Analyze Statement I

Statement I says:

$2$-methylbutane on oxidation with $KMnO_4$ gives $2$-methylbutan-$2$-ol.

$2$-Methylbutane is an alkane: $\mathrm{CH_3-CH(CH_3)-CH_2-CH_3}$

Conversion of an alkane directly into an alcohol is not the usual oxidation reaction with $KMnO_4$. However, in standard hydrocarbon chemistry, branched alkanes having a tertiary hydrogen are considered susceptible to oxidation at that carbon, and the product is represented as the corresponding tertiary alcohol.

In $2$-methylbutane, carbon-2 is a tertiary carbon bearing one hydrogen. Oxidation at this position gives: $\mathrm{CH_3-C(OH)(CH_3)-CH_2-CH_3}$ which is $2$-methylbutan-$2$-ol.

So, Statement I is correct.

Analyze Statement II

Statement II says:

$n$-alkanes can be easily oxidised to corresponding alcohols with $KMnO_4$.

This is incorrect. Alkanes are generally quite inert toward oxidation because they contain only strong $\mathrm{C-C}$ and $\mathrm{C-H}$ sigma bonds. They do not get easily oxidised by $KMnO_4$ to give alcohols.

In fact:

• alkanes resist mild oxidising agents,
• strong oxidation generally leads to cleavage or combustion-type products,
• direct selective conversion of an $n$-alkane into the corresponding alcohol with $KMnO_4$ is not a standard easy reaction.

Hence, Statement II is incorrect.

Choose the correct option

• Statement I: Correct
• Statement II: Incorrect

Therefore, the correct option is: $\boxed{\text{C}}$

Comparison with stored answer

Stored correct answer: $\boxed{\text{C}}$

My derived answer matches the stored answer.