Haloform reaction

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The haloform reaction is a chemical reaction where a haloform (CHX3, where X is a halogen) is produced by the exhaustive halogenation of a methyl ketone (a molecule containing the R–CO–CH3 group) in the presence of a base.[1] R may be alkyl or aryl. The reaction can be used to produce chloroform (CHCl3), bromoform (CHBr3), or iodoform (CHI3).

Haloform reaction scheme


Substrates that successfully undergo the haloform reaction are methyl ketones and secondary alcohols oxidizable to methyl ketones, such as isopropanol. The only primary alcohol and aldehyde to undergo this reaction are ethanol and ethanal, respectively. 1,3-Diketones such as acetylacetone also give the haloform reaction. β-ketoacids such as acetoacetic acid will also give the test upon heating. Acetyl chloride and acetamide don't give this test. The halogen used may be chlorine, bromine, or iodine. Fluoroform (CHF3) cannot be prepared from a methyl ketone by the haloform reaction due to the instability of hypofluorite,[clarification needed] but compounds of the type RCOCF3 do cleave with base to produce fluoroform; this is equivalent to the second and third steps in the process shown above.


In the first step, the halogen disproportionates in the presence of hydroxide to give the halide and hypohalite (example with bromine, but reaction is the same in case of chlorine and iodine; one should only substitute Br for Cl or I):

\mbox{Br}_{2} + 2 \mbox{OH}^{{}-{}} ~ \rightarrow ~ \mbox{Br}^{{}-{}} + \mbox{BrO}^{{}-{}} + \mbox{H}_{2} \mbox{O}

If a secondary alcohol is present, it is oxidized to a ketone by the hypohalite:

Bromoform 1.svg

If a methyl ketone is present, it reacts with the hypohalite in a three-step process:

(1) Under basic conditions, the ketone undergoes keto-enol tautomerization. The enolate undergoes electrophilic attack by the hypohalite (containing a halogen with a formal +1 charge).
Haloform Schritt 1.svg
(2) When the α position has been exhaustively halogenated, the molecule undergoes a nucleophilic acyl substitution by hydroxide, with CX3 being the leaving group stabilized by three electron-withdrawing groups. In the third step the CX3 anion abstracts a proton from either the solvent or the carboxylic acid formed in the previous step, and forms the haloform.
Haloform Schritt 2.svg
Haloform startAnimGif


This reaction was traditionally used as a chemical test for qualitative organic analysis to determine the presence of a methyl ketone, or a secondary alcohol oxidizable to a methyl ketone through the iodoform test. Nowadays, spectroscopic techniques such as NMR and infrared are easy and quick to perform instead of qualitative tests. Iodoform reaction is used as test reaction of ethyl alcohol in given sample of alcohols.

It was formerly used to produce iodoform, bromoform, and even chloroform industrially.[citation needed]

In organic chemistry, this reaction may be used to convert a terminal methyl ketone into the analogous carboxylic acid.

Iodoform test[edit]

Negative and positive iodoform test

When iodine and sodium hydroxide are used as the reagents, a positive reaction gives iodoform. Iodoform (CHI3) is a pale-yellow substance. Due to its high molar mass caused by the three iodine atoms, it is solid at room temperature (cf. chloroform and bromoform). It is insoluble in water and has an antiseptic smell. A visible precipitate of this compound will form from a sample only when either a methyl ketone, ethanal, ethanol, or a methyl secondary alcohol is present.


The haloform reaction is one of the oldest organic reactions known.[2] In 1822, Serullas added potassium metal to a solution of iodine in ethanol and water to form potassium formate and iodoform, called in the language of that time hydroiodide of carbon.[3] In 1831, Justus von Liebig reported the reaction of chloral with calcium hydroxide to form chloroform and calcium formate. The reaction was rediscovered by Adolf Lieben in 1870. The iodoform test is also called the Lieben haloform reaction. A review of the Haloform reaction with a history section was published in 1934.[4]


If NaOH is used as the base, then three equivalents of water and three equivalents of NaX will also be produced.


  1. ^ Chakrabartty, in Trahanovsky, Oxidation in Organic Chemistry, pp 343–370, Academic Press, New York, 1978
  2. ^ László Kürti and Barbara Czakó (2005). Strategic Applications of Named Reactions in Organic Synthesis. Amsterdam: Elsevier. ISBN 0-12-429785-4. 
  3. ^ Georges-Simon Surellas, Notes sur l'Hydriodate de potasse et l'Acide hydriodique. – Hydriodure de carbone; moyen d'obtenir, à l'instant, ce composé triple [Notes on the hydroiodide of potassium and on hydroiodic acid – hydroiodide of carbon; means of obtaining instantly this compound of three elements] (Metz, France: Antoine, 1822). On pages 17–20, Surellas produced iodoform by passing a mixture of iodine vapor and steam over red-hot coals. However, later, on pages 28–29, he produced iodoform by adding potassium metal to a solution of iodine in ethanol (which also contained some water).
  4. ^ Reynold C. Fuson and Benton A. Bull (1934). "The Haloform Reaction". Chemical Reviews 15 (3): 275–309. doi:10.1021/cr60052a001.