Synthesis and X-ray crystallographic studies of Ni (II) and Cu (II) complexes of [5-(4-pyridyl)-1,3,4] oxadiazole-2-thione/thiol formed by transformation of N-(pyridine-4-carbonyl)-hydrazine carbodithioate in the presence of ethylenediamine

The reaction of [M(H2L)2] [M = Ni(II) Cu(II)] (K+H2L− = N-(pyridine-4-carbonyl)-hydrazine carbodithioate) with excess of ethylenediamine (en) gave mixed ligand complexes [Ni(en)2(4-pytone)2] (4-pytone = 5-(4-pyridyl)-1,3,4-oxadiazole-2-thione), and [Cu(en)2](4-pytol)2·H2O (4-pytol = 5-(4-pyridyl)-1,3,4-oxadiazole-2-thiol). The metal complexes have been characterized with the aid of elemental analyses, IR, magnetic susceptibility and single crystal X-ray studies. Complexes (1) and (2) crystallize in monoclinic system, space group P1 21/n1 and C2/c, respectively. The ligand after cyclization is present in the deprotonated thiol form in the Cu(II) complex where it is ionically bonded through sulfur. In the Ni(II) complex (1) bonding of the ligand take place through oxadiazole nitrogen and the ligand exists as the thione form.

Graphical abstractThe reaction of [M(H2L)2] [M = Ni(II) Cu(II)] (K+H2L− = N-(pyridine-4-carbonyl)-hydrazine carbodithioate) with excess of ethylenediamine (en) gave mixed ligand complexes [Ni(en)2(4-pytone)2] and [Cu(en)2](4-pytol)2·H2O. The metal complexes have been characterized with the aid of elemental analyses, IR, magnetic susceptibility and single crystal X-ray studies. The ligand (4-pytone) is bonded through oxadiazole nitrogen in the thione form in 1 and is ionically present in the thioate form in 2.Figure optionsDownload full-size imageDownload as PowerPoint slide