The reaction Ph-CDO xrightarrow{50% text{ aq. | vedclass

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Question

(C) The reaction is a Cannizzaro reaction. In the mechanism, the nucleophilic attack of $OH^-$ on the carbonyl carbon of $Ph-CDO$ forms a tetrahedral

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$Ph-CD_2-OH$

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(C) The reaction is a Cannizzaro reaction. In the mechanism, the nucleophilic attack of $OH^-$ on the carbonyl carbon of $Ph-CDO$ forms a tetrahedral intermediate. In the rate-determining step, a hydride (or deuteride) ion is transferred from this intermediate to another molecule of $Ph-CDO$. Since the intermediate contains a $D$ atom attached to the carbon, a deuteride ion $(D^-)$ is transferred to the second $Ph-CDO$ molecule. This results in the formation of $Ph-CD_2-O^-$ and $Ph-COO^-$. Upon workup (or proton exchange with the solvent), the alkoxide $Ph-CD_2-O^-$ becomes $Ph-CD_2-OH$.

The reaction $Ph-CDO \xrightarrow{50\% \text{ aq. } NaOH, \text{ warm}} Ph-COO^- + \text{an alcohol}$. This alcohol is:

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