Reaction Of Lithium Anilide With (Pph(3))(3)Rhcl And Related Rhodium(I) Complexes

The reaction of LiNHPh with four Rh(I) complexes, (PPh(3))(3)RhCl, [(PPh(3))(2)RhCl](2), (PEt(3))(3)RhCl, and [(PEt(3))(2)-RhCl](2), in THF has been studied by NMR spectroscopy. The reaction of LiNHPh (1.3 equiv/Rh) with (PPh(3))(3)-RhCl or [(PPh(3))(2)RhCl](2) leads to an equilibrium mixture of syn- and anti-[(PPh(3))(2)RhNHPh](2), which is slowly converted to [(PPh(3))(2)Rh](2) mu-Cl)(mu-NHPh). When these reactions are conducted with a large excess (10 equiv/Ph) of LiNHPh, a single complex forms, for which the formula [(PPh(3))(2)Rh(NHPh)(2)](-) Li+ is proposed on the basis of H-1, P-31{H-1}, and Rh-103{H-1} NMR data. This complex can also be generated by treatment of the mixture of syn- and anti-[(PPh(3))(2)RhNHPh](2) with an excess of LiNHPh. Reaction of LiNHPh (1.3 equiv/Rh) with (PEt(3))(3)-RhCl leads to (PEt(3))(3)RhNHPh, which is in equilibrium with the starting complex. In the presence of excess LiNHPh (10 equiv/Rh), (PEt(3))(3)RhCl reacts to give an equilibrium mixture of (PEt(3))(3)RhNHPh and [(PEt(3))(2)Rh(NHPh)(2)]Li--(+). Reaction of LiNHPh (1.3 equiv/Rh) with[(PEt(3))(2)RhCl](2) leads to a mixture of [(PEt(3))(2)-RhNHPh](2) and [(PEt(3))(2)Rh](2)(mu-Cl)(mu-NHPh). When conducted with excess LiNHPh (10 equiv/Rh), this reaction gives [(PEt(3))(2)Rh(NHPh)(2)]-Li+ as the major product. In the presence of excess LiNHPh, the two anionic complexes [(PR(3))(2)Rh(NHPh)(2)]Li--(+) (R = Ph, Et) are stable for days in refluxing THF.