One mole of 1,2-dibromopropane on treatment with X moles of  NaNH2  followed by treatment with ethyl bromide gave a pentyne. The value of  X  is

The reaction of 1,2-dibromopropane with NaNH2 results in the formation of a terminal alkyne through a process known as double dehydrohalogenation.

Here are the steps:

1. The first mole of NaNH2 (sodium amide) removes a hydrogen and a bromine atom from 1,2-dibromopropane, resulting in the formation of a bromopropene and ammonia.

2. The second mole of NaNH2 then removes the remaining hydrogen and bromine atom, resulting in the formation of a propyne.

3. The propyne is then treated with ethyl bromide, which adds an ethyl group to the terminal carbon of the propyne, resulting in the formation of a pentyne.

Therefore, the value of X is 2, as two moles of NaNH2 are required for the reaction to proceed.