See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Ozonolysis of aromatic compounds (von Auwers ozonolysis) cleaves all three double bonds of the benzene ring, producing carbonyl fragments from each double bond. Step 1: Identify the structure. The starting material is o-xylene (1,2-dimethylbenzene), which has a benzene ring with methyl groups on adjacent carbons (C1 and C2). Step 2: Number the ring carbons and label substituents. C1 has CH3, C2 has CH3, and C3 through C6 are unsubstituted. Step 3: Ozonolysis cleaves each C=C double bond of the Kekule structure. The three double bonds in one Kekule resonance form of o-xylene are C1=C2, C3=C4, and C5=C6. Step 4: Determine fragments from each cleavage: - Cleavage of C1=C2 (both carbons bear CH3): gives CH3-C(=O)- from C1 and -C(=O)-CH3 from C2, which recombine to give CH3-C(=O)-C(=O)-CH3 = 2,3-butanedione (1 molecule). - Cleavage of C3=C4 (both unsubstituted, connected to C2 and C5 respectively): gives H-C(=O)- fragments; these are the terminal aldehyde carbons. C3 is connected to C2 (which gives -C(=O)-CH3) and C4 is connected to C5. So C3 fragment: H-C(=O)- and C4 fragment: H-C(=O)-. These join with adjacent fragments. - Cleavage of C5=C6 (both unsubstituted, connected to C4 and C1 respectively): C5 fragment: H-C(=O)- and C6 fragment: H-C(=O)-. Step 5: Assemble the linear chain of fragments from ozonolysis of the ring opened as a chain: C1(CH3)-C2(CH3)-C3(H)-C4(H)-C5(H)-C6(H)-back to C1. Opening gives the dialdehyde chain: OHC-C(CH3)(=O) | C(CH3)(=O)-CHO | CHO-CHO. More carefully: the six carbons in sequence give three dicarbonyl products: - C1-C2 bond cleavage (both methylated): CH3CO-COCH3 = 2,3-butanedione (1 mol) - C3-C4 bond cleavage (both H): OHCCHO = glyoxal (1 mol) - C5-C6 bond cleavage (H and CH3 adjacent): but we must pair C6 with C1(CH3) and C3 with C2... Step 6: Re-approach systematically. The ring opens into three 2-carbon units. The pairs are: {C1,C2} both have CH3 -> 2,3-butanedione; {C3,C4} both have H -> glyoxal; {C5,C6} one is adjacent to C1(CH3) giving C6 as CHO end, C5 as CHO end, but C5 is adjacent to C4(H) and C6 is adjacent to C1(CH3). Actually each carbon becomes one carbonyl. The three products come from pairing adjacent carbons across each cleaved double bond: C1(CH3)+C6(H) -> CH3CO-CHO = methylglyoxal (pyrualdehyde); C2(CH3)+C3(H) -> CH3CO-CHO = methylglyoxal (pyrualdehyde); C4(H)+C5(H) -> OHCCHO = glyoxal. Step 7: Count: Glyoxal = 1 molecule (from C4-C5 cleavage), Pyrualdehyde = 2 molecules (from C1-C6 and C2-C3 cleavages), 2,3-Butanedione = ... wait, this doesn't give butanedione. Step 8: The correct Kekule pairing for o-xylene ozonolysis: double bonds are C1=C6, C2=C3, C4=C5 (alternate). Cleavage: C1(CH3)-C6(H) -> methylglyoxal x1; C2(CH3)-C3(H) -> methylglyoxal x1; C4(H)-C5(H) -> glyoxal x1. Other Kekule: C1=C2, C3=C4, C5=C6: C1(CH3)-C2(CH3)->butanedione x1; C3(H)-C4(H)->glyoxal x1; C5(H)-C6(H)->glyoxal x1. Average of both Kekule structures: glyoxal = (1+2)/2 = 1.5, pyrualdehyde = (2+0)/2 = 1, butanedione = (0+1)/2 = 0.5. Ratio glyoxal:pyrualdehyde = 1.5:1 = 3:2. Step 9: This matches option (c) 3:2. Why other options fail: 1:3 and 3:1 don't reflect the averaged fragment count; 2:3 inverts the correct ratio. Therefore, the correct answer is C.