Welcome to the second part of catalytic organometallics. The subjects of today are the basics of transition metal catalysis. And we will start with the electron configuration of

generally catalytically active complexes. In the main group, chemistry, we have the eight electron rule, which means we have the electron configuration of

we try to adopt the electron configuration of in-air noble gas. And similarly, in transition metal catalysis, we have the 18 electron rule. So, let's start with an example.

Palladium in the oxidation state zero. Well, palladium is the element number 46. And as the electron configuration, we have the electron configuration of krypton and additionally fully occupied d orbitals and

four d orbitals and empty five s orbital and the three five p orbitals also empty.

To get to the electron configuration of xenon, we of course need two electrons for filling up the five s orbital

and another six electrons for filling up the three p orbitals. Well, then we have the electron configuration of xenon. And well, this must not necessarily be very stable,

but it is clearly saturated. Well, okay, so for palladium, this means for palladium zero, this means, well, there is a rather common complex

with phosphines, triphenyl phosphines. Well, with four equivalents of triphenylphosphine, phosphine we get tetrakis triphenylphosphine palladium,

which is regarded as a very useful catalyst in palladium catalyzed reactions. Well, although it is a saturated complex,

coordinatively saturated.

So how do we have to count to notice that this is coordinatively saturated? Well, let's start with a table ligand,

a phosphine ligand, our three substituents, has the charge of zero.

But with a three electron pair at the phosphine, we have two donor electrons, which are donated to the formation of the complex.

So oxidation state zero for phosphine units makes over all eight electrons. That means this complex has then the configuration electron configuration of xenon

is coordinatively saturated. And since this is then saturated, well, it can't interact, for instance, with olefins, because it is saturated. Therefore, this means this is certainly not the catalytically active species.

Well, but no problem. There are, of course, equilibriums in solution with the palladium,

with only three phosphine ligands, a 16 electron complex,

and the one with just only two phosphine ligands, a 14 electron complex,

which is indeed regarded as the active catalytic species, for instance, in the famous Heck reaction.

So let's have a look at other ligands. An olefin, where the metal is coordinated to the central electron density.

This would be called as an eta two alkene complex. You can also call it dihabto.

It's another word for it, dihabto complex. The charge, again, is zero, and it also delivers two electrons. The same is, of course, true for the similar alkyne complex, eta two alkyne.

And now compare it with a complex like this. Here we have an ethanol substituent, or a vinyl substituent,

or even alkyl or aryl metal. Well, in all these cases, the ligand, well, is regarded to have the charge minus one,

and it delivers two electrons.

Other interesting examples. Well, just to add, of course, the same is true for halides, like chloride, bromide, iodide, or other substituent, acetate, for instance.

Also minus one, and delivering two donor electrons. Let's go on with this table.

What about this one here? A cyclopentadienyl ligand. This is eta five, cyclopentadienyl.

The charge, minus one, and it delivers, overall, six electrons.

We also know complexes where an aryl ring, benzoic, aryl ring,

is coordinated by a metal, for instance, chromium complexes, we got to know in stoichiometric organometallics. In this case, we have a charge of zero, and again, six electrons.

Another interesting case is one of alu complexes. This here is the eta three alu complex, charge minus one,

and overall, four electrons. Well, I should add that this is the general scheme for an eta six aryl complex.

Well, let's have a look at this complex, bistriphenylphosphine palladium dichloride.

While the tetrakis palladium, the tetrakis triphenylphosphine palladium complex, has a tetrahedral coordination sphere. In this case, it is square planar.

We have an oxidation state of plus two, since we have the two minus one chloride substituents.

Well, since this is a palladium two complex, and with a palladium zero complex, we need eight electrons to get the noble gas configuration. With palladium two, we need two electrons more.

That means we are in need of ten electrons. Well, we have overall four substituents, all delivering two donor electrons,

and that means overall this is a sixteen electron complex, sixteen which is coordinatively unsaturated. And that means that, for instance, this palladium complex

could coordinate to an olefin, for instance. By that, then a reaction could start.

It could be already catalytically active. Another nice example, such an intermediary complex

we will discuss later on, maybe already next week.

And this intermediary complex has been trapped and stabilized by a phenanthroline ligand.

The presence of phenanthroline, this can be isolated. This has been done by Professor Alan Kente

from the University of Tasmania in Hobart. And he did a nice experiment with that. Well, let's see, what kind of complex is that?

It's also a sixteen electron complex, coordinatively unsaturated. You see, minus one, minus one, that means this has the oxidation state of plus two, the palladium. So, since we have four ligands delivering two electrons each,

this is then a sixteen electron complex, coordinatively unsaturated. That means it can undergo an additional reaction. And it reacts indeed with, for instance, methyl iodide.

And it adds methyl iodide. And this complex is also, well, under moderate conditions to be isolated.

So, what kind of complex do we have here? Now we have four, one, two, three, four substituents, which are regarded as charged minus one.

That means this is a palladium four complex, starting from palladium zero, is in need of four electrons, palladium four.

Sorry, palladium zero is in need of eight electrons, palladium four, in need of twelve electrons. We now have six substituents, all with two donor electrons. That means twelve.

Well, fine, this is then coordinatively saturated. Okay? That's how you could count that quite easily. So, let's go on with key steps of transition metal catalysis.

First of all, there is, of course, the ligand exchange.

One transition metal, plus a ligand, with a ligand there, and a ligand prime, another ligand is offered,

and overall, in some kind of equilibrium, we will get to the other situation, to the situation with the exchanged ligands. So, there are some similarities to, for instance, nucleophilic substitution,

we know from general carbon chemistry. However, there are much more mechanistic options when working with transition metals.

For instance, let's assume we have a coordinatively unsaturated complex, of this general scheme.

So, square planar, then here we have a trans ligand, this one cis, and this one also cis.

A nucleophilic other ligand to be exchanged, this should be exchanged by that Y ligand. So, should be nucleophilic, this is coordinatively unsaturated,

and then it just adds in. So, from a geometry, which is square planar,

we will get to a situation of a square pyramidal geometry.

Now, just a little change, the X ligand moves down

to a position where it is located trans to the Y ligand,

and this ligand just moves a little bit to this position. So, then the geometry has changed to bipyramidal.

Well, now you can imagine that again the change could take place,

that this substituents moves into the plane of those three L ligands.

Okay, then we are again in a square pyramidal geometry, and finally X leaves a complex,

and it's clear that we will have then the situation, the initial situation, but with X exchanged with Y.

So, you see that these ligand exchange reactions, or substitution reactions, some kind of substitution reactions,

could proceed much more complicated as compared to what we are used to in chemistry substitution at carbon centers. Well, and you can imagine that those ligands, of course,

have a significant influence on reaction rate, especially the trans ligand is the one who is determining reaction rate, and by applying the right ligand,

you can indeed trigger reaction rate of such processes. So, in the case of a ligand exchange with a qualitatively saturated complex,

for instance with this tetrahedral, nickel tetra carbonyl complex, well, before you can exchange something, the one carbon monoxide has to leave the complex,

and this is generally the slow reaction step,

then you have a coordination site where another ligand can coordinate,

and well, that second step is generally the fast reaction step. Alternatively, with certain ligands present,

coordination sites can be opened by a reaction you sometimes call a slippage, for instance a ring slippage.

Well, let's imagine we have an allyl, an eta3 allyl complex, as we said already here in that table,

we have a charge of minus one for the ligand and four donor electrons, and here we can sometimes observe an equilibrium with the eta1 allyl complex,

so again, still minus one, but clearly only two donor electrons.

Imagine that this one is, with other ligands, coordinatively saturated as maybe an 18 electron complex, then you have here the situation of an unsaturated 16 electron complex.

Or cyclopentadienyl minus one, six electrons from eta5 cyclopentadienyl complex,

to an eta3 cyclopentadienyl complex, minus one, four electrons. This is the so-called ring slippage, of course thermodynamically disfavored,

since it loses its aromaticity through this step.

Well, but nevertheless it can go on in equilibrium to the eta1 complex,

so here eta5, eta3, eta1, and here we have again minus one and two electrons.

Another example, not a cyclopentadienyl complex, but just a pentadienyl complex of a manganese carbonyl.

Here manganese has the oxidation state plus one, it's a D6 complex, 18 electrons saturated.

There's an equilibrium possible from eta5 to eta3,

and now it is an unsaturated, cognitively unsaturated manganese complex.

If another ligand is offered, it can coordinate to the manganese, and if now in equilibrium a carbon monoxide leaves the complex,

then it can return to the eta5 situation,

now with only two carbon monoxide ligands, but in addition with the ligand which has been introduced by that ligand exchange.

So after ligand exchange reactions,

let's discuss reactions which are called oxidative addition and it's reverse reaction, retroreaction, reductive elimination.

Generally scheme, a metal in the oxidation state N reacts with a molecule AB

and the oxidative addition will lead to a complex with A and B

that has ligands or substituents at the metal and the metal has changed to the oxidation state N plus two. So and the retroreaction is the other way around and it's called reductive elimination.

So generally rather electron rich transition metal complexes, those for instance with phosphine ligands react quite nicely

with carbon halide bonds under oxidative addition reaction. If you have carbon monoxide or cyanide as ligands,

then this diminishes the electron density and hinders oxidative addition reactions of that kind. On the other hand, electron deficient complexes might react with more electron rich single bonds,

well for instance carbon hydrogen bonds. Well, let's have a look for some examples and the mechanisms which can occur for oxidative addition reactions.

For instance, the SN2 type oxidative addition.

Let's assume such a square planar complex with methyl iodide and it should be the slow reaction step,

just adding the methyl iodide via a nucleophilic substitution SN2 type process.

So and the fast reaction step should be the next one. There the iodide then also coordinates to the metal.

Well, actually we already have discussed an example for that. Well, it is the oxidative addition from a palladium II complex

to a palladium IV complex as reported by Alan Canty. So this is SN2 type.

Another type of oxidative addition, once again with the general scheme. So these brackets here symbolize that there is of course not a naked metal.

There are already ligands and substituents. So it reacts with the reagent AB and in equilibrium that unsaturated complex might coordinate

to the electron density of the single bond between A and B. This is called an agostic interaction

and it often leads then via a two electron three center bonded system

to the oxidative addition product, this one.

One example which has been studied in detail is the tungsten complex.

So what kind of geometry does this drawing symbolize?

Well, those four substituents are in a plane and with this one it is square pyramidal.

If molecular hydrogen is offered, then that molecular hydrogen can be coordinated to the tungsten

and indeed this dihydrogen complex was isolated at low temperatures

and was analyzed by x-ray diffraction and it was found that we have a bond length between those two hydrogen atoms

of 0.82 angstroms compared to 0.74 for the molecular hydrogen.

There is further equilibrium to the complex where the hydrogens are clearly separated. Well, I didn't find the distance for that case.

However, this is then not a dihydrogen complex anymore but a dihydride complex

and it should be clear for everyone that these equilibriums are important for hydrogenation reactions under homogeneous catalytic conditions.

Similarly, the oxidative addition of palladium zero to carbon halide bonds where the carbon is sp2 hybridized.

In these cases, we cannot assume some SN2 kind of process.

As you know, we don't have SN2 reactions also in the normal carbon chemistry. When the carbon is sp2 hybridized, then we have always addition elimination processes. So, palladium zero with these four ligands could be four phosphine ligands.

As already discussed earlier, there is then the equilibrium until the 14 electron complex L2 palladium

which is regarded to be the one that is catalytically active for the oxidative addition step. So, an aryl halide, then it is assumed that also in this case we have three centered system.

And finally, from palladium zero, we have changed palladium four, 14 electron palladium zero complex

to a palladium two complex still, coordinatively unsaturated since this is a 16 electron complex.

So, these will be important as intermediates and we will discuss the head reaction in the next lecture. Other mechanisms which might occur for oxidative addition processes are ionic pathways.

For instance, in cases where you have cationic transition metal complexes which are very often highly reactive because they are so electron deficient

and one should keep in mind that also radical reaction pathways can occur and have been indeed proven for some cases. So, next subject are insertion reactions.

For instance, insertion of carbon monoxide.

So, carbon monoxide is offered to a cognitively unsaturated complex with a substituent R,

maybe an alkyl or aryl substituent will coordinate to the metal and then in equilibrium it could insert into the carbon-metal bond forming such an acyl complex.

And indeed, this reaction, it is in equilibrium since it is reversible. Similarly, the insertion of an olefin and that's almost the same as the insertion of an alkyne.

So, an olefin plus again such a complex with a substituent R.

You see the mechanism is in principle all the same.

The olefin is coordinated at the metal and the insertion reaction takes place. So, in this case that insertion reaction we would call a carbometallation

if R is indeed a carbon-centered substituent and this is also a reversible process.

Well, if R just is a hydride then it's a hydrometallation

and this, as you know, is also a reversible process as you have learned for the hydroboration. And this type of process, well, generally is a concerted one which proceeds in a zone fashion.

Let's assume we have hydrogens at the better position.

So, then for a better hydride elimination this carbon-carbon bond should rotate

until a hydrogen is in a soon periplanar position and then a concerted better hydrogen elimination,

a better hydride elimination can occur.

The first step with the metal still coordinated to the olefin while the re-addition could occur and so on is also a typical process which we will discuss again in the next session

with the Heck reaction of course.

So, what remains to be mentioned that is the transmetallation. Of course you already know the transmetallation as a typical step in stoichiometric organometallics.

So, we have some organometallic compound and a transition metal complex with a substituent X,

most of the time halide and this would be then the transmetallation M' symbolizes a transition metal complex

or complexated transition metal and M' can be lithium, magnesium, zinc, barone, tin, zirconium, silicon and some more.

So, with this overview we have discussed today, we have all the instruments we need for discussing transition metal catalyzed reactions in details

and as I said we will start then with the Heck reaction on next Tuesday. Thank you for listening, see you next week.