Question

l-chorobutane on reaction with alcoholic potash gives:
(A) 1-butene
(B) 1-butanol
(C) 2-butane
(D) 2-butanol

Answer 1

Hint: In alkenes, the molecule contains the double bond between the adjacent carbon atoms. The alkenes can be synthesized from the haloalkanes. Haloalkanes are treated with the potassium hydroxide $\text{ KOH }$. The $\text{ }\beta \text{ }$ hydrogen atom shifts its bonds and forms the double bond followed by the removal of the halogen atom at the $\alpha $ position. The reaction contains the removal of hydrogen and halogen, thus the reaction is known as dehydrohalogenation.

Complete step by step answer:
Let us first start the answer by writing the chemical reaction. So the chemical reaction goes as follows:
\[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{Cl }\!\!~\!\!\text{ }\xrightarrow{\text{alcoholic KOH}}\text{ C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CH = C}{{\text{H}}_{\text{2}}}\]

In the above reaction, alcoholic KOH plays the role of a dehydrohalogenation agent. The alcoholic KOH reacts with an alkyl halide and an alkene is formed as the product of the chemical reaction.
The product that is formed consists of 4 carbon atoms and as we can see there is a double bond that is present between the first and the second carbon atoms. So, the name of the compound will be 1-butene.
The mechanism is as shown below:

The answer to the question which is 1-butene is an organic compound that is highly flammable and is a highly condensed gas. 1-butene is soluble in ether and alcohol but is not soluble in water.
So, the correct answer is “Option A”.

Additional Information:
By alcoholic solution, as mentioned in the answer, we mean a solution which is a mixture of water and ethanol, which is used as a solvent. Substances that contain sugar can be fermented using this alcoholic solution.

Note: Alcoholic potash that is used in the chemical reaction mentioned above is a very strong base. Alcoholic potash or as we know KOH dissolves in water to give $\text{R}{{\text{O}}^{-}}$ ions. Alcohol abstracts the hydrogen which results in the formation of elimination in the reaction.