Primary alcohol identification via oxidation product reacting with sodium Steps:

• Reaction with sodium liberating H₂ indicates an -OH group in T, typical of alcohols.
• Hot acidified KMnO₄ oxidizes alcohols: primary to carboxylic acids, secondary to ketones, tertiary/ketones unchanged.
• The query implies the oxidation product reacts with sodium to give H₂, which carboxylic acids do (2RCOOH + 2Na → 2RCOONa + H₂) but ketones do not.
• Thus, T is a primary alcohol, matching option A (CH₃CH₂CH₂OH oxidized to CH₃CH₂COOH).

Why A is correct:

• CH₃CH₂CH₂OH (propan-1-ol) is primary; oxidation gives propanoic acid, which reacts with Na per the reaction 2CH₃CH₂COOH + 2Na → 2CH₃CH₂COONa + H₂.

Why the others are wrong:

• B: CH₃CH(OH)CH₃ is secondary; oxidation gives CH₃COCH₃ (ketone), which does not liberate H₂ with Na.
• C: (CH₃)₂CHOH is secondary; same as B, oxidation product (ketone) does not react with Na to give H₂.
• D: CH₃COCH₃ is a ketone; not oxidized by KMnO₄.

Final answer: A