Organic Chemistry · Alkynes: An Introduction to Organic Synthesis

Reduction of Alkynes

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

of an alkyne means adding hydrogen across the triple bond, but how much hydrogen you add — and which alkene isomer you get — depends entirely on the reagent you choose. An alkyne has two π bonds, so it can be reduced in stages:

Complete reduction with H₂ over a metal catalyst (Pd, Pt, or Ni) adds two equivalents of hydrogen and gives an alkane.
Lindlar (hydrogen gas over a "poisoned" palladium catalyst) stops after adding one equivalent and gives a cis-alkene — both hydrogens add to the same face of the triple bond.
with sodium or lithium in liquid ammonia adds the two hydrogens from opposite faces and gives a trans-alkene.

The practical message: alkynes are the only hydrocarbon family from which you can prepare both geometric isomers of an alkene selectively. That stereochemical control is why alkynes appear so often in synthesis problems and in the preparation of natural products.

Why this matters

Stereochemistry is chemistry in three dimensions, and reduction of alkynes is a textbook demonstration of how reaction conditions — not just starting materials — control the shape of a product. Cis- and trans-alkenes have very different physical properties and biological activities: fatty acids in your body are cis-unsaturated, margarine-style hardened fats contain trans isomers, and insect pheromones often signal with a specific alkene geometry. Being able to build either isomer deliberately is a core skill in organic synthesis, pharmaceutical process chemistry, and the study of lipid biochemistry. On exams, "reduce this alkyne" is a favorite question precisely because the reagent choice (Lindlar vs. Na/NH₃ vs. H₂/Pd) completely changes the answer.

The college version

Core Concepts

Complete reduction to an alkane

With excess H₂ and a metal catalyst such as palladium on carbon, both π bonds are hydrogenated and the alkyne becomes the fully saturated alkane:

$R - C \equiv C - R' + 2 H 2 Pd/C ⟶ R - CH 2 - CH 2 - R'$

The reaction is exothermic, and the catalyst adsorbs hydrogen and the alkyne onto its surface, so both hydrogens add to the same face (). Because the intermediate alkene is also adsorbed and hydrogenated rapidly, the alkene is not isolated — the alkane is the only product. This method is used when the goal is simply to remove the triple bond completely.

Lindlar hydrogenation: syn addition to a cis-alkene

The is palladium deposited on calcium carbonate and "poisoned" with a lead compound and quinoline. The poison slows the catalyst enough that hydrogenation stops after the first equivalent of H₂, leaving the alkene:

$R - C \equiv C - R' + H 2 Lindlar catalyst ⟶ cis - R - CH = CH - R'$

Hydrogen adds to the alkyne's two carbons from the same side of the triple bond (syn addition), and the substituents on the resulting double bond therefore end up on the same side — the cis isomer. For example, 2-butyne gives cis-2-butene. Lindlar hydrogenation is the standard way to prepare a cis-alkene from an alkyne.

Dissolving-metal reduction: anti addition to a trans-alkene

Treating the alkyne with sodium or lithium metal in liquid ammonia (a "dissolving-metal" reduction) delivers electrons one at a time. The mechanism goes through a , then a vinylic radical, then a vinylic anion — each intermediate is protonated by the ammonia solvent. The key stereochemical result is : the two hydrogens end up on opposite faces of the double bond, giving the trans alkene:

$R - C \equiv C - R' + 2 Na + 2 NH 3 liquid NH 3 ⟶ trans - R - CH = CH - R' + 2 NaNH 2$

The trans isomer is favored because the intermediate anions prefer the conformation with the bulky substituents as far apart as possible. For example, 2-butyne gives trans-2-butene. Note that dissolving-metal reduction also stops at the alkene — it does not reduce the double bond further, because alkenes are not reduced by Na/NH₃ under these conditions.

Choosing the reagent: a decision map

Want an alkane? Use H₂ + Pd/C (or Pt, Ni) with excess hydrogen.
Want a cis-alkene? Use H₂ + Lindlar catalyst.
Want a trans-alkene? Use Na (or Li) in liquid NH₃.

The same starting alkyne can give two different alkene isomers, which makes alkyne reduction a powerful, predictable tool for stereoselective synthesis. Terminal alkynes (e.g., 1-butyne, CH₃CH₂C ≡ CH) give 1-alkenes with either reagent — there is no cis/trans distinction at a terminal carbon, so Lindlar and Na/NH₃ give the same 1-alkene.

Common Confusions

Do Not Confuse	With	The Difference
Lindlar catalyst	H₂/Pd	Lindlar is poisoned and stops at the alkene (cis); plain Pd/C goes all the way to the alkane
Na/NH₃	Lindlar	Na/NH₃ adds H's to opposite faces (trans); Lindlar adds them to the same face (cis)
Syn addition	Anti addition	Same face vs. opposite faces of the multiple bond
"Reduction"	"Hydrogenation"	Hydrogenation is reduction by H₂; reduction can also happen via electrons (Na/NH₃) with no H₂ gas
Cis vs. trans for terminal alkynes	Cis vs. trans for internal alkynes	Terminal alkynes (RC≡CH) give 1-alkenes with no geometric isomers; only internal alkynes show cis/trans product differences

Eli explains

The same idea, in plain words

Explain it like I’m 10

A triple bond is like two ropes twisted tightly together. If you pull them apart gently with a special tool (Lindlar), the two ends stay on the same side — that's the cis shape, like your hands meeting in front of you. If you instead use a different tool (sodium in liquid ammonia), the ends flip to opposite sides — that's the trans shape, like your hands spread wide apart. And if you just keep pulling until the ropes are completely loose, you get a plain single bond (an alkane). Same rope, three different results — the tool you pick decides the shape.

Worked example

Worked example 1 (reagent → product). Starting from 3-hexyne, CH₃CH₂C ≡ CCH₂CH₃:

H₂ + Lindlar → cis-3-hexene (CH₃CH₂CH=CHCH₂CH₃, both ethyl groups on the same side).
Na in liquid NH₃ → trans-3-hexene (ethyl groups on opposite sides).
Excess H₂ + Pd/C → hexane, CH₃CH₂CH₂CH₂CH₂CH₃.

Worked example 2 (stoichiometry with dimensional analysis). How many liters of H₂ gas (at STP) are needed to completely reduce 10.0 g of 2-butyne, CH₃C ≡ CCH₃, to butane? Write the conversion chain first, then substitute numbers:

$n(2 - butyne) = 10.0 g 54.09 g/mol = 0.185 mol$

Complete reduction needs 2 mol H₂ per mol alkyne:

$n(H 2) = 0.185 mol \times 2 mol H 2 1 mol alkyne = 0.370 mol$

At STP, 1 mol of gas occupies 22.4 L:

$V(H 2) = 0.370 mol \times 22.4 L/mol = 8.29 L$

So 8.29 L of H₂ gas at STP is required.

Worked example 3 (synthesis planning). A chemist needs cis-2-pentene. Starting from 2-pentyne, CH₃C ≡ CCH₂CH₃, the answer is a single step: hydrogenate with H₂ over the Lindlar catalyst to give cis-2-pentene. If trans-2-pentene were needed instead, the same alkyne would be treated with Na in liquid NH₃. The alkyne is the "universal alkene precursor" because both isomers are accessible from it.

Key takeaways

Three reagents, three products: H₂/Pd → alkane; Lindlar → cis-alkene; Na/NH₃ → trans-alkene.
Lindlar = syn addition (both H's to the same face) → cis.
Na/NH₃ = anti addition (H's to opposite faces) → trans, via radical-anion intermediates.
Both partial reductions stop at the alkene — they do not make the alkane.
Terminal alkynes give 1-alkenes regardless of which partial-reduction reagent is used (no cis/trans isomers possible at a terminal carbon).
Hydrogen stoichiometry: 1 mol alkyne needs 2 mol H₂ for complete reduction, 1 mol H₂ for partial reduction.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

What product results from treating 2-butyne with H₂ and Lindlar catalyst?
Show answer
cis-2-butene — Lindlar hydrogenation is syn addition and stops at the alkene.
What product results from treating 2-butyne with Na in liquid NH₃?
Show answer
trans-2-butene — dissolving-metal reduction is anti addition, giving the more stable trans isomer.
How many moles of H₂ are consumed when 1 mol of an alkyne is completely reduced to an alkane?
Show answer
2 mol H₂ — one equivalent for each of the two π bonds of the triple bond.
Why does Lindlar hydrogenation stop at the alkene instead of continuing to the alkane?
Show answer
The catalyst is "poisoned" with lead and quinoline, which slows hydrogenation so much that the alkene (which would need a second addition) is essentially unreactive under these conditions; the reaction stops after one equivalent.
A student claims Na/NH₃ reduction adds two hydrogens to the same face of the triple bond. Is that correct?
Show answer
No. Na/NH₃ reduction adds hydrogens to opposite faces (anti addition), giving the trans alkene. Same-face addition is Lindlar's job.
What single reagent converts 1-hexyne into hexane?
Show answer
Excess H₂ over Pd/C (or Pt or Ni) — complete hydrogenation to hexane, CH₃(CH₂)₄CH₃.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Reduction: Addition of hydrogen (or gain of electrons) to a molecule
Hydrogenation: Reaction of a compound with H₂, usually over a catalyst
Lindlar catalyst: Poisoned Pd/CaCO₃ that stops at the alkene
Syn addition: Both new atoms add to the same face of the multiple bond
Anti addition: New atoms add to opposite faces
Dissolving-metal reduction: Na or Li in liquid NH₃ that transfers electrons one at a time
Radical anion: Species with an unpaired electron and a negative charge

Sources & references

openstax.org — Organic Chemistry

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