(N/A) The reaction of $2$-methylpropane with $KMnO_{4}$ is an oxidation reaction.
The chemical equation is:
$(CH_{3})_{3}CH \xrightarrow{KMnO_{4}} (CH_{3})_{3}COH$
In this reaction,the tertiary hydrogen atom of $2$-methylpropane is replaced by a hydroxyl group $(-OH)$.
This is classified as an oxidation reaction because the oxidation state of the tertiary carbon atom increases. In the reactant $(CH_{3})_{3}CH$,the tertiary carbon is bonded to one hydrogen atom,while in the product $(CH_{3})_{3}COH$,it is bonded to an oxygen atom,which is more electronegative,leading to an increase in the oxidation state of the carbon atom.