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Explain why acetone (CH3COCH3) gives a positive iodoform test when treated with aqueous iodine and sodium hydroxide. Write the step‑wise mechanism, including all intermediates, and state the final product formed.

hard 5 marks Class 12 · Chemistry · Aldehydes Ketones and Carboxylic Acids
Answer
Acetone yields yellow precipitate of iodoform (CHI₃) and acetate ion (CH₃COO⁻).

The correct answer is Acetone yields yellow precipitate of iodoform (CHI₃) and acetate ion (CH₃COO⁻). — Conclusion: Because acetone possesses the –COCH₃ group (methyl ketone), it undergoes complete iodination and cleavage, giving iodoform as the characteristic yellow solid.

Step-by-step solution

  1. Step 1 – Enolate formation: OH⁻ abstracts an α‑hydrogen from acetone, producing the enolate ion CH₂=C(O⁻)CH₃. This is possible because the α‑hydrogen is acidic.
  2. Step 2 – Iodination: The enolate attacks I₂, giving mono‑iodinated acetone CH₃COCH₂I and regenerating I⁻. The process repeats three times, each time deprotonating the remaining α‑hydrogens, leading to the triiodinated intermediate CH₃COCI₃.
  3. Step 3 – Further deprotonation: OH⁻ removes the remaining α‑hydrogen from CH₃COCI₃, forming the carbanion ⁻C(CI₃)=OCH₃.
  4. Step 4 – Nucleophilic acyl substitution: OH⁻ attacks the carbonyl carbon of the triiodinated intermediate, causing cleavage of the C–C bond between the carbonyl carbon and the C(I)₃ group. This yields iodoform (CHI₃) as a yellow precipitate and acetate ion (CH₃COO⁻).
  5. Step 5 – Overall reaction: 3 I₂ + CH₃COCH₃ + 4 OH⁻ → CHI₃↓ + CH₃COO⁻ + 3 I⁻ + 3 H₂O.
  6. Conclusion: Because acetone possesses the –COCH₃ group (methyl ketone), it undergoes complete iodination and cleavage, giving iodoform as the characteristic yellow solid.
Common mistake: Thinking that only aldehydes give iodoform; the test is specific for methyl ketones and compounds that can be oxidised to them.

Concept tested: Haloform (iodoform) reaction of methyl ketones

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