Haloform Reaction

The haloform reaction is a chemical reaction in which a haloform (CHX₃, where X is a halogen) is formed by exhaustive halogenation of a methyl ketone in the presence of a base (RCOCH₃, where R may be either a hydrogen atom, an alkyl or an aryl group). The procedure may result in the formation of chloroform (CHCl₃), bromoform (CHBr₃), or iodoform (CHBr₃) from acetyl groups (CHI₃). [Note: this method does not work with fluoroform (CHF₃).]

Methyl ketones and secondary alcohols oxidizable to methyl ketones, such as isopropanol, are the most common substrates. Ethanol and acetaldehyde are the only main alcohols and aldehydes that may undergo this reaction. The haloform reaction may also be triggered by 1,3-diketones like acetylacetone. When heated, ketoacids such as acetoacetic acid will also yield the test. This test is not performed with acetyl chloride or acetamide. Chlorine, bromine, iodine, or sodium hypochlorite are examples of halogens. This approach cannot produce fluoroform (CHF₃) because it requires the presence of the exceedingly unstable hypofluorite ion. When ketones with the structure RCOCF₃ are treated with base, they cleave and create fluoroform, which is comparable to the second and third phases of the preceding process.

The iodoform test, which was widely used in the past as a chemical test to assess the presence of a methyl ketone or a secondary alcohol oxidizable to a methyl ketone, is based on this reaction. When iodine and sodium hydroxide are employed as reagents, a positive reaction produces iodoform, which is a solid at room temperature and has a tendency to precipitate out of solution, resulting in a noticeable cloudiness.

This reaction may be used to convert a terminal methyl ketone into an equivalent carboxylic acid in organic chemistry.

Ketone

A ketone is a carbonyl group (C=O) with two carbon atoms flanking it. “Alpha” carbons are those that are immediately attached to the carbonyl carbon, “beta” carbons are those two bonds away, “gamma” carbons are those three bonds away, and so on.

Ketones containing C-H bonds on the alpha carbon are “enolizable”; not only may they achieve equilibrium with an enol tautomer, but they can also deprotonate the alpha carbon to produce its conjugate base, which we call an enolate.

Because the negative charge resulting from deprotonation can be delocalised through resonance to the oxygen atom, which is superior at stabilizing negative charge due to its higher electronegativity, the C-H bonds on the alpha carbons of ketones are unusually acidic (the pKa of the simplest ketone, 2-propanone, is about 19 in water), the C-H bonds on the alpha carbons of ketones are unusually acidic

The Mechanism of the Haloform Reaction

The initial step:

The base (hydroxide ion) removes the alpha hydrogen, resulting in the formation of enolate. After that, the enolate reacts with the halogen, resulting in the synthesis of the halogenated ketone and the halogen’s corresponding anion.

2nd step:

To make a tri-halogenated ketone, perform Step 1 twice more. The whole reaction is held till the tri-halogenated ketone is formed.

3rd step:

As a nucleophile, the hydroxide ion attacks the electrophilic carbon that is doubly bound to oxygen. The double bond between carbon and oxygen is broken, resulting in the oxygen atom’s being anionic. As a result, the carbon oxygen double bond reforms favorably, and the carbon linked to three halogens is displaced, yielding carboxylic acid. The carboxylic acid transfers a proton to the tri-halomethyl anion, resulting in the necessary haloform product in an acid-base reaction.

Iodoform

The iodoform test is used to determine if a given unknown chemical contains carbonyl compounds with the structure R-CO-CH₃ or alcohols with the structure R-CH (OH)-CH₃

A yellow precipitate with an “antiseptic” odor is formed when iodine, a base, and a methyl ketone react. A few secondary alcohols with at the most one methyl group in the alpha position also show up in the test.

Description of the Iodoform Test

A light yellow precipitate of iodoform or triiodomethane is generated when iodine and sodium hydroxide are added to a molecule that includes either a methyl ketone or a secondary alcohol with a methyl group in the alpha position. It may be used to distinguish between aldehydes and ketonesBecause acetaldehyde is the only aldehyde with a CH₃C=O group, it must be acetaldehyde if an aldehyde provides a positive iodoform test. A few examples of positive iodoform test responses are shown below.

Positive Iodoform Test Compounds:

• Acetaldehyde
• Methyl Ketones
• Ethanol
• Secondary Alcohols with Methyl Groups in Alpha Position

Conclusion:

Ketones are normally inert to oxidation, but aldehydes are converted to carboxylic acids, as you may know from your oxidation reactions class. The haloform reaction is a rare case of a ketone being net oxidized to produce a carboxylic acid. (The Baeyer-Villiger oxidation, which produces esters from ketones, is another example.)However, there are certain limits. The haloform reaction only works with methyl ketones for our needs.Other alkyl ketones will undergo halogenation but not carboxylic acid cleavage.