Lesson 11 · Alkyne Reactions
Alkynes have two π bonds, so they often react twice. Almost every reaction in this chapter is either a question of how far you let the addition go, or a piece of chemistry that only terminal alkynes can do. Learn it in those two halves and the reagent list shrinks fast.
Rule to remember
Two π bonds means two chances to react. Always check how many equivalents the question gives you before you draw anything.
Learning goals
Key terms (click to reveal)
Part 01
A C≡C triple bond is one σ bond plus two π bonds. Both π bonds are electron rich and both can attack an electrophile, so an alkyne can run the same addition twice. The first addition converts the alkyne into an alkene. The second converts that alkene into a saturated product.
Two π bonds, two chances to add. Where the reaction stops depends on the reagent and the equivalents.
That means the exam question is usually not just "what adds?" but "how far does it go?" Three things control the answer. Equivalents: one equivalent of HBr stops at the vinyl halide, while excess HBr keeps going. Catalyst: ordinary Pd runs all the way to the alkane, while poisoned Lindlar palladium stops at the alkene. And mechanism: after hydration adds water once, the product tautomerizes into a carbonyl that simply cannot add again.
Big picture
Alkene chemistry asked what adds and where. Alkyne chemistry asks the same questions plus one more: does it happen once or twice?
Part 02
Every alkyne reaction in this course, tagged by bucket. Read the four distinctions that follow before you try to memorize any row: the color tags are the organizing idea, not decoration.
| Reaction | Reagents | Main product | Key point | Bucket |
|---|---|---|---|---|
| Full hydrogenation | H₂ (excess), Pd/C, Pt, or Ni | Alkane | Adds 2 equivalents of H₂, straight through the alkene stage | Reduction |
| Partial hydrogenation | H₂, Lindlar catalyst | cis alkene | Syn addition, stops after one equivalent | Reduction |
| Dissolving-metal reduction | Na, NH₃ (l) | trans alkene | Anti addition through radical intermediates | Reduction |
| Hydrohalogenation | HX (1 eq., then excess) | Vinyl halide, then geminal dihalide | Markovnikov both times, so both X end up on the same carbon | Addition |
| Halogenation | Br₂ or Cl₂ (1 eq., then excess) | Dihaloalkene, then tetrahalide | Anti addition each time; can add twice | Addition |
| Acid hydration | H₂O, H₂SO₄, HgSO₄ | Ketone (methyl ketone if terminal) | Markovnikov, enol tautomerizes to the keto form | Hydration |
| Hydroboration-oxidation | 1. R₂BH 2. H₂O₂, OH⁻ | Aldehyde (terminal) or ketone (internal) | Anti-Markovnikov, enol tautomerizes to the keto form | Hydration |
| Oxidative cleavage | O₃ then H₂O, or hot concentrated KMnO₄ | Two carboxylic acids | A terminal alkyne carbon becomes CO₂ | Cleavage |
| Acetylide formation | NaNH₂, NH₃ (l) | Acetylide anion | Terminal alkynes only, pKa about 25 | Terminal only |
| Acetylide alkylation | 1. NaNH₂ 2. R−X | Longer alkyne (internal, unless you start from acetylene) | New C−C bond; works with methyl and 1° halides only | Terminal only |
Part 03
Start with an internal alkyne, R−C≡C−R′. Three reagent sets take it to three different places, and exams test all three against each other constantly.
H₂ (excess), Pd/C
Full reduction
H₂, Lindlar
Lindlar = cis
Na, NH₃ (l)
Na/NH₃ = trans
Lindlar → cis. Na/NH₃ → trans. Plain H₂/Pd → all the way to the alkane.
The reason behind the split is worth a sentence each. On Lindlar, the alkyne lies flat on the poisoned metal surface and both hydrogens are delivered from that same face, so the two R groups stay on the same side: cis. With Na/NH₃, the alkyne picks up an electron to form a radical anion, and that intermediate settles into the arrangement where the bulky R groups sit as far apart as possible before it grabs a proton: trans.
Part 04
This is the highest-yield comparison in the chapter. Both routes add water across the triple bond. Both first make an enol. Both enols immediately tautomerize. The only difference is regiochemistry, and that difference changes the entire product class.
The enol is never the answer. It tautomerizes to the keto form every time.
H₂O, H₂SO₄, HgSO₄
Mercury hydration → ketone
Markovnikov: OH lands on the more substituted alkyne carbon. On a terminal alkyne R−C≡CH that is C2, so after tautomerization you get the methyl ketone R−CO−CH₃.
1. R₂BH 2. H₂O₂, OH⁻
Hydroboration → aldehyde
Anti-Markovnikov: boron, then OH, lands on the terminal carbon. After tautomerization you get the aldehyde R−CH₂−CHO. A hindered borane such as R₂BH is used so it only adds once.
High-yield rule
Terminal alkyne + Hg²⁺ hydration → methyl ketone. Terminal alkyne + hydroboration-oxidation → aldehyde.
One caution on internal alkynes: both routes give ketones there, and unless the alkyne is symmetric you will get a mixture of two ketones, because neither carbon is clearly favored. Exams that want a clean single product almost always hand you a terminal alkyne.
Part 05
With one equivalent of HBr, R−C≡CH gives a vinyl bromide: Markovnikov puts the Br on the more substituted carbon. With excess HBr the reaction runs again, and here is the part students get wrong: both bromines end up on the same carbon, giving R−CBr₂−CH₃, a geminal dihalide.
Markovnikov twice in a row puts both halogens on the same carbon: geminal, not vicinal.
The reason is the second protonation. It happens in the direction that gives the more stable cation, and the carbon already bearing a bromine can stabilize the positive charge with a bromine lone pair. So the second bromide attacks that same carbon. Geminal is the mechanistic consequence of Markovnikov applying twice, not a separate rule.
Halogenation behaves in a parallel way. One equivalent of Br₂ or Cl₂ gives the dihaloalkene, with anti addition placing the two halogens on opposite sides. Excess halogen adds across the remaining π bond as well, giving the tetrahalide with all four halogens on the two original alkyne carbons.
Oxidative cleavage sits at the far end of this section. O₃ followed by workup, or hot concentrated KMnO₄, snaps the triple bond entirely and gives two carboxylic acids. If the alkyne was terminal, that end carbon has no substituent to keep, so it leaves as CO₂ and you recover only one acid.
Part 06
This is the one thing alkenes cannot do. The C−H of a terminal alkyne has a pKa around 25, which is enormously more acidic than an alkene C−H (about 44) or an alkane C−H (about 50). The reason is hybridization: the alkyne carbon is sp, so its bonding electrons sit closer to the nucleus and the resulting carbanion holds its negative charge much more comfortably.
A strong enough base, NaNH₂, removes that proton and gives the acetylide ion R−C≡C⁻. Note that the base has to be strong: NaOH will not do it, because its conjugate acid, water (pKa 15.7), is a much stronger acid than the alkyne, so the equilibrium sits on the alkyne side. Amide works because its conjugate acid, ammonia (pKa 38), is a much weaker acid than the alkyne.
Deprotonate, then attack. The acetylide is the course's main carbon-carbon bond builder.
The acetylide is a strong carbon nucleophile, so it attacks an alkyl halide and forms a new C−C bond, lengthening the chain. That reaction is SN2, which brings its own restriction: it works with methyl and primary halides and fails on secondary and tertiary ones. With a hindered halide, the acetylide acts as a base instead and you get E2 elimination.
Why this matters later
Acetylide alkylation makes carbon skeletons bigger, and the reduction reactions then set the alkene geometry. Build the chain with an acetylide, then use Lindlar or Na/NH₃ to choose cis or trans. That two-move combination shows up all over the synthesis chapter.
Part 07
When a question hands you an alkyne and a reagent, sort the reagent into one of these four groups first. Most of the work is done at that point.
Bucket 1
Reduction
Bucket 2
Addition
Bucket 3
Hydration
Bucket 4
Terminal alkyne chemistry
Watch out
Stopping at the enol
An enol is an intermediate, not a product. Both hydration routes tautomerize to a carbonyl. If your answer has an OH on an alkene carbon, keep going.
Swapping Lindlar and Na/NH₃
Lindlar gives cis, Na/NH₃ gives trans. Tie each to its mechanism: surface delivery from one face is syn, while radical intermediates relax to the roomier anti arrangement.
Drawing a vicinal dihalide from excess HX
Two equivalents of HX give a geminal dihalide with both halogens on the same carbon. Vicinal (adjacent) is what X₂ gives, not HX.
Trying to make an acetylide from an internal alkyne
NaNH₂ needs an acidic terminal C−H. R−C≡C−R′ has none, so nothing happens. Check for a terminal hydrogen before writing an acetylide.
Using a 2° or 3° halide in acetylide alkylation
The alkylation step is SN2. Hindered halides give E2 elimination instead, since the acetylide is also a strong base. Methyl and 1° only.
Ignoring equivalents
One equivalent versus excess is often the entire question on HX and X₂ additions. Read the conditions line before you decide how far the reaction runs.
Forgetting CO₂ in terminal cleavage
Oxidative cleavage of a terminal alkyne gives one carboxylic acid plus CO₂, not two acids. The terminal carbon has no substituent to carry.
Expecting a clean product from internal alkyne hydration
Unless the internal alkyne is symmetric, hydration gives a mixture of two ketones. Do not force a single answer when the substrate does not allow one.
Checkpoint
You need cis-3-hexene from 3-hexyne. Which reagent, and what would you get if you used Na in liquid NH₃ instead?
H₂ with Lindlar catalyst. Na/NH₃ (l) would give trans-3-hexene instead.
Lindlar is a poisoned catalyst, so the alkyne lies on the surface and picks up two hydrogens from the same face before falling off: syn addition gives cis. Na/NH₃ goes through a radical anion that prefers the less crowded geometry, so the two R groups end up opposite each other: anti, giving trans.
1-hexyne is treated with H₂O, H₂SO₄, and HgSO₄. A classmate answers hex-1-en-2-ol. Why is that wrong, and what is the real product?
The enol tautomerizes. The product is 2-hexanone, a methyl ketone.
Markovnikov hydration puts the OH on the more substituted alkyne carbon (C2), giving an enol. Enols are not stable products: keto-enol tautomerization moves a proton and shifts the π bond to give the far more stable carbonyl. On a terminal alkyne this always lands you at a methyl ketone.
How would you convert 1-octyne into octanal? What single change in reagents would give you 2-octanone instead?
Use 1. R₂BH (a hindered borane such as disiamylborane), 2. H₂O₂/OH⁻. Switching to H₂O/H₂SO₄/HgSO₄ gives 2-octanone.
Hydroboration is anti-Markovnikov, so boron and eventually OH end up on the terminal carbon. That enol tautomerizes to an aldehyde. Mercury-catalyzed hydration is Markovnikov, so the OH goes to C2 and the enol tautomerizes to the methyl ketone. Same net addition of water, opposite regiochemistry, two different carbonyl products.
Propyne reacts with excess HBr. Where do the two bromines end up, and what is that product class called?
Both bromines land on C2, giving 2,2-dibromopropane: a geminal dihalide.
The first addition is Markovnikov, so Br goes to C2 and you get a vinyl bromide. In the second addition, protonation again happens in the direction that gives the more stable cation, which is the one stabilized by the bromine lone pair already on C2. So the second Br joins the first on the same carbon.
You want to make 2-hexyne from acetylene (ethyne). Sketch the sequence, and explain why the alkylation steps must use primary halides.
1. NaNH₂, then CH₃CH₂CH₂Br gives 1-pentyne. 2. NaNH₂ again, then CH₃I gives 2-hexyne.
NaNH₂ deprotonates the terminal alkyne to make an acetylide, which then attacks the alkyl halide in an SN2 reaction and forms a new C−C bond. Because the mechanism is SN2, it needs an unhindered electrophile: methyl and 1° halides work, while 2° and 3° halides mostly undergo E2 instead, since the acetylide is also a strong base.
Practice workbook
Product prediction, reagent selection, cis versus trans control, and multi-step acetylide synthesis problems, organized by the four buckets.
Lesson summary