Heterometallic boride clusters: formation of octahedral [M2Ru4(CO)16B]–(M = Rh or Ir) and gold(I) phosphine derivatives. Crystal structures of [N(PPh3)2][trans-Ir2Ru4(CO)16B], trans-[Rh2Ru4(CO)16B{µ3-AuP(C6H11)3}] and cis-[Ir2Ru4(CO)16B{µ-AuP(C6H11)3}]

Abstract

The reaction of the butterfly cluster [Ru₄H(CO)₁₂BH]^– with [Rh₂(CO)₄Cl₂] led to the octahedral boride [Rh₂Ru₄(CO)₁₆B]^–1. In solution, ¹¹B NMR spectroscopic evidence supports the presence of both cis- and trans-isomers of 1. The trans form is predominant. When treated with [AuCl{P(C₆H₁₁)₃}], 1 yielded [Rh₂Ru₄(CO)₁₆B{AuP(C₆H₁₁)₃}]3 in two isomeric forms 3a and 3b. Similar reactions occur with other phosphine gold(I) derivatives. The crystal structure of compound 3a has been determined and reveals a trans arrangement of Rh atoms and a face capping AuP(C₆H₁₁)₃ unit. The anion [Ru₄H(CO)₁₂BH]^– reacted with [Ir₂L₄Cl₂](L = cycloctene or L₂= cycloocta-1,5-diene) under a stream of CO to give [Ir₂Ru₄(CO)₁₆B]^–2; the crystal structure of 2 has been determined and confirms an octahedral metal framework with trans iridium atoms. Cluster 2 reacted with [AuCl{P(C₆H₁₁)₃}] to yield [Ir₂Ru₄(CO)₁₆B{AuP(C₆H₁₁)₃}]4 and crystallographic data for 4 show that the Ir atoms are mutually cis in the octahedral Ir₂Ru₄ skeleton. The AuP(C₆H₁₁)₃ unit bridges the Ir–Ir edge. In [Ru₄H(CO)₁₂BH]^–, the four ruthenium atoms define a butterfly framework with the boron atom lying in a semi-interstitial position. The addition of two Group 9 metal atoms to close up the metal cage to an octahedral one with an M₂Ru₄ core should initially give a cis isomer. In fact the trans isomer is observed for both 1 and 2, although for 1 both cis and trans isomers are observed by ¹¹B NMR spectroscopy. Geometrical preferences which accompany the addition of a gold(I) phosphine fragment to anions 1 and 2 to give 3 and 4 respectively are discussed.