Haloform reaction

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

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Uses

In analytical chemistry, this reaction was traditionally used to determine the presence of a methyl ketone, or a secondary alcohol oxidizable to a methyl ketone through the the iodoform test. Nowadays, spectroscopic techniques such as NMR and infrared spectroscopy are preferred because they require small samples, may be non-destructive (for NMR) and are easy and quick to perform.

Formerly, it was used to produce iodoform and bromoform and even chloroform industrially.^{[citation needed]}

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

Iodoform test

  The iodoform test or iodoform reaction is a qualitative chemical test for the detection of ketones and aldehydes carrying an alpha methyl group. The reagents are iodine and sodium hydroxide.

Only methyl ketones, or alcohols with the feature:

CH₃CH(OH)-R

may undergo this reaction.

There are three steps to the iodoform reaction when starting from an alcohol. If the reaction starts from a methyl ketone, the first step is not needed.

• Addition of I₂(aq) (usually I₂ dissolved in KI, shown here reacting with 2-propanol)

2 NaOH + I₂ + CH₃CH(OH)CH₃ → 2 NaI + CH₃COCH₃ + 2 H₂O

oxidizes the hydroxyl group to a ketone,

• The ketone formed reacts with the I_2(aq) in a substitution reaction to produce a triiodoketone, for example triiodoacetone.

CH₃COCH₃ + 3I₂ → CH₃COCI₃ + 3HI

(Thus, ketone and aldehyde with structure -COCH₃ also show positive results). Ethanal is the only aldehyde to give the iodoform reaction.

• Finally, OH^- (aqueous) is added, which reacts with, for example, the triiodoacetone.

CH₃COCI₃ + OH^- → CHI₃ + CH₃COO^-

Iodoform, or triiodomethane (CHI₃) is a pale yellow substance with a relatively high molar mass due to the iodine atoms. It is therefore a solid at room temperature. It is insoluble in water and has an antiseptic smell. A visible precipitate of this compound will form from a sample only when a methyl ketone is present.

Reaction mechanism

First, the halogen disproportionates in the basic solution to give halide and hypohalite (= halate (I)) anions. In the case of chlorine, this can be shown:

Cl₂ + 2OH⁻ → Cl⁻ + OCl⁻ + H₂O

The reaction mechanism for the reaction of a methyl ketone with hypochlorite is then a three step process:

(1) R-CO-CH₃ + 3OCl^- → R-CO-CCl₃ + 3 OH⁻

(2) R-CO-CCl₃ + OH⁻ → RCOOH + ⁻CCl₃

(3) RCOOH + ⁻CCl₃ → RCOO⁻ + CHCl₃

an exhaustive chlorination (actually an electrophilic substitution of H by Cl⁺) replacing all the hydrogen atoms on the methyl group with chlorine followed by a nucleophilic acyl substitution of the hydroxide anion with the chloromethyl anion as the leaving group and finally proton transfer from the carboxylic acid to get to chloroform. Note that the latter two steps can be combined as an alkaline hydrolysis of the trihaloketone intermediate.

The graphical depiction of the reaction of a methyl ketone with a hypobromite:

where the first step displays a keto-enol tautomerism.

Haloforms can also be prepared directly from any secondary alcohol containing the CH3-CHOH- group (such as isopropyl alcohol), since this is oxidised under the reaction conditions to the corresponding methyl ketone, which can then undergo the haloform reaction.

With chlorine, it is possible to use sodium hypochlorite as a convenient source of both base and chlorine.

Fluoroform (CHF₃) cannot be prepared from a methyl ketone by the haloform reaction due to the instability of hypofluorite, but compounds of the type RCOCF₃ do cleave with base to produce CHF₃; this is equivalent to the second and third steps in the process shown above. The iodoform test is a laboratory test for the presence of a methyl ketone or the related alcohol, by formation of a pale yellow precipitate of iodoform.

History

The haloform reaction is one of the oldest organic reactions around ^[2]. In 1822 Serullas reacted ethanol with iodine and sodium hydroxide in water to sodium formate and iodoform, called in the language of that time hydroiodide of carbon. In 1831 Justus Liebig reported the reaction of chloral with calcium hydroxide to chloroform and calcium formate. The reaction was rediscovered by A. 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 1934.^[3]

References

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. ^ Reynold C. Fuson and Benton A. Bull (1934). "The Haloform Reaction". Chemical Reviews 15 (3): 275 - 309. doi:10.1021/cr60052a001.