Alkyne Formation via Elimination
Alkyne Formation via Elimination: Alkynes can be made from vicinal or geminal dihalides by removing two molecules of HX.
What happens
Alkynes can be made from vicinal or geminal dihalides by removing two molecules of HX. A very strong base, commonly NaNH₂, is used to perform two elimination reactions. The result is the formation of a C≡C triple bond.
Reaction summary
Vicinal or geminal dihalide + 2 NaNH₂ → alkyne (two E2 eliminations)
Reagents and conditions
- Starting material: 2,3-Dibromobutane (vicinal dihalide)
- Reagents: NaNH₂ (2 equivalents)
- Conditions: Very strong base, commonly NaNH₂; Internal alkyne: two equivalents of NaNH₂ (2 NaNH₂ → two eliminations)
- Products: 2-Butyne (internal alkyne)
Mechanism
- First Elimination (E2): NaNH₂ (the base) removes a hydrogen while a halide leaves. A C=C double bond forms: dihalide → vinyl halide.
- Vinyl Halide Intermediate: The first elimination gives a vinyl (alkenyl) halide: a C=C double bond that still carries one halogen.
- Second Elimination (E2): Another equivalent of NaNH₂ (the base) removes another hydrogen while the second halide leaves. The C≡C triple bond forms.
- Product: Two HX have been removed and the C≡C triple bond is formed: the alkyne.
Common mistakes
- The second elimination is difficult because it occurs from a vinyl halide. That's why a very strong base such as NaNH₂ is required: a weaker base usually isn't strong enough to efficiently perform both eliminations.
- If the product is a terminal alkyne, excess NaNH₂ can remove its terminal H and form an acetylide ion. A final H₂O workup gives the terminal alkyne back.
- How many equivalents of NaNH₂? For formation of an internal alkyne, two equivalents are generally required for the two eliminations (2 NaNH₂ → two eliminations).
Memory trick
Two halogens out → triple bond in. C–C → C=C → C≡C. One elimination = double bond; two eliminations = triple bond. And: NaNH₂ = strong enough to do it twice.