Double elimination to an alkyne: practice and mechanism
Two equivalents of the very strong base NaNH₂ eliminate HBr twice: the vicinal dibromide becomes an alkyne via the vinyl bromide.
Example
With the very strong base sodium amide a vicinal dibromide loses HBr twice (two E2 steps). The triple bond forms via the vinyl bromide. This is how alkenes are turned into alkynes via Br₂ addition.
Mechanism step by step
- 1
The amide ion removes an H from the carbon next to a Br while Br⁻ leaves (E2): a vinyl bromide forms.
- 2
A second amide ion removes HBr again. This step is slower and needs heat; the triple bond forms.
- 3
With terminal alkynes a third equivalent deprotonates the alkyne to the acetylide; aqueous workup returns the alkyne.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- That is only the first elimination. With 2 equivalents of NaNH₂ and heat the second HBr is removed too, giving the alkyne.
- Amide acts as a base here, not as a nucleophile for substitution. It removes protons and eliminates HBr instead of replacing Br by NH₂.
- An alkene forms from a dibromide with zinc (debromination), not with NaNH₂. The base removes HBr twice and the hydrogens do not stay: a triple bond forms.
Often confused with: E2 (Zaitsev product)
Frequently asked questions
Why NaNH₂ and not KOH?
The second elimination at the vinyl bromide is difficult and needs a very strong base; NH₂⁻ (pKa of NH₃ ≈ 38) manages it, OH⁻ only under harsh conditions.
How do I get from an alkene to an alkyne?
In two steps: add Br₂ (vicinal dibromide), then eliminate twice with 2 NaNH₂.