Haloform reaction on symmetric 1,3-diol yielding malonic acid Steps:

• Pentan-2,4-diol oxidizes under alkaline I₂ to 2,4-pentanedione (CH₃COCH₂COCH₃), a bis-methyl ketone.
• Each methyl ketone undergoes haloform reaction, cleaving to CHI₃ and a carboxylate.
• First cleavage yields acetoacetic acid salt (CH₃COCH₂COONa).
• Second cleavage of the remaining methyl ketone gives malonate (⁻OOCCH₂COO⁻).
• Acidification protonates to HOOC-CH₂-COOH (malonic acid), where R = CH₂.

Why C is correct:

• Pentan-2,4-diol has two CH₃CH(OH)- groups separated by CH₂, enabling double haloform cleavage to malonic acid (HOOC-CH₂-COOH) per haloform reaction mechanism.

Why the others are wrong:

• A: Primary-secondary diol; haloform at secondary end gives 4-hydroxybutanoic acid, a monocarboxylic acid.
• B: Both primary alcohols; no methyl carbinol groups, so no haloform reaction or diacid.
• D: Vicinal diol; haloform only at one CH₃CH(OH)- end, yielding 2-hydroxybutanoic acid, a monocarboxylic acid.

Final answer: C