What will happen in the following reaction if -NO2 group is introduced at para position to -Cl?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Reaction will occur at comparatively lower temperatures.
Aryl halides are very unreactive towards nucleophilic substitution. The C–Cl bond has partial double bond character (resonance with the ring), the carbon is sp2, and the electron-rich ring repels the incoming nucleophile. So chlorobenzene gives phenol only under drastic conditions, 623 K and 300 atm.
A para –\( \ce{NO2} \) group delocalises the negative charge of the intermediate onto its own oxygen atoms and so stabilises it.
With –\( \ce{NO2} \) para to –Cl, sodium hydroxide replaces the chlorine at 443 K, and acidification gives 4-nitrophenol:
\( \displaystyle \ce{O2N-C6H4-Cl + 2NaOH} \xrightarrow{443\ \mathrm{K}} \ce{O2N-C6H4-ONa + NaCl + H2O} \)
\( \displaystyle \ce{O2N-C6H4-ONa + HCl -> O2N-C6H4-OH + NaCl} \)
So with –\( \ce{NO2} \) para to –Cl, the reaction occurs at a comparatively lower temperature (443 K instead of 623 K).
Why the other options are incorrect
Option (1) Reaction will not take place — Incorrect. The nitro group does not stop the reaction; it makes it easier.
Option (2) Reaction will occur at comparatively higher temperatures — Incorrect. A higher temperature would be needed for an electron releasing group, not for an electron withdrawing one.
Option (4) Reaction will occur at the same temperature — Incorrect. The para –\( \ce{NO2} \) group stabilises the intermediate, so the conditions change markedly.
NCERT (2026-27 edition):
Reactions of Haloarenes – …
