244761-19-1Relevant articles and documents
Metal tetrahydridoborates and tetrahydridoborato metalates. 23.1 amine solvates of lithium and sodium tetrahydridoborate
Giese, Hans-Hermann,Habereder, Tassilo,N?th, Heinrich,Ponikwar, Werner,Thomas, Steffen,Warchhold, Marcus
, p. 4188 - 4196 (1999)
A series of amine solvates of LiBH4 and NaBH4 have been prepared and characterized by IR and NMR spectroscopy as well as by X-ray single-crystal structure determinations. LiBH4 crystallizes from pyridine as LiBH4·3(py), 1, in which the BH4 anion acts as a bidentate ligand. However, in the structure of LiBH4·3py*, 2 (py* = p-benzylpyridine), a tridentate But group is observed. In contrast, LiBH4·2(coll), 3 (coll = 2,4,6-trimethylpyridine, collidine), possesses only a bidentate tetrahydridoborate group, while a tridentate BH) group is present in monomeric LiBH4·PMDTA, 4 (PMDTA = pentamethyldiethylenetriamine). In contrast, NaBH4·PMDTA, 6, is dimeric in the solid state: three of the four H atoms of each BH4 group coordinate to the Na atoms; two form a double bridge to two Na atoms while the third one is bonded only to one Na center. LiBH4·TMTA, 5 (TMTA = trimethylhexahydrotriazine), is also dimeric; however, only two of the nitrogen atoms of the TMTA ligand coordinate to Li. The BH4 groups bridge the two Li centers each with one H atom coordinating to two Li atoms, and two bind to a single Li atom. A totally different situation exists for NaBH4·TMTCN, 7 (TMTCN = trimethyltriazacyclononane), which is tetrameric in the crystal. Only one hydrogen atom of the BH4 group functions as a hydride bridge and binds to three Na centers. The molecule contains a Na4B4 heterocubane core. Thus, the different modes of the interaction of the BH4 groups with the alkali metal atoms are determined by the number of donor atoms from the neutral amine ligand and the size of the cation. No definitive conclusion as to the structure of the amine solvates can be derived from IR and/or 11B NMR spectra for the solution state. The crystallographic data are as follows. 1: a = 10.9939(5) A?, b = 9.9171(4) A?, c = 14.8260(8) A?, β= 94.721(3)°, V = 1611.0(1) A?3, monoclinic, space group P2(1)/n, Z = 4, R1= 0.0823. 2: a = 10.121(1) A?, b = 12.417(2) A?, c = 13.462(3) A?, α = 83.189(2)°, β = 86.068(3)°, γ = 69.166(4)°, V = 1369.3(5) A?3, triclinic, space group P1?, Z = 2, R1 = 0.0689. 3: a = 28.527(3) A?, b = 10.858(1) A?, c = 11.319(1) A?, V = 3505.7(6) A?3, orthorhombic, space group Fdd2, Z = 8, R1 = 0.0502. 4: a = 7.591(3) A?, b = 15.325(6) A?, c = 8.719(4) A?, β = 99.80(2)°, V = 999.5(7) A?3, monoclinic, space group P2(1)/c, Z = 4, R1 = 0.0416. 5: a = 14.68(1) A?, b = 11.830(7) A?, c = 16.960(8) A?, V = 2946(3) A?, orthorhombic, space group P2(1)2(1)2(1), Z = 8, R1 = 0.0855. 6: a = 9.993(2) A?, b = 10.008(3) A?, c = 14.472(4) A?, β= 93.55(2)°, V = 1444.6(7) A?3, monoclinic, space group P2(1)/n, Z = 4, R1 = 0.0455. 7: cubic, a = b = c = 13.859(5) A?, V = 2662(2) A?3, cubic, space group I4?3m, Z = 8, R1 = 0.0871.
