Substitution reactions of square planar complexes

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

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

nucleophilicity increases with

softness of the donor 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

Hence, for NO2

2log 0.410 3.22 0.62

0.70

k

0.702

-1 -1

10

5.0 M s

k

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

The solvento intermediate has been trapped and isolated in some cases

Pt

N

N

N

I Pt

N

N

N

H2O Pt

N

N

N

Y

+ 2+ +

k1 H2O Y-

Pt

N

N

N

OH

+

base

kinetically inert