We have a thread on making R2AlH by elimination of alkenes from R2AlEt, R2AlPr or R2AliBu, where Pr=n-propyl iBu=isobutyl, and in theory you ought to
be able to make a dialkoxyaluminium hydride and react this with a sodium alkoxide to obtain the trialkoxyaluminium hydride... this is not Red-Al, but
it is close enough for some things.
EDIT: Actually, maybe you can make it all a bit easier if you do it in reverse:
EtAlCl2 + 3 NaOMe >> NaEtAl(OMe)3 + 2 NaCl
NaEtAl(OMe)3 + heat >> NaHAl(OMe)3
[Edited on 28-5-2017 by clearly_not_atara] |