C_6H_5CHO + C_6H_5COCH_3 xrightarrow[293 K]{ | vedclass

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Question

(B) The reaction between benzaldehyde $(C_6H_5CHO)$ and acetophenone $(C_6H_5COCH_3)$ in the presence of a dilute base $(OH^-)$ is a Claisen-Schmidt c

Vedclass · Accepted Answer

$C_6H_5-CH=CH-CO-C_6H_5$

Vedclass · Answer

(B) The reaction between benzaldehyde $(C_6H_5CHO)$ and acetophenone $(C_6H_5COCH_3)$ in the presence of a dilute base $(OH^-)$ is a Claisen-Schmidt condensation reaction.In this reaction,the enolate ion formed from acetophenone attacks the carbonyl carbon of benzaldehyde.The initial product is a $\beta$-hydroxy ketone,which subsequently undergoes dehydration to form an $\alpha,\beta$-unsaturated ketone.The reaction is:$C_6H_5CHO + CH_3COC_6H_5$ $\xrightarrow{OH^-} C_6H_5CH(OH)CH_2COC_6H_5$ $\xrightarrow{-H_2O} C_6H_5CH=CHCOC_6H_5$The final product is benzalacetophenone (also known as chalcone),which corresponds to option $B$.

$C_6H_5CHO + C_6H_5COCH_3 \xrightarrow[293 \ K]{OH^{-}}$ product of reaction is . . . . . . .

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