Alkylation of acetylides: practice and mechanism
Terminal alkynes are weakly acidic (pKa ≈ 25). NaNH₂ deprotonates them to the acetylide, which attacks the bromoalkane via SN2. The chain grows by the alkyl group.
Example
Terminal alkynes are weakly acidic (pKa ≈ 25). Sodium amide deprotonates them to the acetylide anion, which attacks the primary bromoalkane via SN2 as a nucleophile. This extends the carbon chain by the alkyl group.
Mechanism step by step
- 1
NaNH₂ deprotonates the terminal alkyne (sp C–H, pKa ≈ 25): the acetylide anion forms.
- 2
The acetylide attacks the primary bromoalkane from the back (SN2).
- 3
Bromide leaves and an internal alkyne with a longer chain forms.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- Amide does not add to the triple bond. NaNH₂ is the base that removes the terminal H; the acetylide then forms a new C–C bond with bromoethane.
Often confused with: Hydration of alkynes · Addition of HBr to alkynes
Frequently asked questions
Why is the alkyne H acidic?
The sp orbital has 50 % s character, so the anion electrons are close to the nucleus and well stabilised. Alkene H (sp², pKa ≈ 44) and alkane H (sp³, ≈ 50) are far less acidic.
Does it work with tertiary haloalkanes?
No. Acetylides are strong bases; with secondary and tertiary haloalkanes E2 elimination wins.