Lecture 2

Question 1

Factors affecting likelihood of metal-olefin complex formation

Answer 1

Sigma-acceptor properties of the metal (associated with the electron affinity of the metal)
Pi-donor properties of the metal (associated with the promotion energy of the metal)

Question 2

Ni(0)

Answer 2

Electron rich (d10) so can donate electron density
Low electron affinity (poor sigma-acceptor)
Low promotion energy (so good pi-donor)

Question 3

Hg(II)

Answer 3

Electron poor so can accept electron density
High electron affinity (good sigma-acceptor)
High promotion energy (poor pi-donor)

Question 4

Pd(II)

Answer 4

Intermediate electronics (d8) c.f. Ni(0)
Good sigma-acceptor (high electron affinity) and good pi-donor (low promotion energy)

Question 5

Orbital-matched ligand/metal combinations

Answer 5

Sigma bond = L(pi) to M s, pz, dz2, d(x2-y2)
Pi bond = M py, dij to L(pi*)
DRAW

Question 6

Which type of bond dominates the bonding contribution?

Answer 6

M–>L (esp for C2H4)

Question 7

Effect of bonding on bond order of C=C bond

Answer 7

Both L–>M and M–>L bonding reduce the order of the C=C bond so it is much less than 2
i.e. both sigma and pi interactions weaken the C=C bond

Question 8

How can we see the effects of C=C bond weakening

Answer 8

IR spectroscopy

Question 9

Free C2H4 frequency

Answer 9

1623 cm^-1

Question 10

Dewar-Duncanson-Chatt model

Answer 10

3c-2e MC2 sigma- and pi-bonds
‘Donor-acceptor’ type bonding
Alkene should be planar (but is not)

Question 11

Model for bonding in metal-olefin complexes

Answer 11

In between 3c-2e donor-acceptor model and that of an epoxide-type covalent bonding model
(Epoxide = 2c-2e bond, sp3 hybridised (even though geometry/bond angles aren’t sp3))
The best approximation for the bonding is to combine 2 x M-C 2c-2e bonds to give delocalised orbitals, where the electrons are delocalised around the system
This description suggests more p-character (i.e. towards sp3 bonding)
eta2 ethylene is non-planar
M-C bond is short
C=C bond is long

Question 12

Evidence for the bonding model

Answer 12

1. If eta2 coordination of alkenes results in loss of planarity (i.e. pyramidalisation) then strained, non-planar alkenes should be pre-disposed to forming strong bonds to TMs (which is the case)
2. NMR spectroscopy - coordinated alkenes/alkynes display hindered rotation - rotational barrier due to M–>L backbonding. Presence of electronegative groups on alkene increases amount of back-bonding, meaning rotation requires more energy so higher temps are required to reach coalescence in NMR

Question 13

Cp ligand

Answer 13

Good sigma donor
Amazing pi donor
Pi acceptor