Which of the following statement(s) is/are correct with reference to substitution reactions of \(C_6H_5Cl\)?
(A) \(Cl\) acts as an activating and \(o/p\) directing group.
(B) \(Cl\) acts as a deactivating and \(o/p\) directing group.
(C) \(Cl\) acts as an activating and \(m-\) directing group.
(D) \(Cl\) acts as a deactivating and \(m-\)directing group.
(E) Presence of \(NO_2\) at \(p-\) position facilitates the reactions.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → B and E only
(A) Cl acts as an activating and o/p directing group. — Incorrect. The −I effect of chlorine withdraws electrons from the ring, so it deactivates the ring.
(B) Cl acts as a deactivating and o/p directing group. — Correct. In electrophilic substitution the −I effect outweighs the +R effect, so the ring is deactivated on balance; resonance still raises the electron density at the ortho and para positions, so the new group goes ortho/para.
(C) Cl acts as an activating and m- directing group. — Incorrect. Chlorine neither activates the ring nor directs to the meta position.
(D) Cl acts as a deactivating and m-directing group. — Incorrect. Chlorine deactivates, but it directs ortho/para, not meta.
(E) Presence of NO2 at p- position facilitates the reactions. — Correct. In nucleophilic substitution of the Cl atom, a para –\( \ce{NO2} \) group withdraws electron density and stabilises the negative intermediate, so Cl is replaced much more easily (443 K instead of 623 K):
\( \text{p-}\ce{O2NC6H4Cl} \xrightarrow[\text{(ii) }\ce{H+}]{\text{(i) NaOH, 443 K}} \text{p-}\ce{O2NC6H4OH} \)
NCERT (2026-27 edition):
Electrophilic Substitution Reactions – “… Halogen atom besides being slightly deactivating is o, p-directing; therefore, further substitution occurs at ortho- and para-positions with respect to the halogen atom.”