When sulphuric acid (H₂SO₄) is treated with ethanol (C₂H₅OH), sulphuric acid acts as:

When ethanol (C₂H₅OH) is treated with concentrated sulphuric acid (H₂SO₄), the acid does not act primarily as a proton donor but as a powerful dehydrating agent. This means it promotes the removal of a water molecule (H₂O) from ethanol, leading to the formation of ethene (C₂H₄), a gaseous hydrocarbon belonging to the alkene family.

The reaction can be represented as:

C2H5OH→170°Cconc. H2SO4C2H4+H2OC₂H₅OH \xrightarrow[170°C]{conc. \ H₂SO₄} C₂H₄ + H₂OC2H5OHconc. H2SO4170°CC2H4+H2O

Here’s how the reaction proceeds:

1. 
   

   At around 170°C, concentrated H₂SO₄ protonates the ethanol molecule, forming an intermediate oxonium ion (C₂H₅OH₂⁺).

   
   


2. 
   

   This intermediate then loses a water molecule (dehydration), resulting in the formation of a carbocation.

   
   


3. 
   

   The carbocation subsequently loses a proton (H⁺) to yield ethene (C₂H₄) as the main product.

   
   

This process shows that H₂SO₄ facilitates water elimination, hence acting as a dehydrating agent rather than an oxidizing or sulphonating agent. If the reaction were carried out at a lower temperature (around 140°C), the same acid would lead to ether formation (diethyl ether) instead of ethene, showing its dual catalytic role depending on temperature.

The dehydrating property of concentrated sulphuric acid also appears in many reactions, such as charring of carbohydrates (e.g., removing water from sugar to form carbon). This ability arises because sulphuric acid has a strong affinity for water — it readily absorbs and removes it from other compounds.

In summary, during the reaction between ethanol and concentrated sulphuric acid, the acid acts as a dehydrating agent, eliminating water and producing ethene, an important reaction in Organic Chemistry and industrial ethanol dehydration processes