Chapter 13 Lecture 2 More Ligand Types
I.
Ligands with Extended p Systems
A.
Linear p Systems
1) Ethylene (C2H4)
a) Single p-bond composed of two overlapping p-orbitals
b) One bonding and one antibonding p molecular orbital
2)
Allyl Radical (C3H5)
CH
CH2
HC
HC
CH
CH2
3)
1,3-Butadiene
4)
Other extended p systems
H2 C
CH2
CH2
HC
HC
HC
CH
CH
HC
HC
CH2
CH
CH
HC
CH2
CH2
B.
Cyclic p 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 p 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 p Systems
1) Ethylene Complexes
a) Sidebound geometry is most common
b) Bonding: s-donation from p MO, p-acceptance from p* MO
c) Coordination weakens C=C bond (137.5 pm, 1516 cm-1) compared to free
ethylene (133.7 pm, 1623 cm-1)
2)
p-Allyl Complexes
a) Can be trihapto: both s- and p-bonding
b) Can be monohapto: s-bonding only from sp2 hybrid orbital (120o bond angle)
c)
The lowest energy MO provides s-bonding, highest energy MO = p-acceptor
p3 is a p-acceptor
p2 can be donating
or accepting depending
on metal e- distribution
p1 is a s-donor
3)
Other linear p-system coordination
D.
Bonding in Cyclic p Systems
1) Cyclopentadienyl = Cp = C5H5- is the most important cyclic ligand
2)
Ferrocene Synthesis: FeCl2 + 2 NaC5H5
a) Called metallocene or sandwich complex
b) 18-electron complex: Fe2+ = d6 and 2 Cp x 6 ec) Bonding Group Orbitals of 2 eclipsed Cp rings
D5h
0-Node Group Orbitals
(h5-C5H5)2Fe + 2 NaCl
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
Reactivity
i. Follows 18-electron rule, but not inert
ii. Ligand reactions on Cp ring are most common reactions
f)
D5h
II.
M—C Single, Double, and Triple Bonds
A.
Metal Alkyl Complexes
1) Grignard Reagents: X—M—CH2CH2CH2CH3
2) Bonding in Transition Metal Complexes
a) s-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) s-bond from C sp2 hybrid to metal
b) p-bond from C p-orbital(s)
c) Heteroatom delocalizes p-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 Complexes
A.
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 n = 2000 cm-1 [V(CO)6]- n = 1858 cm-1 [Mn(CO)6]+ n = 2095 cm-1
b)
c)
Bonding Mode
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 p 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.
HC
3
S
Examples
N C
1) [(Cp)Mo(CO)3]2 + tds
Product?
H3C
S
tds
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)
S
CH3
C N
S
CH3
CO
OC
2)
OC
Re
CO
Br
C
O
PPh3, toluene, 
II
+
III
O
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
CO
CO
OC
Ph3P
Re
CO
II
OC
Br
C
O
O
Ph3P
Re
Br
C
PPh3 O
III
O