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Organic Chemistry 1 · Alkene and Alkyne Chemistry

Reactions of Alkynes

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On this page 7 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Study tools

In 30 seconds

Terminal alkynes are weakly acidic; strong bases remove the C≡C–H proton to give acetylide ions (R–C≡C⁻), powerful nucleophiles that form new carbon-carbon bonds by SN2 alkylation with methyl or primary alkyl halides. Alkynes add two equivalents of H₂ or X₂/HX, and can be selectively reduced — Lindlar's catalyst gives the cis-alkene (syn addition), sodium in liquid ammonia gives the trans-alkene (anti addition). gives an enol that tautomerizes to a ketone (an aldehyde only from ethyne); instead gives an aldehyde from terminal alkynes. cleaves the triple bond to carboxylic acids.

Why this matters

builds carbon frameworks for pharmaceuticals and advanced materials, and alkyne click chemistry (Cu-catalyzed azide–alkyne cycloaddition) is used in drug discovery and biological labeling. Several reagents here are hazardous — acetylene and organometallic bases are flammable and air-sensitive, and catalytic uses flammable H₂ under pressure — so all handling, quantities, PPE, and disposal must follow approved institutional safety documentation; no operational instructions are given here.

The college version

1. Terminal Alkyne Acidity and Carbon-Carbon Bond Formation

A terminal alkyne's sp-hybridized carbon holds its C–H electrons tightly (more s character), so the proton is weakly acidic (pKa ≈ 25). A strong base (NaNH₂ or an organolithium) deprotonates it to an , R–C≡C⁻, whose lone pair sits in an sp orbital. Acetylides are good nucleophiles and undergo SN2 alkylation with methyl or primary alkyl halides to form new carbon-carbon bonds. : the electrophile must be unhindered — secondary and tertiary halides eliminate instead — and it should bear a good leaving group (I > Br > Cl). This is the key chain-extension reaction (topic 42).

2. Addition Reactions: Hydrohalogenation, Halogenation, and Hydration

  • (HX): adds Markovnikov to give a vinyl halide; excess HX adds a second equivalent to give a geminal dihalide (both X on one carbon). Markovnikov's rule is a prediction tool, not a substitute for mechanism (attack gives the more stable vinylic carbocation).
  • (X₂): one equivalent of Br₂ or Cl₂ adds anti to give a trans-dihaloalkene; excess X₂ gives a tetrahalide.
  • Hydration: acid-catalyzed with a mercury(II) salt (HgSO₄) adds water Markovnikov to give an enol (OH on a double-bond carbon). The enol rapidly undergoes — a proton shifts from O to C while the π bond rearranges — to the more stable ketone. Terminal alkynes give methyl ketones; ethyne is the only alkyne that gives an aldehyde.

3. Reduction and Oxidation: Hydrogenation, Partial Reduction, Hydroboration-Oxidation, Ozonolysis

  • Hydrogenation to alkane: H₂ over Pt, Pd, or Ni reduces the alkyne fully to the alkane (two H₂ equivalents).
  • to cis-alkene: H₂ over Lindlar's catalyst (Pd on CaCO₃ poisoned with lead/quinoline) adds syn, giving the cis-alkene.
  • Partial reduction to trans-alkene: sodium in liquid ammonia (dissolving-metal reduction) adds anti, giving the trans-alkene.
  • Hydroboration-oxidation: a bulky borane (disiamylborane, Sia₂BH) adds anti-Markovnikov (syn); oxidation gives an enol that tautomerizes to an aldehyde from a terminal alkyne (a ketone from an internal alkyne) — the ketone vs aldehyde contrast with hydration.
  • Ozonolysis: O₃ followed by workup cleaves the triple bond to carboxylic acids (one per alkyne carbon).

How it works

  1. Deprotonate a terminal alkyne to an acetylide, then alkylate it (methyl/primary halide) to lengthen the chain.
  2. Reduce with H₂/Pt or Ni (alkane), Lindlar (cis-alkene), or Na/NH₃ (trans-alkene).
  3. Add HX or X₂ (once or twice) for hydrohalogenation/halogenation products.
  4. Add water (hydration → ketone) or hydroborate–oxidize (→ aldehyde/ketone).
  5. Cleave with O₃ → carboxylic acids.

Common confusions

Do not confuseWithDifference
Acetylide alkylation (SN2)EliminationMethyl/primary → SN2; secondary/tertiary → E2
Lindlar (cis)Na/NH₃ (trans)Syn vs anti hydrogen addition
Hydration productHydroboration-oxidation productMarkovnikov → ketone; anti-Markovnikov → aldehyde
EnolKetoneEnol is the fleeting OH-on-C=C form; ketone is the stable tautomer

Memory aids

"cis-Lindlar, trans-Na; hydration ketone, HB aldehyde." Acetylides make C–C bonds, and ozonolysis gives acids.

Quick review

Topic Recap

Terminal alkynes are weakly acidic and are deprotonated to acetylide ions, which build carbon-carbon bonds by SN2 alkylation with methyl or primary halides. Alkynes are reduced fully to alkanes (H₂/catalyst) or partially to cis-alkenes (Lindlar) or trans-alkenes (Na/NH₃). HX or X₂ adds once or twice; hydration gives ketones via keto-enol tautomerization, hydroboration-oxidation gives aldehydes from terminal alkynes, and ozonolysis cleaves the triple bond to carboxylic acids.

Knowledge Check

  1. What base is commonly used to deprotonate a terminal alkyne, and why is that proton acidic?
  2. What reagent converts an internal alkyne to a cis-alkene, and what is the stereochemical basis?
  3. Predict the product of hydrating but-1-yne.
  4. What product forms when a terminal alkyne undergoes hydroboration-oxidation?
  5. Why must the alkyl halide in acetylide alkylation be methyl or primary?

Answers and Rationales

  1. NaNH₂ (sodium amide). The sp carbon holds electrons tightly (more s character), stabilizing the conjugate-base acetylide and making the C–H weakly acidic (pKa ≈ 25).
  2. Lindlar's catalyst (H₂, Pd/CaCO₃, poisoned). Hydrogen adds syn to the same face, giving the cis-alkene.
  3. Butan-2-one (a methyl ketone). Markovnikov water addition gives an enol that tautomerizes to the ketone.
  4. An aldehyde (butanal). Anti-Markovnikov water addition via the organoborane gives an enol that tautomerizes to the aldehyde.
  5. SN2 needs an unhindered carbon. Methyl/primary halides react by SN2; secondary/tertiary are too hindered and eliminate.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Think of an alkyne as a tightly wrapped cable with one slightly loose end. That loose end — the hydrogen on a terminal alkyne — can be popped off by a strong base, leaving a negatively charged "hook" (the acetylide) that snaps onto the carbon of an alkyl halide and lengthens the chain.

The triple bond is also like a triple-zippered seam: you can unzip it partway (to a double bond, "cis-zipped" or "trans-zipped") or all the way (to a single bond), depending on the tool. Different tools — hydrogen over a poisoned catalyst, or sodium metal in ammonia — leave the seam zipped on opposite sides.

Where this stops being exact: "snapping on" hides the strict SN2 requirement that the target carbon be unhindered (methyl or primary), and "unzipping" hides that each reagent adds in a specific direction (Markovnikov vs anti-Markovnikov) with specific stereochemistry, not just "opening" the bond.

Simple Example

  • Propyne + NaNH₂ → CH₃–C≡C⁻ Na⁺; then + CH₃Br → but-2-yne (new C–C bond).
  • But-2-yne + H₂ / Lindlar catalyst → cis-but-2-ene.
  • Propyne + H₂O / H₂SO₄ / HgSO₄ → propan-2-one (acetone) via its enol.

Worked example

Follow acetylide alkylation and hydration, moving electrons before naming products.

  1. Deprotonation (acid–base). A double-headed arrow from the lone pair of NaNH₂ removes the terminal C≡C–H proton of propyne; the C–H bonding pair stays on carbon as a new lone pair, forming acetylide CH₃–C≡C⁻ (Na⁺ counterion). Acetylide is a weaker base than NH₂⁻, so equilibrium favors products.
  2. SN2 alkylation. A double-headed arrow from the acetylide lone pair attacks the back side of bromomethane's C–Br bond while Br⁻ leaves; the primary carbon is unhindered, so SN2 proceeds with inversion. Product: but-2-yne, CH₃–C≡C–CH₃. Charge balances (neutral product + Br⁻); no carbocation forms.
  3. Hydration of propyne. H⁺ adds Markovnikov to the terminal carbon to give a vinylic carbocation on the more substituted carbon; water attacks and deprotonation gives the enol CH₂=C(OH)–CH₃.
  4. Keto-enol tautomerization. A double-headed arrow shows base/water removing the O–H proton while the C=C π pair shifts to form C=O, giving propan-2-one (acetone). The ketone is far more stable, so equilibrium lies on the ketone side.

Key takeaways

  • High yield: Terminal alkynes (pKa ≈ 25) are deprotonated by NaNH₂ to acetylides.
  • High yield: Acetylide + methyl/primary halide = new C–C bond (SN2, inversion).
  • High yield: Lindlar's catalyst → cis-alkene (syn); Na/NH₃ → trans-alkene (anti).
  • High yield: Hydration gives a ketone (aldehyde only from ethyne) via enol tautomerization.
  • High yield: Hydroboration-oxidation of a terminal alkyne gives an aldehyde (anti-Markovnikov).
  • Excess HX or X₂ adds twice (geminal dihalide / tetrahalide); ozonolysis gives carboxylic acids.
  • Markovnikov and Zaitsev rules are prediction tools, not replacements for mechanism analysis.

Keep learning

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Practice Organic Chemistry 1

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Study tools & related lessonsYou’ll learn to · Key vocabulary · Related

You’ll learn to

  • Explain terminal alkyne acidity and how acetylide ions form carbon-carbon bonds by SN2 alkylation.
  • Predict the products of alkyne reduction (full hydrogenation and partial reduction to cis and trans alkenes).
  • Predict the products of electrophilic addition — hydrohalogenation, halogenation, and hydration — including keto-enol tautomerization.
  • Predict the products of hydroboration-oxidation and ozonolysis, and distinguish when a ketone versus an aldehyde forms.

Key vocabulary

Terminal alkyne acidity
Weak acidity (pKa ≈ 25) of C≡C–H
Acetylide ion
Deprotonated terminal alkyne (R–C≡C⁻)
Acetylide alkylation
Acetylide + methyl/primary halide (SN2)
SN2 limitations
Needs an unhindered electrophile
Hydrogenation
H₂ + metal catalyst → alkane
Partial reduction
Selective reduction to an alkene
Hydrohalogenation
Addition of HX (Markovnikov, twice)
Halogenation
Addition of X₂
Hydration
Addition of water (Markovnikov)
Keto-enol tautomerization
H shift that converts enol → ketone
Hydroboration-oxidation
Anti-Markovnikov water addition
Ozonolysis
O₃ cleavage of the triple bond
Hydroboration–oxidation
Two-step anti-Markovnikov hydration via organoboranes

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