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Chapter 13 Lecture 2 More Ligand Types

Chapter 13 Lecture 2 More Ligand Types. Ligands with Extended p Systems Linear p Systems Ethylene (C 2 H 4 ) Single p -bond composed of two overlapping p-orbitals One bonding and one antibonding p molecular orbital Allyl Radical (C 3 H 5 ). 1,3-Butadiene Other extended p systems.

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Chapter 13 Lecture 2 More Ligand Types

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  1. Chapter 13 Lecture 2 More Ligand Types • Ligands with Extended p Systems • Linear p Systems • Ethylene (C2H4) • Single p-bond composed of two overlapping p-orbitals • One bonding and one antibonding p molecular orbital • Allyl Radical (C3H5)

  2. 1,3-Butadiene • Other extended p systems

  3. Cyclic p Systems • Cyclopropene • Similar construction of orbitals as in linear systems • Degenerate orbitals have the same number of nodes • Polygon Method for finding cyclic p system MO’s • Draw molecule as a polygon with vertex down • One MO per vertex gives energy ordering and degeneracy • Number of nodes increases as energy increases

  4. Bonding Between Metals and Linear p Systems • Ethylene Complexes • Sidebound geometry is most common • Bonding: s-donation from p MO, p-acceptance from p* MO • Coordination weakens C=C bond (137.5 pm, 1516 cm-1) compared to free ethylene (133.7 pm, 1623 cm-1) • p-Allyl Complexes • Can be trihapto: both s- and p-bonding • Can be monohapto: s-bonding only from sp2 hybrid orbital (120o bond angle)

  5. c) The lowest energy MO provides s-bonding, highest energy MO = p-acceptor • Other linear p-system coordination p3 is a p-acceptor p2 can be donating or accepting depending on metal e- distribution p1 is a s-donor

  6. Bonding in Cyclic p Systems • Cyclopentadienyl = Cp = C5H5- is the most important cyclic ligand • Ferrocene Synthesis: FeCl2 + 2 NaC5H5 (h5-C5H5)2Fe + 2 NaCl • Called metallocene or sandwich complex • 18-electron complex: Fe2+ = d6 and 2 Cp x 6 e- • Bonding Group Orbitals of 2 eclipsed Cp rings D5h 0-Node Group Orbitals

  7. Matching with metal d-orbitals: dyz orbital example • MO Description • 6 strongly bonding MO’s hold electrons from Cp ligands • 8 antibonding orbitals are empty • 5 mid-range energy orbitals holding metal d-electrons • Reactivity • Follows 18-electron rule, but not inert • Ligand reactions on Cp ring are most common reactions

  8. D5h

  9. M—C Single, Double, and Triple Bonds • Metal Alkyl Complexes • Grignard Reagents: X—M—CH2CH2CH2CH3 • Bonding in Transition Metal Complexes • s-donation from C sp3 hybrid orbital • 2 electron, -1 charge for electron counting • Synthesis • ZrCl4 + 4 PhCH2MgCl Zr(CH2Ph)4 • Na[Mn(CO)5] + CH3I CH3Mn(CO)5 + NaI • Other M—C single bond ligands

  10. Metal Carbene Complexes • M=C counted as 2 electron, neutral ligand in electron counting • Schrock Alkylidenes: only H or C attached to the carbene Carbon • Fisher Carbenes: heteroatom attached to the carbene Carbon (our focus) • s-bond from C sp2 hybrid to metal • p-bond from C p-orbital(s) • Heteroatom delocalizes p-system to 3 atoms, stabilizing it by resonance

  11. Metal Carbyne Complexes • First synthesis in 1973 by Lewis Acid attack on carbene complex • Bonding • 180o bond angle and short bond length confirm triple bond • 3 electron, 0 charge for electron counting

  12. Spectroscopy of Organometallic Complexes • Infrared Spectroscopy • Number of Bands is determined by group theory (chapter 4 procedure) • Monocarbonyl = 1 band only • Dicarbonyl • Linear arrangement = 1 band only • Bent arrangement = 2 bands • 3 or more Carbonyls: table 13.7 in your book • Position of IR Bands • Electron Density determines Wavenumbers Cr(CO)6n = 2000 cm-1 [V(CO)6]-n = 1858 cm-1[Mn(CO)6]+n = 2095 cm-1 • Bonding Mode • Other ligands

  13. NMR Spectroscopy • Proton NMR • Hydride Complexes M—H hydrogen strongly shielded (-5 to –20 ppm) • M—CH3 hydrogens 1-4 ppm • Cyclic p system hydrogens 4-7 ppm and large integral because all the same • 13C NMR • Useful because “sees” all C ligands (CO) and has wide range (ppm) • CO: terminal = 195-225 ppm, bridging slightly larger

  14. tds • Examples • [(Cp)Mo(CO)3]2 + tds Product? • Data: 1H NMR: 2 singlets at 5.48 (5H) and 3.18 (6H) IR: 1950, 1860 cm-1 Mass = 339 • Solution: proton nmr 5.48 = Cp, 3.18 = ½ tds IR: at least 2 CO’s Mass: 339 - (Mo=98) – (Cp=65) – 2(CO) = 120 = ½ tds Product = (Cp)Mo(CO)2(S2CN(CH3)2) I: proton = 4.83 (4H), carbon = 224, 187, 185, 184, 73 II: proton = 7.62-7.41 m (15H), 4.19 (4H) carbon: 231, 194, 189, 188, 129-134,72 III: proton = 7.70-7.32 m (15H), 3.39 s (2 H) carbon: 237, 201,193,127-134, 69 Solution: 224 = M=C; 184-202 = CO; 73 = CH2CH2

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