Start with propyne, CH₃–C≡C–H: three carbons, with a triple bond between the second and the third. A carbon–carbon triple bond is what makes a compound an alkyne. When the triple bond sits at the end of the chain, as in propyne, the last carbon carries a hydrogen, and the compound is a terminal alkyne. When both triple-bond carbons carry carbon groups, as in but-2-yne, CH₃–C≡C–CH₃, it is an internal alkyne.
Reagents add to alkynes just as they add to alkenes, and because an alkyne has two pi bonds, they can add twice. In synthesis, though, alkynes matter more for a different reason. The hydrogen on a terminal alkyne is acidic enough for a strong base to remove, and the anion left behind is one of the very few reliable carbon nucleophiles available at this stage. That anion is how you make a carbon skeleton bigger.
sp hybridization and linear geometry
Each carbon of a triple bond has two groups around it: the other triple-bond carbon and one more atom. So it is sp hybridized. Its two sp orbitals point 180° apart and form its two sigma bonds. That leaves two unhybridized p orbitals on each carbon, at right angles to each other and to the bond axis.
Each p orbital overlaps side by side with the matching p orbital on the other carbon. One pair forms a pi bond above and below the axis; the other pair forms a second pi bond in front of and behind it. A triple bond is therefore one sigma bond plus two separate pi bonds. The Bonding section in Foundations takes ethyne apart the same way.
Because the two sp orbitals point in opposite directions, the two triple-bond carbons and the two atoms attached to them lie on one straight line. In propyne, the CH₃ carbon, both sp carbons and the terminal hydrogen sit in a row. The C≡C bond is about 120 pm long, and breaking all three of its bonds takes roughly 200 kcal/mol.
A straight line is hard to fit into a ring. In a ring of fewer than about eight carbons, a triple bond is severely strained. Cyclooctyne, the smallest cyclic alkyne that can be kept in a bottle, still has its C–C≡C angles bent well short of 180°.
Compare but-1-yne and buta-1,3-diene. Both are C₄H₆: one has a triple bond, the other two double bonds, and the formula cannot tell them apart. That is because a triple bond counts as two degrees of unsaturation (see Alkene structure, earlier in this chapter), the same as two double bonds, or as one ring plus one double bond, as in cyclobutene. Telling such isomers apart is exactly the kind of job IR & Mass Spectrometry, the next chapter, is for.
Terminal alkyne C–H: unusually acidic
Compare three C–H bonds whose carbons differ only in hybridization. The C–H of a terminal alkyne has a pKa of about 25. An alkene C–H sits near 44, and an alkane C–H near 50, which no ordinary base can touch. Each pKa unit is a factor of ten, so the alkyne C–H is nineteen orders of magnitude more acidic than the alkene C–H, from nothing but hybridization.
This is the Orbital factor, the O of the ARIO framework in Acids & Bases: it asks what kind of orbital holds the charge in the conjugate base. When the proton leaves, the bonding pair stays on carbon, in an sp orbital. An sp orbital is 50% s in character, against 33% for sp² and 25% for sp³. More s-character holds the pair closer to the nucleus, which lowers the anion's energy and makes the parent C–H more acidic.
Removing that proton takes a base whose conjugate acid is a weaker acid than the alkyne. An acid–base equilibrium favors the side with the weaker acid, the one with the higher pKa. Sodium amide, NaNH₂, qualifies easily: its conjugate acid, ammonia, has a pKa of 38, so the equilibrium lies about 10¹³ to one toward the anion. Sodium hydride, NaH, works too: its conjugate acid, H₂, has a pKa of about 36.
Acetylide anions: making carbon–carbon bonds
The anion made by deprotonating a terminal alkyne is an acetylide anion. It is a carbanion, a carbon that carries a lone pair and a negative charge, and it is small and strongly nucleophilic. It does what any strong nucleophile does in Substitution & Elimination: an SN2 on an unhindered alkyl halide. The bond it forms is a new carbon–carbon bond, and attaching an alkyl group this way is called alkylation.
Make hex-1-yne from acetylene (ethyne), HC≡CH.
Step 1: NaNH₂ removes one of acetylene's two protons, giving sodium acetylide, HC≡C⁻ Na⁺.
Step 2: add 1-bromobutane. The acetylide carbon attacks the CH₂ that carries the bromine, from the side opposite the bromine, and bromide leaves. The product is HC≡C–CH₂CH₂CH₂CH₃, hex-1-yne.
Two carbons plus four carbons, joined by a C–C bond you made. The product still has a terminal C–H, so you can run the same two steps again on it. That is how you build an internal alkyne with whatever two groups you like on either end.
This matters more than it may look. Most of the reactions you have met so far rearrange or decorate a skeleton that already exists; this one makes the skeleton longer. Acetylide alkylation, together with the enolate and organometallic chemistry of later chapters, is one of a short list of dependable ways to build carbon frameworks.
Making the triple bond: double dehydrohalogenation
Acetylide alkylation lengthens a chain that already contains a triple bond. The first triple bond comes from elimination, run twice. Take 2,3-dibromobutane, a vicinal dihalide: its two halogens sit on adjacent carbons. Treat it with sodium amide, NaNH₂. In an E2, the amide ion removes an H from one carbon while Br⁻ leaves from the next.
Losing HX this way is called dehydrohalogenation. The product is a C=C with a bromine still on one of its carbons, and a compound with a halogen on a C=C carbon is a vinyl halide. A second E2 removes another HBr from the vinyl halide and gives but-2-yne.
A geminal dihalide, with both halogens on the same carbon, gives an alkyne by the same two eliminations. The first again gives a vinyl halide, and the second gives the triple bond.
The second elimination is the hard one. In a vinyl halide the leaving group sits on an sp² carbon, and the hydrogen that must go sits on the other alkene carbon. That makes it a much less willing substrate than an ordinary alkyl halide, which is why the reagent is NaNH₂, usually with heat, and not KOH in ethanol.
Vicinal dihalides come straight from an alkene plus Br₂, as the Addition reactions section of this chapter showed. So the pair of steps Br₂, then excess NaNH₂ turns an alkene into an alkyne.
Addition reactions: partial versus complete
Two pi bonds mean that addition can happen once, giving an alkene, or twice, giving a product with no pi bond left. For HX and X₂, both electrophilic additions you met earlier in this chapter, the number of equivalents you supply usually decides which.
HX. Take but-1-yne and HBr. One equivalent adds across one pi bond with Markovnikov selectivity: the H goes to the end carbon, C1, and the bromine to the more substituted carbon, C2. The product, 2-bromobut-1-ene, is a vinyl halide. A second equivalent adds to that alkene, and Markovnikov again puts the second bromine on C2, the carbon that already has one. The product is a geminal dihalide, 2,2-dibromobutane, not 2,3-dibromobutane.
The reason is the bromine already on C2. When the second H⁺ adds to C1, the positive charge lands on C2, the carbon that already carries that bromine. A lone pair on the bromine can form a pi bond to the cationic carbon, which shares out the charge in a second resonance form. Adding the H⁺ to C2 instead would leave a primary cation on C1 with no such help.
X₂. Take but-2-yne and Br₂. One equivalent gives 2,3-dibromobut-2-ene with the two bromines trans, on opposite sides of the new C=C. The anti addition comes from the same bridged-ion opening that makes alkene bromination anti: bromine bridges both carbons, and bromide attacks from the side opposite the bridge. A second equivalent adds across the remaining pi bond and gives the tetrahalide, 2,2,3,3-tetrabromobutane.
The dihaloalkene is less reactive than the alkyne it came from, because its two halogens withdraw electron density from the remaining pi bond. So one equivalent of halogen, added in the cold, stops cleanly at the dihaloalkene.
Reduction breaks both of those patterns. None of its reagents adds as an electrophile, and the reagent, not the number of equivalents, decides where the reaction stops. The Catalytic hydrogenation section of this chapter covers the three choices in full: H₂ over Lindlar catalyst stops at the alkene and delivers both hydrogens from one metal surface to one face, giving the cis alkene; Na in liquid NH₃ gives the trans alkene; H₂ over ordinary Pd or Pt does not stop at the alkene at all and runs on to the alkane.
Sodium in ammonia releases solvated electrons, electrons held in the solution by the ammonia around them. The alkyne takes one electron, then a proton from ammonia, then a second electron and a second proton. For but-2-yne, the first electron and proton give a vinyl radical. The second electron turns that radical into a vinyl anion. The vinyl radical flips between its two bent shapes very fast, and it spends most of its time in the trans shape, where the two methyls are apart. The vinyl anion flips much more slowly, so it keeps the trans shape until ammonia protonates it. The product is trans-but-2-ene.
So a single internal alkyne can become the cis alkene, the trans alkene or the alkane, purely by reagent choice. That is a large part of why alkynes are so useful as synthetic intermediates, since alkene geometry is otherwise hard to control.
Hydration and the enol
Treat propyne with water, sulfuric acid and a mercury(II) salt such as HgSO₄. Water adds across one pi bond with Markovnikov selectivity: the OH goes to C2, the internal carbon, and the H to C1.
The mercury does here what it did in the oxymercuration of an alkene. The pi bond attacks Hg²⁺, the mercury bridges both carbons, and water opens that bridge at the carbon carrying more positive character, the more substituted one. So the Markovnikov outcome is the bridged ion's doing, not a separate rule for alkynes. Unlike oxymercuration, no borohydride step is needed: the mercury falls off on its own while the product forms.
The first product is an enol, a compound with an OH attached directly to a carbon of a C=C. From propyne it is CH₂=C(OH)–CH₃. Enols are not stable, and this one immediately tautomerizes: a hydrogen moves from the oxygen to C1, and the pi bond moves from C=C to C=O. The product is a ketone, acetone.
The ketone is called the keto form. Keto and enol forms are tautomers (see Resonance, in Drawing Molecules & Moving Electrons). A hydrogen moves between them, so they are two different compounds in equilibrium, drawn with ⇌, not resonance forms of one compound.
The keto form is favored by at least 10⁵ to one. For acetone the equilibrium constant is nearer 10⁸, so a bottle of acetone holds only a few parts per billion of enol. The reason is bond energy. Add up the bonds that differ: the keto form's C=O, C–H and C–C are stronger in total than the enol's C=C, O–H and C–O, mostly because a C=O pi bond is much stronger than a C=C pi bond.
Every terminal alkyne except ethyne itself behaves like propyne here. Markovnikov hydration puts the oxygen on the internal carbon, so the product is a methyl ketone, with the old end carbon finishing as a CH₃. In ethyne both carbons carry only hydrogen, so the carbonyl carbon keeps an H, and the product is an aldehyde, acetaldehyde (CH₃CHO).
The regiochemistry can be flipped. Hydroboration–oxidation, from the Markovnikov section of this chapter, puts the oxygen on the terminal carbon instead. For an alkyne the borane has to be bulky, meaning large carbon groups replace two of its three hydrogens; disiamylborane and 9-BBN are the usual choices. BH₃ itself would add to both pi bonds.
A bulky borane adds only once, and its size steers the boron firmly to the less crowded end carbon. Oxidation then gives the enol with its OH on C1, and that enol tautomerizes to an aldehyde. Propyne gives propanal.
What carries forward
Acetylide alkylation returns among the carbon–carbon bond-forming tools of Synthesis & Retrosynthesis, and acetylides appear again beside organolithium reagents in Organolithiums, Cuprates & Cross-Coupling.