Deprotonation of a Terminal Alkyne
Deprotonation removes the H from a terminal alkyne using a strong base, usually NaNH₂. Step-by-step mechanism, reagents and practice.
What happens
Deprotonation removes the H from a terminal alkyne using a strong base, usually NaNH₂. This forms an acetylide ion, which has a negative charge on carbon. The acetylide ion is important because it is a strong nucleophile and can be used to form new C–C bonds.
Reaction summary
R–C≡C–H + NaNH₂ → R–C≡C⁻ Na⁺ + NH₃
Reagents and conditions
- Starting material: Ethyne (terminal alkyne, R = H)
- Reagents: NaNH₂
- Products: Acetylide ion (H–C≡C⁻); Sodium ion (Na⁺); Ammonia (NH₃)
Mechanism
- Deprotonation: This is a simple acid–base reaction. The strong base NH₂⁻ attacks the H on the terminal alkyne, and the electrons from the C–H bond move onto carbon.
- Product: An acetylide ion (C≡C⁻) and NH₃ are formed.
Common mistakes
- This works with terminal alkynes because they have a hydrogen directly attached to the triple-bond carbon. A terminal alkyne is R–C≡C–H; an internal alkyne, R–C≡C–R, has no terminal H to remove.
- A normal base such as OH⁻ is not strong enough to efficiently deprotonate a terminal alkyne. A much stronger base, such as NH₂⁻, is needed.
Memory trick
Terminal H + strong base → H is gone 👋. R–C≡C–H → R–C≡C⁻. And now that carbon has a negative charge, it's ready to attack and build a new C–C bond.