An alkene X ($\mathrm{C}{}_5\mathrm{H}{}_{10}$) is oxidised by hot, acidified $\mathrm{KMnO}{}_4$ to a mixture of propanone and ethanoic acid. Addition of $\mathrm{HBr}$ (no peroxide) to X gives mainly

Propanone and ethanoic acid come from $(\mathrm{CH}{}_3){}_2\mathrm{C}=$ and $=\mathrm{CH}-\mathrm{CH}{}_3$ ends, so X is $(\mathrm{CH}{}_3){}_2\mathrm{C}=\mathrm{CH}-\mathrm{CH}{}_3$ (2-methylbut-2-ene). Markovnikov addition places H on the carbon with more hydrogens, and Br ends on the tertiary carbon through the more stable tertiary cation: 2-bromo-2-methylbutane.