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: Track carbons
CH3Br has one carbon and CN adds one more, so A has two carbons and B must also have two carbons.

Step 2: Track the functional group
The CN group ($\text{C}\equiv\text{N}$) is reduced to $\text{CH}_2\text{NH}_2$ with Na/ethanol.

Step 3: Product
$\text{CH}_3\text{CN} \to \text{CH}_3\text{CH}_2\text{NH}_2$, ethanamine.

Step 4: Answer
Option (B). Methanol has one carbon and no nitrogen, ethanol has no nitrogen and ethane has no nitrogen, so none can come from a nitrile reduction.

Final Answer:
B is ethanamine. This is option (B). \[ \boxed{\text{(B) }\text{Ethanamine}} \]