Which of the following organic compounds will yield a bright yellow crystalline precipitate of iodoform (CHI3) when warmed with iodine and aqueous sodium hydroxide (the haloform test)?

Answer and explanation

Correct answer: D. Acetophenone (C6H5COCH3) and 2-Propanol (CH3CH(OH)CH3)

  1. Step 1: The haloform (iodoform) reaction is specific to compounds containing either a methyl carbonyl group () or a methyl carbinol group () that can be oxidized in situ by hypoiodite () to a methyl ketone.
  2. Step 2: Acetophenone () contains a methyl ketone group, which undergoes multiple iodination to , cleaved by into and yellow .
  3. Step 3: 2-Propanol () is oxidized by to acetone, which then yields iodoform.
  4. Step 4: Neither benzophenone, 3-pentanone, 1-propanol, methanol, nor benzaldehyde possess the required or structural moiety.
Answer
Acetophenone (C6H5COCH3) and 2-Propanol (CH3CH(OH)CH3)
Why not the others: 'Benzophenone and Methanol' lack methyl carbonyl/carbinol groups; '3-Pentanone and 1-Propanol' are ethyl ketone and primary alcohol respectively, failing the test; 'Benzaldehyde and Formic acid' do not possess methyl groups.

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