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Question

(B) The given reaction involves the reduction of both a ketone group $(-C=O)$ and a carboxylic acid group $(-COOH)$ to their corresponding alcohols. $

Vedclass · Accepted Answer

$LiAlH_4$

Vedclass · Answer

(B) The given reaction involves the reduction of both a ketone group $(-C=O)$ and a carboxylic acid group $(-COOH)$ to their corresponding alcohols. $LiAlH_4$ (Lithium aluminium hydride) is a strong reducing agent that can reduce both carboxylic acids to primary alcohols $(-CH_2OH)$ and ketones to secondary alcohols $(-CH(OH)-)$. $NaBH_4$ is a milder reducing agent that typically reduces ketones but not carboxylic acids. $PCC$ is an oxidizing agent,and $KMnO_4$ is a strong oxidizing agent. Therefore,$LiAlH_4$ is the correct reagent for this transformation.

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