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: Count carbons:
The nitrile carbon becomes the aldehyde carbon, so the product has the same number of carbons as the nitrile: $4$.

Step 2: Name:
$(\text{CH}_3)_2\text{CH}-\text{CHO}$ has a methyl on carbon 2 of a three carbon aldehyde chain: 2-methylpropanal.

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