Iodoform reaction with alkaline I₂(aq)

Steps:

• Alkaline I₂(aq) tests for alcohols with CH₃CH(OH)- group via oxidation to methyl ketone, then haloform cleavage.
• Propan-2-ol (CH₃CH(OH)CH₃) oxidizes to CH₃COCH₃, which reacts to give CH₃COO⁻ and CHI₃ (noted as CH₃I).
• Primary alcohols like propan-1-ol or butan-1-ol lack this structure, yielding no iodoform products.
• Match products to alcohol: only propan-2-ol fits CH₃COO and CH₃I.

Why C is correct:

• Propan-2-ol follows haloform reaction formula: CH₃CH(OH)CH₃ → CH₃COCH₃ → CH₃COO⁻ + CHI₃.

Why the others are wrong:

• A: Propan-1-ol oxidizes to propanal/propanoate, no methyl ketone for iodoform.
• B: Propan-2-ol products are CH₃COO⁻ (not CH₃CO), as cleavage yields carboxylate.
• D: Butan-1-ol oxidizes to butanal/butanoate, no CH₃C(O)- group for reaction.

Final answer: C