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(D) The reaction between benzaldehyde $(Ph-CHO)$ and acetaldehyde $(CH_3-CHO)$ in the presence of a base like $Ca(OH)_2$ followed by heating $(\Delta)

Vedclass · Accepted Answer

$Ph-CH=CH-CHO$

Vedclass · Answer

(D) The reaction between benzaldehyde $(Ph-CHO)$ and acetaldehyde $(CH_3-CHO)$ in the presence of a base like $Ca(OH)_2$ followed by heating $(\Delta)$ is a Claisen-Schmidt condensation,which is a type of cross-aldol condensation. $1$. The $\alpha$-hydrogen of acetaldehyde is removed by the base to form an enolate nucleophile $(CH_2^-CHO)$. $2$. This enolate attacks the carbonyl carbon of benzaldehyde to form a $\beta$-hydroxy aldehyde intermediate. $3$. Subsequent dehydration upon heating $(\Delta)$ gives cinnamaldehyde $(Ph-CH=CH-CHO)$ as the major product.

List-$I$ (Reaction)	List-$II$ (Product)
$A$. $\text{Ketone} + NaBH_4$	$I$. $1^\circ$-alcohol
$B$. $\text{Ester} + LiAlH_4$	$II$. $3^\circ$-alcohol
$C$. $RMgX + \text{Ketone}$	$III$. $2^\circ$-alcohol
	$IV$. $\text{Alkane}$

Find the major product of the following reaction:
$Ph-CHO + CH_3-CHO \xrightarrow[\Delta]{Ca(OH)_2}$ Major product

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Match the following reactions in List-$I$ with their products in List-$II$.
List-$I$ (Reaction) List-$II$ (Product)
$A$. $\text{Ketone} + NaBH_4$ $I$. $1^\circ$-alcohol
$B$. $\text{Ester} + LiAlH_4$ $II$. $3^\circ$-alcohol
$C$. $RMgX + \text{Ketone}$ $III$. $2^\circ$-alcohol
$IV$. $\text{Alkane}$

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Find the major product of the following reaction: $Ph-CHO + CH_3-CHO \xrightarrow[\Delta]{Ca(OH)_2}$ Major product