Consider the skeletal structures of the four hydrocarbons shown...

Alkanes are generally resistant to oxidation by strong oxidizing agents like $\mathrm{KMnO}_4$.

However, alkanes containing a tertiary hydrogen atom can be oxidized to the corresponding tertiary alcohol by $\mathrm{KMnO}_4$.

Let us analyze the given hydrocarbons:
Compound I (n-pentane) contains only primary ($1^\circ$) and secondary ($2^\circ$) hydrogen atoms.
Compound II (2-methylbutane) contains a tertiary ($3^\circ$) hydrogen atom at the second carbon.
Compound III (2,2-dimethylpropane) contains only primary ($1^\circ$) hydrogen atoms.
Compound IV (cyclohexane) contains only secondary ($2^\circ$) hydrogen atoms.

Since only Compound II has a tertiary hydrogen atom, it will be oxidized to 2-methylbutan-2-ol upon treatment with $\mathrm{KMnO}_4$.

Answer: Compound II