Q.Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene?

The key idea is that ozonolysis cleaves each C=C double bond and turns each doubly-bonded carbon into a carbonyl group, so the products reveal where the double bonds were.

For o-xylene (methyls on adjacent carbons C1, C2), a fixed Kekulé structure would give the products of only one arrangement of double bonds. Considering both equivalent Kekulé structures:

• Form I (double bonds 1-2, 3-4, 5-6) → 2 methylglyoxal + 1 glyoxal
• Form II (double bonds 2-3, 4-5, 6-1) → 1 dimethylglyoxal + 2 glyoxal

Since both contribute equally, the combined products are glyoxal : methylglyoxal : dimethylglyoxal = 3 : 2 : 1:

• Glyoxal, $\ce{OHC-CHO}$
• Methylglyoxal, $\ce{CH3CO-CHO}$
• Dimethylglyoxal, $\ce{CH3CO-COCH3}$ …

Ozonolysis cleaves each carbon–carbon double bond and caps each of the two carbons with a carbonyl (C=O) group. If benzene really had three fixed, alternating double bonds, ozonolysis of o-xylene would reveal exactly where they are. The surprising result is one of the classic pieces of evidence that benzene's bonds are not localised.

What a single Kekulé structure predicts

Number the ring carbons 1–6, with the two methyl groups on C1 and C2 (ortho). A Kekulé structure has three fixed double bonds and three fixed single bonds. When ozone cleaves the double bonds, the ring falls apart into three fragments — each held together by one of the surviving single bonds, with a carbonyl at each end.

Kekulé form I (double bonds at C1=C2, C3=C4, C5=C6; single bonds at C2–C3, C4–C5, C6–C1). The fragments are the pairs still joined by a single bond:

• C2–C3 → one methyl, one H → methylglyoxal ($\ce{CH3CO-CHO}$)
• C4–C5 → both H → glyoxal ($\ce{OHC-CHO}$)
• C6–C1 → one H, one methyl → methylglyoxal

So form I gives 2 methylglyoxal + 1 glyoxal.

Kekulé form II (double bonds at C2=C3, C4=C5, C6=C1; single bonds at C1–C2, C3–C4, C5–C6). The fragments are:

• C1–C2 → both methyl → dimethylglyoxal ($\ce{CH3CO-COCH3}$)
• C3–C4 → both H → glyoxal
• C5–C6 → both H → glyoxal

So form II gives 1 dimethylglyoxal + 2 glyoxal.

Combining the two forms

A fixed Kekulé structure would give the products of only one form. But the real molecule is a resonance hybrid in which both patterns of overlap are equally probable, so both sets of fragments form with equal weight. Adding them:

• Glyoxal: 1 (form I) + 2 (form II) = 3
• Methylglyoxal: 2 (form I) = 2
• Dimethylglyoxal: 1 (form II) = 1

giving glyoxal : methylglyoxal : dimethylglyoxal = 3 : 2 : 1. …