Find the total number of alkene

Question: How many numbers of alkenes possible by dehydrobromination of 3-bromo 3-cyclopentylhexane using alcoholic KOH?

Solution:

When any alkyl halide is treated with alcoholic KOH, it removes hydrogen and a halogen atom from it and forms alkene. This is basically a dehydrohalogenation reaction. (You can recall it’s basic from here). The halogen atom removes from the α-carbon and hydrogen atom from β-carbon of the alkyl halide.

 In 3-bromo 3-cyclopentylhexane, α-carbon atom is at position 3 (to which halogen atom attached) and three β-carbon atom which are at 2, 4 and 3 (C of cyclopentane). 

Because of three types of β-hydrogen, there are more than one alkenes are expected.

So let us find out.

11)   Hydrogen remove from cyclopentane i.e. position 3:

2

  2   2) Hydrogen remove from position 2:

Here two alkenes are obtained, one is cis from and other one is trans.

  3) Hydrogen remove from position 4:

Here also two alkenes are obtained, one is cis from and other one is trans.

So the total numbers of alkenes obtained from dehydrobromination of 3-bromo 3- cyclopentylhexane are 5.