Question

Acetonitrile on reduction with sodium in ethanol gives A, which reacts with \(\text{NaNO}_{2}\) and dil. HCl to form B. Identify product B.

• A. \(\text{CH}_{3}\text{-CH}_{2}\text{-NH}_{2}\)
• B. \(\text{CH}_{3}\text{-CH}_{2}\text{-NO}_{2}\)
• C. \(\text{CH}_{3}\text{-CH}_{2}\text{-N}_{2}\text{Cl}\)
• D. \(\text{CH}_{3}\text{-CH}_{2}\text{-OH}\)

Hint:

Na/ethanol reduces the nitrile to a primary amine, which then gives an alcohol with nitrous acid.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Acetonitrile, \(\text{CH}_{3}\text{CN}\), is reduced by sodium in ethanol (Mendius reduction) to a primary amine.

Step 2: Find A
The product A is ethylamine, \(\text{CH}_{3}\text{CH}_{2}\text{NH}_{2}\).

Step 3: Find B
\(\text{NaNO}_{2}\) with dilute HCl produces nitrous acid. A primary aliphatic amine gives an unstable diazonium salt that loses \(\text{N}_{2}\) and forms the alcohol, so B is \(\text{CH}_{3}\text{CH}_{2}\text{OH}\).

Final Answer:
B is ethanol. \[ \boxed{\text{(D) }\text{CH}_{3}\text{-CH}_{2}\text{-OH}} \]