Answer
The reaction is:
$H−C≡C-H$ + $NaNH_2$ --$\gt$ $H−C≡C:^-$ $Na^+$
+ $CH_3CH_2Br$ --$\gt$ $H−C≡C-CH_2CH_3$ + $NaNH_2$ --$\gt$ $^-:C≡C-CH_2CH_3$ $Na^+$ + $CH_3(CH_2)_5Br$ --$\gt$ $CH_3(CH_2)_5−C≡C-CH_2CH_3$
Work Step by Step
$NaNH_2$ is added to alkynes to form the acetylide ion. The acetylide ion attacks on the alkyl bromide to give the alkyne. Same thing happens when $NaNH_2$ is added to $H−C≡C-CH_2CH_3$ which then attacks on the hexyl bromide to give the $CH_3(CH_2)_5−C≡C-CH_2CH_3$
