Welcome to the third part of the lecture on designing organic synthesis. In the preceding lesson, we have analyzed this target structure with the help of having in mind using an oxy-cope rearrangement within our plan for synthesizing this molecule.

We noticed that we have the retron of an oxy-cope rearrangement within this molecule and this is a carbonyl here and alpha, beta, gamma, delta, epsilon and saturation.

So the result of our considerations was that we have to synthesize this type of functionalized bicyclic ring system.

So, and well, then we should check, and we didn't do that in the preceding lesson,

should check what would be the result of the oxy-cope rearrangement. So reaction conditions, potassium hydride to synthesize the potassium alkylate

and then heating it up so that the rearrangement can occur. So what would happen? Well, carbonyl group is formed at this position finally.

This position is connected with that, connected with that.

Well, it should be right, okay. So, and let's translate that into a drawing where we don't have a three-dimensional perspective in mind.

So this structure with an oxy-cope rearrangement would lead to this product and if we compare that to our initial target, we notice, of course, these are mirror images, right?

So that means, well, if we have a racemic resumate as the target, we wouldn't have to care about the absolute stereochemistry, but if we have that as a natural product in mind and want to target it enunciate selectively,

then this here is the wrong enunciomer to perform that rearrangement. So, we need the mirror image.

So in order to get to this substrate, we should consider using, for instance,

chiral Lewis acids for enunciate selective Diels-Alder reaction. That would be the solution to the problem.

Well, next, let us draw again that structure we had a glimpse upon at the first lesson.

And now it's time for retrosynthetic analysis to this target molecule. The problem is, where should we start?

It's often a very good idea to look at the most reactive functional groups because the most reactive functional groups you want to introduce into your molecule at the end of the reaction.

So, if you want to introduce them at the end of your sequence, well, then it would be, of course, the other way around with one of the first steps in your retrosynthetic analysis.

So, for instance, cleaning up or with oxygen and with water, nothing will happen here. It's not possible to hydrolyze here such an ether moiety under moderate reaction conditions.

In this case also this is rather stable. You wouldn't mind about that. On the other hand, well, of course, these functional groups here are the most reactive in the molecule.

So, we should identify what we have here. Certainly, a cyclic carboxylic amide. So, here we have the amide. On the other hand, an enamine here at that position.

So, how is an enamine formed? Simply by a condensation reaction. So, why not focus on that enamine and the condensation reaction? For that condensation reaction you need a ketone, of course.

So, let's directly introduce the ketone functionality without thinking about the synthon. It's directly a substrate which would give then, on the other hand, by condensation reaction, the enamine.

So, next step. Where do we have a strategic bond for cleaving within the retrosynthetic analysis?

And getting to a much more simple structure. On the other hand, a bond which is easy to be formed. Well, it could be this one.

Donator reactivity here. Acceptor reactivity there. And that acceptor reactivity is translated into alphabetic unsaturated carbonyl compound for the synthetic equivalent.

Well, actually this sequence, this acrylic amide, the treatment of the base and deprotonating here,

the overall sequence to this target should remind you of the Robinson undulation. Well, keep in mind that similarity. Here the acrylic amide.

And in the Robinson undulation you are normally dealing with methyl vinyl ketone, abbreviated like that.

Okay. So, nice initial idea. Now we have to check would that really work. So, there might be some problems. First of all, if we deprotonate here, we form an enolate.

So, and the enolate is a base and on the other hand we have this quite acidic amide. So, in equilibrium we would also get a deprotonation here.

That might cause some problems. Might cause occurrence of some competing reactions. Well, we could try that. If it doesn't work, we should improve.

So, it might not be necessary directly having the amino group here. We could think about using acrylonitrile instead.

So, later on, XEMI idolization of the nitrile group will lead to the amide functionality. On the other hand, well, why not using an acrylic acid, aster, then we would have this as an intermediary product and could treat.

But simply, the ammonia should work.

We have to find out the right reaction. Okay.

With these considerations, do we already have solved all the problems connected with that step? I'm afraid no. Because if we treat this with a base, how do we get the selectivity that it is deprotonated here and up there?

That's a problem. Actually, I don't see any chance for any base to get this preferentially deprotonated here or there.

So, maybe we have to play a trick on that and perform a multi-step sequence as a one-pot procedure.

So, for instance, simplifying it in a retrosynthetic analysis to this alpha-beta unsaturated ketone

plus this nucleophile, donate a reactivity here, accept a reactivity there, putting on a metal there.

And, of course, M presumably should be some kind of copper complex.

An organocutrate we should use for that alpha-beta for this conjugate addition reaction to the alpha-beta unsaturated ketone. And, of course, if we perform that in a one-pot reaction, having then already

the enolate in there, well, you have to use some solvents that stabilize your enolate. And then you can trap the enolate directly, well, for instance, with the acrylic acid astra.

Well, okay. So, we are now down to a structure as simple as this one, a methylcyclohexanone.

Well, let us assume that we try a racemic synthesis, don't care about the absolute stereochemistry here.

So, what about this idea, introducing that methyl group also through a conjugate addition reaction with a methyl group rate?

Wouldn't that be simple? No one with any doubts.

So, the problem that you have is, where do you get this one from? At least you can't buy it, and there is a simple reason.

Under acidic as well as slightly basic reaction conditions, which would, of course, readily, I summarize under formation of the aromatic ring system, it summarizes simply to phenol. So, therefore, this is not a good idea, of course.

So, how should one analyze the synthesis of that molecule? Well, this is so simple, you can be sure that someone has synthesized it and has published it,

and so, in that case, I would just have a look in SciFinder and figure out how they did it, and maybe also thinking about improving if the reaction sequence which has been done is not completely satisfying.

Well, so, simply let us have a look how this is synthesized, already synthesized by others. Well, there is acetic acid ester as one component, this acrylic acid ester derivative as the second component,

and this is classical carbonyl chemistry, so under the basic reaction conditions with sodium ethanolate and ethanol,

or what would happen, this is, of course, the most acidic position, double activated by carbonyl groups, here you have the nucleophile, and in a Michael addition reaction, you get that conjugate addition product.

In equilibrium, also, every now and then, that CH3 group will be then deprotonated and making another condensation reaction here.

So, this will result in that cyclohexane dione like that, so, and after hydrolysis of that ester,

so, ethanol is formed and a decarboxylation will occur.

Remember from organic chemistry, too, that better keto-carboxylic acids are not stable.

So, we need now some kind of reduction, we need to double bond there and get rid of one of the carbonyl groups.

Well, with an ortho ester like that, this is an ortho ester, is an acetal of formic acid ester.

Under acidic conditions, you will get this transformation then. Next step is reduce the remaining carbonyl group to give an alcohol,

and then, simple hydrolysis under slightly acidic reaction conditions will give you that OH group as a leaving group

and the hydrolysis will form our target intermediate.

Let us go on with additional strategies and tactics in retrosynthetic analysis.

One very simple but often used and very useful is functional group interconversion.

Often abbreviated as FGI, for instance in the retrosynthesis books of Stuart Warren.

One simple example.

Okay, normally you wouldn't bother about the synthesis of this molecule since you can buy that. And it's not that expensive. However, imagine we have more complex molecule in mind with some additional substituents,

maybe a long chain, a long side chain or something like that. Let's concentrate on these functional groups. So, for instance, we start looking at that disconnection.

What do we need here? Donator reactivity for that amino group which we want to introduce.

And on the other hand here, acceptor reactivity. Okay, looks good on the first glimpse because donator reactivity is just the natural reactivity for an amine like that.

Acceptor reactivity, okay, something like nucleophilic aromatic substitution.

However, we have a problem in translating it into synthetic equivalents simply because, well, let's put on a chloride there as the leaving group using, for instance, simply ammonia solution.

We don't want to use sodium amide. This is much too reactive there. However, this would react, of course, with the acid.

So, we would have to protect the acid. Let's try it as an ester. Well, maybe then the nucleophile would still react preferentially than at the ester functionality.

So, we have a problem there. Let's try the other one. Aniline with a donator reactivity in power position.

And this synthon as a fragment, acceptor reactivity here. Well, again, both are the natural reactivities of these positions. However, if you try some Friedel-Crafts type stuff, you have always

kinetically preferred the reaction of the free electron pair of that amino group. You are always facing some serious problems. Therefore, functional group interconversion is a good idea

and in this case, it's simple changing the oxidation state of the functional group. For instance, here in this case, let's leave the amino group

and we go to the oxidation state of the nitro group, since it's usually no problem for reduction of the nitro group

with hydrogen and an appropriate catalyst. Next step in our retrosynthetic plan. Again, changing the oxidation state.

Now, from the acid functionality, just go back to the simple methyl group.

And para-nitrotoluene is easy to make. No problem in performing this reaction sequence.

So, another good idea is looking out for suitable starting materials. So, identifying suitable starting materials.

For instance, from the chiral pool. You all know the chiral pool consists of all those natural products

we can get hands-on and having some chiral centers. We could make use of them starting the reaction sequence to our target molecule from a chiral pool compound.

Actually, this is then not the pure retrosynthesis anymore. We have our target molecule and we identify a natural product,

something from the chiral pool, we can buy or we can isolate from nature and somehow they are related structurally to each other and we have to connect them within our synthetic plan.

We make retrosynthetic analysis and try to get a structure where we know that we get to that structure from the other side, starting from a natural product. Therefore, it's in that case a bi-directional analysis.

One example you will find in literature is the synthesis of picrotoxinin.

While picrotoxinin has been synthesized from the Coray group

and also from Barry Martin Trow's group at Stanford, Kurt Heffner was the one who finished that synthesis. Let's draw that target molecule.

You will notice this is rather complicated in structure.

So, again Barry Trow's and Kurt Heffner. So, they had to look for a natural product from the chiral pool

which has structural relationship to that and found that there are some similarities with carvone.

It should be minus carvone. So, we have that six-membered central ring with an isopropyl group here.

Here the six-membered ring, the isopropyl group. The same stereochemistry, a methyl group at this position and an oxygen functionalization there. Here the same methyl group and oxygen functionalization.

So, bi-directional analysis. In that case, having in mind some palladium-catalyzed

annihilation reactions developed in the Trow's group and the overall synthesis took about 24 or 25 steps. It's not easy to do, of course.

So, in addition, stereochemical considerations.

Let's explain that with this model compound.

So, in this simple, rather simple model compound, we have three stereogenic centers. And they are all in close proximity.

Therefore, it's a good idea to look for one stereogenic center which could be the control element of a stereochemistry through your reaction sequence.

A bond which is easy to make is this one. So, a malonic acid, ester nucleophile, could be introduced here.

Easily to be deprotonated, so this is the nucleophile then. We need an electrophilic center here. SN2 process. That means we need the opposite configuration

for that leaving group X here.

And this lactone is now the target we have in mind. We analyze the next retrosynthetic step. So, everyone who knows that reaction which is called the iodo-electronization

now has an advantage because you then can simply write down simple structure, this one.

So, now we know this stereogenic center should be a control element of that sequence. It determines the relative configuration here. And if you start from an enantiomer,

then you control the absolute stereochemistry of our target molecule. Just to remind you of the Baldwin rules,

how do we classify this cyclization reaction?

So, what kind of cyclization is that according to the Baldwin rules? Well, a five-membered ring is formed. So, we write down the five.

This bond is formed. The double bond is broken to a single bond. That double bond is located outside the newly formed ring. So, we have to add an exo here. And the reaction takes place at an sp2 hybridized center

that means a trigonal center. So, it's a five exo-trig cyclization.

So, the next good idea in retrosynthetic analysis sounds a bit complicated.

Identifying strategic bonds by topological criteria.

Sounds great, very well educated. Well, actually it's rather easy. We talked already about that. A strategic bond is a bond if you disconnect it,

you get really a simplification. If it's not a simplification, then it's not a strategic bond. Okay, that's it.

Becomes more clear with this example. So, this one is not a strategic bond. It doesn't become more simple.

Well, we could disconnect here, but then we still have that three-membered ring. But maybe it's possible to simultaneously form these two bonds

and then it makes sense in our retrosynthetic analysis to break those simultaneously.

And this looks already much simpler than that. So, the idea is could we use some kind of 2 plus 1 cycloaddition reaction

with a carbine or a carbine complex. I found funny if we analyse that again and disconnect.

So, R1 here, this as R2.

Well, we will notice that both structures are the same. So, imagine we would generate this carbine.

Would this carbine indeed give that cycloaddition reaction?

It doesn't because there is a competing reaction which is much faster. The insertion of the carbine functionality into the neighbouring C-H bond.

This is indeed a very fast reaction.

So, you know, in your undergraduate lab you worked for instance, you made that cyclopropanation reaction with a dichlorocarbine. This will work, but you can't do that with a dimethyl carbine

because of this C-H insertion. So, therefore, we have to modify our plan. We should not have hydrogen atoms here.

Well, okay, let's change the oxidation state. This would give then, is that correct?

Yeah, okay. This would give that cyclopropanation

and some kind of reduction which should be chosen carefully, of course, because of that sensitive cyclopropane unit here, should give then that result.

And where do you get that kind of carbine from? From that type of diazo compound, for instance,

making use of the rhodium or a copper catalyst. Well, of course, one should then further analyse.

This is still not a simple starting material. So, thank you for listening for today and see you tomorrow.