Alkyne preparation and reactions

A carbide lamp works on one simple reaction: drip water onto calcium carbide and it hisses out ethyne gas, bright enough to burn as a lantern flame in old mining tunnels and bicycle lamps. That gas, HC≡CH\text{HC}\equiv\text{CH}, is the simplest member of the alkyne family — hydrocarbons carrying at least one carbon-carbon triple bond, general formula CnH2n−2\text{C}_n\text{H}_{2n-2}. The triple bond packs one σ\sigma bond and two π\pi bonds into a short, linear, electron-rich site, and that extra π\pi bond is the whole story of alkyne chemistry: every addition reagent has to cross the triple bond twice, once for each π\pi bond, on its way to a fully saturated product. This article builds the two laboratory routes to alkynes, the spsp-hybridised structure behind their reactivity, and the two-stage addition chemistry — hydrogenation, halogenation, hydrohalogenation, and Markovnikov hydration to a carbonyl compound — that follows directly from having two π\pi bonds to satisfy instead of one.

The core insight

An alkyne's triple bond is one σ\sigma plus two π\pi bonds between spsp-hybridised, linearly arranged carbons, so every addition reagent adds twice — once per π\pi bond — turning an alkyne first into an alkene-type intermediate and then into the fully saturated product.

The core idea

Two laboratory routes build the triple bond. The classic route to ethyne is calcium carbide plus water, a fast reaction that also generates the Ca(OH)2_2 used to test for the gas:

CaC2+2 H2O⟶HC≡CH+Ca(OH)2CaC_2 + 2\,H_2O \longrightarrow HC\equiv CH + Ca(OH)_2

where: CaC2CaC_2 is calcium carbide (an ionic solid built of Ca2+^{2+} and C22−_2^{2-} carbide ions), and the reaction is simply the carbide ion picking up two protons from water. The second route builds a triple bond out of a double bond by removing hydrogen halide twice — a 1,2-dihaloalkane treated with excess alcoholic KOH and heat eliminates HX, forms a vinyl halide, and eliminates HX again:

CH2Br-CH2Br→alc. KOHCH2=CHBr→alc. KOHCH≡CHCH_2Br\text{-}CH_2Br \xrightarrow{\text{alc. KOH}} CH_2=CHBr \xrightarrow{\text{alc. KOH}} CH\equiv CH

A vicinal or geminal tetrahalide works the same way, losing two molecules of HX to the same excess-base, heat conditions.

Structure explains the reactivity. Each triple-bond carbon is spsp hybridised: the two spsp orbitals point 180∘180^\circ apart (perfectly linear geometry) and overlap head-on to form the σ\sigma bond, while the two unhybridised pp orbitals on each carbon overlap sideways to give two mutually perpendicular π\pi bonds. Four electrons sit exposed in those two π\pi clouds, shorter and stronger than a double bond's single π\pi bond but also a richer target for electrophiles — which is exactly why alkynes undergo the same families of addition reactions as alkenes, only twice over.

Figure
Two stage addition across a triple bond showing an alkyne reacting first at one pi bond to give an alkene intermediate and then at the second pi bond to give the saturated product
Every reagent crosses the triple bond twice HC≡CH alkyne, two π bonds + X-Y CH2=CHY alkene stage, one π left + X-Y CHY2-CH2Y fully saturated product First addition uses up one π bond, second addition uses up the other. Same pattern for H2, X2, and HX; only the final product differs.

Derivation

Derivation AGeneral formula of an alkyne and how many moles of reagent it consumes

The goal is to derive the general molecular formula of an open-chain alkyne from the corresponding saturated alkane by counting exactly how many hydrogen atoms are removed per degree of unsaturation, and then to use that same count of "unsaturation units" to fix how many moles of X2_2 or H2_2 a mole of alkyne consumes when addition runs to completion, which is the number needed for every stoichiometry question on this topic.

Step 1. Start from the fully saturated open-chain hydrocarbon with nn carbons, the alkane CnH2n+2\text{C}_n\text{H}_{2n+2}. Every carbon is sp3sp^3 and every bond is a single bond, so this is the maximum possible number of hydrogens for nn carbons — there is nothing left to remove.

Step 2. Introducing one π\pi bond (one degree of unsaturation, as in an alkene) forces two adjacent carbons to each give up one H atom to make room for the extra bond, so the formula drops by exactly 2H2\text{H}: CnH2n+2→CnH2n\text{C}_n\text{H}_{2n+2} \rightarrow \text{C}_n\text{H}_{2n}.

Step 3. A triple bond is two π\pi bonds stacked on the same pair of carbons, so it is two degrees of unsaturation, removing a further 2H2\text{H} beyond the alkene stage: CnH2n→CnH2n−2\text{C}_n\text{H}_{2n} \rightarrow \text{C}_n\text{H}_{2n-2}. This recovers the alkyne general formula stated at the start of this article, now derived rather than quoted.

Step 4. Each degree of unsaturation is satisfied by exactly one mole of an addition reagent (H2_2 or X2_2) reacting across it. Since an alkyne carries two degrees of unsaturation, complete addition of H2_2 or X2_2 to one mole of alkyne always consumes exactly two moles of that reagent — this is the arithmetic behind the two-stage addition scheme above, and it is what the worked example below uses directly. Check: ethyne, n=2n=2, gives C2H2\text{C}_2\text{H}_2 from 2n−2=22n-2 = 2, matching HC≡CH\text{HC}\equiv\text{CH}; and its two π\pi bonds correctly predict that it takes 2 mol Br2_2 to fully saturate 1 mol ethyne to CHBr2-CHBr2\text{CHBr}_2\text{-CHBr}_2.

Worked example

Problem: What mass of calcium carbide is needed to prepare enough ethyne to completely decolourise 800 g of bromine, Br2_2 (molar mass 160 g/mol), by addition across the triple bond? (Molar mass of CaC2_2 = 64 g/mol.)

Step 1. Find moles of Br2_2 available: n(Br2)=800 g160 g/mol=5 moln(\text{Br}_2) = \dfrac{800\ \text{g}}{160\ \text{g/mol}} = 5\ \text{mol}.

Step 2. Use the stoichiometry from the derivation above: complete addition to an alkyne's triple bond consumes exactly 2 mol of X2_2 per mole of alkyne, so

n(ethyne)=n(Br2)2=52=2.5 moln(\text{ethyne}) = \frac{n(\text{Br}_2)}{2} = \frac{5}{2} = 2.5\ \text{mol}

Step 3. Read the preparation equation, CaC2+2H2O⟶HC≡CH+Ca(OH)2CaC_2 + 2H_2O \longrightarrow HC\equiv CH + Ca(OH)_2: the mole ratio of CaC2_2 to ethyne produced is exactly 1:11:1, so n(CaC2)=2.5 moln(\text{CaC}_2) = 2.5\ \text{mol}.

Step 4. Convert to mass: m(CaC2)=2.5 mol×64 g/mol=160 gm(\text{CaC}_2) = 2.5\ \text{mol} \times 64\ \text{g/mol} = 160\ \text{g}. The lamp needs 160 g of calcium carbide.

Common traps

  • Treating alkyne addition as a single step: it always happens in two stages, through an alkene-type intermediate, because there are two π\pi bonds to satisfy — this is exactly why one mole of alkyne needs two moles of H2_2 or X2_2, not one.
  • Forgetting Markovnikov's rule applies at each stage separately for unsymmetrical HX addition, not just once overall — the second addition follows the regiochemistry set up by the first.
  • Writing the hydration product as the unstable enol instead of its tautomer — the enol is never isolated; it converts immediately to the carbonyl compound.
  • Assuming every alkyne gives an aldehyde on hydration — only ethyne does; every other alkyne gives a ketone, because Markovnikov addition puts OH on the more substituted carbon.

Ask the tutor: if you want to see the two-stage addition mapped out atom by atom for a specific alkyne (say propyne with HBr, or but-2-yne with Br2), ask the AI tutor to walk through both stages with you.

Why does hydration of ethyne give an aldehyde but hydration of every other alkyne gives a ketone?

Markovnikov's rule sends the OH to the more substituted (more stable carbocation-forming) carbon of the triple bond. In ethyne both carbons are identical (each bears one H), so the resulting enol tautomerises to acetaldehyde, an aldehyde. In any other alkyne, at least one triple-bond carbon already carries an alkyl group, so OH lands there, giving a ketone after tautomerisation. Ethyne is the one symmetric exception, not the rule.

Why does the triple bond need excess alcoholic KOH and heat to form from a dihalide, rather than mild conditions?

Removing HX twice means two separate elimination steps must both happen, and each elimination is itself harder to drive than the single elimination that makes an alkene from a monohalide, because forming a π\pi bond next to one that already exists is less favourable. Excess base (so there is always KOH left after the first elimination) and heat (to supply the extra activation energy) push both eliminations to completion; with only mild conditions or limited base, the reaction commonly stalls at the vinyl halide (alkene) stage.

Does the two-stage addition rule mean I can isolate the alkene intermediate from an alkyne reaction?

Sometimes, with the right catalyst choice, yes. Ordinary catalytic hydrogenation over Ni, Pd or Pt does not stop at the alkene — both additions happen readily and the alkane is the product. But Lindlar's catalyst is specifically poisoned to be far less active, so it delivers H2_2 across only the first π\pi bond and halts cleanly at the cis-alkene. For halogenation and hydrohalogenation, the intermediate dihalide can often be isolated simply by controlling the amount of X2_2 or HX added — one equivalent stops at the alkene stage; excess reagent drives full addition.

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