CpM fragment

The five-membered heterocycle Et2C2B2Mc2S reacts with 17 -CpM fragments (M = Ni, Co) from [Tj -CpNi(CO)2]2 and i) -CpCo(CO)2 to form mixed sandwich complexes of the type rj -CpM(Et2C2B2MeS) , whereas Ni(CO>4 yields the bis(di-hydrothiadiborol) complex. The (Et2C2B2Me2S)Ni(CO)2 is formed in the first step. 

Exercise 4.4. Suggest a CpM fragment isolobal with a BH fragment. Systematically replace one and two BH fragments in B5H9 with this metal fragment and enumerate all isomers. 

Answer. A B H fragment possesses three orbitals (two of tt and one of a symmetry) containing two electrons. Hence, a CpM fragment with M = a group-9 metal is the appropriate choice (Figure 4.7). This leads to metal derivatives of B5H9 of the type (CpM) (BH) 5(H)4 for M = Co, Rh or Ir. The square pyramidal mVio-clustcr possesses two types of vertices, one apical and four basal. Hence, for n= 1, two isomers are possible whereas for n = 2, three isomers are possible. The structures... 

A CpM fragment (like CpMn) will have C2 + ax filled. The + 2a set of three levels are empty and because of their hybridization (see 20.6 and 20.7) will form the strongest interactions with extra ligands. 

The reader should carefully compare these valence orbitals with those of a ML3 fragment in Figure 20.1. The ML3 fragment has the same three below, three above level pattern and almost identical atomic composition at the metal. In the CpM fragment it is clear that the cs set is of 8 symmetry and is of tt symmetry. However, c in ML3 is primarily 8 with some tt character and 2e is mainly Trwilh... 

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