In the following reaction,C_6H_5CN xrightarro | vedclass

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(D) The reaction proceeds in two main steps:$1$. The first step is the $Stephen \ reduction$ of benzonitrile $(C_6H_5CN)$ using $SnCl_2/HCl$ followed

Vedclass · Accepted Answer

$X = C_6H_5CHO, Y = 	ext{4-methylchalcone}$ (with correct structure)

Vedclass · Answer

(D) The reaction proceeds in two main steps:$1$. The first step is the $Stephen \ reduction$ of benzonitrile $(C_6H_5CN)$ using $SnCl_2/HCl$ followed by hydrolysis,which yields benzaldehyde $(C_6H_5CHO)$ as product $X$.$2$. The second step is the $Claisen-Schmidt$ condensation (a type of aldol condensation) between benzaldehyde $(X)$ and $4-methylacetophenone$ in the presence of dilute $NaOH$. This reaction produces an $\alpha, \beta$-unsaturated ketone,which is $Y$ (specifically,$1-(4-methylphenyl)-3-phenylprop-2-en-1-one$).

In the following reaction,$C_6H_5CN$ $\xrightarrow[(ii) H_3O^+]{(i) SnCl_2/HCl} X$ $\xrightarrow{Dil. NaOH} Y$. Here,$X$ and $Y$ are:

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