Alkynes are formed by creating a carbon-carbon triple bond, which requires removing two equivalents of HX from a dihalide starting material. This double dehydrohalogenation is one of the most reliable ways to build a triple bond, and it connects directly to the elimination chemistry you already know from E2 reactions.

Alkyne Preparation via Elimination

The core reaction here is eliminating two molecules of hydrogen halide (HX) from a vicinal dihalide, which is a compound with halogens on adjacent carbons. This double dehydrohalogenation requires a strong base such as sodium amide ( $NaNH_2$ ) or potassium tert-butoxide ( $KOC(CH_3)_3$ ).

The reaction proceeds in two stages, both through an E2 mechanism:

The base abstracts a proton from one of the carbons bearing a halogen. One equivalent of HX is eliminated, forming a vinylic halide intermediate.
A second equivalent of base abstracts the remaining proton from the vinylic halide, eliminating the second HX and forming the alkyne product.

Sodium amide ( $NaNH_2$ ) is often the preferred base because it's strong enough to drive both eliminations to completion. Weaker bases may stall at the vinylic halide stage, since vinylic C-H bonds are harder to deprotonate than typical sp3 C-H bonds.

Alkene to Alkyne Conversion

You can also synthesize alkynes starting from alkenes through a two-step sequence: halogenation followed by dehydrohalogenation.

Step 1: Halogenation of the alkene

Treat the alkene with $X_2$ (where X = Cl or Br) to add halogens across the double bond, forming a vicinal dihalide.
This proceeds through electrophilic addition with anti stereochemistry, meaning the two halogens add to opposite faces of the alkene.

Step 2: Double dehydrohalogenation of the vicinal dihalide

Treat the vicinal dihalide with excess strong base (typically $NaNH_2$ ) to eliminate two equivalents of HX, forming the alkyne.
This proceeds through the same two-stage E2 mechanism described above.

The overall transformation converts a double bond into a triple bond. Since the halogenation step is anti and the E2 eliminations are also anti, the geometry works out cleanly to produce the alkyne.

Alkyne preparation via elimination, Organic chemistry 13: Bimolecular beta elimination (E2) - regioselectivity and stereoselectivity

Vinylic Halides in Alkyne Synthesis

A vinylic halide is an alkene with a halogen directly attached to one of the double-bond carbons. These compounds show up in two important contexts:

As intermediates: During double dehydrohalogenation of a vicinal dihalide, the vinylic halide is the halfway point. The first equivalent of base removes one HX to give the vinylic halide; the second equivalent removes the remaining HX to give the alkyne.
As starting materials: If you already have a vinylic halide, you only need one equivalent of strong base to perform a single dehydrohalogenation and form the alkyne directly via an E2 mechanism.

The second elimination (from vinylic halide to alkyne) is typically slower and more difficult than the first. That's because the C-H bond on a vinylic carbon has more s-character (sp2), making the hydrogen less accessible to the base. This is why a particularly strong base like $NaNH_2$ is needed rather than something milder like $NaOH$ .

Additional Alkyne Preparation Methods

Beyond the elimination routes above, alkynes can also be prepared through:

Alkylation of terminal alkynes: A preexisting terminal alkyne can be deprotonated with a strong base (e.g., $NaNH_2$ ) to form an acetylide anion, which then reacts with a primary alkyl halide via an $S_N2$ mechanism to give an internal alkyne. This is covered more in alkyne alkylation sections but is worth noting as a key synthetic tool.
Elimination from geminal dihalides: Compounds with two halogens on the same carbon (geminal dihalides) can also undergo double dehydrohalogenation with strong base to form alkynes, following a similar two-stage E2 pathway.

In all of these preparations, the choice of base matters. Sodium amide ( $NaNH_2$ ) in liquid ammonia or mineral oil is the go-to for driving eliminations all the way to the alkyne, while weaker bases risk stopping at the vinylic halide intermediate.