The three-dimensional configuration of the ester heterocycle is basically the same as that of the carbocycle. Compound: Diiodo(1,5-cyclooctadiene)platinum(II)(SMILESS: I[Pt]I.C1=CCC/C=CCC/1,cas:12266-72-7) is researched.Recommanded Product: 38006-08-5. The article 《3,6-Bis(2′-pyridyl)pyridazine iridium-palladium and iridium-platinum heterodinuclear complexes》 in relation to this compound, is published in Congr. Naz. Chim. Inorg., [Atti], 16th. Let’s take a look at the latest research on this compound (cas:12266-72-7).
[Ir(NO)L(PPh3)2]2+ (I) (L = 3,6-bis(2′-pyridyl)pyridazine) reacted with (PhCN)2PdCl2 to give Ir(NO)L(PPh3)2[PdCl2]2+ which lost PPh3 in CH3CN and added Cl- to give [Ir(NO)L(PPh3)PdCl3]+ (II). I also reacted with halide ions to give [Ir(NO)L(PPh3)X]+ (III) (X = Cl, Br, I) which then reacted with (PhCN)2PdCl2 to give II. I reacted with (COD)PtI2 (COD = cyclooctadiene) to give [Ir(NO)L(PPh3)PtI3]+ but I or III (X = Cl) reacted with (COD)PtCl2 to give [IrL(PPh3)2Cl2]+. The complexes were characterized by IR and 31P NMR spectra and elec. conductivity data.
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