The compound resistant to oxidation is

The correct answer is Option (2) → t-Butanol

Note: The given answer is as per the official NTA Answer Key. However, there is more than one correct option for this question, as explained below.

Oxidation of an alcohol removes a hydrogen from the –OH group and a hydrogen from the carbon carrying it, so that a C=O bond forms. How far an alcohol can be oxidised therefore depends on how many hydrogen atoms that carbon has:

• Primary alcohol, \( \ce{R-CH2OH} \): two such hydrogens; oxidised to an aldehyde and then to a carboxylic acid.

• Secondary alcohol, \( \ce{R2CHOH} \): one such hydrogen; oxidised to a ketone.

• Tertiary alcohol, \( \ce{R3COH} \): no hydrogen on the carbinol carbon, so no C=O can form without breaking C–C bonds.

• Option (1) Carbonic acid — Is also correct. Carbonic acid, \( \ce{H2CO3} \), is not an alcohol; it is the unstable acid formed when carbon dioxide dissolves in water. Its carbon is already in the +4 state, the highest oxidation state of carbon, so it cannot be oxidised further.

• Option (2) t-Butanol — Correct. t-Butanol (2-methylpropan-2-ol), \( \ce{(CH3)3C-OH} \), is a tertiary alcohol, so it resists oxidation. Only strong oxidising agents such as hot \( \ce{KMnO4} \) attack it, and then the carbon chain breaks into smaller acids.

• Option (3) Ethyl alcohol — Incorrect. A primary alcohol; it is oxidised to ethanal and then to ethanoic acid.

      \( \displaystyle \ce{CH3CH2OH} \xrightarrow{[\ce{O}]} \ce{CH3CHO} \xrightarrow{[\ce{O}]} \ce{CH3COOH} \)

• Option (4) Methyl alcohol — Incorrect. Also a primary alcohol; it is oxidised to methanal and then to methanoic acid.

      \( \displaystyle \ce{CH3OH} \xrightarrow{[\ce{O}]} \ce{HCHO} \xrightarrow{[\ce{O}]} \ce{HCOOH} \)

NCERT (2026-27 edition):
Chemical Reactions – …