In terms of thermodynamics, more electron-deficient metal centers increase the likelihood of β-alkyl elimination. For example, β-alkyl elimination is more favorable than β-hydride elimination when it is bonded to electron-deficient early transition metals (Hf, Ti, Zr, Nb, etc.) with d configuration. Computational studies show a thermodynamic preference for β-Me elimination over β-H elimination in these complexes due to additional stability for the metal–alkyl species. The origin of the additional bonding interaction comes from an orbital centered on the CH weakly π-donating to the LUMO of the d of the metal center which is analogous to the hyperconjugation effect (see figure on the right), thus increasing the stability of M−CH over M−H species. Their calculations predict that a more electrophilic metal ion enhances the −CH π-donation, which consequently increases the stability of M−CH over M−H species. Conversely, a more electron-rich metal ion will favor M−H formation (for example, using the more electron-donating Cp* ligand in Cp*MX).
In terms of kinetics, steric effects of ligands could play a role in increasing the energy barrier of β-H elimination relative to β-alkyl elimination, specifically when the ligand is Cp*. A model was proposed to illustrate this effect: In both β-methyl elimination (A) and β-hydride elimination (B), the transferring group aligns perpendicular to the Cp*(centroid)−Zr−Cp*(centroid), allowing the σ or σ bond to overlap with the metal d-orbital. However, to achieve the prerequisite geometry for β-H elimination (B), the adjacent methyl group experiences a significant steric repulsion from the Cp* ligand, thereby elevating the barrier to hydride transfer. By contrast, transition state A for β-Me elimination experiences less steric interaction with the Cp* ligand.