Question:medium

Write the structure of (A), (B) and (C) in the following reaction:
\[ C_6H_5NO_2 \xrightarrow{Fe/HCl} (A) \xrightarrow{NaNO_2 + HCl,\ 273\,K} (B) \xrightarrow{H_2O/H^+,\ \Delta} (C) \]

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Fe/HCl reduces nitrobenzene to aniline; cold NaNO2/HCl diazotises it to benzenediazonium chloride; warm water/H+ replaces the diazonium group with OH to give phenol.
Updated On: Jul 10, 2026
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Solution and Explanation

Follow the nitrogen group as it changes at each arrow.

First arrow (Fe/HCl): This is a classic reduction. The pair Fe + HCl supplies hydrogen that knocks the oxygen off the nitro group and adds hydrogen, turning \(-NO_2\) into \(-NH_2\). Nitrobenzene therefore becomes aniline, \(C_6H_5NH_2\), which is (A).

Second arrow (\(NaNO_2\) + HCl, 273 K): Mixing sodium nitrite with hydrochloric acid makes nitrous acid on the spot. Nitrous acid attacks the amino group and rebuilds it as a diazonium group, \(-N_2^{+}\). The ice-cold temperature is essential, since the salt decomposes if it warms up. The product (B) is benzenediazonium chloride, \(C_6H_5N_2^{+}Cl^{-}\).

Third arrow (water/\(H^+\), heat): On boiling with water in acidic medium, the diazonium group is a very good leaving group; it departs as nitrogen gas and a hydroxyl group takes its place on the ring. That gives phenol, \(C_6H_5OH\), which is (C), with \(N_2\) bubbling off.

So the sequence marches nitrobenzene → aniline → benzenediazonium chloride → phenol.
\(\boxed{A:\ C_6H_5NH_2,\quad B:\ C_6H_5N_2^{+}Cl^{-},\quad C:\ C_6H_5OH}\)
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