Welcome to the lecture on Catalytic Organometallics, part 13. Very often in literature you find the expression CH activation.

We should discuss that term. CH activation is a term used for, well indeed, gaining importance.

A label given to publications and it gains importance. Why is CH activation that important? Well, very simple. You have for instance an alkane, unfunctionalised with its CH bonds.

And you have a reaction which transforms that into a functionalised molecule with a functional group. And the CH activation publications are very often connected with transition metal catalysis.

This transformation is of course important since hydrocarbons are cheap and functional groups in a molecule means that product has far more value.

So it gains the prospect of industrial importance. And there has been, for instance, a suggestion that this label CH

activation should be restricted to transition metal catalysis since, well, also radical chemistry. Radical halogenation is somehow CH activation.

And there are other reactions, for instance, deprotonation for metalation. You can also call CH activation. Therefore, the importance is then a bit ridiculous. However, I think the main problem is that CH activation is used as a label for a certain kind of reactions.

On the other hand, it should be more appropriate for describing a mechanistic step.

For instance, well, let's take cyclohexanone and let's add an acid to it and the acid will protonate.

In equilibrium, the cyclohexanone at the free electron pair, the X minus coordinated by a hydrogen bond.

And of course, by protonating this molecule, you will increase the delta plus at the hydrogens in alpha position. You will activate this CH bond since it's more polarized and you will activate

it by simply increasing the acidity activated for the deprotonation as the next step. So, this is, of course, mechanistically the CH activation. There's no reason why you shouldn't call that CH activation.

So, this is the problem with that term. Very often, you find an alternative expression. This is the CH functionalization.

Of course, CH functionalization is an abbreviation for CH bond functionalization. Well, at least I see a small problem with this expression. You just can't functionalize a CH bond.

After the functionalization, the CH bond is gone. You can't functionalize a molecule, but not a CH bond. I think the better expression is CH transformation.

You transform a CH bond into a functional group.

And if you want to be more precise, well, okay, then just say transition metal catalyzed CH transformation. However, if you want to look up in literature the subject we are talking about, you just have

to look for CH activation and CH functionalization because those two terms are the most common within this context. And, well, clearly it is important chemistry.

So, in palladium-catalyzed reactions, we have also seen a few examples where CH bonds are activated. Well, this was in the context of cyclopallidation processes and sp3 hybridized CH groups, well, tend to be less reactive.

Here, for instance, methoxy anisole with an iodo group.

I think we have discussed that palladium-catalyzed domino process which starts with forming a palladium II intermediate with a palladium II in close proximity to the CH3 group.

And in that case, indeed, a CH activation by the palladium is observed.

A palladium cycle is formed and then lots of more steps will occur since the

reductive elimination to a four-membered ring is at least in this case somewhat inhibited. And, as you already know, the oxygen is not important in this case.

It also works with the tertiary butyl group and again a palladium cycle will be formed which also goes into some domino processes.

You can use these kind of reactions, reaction at a CH group for forming, direct forming cycles.

If you form a six-membered palladium cycle, then it is easy that it undergoes, it will readily undergo better reductive elimination forming than the five-membered hetero or carbon cycle.

I have two examples for this kind of reaction.

For activating this palladium-catalyzed process at CH groups, reaction is performed with 5% palladium acetate.

You have 30% of this pyrolytic acid or corresponding salts while in this case you have cesium carbonate present and of course the cesium salt is then formed.

You need an additional ligand, usually one which is sterically rather crowded, tricyclohexylphosphine, 10%, that means two equivalents to palladium.

You need a solvent, mesotulene and you heat that up 140 degrees, 3 hours and

that five-membered hetero cycle is formed and has been isolated in 65% yield.

Second example with the same reaction conditions, cyclohexyl substituent and methyl substituent, as I said, same reaction conditions as before.

The reaction takes place at the methyl group after the oxidative addition of the palladium zero into this carbon bromide bond.

We will have, as an intermediate, this six-membered palladium cycle and reductive elimination,

then delivers this isoindolyne known with a cyclohexyl group here, 88% was isolated.

Slightly modified substrate, just missing that carbonyl group, so not an amide, just the tertiary amine, at least not the cyclisation product is observed.

Why? All this is simply explained, I think. We will have the palladium here, the electrophilic palladium here and it will be coordinated by the free electron pair of that nitrogen than stabilising, in this case, the palladium 2 intermediate.

Well, after a longer reaction time you will presumably notice better

hydride elimination forming iminium salts as intermediates and then some decomposition reaction. Well, the difference is, here in this case, the electron density here is diminished, therefore not the nitrogen will coordinate

to the palladium, maybe here that oxygen, however, still an electrophilic palladium 2 will remain, which can undergo the insertion reaction.

At the CH3 group and this is different to the situation here where that free electron pair of the nitrogen delivers already electron density to the palladium. What we will see soon in the next examples is the important role of that pivalenic acid, moiety.

This is best illustrated by intermolecular processes. For instance, imagine you

want to synthesise this one phenyl naphthalene. How would you do that?

Well, the idea of course would be using bromo naphthalene as one coupling component plus

that phenyl boronic acid, palladium catalysis, while this is, as you know, the Suzuki coupling reaction.

Well, and where do you get this boronic acid from? Well, from the corresponding halide. So,

the initial CH transformation had been done by a halogenation process, electrophilic substitution at the aromatic system. Of course, it would be much nicer if you could just skip one halogenation and doing a

coupling reaction for instance with bromo naphthalene and just benzene, as you know, having a tether in between.

Then, performing the reaction intramolecularly wasn't a problem at all. The alternative would be, for instance, naphthalene plus bromochloroiodobenzene and essentially the same process, palladium catalyzed.

At least, this one has been made possible after Fan Yu found that pyrolytic acid is an essential ingredient.

So, the reaction is possible with an 80% yield applying 3% palladium acetate, potassium carbonate as the base.

You need a sterically crowded phosphine ligand. In this case, for the 80% yield, it was this one, two cyclohexyl units at the phosphine and a dimethyl aminophenu group as the additional substituent at that phenyl group.

Okay, so as an additive, as I said, pyrolytic acid, abbreviated generally as PIFOH, again 30%, you should have a huge excess of benzene in there.

So, therefore, this combination, one reason why this combination is preferred compared to that one, and naphthalene should be solvent in the other case.

Okay, so you want to have that in high concentration. Best would be having the benzene as the solvent itself. You get some problems then with potassium carbonate and so on.

Well, it's better to have a co-solvent in there. So, they use mixture of benzene and DMA, dimethyl acid amide in the ratio 1 to 1.2 as the solvent and then the reaction works very well with the 80% yield.

Let's have a look at the catalytic cycle of a palladium zero as the initial catalyst, again palladium acetate reduced as usual.

So, an aryl bromide oxidative addition reaction. I will omit additional ligands for clarity.

That pyrolytic acid, of course, reacts the potassium carbonate forming the potassium salt.

Potassium bromide is formed and that pivaloyl substituents will be located at the palladium.

Also suggested by calculations is this transition state with the arrow palladium, still with a phosphine ligand present, of course.

That complex will react with the benzene while that pivaloyl substituent will trap the proton. This is accepted as the preferred transition state.

The pivaloyl acid is formed which will be then deprotonated again

with potassium carbonate and the aryl palladium with the additional benzene substituent. Reductive elimination gives the final product with palladium zero again and it again goes into the catalytic cycle.

So, next subject are palladium catalyzed reactions at C-H bonds where no aryl halides are involved.

Well, of course, we need palladium II intermediates, electrophilic, for reaction with the C-H bond.

Whereas in the examples we discussed here, you have the palladium zero going into the oxidative addition forming then the palladium II.

In the other cases where we don't have an aryl halide present, then we need, we always need some oxidizing agents to reoxidize the palladium zero which is usually formed within the catalytic cycle.

Here we're oxidizing that palladium II. Well, similar to Wacker type processes. But now we don't oxidize the C double bond, the olefin, but we oxidize as a net result a C-H bond.

So, and this of course also works better intramolecularly. So, what we need or what is a nice suggestion and this has of course been done as a general scheme, we need a ligand, a tether and there is then the C-H bond.

A palladium II is coordinated to that ligand by the pre-coordination located in close proximity to the C-H bond.

Then HX is eliminated generating such an alkyl or aryl palladium II intermediate which is intramolecular coordinated here.

This is not a palladium cycle. It's better to call that a cyclopalladated complex per definition.

So, a cyclopalladation is this reaction until that stage.

Then that X might be exchanged to something else or the X itself is transferred to that carbon.

Then we have this product as our final product of the reaction. Palladium 0 is formed.

Well, and of course this should be re-oxidized to the palladium II. So, let's have a look at appropriate examples.

This bansoquinoline seems to be a model compound of choice.

We will have that several times. So, let's call that the model compound, model substrate A.

Here a reaction which has been developed by Melanie Sanford's group. She's very active in this area. 2% palladium acetate, aceto-nitrile as solvent, 75 degrees, 12 hours reaction time.

And, well, the oxidizing agent, you need two equivalents of that, is a eudonium acetate that phenyl eudonium bis acetate.

And in that case you will get an acetate functionality to this remote position.

So, palladium is coordinated here and CH activation takes place there.

86% yield. Another interesting case is the tertiary butyl pyridine.

By the way, very easy to synthesize. Just take tertiary butyl lithium and pyridine and it will react well.

This reaction is a bit similar to the Chi Chi Bobbin reaction which you certainly will know. Same reaction conditions but some more oxidizing agent is necessary to perform this CH transformation three times.

This tris astro was obtained in 63% yield. Some more examples.

Just the product. You see also other functional groups can be applied here in amide.

And the acetoxy group is then introduced in the ortho position. This is preferred to the other ortho position. This would be more crowded, of course.

77%. Azobenzene reacted in the same way.

62% yield. And now two more examples with the reaction taking place at SP3 hybridized CH groups.

This immunogroup here has a ligand.

The acetate is introduced at that position. 78%. And especially interesting is this one. In this case the acetate functionality will end up in this position.

Stereoselectively giving this product 81% yield.

I think it's an interesting problem. Why is the acetate functionality group at this position, the same side at that hydrogen? Well if you draw it like that, it looks like that hydrogen is in between that group and the former hydrogen which has been, which had reacted.

So, you are right. We should draw the appropriate confirmir of the starting compound.

We have two annihilated cyclohexane moieties.

These hydrogens in axial position. Here we have the amino group and then that hydrogen in equatorial position.

The other one in axial position. And now imagine that the palladium II is coordinated here.

Then it's clear that it is coordinated in close proximity to this equatorial hydrogen. And this is activated there.

Once again, substrate A, this benzoquinoline, isopropanol as solvent.

Some other oxidizing agent. For instance, a peroxodisulfate and then an alkoxy group will be introduced 72%.

There are other protocols how to introduce a halide, even a fluoride in this position.

But what about the aerulation here?

Again, substrate A, then an aryl halide, introducing a phenyl group to this position.

This doesn't seem to work that well on the palladium catalysis. The best seems to be catalytic amounts of ruthenium chloride hydrate. You always need to start with a hydrate because the dry ruthenium chloride is an almost insoluble polymer.

Trifenylphosphine, sodium carbonate as a base, 140 degrees, 16 hours.

You need a polar solvent that n-methylpyrrolidone. And then you get an almost quantitative yield of that product.

So in this case ruthenium and also rhodium is definitely better than palladium. So we are almost done with our palladium chemistry. But of course best would be if you could use substrates like A and an arene with no halide present.

Coupling that, well it is possible with catalytic palladium acetate.

Well if I remember it correctly it's then 10%, that's not that nice. But at least it works. You need stoichiometrically silver carbonate, DMSO as the solvent, 130 degrees.

Reoxidizing with Parvacrinone and well the reaction is not bad with 66% yield.

But of course one hopes that this type of reactions having no halide present and combining than these aryl groups will be further improved in future.

As I said this is, well we are finished with our palladium chemistry. In the next lesson we will go on with gold catalysis. Thank you for listening, see you next week.