The product of the following reaction is C_6H | vedclass

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(A) Step $1$: The reaction is a Claisen-Schmidt condensation, which is a type of crossed aldol condensation between benzaldehyde (which has no $\alpha

Vedclass · Accepted Answer

$C_6H_5CH=C(CH_3)CHO$

Vedclass · Answer

(A) Step $1$: The reaction is a Claisen-Schmidt condensation, which is a type of crossed aldol condensation between benzaldehyde (which has no $\alpha$-hydrogens) and propanal (which has $\alpha$-hydrogens).Step $2$: In the presence of dilute $NaOH$, the $\alpha$-carbon of propanal acts as a nucleophile and attacks the carbonyl carbon of benzaldehyde to form an aldol intermediate, $C_6H_5CH(OH)CH(CH_3)CHO$.Step $3$: Subsequent dehydration (loss of $H_2O$) under heating $(\Delta)$ yields the $\alpha,\beta$-unsaturated aldehyde, $C_6H_5CH=C(CH_3)CHO$.Step $4$: Thus, the correct product is $C_6H_5CH=C(CH_3)CHO$.

The product of the following reaction is $C_6H_5CHO + CH_3CH_2CHO \xrightarrow[\Delta]{\text{Dil. NaOH}}$

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