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: Step A:
Methyl iodide reacts with KCN by an SN2 reaction. KCN is mostly ionic, so the carbon end of the cyanide ion attacks and the product is the nitrile \(\text{CH}_3\text{CN}\) (methyl cyanide). So A is \(\text{CH}_3\text{CN}\).

Step 2: Step B:
Na with C2H5OH is nascent hydrogen. It reduces a nitrile to a primary amine (Mendius reduction): \(-\text{CN}\) becomes \(-\text{CH}_2\text{NH}_2\).
\[ \text{CH}_3\text{CN} \xrightarrow{\text{Na/C}_2\text{H}_5\text{OH}} \text{CH}_3\text{CH}_2\text{NH}_2 \]

Step 3: Why the other options are wrong:
Option (A), ethyl cyanide, has an extra carbon and is not formed from CH3I plus KCN. Option (B), propane, would need loss of both the nitrogen and one carbon. Option (D), N-methylethanamine, is a secondary amine, and a nitrile gives a primary amine on this reduction.

Final Answer:
B is ethanamine (ethylamine), option (C). \[ \boxed{\text{CH}_3\text{CH}_2\text{NH}_2} \]