Question

2. Complete the reaction.
CH3CH2CH2CL in presence of alc. KOH​

Answer 1

When a haloalkane is heated with a concentrated alcoholic solution of potassium hydroxide, a hydrogen atom is eliminated from the β-carbon (i.e., the carbon adjacent to the halogenated carbon) and a halogen atom is eliminated from the $\alpha$-carbon. Therefore, an alkene is formed. This reaction is known as Dehydrohalogenation.

CH$_{3}$ –CH$_{2}$ –CH$_{2}$Cl in presence of alcoholic KOH gives CH$_{3}$ –CH=CH$_{2}$ (Propene).

Answer 2

It will result in the formation of CH3CH=CH2.

Explanation:

• Alcoholic KOH is the specific reagent used for the purpose of dehydrohalogenation.
• As a result chlorine will get eliminated along with a hydrogen from the adjacent carbon and result in the formation of a double bond.
• This process therefore helps in the formation of an alkene with the formula CH3CH=CH2.