Suzuki-Miyaura Coupling
Suzuki-Miyaura Coupling: A Pd(0) catalyst inserts into the aryl halide bond (oxidative addition), the boronic acid transfers its aryl group to palladium…
What happens
A Pd(0) catalyst inserts into the aryl halide bond (oxidative addition), the boronic acid transfers its aryl group to palladium after activation by base (transmetalation), and reductive elimination forges the new C–C bond while regenerating the Pd(0) catalyst.
Reaction summary
PhB(OH)₂ + PhBr → Ph–Ph (biphenyl), Pd(0) catalyst, base
Reagents and conditions
- Starting material: Phenylboronic acid; Bromobenzene
- Reagents: Pd(0) catalyst (e.g. Pd(PPh₃)₄); Base (e.g. K₂CO₃)
- Products: Biphenyl
Mechanism
- Oxidative addition: Pd(0) inserts into the C–Br bond of bromobenzene, giving an aryl-Pd(II)-bromide complex.
- Transmetalation: Base activates the boronic acid, forming a more nucleophilic boronate; the aryl group transfers from boron to palladium, displacing bromide.
- Reductive elimination: The two aryl groups on palladium couple to form the new C–C bond, releasing biphenyl and regenerating active Pd(0).
Selectivity
- Chemoselectivity: Highly chemoselective for aryl/vinyl halides (I > Br ≫ Cl reactivity) and tolerant of many other functional groups, which is why it's a workhorse of pharmaceutical synthesis.
Common mistakes
- Forgetting the base is required to activate the boronic acid for transmetalation, without it the catalytic cycle stalls.
- Confusing the boron-bearing partner with the halide partner, they play distinct mechanistic roles even though either aryl group could in principle carry either substituent.
Memory trick
"Suzuki = boron + halide + Pd(0)", three steps: oxidative addition, transmetalation, reductive elimination.