Iodine and NaOH ($I_2 + NaOH$)
The Iodoform Test and Disproportionation Reactions.
By chemca Team • Updated Jan 2026
The mixture of Iodine ($I_2$) and Sodium Hydroxide ($NaOH$) generates Sodium Hypoiodite ($NaOI$) in situ. This reagent is primarily used for the Iodoform Test to detect specific methyl groups and as a mild oxidizing agent.
1. The Iodoform Test (Haloform Reaction)
Detection of Methyl Carbonyls
Compounds containing the Methyl Ketone group ($CH_3-CO-$) or Methyl Carbinol group ($CH_3-CH(OH)-$) react with $I_2/NaOH$ to form a yellow precipitate of Iodoform ($CHI_3$).
A. Reaction with Methyl Ketones:
$$ R-CO-CH_3 + 3I_2 + 4NaOH \rightarrow \underset{\text{Sodium Carboxylate}}{R-COONa} + \underset{\text{Yellow ppt}}{CHI_3 \downarrow} + 3NaI + 3H_2O $$
B. Reaction with Methyl Carbinols (Alcohols):
First, the alcohol is oxidized to a methyl ketone by $NaOI$, which then undergoes the haloform reaction.
$$ R-CH(OH)-CH_3 \xrightarrow{NaOI} R-CO-CH_3 \xrightarrow{NaOI} R-COONa + CHI_3 $$
Compounds Giving Positive Test:
- Aldehydes: Only Acetaldehyde ($CH_3CHO$).
- Ketones: All Methyl Ketones (Acetone, Acetophenone, 2-Pentanone).
- Alcohols: Ethanol ($CH_3CH_2OH$), Propan-2-ol ($CH_3CH(OH)CH_3$), Butan-2-ol.
2. Mechanism of Iodoform Reaction
Step-by-Step
- Oxidation (if Alcohol): $NaOI$ oxidizes $CH_3-CH(OH)-$ to $CH_3-CO-$.
- $\alpha$-Halogenation: The base ($OH^-$) abstracts acidic $\alpha$-hydrogens, followed by reaction with $I_2$. This repeats until $R-CO-CI_3$ is formed.
- Cleavage: The $OH^-$ ion attacks the carbonyl carbon, leading to the cleavage of the $C-C$ bond. The $CI_3^-$ leaving group abstracts a proton to form $CHI_3$.
$$ R-CO-CH_3 \xrightarrow{3I_2/OH^-} R-CO-CI_3 \xrightarrow{OH^-} R-COOH + ^-CI_3 \rightarrow R-COO^- + CHI_3 $$
3. Disproportionation of Iodine
Effect of Temperature
Halogens react with alkalis to undergo disproportionation (simultaneous oxidation and reduction).
A. Cold and Dilute NaOH:
Forms Sodium Hypoiodite ($NaOI$).
$$ 2NaOH + I_2 \xrightarrow{\text{Cold}} NaI + NaOI + H_2O $$
B. Hot and Concentrated NaOH:
Hypohalites are unstable at high temperatures and disproportionate further to form Sodium Iodate ($NaIO_3$).
$$ 6NaOH + 3I_2 \xrightarrow{\text{Hot}} 5NaI + NaIO_3 + 3H_2O $$
4. Distinguishing Organic Compounds
| Pair | Reagent: $I_2 + NaOH$ | Observation |
|---|---|---|
| Methanol vs Ethanol | Ethanol reacts | Yellow ppt ($CHI_3$) with Ethanol only. |
| Propan-1-ol vs Propan-2-ol | Propan-2-ol reacts | Yellow ppt with Propan-2-ol (Methyl Carbinol). |
| Pentan-2-one vs Pentan-3-one | Pentan-2-one reacts | Yellow ppt with Pentan-2-one (Methyl Ketone). |
| Acetophenone vs Benzophenone | Acetophenone reacts | Yellow ppt with Acetophenone. |
Knowledge Check
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