Animated Solution for Chemistry - Organic Chemistry: Given below are two statements. Statement-I 2-methylbutane on oxidation with ${\text{KMnO}}_4$ gives 2-methylbutan-2-ol. Statement-II $n$-alkanes can be easily oxidised to corresponding alcohols with ${\text{KMnO}}_4$. Choose the correct option.

Alkanes are notoriously known as the "paraffins" of organic chemistry, a term derived from Latin meaning "little affinity." This is because their structure consists entirely of strong, non-polar $C-C$ and $C-H$ single bonds. Without any $\pi$-electrons or highly electronegative atoms to act as reactive sites, alkanes generally turn a blind eye to acids, bases, and even strong oxidizing agents like ${\text{KMnO}}_4$ under normal conditions.

When we look at $n$-alkanes (straight-chain alkanes), they are composed exclusively of primary ($1^{\circ }$) and secondary ($2^{\circ }$) carbon atoms. The $C-H$ bonds on these carbons are incredibly robust. Therefore, if you try to treat an $n$-alkane with ${\text{KMnO}}_4$, absolutely nothing happens. The solution remains purple, and the alkane remains untouched. This directly proves that Statement II is incorrect.

However, chemistry is a science of fascinating exceptions. While $1^{\circ }$ and $2^{\circ }$ $C-H$ bonds are stubborn, the tertiary ($3^{\circ }$) $C-H$ bond is the Achilles' heel of branched alkanes.

A tertiary carbon is bonded to three other electron-donating alkyl groups. These groups provide significant inductive stabilization to any intermediate (like a radical) that forms during a reaction. Because of this stabilization, the $3^{\circ }$ $C-H$ bond is noticeably weaker and more reactive than its primary and secondary counterparts.

Let's bring 2-methylbutane onto the stage. If you draw its structure, you will immediately spot a tertiary carbon at the second position, proudly holding onto a single hydrogen atom.

When 2-methylbutane is subjected to oxidation with ${\text{KMnO}}_4$, the strong oxidizing agent specifically targets this vulnerable $3^{\circ }$ $C-H$ bond. The oxygen atom inserts itself right between the carbon and the hydrogen, transforming the alkane into a tertiary alcohol. The product formed is 2-methylbutan-2-ol.

This elegant transformation perfectly aligns with the claim made in the first statement. Thus, Statement I is absolutely correct.

By combining our observations, we can confidently conclude that Statement I holds true due to the unique reactivity of the tertiary hydrogen, while Statement II falls flat because $n$-alkanes lack this reactive site. The correct choice is option (c).