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Pd-catalyzed cross-coupling

Buchwald-Hartwig amination

How the Buchwald-Hartwig amination 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: ArBr, R₂NH, Pd₂(dba)₃ / bulky phosphine (XPhos, BINAP), NaOtBu. Conditions: Toluene or dioxane, 80–110 °C. Bonds formed: Ar–N (C–N σ); bonds broken: Ar–Br; N–H. Stereochemistry: No new stereocentre; retention of aryl geometry. Rate and selectivity: Bulky, electron-rich ligand speeds oxidative addition and C–N reductive elimination.

At a glance

ReagentsArBr, R₂NH, Pd₂(dba)₃ / bulky phosphine (XPhos, BINAP), NaOtBu
Solvent / conditionsToluene or dioxane, 80–110 °C
Bonds formedAr–N (C–N σ)
Bonds brokenAr–Br; N–H
StereochemistryNo new stereocentre; retention of aryl geometry
Rate / selectivityBulky, electron-rich ligand speeds oxidative addition and C–N reductive elimination

Step 1: Oxidative addition (bulky ligand L)

+oxidativeadditionPd(0)LBrPhPdBrL1 · OXIDATIVE ADDITION (BULKY LIGAND L)Pd(0) catalystCATALYSTbromobenzeneC₆H₅BrELECTROPHILEL–Pd(II)(Ph)BrINTERMEDIATEBulky ligand favours a reactive 12-electron mono-ligated Pd(0).

Pd(0), supported by a bulky phosphine ligand, oxidatively adds an aryl halide, forming Ar–Pd(II)–X.

Step 2: Amine binds, base removes N–H

++−tBuOH, −Br⁻PhPdBrLNHOPhPdNL2 · AMINE BINDS, BASE REMOVES N–HL–Pd(II)(Ph)BrINTERMEDIATEdimethylamineNUCLEOPHILEtert-butoxideBASEPd amido complexINTERMEDIATEAmido ligand forms on Pd — the C–N analogue of transmetalation.

An amine coordinates to palladium and is deprotonated by base (an amido ligand forms directly on the metal) — this step plays the same structural role as transmetalation, putting a second group cis to the aryl.

Step 3: Reductive elimination

reductiveelimination+PhPdNLNPd(0)L3 · REDUCTIVE ELIMINATIONcis-aryl Pd amidoINTERMEDIATEN,N-dimethylanilineC₈H₁₁NPRODUCTPd(0) catalystCATALYSTC–N bond forms; Pd(0) returns to the cycle.

Reductive elimination forges the new C–N bond and regenerates Pd(0) — the same three-step logic as Suzuki coupling, but building carbon-nitrogen bonds instead of carbon-carbon ones.

Abbreviations and terms

Ar
Aryl group. Any aromatic ring substituent such as phenyl.
X
Halogen / leaving group. Cl, Br or I (sometimes any leaving group).
L
Ligand. A Lewis-base group (often a phosphine) bound to a metal centre.
Ph
Phenyl. –C₆H₅.
Pd2(dba)3
Tris(dibenzylideneacetone)dipalladium(0). Air-stable source of Pd(0).
dba
Dibenzylideneacetone. Weakly bound alkene ligand on Pd(0).
BINAP
2,2′-Bis(diphenylphosphino)-1,1′-binaphthyl. Chiral bidentate phosphine for asymmetric catalysis.

How it compares with related reactions

Buchwald-Hartwig aminationHeck reactionHiyama coupling
ReagentsArBr, R₂NH, Pd₂(dba)₃ / bulky phosphine (XPhos, BINAP), NaOtBuArBr, alkene (e.g. methyl acrylate), Pd(OAc)₂/PPh₃, Et₃NArX, Ar′Si(OMe)₃ (or ArSiF₃), Pd(0) cat., TBAF (F⁻ activator)
ConditionsToluene or dioxane, 80–110 °CDMF or MeCN, 80–120 °CTHF or DMF, 60–100 °C
StereochemistryNo new stereocentre; retention of aryl geometrySyn insertion, syn β-H elimination; E-alkene favouredRetention at sp² carbons; E/Z alkenes preserved
Rate / selectivityBulky, electron-rich ligand speeds oxidative addition and C–N reductive eliminationAryl adds to the less hindered alkene carbonTransmetalation (needs F⁻ activation) is usually rate-limiting

Test yourself

Try to answer each question before opening it.

Which reagents and conditions are typical for the Buchwald-Hartwig amination?

ArBr, R₂NH, Pd₂(dba)₃ / bulky phosphine (XPhos, BINAP), NaOtBu. Conditions: Toluene or dioxane, 80–110 °C.

Which bonds are formed and which are broken?

Formed: Ar–N (C–N σ). Broken: Ar–Br; N–H.

What is the stereochemical outcome?

No new stereocentre; retention of aryl geometry

What controls the rate or the selectivity?

Bulky, electron-rich ligand speeds oxidative addition and C–N reductive elimination

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