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Substitution reactions of square planar complexes. especially d 8 : Ni(II), Rh(I), Pd(II), Ir(I), Pt(II), Au(III). Reaction:. [ML 3 X] + Y [ML 3 Y] + X. General mechanism. A. k 2 [Y]. P. k 1 +S. k -1 -S. k 3 [Y]. B. General rate law:. [ML 3 X] A. (1).
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especially d8: Ni(II), Rh(I), Pd(II), Ir(I), Pt(II), Au(III)
Reaction: [ML3X] + Y [ML3Y] + X General mechanism A k2 [Y] P k1 +S k-1 -S k3 [Y] B General rate law: [ML3X] A
(1) Assume [B] is in steady state Substituting into (1)
The rate of attack of solvent on A is rate limiting k3[Y] >> k-1 which is in agreement with the experimental rate law Two situations usually arise for the solvent pathway
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
k1 k2 0 Study the rate of the reaction as a function of [Y]
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 k2 pathway The greater Pt, the greater the nucleophilicity of the ligand nucleophilicity parameter
nucleophilicity increases with softness of the donor ligand
where C = log k2o Now generalise for any square planar Pt complex [PtL3X] [PtL3X] + Y [PtL3Y]+ X
S is the nucleophilic discrimination factor and gives the sensitivity of the rate constant to the nucleophilicity of the incoming ligand
trans-[PtCl2(PEt3)2] S is larger SeCN SCN I thiourea SCN [PtCl2(en)] N3 I NH3 Br Cl CH3OH NO2 CH3OH
Usually... The discrimination factor, S, decreases As reactivity towards the common ligand, MeOH, increases
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
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
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.
For the generalised reaction 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
The rate of the reaction depends significantly on the nature of the trans ligand, T, but hardly at all on the cis ligands C
Stronger σ donors Weaker σ donors Stronger π acceptors Weaker π acceptors 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-
Observations consistent with a trigonal bipyramidal transition state in which the cis ligands are axial, and T, X and Y are equatorial
If T is a good donor ligand, it is readily polarisable... ...and this weakens and labilises the MX bond. ...and polarises electron density from M towards it (i.e., the TM bond has significant covalency... i.e., T destabilises the ground state
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
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- 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
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
The intermediate must be shortlived, else scrambling of stereochemistry would be expected
k1k2 S‡ /J K-1 mol-1 -59 -121 V‡ /cm3 mol-1 -67 -63 Activation parameters Both the k1 and the k2 pathways have S‡ and V‡ values that are negative. For example:
The k1 pathway B is a solvento intermediate. The solvento intermediate has been trapped and isolated in some cases
kinetically inert The solvento intermediate has been trapped and isolated in some cases