Question

When 2-bromobutane reacts with alcoholic KOH, the
reaction is called
(a) halogenation (b) chlorination
(c) hydrogenation (d) dehydro-halogenation

Answer 1

Explanation:

I think the correct answer is (c)

Answer 2

Option (d)

dehydro-halogenation

Explanation:

2 bromobutane

The halogen group present in 2 bromobutane.

When a haloalkanes group react with alcoholic KOH. It generally removes the halogen and hydrogen and produced KBr and water.

Here hydrogen and halogen group remove.

This reason the other name of the process is  dehydro-halogenation.

2 CH₃ -CH₂ - CH - CH₃       +    Alc. KOH

                        |

                      Br

→CH₃ - CH₂= CH- CH₃      +   CH₃ - CH₂ - CH = CH₃  + KBr+ H₂O

    But 2-ene (80%)                     But 1- ene (20%)