Ligand exchange – Flashcards

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question
Stoichiometric Mechanism
answer

This distinguishes those mechanisms where a definite intermediate can be detected (by spectroscopy, kinetic or product distribution studies) from those where none can be seen. The latter are termed I (for interchange) mechanisms. The former are termed D (dissociative) if a five–coordinate intermediate is formed (c.f. SN1 mechanism in organic chemistry), or A (associative) if a seven–coordinate intermediate is formed.


DIAGRAM PLEASE  

question
Intimate Mechanism
answer

If experiments are performed where the rate of substitution of X by a range of incoming ligands Y is studied, and it is found that the rate depends on the nature of Y, this shows that the formation of the new bond M—Y is important at the transition state. Such a reaction is said to be associatively activated and is labelled an Ia mechanism. If there is no significant dependence on Y, we can presume that M—X bond breaking is more advanced in the transition state, and the mechanism is said to be dissociatively activated, labelled Id.

 

 

In practice, most ligand substitution reactions of 'classical' octahedral complexes are I in character (Id or Ia). Many organometallic complexes which obey the 18–electron rule find it easier to react by a dissociative than an associative mechanism, and most (e.g. [Cr(CO)6]) react by D mechanisms (i.e. via an intermediate [Cr(CO)5] in the case of [Cr(CO)6]). Most square planar complexes show Ia or A type mechanisms.

question

DH, DS and DV  and how are they measured

answer

These are, respectively, the enthalpy and entropy change as the reactants go from the ground to the activated state, and the volume difference between the ground and activated state. DH and DS can be derived by measuring the rate of reaction k as a function of temperature T, over a limited temperature range. From thermodynamics, 'it can be shown that':

 

   - The Eyring equation. kB is the Boltzmann constant, h is Planck's constant.

(This equation only holds over a limited temperature range).

 

The most useful of these two parameters is probably DS since this can be related to whether a mechanism is dissociative or associative. If there is an increase in disorder as the reactants reach the transition state, this implies that bond–breaking is the important step (Id or D), hence DS will be positive. If a reaction proceeds by an associative mechanism (Ia or A), DS will be negative, as entropy is lost as the new bond forms on going to the transition state or intermediate.

 

Usually more reliable is measurement of DV. This is done by studying the effect of pressure on the rate of a reaction, at constant temperature. If the total volume of the reactants decreases on reaching the transition state (i.e. if DV is negative), the rate should increase, and vice versa. If DV is negative, this implies that bond formation is important in the transition state and the mechanism is Ia or A. If DV is positive, this implies that bond breaking is more important in the transition state and the mechanism is Id or D.

 

 

question

The mechanism of ligand substitution for Pt(II) complexes

The Rate Law 

answer

For the ligand substitution reaction

 LnPt—X + Y ® LnPt—Y + X

in water, it is usually found that the rate law takes the general form:

 

–d[LnPt—X]/dt  =  k1[LnPt—X] + k2[LnPt—X][Y]

 

This indicates that there are two reactions going on here. In the first (rate constant k1), the entering group Y is not involved in the rate determining step. It is very common in ligand substitution reactions carried out in water to find that water itself (a reasonable ligand, present at 55 M concentration!) can play a role, and since it is present in such a huge excess, there is no way of determining whether it plays a role in the rate law. It turns out in this case that it does do so: the reaction whose rate is k1 goes as follows:

                             LnPt—X + H2®  LnPt—OH2 + X   (rate-determining step, rate const k1)

                             LnPt—OH2 + Y  ®  LnPt—Y + H2O   (fast)

Therefore, although ligand Y replaces ligand X in the overall reaction, there is no dependance of the rate of this reaction upon [Y].

 

In the second reaction, incoming ligand Y displaces outgoing ligand X directly:

LnPt—X + Y  ®  LnPt—Y + X   (slow; rate-determining step, rate const k2)

 

 

Note that k1 will vary with solvent; a solvent which is a poorer ligand than H2O will result in a smaller value of k1. If we plot the rate of disappearence of LnPt—X against [Y], we should get a straight line, from the slope of which we can deduce k2 and from the intercept ([Y] = 0) k1.

question

Pt. Effect of varying incoming and leaving ligands

answer

If we react trans–[PtCl2(pyridine)2], a 'model' complex, with a series of ligands Y to replace one Cl, we find big variations in k2. For example:

 

          Y                 k2 (L mol–1 s–1)

          NH3            4.7 ? 10–4

          Br–             3.7 ? 10–3

          I–                0.107

          thiourea     6.00            ({H2N}2C=S, a sulfur donor to Pt(II))

 – a factor of 104 overall. However, if we vary the leaving group, from Cl through Br to I, it makes very little difference to the rate (only a factor of 3.5 overall). Therefore bond formation to Y is important in the transition state, suggesting an associative reaction (A, or Ia).

 

 

question

Pt. Isolation of five–coordinate complexes

answer

A helpful general rule when proposing mechanisms is that a mechanism must be chemically reasonable. Although PtII complexes are nearly always square planar, a few trigonal bipyramidal complexes (e.g. [Pt(SnCl3)5]3–) have been isolated, so a trigonal bipyramidal intermediate or transition state is chemically reasonable.

 

 

question
pt Steric effects
answer

Another piece of evidence comes from varying the steric effects of the other ligands. In the reaction of trans–[Pt(Ar)Cl(PEt3)2] with pyridine in EtOH, the relative rates of reaction varied dramatically with the size of Ar:

question
          The mechanism of ligand substitution for Pt(II) complexes: Conclusion
answer

We can rationalise all these observations by invoking the mechanism above. The incoming ligand Y approaches axially and donates its lone pair to Pt. Ligands X and T (for trans) move downwards to make a trigonal bipyramidal transition state (in brackets). By twisting ligand Y down into the equatorial plane and at the same time moving T back to its former position, ligand X is forced to go axial, from where it can leave.

 

 

question
A or Ia?
answer

It is interesting to consider whether the trigonal bipyramidal species is just a transition state, on the way to the product (an Ia mechanism), or a genuine intermediate which could be detected in solution (an A mechanism; see diagram below). It turns out that, for Pt(II), the reactions are nearly exclusively Ia, but for other d8 square planar complexes (RhI, IrI, PdII, NiII) , they are often A. This is because other d8 square planar ions have a greater tendency to give five–coordinate complexes than does PtII. For instance, with linear tridentate phosphine ligands (L–L–L), it is possible to isolate five–coordinate complexes [Pd(L—L—L)Br2], but the same ligands with Pt(II) give square planar salts [Pt(L—L—L)Br]Br. Nickel(II) has an even greater tendency to form five–coordinate species than Pd(II) (and of course, it also forms many six–coordinate complexes). It is interesting that the greater the tendency to increased coordination number, the faster the reaction rates. For complexes [MCl(o–tolyl)(PEt3)2], the relative rates for M = Ni, Pd and Pt are 5 ? 106:105:1 respectively.

question

Nucleophilicity (versus basicity) of incoming ligands

answer

You are already familiar (or should be!) with the idea of the strength of a Lewis base being measured using the equilibrium constant for its interaction with a Lewis acid; a special example is protonation:

[image]

The stronger the base, the larger is K. Therefore, basicity is a thermodynamic parameter. The kinetics of such a reaction do not play a role as the kinetics are always very fast for protonation.

 

With Pt(II) as the Lewis acid, however, the kinetics are slow, and the outcome is often determined by them, and not by thermodynamics. We can order a series of Lewis bases (ligands) depending on the rates  for reaction with Pt(II). The faster the rate, the more nucleophilic are the ligands to Pt(II). Consider the following standard reaction for Pt(II) chemistry, done in MeOH as solvent:

 

trans–[PtCl2(pyridine)2]  +  Y  ®  trans–[PtCl(Y)(py)2]+   +  Cl

 

This reaction has been done for a wide range of ligands Y. We can express how good a nucleophile is for Pt(II) by measuring how much faster k2 is for that ligand than k1 (i.e. than for the reaction with MeOH alone). This is expressed using the Swain–Scott equation:

 

log (k2/k1) = nPt ? s

 

where nPt is the nucleophilicity of the incoming ligand Y towards PtII and s is called the substrate parameter (defined as 1 for this particular, standard, reaction). Some values for nPt are: MeOH 0 (by definition!), Cl 3.04, NH3 3.07, pyridine 3.19, NO2 3.22, Br 4.18, Et2S 4.52, I 5.46, CN 7.14, PPh3 8.93, PEt3 8.99. (PEt3 attacks [PtCl2(py)2] 108.99 times as fast as MeOH does).

 

 

Clearly, the most nucleophilic ligands (e.g. R3P, I) are not necessarily the strongest bases. The series is remarkably constant for any Pt(II) starting material, not just trans–[PtCl2(pyridine)2]. Since Pt(II) is a soft metal, the best nucleophiles are soft Lewis bases (e.g. soft I > Br > Cl; the hard F ion is the poorest nucleophile; P > N, S > O). Notice that this order is special for PtII, and similar experiments on nucleophilicity for organic reactions (using, for instance, SN2 substitution of Me–I as a standard substrate) would not necessarily give the same order.

question
Trans effect
answer

'the effect of a coordinated group on the rate of substitution of ligands trans to itself.' A rough order of trans effects is: H2O, OH < NH3, py < Cl, Br, NCS < SCN, I, NO2 < C6H5 < tu, Me < H, PR3 < C2H4, CN, CO. Note that ligands which are good at p–backbonding, or which are excellent s–donors, or both, are high in this series. The better the trans ligand, the faster is the rate of substitution. Note: The trans effect is kinetic; we are measuring rates of substitution.

 

 

This may occur in one of two ways: (i) a better trans effect ligand may weaken the bond between PtII and X in the ground state, and/or (ii) it may stabilise the activated complex (or transition state). Either will result in a smaller DG and therefore a faster rate. The reaction coordinate diagram may illustrate this better (above: 1: Bond to X weakened in ground state. 2: Transition state more stabilised by better trans ligand. T1 is a better trans ligand than T2 in these pictures).

question
The trans influence
answer

In theory, we should be able to detect if a given ligand weakens the bond trans to it in the ground state, using structural and spectroscopic methods. This is known as trans influence.  E.g.:

X–Ray crystallography:  In the complex [PtCl3(h–C2H4)], the length of the Pt–Cl bond trans to the ethene, a ligand higher in the trans effect series than Cl, is slightly longer (therefore weaker) than are the two Pt–Cl bonds cis to it. Some other examples:

[image] (Length of Pt–Cl in bold)

Note that the higher the ligand trans to the bold Pt–Cl bond is in the trans effect series, the longer is the Pt–Cl bond. Therefore, here, there must be some ground–state bond weakening (trans influence). Notice that this is much smaller for ethene than for a phosphine.

Infrared spectroscopy:  In the infrared spectrum of [PtCl3(h–C2H4)], the Pt–Cl stretching frequency due to Cl—Pt–trans–C2H4 is at lower energy than the average of the Cl—Pt–trans–Cl modes. Again, this indicates that the Cl—Pt–trans–C2H4 bond is weaker than the Cl—Pt–trans–Cl bond. Also, compare the following data for complexescis–[PtCl2(L)2]:

 

Ligand L

Pt—Cl stretches (cm–1)

pyridine

343, 329

PMe3

303, 300

AsMe3

314, 294

SEt2

330, 318

 

 

This suggests the following order for trans influence: PMe3 ~ AsMe3 > SEt2 > pyridine, which fits in with the idea that 'softer' Lewis bases have a higher trans influence. The greater the trans influence, the weaker the Pt—Cl bond, the lower the Pt—Cl stretching frequencies. Clearly, the weaker the Pt–Cl bond in the ground state, the faster Cl will undergo substitution. 

question
Transition state stabilisation to rationalise the trans effect
answer

Some ligands which are very high in the trans effect series (CO, alkenes) do not have a large trans influence. Therefore, these ligands must mostly act by stabilising the five–coordinate transition state.

[image]How might a good trans effect ligand stabilise the five–coordinate transition state?   Some ligands high in the trans effect series (PR3, CO, C2H4) are known to be good p–acceptors. On going from the ground state to the activated complex, the metal acquires additional electron density from the incoming donor ligand Y. This may be accomodated better by a good p–acceptor ligand, as (above, left).

 

There is still debate about the relative importance of these two effects. But it is generally agreed that where ligands are high in the trans effect series, but are incapable of p–backbonding (e.g. alkyl groups, H), they must act mostly by ground–state bond weakening. Where ligands are high in the trans effect series, are good potential p–acceptors, but have been shown not to have a large trans influence (e.g. alkenes, CO), they must be acting largely by stabilising the transition state. Some ligand types (e.g. phosphines) probably 

question
Charge and size effects on rate
answer

It is immediately obvious that the size of the metal ion plays an important role (e.g. look at alkali metal and alkaline earth metal ions – the larger the ion, the faster the rate of exchange). Secondly, the charge of the ion is important; the larger the positive charge, the slower the rate. The bonding in all these complexes is electrostatic, so the strength of the bond will be proportional to Z2/r {(charge)2/radius}. The weaker the bond, the faster the rate of exchange, regardless of what the mechanism actually is, so a larger r or, even more important, a smaller Z  will lead to faster exchange.

question

Classification according to size of rate constant

answer

It is convenient to classify these reactions according to the rate constants, and this is done as follows:

 

Class I. Very rapid water exchange (? 108 s–1), where we are near the  diffusion–controlled limit. Note that most of the ions in this range bear a low charge (+1, +2) and/or are quite large (Ca2+, Sr2+, Ba2+, Cd2+, Hg2+); the few M3+ ions in this range are exceptionally large (lanthanides Ho3+ ® Gd3+).

 

Class II: The exchange of water is still fast (105 – 108 s–1), but less than the diffusion–controlled limit. Note that in this range are those transition metal 2+ ions with some CFSE; the greater the CFSE, the slower is the rate of exchange (for ions of similar Z2/r); thus, Ni2+ is the slowest here.

 

Class III. (10 – 104 s–1): Tripositive metal ions with some CFSE fall into this category. Also here is Be2+, which, because Be is a first row metal, is exceptionally small (very large Z2/r). Al3+ is also very small.

 

Ions in classes I to III are labile (their lifetime in solution is less than a few seconds).

 

Class IV (10–1 down to ca. 10–9 s–1): These are the only ions which are inert (lifetimes long enough to persist on mixing, for times ranging from a few seconds, to many years). They are distinguished by large CFSE's (d3, low spin d5 [Ru3+] low spin d6 [Ru2+; Co3+, Rh3+, Ir3+ not shown – they are off the scale!] or considerable covalency and CFSE (Pt2+ which we have already considered), and are often more highly–charged (Cr3+, Co3+, Rh3+, Ir3+).

 

 

question
The mechanism of solvent exchange for transition metal ions
answer

The determination of DV for these reactions is quite difficult because the actual numbers, whether positive or negative, are small, and there is experimental uncertainty. Interpretation of DV values for ligand substitution reactions is complicated by the fact that DV represents the total volume change for the whole system, in other words not just the complex ion and the incoming ligand, but also the volume of the solvent 'shell' surrounding the complex ion. Particularly for reactions between a metal cation and an anionic ligand, there can be such large changes in the volume of the solvent 'shell' that this outweighs changes in the volume of the complex. However, for solvent exchange, changes in solvation at the transition state will be relatively small, so although the numbers are small, they are fairly meaningful in this case!

 

As we go across the transition series above, the reactions are probably Ia on the left (negative, if small, DV) and Id on the right (positive DV). Notice also that on the right, there are quite large, positive values of DS, which also indicates a dissociative mechanism.

 

This can be explained by considering two factors, (i) ion size and (ii) population of the t2g orbitals. On the left of the d–block, the ions are quite large, which would favour an associative mechanism (effectively, an increase in coordination number in the activated state). Also on the left, there are fewer d–electrons, and the t2g orbitals are less occupied. In an associative mechanism, the incoming water must approach the metal ion along one of the triangular faces of the octahedron, and since this is where the t2g orbitals are pointing, if they are unoccupied, they are perfectly set up to receive the incoming water lone pair. But if they are occupied, the t2g electrons will tend to repel the incoming ligand

question

          Effect of CFSE on the rate of water exchange

answer

Whichever mechanism applies, the complex will have less CFSE in the transition state (or intermediate) than it does in either the product or the reactant, except in the case of d4 (high spin) or d9 complexes, to which we return later. This is because all possible geometries other than octahedral have less CFSE than an octahedral arrangement with the same type of ligand. So the more CFSE the complex loses as the transition state forms, the slower the rate of reaction will be (because loss of CFSE will make DG bigger). This is illustrated in the diagram above. It is possible to calculate the loss in CFSE as a metal ion goes from Oh to square based pyramidal or trigonal bipyramidal (for a D mechanism) or pentagonal bipyramidal or octahedral wedge–shaped (A mechanism). The calculation is similar to that which is used in CHEM212/214 to rationalise why d3 and d8 ions form so few tetrahedral complexes, by comparing CFSE for Oh and Td arrangements. It is not surprising, then, that for 3d transition metals, d3 (–12Dq), d5 low spin and d6 low spin (–20 Dq and –24 Dq) and octahedral d8 ions like Ni2+ (–12Dq), undergo slower ligand exchange than would be predicted for their charge/radius ratio alone. Notice that 4d and 5d metal ions all undergo slow ligand exchange; remember that CFSE's are bigger for 4d and 5d metal ions, low–spin ions dominate even with weak field ligands, and so loss of CFSE in the transition state is even more pronounced for these.

 

These effects, together with the repulsion to incoming ligands offered by the full or half–full t2g orbitals and the high charge/radius ratio, explains why the 'inert' metal ions (Ru3+, Rh3+, Ir3+, Cr3+) undergo ligand exchange reactions so slowly.

 

 

Finally, notice that high–spin d4 (Cr2+) and d9 (Cu2+) ions have unusually high rates of water exchange, close to the diffusion controlled limit. This is easily rationalised. Both these configurations are subject to the Jahn–Teller effect and, as we saw in CHEM212/214, have two weakly–bound axial ligands. These are very easily exchanged; the activation barrier is low as these M–OH2 bonds are extremely weak. A low–energy vibrational mode interconverts all six coordination sites in these complexes, so all six water ligands are very readily exchanged, and not just the two axial ones.

question
the Eigen–Wilkins mechanism
answer

Eigen and Wilkins attempted to describe in more detail the events that occur in octahedral ligand substitution reactions, and to describe a general mechanism applicable to all such reactions. The first step in the Eigen–Wilkins mechanism is that the entering ligand, Y, and the metal complex LnM—X meet in a diffusional encounter. Since they can also diffuse apart again, and since these events are likely to happen with very fast rates (much faster than the actual substitution reactions) we can write the following pre–equilibrium for the formation of this encounter complex

[image]

kobs can be measured. KE can be estimated, taking into account factors such as the charge on the complex ion, the charge on the ligand, the size of the complex LnM—X and ligand Y, the dielectric constant of the solvent, etc., using an equation called the Fuoss–Eigen equation (not outlined here). This equation says that (i) larger Y and (ii) Y with opposite charge to the LnM—X complex, will give bigger KE's.

 

What this enables us to do is, essentially, to correct the observed rate constant for the tendency of the incoming ligand to form an encounter complex. It therefore lets us compare reactions of a given metal complex (e.g. [Ni(H2O)6]2+) with different ligands, which might have very different tendencies to give an encounter complex. The following values have been determined for [Ni(H2O)6]2+:

 

Complex formation by [Ni(H2O)6]2+

Ligand Y

kobs(dm3mol–1s–1)

KE 

(kobs/KE)

CH3CO2

1 ? 105

3

3 ? 104

F

8 ? 103

1

8 ? 103

HF

3 ? 103

0.15

2 ? 104

H2O

 

 

3 ? 103

NH3

5 ? 103

0.15

3 ? 104

H2NCH2CH2NH3+

4 ? 102

0.02

2 ? 103

SCN

6 ? 103

1

6 ? 103

 

Take–home message from this section:

Note that while kobs varies substantially, which might have led us to suppose an Ia mechanism for Ni(II) (i.e. that bond formation is important in the activation step), when we take into account that (i) larger Y and (ii) anionic Y have a much larger tendency to give an encounter complex, the rates corrected for this are actually very similar (only one order of magnitude difference between the smallest and largest).

 

 

This ties in well with the DV evidence for water self–exchange and lends support to an Id mechanism for Ni2+ (a later transition metal ion).

question

Effects of other ligands upon substitution in inert complexes

 

4.3.1 Leaving group effect on direct substitution at Co(III) and Rh(III)

answer

It is found that DV for reactions such as

[CoX(NH3)5]2+  +  H2®  [Co(NH3)5(H2O)]3+   +   X

are positive, indicating an Id (or D) mechanism. However, this is not as useful as measuring DV for solvent exchange, because in these reactions, we are generating an anion, and the charge on the metal complex is also increasing from 2+ to 3+. This means that we could expect big changes in the volume of the solvation shell at the activated state simply due to the increasing charge. Therefore, additional evidence is required.

[image]

Interestingly, there is further, very good, evidence for this, which comes from determining the dependence of the reaction on the nature of X. As we have already seen, for an Id (or D) mechanism, the rate should depend upon the nature of X (leaving group), but not much on the nature of Y (entering group), as in the activated complex, bond breaking is more important than bond making. The rates of these reactions do indeed depend upon X, but in an interesting fashion.

 

It is found that if the logs of the rate constants for these reactions are plotted against the logs of the equilibrium constants, a straight line is obtained, with a gradient close to 1 (see above), i.e.

 

          ln k = p ln K + c (c is a constant, and p  is approximately 1)

 

This is called a linear free energy relationship and is strong evidence for a D (or very strongly Id) mechanism. Think about what is going on using a reaction coordinate diagram (above). For a dissociative mechanism, changing X changes DG for the conversion of Co—X to the activated complex, because bond–breaking is important in the transition state. It will also change DG for the complete elimination of X to give Co—Y by the same amount..

 

The same exercise for Rh(III) complexes [Rh(NH3)5X]2+ gives a similar result, except that (i) the slope of the log–log plot is somewhat less than 1; the reaction is not as completely dissociative in character, as Rh(III) is bigger than Co(III), and therefore more prone to associative character, and (ii) the order of leaving group reactivity is reversed (i.e. the reaction rates are in the order I < Br < Cl). Rh(III) is a 'soft' Lewis acid, and will therefore prefer to bind I than Cl, and conversely it will be less prone to lose I than Cl. Co(III) is usually a 'hard' Lewis acid and (certainly with 5 hard NH3 ligand already bound) prefers Cl to I.

 

 

Interestingly, it is possible to change the character of a metal centre by altering the co–ligands. Much work has been done on the substitution reactions of complexes [Co(CN)5X]3–. The substitution of X here is almost certainly via a dissociative (D) mechanism. The order of reactivity is now I < Br < Cl << F (that the 'reactivity' of the fluoride is very high can be guessed from the fact that no–one has yet succeeded in isolating [Co(CN)5F]3–!). This order comes about because when Co(III) is bound to five 'soft' CN ligands, its character is made 'soft' also, and it prefers to bind to another 'soft' ligand.  This is called a 'synergistic' effect.

question
Co–ligand effects on substitution
answer

For a mechanism of dissociative character, remember that we predicted that increasing the size of the co–ligands should increase the rate of substitution, whereas if the reaction is of associative character, larger co–ligands will decrease the rate.

 

For Co(III), the rate of hydrolysis of [CoCl(NH2Me)5]2+ is 22 times as great as the rate of hydrolysis of [CoCl(NH3)5]2+. This is good evidence for a dissociative mechanism.

 

A series of complexes where there has been extensive study of the effects of the co–ligands upon hydrolysis is cis– and trans–[Co(A)(X)(en)2]+ (A = OH, NCS, Cl; X is the leaving group and is Cl or Br; en = 1,2–ethanediamine). The cis complexes are more labile than the trans . Moreover, the reactions are 100% stereospecific, i.e.

 

cis–[Co(A)(X)(en)2]+ + H2®  X + cis–[Co(A)(H2O)(en)2]2+ only.

 

The reactions of the trans isomers (i) are slower and (ii) are not stereospecific:

trans–[Co(A)(X)(en)2]+ + H2®  X + trans–[Co(A)(H2O)(en)2]2+  + cis–[Co(A)(H2O)(en)2]2+

 

Product(s) of hydrolysis of [Co(A)(X)(en)2]+

Isomer

A

X

% cis in product

cis

OH

Cl

100

cis

Cl

Cl

100

cis

NCS

Cl

100

cis

Cl

Br

100

trans

NO2

Cl

0

trans

NCS

Cl

50 – 70

trans

Cl

Cl

35

trans

OH

Cl

75

 

How can we account for this?

 

The mechanism of substitution in these reactions is Id. The five–coordinate metal centre in the activated complex could resemble either a square–based pyramid or a trigonal bipyramid. Reaction from a square–based pyramid activated complex would result in retention of stereochemistry. Reaction from a trigonal bipyramidal activated complex would give some scrambling because there are different points from which the incoming water can attack.

[image]

The trigonal bipyramidal activated complex must be a higher energy state than the square based pyramid, to rationalise the difference in rates.

 

It seems that a good pdonor ligand trans to the leaving group stabilises the trigonal bipyramidal activated complex, and this promotes isomerisation (we will see another effect of this when we consider base catalysis later). Anions like Cl, OH or NCS are p–donors, but RNH2 groups are not. Therefore isomerisation is seen for the trans isomers (X trans to A), but not for the cis

question

Acid–catalysed reactions

answer

If the leaving group is the salt of a weak acid, and it has an additional lone pair, it is possible for acid to catalyse the reaction as follows:

 

[image]

 

The M—X bond is weakened by donation of the second lone pair to H+, and loss of X is therefore easier. An example of this is the hydrolysis of [CrF(H2O)5]2+ to [Cr(H2O)6]3+. In neutral solution, the observed rate constant is 6.2 ? 10–10 s–1, but at pH 2 the rate constant is 1.4 ? 10–8 s–1.

question

Base–catalysed hydrolysis of Co(III) complexes; the Dcb mechanism

answer

Octahedral substitution is often greatly accelerated by OH. For example, the solvolysis reaction [CoCl(NH3)5]2+  +  H2®  [Co(H2O)(NH3)5]3+ is one million times faster in basic conditions than in acid. It might be thought that perhaps OH acts as a nucleophile to Co(III). However, this is not the case. If the reaction is examined in the presence of a competing nucleophile Y, the ratio of the products [CoY(NH3)5]2+/[Co(H2O)(NH3)5]3+ increases linearly as a function of [Y] but does not vary with [OH]. This indicates that it is H2O that is the nucleophile in the hydrolysis reaction, not OH. Although the ratio does not change, the rates of both reactions increase as a function of [OH], to the same extent. Therefore, this tells us that there is a common intermediate giving rise to both products, that this involves attack of OH on some part of the complex, and that this gives rise to a much faster reaction than usual.

 

Further evidence as to the actual mechanism comes from consideration of the fact that only complexes which have co–ligands with acidic protons show this rate increase in the presence of base. For instance, while cis–[CoCl2(en)2]+ shows this effect, cis–[CoCl2(pyridine)4]+ does not. En has weakly acidic NH protons (which become more acidic when it is coordinated to a cationic metal centre) whereas pyridine does not.

 

[image]

The most widely–supported explanation for these effects is as follows. The OH removes a proton from a coordinated NH3 or en co–ligand, generating a CoIII–(deprotonated amine) species. We have already seen that good p–donor ligands are capable of stabilising a trigonal bipyramidal transition state in CoIII chemistry. This is probably because the additional electron density donated via the p–bonding helps compensate the metal centre in the five–coordinate activated state for the loss of one s–donor ligand. A deprotonated amine should be an excellent p–donor, much better than Cl for instance, simply because nitrogen is less electronegative, therefore more electron-releasing, than chlorine. Therefore, it greatly stabilises the transition state, and thereby speeds the rate of reaction.

 

This mechanism is often referred to in older texts as the Dcb (or even SN1cb) mechanism, as it is assumed from rate law and other data that the mechanism involves a definite five–coordinate intermediate. However, the supposed five–coordinate intermediate (required for a truly D mechanism) has not been conclusively established. Moreover, none of the data excludes an Id process. Therefore, until anyone comes up with better evidence, it is probably correct to refer to this mechanism as Id cb.

 

 

 

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