Question

1-Chlorobutane on reaction with alcoholic potash gives:
(A). $1 - \operatorname{butene} $
(B). $1 - \operatorname{butanol} $
(C). $2 - \operatorname{butene} $
(D). $2 - \operatorname{butanol} $

Answer 1

Hint: Dehydrohalogenation reaction is an elimination reaction in which a hydrogen halide is being removed from a substrate. Generally, the alkyl halide compounds are being served as the substrates for the Dehydrohalogenation reaction.

Complete step by step answer:
1- Chloro-butane is an alkyl halide with the chemical formula as $C{H_3}{\left( {C{H_2}} \right)_3}Cl$. It is a colorless and flammable liquid. On the other hand, alcoholic potash is an inorganic of$KOH$. It Is chemically called potassium hydroxide. It is a colorless solid.
The alcoholic potash $\left( {alc.KOH} \right)$ is a dehydrohalogenation agent which when reacted with an alkyl halide, gives an alkene as a product. Thus when the alcoholic potash reacts with 1-chlorobutane the reaction product is butane.
This reaction can be written as:
$C{H_2} - C{H_2} - C{H_2} - C{H_2}Cl\xrightarrow{{alc.KOH}}C{H_3} - C{H_2} - CH = C{H_2} + HCl$
$\text{(1-Chlorobutane) (1-butene) (Hydrogen chloride)}$
In the above reaction, $1 - \operatorname{butene} $ is formed as a product. Due to the formation of double bond, the chlorine atom and a hydrogen atom from the adjacent carbon have been removed and hydrogen chloride is formed as a byproduct.

So, the correct answer is Option A.

Additional information:
In this case the position double bond is at first position, depending upon the position of the chlorine atom in 1-chlorobutane. In other words, we can say that the position of the double bond in the reaction product i.e. alkenes, depends on the position of halide in the alkyl group.
Moreover, if a halo-alkane, used for this process has the ability to form more than one alkene as the reaction product, depending on the position of the halogen atom, then the alkene which is most substituted is the main product of the reaction.

Note: The order of reaction of halo-alkanes in the dehydrohalogenation reaction is: $3^\circ > 2^\circ > 1^\circ $ halo-alkane. This is because in the $3^\circ $ halo-alkanes, there are more possibilities for the formation of double bonds than in $1^\circ $halo-alkanes.