Role of o-Methylene-Bridged Bis(aniline) as a Spectroscopic Probe for the Metal Coordination Environment

• Published in: Inorganic chemistry Vol. 37; no. 1; pp. 44 - 49
• Main Authors: Lesueur, William, Rogers, Andrew J, Floriani, Carlo, Chiesi-Villa, Angiola, Rizzoli, Corrado
• Format: Journal Article
• Language: English
• Published: United States American Chemical Society 12.01.1998
• ISSN: 0020-1669; 1520-510X
• DOI: 10.1021/ic9706010

Reaction of the bidentate ligand 2,2‘-methylenebis(5-(dimethylamino)aniline) (2, L) with PdCl2 and K2PtCl4 and the bidentate ligand 2,2‘-methylenebis(5-(acetylamino)aniline (5, L‘) with K2PtCl4 in polar solvents gave the complexes [PdCl2(L)] (6), [PtCl2(L)] (8), and [(L‘)PtCl2] (7), respectively. Treatment of 2 with K2PtCl4 in the presence of DMSO resulted in the solvolysis of a chloride ion to give [(L)Pt(Cl)(DMSO)]Cl (9). The resulting eight-membered metallocyclic rings have rigid elongated chair conformations forcing one of the two bridging methylene hydrogens (Hendo) in close proximity to the metal centre. An 1H NMR study revealed that they are now diastereomeric (Hendo, Hexo); the higher field signal position does not vary irrespective of the ligand or metal, whereas the proton corresponding to the lower field signal is apparently more sensitive to the metal coordination environment. The ability of this spectroscopic probe to predict the bonding mode of the metal was also studied by reacting 2 with K[PtCl3(C2H4)]·H2O to give [Pt2(μ-L)Cl4(C2H4)2] (10), which occurred irrespective of the L/Pt ratio. The corresponding methylene hydrogens were not diastereomeric, showing only one singlet in the 1H NMR spectrum. Crystal structures of 6 and 9 have been determined:  6 is triclinic, space group P1̄, a = 10.179(2) Å, b = 12.136(2) Å, c = 9.686(2) Å, α = 97.79(1)°, β = 95.98(2)°, γ = 78.02(1)°, V = 1156.1(4), Z = 2, and R = 0.034; 9 is triclinic, space group P1̄, a = 12.348(1) Å, b = 12.407(1) Å, c = 10.159(1) Å, α = 103.02(1)°, β = 104.87(2)°, γ = 117.09(1)°, V = 1313.0(3), Z = 2, and R = 0.035.