Question

The final product formed when ethyl bromide is treated with excess of alcoholic KOH is
A) Ethylene
B) Ethane
C) Ethyne
D) Vinyl bromide

Answer 1

Hint: The reaction of an alkyl halide with an excess potassium hydroxide of the alcoholic nature gives the elimination reaction. This reaction is also termed dehydrohalogenation reaction.

Complete Step by Step Answer:
Let's understand the dehydrohalogenation reaction in detail. In this reaction, alkyl halide undergoes a reaction with a strong base, that is, potassium hydroxide(alcoholic). In this reaction, OH, a strong base, attracts one proton from the beta carbon. This causes the release of one hydrogen atom and one atom of chlorine from the alkyl halide and the formation of alkene takes place. And this reaction is called \[\beta \] elimination because of the removal of a proton from the \[\beta \] carbon. A beta carbon indicates the carbon atom beside the carbon atom to which the halogen atom is bonded.

Here, the reaction of ethyl bromide with excess alcoholic potassium hydroxide (KOH) happens. The strong base OH attracts a proton and the Br atom leaves the compound. This results in the formation of an unsaturated compound, that is, ethene or ethylene.
Therefore, option A is right.

Note: Always remember that, when an alkyl halide undergoes a reaction with potassium hydroxide whose nature is aqueous, the product obtained is alcohol. That means alcoholic potassium hydroxide gives the substitution reaction when it undergoes a reaction with an alkyl halide.