Answers are grounded in this paper and cite their sources.
The reaction of 2,6-[ArN(SiMe 3 )CH 2 ] 2 NC 5 H 3 ( 1, BDPP(SiMe 3 ) 2, Ar = 2,6- i Pr 2 C 6 H 3 ) with TaCl 5 yields the complex mer -(BDPP)TaCl 3 ( 2 ) and 2 equiv of ClSiMe 3 . The reduction of compound 2 with excess Na/Hg in the presence of alkynes yields the pseudo 5-coordinate Ta(III) derivatives (BDPP)Ta(η 2 -RC⋮CR‘)Cl ( 3a, R = R‘ = Pr; 3b, R = R‘ = Et; 3c, R = R‘ = Ph; 3d, R = Ph, R‘ = H). An X-ray study of 3a revealed a distorted square pyramidal geometry with the chloride occupying the apical position. Compound 3a reacts with LiC⋮CR to give the acetylide octyne derivatives (BDPP)Ta(η 2 -PrC⋮CPr)(C⋮CR) ( 5a, R = Ph; 5b, R = Bu; 5c, R = SiMe 3; 5d, R = o -Me 3 SiC 6 H 4 ). Compound 5a reacts with phenylacetylene to give the metallacyclic derivative (BDPP)Ta[(η 2 -PhC⋮C)PrC CPrHC CPh] ( 6a ). An X-ray study of 6a again revealed a square pyramidal geometry with the alkenyl carbon occupying the apical position. Similarly, compounds 5b, c react with HC⋮CBu and HC⋮CSiMe 3 to give the metallacycles (BDPP)Ta[(η 2 -BuC⋮C)PrC CPrHC CBu] ( 6b ) and (BDPP)Ta[(η 2 -Me 3 SiC⋮C)PrC CPrHC CSiMe 3 ] ( 6b ), respectively. Compound 5b reacts with HC⋮CPh to give (BDPP)Ta[(η 2 -BuC⋮C)PrC CPrHC CPh] ( 7b ) only, establishing that the starting acetylide is retained in the final product.
No takes yet. Share an insight, caveat, or question.
Guérin et al. (1998) studied this question.
Synapse has enriched 10 related papers. Consider them for comparative context: