Question: Which of the following alcohols will not undergo oxidation? (a) Butanol (b) Butan-2-ol (c) 2-Methylbutan-2-ol (d) 3-Methylbutan-2-ol

Step 1 (Governing Principle): [0.5 Mark]
The oxidation of alcohols depends on the presence of at least one $\alpha$-hydrogen atom attached to the carbon bearing the hydroxyl ($-OH$) group.

• Primary alcohols ($R-CH_2OH$) have two $\alpha$-hydrogens.
• Secondary alcohols ($R_2CHOH$) have one $\alpha$-hydrogen.
• Tertiary alcohols ($R_3C-OH$) have zero $\alpha$-hydrogens.

Since tertiary alcohols lack an $\alpha$-hydrogen, they do not undergo oxidation under normal conditions.

Step 2 (Structural Analysis): [0.25 Mark]
Let us examine the structure of 2-Methylbutan-2-ol: $$CH_3 - CH_2 - \underset{\overset{\displaystyle |}{CH_3}}{\overset{\displaystyle |}{C}} - OH$$ The carbon atom bonded to the $-OH$ group is attached to three other carbon atoms (two methyl groups and one ethyl group). Therefore, it is a tertiary ($3^\circ$) alcohol. It has no hydrogen atom attached to the $\alpha$-carbon.

Step 3 (Final Concluded Answer): [0.25 Mark]
Comparing the options:

• (a) Butanol: Primary ($1^\circ$) - Oxidizes to butanal.
• (b) Butan-2-ol: Secondary ($2^\circ$) - Oxidizes to butanone.
• (c) 2-Methylbutan-2-ol: Tertiary ($3^\circ$) - No oxidation.
• (d) 3-Methylbutan-2-ol: Secondary ($2^\circ$) - Oxidizes to 3-methylbutan-2-one.

Correct Option: (c) 2-Methylbutan-2-ol

💡 EL10 Concept Insight: Think of the $\alpha$-hydrogen as a "handle" that an oxidizing agent needs to grab onto to pull the molecule apart; because a tertiary alcohol is completely surrounded by carbon "walls" and has no hydrogen handle, the oxidizing agent simply can't react with it.