See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: When o-phenylenediamine reacts with NaNO2 and HCl, one of the two amino groups undergoes diazotization to form a diazonium salt. The diazonium group (N2+) then couples with the adjacent free amino group (–NH2) on the same molecule in an intramolecular reaction. Step-by-step reasoning: 1. o-Phenylenediamine has two –NH2 groups on adjacent carbons of a benzene ring. 2. NaNO2 + HCl generates HNO2 (nitrous acid) in situ. 3. One –NH2 group is diazotized by HNO2 to give an aryl diazonium salt (–N2+Cl–) at one position. 4. The diazonium nitrogen (electrophilic N) couples with the neighboring –NH2 nitrogen (nucleophilic N) on the same molecule. 5. This forms an N–N bond intramolecularly, closing the five-membered triazole ring fused to the benzene ring, yielding 1H-benzotriazole. 6. The bond formed is N–N (between the diazonium nitrogen and the amino nitrogen), not C–N. 7. Since both reacting groups are on the same molecule, the coupling is intramolecular. Why other options fail: - (a) Intermolecular C–N coupling: incorrect because the bond formed is N–N, not C–N, and it is intramolecular. - (b) Intramolecular C–N coupling: incorrect because no new C–N bond is formed; the ring closure involves N–N bond formation. - (c) Intermolecular N–N coupling: incorrect because both nitrogen atoms involved belong to the same o-phenylenediamine molecule, making it intramolecular. Therefore, the correct answer is D.