What is the product formed when ethanol is dehydrated with concentrated sulphuric acid at 443 K?

The correct option is Ethene.

Explanation

The reaction involves the acidic dehydration of alcohols. When alcohols are heated with a protic acid such as concentrated sulphuric acid ($H_2SO_4$) or phosphoric acid ($H_3PO_4$), they lose a water molecule to form alkenes or ethers depending on the reaction conditions.

Detailed Analysis:

• Dehydrating Agent: Concentrated sulphuric acid acts as a strong dehydrating agent.
• Temperature Dependence: The product of the reaction between ethanol and sulphuric acid is strictly dependent on temperature.
  • At 443 K ($170^\circ$C): Ethanol undergoes intramolecular dehydration. A water molecule is removed from a single molecule of ethanol, resulting in the formation of Ethene ($C_2H_4$).
  • Reaction:
    $CH_3CH_2OH \xrightarrow[\text{443 K}]{\text{Conc. } H_2SO_4} CH_2=CH_2 + H_2O$
• Comparison (for clarity): At a lower temperature of 413 K ($140^\circ$C), intermolecular dehydration occurs, leading to the formation of Ethoxyethane (Diethyl ether). Since the question specifies 443 K, the product is Ethene.

Key Takeaway:

The dehydration of ethanol with concentrated sulphuric acid yields ethene at 443 K via elimination, while it yields ether at 413 K via substitution.