The mechanism of cleavage of 3-phenylallyl-silicon and -tin bonds by methanolic sodium methoxide

Abstract

Summary The compounds trans -PhCH CHCH 2 MR′ 3 with M Si or Sn and R′ Me or Et have been cleaved in methanolic sodium methoxide at 50°C to give the products trans -PhCH CHCH 3 (I) and PhCH 2 CH CH 2 (II). (Small amounts of cis -PhCH CHCH 3 sometimes obtained are thought to come mainly or exclusively from cis -PhCH CHCH 2 MR′ 3 in the starting material.) At base concentrations of 0.02–0.40 M the I/II ratio is 1.53 for MR′ 3 SiMe 3 and 1.04 for MR′ 3 SnMe 3 , but the ratio rises somewhat in both cases at higher base concentrations. The values of the product isotope effect, PIE, given by the product ratio PhCH CHCH 3 /PhCH CHCH 2 D obtained on cleavage in 1 : 1 MeOH/MeOD at 50°C, for formation of product I are as follows: (MR′ 3 ) SiMe 3 , 1.26; SiEt 3 , 1.75; SnMe 3 , 2.1; SnEt 3 , 2.2. The corresponding values for the formation of II are 1.45, 2.0, 3.0 and 3.2. For formation of I the values of the rate isotope effect, RIE, given by the ratio of the rate in MeOH to that in MeOD, are 0.60 and 0.77 for MR′ 3 SiMe 3 and SnMe 3 , respectively, and the corresponding values for formation of II are 0.65 and 1.14. The RIE: PIE ratios for MR′ 3 SiMe 3 are 0.48 and 0.45 for formation of I and II, respectively, and the corresponding ratios for MR′ 3 SnMe 3 are 0.37 and 0.38. It is suggested that in all cases the cleavage proceeds by an electrophically-assisted mechanism in which the proton transfers from the solvent to the C(1) or C(3) carbon atoms of the phenylallyl group are concerted with the cleavage of the metal-carbon bond.