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Haloform Reaction and the Iodoform Test: Mechanism, Which Compounds Pass and Exam Traps

ChemistryClass 12, Aldehydes, Ketones and Carboxylic AcidsFor JEE and NEET aspirants

The haloform reaction turns a methyl ketone into a carboxylate with one carbon fewer, and the methyl group into a haloform, . The reagent is sodium hypohalite, NaOX, made from the halogen and sodium hydroxide. With iodine it gives iodoform, , a yellow solid with a strong smell, and that is the iodoform test for a group or a group. The reaction is in the Class 12 NCERT chapter on aldehydes, ketones and carboxylic acids. JEE Main and NEET most often ask which compounds give the iodoform test.

At a glance
  1. General:
  2. Iodoform test, acetone:
  3. Ethanol also gives it:
  4. Needs: a group whose C=O carbon is also bonded to H or C (not OH, OR or ), or a group, which NaOX first oxidises to .
  5. Products: the haloform, (X = Cl, Br, I), and , with one carbon fewer.
  6. Left alone: a C=C elsewhere in the molecule.

1. Mechanism

The reaction has two parts. First the base halogenates the methyl group three times through the enolate. Then hydroxide attacks the C=O, and the group leaves as an anion, which is what makes this a cleavage reaction. In the alkali the halogen and the hypohalite are in equilibrium, and the enolate takes X from (or from HOX):

1.1 Three halogens on the methyl group

Haloform reaction, part 1: the methyl group is halogenated three times A methyl ketone, R-CO-CH3, is drawn with one hydrogen of the methyl group shown. Hydroxide removes that hydrogen; the carbon-hydrogen electrons form a carbon-carbon double bond and the carbon-oxygen pi electrons move onto oxygen, giving the enolate. The enolate's negative oxygen reforms the carbon-oxygen double bond, the carbon-carbon pi electrons bond the carbon to one halogen atom of X2, and the halogen-halogen electrons leave with the other halogen as halide. This gives R-CO-CH2X. The same two steps happen twice more, faster each time, giving R-CO-CX3. Hydroxide removes an α-H: the enolate R C O CH2 H OH The enolate takes X from X2 R C O CH2 X X R–CO–CH2X + X− twice more: R–CO–CHX2, then R–CO–CX3 each X makes the next H easier to remove
Figure 1: In base the methyl group of a methyl ketone is halogenated three times. Each halogen pulls electrons from the carbon, so the remaining H is lost faster; the reaction does not stop at one X.

1.2 The C–C bond breaks

Haloform reaction, part 2: the C-CX3 bond breaks Hydroxide adds to the carbonyl carbon of R-CO-CX3 and the carbon-oxygen pi electrons move onto oxygen. In the intermediate, the negative oxygen's lone pair reforms the carbon-oxygen double bond and the bond to the CX3 group breaks, its electrons leaving with the CX3 group as a trihalomethyl anion. That anion takes a proton from the carboxylic acid, giving the haloform CHX3 and the carboxylate ion. With iodine the haloform is iodoform, a yellow solid. Hydroxide adds to the C=O R C O CX3 OH C=O reforms and the CX3 anion leaves R C O OH CX3 R–COOH + −CX3 → R–COO− + CHX3 haloform: CHI3 is a yellow solid
Figure 2: Three halogens make stable enough to leave, so the C–C bond breaks. The carboxylate has one carbon fewer than the ketone; the lost carbon ends up in the haloform.
  1. Enolate. removes an H from the next to the C=O; the negative charge is shared with the oxygen.
  2. Halogenation. The enolate takes an X from , giving R–CO– and . This happens twice more, faster each time, until the group is .
  3. Addition. adds to the C=O of R–CO–.
  4. Cleavage. The C=O reforms and leaves; it then takes the acid's proton, giving and .
Exam Trick

"Methyl next door." The test needs a on the carbon that carries the oxygen: C=O, or CH–OH (NaOI oxidises CH–OH to C=O first), and that carbon must be bonded to H or C, not part of an acid, ester or amide group. Among compounds with only one functional group, ethanol is the only primary alcohol and ethanal the only aldehyde that pass.

2. The iodoform test: which compounds pass

CompoundIodoform testWhy
Ethanol, YesOxidised to ethanal,
Methanol, NoGives , which has no
Propan-2-ol, YesOxidised to propanone
Propan-1-ol, NoGives propanal; its is not on the C=O carbon
Butan-2-ol, 1-phenylethanolYesBoth have the group
Ethanal, YesThe only simple aldehyde with on the CHO carbon (a keto-aldehyde such as passes through its part)
Propanone, butan-2-one, acetophenoneYesMethyl ketones
Pentan-3-one, NoEthyl groups on both sides of C=O
Ethanoic acid, methyl ethanoate, ethanamideNoThe is part of an acid, ester or amide group

The test is done by warming the compound with iodine and sodium hydroxide (or with KI and NaOCl, which together give KOI). A yellow precipitate of iodoform, which smells like an antiseptic, means a pass. With chlorine or bromine the same reaction gives chloroform or bromoform, which are liquids, so only iodine gives an easily seen yellow precipitate.

JEE Advanced

Why base keeps going. In base, each halogen added makes the remaining α-H more acidic, so the second and third halogenations are faster than the first: the methyl group ends as . In acid, each X makes the C=O oxygen less basic, so enol formation (the slow step) gets slower and the reaction can be stopped after one X.

Why can leave. A carbanion is normally a very poor leaving group, but three halogens pull electron density away from the carbon and stabilise the anion. That is why only a group, not , is split off.

Making acids. The reaction shortens a chain by one carbon: acetophenone with NaOCl gives benzoate, and pent-3-en-2-one gives but-2-enoate with its C=C intact. Acetone chlorinated three times and then treated with gives chloroform and calcium ethanoate.

3. Worked examples

Solved Example 1
Which of these give a yellow solid with NaOI/NaOH: propan-2-ol, propan-1-ol, acetophenone, ethanoic acid, pentan-3-one?
Solution:

Propan-2-ol (it has the group) and acetophenone (a methyl ketone). Propan-1-ol has no on the carbon bearing the oxygen, pentan-3-one has ethyl groups on both sides of its C=O, and in ethanoic acid the is part of the acid group.

Solved Example 2
Pent-3-en-2-one, , is warmed with NaOI. Name the products.
Solution:

The methyl ketone gives iodoform, , and the rest becomes the carboxylate with one carbon fewer: sodium but-2-enoate, . The C=C is not touched.

Solved Example 3
A compound gives a yellow precipitate with NaOI. Which of these is it: 4-methylbenzyl alcohol, 1-phenylethanol, 2-phenylethanol?
Solution:

1-Phenylethanol, . It has the group, which NaOI oxidises to acetophenone, a methyl ketone. The other two alcohols carry their OH on a group.

4. Exam traps

Watch out
  • Ethanoic acid, methyl ethanoate and ethanamide do not give the test, although they contain (an ethyl or isopropyl ester can hydrolyse in warm NaOH to an alcohol that passes).
  • Ethanol passes (it is oxidised to ethanal); methanol and methanal fail.
  • Propan-1-ol and propanal fail: the must sit on the carbon that carries the oxygen.
  • Pentan-3-one fails, butan-2-one passes.
  • The carboxylate has one carbon fewer than the methyl ketone.
  • NaOX (or KOI) is an oxidising agent; NaOCl with KI makes KOI.
  • A C=C elsewhere in the molecule survives.
  • Only iodoform is a solid; chloroform and bromoform are liquids.

Past-paper questions on Haloform reaction

Our past papers: JEE Main 2025-2026, JEE Advanced 2022-2026, NEET 2018-2019, 2022-2026. We found 15 questions in them that test this:

  • JEE Main 13 (2025-2026)
  • JEE Advanced 1 (2024)
  • NEET 1 (2018)

Each question was read and checked by hand (list checked 2026-10-15). A question that shows its structures only as pictures can be missed by our search. Each opens in a new tab, with the question to try and our step-by-step solution.

Learn it with the chapter

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