Organic Chemistry · Alkynes: An Introduction to Organic Synthesis

Alkyne Acidity: Formation of Acetylide Anions

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

Terminal alkynes — alkynes with a hydrogen on the triple-bond carbon, R-C ≡ CH — are the most acidic hydrocarbons. Their pK_a is about 25, compared with roughly 44 for the sp² C–H of an alkene and about 50 for the sp³ C–H of an alkane. That is still weak compared with water (pK_a 15.7), but it is acidic enough that a sufficiently strong base can remove the terminal proton and generate an , R-C ≡ C^-:

$R - C \equiv CH + NaNH 2 \to R - C \equiv C - Na + + NH 3$

The key to this reaction is choosing the right base. , NaNH₂, works because its conjugate acid, ammonia, has a pK_a of about 38 — higher than 25 — so the equilibrium favors the acetylide. Sodium hydroxide does not work, because water's pK_a (15.7) is lower than the alkyne's; the equilibrium lies the other way. This topic explains why alkynes are unusually acidic, how hybridization controls acidity, and how to form the acetylide anions that power the carbon–carbon bond-forming reactions of the next topic.

Why this matters

The acetylide anion is one of the most important carbon nucleophiles in organic synthesis — a negatively charged carbon that attacks an electrophile to form a new carbon–carbon bond. That is the fundamental way chemists build larger skeletons, and it is exactly what the next topic exploits to convert simple alkynes into complex internal alkynes. Beyond synthesis, the acidity trend sp > sp² > sp³ is a cornerstone of organic theory: it shows how hybridization, electronegativity, and the stability of the are connected. Understanding why acetylene is acidic also explains real-world chemistry, from biochemical deprotonations to industrial metal acetylides.

The college version

Core Concepts

Why terminal alkynes are acidic: the role of s-character

Acidity is governed by the stability of the conjugate base. When the proton is removed from a , the negative charge sits on a carbon that is sp-hybridized. An sp hybrid orbital has 50% (one s orbital mixed with one p orbital), and s orbitals hold electrons closer to the nucleus than p orbitals do. The closer the electron pair sits to the nucleus, the better the carbon can stabilize the negative charge:

sp carbon: 50% s-character → most electronegative → most stable anion → most acidic (pK_a ≈ 25)
sp² carbon: 33% s-character → intermediate (pK_a ≈ 44)
sp³ carbon: 25% s-character → least electronegative → least stable anion → least acidic (pK_a ≈ 50)

This is a clean example of a general rule: increasing s-character increases acidity of the C–H bond. The same logic explains why the C–H bonds of alkenes and arenes are more acidic than those of alkanes.

Acetylide ions: structure and properties

Removing the terminal proton leaves the acetylide ion, R-C ≡ C^-, with a full negative charge on the linear sp carbon, stabilized by s-character and the adjacent triple bond. Acetylides are both strongly basic and strongly nucleophilic — powerful but selective reagents. Because they react with water and even air's moisture, they are generated and used in anhydrous conditions (liquid ammonia or ether solvents). As a general laboratory principle, strong-base work requires proper PPE, a fume hood where appropriate, and exclusion of water.

Choosing the base: a pKa comparison

The rule for deprotonation is that a base whose conjugate acid has a higher than the acid being deprotonated will drive the equilibrium toward products. Compare:

NaNH₂ (conjugate acid NH₃, pK_a ≈ 38): works — 38 > 25, so the acetylide forms.
NaH (conjugate acid H₂, pK_a ≈ 35): works.
NaOH or KOH (conjugate acid H₂O, pK_a ≈ 15.7): does not work — 15.7 < 25, so the alkyne stays protonated.
Alcohols/alkoxides (conjugate acid pK_a ≈ 16–18): do not work for the same reason.

Sodium amide in liquid ammonia is the standard reagent: strong enough to deprotonate the alkyne, while the ammonia solvent keeps the acetylide stable and manageable.

Terminal versus internal alkynes

Only terminal alkynes have an acidic hydrogen. An internal alkyne such as 2-butyne, CH₃C ≡ CCH₃, has no hydrogen on either triple-bond carbon, so it cannot form an acetylide under these conditions. This distinction matters enormously in synthesis: the ability to deprotonate is what allows selective functionalization of one end of a molecule, and it is why terminal alkynes are the workhorses of alkyne chemistry.

Common Confusions

Do Not Confuse	With	The Difference
Terminal alkyne acidity	Internal alkyne "acidity"	Only RC≡CH has an acidic proton; internal alkynes cannot form acetylides
pKa	pH	pKa is an intrinsic property of a molecule (acidity); pH describes a solution's H⁺ level
NaNH₂	NaOH	NaNH₂'s conjugate acid (NH₃, pKa 38) is weaker than the alkyne, so it works; NaOH's conjugate acid (H₂O, pKa 15.7) is stronger, so it fails
"Acidic alkyne"	Strong acid	pKa ≈ 25 is still very weak in everyday terms; it means "acidic for a hydrocarbon," not like HCl
Acetylide as base	Acetylide as nucleophile	It is both; which role it plays depends on the reaction partner (proton donors vs. electrophilic carbons)

Eli explains

The same idea, in plain words

Explain it like I’m 10

Think of a hydrogen atom on the end of a molecule as a ball sitting on a hill. On a normal alkane the hill is very steep, so the ball stays put — the hydrogen almost never comes off. On a terminal alkyne the hill is gentler, so with a strong "shove" (a strong base like sodium amide) the ball rolls off, leaving the molecule with a negative charge — that's the acetylide ion. The steeper the hill, the harder it is to knock the ball off; the gentler the hill, the easier. Triple-bond carbons make the gentlest hills, which is why alkynes are the most acidic hydrocarbons.

Worked example

Worked example 1 (base selection). Can NaOH deprotonate 1-hexyne, CH₃(CH₂)₃C ≡ CH? Compare pKa values. The alkyne's pK_a is 25; water's pK_a is 15.7. Since 15.7 < 25, the equilibrium

$R - C \equiv CH + OH - ⇌ R - C \equiv C - + H 2 O$

lies to the left — the hydroxide cannot pull the proton off. With sodium amide instead, the conjugate acid (NH₃) has pK_a ≈ 38 > 25, so the equilibrium lies to the right and the acetylide forms:

$R - C \equiv CH + NaNH 2 \to R - C \equiv C - Na + + NH 3$

Worked example 2 (stoichiometry with dimensional analysis). How many grams of NaNH₂ are needed to deprotonate 10.0 g of 1-hexyne completely? The reaction is 1:1 in moles. First the moles of alkyne, then the moles and mass of amide:

$n(1 - hexyne) = 10.0 g 82.15 g/mol = 0.122 mol$

$n(NaNH 2) = 0.122 mol \times 1 mol NaNH 2 1 mol alkyne = 0.122 mol$

$m(NaNH 2) = 0.122 mol \times 39.01 g/mol = 4.75 g$

So 4.75 g of NaNH₂ is required for complete deprotonation (theoretical; practice often uses a slight excess to drive the equilibrium).

Worked example 3 (conceptual). Rank these C–H bonds by acidity: ethane, ethylene, acetylene. Answer: acetylene (sp, pKa ≈ 25) > ethylene (sp², pKa ≈ 44) > ethane (sp³, pKa ≈ 50). The reason is the increasing s-character of the C–H carbon along that series, which stabilizes the negative conjugate base.

Key takeaways

Terminal alkyne pK_a ≈ 25 — the most acidic hydrocarbon C–H.
Acidity trend: sp (25) > sp² (44) > sp³ (50), driven by s-character and conjugate-base stability.
NaNH₂ works; NaOH does not. Base selection is a pKa comparison: the base's conjugate acid must have a higher pKa than the alkyne (38 > 25 for NH₃; 15.7 < 25 for H₂O).
Only terminal alkynes (RC ≡ CH) form acetylides; internal alkynes have no acidic proton.
Product: acetylide anion R-C ≡ C^- — a carbon nucleophile and strong base.
Lab principle: acetylides react with water/moisture; generate and use them in anhydrous conditions with standard PPE and hood use where required.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

What is the approximate pKa of a terminal alkyne?
Show answer
About 25 — far more acidic than alkenes (~44) or alkanes (~50), though still weak compared with water (15.7).
Why is an sp-hybridized C–H bond more acidic than an sp³ C–H bond?
Show answer
The sp carbon has 50% s-character, so its hybrid orbitals hold electrons closer to the nucleus, making the carbon more electronegative and better able to stabilize the negative charge of the conjugate base.
Which reagent successfully converts 1-butyne into its acetylide: NaOH or NaNH₂? Why?
Show answer
NaNH₂. Its conjugate acid (NH₃) has pKa ≈ 38, which is higher than the alkyne's 25, so the equilibrium favors the acetylide. NaOH's conjugate acid (H₂O, pKa 15.7) is a stronger acid than the alkyne, so the equilibrium favors the protonated alkyne.
Can an internal alkyne such as 2-butyne form an acetylide anion with NaNH₂?
Show answer
No. 2-butyne has no hydrogen on either triple-bond carbon — there is no acidic proton to remove.
Write the acid–base reaction between 1-pentyne and sodium amide, showing the products.
Show answer
CH₃CH₂CH₂C ≡ CH + NaNH₂ → CH₃CH₂CH₂C ≡ C^-Na⁺ + NH₃
Arrange ethane, ethylene, and acetylene in order of increasing acidity.
Show answer
Increasing acidity: ethane (pKa ≈ 50) < ethylene (≈ 44) < acetylene (≈ 25). Acetylene is the most acidic.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Terminal alkyne: Alkyne with a hydrogen on the triple-bond carbon (RC ≡ CH)
Acetylide anion: The conjugate base R-C ≡ C^-
pKa: A number measuring acid strength (lower = stronger acid)
s-character: Fraction of s-orbital character in a hybrid orbital
Conjugate base: What remains after an acid loses its proton
Sodium amide: NaNH₂, a strong base whose conjugate acid is NH₃ (pKa 38)

Sources & references

openstax.org — Organic Chemistry

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