The complex ppn[Au(acac)2]1 can be obtained by reacting ppn[AuCl2] with [TI(acac)] in 1:2 molar ratio, and complex 1 reacts with compounds of the type RH in which H has some acidic character to give the corresponding [AuR2]– species; RH = CH2(CN)22, CH2(CO2Me)23, CH2(CN)CO2Me 4, Me3SiCCH 5, 2-SH-C5H4N 6, AuR RH = CH2(PPh2)2, 7, or [AuR2]+ RH =[(4-MeC6H4)3PCH2C5H4N-2]ClO48.
ppn[Au(acac)2]1的合成可通过ppn[AuCl2]与[TI(acac)]以1:2的摩尔比反应获得,而1与具有某种酸性特征的RH类化合物反应,可得到相应的[AuR2]–物种;RH =
CH2(CN)22、 (CO2Me)23、 (CN)CO2Me 4、Me3SiCCH 5、2-SH-
C5H4N 6、AuR RH = (PPh2)2、7或[AuR2]+ RH =[(4-MeC6H4)3PCH2 -2]ClO48。