形成一个双核亚氨基配合物从反应的钌(VI)氮化物与钌(II)氢化。

Formation of a dinuclear imido complex from the reaction of a ruthenium(VI) nitride with a ruthenium(II) hydride.

机构信息

Department of Chemistry, The Hong Kong University of Science and Technology, Clear Water Bay, Kowloon, Hong Kong, People's Republic of China.

出版信息

Inorg Chem. 2011 Feb 21;50(4):1161-3. doi: 10.1021/ic101520g. Epub 2011 Jan 21.

DOI:10.1021/ic101520g

PMID:21265579

Abstract

The treatment of [Ru(L(OEt))(N)Cl(2)] (1; L(OEt)(-) = Co(η(5)-C(5)H(5)){P(O)(OEt)(2)}(3)) with Et(3)SiH affords [Ru(L(OEt))Cl(2)(NH(3))] (2), whereas that with [Ru(L(OEt))(H)(CO)(PPh(3))] (3) gives the dinuclear imido complex [(L(OEt))Cl(2)Ru(μ-NH)Ru(CO)(PPh(3))(L(OEt))] (4). The imido group in 4 binds to the two ruthenium atoms unsymmetrically with Ru-N distances of 1.818(6) and 1.952(6) Å. The reaction between 1 and 3 at 25 °C in a toluene solution is first order in both complexes with a second-order rate constant determined to be (7.2 ± 0.4) × 10(-5) M(-1) s(-1).

摘要

用三乙硅烷处理[Ru(L(OEt))(N)Cl(2)]（1；L(OEt)(-) = Co(η(5)-C(5)H(5)){P(O)(OEt)(2)}(3)）可得到[Ru(L(OEt))Cl(2)(NH(3))]（2），而与[Ru(L(OEt))(H)(CO)(PPh(3))]（3）反应则生成双核亚胺配合物[(L(OEt))Cl(2)Ru(μ-NH)Ru(CO)(PPh(3))(L(OEt))]（4）。4 中的亚胺基团不对称地与两个钌原子结合，Ru-N 距离分别为 1.818(6) 和 1.952(6) Å。在 25°C 的甲苯溶液中，1 和 3 之间的反应在两个配合物中均为一级反应，确定二级速率常数为(7.2 ± 0.4) × 10(-5) M(-1) s(-1)。