Question

Alkene can be prepared from alkyl halide by the following reagent \[{\rm{R}} - {\rm{X}} + N{u^ - } \to {\rm{Alkene}} + NuH\]
A) Alc. KOH + Heat
B) Aq KOH + Cold water
C) NaOH
D) LiOH

Answer 1

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

Complete Step by Step Solution:
Let's understand the dehydrohalogenation reaction in detail. In this reaction, an alkyl halide undergoes a reaction with a base whose nature is strong, 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 forms a bond.

For example, in the reaction of ethyl bromide with alcoholic KOH, 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.

Image: Dehydrohalogenation reaction

Here, OH acts as a nucleophile.
Therefore, option A is right.

Note: It is the point of confusion among students about the reaction of an alkyl halide with alcoholic KOH and aqueous KOH. As stated above, a reaction with alcoholic KOH of a haloalkane gives an elimination reaction. And the reaction with aqueous KOH gives a substitution reaction, so the product obtained is alcohol.