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Mechanistic details pertaining to the Pd(0)/PCy3-catalyzed intermolecular arylation of a terminal β-C(sp(3))-H bond aryl amide substrate (SM = EtCONH-Ar, where Ar = C6H5, C6F5 and CONH-Ar is a directing group (DG)) in the presence of CsF base were elucidated. Key mechanistic features of this reaction are (1) oxidative addition of the aryl halide PhI to Pd(0)/PCy3, (2) deprotonation of SM by CsF to form DG' = [EtCON-Ar]Cs(+) for subsequent coordination to intermediate I-Pd(II)(PCy3)Ph (the substantially lower pKa of the EtCONHC6F5 in comparison to EtCONHC6H5 is instrumental for the presence of a larger population of the reactive deprotonated amides for Ar = C6F5), (3) "Cs2-I-F" cluster formation upon external (the second) CsF molecule approach to the active site of the I-Pd(II)(PCy3)Ph(DG') intermediate, (4) "Cs2-I-F cluster" assisted β-C(sp(3))-H bond activation via a concerted metalation-deprotonation (CMD) mechanism, and (5) reprotonation of the amide directing group to facilitate the C(sp(3))-Ph reductive elimination. The energy barriers, ΔG(‡) (ΔG(‡disp), associated with the "Cs2-I-F cluster" mediated β-C(sp(3))-H bond activation transition state are 6.5 (8.7) and 10.2 (12.9) kcal/mol when DG = CONHC6H5, CONHC6F5, respectively. It was shown that (a) the PCy3 ligand only semidissociates upon β-C(sp(3))-H bond cleavage and (b) the I-to-F substitution in I-[Pd(II)](Ph)(PCy3)(DG') is a facile process that makes the "direct-halide" assisted β-C(sp(3))-H bond activation relatively less energy demanding and opens the possibility for a competing Ph-F bond formation reaction. It was shown that the "direct-I" assisted C-H bond activation TS, which associates with a relatively large energy barrier, is an H-atom insertion transition state into the Pd-I bond, while the "direct-F" assisted C-H bond activation TS, which occurs with a relatively low energy barrier (but still is much larger than that required for the "Cs2-I-F cluster" assisted pathway), is a direct proton abstraction transition state.
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Figg et al. (2013) studied this question.
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