Question

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

Hint:

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.

Answer 1

Step 1: Nitrobenzene to (A) using Fe/HCl. Iron with hydrochloric acid is a reducing agent. It reduces the nitro group \(-NO_2\) to an amino group \(-NH_2\).
(A) = aniline, \(C_6H_5NH_2\).
\[ C_6H_5NO_2 \xrightarrow{Fe/HCl} C_6H_5NH_2 \]

Step 2: Aniline to (B) with NaNO₂ + HCl at 273 K. \(NaNO_2 + HCl\) gives nitrous acid \((HNO_2)\), which converts the \(-NH_2\) group into a diazonium group. This step (diazotisation) is done in the cold (273-278 K) because the diazonium salt breaks down when warm.
(B) = benzenediazonium chloride, \(C_6H_5N_2^{+}Cl^{-}\).
\[ C_6H_5NH_2 + HNO_2 + HCl \xrightarrow{273\,K} C_6H_5N_2^{+}Cl^{-} + 2H_2O \]

Step 3: (B) to (C) with warm water / H⁺. When the diazonium salt is heated with water (dilute acid), the \(-N_2^{+}\) group is replaced by \(-OH\), releasing nitrogen gas.
(C) = phenol, \(C_6H_5OH\).
\[ C_6H_5N_2^{+}Cl^{-} + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2\uparrow + HCl \]

\[\boxed{A = \text{aniline},\ B = \text{benzenediazonium chloride},\ C = \text{phenol}}\]