Question

Aniline on treatment with \(\text{NaNO}_2\) / HCl at about \(0^{\circ}\text{C}\) gives a compound A, which on further treatment with CuCN / KCN gives a compound B, which when reacts with \(\text{H}_2\) / Ni gives a compound C. The compound C on reaction with \(\text{HNO}_2\) gives a compound D. The structure of D is:

• A. \(\text{C}_6\text{H}_5\text{-CH}_2\text{-OH}\)
• B. \(\text{C}_6\text{H}_5\text{-NH-CH}_2\text{-CH}_3\)
• C. \(\text{C}_6\text{H}_5\text{-NH-OH}\)
• D. \(\text{C}_6\text{H}_5\text{-CH}_2\text{-NH}_2\)

Hint:

Follow the chain: diazonium salt, cyanide, amine by reduction, alcohol with nitrous acid.

Answer 1

The Correct Option is A

Step 1: Approach
Track the carbon skeleton and the nitrogen group across the four compounds.

Step 2: Chain
Aniline ($\text{C}_6\text{H}_5\text{NH}_2$) becomes the diazonium salt (A). Sandmeyer cyanation adds one carbon, making $\text{C}_6\text{H}_5\text{CN}$ (B). Catalytic hydrogenation turns $\text{-CN}$ into $\text{-CH}_2\text{NH}_2$ (C). The last step is deamination by nitrous acid: the $\text{-NH}_2$ on the aliphatic carbon is swapped for OH.

Step 3: Result
D is $\text{C}_6\text{H}_5\text{CH}_2\text{OH}$. The other options retain nitrogen or have no extra carbon, so (A) is correct.

Final Answer:
The sequence ends in benzyl alcohol, option (A). \[ \boxed{\text{C}_6\text{H}_5\text{CH}_2\text{OH}} \]