Alkynes

Every alkyne is built around a carbon-carbon triple bond, the most electron-rich and most reactive of the three aliphatic C-C linkages in this chapter. That triple bond is one sigma bond plus two pi bonds between two spsp-hybridised carbons held rigidly at 180∘180^\circ, and it does two things a double bond cannot: it lets a terminal hydrogen ionise as a weak acid, and it lets an addition reagent add on twice instead of once. Both quirks trace to the same cause, the unusually high s-character of the spsp orbital. This section builds the general formula and unsaturation count, works through bromination and the silver-acetylide test with full numbers, and flags the traps examiners love: assuming every alkyne is acidic, or forgetting that excess reagent means two additions, not one.

The core insight

A carbon-carbon triple bond is one sigma plus two pi bonds between linear, spsp-hybridised carbons, and the high s-character of that spsp orbital makes a terminal alkyne's C-H weakly acidic while letting the triple bond accept an addition reagent twice.

The core idea

Formula and hybridisation. Alkynes follow the general formula CnH2n−2\text{C}_n\text{H}_{2n-2}, two hydrogens short of the corresponding alkane for every triple bond present. Each triple-bonded carbon is spsp hybridised: two spsp orbitals point in opposite directions along the molecular axis, forming linear sigma bonds (one to a substituent, one to the other alkyne carbon), while two unhybridised, mutually perpendicular 2p2p orbitals on each carbon overlap sideways to give two pi bonds. The result is a rigid, linear 180∘180^\circ arrangement at every triple-bonded carbon — the triple bond itself and the two atoms directly attached to it all lie on one straight line.

Why terminal alkynes are acidic. A hydrogen sitting directly on an spsp-hybridised carbon, as in HC≡C-R\text{HC}\equiv\text{C-R}, is weakly acidic — strong enough to be pulled off by sodium amide or precipitated by ammoniacal Ag+\text{Ag}^+/Cu+\text{Cu}^+:

HC≡CH+NaNH2→HC≡C−Na++NH3\text{HC}\equiv\text{CH} + \text{NaNH}_2 \rightarrow \text{HC}\equiv\text{C}^-\text{Na}^+ + \text{NH}_3

where: the acetylide ion HC≡C−\text{HC}\equiv\text{C}^- is stabilised because the lone pair it carries sits in an spsp orbital, which holds 50%50\% s-character (versus 33%33\% in sp2sp^2 and 25%25\% in sp3sp^3). Since s orbitals draw electron density in closer to the nucleus, that negative charge is held tighter and lower in energy on an spsp carbon than on an sp2sp^2 or sp3sp^3 one — so the parent C-H bond gives up its proton more readily. This is exactly why alkanes and non-terminal alkenes show no such acidity: their C-H bonds sit on sp3sp^3 or sp2sp^2 carbons with far less s-character to stabilise the resulting anion.

Addition, and why it can happen twice. Like alkenes, alkynes add reagents across their unsaturation, but each triple bond carries two pi bonds, so with excess reagent it can add twice, first to an alkene-stage intermediate and then on to the fully saturated product:

  • Hydrogenation (Ni/Pt/Pd\text{Ni/Pt/Pd}): full addition gives an alkane; a partially poisoned catalyst can be stopped at the alkene stage.
  • Halogenation (X2\text{X}_2): first addition gives a dihaloalkene; excess X2\text{X}_2 carries it on to a tetrahaloalkane.
  • Hydrohalogenation (HX): adds by Markovnikov's rule (H to the carbon already richer in H, X to the carbon poorer in H), and can add twice with excess HX.
  • Hydration (H2O\text{H}_2\text{O}, Hg2+/H2SO4\text{Hg}^{2+}/\text{H}_2\text{SO}_4): Markovnikov addition of water gives an unstable enol that rearranges to a carbonyl compound — acetaldehyde specifically from ethyne, a ketone from any other alkyne.
Figure
Orbital picture of ethyne showing two sp hybrid orbitals along the molecular axis for the sigma framework and two perpendicular p orbital pairs forming two pi bonds around the triple bond
The C≡C triple bond — one sigma bond plus two pi bonds C (sp) C (sp) H H pi bond (vertical p-p overlap) second pi bond (horizontal p-p overlap, drawn as outline) 180 degree linear sigma framework: H-C≡C-H
Derivation AGeneral formula and degree of unsaturation for an alkyne

The goal is to show, starting only from the saturated alkane formula and the definition of degree of unsaturation, why every alkyne obeys CnH2n−2\text{C}_n\text{H}_{2n-2} and why that formula always signals exactly one triple bond (two degrees of unsaturation) once ring closures are ruled out, so that a student can read the molecular formula of an unknown hydrocarbon and immediately know how much unsaturation it must contain before drawing a single structure.

Step 1. Start from the fully saturated reference. An open-chain (acyclic) alkane with nn carbon atoms has the maximum possible number of hydrogens, CnH2n+2\text{C}_n\text{H}_{2n+2} — every carbon uses all four bonds to either another carbon or a hydrogen, with no double or triple bonds and no rings.

Step 2. Recall what one pi bond costs. Forming a carbon-carbon double bond (as in an alkene) converts one C-C single bond into a double bond, which removes exactly two hydrogens from the saturated count: each carbon now uses one bond for the extra pi overlap instead of an extra C-H bond. This gives the alkene formula CnH2n\text{C}_n\text{H}_{2n}, one degree of unsaturation below the alkane.

Step 3. Add the second pi bond. A triple bond is a double bond plus one more pi bond, so it costs a second pair of hydrogens on top of the first. Removing two more H atoms from CnH2n\text{C}_n\text{H}_{2n} gives CnH2n−2\text{C}_n\text{H}_{2n-2} — the general formula for an acyclic alkyne, carrying two degrees of unsaturation (degree of unsaturation, or index of hydrogen deficiency, IHD=(2n+2)−h2\text{IHD} = \dfrac{(2n+2) - h}{2}, where hh is the actual hydrogen count).

Step 4. Verify against known alkynes. For ethyne, n=2n = 2: the formula predicts C2H(2×2)−2=C2H2\text{C}_2\text{H}_{(2\times2)-2} = \text{C}_2\text{H}_2, which matches HC≡CH\text{HC}\equiv\text{CH} exactly, and IHD=(2×2+2)−22=2\text{IHD} = \dfrac{(2\times2+2)-2}{2} = 2, confirming two degrees of unsaturation from the single triple bond. For but-1-yne, n=4n = 4: predicted formula is C4H(2×4)−2=C4H6\text{C}_4\text{H}_{(2\times4)-2} = \text{C}_4\text{H}_6, matching HC≡C-CH2-CH3\text{HC}\equiv\text{C-CH}_2\text{-CH}_3 (count: 4 carbons, 6 hydrogens), with IHD=(2×4+2)−62=2\text{IHD} = \dfrac{(2\times4+2)-6}{2} = 2 again — the same two degrees, now assigned entirely to the one triple bond rather than to a ring plus a double bond, since the molecule is confirmed acyclic and has no other unsaturation.

Worked example

Q. Ethyne, HC≡CH\text{HC}\equiv\text{CH}, is treated with an excess of bromine, Br2\text{Br}_2. Give the full reaction sequence, name the final product, and confirm its molecular formula matches the hydrogen count expected for full saturation.

Step 1. First addition. One pi bond of the triple bond breaks as Br2\text{Br}_2 adds across it, converting the triple bond to a double bond: HC≡CH→Br2BrCH=CHBr\text{HC}\equiv\text{CH} \xrightarrow{\text{Br}_2} \text{BrCH=CHBr} (1,2-dibromoethene).

Step 2. Second addition. With excess Br2\text{Br}_2 present, the remaining pi bond — now a simple C=C double bond — also undergoes addition: BrCH=CHBr→Br2CHBr2-CHBr2\text{BrCH=CHBr} \xrightarrow{\text{Br}_2} \text{CHBr}_2\text{-CHBr}_2 (1,1,2,2-tetrabromoethane).

Step 3. Name the final product. 1,1,2,2-tetrabromoethane. This confirms the general rule from the derivation above: a triple bond, carrying two pi bonds (two degrees of unsaturation), accepts two successive molecules of an addition reagent when the reagent is in excess, whereas an alkene's single pi bond (one degree of unsaturation) only ever adds once.

Step 4. Sanity-check against the formula. Ethyne is C2H2\text{C}_2\text{H}_2 (IHD=2\text{IHD} = 2, matching Step 4 of the derivation). Fully saturating both degrees of unsaturation with two Br2\text{Br}_2 molecules (4 Br atoms total, no H change) gives C2H2Br4\text{C}_2\text{H}_2\text{Br}_4 — exactly 1,1,2,2-tetrabromoethane, with every carbon now singly bonded to four other atoms. Final answer: 1,1,2,2-tetrabromoethane, C2H2Br4\text{C}_2\text{H}_2\text{Br}_4, formed by two successive additions of Br2\text{Br}_2.

Common traps

  • Assuming every alkyne is acidic — only terminal alkynes (H directly on an spsp-carbon) react with NaNH2\text{NaNH}_2 or give a precipitate with ammoniacal AgNO3\text{AgNO}_3/Cu2Cl2\text{Cu}_2\text{Cl}_2; internal alkynes like but-2-yne have no such H and give no reaction.
  • Forgetting that excess reagent means two additions across a triple bond, not one — each pi bond adds separately, unlike a double bond's single pi bond.
  • Mixing up hybridisation and bond angle — alkyne carbons are spsp with 180∘180^\circ linear geometry, not sp2sp^2/120∘120^\circ like alkenes.
  • Misapplying Markovnikov's rule direction, or forgetting the mechanism, during hydrohalogenation or hydration of an unsymmetrical alkyne — the same "more stable intermediate wins" logic as for alkenes still governs where H and X (or H and OH) land.

Ask the tutor: Want to work through a multi-step alkyne addition sequence, or double-check which of two isomers gives the silver acetylide precipitate?

Why does a triple bond make the attached hydrogen acidic, when a double bond does not really do the same?

It comes down to how much s-character the carbon's bonding orbital carries. An spsp carbon (triple bond) is 50%50\% s, an sp2sp^2 carbon (double bond) is only 33%33\% s, and an sp3sp^3 carbon (single bond) is 25%25\% s. Higher s-character pulls the bonding electrons — and, after the proton leaves, the lone pair on the anion — closer to the nucleus, which lowers the anion's energy and stabilises it. That extra stabilisation is enough to make a terminal alkyne's C-H weakly acidic (removable by strong bases like NaNH2), while an alkene's or alkane's C-H is essentially non-acidic under the same conditions.

If but-2-yne has a triple bond too, why does it not react with ammoniacal AgNO3?

The acidic-hydrogen test only works if there is an acidic hydrogen to remove in the first place. But-2-yne, CH3-C≡C-CH3, has both triple-bond carbons capped with methyl groups — there is no H sitting directly on either sp-carbon. Ammoniacal AgNO3 needs exactly that H, on an sp-carbon, to replace with Ag+ and form the insoluble acetylide precipitate. Since but-2-yne is an internal alkyne with no such H, it simply gives no reaction, even though its triple bond is otherwise just as real and just as capable of undergoing addition reactions like bromination or hydrogenation.

How do I know when an alkyne will add a reagent once versus twice?

It comes down to how much of the reagent is available. Each triple bond has two pi bonds, so in principle it can accept two separate addition events — first converting the triple bond to a double bond, then converting that double bond to a single bond. With a controlled, limited amount of reagent (or a suitably poisoned catalyst for hydrogenation), the reaction can be stopped after the first addition, leaving an alkene-stage product. With excess reagent, both pi bonds react and you end up at the fully saturated product. Always check the question for the word excess, or for a specified stoichiometric ratio, before deciding how far the addition goes.

Before this

Where this leads