Yves Jean - Molecular Orbitals of Transition Metal Complexes (793957), страница 40
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Riehl, Y. Jean, O. Eisenstein, M. Pélissier Organometallics 11, 729(1992) (§ 4.2.2).J.-Y. Saillard, R. Hoffmann J. Am. Chem. Soc. 106, 2006 (1984) (§ 4.1.4).S. Shaik, R. Hoffmann, C. R. Fisel, R. H. Summerville J. Am. Chem.Soc. 102, 1194 (1980) (§ 4.4).D. C. Smith, W.
A. Goddard III J. Am. Chem. Soc. 109, 5580 (1987)(§ 4.4).K. Tatsumi, R. Hoffmann, A. Yamamoto, J. K. Stille Bull. Soc. Chim. Japan54, 1857 (1981) (§ 4.5)ExercisesConformation of d10 -[(η2 -C2 H4 )ML2 ] complexes4.11. Which is the more stable conformation for an [(η2 -C2 H4 )ML2 ]complex with a d10 electronic configuration, 1 or 2?ApplicationsLLMMLL122. How does the barrier to olefin rotation change when the substituents R are varied: R==CN (tetracyanoethylene), R==H (ethylene),R==Cl (tetrachloroethylene)?3.
Repeat question 2, replacing the two non-ethylene ligands on the−M−−L angle tometal by a bidentate ligand that constrains the L−◦be much larger than 120 .ML3 complexes4.21. Starting from the d orbitals of an octahedron, construct thed orbitals (shapes and relative energies) for a ‘T-shaped’ ML3−M−−L3 = 180◦ , L1 −−M−−L2 = L1 −−M−−L3 = 90◦ ).fragment (L2 −L3L1L2Mzyx2. How do the orbital energies of this fragment change on movingfrom the T-shape to a trigonal geometry (120◦ bond angles)?L3L3L1L2ML1ML23.
From the results you obtain in 2 above, predict the geometriesadopted by the complexes [Rh(PPh3 )3 ]+ and [Pt(PPh3 )3 ].Octahedral dioxo complexes4.3We wish to compare the relative stabilities of the cis and trans isomersof d0 and d2 octahedral dioxo complexes; remember that the oxogroup is an X2 -type ligand.ExercisesOzyxMMOOOCisTrans1. Give the shapes and relative energies of the d orbitals for an octahedral complex in the absence of any π-type interactions, in thetwo orientations shown above.
In what follows, we shall only considerthe nonbonding d orbitals.2. We now turn to dioxo complexes. If the four non-oxo ligands arepure σ donors, give the interaction schemes d (metal) ↔ p (oxo)for each isomer.3. Which isomer is the more stable for a d0 complex? For a d2complex?4. Deduce the structures of the complexes [OsO2 F4 ]2− and[MoO2 Cl4 ]2− .Carbene–dihydrogen coupling in an octahedral complex4.4In the molecular hydrogen complex [Os(NH3 )4 (CR 2 )(H2 )]2+ , aFischer carbene and a dihydrogen molecule are in cis positions.
Weshall consider four limiting structures (0, 0), (0, 90), (90, 0), and (90,90) that are characterized by the orientations of the carbene (firstangle) and of the dihydrogen molecule (second angle).HOsOsHHHCC(0, 0)(0, 90)HHHC(90, 0)HOsOsC(90, 90)1. What is the electronic configuration dn for this complex?2.
For each of the four conformations above, analyse the backdonation interactions towards the carbene and towards thedihydrogen.3. From the results of this analysis, suggest the favoured structure(s)for this complex.4. Which limiting conformations are possible for the trans isomer?Which should be the most stable a priori?ApplicationsCoupling of two trans carbenes in an octahedral complex4.5Consider the conformations 1 and 2 of octahedral complexes inwhich two Fischer carbenes are in trans positions. The non-carbeneligands are pure σ donors.1. Give the shapes and relative energies of the three lowest-energyd-block MOs for the two conformations.CCMMCC122. Deduce the conformations for a d2 , d4 , or d6 diamagnetic complex.3.
How is the barrier to rotation modified if the non-carbene ligandsare π acceptors?The reductive elimination reaction4.6We wish to study the reductive elimination of H2 from an octahedral complex with a d6 electronic configuration, following themechanism shown in 4-56, which leads to a square-planar d8 ML4complex.L1L1L4L3MHHML3L2L4HHL24-561. Sketch the correlation diagram for the MO associated with thisreaction. For the reactant, consider the nonbonding d MO and the−H bonds which will be broken;MO that characterize the two M−for the products, the MO of H2 , the occupied d orbitals, and thenonbonding p orbital of the ML4 complex. Two planes should beused to characterize the symmetries of the MO.2. Is this reaction allowed by symmetry?3. How does the correlation diagram show the reduction of themetal? The isolobal analogy1R.
Hoffmann ‘Building bridges betweeninorganic and organic chemistry (Nobellecture)’, Angew. Chem. Int. Ed. 21, 711 (1982).2J. Halpern in Advances in Chemistry;Homogeneous Catalysis, No 70, AmericanChemical Society, Washington, DC (1968)pp. 1–24.In Chapter 2, the d orbitals (and, in some cases, the s, p, or s–p hybridorbitals on the metal) were constructed for many types of MLn complexes in which the ligands have σ -type interactions with the metal. Theshapes and relative energies of these orbitals depend on the number ofligands and their geometrical arrangement around the metal. Naturally,any π-type interactions that are present also play a role. Thus, the dorbitals in the octahedral complex [WCl6 ] (six π -donor ligands) are different from those in the octahedral complex [W(CO)6 ] (six π-acceptorligands).
But ‘there is a time for detail and there is a time for generality’1 : inboth cases, the d orbitals are split 3 + 2, with three degenerate orbitals (the t2g block) lower in energy than the two degenerate orbitalsof the eg block, and this result is characteristic of all octahedral ML6complexes. In the same way, the orbital structure of AHn molecules orfragments (or more generally AR n ), in which A is a main-group element, depends essentially on the number and geometrical arrangementof the substituents on the central atom. But beyond the obvious difference created by the presence of d orbitals in organometallic complexes,there are resemblances between the molecular orbitals (MO) of MLncomplexes and those of AHn molecules. These similarities can lead torelated properties: thus J.
Halpern had noted in 1968 the similarity inbehaviour of organic radicals and of d7 ML5 complexes.2 But it is chieflyR. Hoffmann who developed this concept and showed its remarkablefruitfulness.15.1. The analogy between fragments of octahedralML6 and of tetrahedral CH4In many cases, we established the orbitals of MLn fragments earlier inthis book by starting from a complex in which one or several ligands wereremoved. For example, the orbitals of ML5 complexes (SBP, C4v ), ML4(‘butterfly’, C2v ), and ML3 (pyramidal, C3v ) were obtained from thoseof an octahedral ML6 complex, by removing one, two, or three ligands,respectively (Chapter 2, § 2.3.1, 2.8.1, and 2.8.3). In certain respects,each of these complexes was therefore considered as a fragment of anoctahedron (5-1).The isolobal analogyML5 (SBP)ML4 (‘butterfly’)ML6ML3 (pyramidal)5-1The same procedure can be applied to molecules of the type AHn ,where A is an element from the second or third row of the periodic table.In particular, starting from tetrahedral methane, CH4 , the archetype oforganic molecules, one can obtain the organic fragments pyramidalCH3 (C3v ), bent CH2 (C2v ), and CH(C∞v ), by removing one, two, orthree atoms of hydrogen (5-2).HHHCCH3 (pyramidal)CCH2 (bent)CCHHHHHHCH4CHHHCHHHHHCHH5-2Even though these comments are purely geometrical in nature, itturns out that they have profound chemical consequences.
Thus, anML5 complex (SBP) possesses a vacant site, and it is likely to bind anadditional ligand to form an octahedral complex. In the same way, theML4 (C2v ) and ML3 (C3v ) fragments have two and three available sites,respectively, with respect to the initial octahedron. In the organic series,The analogy between fragments of octahedral ML6 and of tetrahedral CH4the CH3 , CH2 , and CH fragments possess the same characteristics, withthe potential to form one, two, or three new bonds.This similarity in the capacities of inorganic and organic fragmentsto form new bonds can be expressed in orbital terms, particularly if weconsider the number, the shape, and the electronic occupation of the orbitalsthat are available to form these new bonds.
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