Mechanism of Palladium-Catalyzed Alkylation of Aryl Halides with Alkyl Halides through C–H Activation: A Computational Study

Pd-catalyzed C­(sp³)–H activation/alkylation of 2-<i>tert</i>-butylaryl halides with alkyl halides and CH₂Br₂ represents an advantageous strategy for the C–H functionalization with halogens as traceless directing groups. Several possible mechanisms were proposed for the reactions, but no further evidence was available to judge their relative feasibilities. Herein, a mechanistic study was performed with the aid of density functional theory (DFT) methods. Calculations indicate that the coupling of aryl bromides with alkyl chlorides is likely to generate alkylated benzocyclobutenes via aryl–Br oxidative addition on Pd(0) catalysts, C­(sp³)–H activation, alkyl–Cl oxidative addition, aryl–alkyl reductive elimination, aryl–H activation, and aryl–C­(sp³) reductive elimination. The coupling of aryl iodides with CH₂Br₂ is likely to generate indane derivatives via aryl–I oxidative addition, C­(sp³)–H activation, alkyl–Br oxidative addition, aryl–CH₂Br reductive elimination, alkyl–Br oxidative addition, C­(sp³)–alkyl reductive elimination, and reduction of palladium dibromide complexes by amines. By comparison, the metathesis of alkyl chlorides on Pd­(II) intermediates and the pathway involving palladium carbene intermediates are found to be less favored. Meanwhile, the coordination of <i>in situ</i> generated salts KI, KBr, and KHCO₃ to palladium complexes, which has been less considered in previous mechanistic studies, is found to lead to more energetically favored pathways in most of the steps. Finally, the oxidative addition of alkyl halides generating Pd­(IV) intermediates or the reduction of palladium dibromide complexes by amines, rather than the previously proposed C­(sp³)–H activation, is found to be the rate-determining step in the two types of coupling reactions. This result does not go against the reported primary kinetic isotope effect (KIE) based on intramolecular competition reactions because the C­(sp³)–H activation is irreversible according to our calculations