| Abstract
| - The reaction of [Me2Al(μ-OEPh3)]2 with pyridine yields the expected acid−base complexesAlMe2(OEPh3)(py) [E = C (1) and Si (2)]. In contrast, the reaction with THF yields AlMe(OEPh3)2(THF) [E = C (5) and Si (6)], although the dimethyl compounds, AlMe2(OEPh3)(THF) [E = C (3) and Si (4)], are observed in THF-d8 solution. The reaction of [Me2Al(μ-OCPh3)]2 with THF was followed by 1H NMR and found to occur by a two-step process. First,the Al2O2 core of [Me2Al(μ-OEPh3)]2 is cleaved by THF to form compound 3. Second, twomolecules of AlMe2(OCPh3)(THF) react with each other, with prior dissociation of THF fromat least one complex, resulting in the ligand redistribution and the formation of 5 and AlMe3(THF). The conversion of [Me2Al(μ-OCPh3)]2 into compound 3 is exothermic, and thesubsequent formation of 5 and AlMe3(THF) is endothermic. The rate equations for theformation of 3 and its conversion to 5 have been determined. The observation of both alkoxidecleavage and alkyl/alkoxide exchange requires a fine balance between a Lewis base that isof sufficient strength to cleave the dimeric alkoxide, [R2Al(μ-OR‘)]2, while being sufficientlyweak to allow dissociation from the monomeric complex, AlR2(OR‘)(L).
- The observation of both alkoxide cleavage and alkyl/alkoxide exchange for the reaction of [R2Al(μ-OR‘)]2 with a Lewis base requires a fine balance between a Lewis base that is of sufficient strength to cleave the dimeric alkoxide, while being sufficiently weak to allow dissociation from the monomeric complex. In the case of [Me2Al(μ-OCPh3)]2, THF is found to meet these requirements. The conversion of [Me2Al(μ-OCPh3)]2 into AlMe(OCPh3)2(THF) is exothermic, and the subsequent formation of AlMe2(OCPh3)(THF) and AlMe3(THF) is endothermic.
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