Abstract
Abstract In the reaction of diazomethane with four aluminum haloalkoxides, Al(OCH2CF3)3 (I), Al(OCH2CCl3)3 (II), Al(OCH2CBr3)3 (III), and Al[OCH(CH2Cl)2]3 (IV), a methylene insertion into the aluminum-alkoxyl linkage was observed along with the polymethylene formation. By the methylene insertion, an organoaluminum compound with a haloalkoxymethyl-aluminum linkage is formed. Among aluminum alkoxides, these four haloalkoxides are distinguished by this methylene insertion. The usual aluminum alkoxides of unsubstituted alcohols do not undergo a methylene insertion reaction, but only the polymethylene formation. The difference in the reactivity towards diazomethane between the four haloalkoxides and usual alkoxides has been explained by the difference in acid strength.
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