The whole story: a strong base removes the acidic sp C–H, and the acetylide alkylates a halide to forge a new C–C bond.
1. A Terminal Alkyne's C–H Is Unusually Acidic (pKa ≈ 25)
A terminal alkyne (R–C≡C–H) has one acidic sp C–H at pKa ≈ 25, extraordinary for a C–H bond; internal alkynes (R–C≡C–R) have none.
2. Acidity Tracks s-Character: sp > sp2 > sp3
Acidity follows only the hybridization of the carbon bearing the hydrogen:
An s orbital hugs the nucleus, so more s character (sp 50% > sp2 33% > sp3 25%) holds electrons tighter and lower in energy.
3. The Conjugate Base Is an sp-Hybridized Acetylide Anion
Removing the proton leaves the lone pair in that sp orbital, giving an acetylide anion (R–C≡C:–) whose charge is stabilized far better than a vinyl or alkyl anion — more stable anion, stronger acid, lower pKa.
4. Your Base Works Only If Its Conjugate Acid Has a pKa Above 25
Full deprotonation needs a base whose conjugate acid has pKa > 25, since equilibrium runs toward the weaker acid.
- NaNH2 works — its conjugate acid NH3 (pKa 38) is weaker, so equilibrium lies fully toward the acetylide. n-BuLi and LDA also work.
- NaOH fails — water (pKa 15.7) is a stronger acid, so only a trace deprotonates.
5. Acetylides Alkylate Primary Alkyl Halides (SN2, New C–C Bond)
The acetylide is a strong carbon nucleophile that displaces a halide by SN2, forming a new C–C bond; it needs a methyl or primary halide, since secondary/tertiary halides give E2 elimination instead.
1-Butyne → acetylide → SN2 on a primary halide → 3-hexyne.
6. Acetylides Add to Carbonyls to Give Alcohols
The same nucleophile adds to a carbonyl C=O to give an alkoxide; H3O+ workup then yields an alcohol with a new C–C bond.
Acetylide + formaldehyde, then acid workup → propargyl alcohol (a propargylic alcohol with the triple bond intact).
7. Summary
Terminal ≡C–H, pKa ≈ 25 · stabilized by the sp conjugate base (acidity tracks s character, sp > sp2 > sp3) · deprotonate with NaNH2 not NaOH · the acetylide alkylates primary halides (SN2) and adds to carbonyls, both forming new C–C bonds.
Quiz yourself
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Because of the conjugate base. Removing the terminal proton leaves the lone pair in an sp orbital (50% s character), which holds the electrons close to the nucleus and stabilizes the negative charge. An alkane's anion would sit in an sp3 orbital (only 25% s), far less stabilized. More stable conjugate base = stronger acid = lower pKa.
No. Deprotonation runs toward the weaker acid. Hydroxide's conjugate acid is water (pKa 15.7), which is a stronger acid than the alkyne (pKa 25), so equilibrium lies on the alkyne side — only a trace is deprotonated. You need a base whose conjugate acid has pKa > 25, such as NaNH2 (NH3, pKa 38).
The alkylation is an SN2 reaction, which requires an unhindered electrophilic carbon. With secondary or tertiary halides, the acetylide — which is also a strong base — instead causes E2 elimination, so substitution fails. Methyl and primary halides give clean SN2 and the desired longer internal alkyne.
A propargylic alcohol. The acetylide carbon attacks the carbonyl carbon of formaldehyde to give an alkoxide; H3O+ workup protonates it to an –OH. From acetylene + formaldehyde you get propargyl alcohol (HC≡C–CH2OH), with the triple bond untouched and a new C–C bond formed.
Draw this on the whiteboard
Open the OChem Board whiteboard — benzene rings, curved arrows, wedge/dash bonds and a clickable periodic table built in. No account needed.