Haloform reaction

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. [Chakrabartty, in Trahanovsky, "Oxidation in Organic Chemistry", pp 343-370, Academic Press, New York, 1978] R may be Hydrogen, alkyl or aryl. The reaction can be used to produce CHCl3, CHBr3 or CHI3.

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Substrates which successfully undergo the haloform reaction are methyl ketones and secondary alcohols oxidizable to methyl ketones, such as isopropanol. The halogen used may be chlorine, bromine, and 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 (CHF3); 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 hypohalide:

: X2 + OH→ XO+ X + H+ (X = Cl, Br, I)

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

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

:(1) R-CO-CH3 + 3 OX- → R-CO-CX3 + 3 OH

:(2) R-CO-CX3 + OHRCOOH + CX3

:(3) RCOOH + CX3RCOO + CHX3

The detailed reaction mechanism is as follows:

Under basic conditions, the ketone undergoes keto-enol tautomerization. The enolate undergoes electrophilic attack by the hypohalide (containing a halide with a formal +1 charge). When the alpha 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 carboxylic acid formed, or the solvent, 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 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.Fact|date=September 2007

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

It also use for test priamry, secondary alochol.

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 due to the three iodine atoms, it is solid at room temperature (c.f. 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 a methyl ketone is present.

History

The haloform reaction is one of the oldest organic reactions around [cite book | title = Strategic Applications of Named Reactions in Organic Synthesis | author = László Kürti and Barbara Czakó | publisher = Elsevier | location = Amsterdam | year = 2005 | isbn = 0-12-429785-4] . 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 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 1934. [cite journal|title =The Haloform Reaction|author=Reynold C. Fuson and Benton A. Bull|journal=Chemical Reviews
volume= 15|issue= 3|year= 1934|pages= 275–309|doi= 10.1021/cr60052a001
]

References


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