1. Substitution reactions of square planar complexes
Chapter 7

2. especially d8: Ni(II), Rh(I), Pd(II), Ir(I), Pt(II), Au(III)

3. A P B [ML3X] + Y  [ML3Y] + X Reaction: [ML3X]  A General rate law:
General mechanism A P B
k2 [Y]
k3 [Y] k1 +S
k-1 -S

4. (1)
Assume [B] is in steady state
Substituting into (1)

5. Two situations usually arise for the solvent pathway
The rate of attack of solvent on A is rate limiting
k3[Y] >> k-1
which is in agreement with the experimental rate law

6. Two situations usually arise for the solvent pathway
The rate of attack of solvent on A is much faster than attack of Y on the intermediate B
k3[Y] << k-1

7. Study the rate of the reaction as a function of [Y]k2 k1

8. k2o as the 2nd order rate constant when Y = MeOH in the reaction
The k2 pathway
Define
k2o as the 2nd order rate constant when Y = MeOH
in the reaction
trans-[PtCl2(py)2] + Y  trans-[PtClY(py)2] + Cl
then compare the rate for any other ligand Y to the rate when Y = MeOH
The greater Pt, the greater the nucleophilicity of the ligand
nucleophilicity parameter

9. nucleophilicity increases with softness of the donor ligand

10. where C = log k2o
Now generalise for any square planar Pt complex [PtL3X]
[PtL3X] + Y  [PtL3Y]+ X

11. S is the nucleophilic discrimination factor and gives the sensitivity of the rate constant to the nucleophilicity of the incoming ligand

12. trans-[PtCl2(PEt3)2] S is larger [PtCl2(en)] SeCN SCN I thiourea
NH3 I
Br
Cl
CH3OH
NO2
CH3OH

13. The discrimination factor,
Usually...
The discrimination factor,
S, decreases
ligand, MeOH, increases
As reactivity towards
the common

14. All values are significantly > 0, i. e
All values are significantly > 0, i.e., all complexes undergo substitution reactions that are quite sensitive to the nucleophilicity of the entering ligand
This sensitivity is expected for reactions under associative activation

15. 2 Cl, 2 P 2Cl, 2 aromatic N Cl, 3 aliphatic N 2Cl, 2 aliphatic N
As softness of ligands on Pt increases, S increases – the complexes becomes less reactive and more discriminating

16. Example
Calculate the second-order rate constant for the reaction of trans-[PtCl(CH3)(PEt3)2] with NO2, for which Pt = For this complex, I (Pt = 5.42) and N3 (Pt = 3.58), react at 30 oC with k = 40 M-1 s-1 and 7 M-1 s-1, respectively.

17. Hence, for NO2

18. Two important observations:
The nature of the transition state
Two important observations:
For the generalised reaction
whether the predominant product is A or B depends on the relative trans effect of the spectator ligands L1 and L2

19. The rate of the reaction
depends significantly on the nature of the trans ligand, T, but hardly at all on the cis ligands C

20. The trans effect order is
For  donor ligands
H- > PR3 > SCN- > I- > CH3-, CO, CN- > Br-, Cl- > NH3, py > OH-, H2O
Stronger σ donors Weaker σ donors
Stronger π acceptors Weaker π acceptors
For  acceptor ligands
CO, C2H2 > CN- > NO2- > NCS- > I- > Br-

21. Observations consistent with a trigonal bipyramidal transition state in which the cis ligands are axial, and T, X and Y are equatorial

22. If T is a good  donor ligand, it is readily polarisable...
...and this weakens and labilises the MX bond.
...and polarises electron density from M towards it (i.e., the TM bond has significant covalency...
i.e., T destabilises the ground state

23. If T is a good π acceptor ligand…
as X departs in the transition state, there is a build-up of electron density on the metal...
...which can be accommodate by  donation to the T ligand.
i.e., T stabilises the transition state

24. Example AN ASIDE The trans effect order can be exploited in synthesis
Given that the trans effect order is PPh3 > Cl- > NH3, explain how to synthesise trans-[PtCl2(NH3)(PPh3)] starting from [PtCl4]2-
How would you synthesise the cis complex?

25. Steric Effects
Steric crowding at a metal centre will retard an associative reaction, but speed up a dissociative reaction

28. Stereochemistry
The stereochemistry at the metal centre is preserved, consistent with a transition state in which the entering (Y), leaving (X) and trans (T) ligands are in the plane of a trigonal bipyramidal complex

29. The intermediate must be shortlived, else scrambling of stereochemistry would be expected

30. Activation parameters
Both the k1 and the k2 pathways have S‡ and V‡ values that are negative. For example:
k1 k2
S‡ /J K-1 mol
V‡ /cm3 mol

31. The k1 pathway
B is a solvento intermediate.
The solvento intermediate has been trapped and isolated in some cases

32. The solvento intermediate has been trapped and isolated in some cases
kinetically inert