What happens when ethyl chloride is treated with alcoholic potassium hydroxide?
Ans.
When ethyl chloride (C2H5Cl) is treated with alcoholic potassium hydroxide (KOH dissolved in alcohol), it undergoes an elimination reaction known as dehydrohalogenation or elimination of hydrogen halide.
The reaction proceeds as follows:
C2H5Cl+KOH→C2H4+KCl+H2O
In this reaction, the base (KOH) abstracts a proton (H+) from the ethyl chloride molecule, leading to the formation of ethylene (C2H4) as the major product, along with potassium chloride (KCl) and water (H2O) as byproducts.
The mechanism of this reaction involves the removal of a proton from the beta carbon (carbon adjacent to the carbon bonded to the halogen) by the strong base (OH-) to form an alkene (ethylene). This process results in the formation of a carbon-carbon double bond and the elimination of hydrogen chloride (HCl).
Overall, the treatment of ethyl chloride with alcoholic potassium hydroxide results in the formation of ethylene, which is an important reaction used in organic synthesis for the preparation of alkenes.
