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Substitution reactions of square planar complexes

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Substitution reactions of square planar complexes Chapter 7 * especially d8: Ni(II), Rh(I), Pd(II), Ir(I), Pt(II), Au(III) General mechanism A P B k2 [Y] k3 [Y] k1 +S ... – PowerPoint PPT presentation

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Title: Substitution reactions of square planar complexes


1
Substitution reactions of square planar complexes
2
especially d8 Ni(II), Rh(I), Pd(II), Ir(I),
Pt(II), Au(III)
3
ML3X ? A
4
(1)
Assume B is in steady state
Substituting into (1)
5
Two situations usually arise for the solvent
pathway
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
k3Y ltlt k-1
7
Study the rate of the reaction as a function of
Y
8
The k2 pathway
The greater ?Pt, the greater the nucleophilicity
of the ligand
nucleophilicity parameter
9
nucleophilicity increases with softness of the
donor ligand
10
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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
SeCN?
SCN?
I?
thiourea
SCN?
PtCl2(en)
N3?
I?
NH3
Br?
Cl?
CH3OH
NO2?
CH3OH
13
Usually...
The discrimination factor, S, decreases
As reactivity towards the common ligand, MeOH,
increases
14
All values are significantly gt 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
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 3.22. 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
The nature of the transition state
Two important observations
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- gt PR3 gt SCN- gt I- gt CH3-, CO, CN- gt Br-, Cl- gt
NH3, py gt OH-, H2O
For ? acceptor ligands
CO, C2H2 gt CN- gt NO2- gt NCS- gt I- gt 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 p 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
AN ASIDE
The trans effect order can be exploited in
synthesis
Example
Given that the trans effect order is PPh3 gt Cl- gt
NH3, explain how to synthesise trans-PtCl2(NH3)(P
Ph3) starting from PtCl42-
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
26
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27
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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
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
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