JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Hydrocarbons

Reductive ozonolysis cleaves every carbon-carbon double bond and caps the two freed carbons with carbonyl groups, so the fragments reveal exactly where the double bonds sat in the parent ring. For A the two products are ethane-1,2-dicarbaldehyde, $\mathrm{OHC}-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{CHO}$, and glyoxal, $\mathrm{OHC}-\mathrm{CHO}$; rejoining these carbonyl carbons back into a six-membered ring places the double bonds at the 1- and 3-positions, which identifies A as cyclohexa-1,3-diene, since cleaving its $C1=C2$ and $C3=C4$ bonds detaches the two-carbon $C2$-$C3$ piece as glyoxal and leaves the $C4$-$C5$-$C6$-$C1$ piece as the four-carbon dialdehyde. For B a single product, 5-oxohexanal $\mathrm{C}{\mathrm{H}}_3-\mathrm{CO}-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{CHO}$, signals one ring double bond carrying a methyl group: the methyl-bearing carbon becomes the ketone and its partner becomes the aldehyde, which is exactly what 1-methylcyclopent-1-ene gives on cleavage. Hence A is cyclohexa-1,3-diene and B is 1-methylcyclopent-1-ene.