An alkene A ($\mathrm{C}{}_6\mathrm{H}{}_{12}$) gives only one carbonyl compound B on reductive ozonolysis. B gives a yellow precipitate with $\mathrm{I}{}_2$/$\mathrm{NaOH}$. A is

One carbonyl product means the alkene is symmetrical. Iodoform positive means B has a $\mathrm{CH}{}_3\mathrm{CO}$ unit, so B is $(\mathrm{CH}{}_3){}_2\mathrm{C}=\mathrm{O}$ and A is $(\mathrm{CH}{}_3){}_2\mathrm{C}=\mathrm{C}(\mathrm{CH}{}_3){}_2$. Hex-3-ene gives propanal (no iodoform), 2-methylpent-2-ene gives two carbonyls, and cyclohexene gives a dialdehyde.