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Cu-catalyzed unsymmetrical coupling

Cadiot-Chodkiewicz coupling

How the Cadiot-Chodkiewicz coupling works, in 3 steps. Each step shows the structures and the curved arrows that move the electrons.

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In short. A 3-step mechanism. Typical reagents: terminal alkyne, 1-bromoalkyne, cat. CuCl, amine base (EtNH₂ or BuNH₂), NH₂OH·HCl. Conditions: MeOH or aq. amine, 0–25 °C, under N₂. Bonds formed: C(sp)–C(sp) bond (unsymmetrical 1,3-diyne); bonds broken: C–H (alkyne), C–Br. Stereochemistry: None (linear sp carbons). Rate and selectivity: Cross-selective; NH₂OH keeps copper as Cu(I).

At a glance

Reagentsterminal alkyne, 1-bromoalkyne, cat. CuCl, amine base (EtNH₂ or BuNH₂), NH₂OH·HCl
Solvent / conditionsMeOH or aq. amine, 0–25 °C, under N₂
Bonds formedC(sp)–C(sp) bond (unsymmetrical 1,3-diyne)
Bonds brokenC–H (alkyne), C–Br
StereochemistryNone (linear sp carbons)
Rate / selectivityCross-selective; NH₂OH keeps copper as Cu(I)

Step 1: Cu(I) acetylide forms

+++RHCu⁺NH₂RCu(I)NH₃1 · CU(I) ACETYLIDE FORMSterminal alkyneSUBSTRATEcopper(I)Cu⁺CATALYSTamine baseC₂H₇NBASEcopper acetylideINTERMEDIATEammoniumC₂H₈N⁺BYPRODUCTSame Cu–acetylide formation as in Glaser coupling.

Cu(I) deprotonates one terminal alkyne to form a copper acetylide, exactly as in Glaser coupling — but here, a haloalkyne (rather than a second terminal alkyne) is the coupling partner.

Step 2: Acetylide displaces bromide

+RCu(I)R'BrRCu(III)BrR'2 · ACETYLIDE DISPLACES BROMIDEcopper acetylideNUCLEOPHILE1-bromoalkyneELECTROPHILEdialkynylcopper(III)INTERMEDIATEOxidative-addition-like step: both alkynyl groups end up on copper.

The copper acetylide's nucleophilic carbon performs an oxidative-addition-like displacement of halide from the haloalkyne, joining the two alkynyl fragments on copper.

Step 3: Reductive elimination

reductiveelimination+RCu(III)BrR'RR'BrCu3 · REDUCTIVE ELIMINATIONdialkynylcopper(III)INTERMEDIATEunsymmetrical 1,3-diynePRODUCTCuBr (Cu(I) regenerated)CATALYSTHaloalkyne as partner gives cross-selectivity: no statistical Glaser mixture.

Reductive elimination releases the unsymmetrical 1,3-diyne and regenerates Cu(I) — using a haloalkyne instead of a second terminal alkyne is what makes this cross-selective, avoiding the statistical mixture of products that plain Glaser homocoupling would give with two different alkynes.

Abbreviations and terms

R
R group. Any carbon substituent (alkyl, aryl…) whose identity does not matter for the mechanism.
MeOH
Methanol. Polar protic solvent (also a nucleophile in solvolysis).
H2
Hydrogen gas. Reductant for catalytic hydrogenation.
HCl
Hydrochloric acid / hydrogen chloride. Strong acid; chloride is the nucleophile in additions/substitutions.

How it compares with related reactions

Cadiot-Chodkiewicz couplingBuchwald-Hartwig aminationBuchwald-Hartwig etherification
Reagentsterminal alkyne, 1-bromoalkyne, cat. CuCl, amine base (EtNH₂ or BuNH₂), NH₂OH·HClArBr, R₂NH, Pd₂(dba)₃ / bulky phosphine (XPhos, BINAP), NaOtBuArBr, ROH, Pd₂(dba)₃ or Pd(OAc)₂ / bulky biaryl phosphine (RockPhos, tBuBrettPhos), Cs₂CO₃ or NaOtBu
ConditionsMeOH or aq. amine, 0–25 °C, under N₂Toluene or dioxane, 80–110 °CToluene or dioxane, 80–100 °C
StereochemistryNone (linear sp carbons)No new stereocentre; retention of aryl geometryNo new stereocentre; aryl geometry retained
Rate / selectivityCross-selective; NH₂OH keeps copper as Cu(I)Bulky, electron-rich ligand speeds oxidative addition and C–N reductive eliminationC–O reductive elimination is slow, so very bulky ligands are needed

Test yourself

Try to answer each question before opening it.

Which reagents and conditions are typical for the Cadiot-Chodkiewicz coupling?

terminal alkyne, 1-bromoalkyne, cat. CuCl, amine base (EtNH₂ or BuNH₂), NH₂OH·HCl. Conditions: MeOH or aq. amine, 0–25 °C, under N₂.

Which bonds are formed and which are broken?

Formed: C(sp)–C(sp) bond (unsymmetrical 1,3-diyne). Broken: C–H (alkyne), C–Br.

What is the stereochemical outcome?

None (linear sp carbons)

What controls the rate or the selectivity?

Cross-selective; NH₂OH keeps copper as Cu(I)

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