Question

Aniline undergoes diazotization with \(\text{NaNO}_{2}\)/HCl to give compound A. Compound A reacts with fluoroboric acid to give compound B. When B is heated with aq. \(\text{NaNO}_{2}\) in the presence of copper, it forms compound C. Identify C.

• A. \(\text{Ar}-\text{F}\)
• B. \(\text{Ar}-\text{Cl}\)
• C. \(\text{Ar}-\text{H}\)
• D. \(\text{Ar}-\text{NO}_{2}\)

Hint:

Follow each step: diazonium chloride, then diazonium fluoroborate, then replacement of the diazonium group by a nitro group using nitrite and copper.

Answer 1

The Correct Option is D

Step 1: Compound A
Aniline with \(\text{NaNO}_2\)/HCl at \(273-278\) K gives benzenediazonium chloride, \(\text{C}_6\text{H}_5\text{N}_2^{+}\text{Cl}^{-}\).

Step 2: Compound B
With fluoroboric acid, \(\text{HBF}_4\), it gives the insoluble benzenediazonium fluoroborate, \(\text{C}_6\text{H}_5\text{N}_2^{+}\text{BF}_4^{-}\).

Step 3: Compound C
Warming the fluoroborate with aqueous \(\text{NaNO}_2\) and copper replaces the diazonium group by \(\text{NO}_2\), with loss of \(\text{N}_2\).

Step 4: Conclusion
So C is nitrobenzene, \(\text{C}_6\text{H}_5\text{NO}_2\), written as \(\text{Ar-NO}_2\). Heating the dry fluoroborate alone would give fluorobenzene, but nitrite is present here.

Final Answer:
Compound C is the nitro compound, Ar-NO2. \[ \boxed{\text{(D) }\text{Ar}-\text{NO}_{2}} \]