Ethanol is treated with concentrated sulphuric acid. The product formed is:

$C_2H_4$

The correct answer is Option (2) → $C_2H_4$

When ethanol \((C_2H_5OH)\) is treated with concentrated sulfuric acid \((H_2SO_4)\), a dehydration reaction occurs. Here's how the process works:

The concentrated sulfuric acid protonates the hydroxyl group \((-OH)\) of ethanol, forming an oxonium ion. This step increases the leaving group ability of the \(-OH\) group.

The protonated ethanol then loses a water molecule \((H_2O)\), leading to the formation of an alkene. Specifically, this reaction eliminates a hydrogen atom from the adjacent carbon, resulting in the formation of ethylene \((C_2H_4)\).

The overall reaction can be summarized as:

Mechanism:

Why Other Options Are Incorrect:

\(C_2H_6\) (Ethane): This is a saturated hydrocarbon, and the reaction with sulfuric acid does not produce ethane. Instead, it promotes dehydration to form an alkene.

\(C_2H_2\) (Acetylene): This alkyne cannot be produced directly from ethanol through this reaction with concentrated sulfuric acid.

\(CH_3CH_2.OSO_3OH\) (Ethanol Sulfonate): While this compound may form as an intermediate in some reactions involving sulfuric acid, it is not the main product of the dehydration process under these conditions.

Conclusion:

The primary product of the reaction of ethanol with concentrated sulfuric acid is \(C_2H_4\) (Ethylene) through the dehydration process.