The hydrolysis of optically active $ 2 $-bromobutane with aqueous $ \mathrm{NaOH} $ result in the formation of

Question

The hydrolysis of optically active $ 2 $-bromobutane with aqueous $ \mathrm{NaOH} $ result in the formation of, $ (+) $ butan-2-ol, (-) butan-2-ol, ( $ \pm $ ) butan-1-ol, ( $ \pm $ ) butan-2-ol

MARKS App · Accepted Answer

2-Bromobutane is a secondary alkyl halide, therefore, hydrolysis will proceed through $ S_{N} 1 $ mechanisms, hence, racemic mixture will be formed, that is, (±) butan-2-ol. (+) Butan-2-ol (-) Butan-2-ol

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