Identify the product 'B' in the following rea | vedclass

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(C) $1$. The reaction is the Stephen reduction of a nitrile. Isopropyl cyanide is $(CH_3)_2CH-CN$.$2$. In the first step, the nitrile is reduced by $S

Vedclass · Accepted Answer

$2-$Methylpropanal

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(C) $1$. The reaction is the Stephen reduction of a nitrile. Isopropyl cyanide is $(CH_3)_2CH-CN$.$2$. In the first step, the nitrile is reduced by $SnCl_2/HCl$ to form an imine intermediate $(A)$, which is $(CH_3)_2CH-CH=NH$.$3$. In the second step, the acid hydrolysis of the imine $(A)$ yields the corresponding aldehyde $(B)$.$4$. The reaction is: $(CH_3)_2CH-CN \xrightarrow{SnCl_2, HCl} (CH_3)_2CH-CH=NH \xrightarrow{H_3O^+} (CH_3)_2CH-CHO + NH_4Cl$.$5$. The product '$B$' is $(CH_3)_2CH-CHO$, which is $2-	ext{Methylpropanal}$.

List-$I$ (Reagents)	List-$II$ (Reaction Name)
$A$. $H_2, Pd-BaSO_4$	$I$. Etard Reaction
$B$. $SnCl_2, HCl$	$II$. Rosenmund Reduction
$C$. $CrO_2Cl_2, CS_2$	$III$. Gatterman-Koch Reaction
$D$. $CO, HCl, \text{Anhyd. } AlCl_3$	$IV$. Stephen Reaction

Identify the product '$B$' in the following reaction: $\text{Isopropyl cyanide} \xrightarrow{SnCl_2, HCl} A \xrightarrow{H_3O^+} B + NH_4Cl$

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