Bis(diethyl sulfide)bis(2,4,6-trimethylphenyl)iridium(II) and related compounds

Abstract

The interaction of mer-IrCl3(SEt2)3 with Mg(mes)2(thf)2[mes = mesityl (2,4,6-trimethylphenyl), thf = tetrahydrofuran] in Et2O-thf gives trans-Ir(mes)2(SEt2)21. With PMe31 gives trans-Ir(mes)2(PMe3)22; both compounds are paramagnetic. With NO, 2 reacts to form the formally iridium(III) complex Ir(mes)2(NO)(PMe3)24 that has a bent IrNO group, while H2 gives the known hydride IrH5(PMe3)2. With CO 1 forms the square iridium(I) complex Ir(mes)(CO)2(SEt2)3. The trans-square planar molecules of compound 1 lie on centres of symmetry; Ir–S and Ir–C 2.298(4) and 2.09(1)Å, respectively, with a unique S–Ir–C angle of 88.2(3)°. Molecules of 2 are also trans square planar, but lie in general positions; Ir–P 2.302(4), 2.305(4)Å, Ir–C 2.097(9), 2.112(9)Å and P–Ir–C 89.6(3)–90.5(3)°. The crystal structure of 4 contains two crystallographically independent molecules, both of which lie on two-fold axes, coincident with the Ir–N bond. The geometry is slightly distorted square pyramidal; N–Ir–P and N–Ir–C 91.7(2)(× 2) and 99.1(3), 99.3(3)°, respectively. The Ir–N, Ir–P and Ir–C bond lengths are 1.90(1), 1.91(1); 2.324(5), 2.328(4); and 2.186(9), 2.182(12)Å, respectively. The bent Ir–N–O system [133.7(10) and 133.4(8)°] is disordered over two sites related by the two-fold axes, but in a single molecule is fixed in one position and believed to associate, via O ⋯ P interaction, with the phosphorus atom of one phosphine.