S_N1$ and retention in $S_N2$
  ❌ Wrong, because $S_N2$ gives inversion, not retention.

• C: Racemisation in both $S_N1$ and $S_N2$
  ❌ Wrong, because $S_N2$ gives inversion.

• D: Racemisation in $S_N1$ and inversion in $S_N2$
  ✅ Correct.

Final answer

The correct statement is:

$\boxed{\text{D}}$