Explain the difference in the following reactions.
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Explanation

When the concentration of the nucleophile `(overset (Ө) O H)` is high `(5N NaOH)`, it follows `SN^2` reaction and inversion takes place.

With dil. `NaOH, SN^1` mechanism is followed when `(C=O)` bond at chircal carbon is cleaved forming a carbonium. Hence, attack by nucleophile `(overset (Ө) O H)` takes place from both directions forming inversion and retention products resulting in `(+-)` forms, i.e., inactive forms.
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