Iodoform Test: Identification of Methyl Ketones & Alcohols | Chemca

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Iodoform Test: Identification & Mechanism

By Chemca Editorial Team • Last Updated: January 2026 • 8 min read

The Iodoform Test is a specific chemical test used to detect the presence of Methyl Ketones ($CH_3-C=O$) or Methyl Carbinols ($CH_3-CH(OH)-$). A positive test is indicated by the formation of a Yellow Precipitate of Iodoform ($CHI_3$).

1. The Reaction

The compound is heated with Iodine ($I_2$) and a strong base like Sodium Hydroxide ($NaOH$) or Potassium Hydroxide ($KOH$).

$$ R-CO-CH_3 + 3I_2 + 4NaOH \xrightarrow{\Delta} \underbrace{CHI_3 \downarrow}_{\text{Yellow Ppt.}} + R-COONa + 3NaI + 3H_2O $$

Active Reagent: The reaction uses Sodium Hypoiodite ($NaOI$), formed in situ:

$$ I_2 + 2NaOH \rightarrow NaOI + NaI + H_2O $$

2. Structural Requirement (Who gives the test?)

The test is specific for compounds containing the following structural units:

1. Methyl Ketones

Structure: $CH_3-C(=O)-R$

• Acetaldehyde ($CH_3CHO$) - Only Aldehyde
• Acetone ($CH_3COCH_3$)
• Acetophenone ($Ph-COCH_3$)
• 2-Pentanone

2. Methyl Carbinols

Structure: $CH_3-CH(OH)-R$

(Oxidized to Methyl Ketones during reaction)

• Ethanol ($CH_3CH_2OH$) - Only Primary Alcohol
• Isopropanol ($(CH_3)_2CHOH$)
• 2-Butanol

3. Step-by-Step Mechanism

Step 1: Oxidation (If Alcohol)

If the starting material is an alcohol ($CH_3-CH(OH)-R$), it is first oxidized by the Hypoiodite ion ($IO^-$) to the corresponding methyl ketone ($CH_3-CO-R$).

Step 2: $\alpha$-Iodination

The base ($OH^-$) removes the acidic $\alpha$-hydrogens of the methyl group. The resulting enolate reacts with $I_2$. This repeats three times to form a Tri-iodo ketone ($R-CO-CI_3$).

$$ R-CO-CH_3 \xrightarrow{3I_2, \ 3OH^-} R-CO-CI_3 $$

Step 3: Nucleophilic Acyl Substitution

The hydroxide ion ($OH^-$) attacks the carbonyl carbon. The $CI_3^-$ group is a good leaving group due to the electron-withdrawing nature of the three iodine atoms.

$$ R-CO-CI_3 + OH^- \rightarrow R-COOH + ^-:CI_3 $$

Step 4: Acid-Base Reaction (Precipitation)

The $CI_3^-$ anion instantly abstracts a proton from the carboxylic acid to form insoluble Iodoform.

$$ R-COOH + ^-:CI_3 \rightarrow R-COO^-Na^+ + CHI_3 \downarrow \text{ (Yellow)} $$

4. Distinguishing Compounds (Exam Favorites)

Pair	Iodoform Positive (+)	Iodoform Negative (-)
Methanol vs Ethanol	Ethanol ($CH_3CH_2OH$)	Methanol ($CH_3OH$)
1-Propanol vs 2-Propanol	2-Propanol	1-Propanol
2-Pentanone vs 3-Pentanone	2-Pentanone ($CH_3-CO-$)	3-Pentanone ($Et-CO-Et$)
Acetophenone vs Benzophenone	Acetophenone ($Ph-CO-CH_3$)	Benzophenone ($Ph-CO-Ph$)
Acetaldehyde vs Formaldehyde	Acetaldehyde ($CH_3CHO$)	Formaldehyde ($HCHO$)

5. Exceptions and Notes

• Acetic Acid ($CH_3COOH$): Does NOT give the test. Although it has a $CH_3-C=O$ group, the $OH$ of the acid reacts with the base ($NaOH$) to form salt, killing the nucleophilic attack mechanism.
• Acid Chlorides/Amides/Esters: Generally do not give the test because the leaving group ($Cl, NH_2, OR$) leaves before the $CI_3$ group can form or cleave properly.

Iodoform Quiz

Test your ability to distinguish compounds. 10 MCQs with explanations.

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