The hydrolysis of optically active 2-bromobut | vedclass

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Question

(D) $2-$Bromobutane is a secondary alkyl halide. In the presence of aqueous $NaOH$,the reaction proceeds primarily through the $S_N1$ mechanism.This m

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$(\pm)$ butan$-2-$ol

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(D) $2-$Bromobutane is a secondary alkyl halide. In the presence of aqueous $NaOH$,the reaction proceeds primarily through the $S_N1$ mechanism.This mechanism involves the formation of a planar carbocation intermediate.The nucleophile $(OH^-)$ can attack the carbocation from either side with equal probability.This leads to the formation of both enantiomers in equal amounts,resulting in a racemic mixture of $(\pm)$ butan$-2-$ol.

The hydrolysis of optically active $2-$bromobutane with aqueous $NaOH$ results in the formation of:

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