Question

Identify the product 'B' in the following reaction
\(\text{Isopropyl cyanide}\overset{\text{SnCl}_2,\text{HCl}\,}{\rightarrow }\text{A}\overset{\text{H}_3\text{O}^+\,}{\rightarrow }\text{B}+\text{NH}_4\text{Cl}\)

• A. Propanal
• B. Propanone
• C. 2-Methylpropanal
• D. 2-Methylpropanoic acid

Hint:

Stephen reduction turns a nitrile into an aldehyde with the same carbon skeleton.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In the Stephen reaction, a nitrile is reduced by \(\text{SnCl}_2\) and HCl to an imine hydrochloride, which hydrolyses to an aldehyde.

Step 2: Key Formula or Approach:
\(\text{RCN} + 2[\text{H}] + \text{HCl} \xrightarrow{\text{SnCl}_2} \text{RCH=NH} \cdot \text{HCl}\), then \(\text{RCH=NH}\cdot\text{HCl} + \text{H}_2\text{O} \to \text{RCHO} + \text{NH}_4\text{Cl}\).

Step 3: Detailed Explanation:
Isopropyl cyanide is \((\text{CH}_3)_2\text{CHCN}\), so \(\text{R} = (\text{CH}_3)_2\text{CH}-\).
A is the imine \((\text{CH}_3)_2\text{CHCH=NH}\).
Hydrolysis gives B \(= (\text{CH}_3)_2\text{CHCHO}\), which is 2-methylpropanal, with \(\text{NH}_4\text{Cl}\) as the by-product.
Propanal has one carbon too few and the acid would be the product of a full hydrolysis, which does not occur here.

Final Answer:
B is 2-methylpropanal, option (C). \[ \boxed{\text{2-Methylpropanal}} \]