Source: Organic Chemistry I, University of Minnesota Twin Cities
Tags: alkyne reactions, hydrohalogenation, halogenation, hydrogenation, Lindlar catalyst, hydration, keto-enol tautomerization, ozonolysis, alkylation, elimination, alkyne synthesis
Difficulty: Intermediate | Prerequisites: Part 1 of these notes (alkyne structure, sp hybridisation, acidity), Ch. 8 (alkene addition reactions, Markovnikov’s rule, stereoselectivity).
Alkyne reactions are largely the triple-bond versions of the alkene reactions from Ch. 8. Because a triple bond can undergo one or two successive additions, most reactions here come with a "1 eq vs. 2 eq" decision that controls whether you stop at an alkene intermediate or push through to a fully saturated product. The chapter also introduces two pieces of chemistry that are unique to alkynes: keto-enol tautomerization (hydration gives a ketone, not an alcohol) and alkylation (terminal alkynes form new C–C bonds). These make alkynes powerful building blocks in synthesis. If the alkene addition reactions from Ch. 8 are solid, this is a manageable step up.
Alkynes undergo addition reactions (hydrohalogenation, halogenation, hydrogenation, hydration, ozonolysis) that parallel alkene chemistry, with the key twist that you can often stop at one equivalent to get a substituted alkene or push to two equivalents for the fully saturated product. Hydration of alkynes produces ketones through keto-enol tautomerization rather than alcohols. Alkynes can also be synthesised by double E2 elimination, and terminal alkynes can be alkylated to build longer carbon chains via Sₙ₂.
Hydrohalogenation (of Alkynes)
Addition of HX (HCl, HBr, HI) across the triple bond. One equivalent gives a vinyl halide; two equivalents (or excess) gives a geminal dihalide. Follows Markovnikov’s rule.
Halogenation (of Alkynes)
Addition of X₂ (Cl₂, Br₂, I₂) across the triple bond. One equivalent gives a trans-dihaloalkene (anti addition); two equivalents gives a tetrahalide.
Lindlar Catalyst
Pd/CaCO₃ poisoned with Pb. Specifically deactivated so it catalyses only the first hydrogenation step (triple bond to cis-alkene), then stops. Other catalysts (Pd/C, Pt, Ni, Rh) reduce all the way to the alkane.
Keto-Enol Tautomerization
The acid-catalysed equilibrium between an enol (C=C–OH) and a ketone (CH–C=O). The ketone form is thermodynamically favoured. Think of it as a proton shuffle: the OH proton migrates so the double bond shifts from C=C to C=O.
Enol
An intermediate with an –OH group on a carbon–carbon double bond. Unstable relative to the corresponding ketone and tautomerizes rapidly under acidic conditions.
Alkylation (of Terminal Alkynes)
A two-step sequence: (1) deprotonation of the terminal alkyne with NaNH₂ to form the acetylide anion, then (2) Sₙ₂ reaction with a methyl or primary alkyl halide. Builds a new C–C bond.
E2 Elimination (to Form Alkynes)
Double dehydrohalogenation using NaNH₂. A dihalide (vicinal or geminal) undergoes two E2 eliminations in succession. If a terminal alkyne forms, NaNH₂ deprotonates it; H₂O is added at the end to re-protonate.
C≡C → H–C=C–X → CH₂–CX₂
Reagent: HCl, HBr, or HI (1 eq or 2 eq / excess)
Selectivity: Markovnikov, not stereoselective
1 eq gives a vinyl halide (one addition only)
2 eq or excess pushes through to a geminal dihalide
The second addition is slower than the first
Similar to hydrohalogenation of alkenes
First step forms an sp-hybridised carbocation intermediate, which is less stable than the sp² carbocation from alkenes, so the reaction is slower
The first addition is Markovnikov-selective
The second carbocation intermediate sits next to the halogen, which destabilises it by electronegativity but stabilises it by resonance. Net result: stabilised overall. Also Markovnikov-selective.
C≡C → X–C=C–X → CX₂–CX₂
Reagent: Cl₂, Br₂, or I₂ (1 eq or 2 eq / excess)
Selectivity: anti (via halonium intermediate). Regioselectivity is not applicable since both atoms adding are the same.
1 eq gives a trans-dihaloalkene (one addition, anti)
2 eq or excess gives a tetrahalide
The second addition is slower than the first, so you can reliably stop at one equivalent
Mechanism is analogous to halogenation of alkenes, proceeding through a halonium ion intermediate
C≡C → H–C=C–H → CH₂–CH₂
Reagent: H₂ with a metal catalyst
Selectivity: syn addition. Regioselectivity not applicable (both atoms are H).
Lindlar catalyst (Pd/CaCO₃ + Pb): specifically deactivated so it performs only the first addition. Reduces a triple bond to a cis-alkene and stops.
Pd/C, Pt, Ni, Rh: these catalyse both additions and cannot be stopped after the first. The product is a fully saturated alkane.
This distinction is one of the most testable points in the chapter. If the question asks you to make a cis-alkene from an alkyne, the answer is H₂ with Lindlar catalyst. If it asks for the alkane, use Pd/C or Pt.
Mechanism is not covered in this course.
C≡C → [H–C=C–OH] → CH₂–C=O
Reagent: H₂O, Hg²⁺, H₂SO₄
Hg²⁺ and H₂SO₄ are catalytic (not consumed)
Selectivity: Markovnikov, not stereoselective
The initial product is an enol (vinyl alcohol, H–C=C–OH), which is unstable
The enol immediately tautomerizes to a ketone (CH–C=O) under the acidic conditions
Catalysed by any acid (e.g., H₂SO₄)
The equilibrium strongly favours the ketone (more stable)
Proceeds through a carbocation intermediate positioned next to the oxygen
This carbocation is destabilised by the electronegativity of oxygen but very effectively stabilised by resonance donation from the oxygen lone pairs
Net result: a well-stabilised intermediate
This is a key "twist" compared to alkene hydration: alkene hydration gives an alcohol, but alkyne hydration gives a ketone.
C≡C → HO–C=O + O=C–OH
Reagent: O₃ (step 1), then H₂O (step 2)
Selectivity: not applicable (the triple bond is cleaved entirely)
Analogous to ozonolysis of alkenes, but no reductant (e.g., Zn or DMS) is needed in the second step
The triple bond is broken completely, producing two carboxylic acids
Mechanism is not covered in this course
XHC–CHX or X₂C–CH₂ → [X–C=C–H] → C≡C
Reagent: NaNH₂, then H₂O
Selectivity: not applicable
This is a double E2 elimination: two consecutive eliminations in one pot
NH₂⁻ is a very strong base, so E2 is fast
The reaction does not stop after the first elimination; both E2 steps proceed
If a terminal alkyne forms, NaNH₂ deprotonates it (recall pKa ~26 vs. ~38 for NH₃), pulling the terminal alkyne out of the reaction as the acetylide anion
Water (H₂O) is added at the end to re-protonate the acetylide back to the terminal alkyne
Typically used to make terminal alkynes from vicinal or geminal dihalides
This reaction works on both vicinal dihalides (X on adjacent carbons) and geminal dihalides (both X on the same carbon).
C≡C–H + X–CH₂R → C≡C–CH₂R
Reagent: (1) NaNH₂, then (2) alkyl halide (R–X)
Only works with methyl or primary (1°) alkyl halides
Step 1 (acid–base): NaNH₂ deprotonates the terminal alkyne to form the acetylide anion (⊖C≡C–)
Step 2 (Sₙ₂2): the acetylide anion acts as a nucleophile, attacking the methyl or primary alkyl halide in a backside displacement
The acetylide anion is both a very good nucleophile and a very strong base. With methyl and 1° substrates, Sₙ₂2 dominates. With 2° and 3° substrates, E2 elimination outcompetes Sₙ₂2 because the acetylide’s basicity takes over, and you get an alkene instead of the desired alkylation product.
Alkynes are versatile synthetic intermediates. The chapter highlights three unique things alkynes let you do.
New C–C bonds: via alkylation, adding carbons to the chain
Cis-alkenes: via hydrogenation with Lindlar catalyst (this is significant because standard E2 elimination preferentially makes trans-alkenes)
1,1-Dihaloalkanes and ketones: via hydrohalogenation (2 eq) or hydration, respectively
From dihalides: double E2 elimination with NaNH₂, then H₂O
From shorter alkynes: alkylation extends the carbon chain
From alkenes: bromination (Br₂ addition to give a vicinal dibromide), followed by double E2 elimination with NaNH₂
Synthesis problems on exams typically combine these tools: make an alkyne, extend it by alkylation, then convert it to a cis-alkene or a ketone.
The Lindlar reduction (alkyne to cis-alkene) is used in the industrial synthesis of vitamin A and other compounds where cis-alkene geometry is required. Alkylation of terminal alkynes is a staple of pharmaceutical retrosynthesis: the acetylide anion’s ability to form C–C bonds makes it a workhorse for building carbon skeletons. Keto-enol tautomerization is not limited to alkyne hydration; it underlies many biological and synthetic processes, including the chemistry of enolates in advanced organic chemistry.
"Hydration of an alkyne gives an alcohol, just like hydration of an alkene." It does not. The initial enol product tautomerizes to a ketone. You end up with a C=O, not a C–OH.
"Lindlar catalyst fully reduces a triple bond to an alkane." Lindlar catalyst is specifically deactivated to stop at the cis-alkene stage. To get the alkane, you need Pd/C, Pt, or a similar unmodified catalyst.
"Alkylation works with secondary and tertiary alkyl halides." It does not. The acetylide is such a strong base that E2 elimination dominates with 2° and 3° halides. Only methyl and 1° halides give the Sₙ₂2 product.
"The double elimination with NaNH₂ stops after one E2." It does not stop. NaNH₂ is a strong enough base to drive both E2 eliminations in sequence. If a terminal alkyne forms, it even gets deprotonated, requiring water at the end to re-protonate it.
⚠️ Know the reagent and selectivity for every reaction: hydrohalogenation (HX, Markovnikov), halogenation (X₂, anti), hydrogenation (H₂ + catalyst, syn), hydration (H₂O/Hg²⁺/H₂SO₄, Markovnikov), ozonolysis (O₃ then H₂O)
⚠️ Distinguish 1 eq vs. 2 eq outcomes for hydrohalogenation and halogenation
⚠️ Lindlar catalyst vs. other catalysts: this is a classic exam question. Lindlar = cis-alkene, Pd/C or Pt = alkane.
⚠️ Hydration gives a ketone (via tautomerization), not an alcohol. Exam problems love to test this.
⚠️ Alkylation only works with Me/1° halides; be prepared to explain why (Sₙ₂2 vs. E2 competition)
⚠️ Multi-step synthesis problems: expect to chain together elimination → alkylation → Lindlar reduction or hydration
True or False: Hydration of a terminal alkyne with H₂O/Hg²⁺/H₂SO₄ gives an alcohol. (False – gives a ketone via tautomerization)
Lindlar catalyst reduces an alkyne to a _____ with _____ stereochemistry. (cis-alkene, syn)
True or False: Alkylation of a terminal alkyne works with 2° alkyl halides. (False – E2 dominates)
Halogenation of an alkyne with 1 eq Br₂ gives _____ addition. (anti)
True or False: Double E2 elimination with NaNH₂ stops after one elimination. (False – both eliminations proceed)
Q: What product results from treating propyne with 1 eq of HBr? What about with excess HBr?
A: With 1 eq: 2-bromopropene (Markovnikov addition, Br goes to the more substituted carbon). With excess: 2,2-dibromopropane (geminal dihalide, both Br on the same carbon).
Q: You need to convert an internal alkyne into a cis-alkene. What reagent and catalyst do you use?
A: H₂ with Lindlar catalyst (Pd/CaCO₃/Pb). This gives syn addition and stops at the cis-alkene.
Q: Why does hydration of an alkyne produce a ketone rather than an alcohol?
A: The initial addition follows Markovnikov’s rule, placing the OH on the more substituted carbon of the resulting double bond. This enol intermediate (C=C–OH) is unstable and undergoes keto-enol tautomerization to give the more stable ketone (C=O).
Q: Propose a two-step synthesis of pent-2-yne from propyne.
A: (1) Treat propyne with NaNH₂ to deprotonate the terminal C–H and form the acetylide anion. (2) React the acetylide with bromoethane (CH₃CH₂Br, a primary alkyl halide) via Sₙ₂2. The product is pent-2-yne.
Q: Explain why you cannot alkylate a terminal alkyne with 2-bromopropane.
A: 2-Bromopropane is a secondary (2°) alkyl halide. The acetylide anion is a strong base as well as a good nucleophile. With a 2° substrate, E2 elimination outcompetes Sₙ₂2, so you get propene as the major product instead of the desired alkylation product.
Q: Starting from cyclohexene, how would you make an alkyne? Outline the steps.
A: (1) Add Br₂ to cyclohexene to get trans-1,2-dibromocyclohexane (anti addition). (2) Treat with NaNH₂ (excess) to perform double E2 elimination, then add H₂O to protonate. The product is cyclohexyne (though note: small-ring cycloalkynes are very strained; this sequence works better with larger or acyclic substrates).
Every addition reaction here parallels its alkene counterpart from Ch. 8, so mastering the comparison table (same reagent, similar mechanism, but one vs. two additions possible) ties the two chapters together. Keto-enol tautomerization reappears heavily in later chapters on carbonyl chemistry (enolate reactions, aldol condensation). Alkylation via the acetylide anion is one of the first C–C bond-forming reactions in the course, and the logic of deprotonation followed by nucleophilic attack will return with enolates and organometallic reagents (Grignard, organolithium) later in the course.
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