具三甲基矽基之磷硫多牙基的新穎三價銅錯合物合成及其性質探討

Abstract

我們利用去質子化的三腳四牙配位基P(C6H3-3-SiMe3-2-SH)3 (PS3TMS)與CuCl2和PPNCl在乙腈與四氫呋喃混合溶劑下反應，會發生自身氧化還原反應得到[PPN][CuClPS3TMS] (1)和PPNCuCl2，利用通入氧氣來氧化正二價銅離子可提高錯合物1的產率。如果使用類似之去質子化配位基PS3進行反應，就沒有辦法觀察到正三價銅錯合物之產生。 將錯合物1和NaOMe進行反應會得到Cu(NHPPh3)PS3TMS (3)，因此推測反應過程中可能是先經由β-hydride elimination生成[PPN] [CuHPS3TMS]的中間態，進而與[PPN]+反應生成NHPPh3並配位到正三價銅離子中心，才會得到錯合物3。為了證明此反應過程會經過[PPN][CuHPS3TMS]中間態，因此將錯合物1直接與KBEt3H反應，也成功地得到錯合物3。若將錯合物1的陽離子置換成[PPh4]+生成[PPh4] [CuClPS3TMS] (2)，再與KBEt3H反應，會得到K2[(CuPS3TMS)2] (4)和氫氣，這個結果告訴我們反應過程中可能生成[PPh4][CuHPS3TMS]。

The tripodal tetradentate ligand P(C6H3-3-SiMe3-2-SH)3 (PS3TMSH3) was deprotonated by NaH, and further reacted with CuCl2 in CH3CN and THF. The disproportionation of CuCl2 was occurred to form [PPN][CuClPS3TMS] (1) and PPNCuCl2. The yield of complex 1 in previous reaction could be improved by using dry O2 to oxidize Cu(II) ion. A similar ligand, P(C6H4- 2-SH)3 (PS3H3) was also employed to reacted with CuCl2. However, no CuIII complex was observed. Reaction of 1 with NaOMe afforded complex, [Cu(NHPPh3)PS3TMS] (3). This results suggested that an intermediates, [PPN][CuHPS3TMS], was generated through β-hydride elimination. The hydride attacted the cation, [PPN]+, to form NHPPh3, which coordinated to the CuIII center. To prove the formation of the [PPN][CuHPS3TMS] intermediates, complex 1 was reacted with KBEt3H to afford complex 3. If the counter ion of 1 changed to [PPh4]+ to form [PPh4][CuClPS3TMS] (2). Reaction of 2 with KBEt3H gave a new dinuclear complex, K2[(CuPS3TMS)2] (4) and H2 suggesting the possibility of [PPh4][CuHPS3TMS] formation.