Haloform reaction

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 H, alkyl or aryl. The reaction can be used to produce chloroform (CHCl₃), bromoform (CHBr₃), or iodoform (CHI₃).

Scope

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. 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, 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.

Mechanism

In the first step, the halogen disproportionates in the presence of hydroxide to give the halide and hypohalite:

X₂ + OH⁻→ XO⁻+ X⁻ + H⁺ (X = Cl, Br, I) (Unbalanced)

If a secondary alcohol is present, it is oxidized to a ketone by the hypohalite (hydroxide depicted below should be hypohalite and the water≡H₂O should be hypohalic acid≡HXO):

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

(1) R–CO–CH₃ + 3 OX⁻ → R–CO–CX₃ + 3 OH⁻

(2) R–CO–CX₃ + OH⁻ → RCOOH + ⁻CX₃

(3) RCOOH + ⁻CX₃ → RCOO⁻ + CHX₃

The detailed reaction mechanism is as follows:

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). 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. The ⁻CX₃ anion abstracts a proton from either the solvent or the carboxylic acid formed in the previous step, and forms the haloform.

Uses

This reaction was traditionally used 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.

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 appropriate carboxylic acid.

Iodoform test

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.

History

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]

References