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Haloform Reaction: Iodoform Test & Mechanism

By Chemca Editorial Team • Last Updated: January 2026 • 10 min read

The Haloform Reaction describes the exhaustive halogenation of a methyl ketone (or a compound oxidizable to a methyl ketone) in the presence of a base, followed by cleavage to produce a Haloform ($CHX_3$) and a carboxylate ion.

1. Structural Requirement

The reaction is given only by compounds containing the Methyl Ketone group ($CH_3-C=O$) or Methyl Carbinol group ($CH_3-CH(OH)-$).

• Ketones: $R-CO-CH_3$ (e.g., Acetone, Acetophenone).
• Aldehydes: Only Acetaldehyde ($CH_3CHO$).
• Alcohols: $R-CH(OH)-CH_3$ (e.g., Ethanol, Isopropanol).

2. General Reaction

Reaction of a methyl ketone with Halogen ($X_2$) and Sodium Hydroxide ($NaOH$).

$$ R-CO-CH_3 + 3X_2 + 4NaOH \rightarrow \underbrace{CHX_3}_{\text{Haloform}} + R-COONa + 3NaX + 3H_2O $$

Common Products:

• Chloroform ($CHCl_3$): Colorless liquid.
• Bromoform ($CHBr_3$): Colorless liquid.
• Iodoform ($CHI_3$): Yellow solid precipitate (The Iodoform Test).

3. Detailed Mechanism

The reaction proceeds in two stages: Halogenation and Cleavage.

Step 1: $\alpha$-Halogenation (repeated 3 times)

The base abstracts an acidic $\alpha$-proton. The resulting enolate reacts with the halogen. This repeats until all three $\alpha$-hydrogens are replaced by halogens.

$$ R-CO-CH_3 + 3X_2 + 3OH^- \rightarrow R-CO-CX_3 + 3X^- + 3H_2O $$

Note: The introduction of each halogen makes the remaining $\alpha$-hydrogens more acidic, speeding up subsequent halogenations.

Step 2: Nucleophilic Acyl Substitution

The hydroxide ion ($OH^-$) attacks the carbonyl carbon of the trihalo ketone.

$$ R-CO-CX_3 + OH^- \rightarrow [R-C(OH)(O^-)-CX_3] $$

Step 3: Cleavage ($C-C$ Bond Breaking)

The intermediate collapses, reforming the double bond and expelling the trihalomethyl anion ($^-CX_3$). This is possible because the three halogen atoms stabilize the negative charge inductively.

$$ [R-C(OH)(O^-)-CX_3] \rightarrow R-COOH + ^-CX_3 $$

Step 4: Acid-Base Reaction

The strongly basic trihalomethyl anion instantly abstracts a proton from the carboxylic acid.

$$ R-COOH + ^-CX_3 \rightarrow R-COO^- \text{ (Salt)} + CHX_3 \text{ (Haloform)} $$

4. The Iodoform Test

This is a specific diagnostic test for methyl ketones or methyl carbinols.

Reagents: $I_2 + NaOH$ (or $NaOI$ - Sodium Hypoiodite).
Observation: Formation of a Yellow Precipitate of Iodoform ($CHI_3$) with a characteristic antiseptic smell.

5. Examples

A. Acetone (Propanone)

$$ CH_3COCH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 \downarrow + CH_3COONa $$

B. Ethanol (Oxidizable to Acetaldehyde)

$$ CH_3CH_2OH \xrightarrow{[O], I_2} CH_3CHO \xrightarrow{I_2, NaOH} CHI_3 + HCOONa $$

C. Acetophenone

$$ C_6H_5COCH_3 \xrightarrow{NaOCl} CHCl_3 + C_6H_5COONa $$

Haloform Quiz

Test your concepts on the Iodoform Test. 10 MCQs with explanations.

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