Reduction by HI & Red Phosphorus: The Ultimate Reducer | Chemca

Organic Chemistry

Reduction by HI & Red Phosphorus

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

Hydrogen Iodide (HI) in the presence of Red Phosphorus is one of the strongest reducing systems available in organic chemistry. It is capable of reducing alcohols, aldehydes, ketones, carboxylic acids, and their derivatives directly into their corresponding Alkanes.

1. General Principle

The reagent combination (HI + Red P) effectively replaces Oxygen functional groups with Hydrogen.

$$ \text{Functional Group} + HI \xrightarrow{\text{Red P}, \ \Delta (150^\circ C)} \text{Alkane} + I_2 + H_2O $$

Conditions:

Reagent: Concentrated HI (57%).
Catalyst: Red Phosphorus.
Temperature: Heated in a sealed tube (approx 423 K).

2. The Crucial Role of Red Phosphorus

Why is Phosphorus needed?

During the reduction, HI is oxidized to Iodine ($I_2$). Iodine can react with the formed alkane to reverse the reaction (forming Alkyl Iodides).

Red Phosphorus reacts with the liberated Iodine to form Phosphorus Triiodide ($PI_3$), which is hydrolyzed back to $H_3PO_3$ and HI, or simply removes the iodine from the equilibrium, driving the reaction forward.

$$ 2P + 3I_2 \rightarrow 2PI_3 $$

3. Reactions with Functional Groups

A. Alcohols to Alkanes

Alcohols are reduced to alkanes. It consumes 2 moles of HI.

$$ R-OH + 2HI \xrightarrow{\text{Red P}} R-H + I_2 + H_2O $$

Example: Ethanol $\to$ Ethane.

B. Aldehydes/Ketones to Alkanes

The carbonyl oxygen is removed completely. It consumes 4 moles of HI.

$$ R-CO-R' + 4HI \xrightarrow{\text{Red P}} R-CH_2-R' + 2I_2 + H_2O $$

Example: Acetone $\to$ Propane.

C. Carboxylic Acids to Alkanes

Even the stable carboxyl group is reduced to a methyl group. It consumes 6 moles of HI.

$$ R-COOH + 6HI \xrightarrow{\text{Red P}} R-CH_3 + 3I_2 + 2H_2O $$

Example: Acetic Acid $\to$ Ethane.

4. Historical Significance: Structure of Glucose

This reaction played a pivotal role in determining the straight-chain structure of Glucose. When Glucose is heated with HI/Red P, it yields n-Hexane.

$$ \text{Glucose } (C_6H_{12}O_6) \xrightarrow{HI/P, \ \Delta} \text{n-Hexane } (CH_3-CH_2-CH_2-CH_2-CH_2-CH_3) $$

This proved that all 6 carbon atoms in glucose are linked in a straight chain without branching.

5. Summary Table

Functional Group	Moles of HI consumed	Product
Alcohol ($R-OH$)	2	Alkane ($R-H$)
Halide ($R-X$)	1	Alkane ($R-H$)
Carbonyl ($>C=O$)	4	Alkane ($>CH_2$)
Acid ($R-COOH$)	6	Alkane ($R-CH_3$)

HI Reduction Quiz

Test your concepts on Reductive Chemistry. 10 MCQs with explanations.

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