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JEE Advanced 2019 Paper 1 · Chemistry · Hydrocarbons · medium

Consider electrophilic aromatic substitution on substituted benzenes.

One or more options correct

Consider electrophilic aromatic substitution on substituted benzenes. The correct statement(s) is(are)
  1. A. Nitration of chlorobenzene gives mainly the meta isomer because chlorine withdraws electrons by the inductive effect.
  2. B. Friedel-Crafts alkylation of benzene with 1-chloropropane and anhydrous AlCl3\mathrm{AlCl_3} gives mainly nn-propylbenzene.
  3. C. Nitration of acetanilide gives mainly the para isomer, as the NHCOCH3\mathrm{NHCOCH_3} group is ortho/para directing but bulkier and less activating than NH2\mathrm{NH_2}.
  4. D. Nitrobenzene undergoes Friedel-Crafts acylation readily to give mm-nitroacetophenone.
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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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