Question

Identify the product B in following reaction.
\(\text{CH}_3\text{Br}\overset{\text{KCN}\,}{\rightarrow }\text{A}\rightarrow _{\text{C}_2\text{H}_5\text{OH}\,}^{\text{Na}\,}\text{B}\)

• A. Ethane
• B. Ethanamine
• C. Ethanol
• D. Methanol

Hint:

KCN gives the nitrile and Na with ethanol reduces it to a primary amine with one more carbon.

Answer 1

The Correct Option is B

Step 1: First step
\(\text{CH}_3\text{Br} + \text{KCN} \to \text{CH}_3\text{CN} + \text{KBr}\). The cyanide ion replaces bromide by \(S_N2\), so A is methyl cyanide (acetonitrile).

Step 2: Second step
Sodium in ethanol supplies nascent hydrogen, which reduces the nitrile group (Mendius reduction): \(\text{CH}_3\text{CN} + 4[\text{H}] \to \text{CH}_3\text{CH}_2\text{NH}_2\).

Step 3: Identify B
B is \(\text{CH}_3\text{CH}_2\text{NH}_2\), ethanamine, with two carbons.

Step 4: Why not others
Ethane would need loss of the CN group, and ethanol and methanol have no nitrogen, so the amine nitrogen could not have been lost during a reduction with Na and ethanol.

Final Answer:
Reduction of the nitrile gives ethanamine. This is option (B). \[ \boxed{\text{(B) }\text{Ethanamine}} \]