Theoretical Study of Platinum(0)-Catalyzed Hydrosilylation of Ethylene. Chalk−Harrod Mechanism or Modified Chalk−Harrod Mechanism

A detailed theoretical investigation was performed on all of the transition states and intermediates involved in the Si−H oxidative addition of H−SiR3 (R = H, Cl, or Me) to Pt(PH3)2, ethylene insertion into Pt−H and Pt−SiR3 bonds, isomerization of the ethylene insertion product, and Si−C and C−H reductive eliminations. In a Chalk−Harrod mechanism, the rate-determining step is the isomerization of Pt(SiR3)(C2H5)(PH3) formed by ethylene insertion into the Pt−H bond and its activation barrier is 22 kcal/mol for R = H, 23 kcal/mol for R = Me, and 26 kcal/mol for R = Cl (MP4SDQ values are given hereafter). In a modified Chalk−Harrod mechanism, the rate-determining step is the ethylene insertion into the Pt−SiH3 bond and its barrier is 44 kcal/mol for R = H and Me and 60 kcal/mol for R = Cl. Thus, it should be reasonably concluded that platinum(0)-catalyzed hydrosilylation of ethylene proceeds through a Chalk−Harrod mechanism. Chlorine substituent on Si causes significant and interesting effects on the reaction, since the Pt−SiCl3 bond is much stronger than Pt−SiH3 and Pt−SiMe3 bonds. A methyl substituent influences the activation barrier of Si−H oxidative addition, ethylene insertion, and Si−C reductive elimination little. Detailed analysis is presented to clarify the reason that ethylene is inserted into the Pt−SiR3 bond with much more difficulty than into the Pt−H bond, since a modified Chalk−Harrod mechanism is difficult due to the high activation barrier of the ethylene insertion into the Pt−SiR3 bond. The Si−C reductive elimination was also investigated in detail, since this is a candidate for the rate-determining step of a Chalk−Harrod mechanism. The present calculations indicate that the Si−C reductive elimination occurs more easily in Pt(SiR3)(CH3)(PH3)(C2H4) than in Pt(SiR3)(CH3)(PH3)2, which agrees well with the experimental finding of Ozawa et al. (J. Am. Chem. Soc. 1995, 117, 8873). The reason is clearly interpreted in terms of a π-back-bonding interaction between Pt d and C2H4 π* orbitals. The ethylene coordination to Pt significantly accelerates the Si−C reductive elimination, and as a result, the Si−C reductive elimination can be clearly excluded from the rate-determining step in a Chalk−Harrod mechanism.