Post on 03-Jan-2016
General mechanism
A
P
B
k2 [Y]
k3 [Y]
k1
+Sk-1
-S
General rate law:1 2Rate ( [Y])[A]k k
Reaction:[ML3X] + Y [ML3Y] + X
[ML3X] A
A
P
B
k2 [Y]
k3 [Y]
k1
+Sk-1
-S
A
P
B
k2 [Y]
k3 [Y]
k1
+Sk-1
-S2 3
[P][A][Y] [B][Y]
dk k
dt
Assume [B] is in steady state
1 1 3
1
1 3
[A] [B] + [B][Y]
[A][B] =
( [Y])
k k k
k
k k
Substituting into (1)
(1)
12 3
1 3
[A][P][A][Y] + [Y]
( [Y])
kdk k
dt k k
1 2Rate ( [Y])[A]k k
A
P
B
k2 [Y]
k3 [Y]
k1
+Sk-1
-S
A
P
B
k2 [Y]
k3 [Y]
k1
+Sk-1
-S
12 3
1 3
[A][P][A][Y] + [Y]
( [Y])
kdk k
dt k k
Two situations usually arise for the solvent pathway
The rate of attack of solvent on A is rate limiting
k3[Y] >> k-1
1 2Rate ( [Y])[A]k k
12 3
3
2 1
1 2
[A][P][A][Y] + [Y]
( [Y])
[A][Y] + [A]
( [Y])[A]
kdk k
dt k
k k
k k
which is in agreement with the experimental rate law
A
P
B
k2 [Y]
k3 [Y]
k1
+Sk-1
-S
A
P
B
k2 [Y]
k3 [Y]
k1
+Sk-1
-S
12 3
1 3
[A][P][A][Y] + [Y]
( [Y])
kdk k
dt k k
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
1 2Rate ( [Y])[A]k k
12 3
1
1 32
1
[A][P][A][Y] + [Y]
+ [A][Y]
'[A][Y]
kdk k
dt k
k kk
k
k
1 32
1
[P]+ [A][Y]
'[A][Y]
k kdk
dt k
k
1 2
[P]( [Y])[A]
dk k
dt
Study the rate of the reaction as a function of [Y]
ko b s
[Y ]
ko b s
[Y ]
k1
k2
1 32
1
+ k k
kk
0
The k2 pathwayA
P
B
k2 [Y]
k3 [Y]
k1
+Sk-1
-S
A
P
B
k2 [Y]
k3 [Y]
k1
+Sk-1
-SDefine
k2o as the 2nd order rate constant when Y = MeOH
in the reactiontrans-[PtCl2(py)2] + Y trans-[PtClY(py)2] + Cl
then compare the rate for any other ligand Y to the rate when Y = MeOH
2Pt o
2
(Y)log
k
k
nucleophilicity parameter
The greater Pt, the greater the
nucleophilicity of the ligand
2Pt o
2
(Y)log
k
k
oPt 2 2
o2 Pt 2
2 Pt
log (Y) log
log (Y) log
log (Y)
k k
k k
k C
where C = log k2o
Now generalise for any square planar Pt complex [PtL3X]
[PtL3X] + Y [PtL3Y]+ X
2 Ptlog (Y)k S C
S is the nucleophilic discrimination factor and gives the sensitivity of the
rate constant to the nucleophilicity of the
incoming ligand
lo g k 2
P t
CH3OH
CH3OH
NO2
Cl
Br
I
SCN
SeCN
NH3
N3
I
SCN
thiourea
trans-[PtCl2(PEt3)2]
[PtCl2(en)]
S is larger
lo g k 2
P t
Usually...
As
reac
tivi
ty t
owar
ds
the
com
mon
liga
nd
, MeO
H, i
ncr
ease
s
Th
e discrim
ination
factor,S
, decreases
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
As softness of ligands on Pt increases, S increases
– the complexes becomes less reactive
and more discriminating
2Cl, 2 aliphatic N
Cl, 3 aliphatic N
2Cl, 2 aromatic N
2 Cl, 2 P
ExampleCalculate 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.
2 Ptlog (Y)k S C
log 40 5.42 (1)
log 7 3.58 (2)
S C
S C
(1) (2)
0.755 1.84
0.410
S
S
1.60 5.42 0.410
0.62
C
C
For the generalised reaction
L1
PtL2 Cl
Cl
L1
PtL2 Cl
Y
L1
PtL2 Y
Cl
Y
Y
A
B
whether the predominant product is A or B depends on the relative trans effect of the spectator ligands L1 and L2
Two important observations:
The nature of the transition state
The rate of the reaction
depends significantly on the nature of the trans ligand, T, but hardly at all on the cis ligands C
C
PtT Cl
C
Y
C
PtT Y
C
The trans effect order is
For donor ligands
H- > PR3 > SCN- > I- > CH3-, CO, CN- > Br-, Cl- > NH3, py > OH-, H2O
For acceptor ligands
CO, C2H2 > CN- > NO2- > NCS- > I- > Br-
Stronger σ donors Weaker σ donors
Stronger π acceptors Weaker π acceptors
Observations consistent with a trigonal bipyramidal transition state in which the cis ligands are axial, and T, X and Y are equatorial
‡
‡
MT
C X
C
T M
S
X
C
C
T M
Y
X
C
C
T M
Y
X
C
C
MT
C Y
C
S
Y
Y
S
‡
T M
Y
X
C
C
If T is a good donor ligand, it is readily polarisable...
...and polarises electron density from M towards it (i.e., the TM bond has significant covalency...
...and this weakens and labilises the MX bond.
i.e., T destabilises the ground state
‡
T M
Y
X
C
C
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
If T is a good π acceptor ligand…
AN ASIDE
The trans effect order can be exploited in synthesis
Example
Given that the trans effect order is PPh3 > Cl- > NH3, explain how to synthesise trans-[PtCl2(NH3)(PPh3)] starting from [PtCl4]2-
Pt
Cl
Cl
Cl
Cl
2-
PPh3Pt
PPh3
Cl
Cl
Cl
2-
NH3Pt
PPh3
Cl
NH3
Cl
2-
How would you synthesise the cis complex?
Steric Effects
Steric crowding at a metal centre will retard an associative reaction, but speed up a dissociative reaction
Pt
L
Cl PEt3
PEt3
+
Pt
L
H2O PEt3
PEt3
2+
H2O N N N
L =
k = 8 x 10-2 2 x 10-4 1 x 10-6 s-1
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
C
T C
XC
CX
Y
T
C
T C
YC
T C
X
Y
X
The intermediate must be shortlived, else scrambling of stereochemistry would be expected
C
T C
X C
CX
Y
T
Y
T C
C
Y
C
C
T
X X
T
CC
Y
Berry pseudo rotation througha square pyramidal intermediate
Activation parameters
Both the k1 and the k2 pathways have S‡ and V‡ values that are negative. For example:
PtEt2P
Br PEt2Pt
Et2P
I PEt2I-
k1 k2
S‡ /J K-1 mol-1 -59 -121V‡ /cm3 mol-1 -67 -63
+ +
The k1 pathwayA
P
B
k2 [Y]
k3 [Y]
k1
+Sk-1
-S
A
P
B
k2 [Y]
k3 [Y]
k1
+Sk-1
-S
B is a solvento intermediate.
The solvento intermediate has been trapped and isolated in some cases