Which among the following pairs of the structures will give different products on ozonolysis? (Consider the double bonds in the structures are rigid and not delocalized.)

Solution

Chemistry ma’am · 4 steps

Ozonolysis cuts every C=C, so each single bond left in the ring becomes one two-carbon dicarbonyl piece.

1. Cut every C=C. Ozone breaks each C=C and puts a C=O on both carbons. In a rigid Kekulé ring the three single bonds are not touched, so the ring falls into three O=C-C=O pieces, one for each single bond.
2. Name the pieces. A single bond joining two ring carbons that carry H gives glyoxal, OHC-CHO. If one of the two carbons carries a methyl, you get methylglyoxal, CH₃COCHO. If both do, you get dimethylglyoxal, CH₃COCOCH₃.

   Pair	Left structure gives	Right structure gives	Same?
   (a) m-xylene	2 CH₃COCHO + OHC-CHO	2 CH₃COCHO + OHC-CHO	same
   (b) p-xylene	2 CH₃COCHO + OHC-CHO	2 CH₃COCHO + OHC-CHO	same
   (c) o-xylene	2 CH₃COCHO + OHC-CHO	CH₃COCOCH₃ + 2 OHC-CHO	different
   (d) toluene	CH₃COCHO + 2 OHC-CHO	CH₃COCHO + 2 OHC-CHO	same

3. Why o-xylene differs. Only in o-xylene are the two methyl-bearing carbons neighbours. In the right structure the bond between them is single, so both methyls stay together in one CH₃COCOCH₃. In the left structure that bond is double, so it is cut and the methyls go into two separate CH₃COCHO pieces.
4. So. Only pair (c) gives different products on ozonolysis.

The idea to keep. When you ozonolyse a Kekulé ring, look at the single bonds: each one becomes a dicarbonyl. Two Kekulé forms give different products only when substituents sit on neighbouring carbons.

Answer from the NTA final answer key. Solution by PrepWiser.