Difficulty: Intermediate | Prerequisites: Alkene reactions, Markovnikov's rule, acid-base chemistry (pKa of terminal alkynes), E2 elimination
Alkyne chemistry builds on everything you have learnt about alkenes and extends it to triple bonds. The reactivity patterns are similar in principle but differ in important details, particularly around regiochemistry and the number of equivalents of reagent needed. This worksheet also introduces alkyne preparation from alkenes via dihalides. If you are comfortable with alkene additions and E2 eliminations, you are ready.
Alkynes are named using IUPAC rules similar to alkenes but with the "-yne" suffix. They can be prepared from alkenes via bromination followed by double elimination with NaNH2. Their key reactions include acid-catalysed hydration (Markovnikov, giving ketones via enol intermediates), dissolving metal reduction (Na/NH3, giving trans-alkenes), HX addition, ozonolysis, and catalytic hydrogenation.
Alkyne
A hydrocarbon containing a carbon-carbon triple bond (one sigma bond and two pi bonds). Alkynes are linear at the triple bond carbons (sp hybridisation, 180° bond angles).
In simple terms, the triple bond makes the molecule rigid and rod-like at those two carbons.
Terminal alkyne
An alkyne with the triple bond at the end of the chain (one hydrogen attached to a triple-bond carbon). Terminal alkynes have a weakly acidic C–H (pKa ~25), which matters for acetylide chemistry.
Internal alkyne
An alkyne with the triple bond between two carbon substituents, not at the end of the chain.
Dissolving metal reduction (Na/NH3 or Li/NH3)
Reduction of an alkyne to a trans-alkene using an alkali metal in liquid ammonia. The mechanism involves sequential electron additions and protonations, and the geometry is controlled by the preference for the more stable trans-vinyl anion intermediate.
Think of it as the one reliable way to get a trans-alkene from an alkyne.
Markovnikov hydration of alkynes
Addition of water across a triple bond using H2O/H2SO4/HgSO4 (mercury(II) sulfate catalyst). Water adds with Markovnikov regiochemistry, placing the OH on the more substituted carbon. The initial product is an enol, which tautomerises to a ketone.
Enol
An unstable intermediate in which a hydroxyl group is attached to a doubly bonded carbon ("ene" + "ol"). Enols rapidly tautomerise to the more stable keto form under acidic or basic conditions.
Keto-enol tautomerisation
The equilibrium interconversion between an enol and a ketone (or aldehyde). The keto form is heavily favoured at equilibrium for simple cases.
Vinyl halide
A halogen bonded directly to a doubly bonded carbon (C=C–X). Formed by addition of one equivalent of HX to an alkyne.
Double dehydrohalogenation
Two successive E2 eliminations from a vicinal dihalide (or geminal dihalide) using a strong base such as NaNH2. The first elimination gives a vinyl halide; the second gives the alkyne. This is the standard method for preparing alkynes from alkenes.
Naming alkynes follows the same logic as alkenes, with "-yne" replacing "-ene":
Find the longest carbon chain that includes the triple bond
Number from the end that gives the triple bond the lowest locant
Name substituents with their position numbers
If both a double and triple bond are present, use "-en-...-yne" and give the triple bond the lower number when there is a tie
Examples from the worksheet:
4-methyl-1,5-octadiyne: an eight-carbon chain with triple bonds at positions 1 and 5, and a methyl group at position 4
3-butyl-2-octyne: an eight-carbon chain with a triple bond at position 2 and a butyl group at position 3
2,6-dimethyl-3-octyne: an eight-carbon chain with a triple bond at position 3 and methyl groups at positions 2 and 6
The standard route from an alkene to an alkyne:
Brominate the alkene with Br2 to give a vicinal dibromide (anti addition)
Treat the dibromide with 2 equivalents of NaNH2 (sodium amide, a very strong base)
Two successive E2 eliminations remove both HBr molecules, first giving a vinyl bromide, then the alkyne
From the worksheet (Q4): an alkene is brominated to give a vicinal dibromide, then treated with NaNH2 to produce 2,6-dimethyl-3-octyne.
Key point: NaNH2 is strong enough to deprotonate the terminal alkyne (pKa ~25) and also to drive the second elimination. Weaker bases like KOH or NaOEt typically stop at the vinyl halide stage.
Addition of HCl (one equivalent):
Follows Markovnikov's rule: Cl adds to the more substituted carbon
Product is a vinyl chloride (a haloalkene)
With two equivalents, a geminal dihalide forms
Dissolving metal reduction (Na/NH3):
Converts an internal alkyne to a trans-alkene
The mechanism proceeds through a vinyl radical and then a vinyl anion; the trans geometry is preferred because the bulky groups adopt an anti arrangement in the anion
This is the complement of catalytic hydrogenation, which gives the cis-alkene
Ozonolysis of alkynes (O3, then (CH3)2S):
Cleaves the triple bond to give carboxylic acids (with oxidative workup) or, with the milder DMS workup shown in the worksheet, can give carbonyl fragments
From the worksheet: an internal alkyne treated with O3/(CH3)2S gives a carbonyl product (aldehyde)
Catalytic hydrogenation (H2/Pd):
Reduces the alkyne to a cis-alkene (syn addition of H2 from the metal surface)
With excess H2 or a more active catalyst, full reduction to the alkane occurs
Lindlar's catalyst (Pd/CaCO3, poisoned with lead acetate and quinoline) stops cleanly at the cis-alkene
Acid-catalysed hydration (H2O/H2SO4/HgSO4):
Markovnikov addition of water across the triple bond
The initial product is an enol (OH on a doubly bonded carbon)
The enol tautomerises to the more stable ketone
For a terminal alkyne, the ketone is always a methyl ketone (the oxygen ends up on C2)
This is the mechanism the worksheet asks you to draw (the rimantadine synthesis step):
The Hg2+ ion coordinates to the triple bond, activating it toward nucleophilic attack
Water attacks the more substituted carbon (Markovnikov), forming a vinyl alcohol (enol) bound to mercury
Proton transfer and loss of the mercury catalyst regenerate Hg2+ and give the free enol
Keto-enol tautomerisation: the enol's OH proton is lost, electrons shift, and a proton is picked up at the terminal carbon, converting C=C–OH into C–C=O (the ketone)
The net transformation: alkyne + H2O → enol → ketone. Mercury acts as a catalyst and is regenerated.
Students often write the hydration product of a terminal alkyne as an aldehyde. Markovnikov addition places the OH on the internal carbon (C2), so after tautomerisation you always get a methyl ketone, not an aldehyde.
Students confuse dissolving metal reduction (Na/NH3 → trans-alkene) with catalytic hydrogenation (H2/Pd or Lindlar → cis-alkene). The stereochemical outcomes are opposite.
When naming alkynes, students sometimes number the chain from the wrong end. The triple bond must receive the lowest possible locant, just like the double bond in alkene naming.
Students forget that double dehydrohalogenation with NaNH2 requires two equivalents of base. One equivalent removes the first HBr to give a vinyl halide; the second removes the second HBr to give the alkyne.
⚠️ The acid-catalysed hydration mechanism (H2O/H2SO4/HgSO4) is a favourite exam question. Be ready to draw the full arrow-pushing mechanism, including the enol intermediate and the tautomerisation step.
⚠️ Know the difference between cis and trans alkene products from alkynes: H2/Lindlar gives cis; Na/NH3 gives trans. This is tested constantly.
⚠️ Alkyne preparation questions typically give you a starting alkene and ask you to show the full sequence: bromination, then double elimination with NaNH2. Be able to draw the vicinal dibromide intermediate.
⚠️ IUPAC naming of alkynes appears regularly. Pay attention to longest chain selection when branching is present, as in 3-butyl-2-octyne (the longest chain through the triple bond is eight carbons, not the longest chain overall).
⚠️ The rimantadine synthesis (Q6) is a real-world application. Exam questions sometimes frame a reaction in a pharmaceutical context and ask for the mechanism of one step.
Fill in the blank: Acid-catalysed hydration of a terminal alkyne produces a ______ after tautomerisation.
Answer: methyl ketone
True or False: Na/NH3 reduction of an internal alkyne gives a cis-alkene.
Answer: False. Na/NH3 gives a trans-alkene. Lindlar's catalyst (or H2/Pd-BaSO4) gives cis.
Fill in the blank: The first step in preparing an alkyne from an alkene is ______ with Br2.
Answer: bromination (forming a vicinal dibromide)
True or False: In IUPAC naming, the triple bond in an alkyne receives the lowest possible locant.
Answer: True.
Fill in the blank: The unstable intermediate formed during Markovnikov hydration of an alkyne, before tautomerisation, is called an ______.
Answer: enol
Q: Show how you would convert 2-methylbut-2-ene into an alkyne. Name the product.
A: Treat the alkene with Br2 to form the vicinal dibromide (anti addition). Then add 2 equivalents of NaNH2 to perform double dehydrohalogenation. The product is 2-methylbut-2-yne (an internal alkyne).
Q: What is the product of treating 1-hexyne with H2O/H2SO4/HgSO4?
A: Markovnikov hydration places the OH on C2 (the internal carbon). The resulting enol tautomerises to 2-hexanone (a methyl ketone).
Q: Draw the mechanism for the acid-catalysed hydration of a terminal alkyne to form a ketone.
A: (1) Hg2+ coordinates to the triple bond. (2) Water attacks the more substituted carbon (Markovnikov). (3) Proton transfer and loss of Hg2+ gives the enol. (4) Tautomerisation: protonation at the terminal carbon and loss of the OH proton converts the enol to the ketone.
Q: You want a trans-alkene from 3-hexyne. What reagents do you use?
A: Na (or Li) in liquid NH3 (dissolving metal reduction). This gives exclusively the trans-3-hexene.
Q: An alkyne is treated with O3 followed by (CH3)2S. The products are an aldehyde and a ketone fragment. What can you deduce about the original alkyne?
A: The alkyne was internal (not terminal, since both fragments are carbonyl compounds). One side of the triple bond had at least one hydrogen (giving the aldehyde), and the other was fully substituted at the triple-bond carbon (giving the ketone).
Q: Give the IUPAC name for an eight-carbon alkyne with the triple bond at C3 and methyl groups at C2 and C6.
A: 2,6-dimethyl-3-octyne.
Alkyne reactions, triple bond chemistry, IUPAC alkyne naming, terminal alkyne, internal alkyne, hydration of alkynes, Markovnikov hydration, HgSO4 catalyst, mercury-catalysed hydration, enol, keto-enol tautomerism, tautomerisation, dissolving metal reduction, Na NH3 reduction, Birch reduction (related), trans-alkene from alkyne, Lindlar catalyst, cis-alkene from alkyne, ozonolysis of alkynes, double dehydrohalogenation, NaNH2 elimination, vinyl halide, vinyl chloride, vicinal dibromide, alkyne preparation, rimantadine synthesis, adamantane, CHM 255 Purdue, PSO Worksheet 7
Connections to other topics: Alkyne hydration connects to carbonyl chemistry later in the course, since the ketone products are the starting materials for nucleophilic addition, aldol reactions, and more. The enol intermediate reappears in enolate chemistry. Dissolving metal reduction and catalytic hydrogenation connect back to alkene stereochemistry (covered in the companion alkene notes). Acetylide anion chemistry (from terminal alkynes deprotonated by NaNH2) is a key tool in carbon-carbon bond formation and will appear in synthesis problems throughout the rest of the course.