The number of molecules which gives the haloform test among the following molecules is ________.

Solution

Chemistry ma’am · 3 steps

Hunt for a CH₃ sitting right on the C=O, or a CH₃CH(OH) that oxidises into one.

1. What the test needs. The haloform test (I₂ or NaOI with base) is positive for a methyl ketone CH₃CO⁻. It is also positive for a CH₃CH(OH)⁻ alcohol, because the hypohalite first oxidises it to CH₃CO⁻. The methyl must be bonded directly to the carbonyl carbon. A methyl one carbon away does not count.

   Molecule	Group next to C=O (or C–OH)	Haloform test
   2,4-dimethylacetophenone, (CH₃)₂C₆H₃COCH₃	CH₃CO⁻ present	positive
   Ethyl 3-oxopentanoate, CH₃CH₂COCH₂COOC₂H₅	ketone carries C₂H₅ and CH₂, no CH₃	negative
   Acetophenone, C₆H₅COCH₃	CH₃CO⁻ present	positive
   Pentan-3-ol, CH₃CH₂CH(OH)CH₂CH₃	oxidises to C₂H₅COC₂H₅, no CH₃CO⁻	negative
   Butan-2-ol, CH₃CH(OH)CH₂CH₃	has CH₃CH(OH)⁻, oxidises to CH₃COC₂H₅	positive
   2,5,5-trimethylcyclohexane-1,3-dione	both C=O are ring carbons; the methyl sits on C-2, not on a C=O	negative

2. The tricky ones. Butan-2-ol has no carbonyl at all, but it still passes because it is oxidised to butan-2-one in the test mixture. In the cyclic dione, each carbonyl carbon is bonded only to ring carbons, so there is no CH₃CO⁻ unit.
3. So. 2,4-dimethylacetophenone, acetophenone and butan-2-ol give the haloform test, so the answer is 3.

The common slip: 2. Checking only the ketones and skipping butan-2-ol gives 2. The hypohalite oxidises CH₃CH(OH)⁻ to CH₃CO⁻, so it counts.

The idea to keep. Haloform needs CH₃CO⁻ or CH₃CH(OH)⁻, with the methyl bonded directly to the carbon carrying the O. Always check alcohols too, since the reagent oxidises them first.

Answer from the NTA final answer key. Solution by PrepWiser.