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(C) The reaction between $CH_3CHO$ (acetaldehyde) and $PhCOCH_3$ (acetophenone) in the presence of dilute $NaOH$ is a cross-aldol condensation reactio

Vedclass · Accepted Answer

$CH_3CH(OH)CH_2COPh$

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(C) The reaction between $CH_3CHO$ (acetaldehyde) and $PhCOCH_3$ (acetophenone) in the presence of dilute $NaOH$ is a cross-aldol condensation reaction.$1$. Acetophenone has an $\alpha$-hydrogen atom,so it can form an enolate ion in the presence of a base $(NaOH)$.$2$. The enolate ion formed from acetophenone acts as a nucleophile and attacks the carbonyl carbon of the acetaldehyde $(CH_3CHO)$,which is more electrophilic than acetophenone.$3$. The nucleophilic attack leads to the formation of a $\beta$-hydroxy ketone.$4$. The reaction proceeds as follows:$PhCOCH_3 + OH^- \rightarrow [PhCOCH_2]^- + H_2O$$[PhCOCH_2]^- + CH_3CHO \rightarrow CH_3-CH(O^-)-CH_2-COPh$$CH_3-CH(O^-)-CH_2-COPh + H_2O \rightarrow CH_3-CH(OH)-CH_2-COPh + OH^-$Thus,the major product is $CH_3-CH(OH)-CH_2-COPh$.

The major product formed in the following reaction is:
$CH_3CHO + PhCOCH_3 \xrightarrow{\text{dil. } NaOH}$

Similar Questions

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Which is the slowest step in the above Cannizzaro reaction?

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The major product formed in the following reaction is: $CH_3CHO + PhCOCH_3 \xrightarrow{\text{dil. } NaOH}$