2-Methylbut-2-ene, $(\mathrm{CH}{}_3){}_2\mathrm{C}=\mathrm{CH}-\mathrm{CH}{}_3$, is heated with alkaline $\mathrm{KMnO}{}_4$ and then acidified. The organic products are

Hot alkaline permanganate cleaves the double bond. A carbon bearing two alkyl groups ends up as a ketone, and a carbon bearing one alkyl group and one H ends up as a carboxylic acid. $(\mathrm{CH}{}_3){}_2\mathrm{C}=$ gives propanone and $=\mathrm{CH}-\mathrm{CH}{}_3$ gives ethanoic acid.