Question

How many alkaline could you treat with ${H_2}$ $Pd/C$ in order to prepare methylcyclopentane?

Answer 1

Hint : ${H_2}$ and $Pd/C$ will reduce alkynes to alkanes it means two hydrogen atoms get added to it . Palladium catalyses the addition of hydrogen to multiple carbon-carbon bonds. This reaction will take place twice while reducing alkynes to alkanes.

Complete Step By Step Answer:
To prepare methylcyclopentane by using ${H_2}$ , $Pd/C$ the required alkaline are $4 -$ methylcyclopent $- 1 -$ ene, $1 -$ methylcyclopent $- 1 -$ ene, methylenecyclopentane and $3 -$ methylcyclopent $- 1 -$ ene.

Methylenecyclopentane

$3 -$ methylcyclopent $- 1 -$ ene.

$1 -$ methylcyclopent $- 1 -$ ene

$4 -$ methylcyclopent $- 1 -$ ene

Note :
This hydrogenation reaction catches fire very easily as it has flammable reagents and solvents as it includes palladium and carbon; they are highly flammable in nature and can ignite fire very easily. The presence of hydrogen gas adds risk of explosion.