1-Chloro-4-nitro-benzene reacts with NaOH at 443K to form

The correct answer is Option (1) → 4-nitrophenol

Aryl halides are very unreactive towards nucleophilic substitution. The C–Cl bond has partial double bond character from resonance with the ring and the carbon is sp² hybridised, so the bond is hard to break. Chlorobenzene itself needs drastic conditions:

    \( \displaystyle \ce{C6H5Cl} \xrightarrow[\text{(ii) }\ce{H+}]{\text{(i) NaOH, 623 K, 300 atm}} \ce{C6H5OH} \)

An electron withdrawing –NO₂ group at the ortho or para position makes the reaction much easier. It pulls electron density out of the ring, helping the attack of \( \ce{OH^{-}} \), and it stabilises the negatively charged intermediate by resonance. The more nitro groups at these positions, the milder the conditions: 443 K for one, 368 K for two, and only warm water for three.

In 1-chloro-4-nitrobenzene the nitro group is para to Cl, so \( \ce{OH^{-}} \) replaces Cl at 443 K. The sodium phenoxide formed is then acidified:

    \( \displaystyle p\text{-}\ce{O2NC6H4Cl + 2NaOH} \xrightarrow{443\ \text{K}} p\text{-}\ce{O2NC6H4O^-Na+ + NaCl + H2O} \)

    \( \displaystyle p\text{-}\ce{O2NC6H4O^-Na+ + H+ -> }\,p\text{-}\ce{O2NC6H4OH + Na+} \)

Only the chlorine is replaced; the nitro group stays where it was. The product is 4-nitrophenol.

Why the other options are incorrect

• Option (2) 4-chlorophenol — Incorrect. Here the nitro group has been replaced and the chlorine kept. In this reaction the halogen is the group that leaves; the –\( \ce{NO2} \) group is what activates the ring.

• Option (3) 2-chloro-5-nitrophenol — Incorrect. This would need an –OH group to replace a ring hydrogen next to Cl. Hydroxide ion does not displace hydrogen; it replaces the halogen.

• Option (4) 2-chlorophenol — Incorrect. This has lost the nitro group and gained an –OH at a ring carbon that carried hydrogen; neither change takes place.

NCERT (2026-27 edition):
Replacement by Hydroxyl Group – …