Sodium amide induces the loss of two equivalents of hydrogen bromide from a vicinal dibromoalkane to give a carbon–carbon triple bond, as in a preparation of phenylacetylene.
Usually two equivalents of sodium amide yields the desired alkyne. Three equivalents are necessary in the preparation of a terminal alkynes because the terminal CH of the resulting alkyne protonates an equivalent amount of base.
Hydrogen chloride and ethanol can also be eliminated in this way, as in the preparation of 1-ethoxy-1-butyne.