See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Diaryliodonium salts undergo nucleophilic substitution (SNAr-type or ligand-coupling mechanism). The nucleophile attacks the ipso carbon of one of the aryl rings attached to the positively charged iodine, displacing the other aryl group as an iodoarene leaving group. Step 1 – Nature of the reagent: The compound is a diaryliodonium salt [(4-NO2-C6H4)(Ph)I]+. The I+ center is hypervalent and electrophilic at the two ipso carbons (positions iii and iv) directly bonded to iodine. Step 2 – Mechanism of nucleophilic attack: ArO- (a nucleophile/aryloxide) attacks one of the two ipso carbons attached to I+. Positions i and ii are not ipso carbons of the iodonium and are not electrophilic in this context; attack there is not the operative mechanism. Step 3 – Regioselectivity (ortho effect): In diaryliodonium chemistry, the nucleophile preferentially attacks the more electron-deficient aryl ring. The ring bearing the -NO2 group is more electron-deficient (electron-withdrawing group). The ipso carbon of the nitrophenyl ring (position iii, the carbon directly bonded to I+) is therefore more electrophilic than the ipso carbon of the plain phenyl ring (position iv). Additionally, the well-known 'ortho effect' in diaryliodonium salts states that nucleophiles attack the aryl group that has an ortho substituent preferentially. Here, the NO2 group on the nitrophenyl ring acts as an activating substituent (electron withdrawal making ipso more electrophilic), reinforcing attack at position iii. Step 4 – Why other positions fail: - Position i (para to NO2 on nitrophenyl ring): Not bonded to I+; electrophilic substitution mechanism doesn't apply here. - Position ii (ortho to I on nitrophenyl ring): Not the ipso carbon; not the site of nucleophilic attack in iodonium chemistry. - Position iv (ipso of plain phenyl ring): Less electrophilic than position iii because the phenyl ring lacks electron-withdrawing substituents. Step 5 – Conclusion: ArO- attacks position iii, the ipso carbon of the 4-nitrophenyl group attached to I+, because it is the most electrophilic site due to the electron-withdrawing NO2 group. Therefore, the correct answer is C.