Question

Ethyl Chloride to Ethene conversions can be carried out by ${C_2}{H_5}Cl + KOH$
(alcoholic solution) $\to C{H_2} = C{H_2} + HCl$ :
A. True
B. False

Answer 1

When alcoholic $KOH$ will react with an alkyl halide, elimination reaction will take place. Elimination reaction is the process by which organic compounds containing only single bonds (saturated compounds) are transformed to compounds containing double or triple bonds (unsaturated compounds). There are two types of elimination reaction: - ${E_1}\& {E_2}$ . Here we see that ${E_2}$ elimination or $\beta$ elimination is taking place.

Complete Step by step answer: This is a dehydrohalogenation reaction as in this reaction removal of hydrogen and a halogen takes place. We see that hydrogen is eliminated from the $\alpha$ carbon and halogen is eliminated from the $\beta$ carbon hence this reaction is termed as $\beta$ elimination.
Let us have a look on the reaction: -

$C{H_3}C{H_2}Cl + alc.KOH \to C{H_2} = C{H_2} + HCl$

Here we see that after the elimination of hydrogen and chlorine from $\alpha$ and $\beta$ carbons respectively formation of ethene and hydrochloric acid takes place. Here $alc.KOH$ is acting as a Dehydrohalogenation agent.
Let us see which is the $\alpha \beta$ carbon: -
$C{H_3}C{H_2}Cl$
$\beta$ $\;\;\;\;\;\;$ $\alpha$
Hence it is true that ethyl chloride to ethene conversion can be followed by using alcoholic $KOH$ .

So, the correct option will be option A i.e. True.

Note: We must know that $alc.KOH$ and $aq.KOH$ have different mechanisms of the reaction. We see that in an aqueous solution, $KOH$ almost completely ionizes to give $O{H^ - }$ ions. It leads to the formation of alcohol as $O{H^ - }$ ion is a weak base, while an alcoholic solution of $KOH$ contains alkoxide $(R{O^ - })$ ion, which is a strong base. Thus, it can abstract a hydrogen from the $\beta$- carbon and form alkene by eliminating a molecule of $HCl$.