Welcome to the sixth part of the lecture on designing organic synthesis. The subjects of today are dicarbonyl compounds, the 1N dicarbonyl compounds,

which means we will start with 1,2-dicarbonyl, then 1,3-dicarbonyl, 1,4. And on the next lesson, next Friday, then we will focus on 1,5 and 1,6-dicarbonyl compounds. So, 1,2-dicarbonyl first, of course, general structure.

And, of course, first idea for a disconnection is that central CC bond between those two carbonyls,

leading us to the idea having one carbonyl compound with an acceptor reactivity,

and the other one with a donator reactivity. Acceptor reactivity, as you know, is a natural reactivity that fits that one with a donator reactivity,

while is a disadvantage, of course, since we need a reactivity on polo. Well, we can translate it into synthetic equivalents.

Here, an acyl group with a leaving group, X, a chloride, or maybe an ester is sufficient. And, well, the other one, for instance, this dithioketal, dithiocetal, which we can deprotonate.

However, this is uncomfortable, tedious, and maybe the reactivity we have here is unnecessary for achieve this transformation. Therefore, it's certainly not a preferred solution to that synthesis problem,

and therefore we should think about other ways to get to 1,2-dicarbonyls. Well, and there are more attractive ways to synthesize that.

For instance, starting from a ketone and performing some kind of oxidation in the alpha position.

This is, for instance, possible with, well, reaction with nitrozyl cation. It reacts with the corresponding enolate, forming the alpha nitrozo ketone, which will tautomerize to the corresponding oxime.

Hydrolysis, then, will deliver our target compound.

There is, of course, the possibility for the direct introduction of that

oxygen with the appropriate oxidation agent, selenium dioxide will do the job. On the other hand, we could, for instance, assume that we get to the dione from a pinacol, sorry, not a pinacol, from a silylene.

Well, of course, consequently, you could think about a pinacol as the starting material, oxidation, and here, again, oxidation.

However, as you certainly know, we have various interesting ways for synthesizing such a silylene, combining two aldehydes.

Well, you have the problem with selectivity in cross-coupling.

However, there are various ways. For instance, catalysis with cyanides. This is chemistry from the basic lecture, organic chemistry 2, or very similar, the Stetta catalysis.

And there is another way for synthesizing this type of moiety, that is, you have two esters and then treating that with sodium sand.

I hope you remember that method. Let's have a look at this interesting small molecule, that cyclohexane 1,2-dione.

And indeed, there are various ways which have been applied for synthesizing that.

It is indeed possible to start from adipenic acid, esters. So, sodium sand. Secondly, hydrolysis.

Also, it is known that this type of hydrolysis works. And therefore, you can start from cyclohexanone with the introduction of that oxime group via the nitroso intermediate, as I explained before.

For this transformation, from that ketone to that dione, a 63% overall yield is reported.

Whereas, the oxidation with selenium dioxide has also been tested and essentially it's the same yield, 65%.

So, I didn't find the overall yield for the sodium sand method. While, in terms of cost efficiency, I think I would prefer this way of synthesizing that dione.

I would try to avoid the selenium dioxide and certainly working with sodium sand is also somewhat more expensive, that's for sure, than that way.

That dione is an interesting substrate for further annihilation processes.

For instance, condensation reaction with ortho phenylene diamine will lead to this interesting heterine.

While this is interesting in terms of medicinal chemistry.

Let's discuss 1,3-dicarbonyl compounds.

We disconnect either here or there. Well, for that discussion it is essentially the same. Donator reactivity alpha to the carbonyl group. This fits to the natural reactivity since we can easily deprotonate in alpha position.

And on the other hand here, acceptor reactivity, the carbonyl group, also natural reactivity means this combination is favorable.

As an interesting example, let us discuss how to synthesize this target molecule.

Well, okay, first idea should be to disconnect here.

And I will directly translate the synthons into their synthetic equivalents. A phenyl acetic acid derivative and a benzylic organometal species.

Well, not a bad combination.

However, that methylation is not, well, cheap if you want to run that, for instance, on a large scale industrial process.

You would try to avoid, for instance, a benzium magnesium halide. The problem with the benzium magnesium halide is, you might know, treating benzyl chloride or benzyl bromide with magnesium.

And single electron transfer processes produces very likely benzylic radicals which recombine. And 1,2-diphenyl ethane is then a major product.

And not, unfortunately not, the desired benzium magnesium halide. Lithiation can be achieved starting from toluene and deprotonating with very basic organometallic bases, butyl lithium.

But for a large scale process, industrial purposes, you would also try to avoid working with butyl lithium, of course.

Well, there is another solution to the problem. A 1,3-dicarbonyl compound is favorable in synthesis.

So, let's transform that by adding a functional group, master functionality here.

Then we have a 1,3-dicarbonyl situation. Okay, this looks more complicated than that. So, we go up in complexity. However, the disconnection now, again at the same bond, allows us to get to two identical fragments, phenyl acetic acid ester.

So, and indeed, no problem treating that with a sodium alcoholate, in the same alcohol, of course, will give the classical ester condensation.

This product, and now very simple, treating it with an aqueous bronsted acid, heating it up a

little bit, will hydrolyze the ester, and then as a better keto carboxylic acid, it will decarboxylate.

Immediately. So, I think you will agree, starting from here, first, a very reliable step and a second simple hydrolysis with decarboxylation is the best way to get to that type of compound.

In addition, it might be interesting to think about the synthesis of structures like that. Of course, phenyl acetic acid ester, you can buy, it's rather cheap, but

imagine you are interested in a structure like that with additional substituents here. For instance, bromide, bromine in the alpha position.

So, then you have to target that, since you can't buy it, and indeed, the standard procedure to get to that is, well, the transformation of the corresponding acetonitrile,

which you get starting from benzyl halide, a chloride, or a bromide, a nucleophilic substitution with a cyanide.

Next example, in this analysis, the first retrosynthesis step is fairly easy.

You would introduce that allylic side chain, starting with that 1,3-dicarbonyl with the

especially acidic hydrogen double-activated by the two carbonyl groups, deprotonating with sodium ethanolate and ethanol.

And then trapping the enolate with the propagolic bromide.

Now we have a choice. Let's test both the disconnection here within the ring as R1 and the disconnection as R2 of that aldehyde functionality there.

So, disconnecting here leads us to the suggestion that we have acceptor reactivity here, translated into a leaving group at that position.

So, some carboxylic acid derivative with that leaving group, and donator

reactivity here we get simply by deprotonation of the aldehyde functionality. Well, okay, correct for the retrosynthetic analysis. However, the next step to set up, how to set up this 1,7-dicarbonyl situation does make problems.

Therefore, let's have a look at suggestion 2. Donator reactivity here,

well, easily achieved through the corresponding enolate and acceptor reactivity here. So, what do we have here? This is a derivative of formic acid and, well, X, let us choose an ethoxy group.

So, for X, the leaving group, so it's formic acid, ethyl ester. And we should treat this combination again with sodium ethanolate and ethanol.

Secondly, hydrolysis, moderate reaction conditions, but we don't get some additional condensation or aldol addition processes.

So, in this situation of course the question arises, how do we get the selectivity here?

Why is this position deprotonated preferentially compared to that one? Well, we could use, for instance, LDA as discussed in the lecture on stoichiometric organometallics.

The sterically hindered base will preferentially deprotonate the sterically less hindered position. Here, the deprotonation at this position is thermodynamically favored.

And under these reaction conditions, sodium ethanolate, heating it up a bit, we should have preferentially the deprotonation there. However, deprotonation reactions are equilibrium reactions when this sodium ethanolate is trapped by that formic acid,

after addition elimination process as mechanism, we get to this result here with the same reagent present in solution of course.

We get to this situation in analogy to that. But here we have that especially acidic proton will be deprotonated immediately

forming this stabilized enolate and this is thermodynamically indeed preferred compared to that.

Since this one is trapped at this enolate, it won't react retro.

On the other hand, here it very well seems to be possible that

this remains still in equilibrium since the sodium ethanolate can attack addition elimination process. And then we are back again there. Remember, aldol addition and aldol condensation processes are very well equilibrium reactions.

Now let's try an exercise. Please think about how would you synthesize this target molecule. Unfortunately, in this case, the very simple solution is not a good one.

Disconnecting here and combining the alpha acidic cyclopentanone with that benzylic halide.

The problems you would have to face when treating that with a base is, well, that base could also react as a nucleophile substituting the chloride.

As soon as you have deprotonated here, well, you wish that it attacks as a nucleophile with benzylic chloride, but you could get aldol addition and aldol condensation as competing reactions.

As soon as you have the first benzo group attached, you want to avoid that the second benzo group is attached. That is a problem. With the first benzo substituent there, the proton in alpha position becomes even more acidic.

You have lots of selectivity problems. There is one way to avoid these selectivity problems with a reaction you already know from our organic, basic organic chemistry lab.

Some of you have synthesized an enamine.

I hope you remember, morpholine combined with cyclopentanone, well, in our undergraduate lab we have cyclohexanone since that is cheaper.

You can form that enamine at a so-called Dean-Stark trap. In German language, that's a Wasserabscheide.

This enamine reacts with benzo chloride via a nucleophilic substitution, alkylation at this position.

Again, an enamine is formed which you can then hydrolyze and get to that structure.

This is possible. But maybe there is another way to get to this product which is connected to the subject we are talking about, 1,3-dicarbonyl compounds.

So, let's add a second carbonyl functionality. You can get rid of

the ester functionality just by hydrolysis and subsequent decarboxylation as discussed earlier.

We get to this cyclopentanone with the ester functionality in alpha position.

Now the deprotonation takes only place once at the especially acidic position here and no problem to get that alkylation done. So, simply with sodium ethanolate and the synthesis of that is also fairly easy.

Again, adipinic acid, ester, but in that case not an acetylene condensation, but simply treating it with sodium ethanolate in ethanol.

This again is an ester condensation, this time intramolecularly and as you know this than is a name reaction called the Diekmann condensation.

Another argument why the synthesis of that target with this as intermediate is far better than starting from that ketone.

The standard procedure for synthesizing that ketone is starting from adipinic

acid, ester via that molecule and then hydrolysis under acidic conditions. As an alternative to this solution, one of you has found an interesting idea.

It's this transformation. Let's add a double bond to that target that we get to an alpha better unsaturated ketone.

And this is indeed a nice idea because you can get rid of that double bond simply by hydrogenation with an appropriate catalyst.

Additionally, if a catalyst is the chiral one in homogeneous catalysis, you might be able to synthesize that and then see you selectively. And while the same importance as the fact that you can easily set up

that, these double bonds alpha better unsaturated carbonyls are easily synthesized by aldol condensation.

In this case of a cyclopentanone with benz aldehyde. And here we don't have a selectivity problem with the initial deprotonation because that aldehyde has no alpha acidic hydrogens.

If it would be an aliphatic aldehyde, then you have again a selectivity problem. Next exercise, the target molecule is a tricarbonyl compound with several 1,3 connections of the carbonyl groups.

So please think about how to synthesize that target. So for analyzing the synthesis or retrosynthetic approach to that target, let us first disconnect here as R1.

Having that intermediate plus this pivaloyl ester, in the first glimpse this should be rather acidic, well I think it is.

However, there is a problem connected with that enolate.

You might notice that this system has a somewhat anti-aromatic character.

You can't synthesize that molecule and keeping it at room temperature at your flask. It is highly reactive because of that somewhat anti-aromatic character.

So before taking that for granted, one should look it up for its reactivity.

To my knowledge it works, you can get that as a nucleophile but it might be some problems connected with it. I am not quite sure about it.

So nevertheless, with that idea you would have to analyze how to get this one. It might become more complicated. Therefore, let's return to this structure and discuss the alternative R2, disconnecting here.

Of course, preferred donator reactivity at this position, double activated CH groups here, methylene group and the acceptor reactivity there.

It is not necessary to have there an especially high reactivity, a chloride is not necessary, an ester is just fine. So, donator reactivity, acceptor reactivity and next disconnection we would have to consider at this bond.

Again, acceptor reactivity here, ester functionality the same as that group before.

So, we simply have phthalic acid diethyl ester, simple and cheap and the tertiary butyl methyl ketone as the second reaction component.

And if we treat that together with sodium ethanolate in ethanol it will directly react to get to the target molecule.

Of course, after hydrolysis since here at this position you have an especially acidic CH group.

I hope you know how to synthesize molecule like that. Yes, you can think about having pivaloyl chloride and a methyl grigner, ok, or on the other hand acetic acid chloride and tertiary butyl lithium.

Well, but if you want to avoid these reagents then you should remember making use of

the pinacol of acetone and performing a pinacol-pinacolone rearrangement which will lead us to that ketone.

Next subject, of course, the 1,4-dicarbonyl compounds.

The bad news in this case we have an unfavorable situation for our retrosynthetic analysis.

Well, disconnecting here has of course the disadvantage that natural reactivity in better position to the carbonyl group is an acceptor reactivity.

Here directly at the carbonyl group also acceptor reactivity. We don't want to have acceptor and acceptor reactivity side by side. So, disconnecting here leads to two similar symptoms.

Unfortunately, one with the acceptor reactivity, one with the donator reactivity. We would need a reactivity unprolung for that partial structure.

Well, we can do that simply adding a leaving group but with bromide or chloride these structures usually are lacrimators as discussed before.

Here we can deprotonate. We have that problem. Let's discuss this example, cyclopentanone and bromoacetic acid ester.

Well, the presence of the base we could hope for deprotonating here and let that attack as a nucleophile.

But maybe you remember one remark from organic chemistry too, but usually it doesn't work.

Because here is also an activated methylene group.

The bromide also activates the acidity here and you can deprotonate here as a competing reaction. Maybe it works but it has a competing reaction. Let me clarify that.

The competing reaction is deprotonation here attacking that as a nucleophile resulting

in an alkylate as the intermediate stir with the bromide attached there.

But then intramolecularly nucleophilic substitution ending with this epoxide alpha 2v ester.

So the prominent example from organic chemistry too is this one. Let that react under basic conditions with that, with the benzaldehyde.

Then you result in this product and this is a glucidic acid ester and it is a Datsen glucid glycid ester synthesis as a name reaction.

Sorry, ester synthesis.

So thinking about that I'm not sure what will happen here in this case. That Datsen's glycidic ester synthesis is a rather nice reaction since with that benzaldehyde you don't have that alpha acidic protons.

And of course the carbonyl group, the aldehyde has a higher carbonyl activity than the ketone.

Here you have that competing situation of either deprotonating alpha 2v ester but additional activated by the bromide or here. Well it could cause a lot of problems. Therefore we should think about an alternative.

Well okay, maybe again making use of that enamine or for instance TMS enolate will avoid that competing reaction pathway.

A related example, now with this aldehyde functionality attached, disconnecting here will lead us either to this combination,

donator reactivity here, acceptor reactivity there or acceptor reactivity here and donator reactivity there.

Both situations are rather bad. If you add a base in this case it's for sure that the position alpha to the aldehyde will be more acidic and deprotonation will take place here.

With these synthetic equivalents in mind adding a base will for sure lead to a deprotonation. Again alpha to the aldehyde, that's what we want but presumably it won't preferentially

attack that C halide bond but the next aldehyde in an aldol addition process.

So both ideas have huge problems. A clever solution to that problem is this one.

Having an allylic sidechain here and simply doing an ozonolysis will lead us to that aldehyde functionality.

Last example for today, also a nice target for an exercise, this one.

So every one of you noticed that this somehow reminds you of the so-called Robinson annihilation.

For Robinson annihilation we would need an additional methylene group here, two methylene groups. Next lesson on Friday.

in connection with 1,5-dicarbonyl compounds, we will talk about Robinson annihilation again. However, the idea of that this is connected somehow with the Robinson annihilation is not bad because we can make use of that

in the first step of our retro-synthetic analysis. Disconnecting here leads to that cyclohexane dione derivative.

And then you should, of course, think of disconnecting there

in analogy to the Robinson annihilation. But in the Robinson annihilation, we will have methyl vinyl ketone as the reaction, second reaction component. And here in this case, we are in need of that bromoacetone,

that infamous lacrymator, we, of course, would like to avoid. There are ways to avoid that.

If you think about functional group interconversion here,

for instance, let us assume that this derives from a propagolic side chain.

Then we just have to add water with Makovnikov selectivity to that acetylene. So formally, it's simple water addition.

You can achieve that relative selectively with an oxymercuration might remember that type of process.

While maybe also just water and an acid could do the job. However, I would prefer another solution,

certainly cheaper, just going to an allylic side chain, easy to introduce, cheaper than the propagolic system, of course.

And now you need the addition of water, Makovnikov style to the double bond and an oxidation to the ketone.

Best would be in a one pot reaction. And indeed, this is known. There is an important industrial process, the Wacker process. So, Wacker process,

making use of palladium chloride and a copper chloride catalyst combination plus molecular oxygen as the oxidizing agent

and this in an aqueous system. We talked about the Wacker process in the lecture on catalytic organometallics.

So for the 1,4-dicarbonyles, please keep in mind that we have initially an unfavorable situation for setting up the 1,4 relationship.

On the other hand, there are some tricks like ozonolysis or Wacker process and so on to avoid these initial problems. And you should also keep in mind

that 1,4-dicarbonyles are especially important just to illustrate that from this hexanedione.

You can easily get to the corresponding furan,

to the corresponding pyroles and to the corresponding thiophene. So, that should be enough for today. Thank you for listening.

We will meet again next Friday at 1 p.m. in the neighboring lecture hall HNC30. See you on Friday.