Propene is treated with $\mathrm{Br}{}_2$ in $\mathrm{CCl}{}_4$, then with excess $\mathrm{NaNH}{}_2$ (3 equivalents) in liquid ammonia, and finally with $\mathrm{CH}{}_3\mathrm{I}$. The final product is:

Propene gives 1,2-dibromopropane. Two equivalents of $\mathrm{NaNH}{}_2$ remove 2 HBr to give propyne, and the third deprotonates the terminal alkyne to sodium propynide. Alkylation with $\mathrm{CH}{}_3\mathrm{I}$ gives $\mathrm{CH}{}_3\mathrm{C}\equiv \mathrm{C}-\mathrm{CH}{}_3$, but-2-yne.