An alkene of formula $\mathrm{C}{}_4\mathrm{H}{}_8$ is treated with cold, dilute $\mathrm{KMnO}{}_4$ and the resulting diol is cleaved by $\mathrm{HIO}{}_4$ to give only methanal and propanone. The alkene is

The cleavage products reveal the diol, $\mathrm{HOCH}{}_2-\mathrm{C}(\mathrm{OH})(\mathrm{CH}{}_3){}_2$, with a $\mathrm{CH}{}_2\mathrm{OH}$ end giving $\mathrm{HCHO}$ and a $\mathrm{C}(\mathrm{OH})(\mathrm{CH}{}_3){}_2$ end giving propanone. The diol comes from syn dihydroxylation of $\mathrm{CH}{}_2=\mathrm{C}(\mathrm{CH}{}_3){}_2$, 2-methylpropene. But-1-ene would give methanal and propanal, and but-2-ene would give ethanal only.