Difficulty: Intermediate | Prerequisites: Alkene reactions (Ch. 7–8), acid-base chemistry, Markovnikov's rule, syn/anti addition.
Alkynes are hydrocarbons containing a carbon-carbon triple bond. This unit extends your knowledge of addition reactions from alkenes to alkynes, where the triple bond can react once (stopping at an alkene) or twice (going all the way to a saturated product). Understanding how to control selectivity, whether you get a cis-alkene, a trans-alkene, a ketone, or a geminal dihalide, is the centrepiece of this chapter. You should already be comfortable with alkene addition mechanisms, Markovnikov regiochemistry, and curly-arrow notation.
Terminal alkynes are weakly acidic (pKa ~25) and can be deprotonated by strong bases such as NaNH₂ to form acetylide anions, which are excellent carbon nucleophiles. Alkynes undergo the same families of addition reactions as alkenes (hydrogenation, hydrohalogenation, halogenation, hydration, hydroboration-oxidation), but can stop at the alkene stage with the right reagents or proceed to doubly-added products with excess reagent.
Terminal alkyne
An alkyne in which the triple bond is at the end of the carbon chain (R–C≡C–H). The sp-hybridised C–H bond makes the terminal hydrogen mildly acidic (pKa ~25). Think of it as: the triple bond has an exposed hydrogen that can be plucked off by a strong enough base.
Internal alkyne
An alkyne in which the triple bond sits between two carbon groups (R–C≡C–R'). These lack the acidic terminal hydrogen and are generated from terminal alkynes by alkylation. In simple terms, both ends of the triple bond are connected to carbon, not hydrogen.
Acetylide anion
The conjugate base formed when a terminal alkyne is deprotonated (R–C≡C:⁻). It acts as a strong nucleophile and base, and is central to carbon-carbon bond-forming reactions. Think of it as: a carbon with a negative charge and a triple bond, ready to attack electrophiles.
Lindlar's catalyst
A "poisoned" palladium catalyst (Pd on CaCO₃ with lead acetate and quinoline) that reduces an alkyne to a cis-alkene and stops there. The catalyst surface is deactivated enough that it cannot reduce the resulting double bond further. In simple terms, it is a gentle enough catalyst to do half the job and then quit.
Tautomerisation
The interconversion of an enol (vinyl alcohol) and a keto form (carbonyl compound) through proton migration. In water with acid or base, the keto form is overwhelmingly favoured. Think of it as: the enol "flips" a proton from oxygen to carbon, landing on the more stable ketone.
Ozonolysis (of alkynes)
Cleavage of a triple bond using excess ozone (O₃), followed by a reductive workup (e.g. CH₃SCH₃), to produce carboxylic acids (or CO₂ from a terminal alkyne). In simple terms, ozone chops the triple bond and oxidises each fragment to its most oxidised carbonyl form.
Terminal alkynes have a pKa of ~25; compare with alkenes (~44) and alkanes (~50). The acidity comes from the sp-hybridised carbon holding electrons closer to the nucleus.
NaNH₂ (pKa of NH₃ ~38) is strong enough to fully deprotonate a terminal alkyne. NaOH and NaH are not strong enough in practice for clean deprotonation.
The resulting acetylide anion (R–C≡C:⁻ Na⁺) is a powerful carbon nucleophile.
From geminal or vicinal dihalides: Double elimination using NaNH₂ (2 equivalents) removes two molecules of HX to form an alkyne.
Alkylation of acetylides: Deprotonate a terminal alkyne with NaNH₂, then treat the acetylide with a primary alkyl halide (SN2). This builds new C–C bonds and converts terminal alkynes into internal alkynes.
Only primary (and methyl) halides work well. Secondary and tertiary halides undergo elimination instead.
Full reduction (alkyne → alkane): H₂ (excess) with Pd or Pt catalyst. Two equivalents of H₂ add across the triple bond via syn addition in two rounds.
Partial reduction to cis-alkene: H₂ with Lindlar's catalyst (Pd/CaCO₃, poisoned). Syn addition of one equivalent of H₂ stops at the cis (Z) alkene.
Partial reduction to trans-alkene: Dissolving metal reduction using Na⁰ in liquid NH₃. This proceeds through a radical anion intermediate and delivers the trans (E) alkene.
First equivalent of HBr adds across the triple bond following Markovnikov's rule, giving a vinyl halide.
Second equivalent of HBr adds to the vinyl halide (again Markovnikov), giving a geminal dihalide (both halogens on the same carbon).
Mechanism: protonation of the triple bond forms a vinyl carbocation; halide attacks.
First equivalent of Br₂ adds to give a trans-dibromoalkene (anti addition).
Second equivalent of Br₂ adds to give a tetrahalide.
Acid-catalysed hydration (H₂SO₄, HgSO₄): Water adds across the triple bond following Markovnikov's rule. The initial enol product tautomerises to a ketone (for internal alkynes) or a methyl ketone (for terminal alkynes).
Mechanism: mercury-assisted electrophilic addition of water, then tautomerisation.
Hydroboration-oxidation (1. BH₃·THF, 2. NaOH/H₂O₂): Anti-Markovnikov addition of water. For a terminal alkyne, the initial enol tautomerises to an aldehyde.
This is the only straightforward route from a terminal alkyne to an aldehyde.
Treatment with O₃ followed by reductive workup (e.g. CH₃SCH₃) cleaves the triple bond entirely, producing carboxylic acids from each fragment.
Useful for determining the position of the triple bond by identifying the cleavage products.
Starting from R–C≡C–H:
H₂, Pd → R–CH₂–CH₃ (alkane)
H₂, Lindlar's catalyst → cis (Z) alkene
Na⁰, NH₃ → trans (E) alkene
NaNH₂ then R'–X (1° only) → internal alkyne R–C≡C–R'
HBr (1 eq) → vinyl bromide; HBr (2 eq) → geminal dibromide
Br₂ (1 eq) → trans-dibromoalkene; Br₂ (2 eq) → tetrabromide
H₂SO₄/HgSO₄/H₂O → methyl ketone (Markovnikov)
BH₃·THF then NaOH/H₂O₂ → aldehyde (anti-Markovnikov)
O₃ then CH₃SCH₃ → carboxylic acids
NaNH₂ then epoxide, then H₃O⁺ → alcohol (nucleophilic ring opening)
pKa of terminal alkyne C–H: ~25
pKa of NH₃: ~38
C≡C bond length: ~1.20 Å
C=C bond length: ~1.33 Å
C–C bond length: ~1.54 Å
Acetylene (HC≡CH) is one of the simplest alkynes and is used industrially in oxyacetylene welding torches because of the enormous energy released when the triple bond burns. The acetylide alkylation strategy is used in pharmaceutical synthesis to extend carbon chains in a controlled, stereospecific way.
Students often think Lindlar's catalyst gives a trans-alkene. It does not. Lindlar's catalyst delivers syn addition, so the product is always cis (Z). For trans (E), you need dissolving metal reduction (Na/NH₃).
Students frequently confuse the hydration products: acid-catalysed (Markovnikov) hydration of a terminal alkyne gives a methyl ketone, not an aldehyde. Only hydroboration-oxidation (anti-Markovnikov) gives the aldehyde.
Students sometimes attempt to alkylate an acetylide with a secondary or tertiary halide. This fails because the acetylide is also a strong base, and elimination (E2) dominates over substitution with bulky electrophiles.
Tautomerisation is an equilibrium, not a resonance. The enol and keto forms are different molecules (constitutional isomers), not different drawings of the same molecule.
⚠️ Know which reagent set gives which alkene geometry: Lindlar's → cis (Z), Na/NH₃ → trans (E). This is a favourite exam question.
⚠️ Be able to draw the full mechanism for acid-catalysed hydration of a terminal alkyne, including the tautomerisation step.
⚠️ Acetylide alkylation is a critical C–C bond-forming reaction. Expect multi-step synthesis problems that require you to build a carbon skeleton using this tool.
⚠️ Ozonolysis products can be used "in reverse" to deduce the structure of an unknown alkyne. Practice working backwards from carboxylic acid fragments.
True or False: The pKa of a terminal alkyne (~25) is lower than that of an alkene (~44), making alkynes more acidic.
Fill in the blank: Hydroboration-oxidation of a terminal alkyne yields an __________ (aldehyde / ketone).
True or False: Lindlar's catalyst reduces an alkyne all the way to an alkane.
Fill in the blank: To convert a terminal alkyne into an internal alkyne, treat with NaNH₂ followed by a __________ alkyl halide.
True or False: The enol form of a carbonyl compound is the thermodynamically favoured tautomer in water.
Q: What product results from treating 1-pentyne with H₂SO₄/HgSO₄/H₂O?
A: 2-pentanone (a methyl ketone), via Markovnikov addition of water to form an enol that tautomerises to the ketone.
Q: Propose a two-step synthesis of 2-butyne from 1-butyne.
A: 1) Deprotonate with NaNH₂ to form the sodium acetylide. 2) Alkylate with CH₃I (methyl iodide) to give CH₃–C≡C–CH₃.
Q: What reagents convert an internal alkyne to a trans-alkene?
A: Na⁰ in liquid NH₃ (dissolving metal reduction).
Q: Draw the products of treating 1-hexyne with (a) one equivalent of HBr and (b) two equivalents of HBr.
A: (a) 2-bromo-1-hexene (Markovnikov vinyl bromide). (b) 2,2-dibromohexane (geminal dibromide, both bromines on C2).
Q: Why can you not use tert-butyl bromide to alkylate an acetylide?
A: The acetylide is a strong base. With a tertiary halide, E2 elimination dominates and you get an alkene instead of the desired C–C bond.
This material connects directly to alkene addition reactions (Ch. 7–8); every reaction here has a close alkene parallel, and exam questions often ask you to compare or chain them. Acetylide alkylation connects to SN2 reactivity and the steric requirements you learned in substitution/elimination (Ch. 6). The tautomerisation concept returns in carbonyl chemistry (Ch. 18–19) where keto-enol equilibria become a major theme.
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