Question

When ethyl chloride and alcoholic KOH are heated, the compound obtained is
A. \[{C_2}{H_4}\]
B. \[{C_2}{H_2}\]
C. \[{C_6}{H_6}\]
D. \[{C_2}{H_6}\]

Answer 1

Hint: The haloalkane with \[\beta \]-hydrogen on reacting with alcoholic solution of potassium hydroxide results in the formation of alkene with the elimination of hydrogen halide.

Complete Step by Step Solution:
The alkene is formed from alkyl halide by reacting with alcoholic potassium hydroxide and eliminating hydrogen halide.The reaction involved is elimination reaction or dehydrohalogenation reaction. Mono Haloalkanes are the compound where only one halogen group is attached to the alkyl group.

In the dehydrogenation reaction, the hydrogen atom is removed from the reactant whereas in dehalogenation, the halogen group is removed from the reactant. Both the reactions taking place are known as dehydrohalogenation reactions.

The dehydrohalogenation reaction means the removal of hydrogen and halogen atoms.
Dehydrohalogenation is the method used for the preparation of alkenes from alkyl halides. In this reaction,hydrogen halide is removed. This reaction takes place by heating in presence of alcoholic potassium hydroxide or alcoholic potash.

The reaction of dehydrohalogenation of ethyl chloride is shown below:
\[C{H_3}C{H_2}Cl + alc.KOH \to C{H_2} = C{H_2} + HCl\]
In the above reaction ethyl chloride is heated with alcoholic potassium hydroxide to form ethene as the main product by eliminating hydrochloric acid.
Therefore, the correct option is B.

Note: It should be noted that when ethyl chloride is reacted with aqueous KOH then the product formed is ethanol and hydrochloric acid. Here, nucleophilic substitution takes place as chloride is a good leaving group and the nucleophile attacks (-OH) the carbocation to form the product.