Reduction of Ir 4 (CO) 12 with Na in THF yields the salt Na Ir(CO) x (A) which has a strong absorption in its IR spectrum (THF solution) at 1892 cm 1 Reduction of A with Na in liquid NH 3 , followed by addition of Ph 3 SnCl and Et 4 NBr, gives Et 4 N B as the iridium containing product CO is lost d...

To determine the structures of A and B lets go step by step Reduction of Ir4CO12 with Na in THF Ir4CO12 is a tetranuclear cluster of iridium with 12 CO ligands When it is reduced with Na in THF it los ...

Reduction of Ir 4 (CO) 12 with Na in THF yields the salt Na[Ir(CO) x ] (A)

Question:

Reduction of Ir₄(CO)₁₂ with Na in THF yields the salt Na[Ir(CO)_x] (A) which has a strong absorption in its IR spectrum (THF solution) at 1892 cm⁻¹. Reduction of A with Na in liquid NH₃, followed by addition of Ph₃SnCl and Et₄NBr, gives [Et₄N][B] as the iridium containing product; CO is lost during the reaction. Elemental analysis of [Et₄N][B] shows that it contains 51.1% C, 4.55% H and 1.27% N. The IR spectrum of [Et₄N][B] shows one strong absorption in the carbonyl region at 1924 cm⁻¹, and the solution ¹H NMR spectrum exhibits multiplets between δ 7.1 and 7.3 ppm (30H), a quartet at δ 3.1 ppm (8H) and a triplet at δ 1.2 ppm (12H). Suggest structures for A and [B]⁻. Comment on possible isomerism in [B]⁻ and the preference for a particular isomer.