Answers are grounded in this paper and cite their sources.
Fluoride ions play three roles in the Suzuki–Miyaura reaction. They favor the reaction by formation of trans-[ArPdF(PPh3)2], which reacts with Ar′B(OH)2 in an unprecedented rate-determining transmetalation, and by promoting the reductive elimination from the trans-[ArPdAr′(PPh3)2] intermediate. Conversely, F− disfavors the reaction by formation of unreactive anionic Ar′B(OH)n−3Fn− (n=1–3), leading to two antagonistic effects of F− in the transmetalation.
No takes yet. Share an insight, caveat, or question.
Amatore et al. (2011) studied this question.
Synapse has enriched 10 related papers. Consider them for comparative context: