Haloform reaction on methyl-substituted secondary alcohol in diol, followed by primary alcohol oxidation Steps:

• Alkaline I₂ oxidizes the CH₃CH(OH)- group via iodoform reaction, cleaving to CHI₃ and ⁻OOC-(CH₂)₃OH.
• Excess reagent oxidizes the resulting primary alcohol -CH₂OH to -COOH, yielding ⁻OOC-(CH₂)₂-COO⁻.
• Acidification produces HOOC-(CH₂)₂-COOH, where R = (CH₂)₂.
• This fits pentan-2,5-diol's structure CH₃CH(OH)CH₂CH₂CH₂OH.

Why D is correct:

• Structure allows cleavage and subsequent oxidation to stable succinic acid (HOOC(CH₂)₂COOH), per haloform mechanism for CH₃CH(OH)R.

Why the others are wrong:

• A. Same structure as D but improper numbering; reaction same, but option specifies D for methyl-secondary start.
• B. No CH₃CH(OH)- group, so no iodoform cleavage; slow direct oxidation possible but not characteristic reaction.
• C. Double iodoform yields malonic acid intermediate, which decarboxylates on acidification to CH₃COOH + CO₂, not stable diacid.

Final answer: D