See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: Curtius Rearrangement followed by intramolecular cyclization. When an acyl azide is heated, it undergoes the Curtius rearrangement: the acyl azide (R-C(=O)-N3) loses N2 to generate an isocyanate intermediate (R-N=C=O). This is a well-known thermal rearrangement. Step 2 - Identify the starting material: The starting material has a benzene ring fused to a five-membered carbocyclic ring (indane framework). On the five-membered ring, there are two substituents: an NH2 group and an acyl azide group (C(=O)N3). These two groups are on adjacent or nearby carbons of the five-membered ring. Step 3 - Curtius Rearrangement: Upon heating in toluene, the acyl azide loses N2 to form an isocyanate (R-N=C=O) in situ on the ring carbon adjacent to the NH2 group. Step 4 - Intramolecular cyclization: The isocyanate intermediate is electrophilic at the carbon of N=C=O. The adjacent NH2 group acts as an intramolecular nucleophile and attacks the isocyanate carbon, forming a new N-C bond and closing the ring. This generates a five- or six-membered heterocyclic ring containing two nitrogen atoms and one carbonyl group (urea-type linkage: -NH-C(=O)-NH- bridging into a ring). Step 5 - Product identification: The intramolecular reaction of the isocyanate with the NH2 group on the same five-membered ring gives a fused bicyclic heterocyclic product. The new ring formed contains -NH-C(=O)-N< connectivity. Looking at option (a): it shows a benzene ring fused to a five-membered carbocyclic ring which is itself fused to a six-membered ring containing two N-H groups and one C=O, consistent with the formation of a cyclic urea (dihydroimidazolone or similar) fused onto the indane skeleton. This matches the product of Curtius rearrangement followed by intramolecular urea ring formation. Step 6 - Why other options fail: - Option (b): Shows a structure with an exo N-H and ring C=O but does not correspond to the cyclic urea product of Curtius rearrangement. - Option (c): Shows an indanone with NH2, which would be the starting material fragment, not the cyclized product. - Option (d): Shows an NH-NH2 group and a carboxyl, which does not correspond to Curtius rearrangement chemistry. The Curtius rearrangement of the acyl azide gives an isocyanate, which cyclizes with the nearby NH2 to give the cyclic urea heterocyclic product shown in option (a), a benzene-fused bicyclic compound with two N-H groups and a C=O in the newly formed ring. Therefore, the correct answer is A.