摘要:
配合物Ti(NRAr F)2(NMe 2)2(4,R = C(CD 3)2 CH 3,Ar F = 2,5-C 6 H 3 FMe),Ti(NRAr F)2(NMe 2)(I)(5),Ti(NRAr F)2(NMe 2)(CH 2 SiMe 3)(6)和Ti(NRAr F)2(I)(CH 2 SiMe 3)(7通过交替的除盐和二甲酰胺脱保护步骤的四个步骤,分别以77%,71%,70%和84%的收率合成了α-己内酰胺。配成Ti(NRAr)(I)2(OAr')(11,Ar = 3,5-C 6 H 3 Me 2,Ar'= 2,6-C 6 H 3 i Pr 2)(65%)通过配合物Ti(NRAr)(NMe 2)3(9)和Ti(NRAr)(NMe 2)2(OAr')(10)产生,这些配合物是原位生成的,经光谱表征且未分离。Ti(NR'Ar)(NMe 2)3(12,R'= C(CD 3)2 Ph,Ti(NR'Ar)(NMe