Solvent isotope effects in the cleavage of benzyltrimethyl-silanes and -stannanes by alkali

Abstract

Product isotope effects, given by the product ratio XC₆H₄·CH₃/XC₆H₄·CH₂D, have been measured for cleavage at ca. 21° of some compounds of the type XC₆H₄·CH₂·MMe₃(M = Sn or Si) in 1 : 1 MeOH–MeOD containing sodium methoxide (2 mol I^–1). Values for various substituents X and M = Sn are: p-Me, 2·4; H, 2·1; p-Cl, 2·1; m-CF₃, 2·1 (all ±0·15); and for M = Si; H, 1·17; m-OMe, 1·25; p-Cl, 1·07; m-CF₃, 1·19 (all ±0·05). The corresponding value for C₆F₅·CH₂·SiMe₃ is 1·16. Overall solvent isotope effects, (k_SH/k_SD)_s given by the individual rates in non-deuteriated and deuteriated media at 50° have been determined for the compounds with X =m-CF₃; the values are: for M = Si, 0·50 in MeOH or in MeOH containing 18·6 mole % H₂O; for M = Sn, 0·95 in MeOH and 0·97 in MeOH containing 20 mole % H₂O.

It is concluded that, in the cleavage of the tin compounds certainly, and probably also in that of the silicon compounds, a free carbanion is not formed. It is suggested that the rate-determining step involves proton transfer from the solvent to the carbon atom of the breaking M–CH₂·C₆H₄·X bond, with the MeO–M bond fully or almost fully formed in the transition state. Initial formation of the M^V intermediates [(MeO)Me₃M·CH₂·C₆H₄X]^– is favoured.