Question

What happens when ethyl chloride is treated with alcoholic potassium hydroxide?

Hint:

In alcoholic KOH, ethyl chloride undergoes an elimination reaction (E2 mechanism) to form ethene by the removal of HCl.

Answer 1

When ethyl chloride (C$_2$H$_5$Cl) is treated with alcoholic potassium hydroxide (KOH), it undergoes an elimination reaction. The result is the formation of ethene (C$_2$H$_4$). The reaction follows the E2 mechanism, where the base abstracts a proton from a carbon atom adjacent to the one bonded to the halogen (chlorine), resulting in the elimination of HCl and the formation of a double bond.

Step 1: The KOH deprotonates the β-carbon (carbon adjacent to the one with the chlorine), leading to the formation of a double bond between the \alpha and \beta carbons.


Step 2: The product formed is ethene (C$_2$H$_4$), and HCl is eliminated.
Reaction: \[ \text{C}_2\text{H}_5\text{Cl} + \text{KOH} \xrightarrow{\text{alcoholic}} \text{C}_2\text{H}_4 + \text{KCl} + \text{H}_2\text{O} \]