A hydrocarbon A (C7H12) decolourises bromine water. On ozonolysis followed by Zn/H2O it gives a single compound B (C7H12O2) which gives a positive…

A has two degrees of unsaturation, and since ozonolysis gives one product containing both carbonyls, A is a cyclic alkene: 1,2-dimethylcyclopentene. Its reductive ozonolysis opens the ring to heptane-2,6-dione, $\mathrm{CH_3COCH_2CH_2CH_2COCH_3}$ (B), which has the $\mathrm{CH_3CO-}$ group needed for the iodoform test. In base, the terminal $\mathrm{CH_3}$ (C-1) enolate attacks the other carbonyl (C-6), closing a six-membered ring (C1 to C6); dehydration gives 3-methylcyclohex-2-en-1-one, $\mathrm{C_7H_{10}O}$. Attack from C-3 would need a strained four-membered ring.