Sodium's reactivity with both alcoholic and carboxylic OH groups

Steps:

• Lactic acid, CH₃CH(OH)COOH, contains an alcoholic -OH (secondary) and a carboxylic -OH.
• Sodium metal reduces to Na⁺ and releases H₂ when reacting with acidic hydrogens.
• It deprotonates the alcoholic -OH to form CH₃CH(ONa)COOH + ½H₂.
• It also deprotonates the carboxylic -OH to form CH₃CH(OH)COONa + ½H₂, consuming both.

Why C is correct:

• Sodium reacts with alcohols (ROH + Na → RONa + ½H₂) and carboxylic acids (RCOOH + Na → RCOONa + ½H₂), engaging both -OH types via redox with H.

Why the others are wrong:

• A: Oxidizes the alcoholic -OH to a carbonyl (CH₃COCOOH) but leaves carboxylic -OH unchanged.
• B: Esterifies only the carboxylic -OH (to ethyl lactate) under catalysis, ignoring alcoholic -OH.
• D: Neutralizes only the carboxylic -OH (to CH₃CH(OH)COO⁻ Na⁺), sparing alcoholic -OH.

Final answer: C