See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Identify the starting material. The starting material is 1-methyl-1H-indene (more precisely, 3-methylindene), which has a benzene ring fused to a cyclopentadiene ring, with a methyl group on the C3 carbon (the exocyclic double bond carbon of the five-membered ring). The double bond in the five-membered ring is between C1 and C3 (or C2-C3 depending on numbering), adjacent to the methyl group. Step 2: Reaction with OsO4/NaHSO3 (dihydroxylation). OsO4 performs a syn-dihydroxylation across the double bond of the five-membered ring. The double bond in the indene system (C1=C2 of the five-membered ring, with the methyl at C3) gets dihydroxylated to give a cis-diol. Product A is the cis-1,2-diol of the five-membered ring of the indene framework (specifically a 1,2-indandiol derivative with methyl substituent). Step 3: Reaction with HIO4 (periodate cleavage). HIO4 cleaves vicinal diols (1,2-diols) oxidatively. The cis-diol in A is cleaved by periodate to give a dialdehyde (or keto-aldehyde if one carbon bears a methyl or other substituent). Since the double bond was between C1 and C2 of the cyclopentene ring (with methyl at C3 on the ring), after dihydroxylation and periodate cleavage, the five-membered ring opens. This cleavage produces a ring-opened dialdehyde — specifically, since the diol carbons are part of the five-membered ring fused to benzene, cleavage gives a 2-formylbenzaldehyde derivative (an ortho-dialdehyde on the benzene ring tether). The product of periodate cleavage is 2-(2-oxopropyl)benzaldehyde or a related ketoaldehyde — considering the methyl group at C3, one end gives an aldehyde and the other gives a methyl ketone (CH3CO-). So the intermediate after HIO4 is an ortho-(2-oxopropyl)benzaldehyde: a benzene ring with CHO at one position and -CH2-CO-CH3 at the adjacent position. Step 4: Reaction with NaOH/heat (intramolecular aldol condensation). The diketone/ketoaldehyde intermediate undergoes intramolecular aldol condensation under basic conditions with heat. The alpha carbon of the methyl ketone attacks the aldehyde intramolecularly, forming a new C-C bond and then dehydrating to give an alpha,beta-unsaturated carbonyl compound. This intramolecular aldol gives a cyclopentenone ring fused back to the benzene ring — specifically 1-indanone or a methylated cyclopentenone fused to benzene. Given the methyl group, the product B is 2-methyl-1H-inden-1-one or more precisely 2-methylindanone/indenone derivative. The intramolecular aldol of 2-(2-oxopropyl)benzaldehyde under NaOH/heat yields 2-methyl-1-indanone after aldol and dehydration, giving an alpha,beta-unsaturated system: 2-methylinden-1(2H)-one, or upon full dehydration, 2-methylinden-1-one (2-methyl-1H-inden-1-one). This is a cyclopentenone ring fused to benzene with a methyl group — consistent with answer choice C. Why other options fail: Other options would differ in ring size, position of methyl, or degree of unsaturation, none of which match the mechanistic outcome of syn-dihydroxylation followed by periodate cleavage and intramolecular aldol condensation of the specific substrate. Therefore, the correct answer is C.