Given below are two statements:
Statement-I: 2-methylbutane on oxidation with KMnO$$_4$$ gives 2-methylbutan-2-ol.
Statement-II: n-alkanes can be easily oxidised to corresponding alcohol with KMnO$$_4$$.
Choose the correct option:

Statement I says that 2-methylbutane on oxidation with $$\text{KMnO}_4$$ gives 2-methylbutan-2-ol. In 2-methylbutane (isopentane), the carbon at position 2 is a tertiary carbon bonded to three other carbons. Oxidation with $$\text{KMnO}_4$$ preferentially attacks tertiary C$$-$$H bonds because they have the lowest bond dissociation energy. The tertiary hydrogen at C-2 is abstracted, and the resulting radical is hydroxylated, forming 2-methylbutan-2-ol (a tertiary alcohol). This is a known reaction specific to branched alkanes with tertiary carbons. So Statement I is correct.

Statement II says that n-alkanes can be easily oxidised to corresponding alcohols with $$\text{KMnO}_4$$. Straight-chain alkanes have only primary and secondary C$$-$$H bonds, which are much more resistant to oxidation. Under mild conditions, $$\text{KMnO}_4$$ cannot oxidise n-alkanes to alcohols. Under harsh conditions, n-alkanes undergo uncontrolled oxidation leading to a mixture of carboxylic acids and other products rather than cleanly forming alcohols. So Statement II is incorrect.

Therefore, Statement I is correct but Statement II is incorrect.