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, (kSH/kSD)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 MV intermediates [(MeO)Me3M·CH2·C6H4X]– is favoured.