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: Understanding the Concept
We follow the sequence one step at a time.

Step 2: Step to A
Aniline with \(\text{NaNO}_2\)/HCl at 0 C gives benzenediazonium chloride, \(\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-\).

Step 3: Step to B
CuCN/KCN replaces the diazonium group by CN (Sandmeyer reaction) and gives benzonitrile, \(\text{C}_6\text{H}_5\text{CN}\).

Step 4: Step to C
\(\text{H}_2\)/Ni reduces the nitrile to a primary amine, benzylamine, \(\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2\).

Step 5: Step to D
A primary aliphatic amine reacts with nitrous acid to form an alcohol with the release of nitrogen:
\[ \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2+\text{HNO}_2\to\text{C}_6\text{H}_5\text{CH}_2\text{OH}+\text{N}_2+\text{H}_2\text{O} \]

Step 6: Conclusion
D is benzyl alcohol, option (A).

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