Welcome to part 11 of the lecture on designing organic synthesis. Today we will discuss two examples of some more complex target molecules and of course again

stereo selectively. Stereo selectivity will be of importance. So the first target structure is this one.

It is the so-called Preloq Gerasilactone and it is a benchmark target

molecule for newly developed methods for stereo selective synthesis.

So it is not too hard to find these bonds we can see as crucial bonds within the retrosynthetic analysis.

So please try some steps into the retrosynthetic analysis with this target molecule. Most of you realized that this CO bond is the crucial strategic bond we should target upon first.

Well, okay. We disconnect this strategic bond. We have donated reactivity at the oxygen, need acceptor reactivity here.

Okay, we could add a leaving group X at that position. But this would not represent a significant simplification. An acceptor reactivity in better position of a carbonyl group well is easy achieved if we assume that we have

an alpha-beta unsaturated carbonyl group there. Okay, so and now the alpha-beta unsaturation and we will get to this

structure. This is

much simpler as the one before since here we have only two cryosenters. Well, we have more stereo information in there because we have a double bond and there are cis and trans isomers of course. So if we

treat that under the right reaction conditions this will act as a nucleophile here. And hopefully will selectively give that product.

The problem is now, of course, we should think about would this cyclization lead to this product?

Presumably yes because this well, it's not directly a natural product, but it derives from natural products and we can assume that if nature produces that this structure has an advantage

compared to other isomers. But with your knowledge in organic chemistry you can prove that. So let's have a closer look at this structure and

therefore we should write it down as a six-membered ring than in the chair conformation. And then we will notice

that for this structure all substituents, the two methyl substituents and the bigger one here are in equatorial position. And this is of course

thermodynamically preferred. And indeed in the thermodynamic equilibrium we have this structure

with a selectivity of four to one concerning its C2 epimir. This all the other three stereogenic centers are highly selective.

Almost 100%. Here four to one. Okay, it's the preferred one. So and since this reaction of course should be an equilibrium with

here lying on that side of the equilibrium then it can react back and forth and as the equilibrium then finally we will have that one. The small part of that might be present. So this is no problem therefore.

Then we could go on. Double bonds are strategic bonds. Some kind of

aldol condensation. Well we have to protect that one and you have to protect this one and with an ester enolate and here the aldehyde is somehow we will make that.

So we get the simplification to this structure acid and an aldehyde functionality.

In this case we might already have present the cyclic heme acetal. This makes clear

that we will get to this

heme acetal by selectively reducing this cyclic acid anhydride

with maybe a chiral complex hydride would be an idea. Reduction with an chiral reducing agent. So this should be enough of that structure. Here

one would have a look in SciFinder of course.

This is small enough that we have good chances that others have already worked out a nice protocol for synthesizing this molecule. Let's go to another interesting target

a sesquiterpenoid.

Its name is aroma dendrine.

The first step in retrosynthetic analysis is fairly easy. Where is the most sensitive functional group? Well, actually it's not the cyclopropane unit but the

exocyclic double bond. One has to be careful with that because under somewhat acidic conditions it will presumably isomerize in. Therefore it's a good idea to transform that into the corresponding ketone since

the corresponding vitic reaction shouldn't be a problem.

So now for this target new target structure we should summarize some preliminary considerations.

So we have one, two, three, four, five stereogenic centers.

All are adjacent therefore, we don't need

some step of remote induction or introduce the chiral information twice into our synthesis. Maybe it's okay if we start from a compound from the chiral pool with just one stereogenic center and this is then the stereogenic control element

for all the other centers generated within our synthesis. So

we have this or not unusual combination of a seven and a five membered ring. You know these seven and five membered ring combination from

the natural product azulene that isomer of naphthalene.

Well, but setting up that not as a saturated system all you need therefore a good idea. And this idea could for instance be if we have a seven and a five membered ring.

So five plus seven membered might derive

from the combination of two six membered ring from two six membered rings.

Of course, annihilated six membered rings. So that means we have two annihilated six membered rings and we have a rearrangement enlarging one of those rings making the other one smaller. That's what we could look for.

In addition we have a ketone directly alpha to the annihilation position and we have seen situations

like that. Maybe you remember from organic chemistry two

that pinacol, pinacolone rearrangement

two five membered rings by enlargement gave a six membered and a five membered spirofused. Now one should think about

well this situation we have two six membered rings and want to get want to achieve

the transformation to seven and five membered enlarging this one and well making that one smaller. How could we achieve that? We need a leaving group here.

Well, let me put on first diol, a pinacol. The pinacol, pinacolone rearrangement in that case wouldn't

work. Since by protonation we would get the stable rather stable carbocation and tertiary carbocation. We need that as a leaving group somehow generating

a cation at this position and then that rearrangement would occur. I think this is an interesting motive

we could work with for designing while designing the synthesis of that target. C another consideration What about the construction of the cyclopropane unit?

Or we could think about having an olefin and the cycloproponation with a corresponding carbine. So how about

this one? Unfortunately the reaction to this won't work. Why? I think you already should know.

The CH insertion is an ultra-fast reaction and this carbine

will transfer to this propene immediately. So that doesn't work. Well maybe one could think about

cyclopropanation with dichlorocarbene carbene and then exchange those chloride against methyl groups. Ah, this is not that easy. Maybe here with a question mark

it's the same oxidation state and there are reactions known. We talked about that in the lecture about stoichiometric organometallics connection with TABOR regions and so on. You can

transform a carbonyl group into a dimethylmethylene unit. But again question marks. So however one could keep in mind that you can set up a three-member ring

having a nucleophilic reactivity here and an acceptor reactivity there. That means a leaving group. Another motive

for taking that into consideration for targeting this aroma dendrine. So now D. We already

noticed that we have five adjacent stereogenic centers that led us to the idea. Let's look for a chiral starting material from the chiral pool.

So and here we have such a structure this aldehyde. This is

a tepanoide called pyrrol aldehyde. Might be interesting to know that the corresponding oxime. Its name is pyrrolatine.

This is two thousand times more sweet than sugar and is applied as a sweetener mainly in Japan.

In pyrrol aldehyde we have here that chiral center and we have those three carbons here.

Maybe it's possible to construct the three-membered ring. Well, and indeed that has been done. So again let's draw the structure of pyrrol aldehyde.

First step hydrogen bromide addition. Markovnikov addition to that double bond

an acetic acid as the solvent leads to

this tertiary bromide. And now the most acidic position is this one where you get that conjugated enol

and an intramolecular

substitution will lead to this structure

with that dimethyl cyclopropane moiety. Let's call that A. And now we have to think about

how do we get from A to that ketone making use of the idea of a pinacol pinacolone rearrangement.

So this now is a bidirectional analysis. Well, please have a try. So let's draw once again that ketone which we are targeting for.

So once again

we would like to bring this structure and that one somehow together. Okay. And we would like to try the idea to apply a pinacol pinacolone rearrangement for the construction of that

adjacent or annihilated seven and five membered ring. And therefore we have to think about a retro pinacol pinacolone rearrangement in our retrosynthetic analysis. So and well

six membered ring six membered ring here the OH group. You don't care about the stereochemistry at this and that center. Here the leaving group is whatever that is. We could think about that later on.

Then, sorry, here's the CH3 group. Well, the stereochemistry here

should be also rather clear. All those will be preferentially on the same side as the others here, the other substituents. So some of you thought about

this bond as strategic bond. Yes, it is not wrong. However, it doesn't lead to a simplification in the next step. But while assuming that we can add

the OH group and that X to these two positions, maybe simultaneously, it's this addition to an olefin that would lead

to a much simplified structure. Intermediate, intermediary structure. So what do we have here? A cyclohexene

moiety, the retron of a Diels-Alder reaction.

A Diels-Alder reaction doesn't work well if we take propene as the olefin. Should be much better if we have

well an electron deficient dienophile. So let's change the oxidation state of that CH3 group. Put in a carbonyl group there.

So, retron Diels-Alder within our retrosynthetic analysis.

Acrylic aldehyde as the dienophile. The Diels-Alder reaction will proceed somewhat stereoselectively. And now getting from here

to there is rather straightforward since again Wittig reaction could provide that olefin.

So let's have a look at the real synthesis that then has been performed based on this retrosynthetic analysis. So as you're told, as I told you, these steps

work quite nicely. Well, the overall yield of this aldehyde with cyclopropane moiety was 41%. Well, that's not excellent, but it is

through several steps with a relatively cheap starting material. And it is the beginning of your synthetic procedure.

The next step, Wittig reaction to this structure worked with almost quantitative yield 98%. Now things became a bit more complicated.

The Diels-Alder reaction was performed at 100 degrees. So the 100 degrees are not necessary for the Diels-Alder reaction to

occur, but should help that more selectively the exo Diels-Alder product is formed. With the ando Diels-Alder product we have the wrong stereochemistry at this center.

So even though the reaction was performed at 100 degrees not the

thermodynamically favored product was the predominant, but the kinetically favored product um

in an 15 percent to 75 no, not percent it was our 15 percent yield of this one

75 yield of this was obtained and a couple of byproducts. So they can be separated and well this one

was then treated with a base or a base similar to potassium t-butoxide

just five minutes at 20 degrees then it is deprotonated and protonated again under equilibrium conditions at this position is the most acidic proton

which will lead to an epimerization here favoring the exo Diels-Alder product. So in equilibrium we have the ratio

of ando to exo 17 to 83 and the one with the 83 that is the thermodynamically favored one

we are in need of Okay So then reduction They reduced that with lithium aluminum hydride you get the tertiary alcohol. They transferred it transformed it into

a leaving group there and then treated tosylate or so and then treated again with a complex hydride for the complete reduction to the CH3 group while this is an alder synthesis in jacks I think from the late 60s

meanwhile there are better straightforward methods for directly getting the reduction done from here to there not the three steps nevertheless these three steps were performed with almost

quantitative yield and they are here Okay So next step osmium tetroxide

oxidation We get the diol the diol was obtained with 84 percent yield the problem now is

If you have your OH group here as I told you this is the badder leaving group We have to transform that selectively into a leaving group or with a tosylation

So first osmium tetroxide Secondly a tosylation since form that into tosyl group and This is selectively possible, of course with the secondary alcohol

Because the tertiary alcohol is sterically more hindered So and if you have this tosylate there the next step again is Works very well

better 80% yield Either adding strong base for instance potassium t-butoxide or

just the chromatography through aluminum oxide Which activates the leaving group, that's it. And then the we are here and

The final step was Wittig reaction Which worked with 79 percent Well, okay tomorrow we will go on with synthesis construction of

heterocycles Starting with three membered heterocycles and then going on to four membered five membered six membered and so on Thanks for listening. See you tomorrow