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JEE Advanced 2022 Paper 2 · Chemistry · Hydrocarbons · medium

Regarding electrophilic aromatic substitution, choose the correct statement(s).

One or more options correct

Regarding electrophilic aromatic substitution, choose the correct statement(s).
  1. A. Nitration of acetanilide with HNO3/H2SO4\mathrm{HNO_3/H_2SO_4} gives mainly pp-nitroacetanilide.
  2. B. Direct nitration of aniline with concentrated HNO3/H2SO4\mathrm{HNO_3/H_2SO_4} gives a substantial amount of mm-nitroaniline.
  3. C. Nitrobenzene undergoes Friedel-Crafts methylation with CH3Cl/AlCl3\mathrm{CH_3Cl/AlCl_3} to give mm-nitrotoluene as the major product.
  4. D. Sulfonation of naphthalene with concentrated H2SO4\mathrm{H_2SO_4} at 80 ∘C80\,^\circ\mathrm{C} gives mainly naphthalene-1-sulfonic acid, whereas at 160 ∘C160\,^\circ\mathrm{C} the major product is naphthalene-2-sulfonic acid.
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Answer: A, B, D

(A) True. The NHCOCH3\mathrm{NHCOCH_3} group is a moderate activator and ortho/para director; the bulk of the acetamido group favours the para product.
(B) True. In the strongly acidic mixture aniline is protonated to anilinium, −NH3+\mathrm{{-}NH_3^+}, a meta director, so a large fraction of the meta product forms alongside para (and oxidation products).
(C) False. Nitrobenzene is strongly deactivated and, in addition, its lone pairs complex with AlCl3\mathrm{AlCl_3}; Friedel-Crafts alkylation does not proceed at all.
(D) True. At low temperature, attack at C-1 is faster (kinetic control, more stabilised arenium ion), but the 1-sulfonic acid suffers peri-strain with the C-8 hydrogen and the reaction is reversible; at higher temperature thermodynamic control gives the more stable 2-isomer.
Correct options: (A), (B), (D).
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