文摘
A series of symmetric divalent Sn(II) hydrides of the general form [(4-X-Ar')Sn(-H)]2 (4-X-Ar'= C6H2-4-X-2,6-(C6H3-2,6-iPr2)2; X = H, MeO, tBu, and SiMe3; 2, 6, 10, and 14), along with the more hinderedasymmetric tin hydride (3,5-iPr2-Ar*)SnSn(H)2(3,5-iPr2-Ar*) (16) (3,5-iPr2-Ar* = 3,5-iPr2-C6H-2,6-(C6H2-2,4,6-iPr3)2), have been isolated and characterized. They were prepared either by direct reduction of thecorresponding aryltin(II) chloride precursors, ArSnCl, with LiBH4 or iBu2AlH (DIBAL), or via a transmetallationreaction between an aryltin(II) amide, ArSnNMe2, and BH3·THF. Compounds 2, 6, 10, and 14 were obtainedas orange solids and have centrosymmetric dimeric structures in the solid state with long Sn···Sn separationsof 3.05 to 3.13 Å. The more hindered tin(II) hydride 16 crystallized as a deep-blue solid with an unusual,formally mixed-valent structure wherein a long Sn-Sn bond is present [Sn-Sn = 2.9157(10) Å] and twohydrogen atoms are bound to one of the tin atoms. The Sn-H hydrogen atoms in 16 could not be locatedby X-ray crystallography, but complementary Mössbauer studies established the presence of divalent andtetravalent tin centers in 16. Spectroscopic studies (IR, UV-vis, and NMR) show that, in solution, compounds2, 6, 10, and 14 are predominantly dimeric with Sn-H-Sn bridges. In contrast, the more hindered hydrides16 and previously reported (Ar*SnH)2 (17) (Ar* = C6H3-2,6-(C6H2-2,4,6-iPr3)2) adopt primarily the unsymmetricstructure ArSnSn(H)2Ar in solution. Detailed theoretical calculations have been performed which includecalculated UV-vis and IR spectra of various possible isomers of the reported hydrides and relevant modelspecies. These showed that increased steric hindrance favors the asymmetric form ArSnSn(H)2Ar relativeto the centrosymmetric isomer [ArSn(-H)]2 as a result of the widening of the interligand angles at tin,which lowers steric repulsion between the terphenyl ligands.