A Comparative Assessment of the Effect of the Lewis Acidity of the Central Tin Atom on Intramolecular Coordination of (3‐Methoxypropyl)stannanes

Abstract

AbstractThe recently studied set of (3‐methoxypropyl)stannanes of general formula RxSnCl4–x [R = (CH2)3OCH3, x = 4 (1), x = 3 (2), x = 2 (3) and x = 1 (4)] is extended with two new sets of (3‐methoxypropyl)stannanes of general formula RPh3–xSnClx [x = 0 (5), x = 1 (6) and x = 2 (7)] and RSn(S2CNEt2)3–xClx [x = 0 (8), x = 1 (9) and x = 2 (10)]. The molecular structures of 6, 7, 9 and 10 in the solid state were determined by X‐ray diffraction. In compounds 6 and 7, the 3‐methoxypropyl ligand forms a C,O‐chelate and the tin atom is coordinated as a distorted trigonal bipyramid with the oxygen and one chlorine atom in the axial positions. Compounds 9 and 10 are hexacoordinate with a distorted octahedral ligand arrangement. While the 3‐methoxypropyl ligand in 10 also forms a C,O‐chelate, the oxygen atom in 9 is not coordinated to the central tin atom. The structures in a non‐coordinating (CDCl3) and a coordinating solvent ([D6]DMSO) were studied by multinuclear 1H, 13C and 119Sn NMR spectroscopy. It is proposed that the structures found in the solid state are retained upon dissolution in CDCl3. Compounds 5, 8 and 9 also preserve their structures in [D6]DMSO. On the other hand, an equilibrium, in which one or more molecules of solvent enter the coordination sphere of the central tin atom and cleave the oxastannacycle, is established in [D6]DMSO solutions of 6, 7 and 10. A comparison of the J coupling constants obtained from 1D 1H,119Sn HMQC and 2D 1H,119Sn J‐HMBC spectra and the J coupling constants derived from conventional 1D 13C NMR spectra shows that the best indicator of the O→Sn donor–acceptor interaction in 3‐methoxypropylstannanes is the 3J value. (© Wiley‐VCH Verlag GmbH & Co. KGaA, 69451 Weinheim, Germany, 2005)