Compound B is ? | vedclass

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(C) The reaction with $NaBH_4$ reduces the carbonyl group $(C=O)$ to a hydroxyl group $(-OH)$ while leaving the carboxylic acid $(-COOH)$ group unaffe

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Vedclass · Answer

(C) The reaction with $NaBH_4$ reduces the carbonyl group $(C=O)$ to a hydroxyl group $(-OH)$ while leaving the carboxylic acid $(-COOH)$ group unaffected. This indicates that compound $A$ contains both a ketone and a carboxylic acid group. $LiAlH_4$ is a stronger reducing agent than $NaBH_4$ and it reduces both the ketone and the carboxylic acid groups. Specifically,the ketone is reduced to a secondary alcohol and the carboxylic acid is reduced to a primary alcohol $(-CH_2OH)$. Therefore,compound $B$ is the product formed by the reduction of both functional groups,which corresponds to the structure with a bridgehead hydroxyl group and a $-CH_2OH$ group.

Compound $B$ is ?

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