Welcome to part 15 of the lecture on designing organic synthesis. The subjects of today are heteroarenes with two or more heteroatoms. First of all, let's have a look at the synthesis of oxazoles having an oxygen and a nitrogen heteroatom.

And let us try the retrosynthetic analysis of this type of target molecules.

Well, first idea could be just following the general scheme for the synthesis of furans.

We just ignore having a nitrogen here and assume that it should be possible to achieve this transformation.

So, forming the heteroarene in the case of a furan, it's just treating with an acid catalyst.

This will not be sufficient in this case because such an amide is through that conjugation a relatively stable and less reactive moiety.

Insofar, it's better to transform this in something that has more electrophilic reactivities, more electrophilic at that center.

And we can achieve that just by treating this amide with PCl5 or phosphor oxychloride.

In that case, imidoyl chlorides will be formed. And imagine now that the enol forms here from the tautomer of this ketone.

Then the nucleophilic oxygen will react there. Then an addition, elimination process, finally giving rise to the aromatic system. This then works very well.

Just an example, so for instance, treatment with phosphor oxychloride, heating it up a bit, will deliver this oxazole in very high yield.

Nevertheless, we should think about how do we synthesize that. Now, we start the retrosynthetic analysis of this structure.

For instance, disconnecting here having an amino ketone and, well for instance, acetic anhydride for the N-acylation.

Should work. However, there is a problem, the sensitivity of amino ketones.

They tend to condensate intermolecularly, of course. So, you can't keep that very long on your shelf.

This is just perfectly stable. So, for getting there, we should generate this and soon trapping the amino ketone with the acetic anhydride.

And indeed, the normal process which is often chosen is this. Having this oxime and under basic conditions, what is basic?

No, it is, in this case, it's acetic conditions. Acetic acid, zinc, well you might remember the chlorpyrrole synthesis.

In that case, we had the same reaction conditions. And also an oxime present. The oxime was reduced to the amino group. So, and here in this case, the presence of acetic anhydride and indeed this reaction works almost quantitatively.

Around 90% and I found a publication that if you use indium instead of zinc, you get quantitative yield.

Well, okay, I wouldn't spend my indium for that. And how do you get the oxime?

From aceto-phenone and treating that with nitrozyl chloride. So, the normal nitro, generating the normal nitrozyl cation.

Well, why not disconnecting here? Let's have a look at the suggestion R2.

Accept a reactivity here. Donate a reactivity there. And these two synthons we can translate into synthetic equivalents.

Just a leaving group here. Well, let's choose chloride for instance. And on the other hand, well, okay.

Just let us apply this acetic amide. Three electron pair there should go directly. Isn't that a more straightforward route compared to this oxime reducing chemistry and then trapping with acetic anhydride?

Well, there must be a reason why you don't find this transformation. These two giving that compound.

Well, very simple. If we would test this reaction, put those together with, well, maybe directly, then we would get a nice product.

And indeed, it is an oxazole, but this one is the wrong isomer.

So, why is that? Well, please remember, if you try to protonate a carboxylic amide, never protonate at the nitrogen.

You will protonate at the oxygen. The oxygen is the more basic center. And at the same time, it's also the more nucleophilic center. Because the corresponding cation is mesomerically stabilized in this case.

So, the oxygen will attack that electrophilic center forming this product of the nucleophilic substitution.

And then, intramolecularly, the emit nitrogen will interact with the carbon group.

So, and, let's draw this hemiozimaminal.

And, well, just eliminating water leads to the final oxazole. And indeed, this reaction works very well.

Directly from these starting materials in more than 60% or 70% yield.

Another oxazole synthesis I found rather interesting. The von Lausen oxazole synthesis.

And actually, it's not that easy to get that idea by retrosynthetic analysis. But it might help to understand this reaction. If you want to get to the von Lausen oxazole synthesis, you would have to, well, assume that in the last step an elimination occurred.

And that this oxazole derives from something like that.

From an oxazoline with a leaving group here. And now, we could get to the idea that we need donator reactivity there,

and acceptor reactivity there, acceptor here, donator there, and here an X.

So, translating these two symphons into synthetic equivalents is easy for this case,

since, well, this is simply an aldehyde. Here, this leaving group X needs to activate its alpha position to be acidic.

So, if we have a leaving group X, which cares for the acidity of its alpha position, then everything would be fine. Okay, and there is a structure, a sulfone.

Sulfone can be a leaving group, and of course, the alpha protons become acidic.

On the other hand, here we need nitrogen, and something with an acceptor reactivity there, and most interestingly, such an isonitrile would resemble a synthon like that.

This is a rather important reagent called tosmic for tosyl methyl isocyanide.

And indeed, if you assume having a phenyl group there, just bands aldehyde,

you need just potassium carbonate as a base, methanol reflux for two hours, and you will get nicely this oxazole in a 91% yield.

Same reaction conditions, but 20 degrees, then one is able to isolate the oxazoline

with the sulfone group still present.

Let's talk about the synthesis of the tosmic itself.

I think this is also rather interesting. Start from para toluenesulfonic acid chloride. Treat that under basic conditions, sodium hydroxide,

sodium carbonate in the presence of zinc as reducing agent,

and you will get a reported 64% yield to the deprotonated sulfonic,

it's not the sulfonic acid, sulfinic acid, with indeed a nucleophilic sulfur.

In combination with simply form amide and form aldehyde,

you will achieve as a one, preparative one-step process,

the CN bond formation and the CS bond formation.

So if you would classify this reaction, how would we classify that? So this is a Mannich reaction.

Well, we must admit rather unusual Mannich reaction, because normally we would apply primary or secondary amine, not an amide.

Yes, we would apply an aldehyde, maybe also form aldehyde, and as nucleophile in a normal Mannich process, we have normally something like an enol or enolate. So, but again, this will form the electrophile.

Here we have by condensation, here we have the nucleophile. Overall, this is close to a Mannich reaction. Well, and here this reaction works with 81% yield,

and the next step forming the tosmic, 82% yield, and you just need phosphooxychloride, triethyl amine as base,

and the reaction takes place in dichloromethane as a solvent. From oxazoles to their isomers, these oxazoles,

actually we already discussed those.

A nice straightforward route is the 1,3-dipolar cycloaddition reaction,

of an alkyne with a nitrile oxide, which reacts as a 1,3-dipole.

How do you produce those nitrile oxides? Well, let's turn that around.

You just treat these imidobromides with a base.

Well, this might be potassium carbonate, methanol. The transformation is quantitatively.

In the presence, well, not quantitatively, this isolated, you generate that in the presence of the alkyne, and this is then trapped. So, and these, well, are easy to get.

Just let that oxime of an aldehyde react with n-bromosocynamide

in the presence also of a base and room temperature, and that works. Under the certain moderate reaction conditions, you can isolate that. You don't want to make this step with the n-b-s

already in the presence of the electron-rich acetylene, of course. So, thiazoles, that's easy.

Let's take an example similar to this one,

just the oxygen exchanged to a sulfur, and then we already have a good idea,

leaving the repair chloride, for instance, and using a thioamide as the bis nucleophile.

Concerning those thiazoles, we should talk about an important one, vitamin B1.

So, its structure, we have a pyrimidine ring,

and it's connected to a thiazolium unit by this methylene bridge,

X minus, generally, a chloride. So, why is this vitamin B1 important?

Well, of course, vitamins are important, and in addition, it's very interesting for us chemists, since these are active as catalysts.

I hope you remember Stetta, catalysis for a silylene formation. So, if you don't remember that, please look that up. Well, essentially, we might deprotonate here,

forming carbene intermediate, this interacts then with aldehydes, and then a domino process goes on.

Domino process behind this transformation.

How would we synthesize this one? The first step of a retrosynthetic analysis is, of course, straightforward, just disconnecting somewhere in the middle of that molecule

at a carbon-heteroatom bond, a strategic bond, having a leaving group here,

and I already add one proton to the amino group, simply because this X-methylene group would be rather sensitive

if you have an amino group here, because the formation of a stabilized cation would cause some troubles. Protonating this amino group will stabilize this as a salt.

We should use this thiazole as a nucleophile reacting with that.

How to synthesize this pyrimidine derivative?

Well, I'm sure you noticed that this connection would be attractive here, starting from acetamidine and as electrophilic, this electrophilic unit.

One could choose something like that, having an aldehyde functionality there, an ester functionality at this position. Of course, later on, you have to get rid of the alcohol late moiety

by replacing that nucleophilic with ammonia. Okay, so it's certainly several steps. Now analyzing that synthesis of thiazole,

similar to what we have discussed before. In this case, we need the thioamide of the form amide,

condensating with the carbonyl group here, and a nucleophilic substitution will connect the sulfur to that carbon.

This will work very well. So now we have to analyze how to synthesize that one.

I found this idea very interesting. This has been successfully applied in several labs with almost quantitative yield.

So how is this transformation achieved?

Very simple. It's just the hydrolyzation of the lactone. Then we have a better keto carboxylic acid which will readily decarboxylate.

That's it. So, and to get to this structure, it's just the chlorination with sulfuric chloride.

Well, I think the conditions of a radical chain reaction, I assume.

And this one. Donator reactivity here, acceptor reactivity there,

acceptor reactivity here, and donator reactivity at that oxygen. So these symptoms are easily translated to acetoacetic acid,

and ester as one reaction component, and ethylene oxide as the other one.

Next, heterines, imidazoles.

While following essentially the same route as with the thiazoles, we can disconnect here, disconnect there,

having a carbonyl component with the leaving group in alpha position,

and those amidines as the bis nucleophilic.

There's an interesting alternative. You could start from these one, two diketones.

One equivalent of an aldehyde, an excess of ammonium acetate.

You will form a bisamine here, monoamine there. These will also condensate.

Then you get to this amino.

And now the transformation to the final aromatic product is just the tautomerization.

Not as simple as an enamine tautomerization usually is, but this tautomerization profits, of course, from thermodynamically since that aromatic system is formed.

Pyrazoles, that would be too easy as a task for an exercise.

You just disconnect here and there.

This pentane 2,4-dione is treated with this hydrazine.

That's it. Let us introduce a slight modification. Having an amino group here, following the same idea,

then we would try this amide of the acetic acid.

So what's the problem?

Well, first of all, we would form the imine here, maybe also then in equilibrium having the enamine, the hydrazino compound.

So first of all, we would form the hydrazone. It could tautomerize to the hydrazino compound.

Just an R. So, but unfortunately, that carbonyl group here is not that electrophilic

since it is embedded in this mesomeric system. To get the CN bond formation done here, we would have to go to rather high temperature, vigorous reaction conditions

or, better, choose a more reactive group here that should be readily available. Well, not an acid chloride.

In case of an acid chloride here, the more nucleophilic or more accessible, less directly hindered amino group will react at the acid chloride. So more electrophilic is this amino group if we have a phenyl group there.

And the free electron pair is conjugated with the phenyl ring. So, better than is just having a nitrile functionality here.

So then, clearly, the terminal amino group will react at that carbonyl group

and intramolecularly, here we have the electrophilic center.

This reaction with acetoacetic acid ester and phenylhydrazine works very well.

Just a bit of acidic catalysis, then the hydrazone will be formed

and actually you have in the reaction mixture this hydrazone,

preferred one since it has a bridging hydrogen bond here, then the stereoisomer, there this nitrogen is on the other side and the free electron pair is closer to the carbonyl group,

then of course the tautomer, the enamine tautomer. However, this is the one which then can nicely react intramolecularly,

closing the ring and tautomer rising, then to this pyrazolone.

So, as a final subject of today, we should talk about triazoles.

First of all, the one, two, three triazoles you are presumably very well aware of.

You let azides react with alkynes, actually a quite old reaction found by Hüssgen.

It's one classical one, three dipolar cycloaddition reaction.

Normally the term, or often the term Hüssgen, cyclization or Hüssgen reaction is used. Nowadays this runs more under the label click reaction, click chemistry,

since it's one example of those reactions that work very, very well with sometimes almost quantitative yield and this reaction has been improved by copper catalysis

and the copper catalysis helps to achieve a nice yield regioselectively because in the classical one, three dipolar cycloaddition reaction you normally obtained a mixture of the two regioisomers.

The other regioisomer is the one where the R prime group is at this position.

A nice example is this one where you get this, well, mono substituted one, two, three triazole.

One could guess what's the alkyne component. Well, it should be simply acetylene but it is more convenient to apply calcium carbide.

Just a hint, I think this is interesting. So these one, two, three triazoles are very often used to just connect two moieties within one molecule

which could maybe electronically interact or the two moieties have both some application

and the combination of those are very interesting in bio organic chemistry this is used and so on. So as one of you noticed we have to make a few corrections here. So we have the phenyl group suddenly occurring here.

Well, it's just CH3 group that means we have to wipe out this one and now here the methyl group is missing but should be at this position. Now this should be correct.

What about the other triazoles, the one, two, four triazoles?

They are in academics less renowned than the one, two, three triazoles but these are indeed far more important in applications.

Medicinal applications as important fumicides.

Actually these triazoles are the most important fumicides in medicinal chemistry as well as in agricultural chemistry. Found that very funny somehow.

Well, let's analyze this structure. I would like to suggest let us analyze this as an exercise. Well, several ways to disconnect just one idea.

Well, it's clear we have hydrazine as one component and on the other hand we could try this emit.

Well, it's not wrong but the question is will it work in this direction?

So another idea would be acetic amide here and acetic hydrazide on the other hand.

Well, it will turn out that in terms of again carbonyl activity

those amides are not very active. If the nucleophile attacks, well then you have always the deprotonation of the amide as the competing reaction. So we should increase the reactivity of one of those components.

Maybe this one is more easy. So more reactive would be just acetonitrile.

Again a similar solution to things we have discussed before today. And if this is not reactive enough then we have an even more reactive reagent

with thioamide.

Advantage of being even more reactive, disadvantage of being rather smelly. So I think this is enough for today. Thank you for listening. See you next week Tuesday.