作者:Jack M. Miller、Raj K. Chadha
DOI:10.1016/s0022-328x(00)85758-7
日期:1981.8
RTeCl3 (R C6H5, p-CH3OC6H4 or p-C6H5OC6H4) reacts with Me3SiNR′2 (R′2 Et2, C4H8) under dry nitrogen atmosphere to give R(R′N)TeCl2 and Me3- SiCl. The products readily decompose to give (R′2NH2)+(RTeCl4). The prod ucts have been characterized by 1H NMR, IR and mass spectra. R2TeCl2 does not react with Me3SiNR'2 even on refluxing for 6 h. Et2NLi, however, reduces R2TeCl2 to R2Te.
RTeCl 3(Rç 6 ħ 5,p -CH 3 OC 6 H ^ 4或p -C 6 H ^ 5 OC 6 H ^ 4)反应与我3 SINR' 2(R' 2 的Et 2,C 4 H ^ 8)在干燥的氮气气氛下,得到R(R'N)TeCl 2和Me 3- SiCl。该产品容易分解,得到(R' 2 NH 2)+(RTeCl4)。产品已通过1 H NMR,IR和质谱表征。R 2 TeCl 2即使回流6小时也不会与Me3SiNR'2反应。但是,Et2NLi会将R 2 TeC 12还原为R 2 Te。