From bismuth amides

Monomeric cyclopentadiene (10 g, 150 mmol) was added to a stirred solution of Bi(NMe2)j (9.00 g, 26.4 mmol) in absolute THF (60 ml) at -70°C. All manipulations need to be done in the dark. After 2 h, the mixture was allowed to warm to room temperature and was stirred for an additional 2 h. The mixture turned from deep purple to red, and a dark brown solid precipitated. The solution was filtered through Celite and the red colored filtrate was concentrated to about 6 ml, cooled to -30°C, and diluted with hexane (6 ml). After drying at 0°C in vacuo, tris(cyclo-pentadienyl)bismuthine was obtained as deep red, transparent crystals (7.40 g, 73.2%). The filtered brown solid was dried in vacuo to give (BiCsHs), as a brown, insoluble powder [95JOM(485)149]. 

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The tin(II) and lead(II) tellurides result from the amide M N(SiMe3)2k and tellurol. The tin derivative is dimeric with the Sn2Te2 unit adopting a butterfly-like structure with the terminal tellurol groups cis109. The antimony and bismuth derivatives result from amides and the tellurol96. 

A solution of bismuth tris[bis(trimethylsilyl)amide] BifNlSiMetfilt (0.98 g, 1.42 mmol) in light petroleum (10 ml) was added to 2,4,6-trimethylbenzenethiol (1.0 g, 4.25 mmol) in the same solvent (40 ml) at room temperature. The resulting red solution was stirred for 1 h and then the solvent was removed in vacuo. The residue was recrystallized from MeCN-ethyl acetate (3 1 v/ v) to give bismuth tris(2,4,6-trimethylbenzenethiolate) as orange crystals (1.1 g, 86%) [95JCS(D)1649]. 

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