I.Ligands with Extended Systems A.Linear Systems 1)Ethylene (C 2 H 4 ) a)Single -bond composed of two overlapping p-orbitals b)One bonding and one.

I. Ligands with Extended SystemsA. Linear Systems

1) Ethylene (C2H4)

a) Single -bond composed of two overlapping p-orbitals

b) One bonding and one antibonding molecular orbital

2) Allyl Radical (C3H5)

Chapter 13 Lecture 2 More Ligand Types

CH

HC

CH2 CH

HC

CH2

3) 1,3-Butadiene

4) Other extended systems

CH2

HC

CH

HC

CH2 CH2

HC

CH

HC

CH2

CH2

HC

CH

HC

CH

H2C

B. Cyclic Systems

1) Cyclopropene

a) Similar construction of orbitals as in linear systems

b) Degenerate orbitals have the same number of nodes

2) Polygon Method for finding cyclic system MO’s

a) Draw molecule as a polygon with vertex down

b) One MO per vertex gives energy ordering and degeneracy

c) Number of nodes increases as energy increases

C. Bonding Between Metals and Linear Systems

1) Ethylene Complexes

a) Sidebound geometry is most common

b) Bonding: -donation from MO, -acceptance from * MO

c) Coordination weakens C=C bond (137.5 pm, 1516 cm-1) compared to free ethylene (133.7 pm, 1623 cm-1)

2) -Allyl Complexes

a) Can be trihapto: both - and -bonding

b) Can be monohapto: -bonding only from sp2 hybrid orbital (120o bond angle)

c) The lowest energy MO provides -bonding, highest energy MO = -acceptor

3) Other linear -system coordination

2 can be donatingor accepting dependingon metal e- distribution

1 is a -donor

3 is a -acceptor

D. Bonding in Cyclic Systems

1) Cyclopentadienyl = Cp = C5H5- is the most important cyclic ligand

2) Ferrocene Synthesis: FeCl2 + 2 NaC5H5 (5-C5H5)2Fe + 2 NaCl

a) Called metallocene or sandwich complex

b) 18-electron complex: Fe2+ = d6 and 2 Cp x 6 e-

c) Bonding Group Orbitals of 2 eclipsed Cp rings

0-Node Group Orbitals

D5h

d) Matching with metal d-orbitals: dyz orbital example

e) MO Description

i. 6 strongly bonding MO’s hold electrons from Cp ligands

ii. 8 antibonding orbitals are empty

iii. 5 mid-range energy orbitals holding metal d-electrons

f) Reactivity

i. Follows 18-electron rule, but not inert

ii. Ligand reactions on Cp ring are most common reactions

D5h

II. M—C Single, Double, and Triple BondsA. Metal Alkyl Complexes

1) Grignard Reagents: X—M—CH2CH2CH2CH3

2) Bonding in Transition Metal Complexes

a) -donation from C sp3 hybrid orbital

b) 2 electron, -1 charge for electron counting

3) Synthesis

a) ZrCl4 + 4 PhCH2MgCl Zr(CH2Ph)4

b) Na[Mn(CO)5] + CH3I CH3Mn(CO)5 + NaI

4) Other M—C single bond ligands

B. Metal Carbene Complexes

1) M=C counted as 2 electron, neutral ligand in electron counting

2) Schrock Alkylidenes: only H or C attached to the carbene Carbon

3) Fisher Carbenes: heteroatom attached to the carbene Carbon (our focus)

a) -bond from C sp2 hybrid to metal

b) -bond from C p-orbital(s)

c) Heteroatom delocalizes -system to 3 atoms, stabilizing it by resonance

C. Metal Carbyne Complexes

1) First synthesis in 1973 by Lewis Acid attack on carbene complex

2) Bonding

a) 180o bond angle and short bond length confirm triple bond

b) 3 electron, 0 charge for electron counting

III. Spectroscopy of Organometallic ComplexesA. Infrared Spectroscopy

1) Number of Bands is determined by group theory (chapter 4 procedure)

a) Monocarbonyl = 1 band only

b) Dicarbonyl

i. Linear arrangement = 1 band only

ii. Bent arrangement = 2 bands

c) 3 or more Carbonyls: table 13.7 in your book

2) Position of IR Bands

a) Electron Density determines Wavenumbers

Cr(CO)6 = 2000 cm-1 [V(CO)6]- = 1858 cm-1 [Mn(CO)6]+ = 2095 cm-1

b) Bonding Mode

c) Other ligands

B. NMR Spectroscopy

1) Proton NMR

a) Hydride Complexes M—H hydrogen strongly shielded (-5 to –20 ppm)

b) M—CH3 hydrogens 1-4 ppm

c) Cyclic system hydrogens 4-7 ppm and large integral because all the same

2) 13C NMR

a) Useful because “sees” all C ligands (CO) and has wide range (ppm)

b) CO: terminal = 195-225 ppm, bridging slightly larger

C. Examples

1) [(Cp)Mo(CO)3]2 + tds Product?

a) Data: 1H NMR: 2 singlets at 5.48 (5H) and 3.18 (6H)

IR: 1950, 1860 cm-1

Mass = 339

b) 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)

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

S SC

S

NCH3

CH3

CS

N

H3C

H3C

tds

CO

Re

CO

OC Br

COCO

O

PPh3, toluene, II + III

CO

Re

CO

OC Br

CPh3PO

OII

CO

Re

PPh3

OC Br

CPh3PO

O

III