Sulphuric acid is not used during the reaction of alcohols with potassium iodide (KI) because it oxidizes the crucial intermediate hydrogen iodide (HI) into iodine (I₂), preventing the formation of the desired alkyl iodide. This makes sulphuric acid unsuitable for converting alcohols into alkyl iodides effectively.
Understanding the Reaction Mechanism
To convert an alcohol into an alkyl iodide (like iodoalkane), the alcohol (ROH) first needs to react with hydrogen iodide (HI). Hydrogen iodide is typically generated in situ (within the reaction mixture) by reacting potassium iodide (KI) with a strong, non-oxidizing acid.
The general steps are:
- Generation of HI: KI + H⁺ (from acid) → HI + K⁺
- Reaction with Alcohol: ROH + HI → RI (Alkyl Iodide) + H₂O
The hydrogen iodide acts as a nucleophile, replacing the hydroxyl group (-OH) of the alcohol, which is a poor leaving group, after it is protonated by the acid.
The Problem with Sulphuric Acid
When sulphuric acid (H₂SO₄) is used as the acid, a critical issue arises due to its strong oxidizing properties. Instead of solely facilitating the production of HI for the alcohol reaction, sulphuric acid reacts with the newly formed hydrogen iodide:
Sulphuric Acid Oxidizes HI:
The nascent hydrogen iodide (HI) is a potent reducing agent. Concentrated sulphuric acid, being a strong oxidizing agent, readily oxidizes HI to elemental iodine (I₂), while itself being reduced to sulphur dioxide (SO₂) or other reduced sulphur species.
$2\text{HI} + \text{H}_2\text{SO}_4 \text{ (conc.)} \rightarrow \text{I}_2 + \text{SO}_2 + 2\text{H}_2\text{O}$
Consequences of Using Sulphuric Acid
- Depletion of HI: The primary reactant needed to convert the alcohol, HI, is consumed in the oxidation reaction with sulphuric acid.
- Reduced Yield of Alkyl Iodide: Since HI is used up, there is very little or no HI left to react with the alcohol (ROH). Consequently, the action of alcohol on acid to produce alkyl iodide cannot occur, leading to a significantly reduced yield or even no formation of the desired alkyl iodide.
- Formation of Undesirable By-products: The appearance of elemental iodine (I₂, often seen as a brownish color in the reaction mixture) indicates this side reaction, and sulphur dioxide (SO₂) is a pungent gas by-product.
Preferred Alternatives
To successfully synthesize alkyl iodides from alcohols using KI, a non-oxidizing or weakly oxidizing acid is essential. The most commonly preferred acid for this purpose is phosphoric acid (H₃PO₄).
- Phosphoric Acid (H₃PO₄): This acid is a strong acid but a very weak oxidizing agent. It efficiently protonates KI to generate HI without oxidizing it to I₂, ensuring that HI is available to react with the alcohol.
| Feature | Sulphuric Acid (H₂SO₄) | Phosphoric Acid (H₃PO₄) |
|---|---|---|
| Role in Reaction | Generates HI, but then oxidizes HI to I₂ | Generates HI, which then reacts with alcohol |
| Oxidizing Property | Strong oxidizing agent | Weak/non-oxidizing agent |
| Suitability for Alkyl Iodide Synthesis | Not suitable; leads to poor/no yield | Highly suitable; ensures good yield of alkyl iodide |
| By-products | I₂ (iodine), SO₂ (sulphur dioxide) | H₂O (water) |
Key Takeaways
- The goal is to generate hydrogen iodide (HI) to react with the alcohol.
- Sulphuric acid actively destroys the HI by oxidizing it to I₂.
- This prevents the desired reaction of alcohol + HI → alkyl iodide.
- Phosphoric acid is preferred because it generates HI without oxidizing it, allowing the synthesis of alkyl iodides.
Understanding the specific properties of reagents, such as the oxidizing nature of sulphuric acid, is crucial for selecting appropriate reaction conditions in organic synthesis. For more information on alkyl halide synthesis, you can refer to resources like LibreTexts Chemistry.
