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Anionic and Neutral Aluminum Bis(N,N'-di-tert-butylethylenediamide) Complexes: [Al{[N(t-Bu)CH(2)](2)}(2)](-) and

Michael G. Gardiner¹, Colin L. Raston, Brian W. Skelton

¹Departments of Chemistry, Monash University, Clayton, Melbourne, Victoria 3168, Australia, and University of Western Australia, Nedlands, Western Australia 6709, Australia.

Inorganic Chemistry

|June 18, 1997

Abstract:

The reaction of the aluminum triamide [{EtaN(t-Bu)CH(2)CH(2)N-t-Bu}Al{N(t-Bu)CH(2)}(2)] (1) with n-BuLi in hexane and diethyl ether gives the monomeric solvent-free lithium aluminum tetraamide [Li{N(t-Bu)CH(2)CH(2)N-t-Bu}(2)Al] (2), and the diethyl ether adduct [Et(2)O.Li{N(t-Bu)CH(2)CH(2)N-t-Bu}(2)Al] (3), respectively. The reaction of the diethyl ether adduct 3 with 2 equiv of the chelating nitrogen Lewis bases TMEDA ( identical withN,N,N',N'-tetramethylethylenediamine) and t-Bu(2)DAB ( identical withN,N'-di-tert-butyl-1,4-diazabutadiene) in diethyl ether affords the ionic complex [Li(TMEDA)(2)][Al{[N(t-Bu)CH(2)](2)}(2)] (4) and a mixture of the paramagnetic species [Li(t-BuDAB)(2)] and [Al{[N(t-Bu)CH(2)](2)}(2)], which cocrystallize 5, respectively. The reaction of the diethyl ether adduct 3 with 0.5 equiv of HgCl(2) yields the neutral paramagnetic species [Al{[N(t-Bu)CH(2)](2)}(2)] (6). Crystals of 2 are triclinic, of space group P&onemacr; (No. 2), with a = 15.894(5) Å, b = 15.473(5) Å, c = 9.975(3) Å, alpha = 93.74(2) degrees, beta = 91.97(2) degrees, gamma = 102.07(3) degrees, and Z = 4. Crystals of 3are monoclinic, of space group C2/c (No. 15), with a = 15.835(5) Å, b = 12.054(3) Å, c = 16.456(5) Å, beta = 111.56(3) degrees, and Z = 4. Crystals of 4are orthorhombic, of space group Pbca (No. 61), with a = 19.349(4) Å, b = 22.227(7) Å, c = 19.070(3) Å, and Z = 8. Crystals of 5 are orthorhombic, of space group Pmmn (No. 59), with a = 13.874(4) Å, b = 13.473(6) Å, c = 12.717(3) Å, and Z = 2. Crystals of 6 are monoclinic, of space group C2/c (No. 15), with a = 16.721(4) Å, b = 9.454(4) Å, c= 16.460(8) Å, beta = 113.27(3), and Z = 4.

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