Question

Identify the product 'B' in the following reaction
\(\text{CH}_3-\text{I}\overset{\text{KCN}\,}{\rightarrow }\text{A}\overset{\text{Na/C}_2\text{H}_5\text{OH}\,}{\rightarrow }\text{B}\)

• A. \(\text{CH}_3\text{-CH}_2\text{-CN}\)
• B. \(\text{CH}_3\text{-CH}_2\text{-CH}_3\)
• C. \(\text{CH}_3\text{-CH}_2\text{-NH}_2\)
• D. \(\text{CH}_3\text{-NH-C}_2\text{H}_5\)

Hint:

KCN replaces I with CN, and Na/C2H5OH reduces the nitrile to a primary amine with one extra CH2.

Answer 1

The Correct Option is C

Step 1: Count carbons:
CH3I has one carbon. KCN adds one more carbon through the cyanide group, so A has two carbons: $\text{CH}_3\text{CN}$.

Step 2: Reduction:
Reducing the $\text{C}\equiv\text{N}$ group with Na/alcohol adds four hydrogens and gives $-\text{CH}_2\text{NH}_2$. So B has two carbons and one NH2 group: $\text{CH}_3\text{CH}_2\text{NH}_2$.

Step 3: Eliminate:
Only option (C) is a two-carbon primary amine. (A) is still a nitrile, (B) has no nitrogen, and (D) is a secondary amine with three carbons.

Final Answer:
B is ethylamine. \[ \boxed{\text{CH}_3\text{CH}_2\text{NH}_2} \]