One or more options correct
Consider electrophilic aromatic substitution on substituted benzenes. The correct statement(s) is(are)
- A. Nitration of chlorobenzene gives mainly the meta isomer because chlorine withdraws electrons by the inductive effect.
- B. Friedel-Crafts alkylation of benzene with 1-chloropropane and anhydrous gives mainly -propylbenzene.
- C. Nitration of acetanilide gives mainly the para isomer, as the group is ortho/para directing but bulkier and less activating than .
- D. Nitrobenzene undergoes Friedel-Crafts acylation readily to give -nitroacetophenone.
Show answer and solution
Answer: C
(A) Chlorine withdraws electrons inductively (deactivating) but donates a lone pair by resonance, which stabilises the ortho and para sigma complexes; it is therefore deactivating yet ortho/para directing, and the para product dominates. Option A is false. (B) The primary carbocation or its AlCl3 complex undergoes a 1,2-hydride shift to the more stable secondary cation, so the main product is cumene (isopropylbenzene) and not the n-propyl compound; false. (C) In acetanilide the nitrogen lone pair is partly delocalised into the carbonyl, so the group is only moderately activating and is sterically larger than amino. The nitronium ion attacks mainly para, giving p-nitroacetanilide; true. (D) The nitro group strongly deactivates the ring, and the acylium ion is too weak an electrophile, so Friedel-Crafts acylation fails on nitrobenzene (nitrobenzene is even used as a solvent for it); false.
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