When an optically pure sample of -bromobutane undergoes an idealized pure hydrolysis, the resulting product mixture is virtually optically inactive. What is the fundamental geometric reason for this p

Question

When an optically pure sample of $(R)-2$ -bromobutane undergoes an idealized pure $S_N1$ hydrolysis, the resulting product mixture is virtually optically inactive. What is the fundamental geometric reason for this precise phenomenon?

Prashant Kotian · Accepted Answer

Correct answer: B. In an idealized reaction, the rate-determining loss of the leaving group generates a free, hybridized carbocation. Because the empty -orbital is planar, the incoming nucleophile () has an equal statistical probability of attacking from the top face (yielding one enantiomer) or the bottom face (yielding the opposite enantiomer). This 50:50 attack intrinsically generates a racemic mixture, completely neutralizing the net optical activity.

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