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首页> 外文期刊>Inorganica Chimica Acta >PALLADIUM(II) COMPLEX FORMATION BY INDOLE-3-ACETATE - MIXED LIGAND COMPLEXES INVOLVING A UNIQUE SPIRO-RING FORMED BY CYCLOPALLADATION
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PALLADIUM(II) COMPLEX FORMATION BY INDOLE-3-ACETATE - MIXED LIGAND COMPLEXES INVOLVING A UNIQUE SPIRO-RING FORMED BY CYCLOPALLADATION

机译:吲哚-3-乙酸酯混合配体配合物形成的钯(II)配合物,并经环加成反应形成独特的螺环

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Formation of mixed ligand paladium(II) complexes involving indole-3-acetate (IA) has been studied by synthetic spectroscopic and X-ray crystallographic methods. Reaction of IA with Na2PdCl4 in methanol gave NaPd(IAH(-1))Cl (1) (IAH(-1)=IA deprotonated from the indole ring), which reacted with pyridine (py) to give ?Pd(IAH(-1))(py)? (2) as orange crystals. Similar reactions carried out in the presence of water gave Pd(IAH(-1)) center dot 1.5H(2)O (1'), which further gave Pd(IAH(-1))(py)center dot 0.5H(2)O (2') by the reaction with py. X-ray crystal structure analysis of 2 revealed a unique dimeric structure, where IAH(-1) coordinates to Pd(II) through the carboxylate oxygen atom and the tetrahedral C3 atom of the indole nucleus, forming a unique spiro-ring. The two complex units are bridged by the indole nitrogens in the 3H-indole form, and there is a clese contact (2.75 Angstroms) around the nitrogen-C2 bonds of the five-membered rings of the indole nuclei positioned in parallel with each other. IA in the neutral form was liberated upon refluxing 1 in methanol containing 10 percent acetic acid, showing that IAH(-1) and IA are interconvertible under proper conditions. The H-3 and C-13 NMR spectra in CDCl3-CD3OD indicated that the C3 atom of IA in 1 and 2 is tetrahedral. Large shift differences of the of the C2 proton signals were observed between 1 and 1' and between 2 and 2', which indicates that 1 and 2 are dimers in solution whereas 1' and 2' are monomers and that the differences are due to the close contact between the two indole rings in 2 as detected in the solid state and probably in 1. [References: 37]
机译:已经通过合成光谱和X射线晶体学方法研究了涉及3-吲哚乙酸酯(IA)的混合配体钯(II)配合物的形成。 IA与Na2PdCl4在甲醇中的反应生成NaPd(IAH(-1))Cl(1)(IAH(-1)=从吲哚环去质子化的IA),与吡啶(py)反应生成αPd(IAH(- 1))(py)? (2)为橙色结晶。在水的存在下进行的类似反应得到Pd(IAH(-1))中心点1.5H(2)O(1'),进一步得到Pd(IAH(-1))(py)中心点0.5H( 2)O(2')与py反应。 X射线晶体结构分析显示2,独特的二聚体结构,其中IAH(-1)通过吲哚核的羧酸氧原子和四面体C3原子与Pd(II)配位,形成独特的螺环。这两个复杂的单元被3H-吲哚形式的吲哚氮桥接,并且在相互平行排列的吲哚核五元环的N-C2键周围存在一个小分子接触(2.75埃)。在含有10%乙酸的甲醇中回流1时,释放出中性形式的IA,表明IAH(-1)和IA在适当条件下可相互转化。 CDCl3-CD3OD中的H-3和C-13 NMR光谱表明1和2中IA的C3原子为四面体。在1和1'之间以及2和2'之间观察到C2质子信号的较大位移差异,这表明1和2是溶液中的二聚体,而1'和2'是单体,并且差异是由于在固态状态下检测到的两个吲哚环之间的紧密接触,可能在1状态下紧密接触。[参考文献:37]

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