Welcome to part 23 of the lecture on designing organic synthesis, subject of today are radical, free radical cyclizations where CC bonds are formed by a radical chain process.

For instance, if you have a hexene moiety and you succeed in generating a

radical center here, then it can cyclize, forming either this type of intermediary radical

and you know this represents a five exo-trick cyclization according to Boltzmann or a

six endo-trick cyclization will occur and you might be able to trap then this radical.

One possibility to trap that is with tributyltin hydride, so in this case it

will lead to the hydrocarbons, so an expensive way to synthesize cyclohexane of course.

Nevertheless it is a very interesting process. First of all you have a very high selectivity for the five exo-trick process.

The product ratio will be in the range of 50 to 1, although it is clear that thermodynamically this is the more stable intermediate, so this is kinetically favored.

Moreover this type of process is highly interesting since also very sterically hindered CC bonds can be formed.

For instance with additional alkyl groups at these positions, quaternary centers then can be formed and it has a high functional group tolerance.

So hydroxyl groups, NH groups, amino groups, cyano groups, all are tolerated

and that means we don't need protecting groups, the presence of alcohols.

If we intend to perform this radical chain process for the construction of our ring system. Additionally frequently a high diastereoselectivity is observed. Let's assume we

have an additional substituent R here, another alkyl group for instance. Again trapping with a tin hydride, preferentially again the five exo-trick cyclization.

This product will be preferentially formed, of course, diastereoselectively, not enunciate selectively, you have a resomate here. Also this product will be observed if you have the substituent here at the three position counting from the radical center with this result.

In contrast having the substituent R in the two position then you will have a trans product as a

result and also with this intermediate substituent R in four position you will also get to a trans diastereoselectivity.

These are findings of Beckwith's group, therefore this is called Beckwith's rules.

Let's have a look at an early example, what worked out by a pioneer of this radical chain processes, Gilbert Stork from Columbia University, New York.

We've exercised at a terpenoid, here we have a bicyclic system first and well it's a tricycle because

here we have another six membered ring, there an olefin, there a methyl group, there also a methyl group and here a methyl group and hydrogen.

Constructing this type of skeleton Stork decided to try a radical process.

Well, but first of all if we think about the retro-synthetic analysis then it's clear again such

an exocyclic double bond first idea should be this is formed in the last step of our synthesis just a wittish reaction with a corresponding carbonyl compound.

So, put on the carbonyl here, yes should be correct.

So, the last step as a wittish type reaction works with a 86% yield.

And in the retro-synthetic analysis they had obviously the idea to form this C-C bond by a radical process.

Of course this is indeed a strategic bond, if you break that you have

that bicyclic skeleton you would hopefully build up by some kind of Diels-Alder process. Here an olefin, there a radical center. But they were also of course facing the problem that this CH3 group has to be somehow ando to this kind of cavity here.

Do you notice that? I hope so.

How do you get the CH3 group into an ando position? Well, the simple idea is getting this done by a hydrogenation step.

Having an olefin here, the hydrogen will approach from the exo phase and that means the CH3 group will get to an ando position of course.

So, simply hydrogenation with a palladium catalyst and this works out just fine at about an 80% yield.

Now one crucial step within the synthesis.

So, homolytic ring cleavage for radical reaction within retro-synthetic analysis.

Radical center here, radical center there.

But in reality, well, we could translate that into this situation having a double bond there.

So, radical reacting with that double bond forming radical center here, this should be trapped by a hydrogen radical. So, now we have to translate that as sunfins into synthetic equivalents.

So, the synthetic equivalents are having a vinyl bromide here and tributyltin hydride as the synthetic equivalent for that radical hydrogen.

So, please keep in mind the stereochemistry at that olefin is not important at all because this radical or the radical electron should be in a p orbital.

So, if you form radical, well, it doesn't care, it of course can react with this position.

Just that this becomes clear, you then have this situation.

And indeed, the reaction conditions are having radical reaction chain starter, AIBN, well known isobisisobutyronitrile, the tin hydride.

And you can start that either by heating it up or photochemically.

And this is a process that works very well, the transformation from here to that one succeeded in a 70% isolated yield.

Sorry, we are missing CH3 groups here, should now be correct.

So, let's go on with the retro-synthetic analysis. Of course, disconnecting there, heterolytically, donator, acceptor.

So, this is the nucleophilic starting compound after deprotonation. We don't have to care about the stereochemistry here.

Deprotonating means we form the enolate and then the electrophile should be attached from the sterically less hindered phase.

And this is of course the one with that olefin because here you have these sterically more hindering hydrogens.

So, nucleophilic component and this the electrophilic component. So, first lithiation, then trapping with this one, okay, 80% yield was achieved.

So, rather good yields in all of those steps leading to the completion of the synthesis.

So, how could we synthesize this one? Well, okay, let's draw the starting compound.

This is the starting compound for that synthesis. Forming the enolate at this position and trapping that with a cellulating agent gives with this silo-enol ether.

This is obviously an electron-rich diene. Now the reaction with the dienophile, maleinic anhydride.

First, maleinic anhydride in toluene. Then secondly, water and heating it up.

So, let's draw. Then we use older product.

With the water, the anhydride will be hydrolyzed. Also, the TMS enolate will be hydrolyzed and will tautomerize to the ketone.

So, and the 77% yield was isolated of this one.

The next step from here to there.

So, is an oxidative, a double oxidative decarboxylation. And at that time, lead tetraacetate in pyridine was applied.

And unfortunately, that gave a yield of about 30%. So that an overall yield from here to this was given 25%.

If one has a better idea than this lead tetraacetate reagent improving that step, then it is a real nice synthesis.

But nevertheless, this was one of the pioneering works for establishing radical chain process for the construction of a framework of rather complex natural products. From the same group, also, Gilbert Stork group is prostaglandin synthesis, which is really very impressive.

Typically, typical for a lot of these syntheses are cyclopentene moieties with oxygens already attached.

It's clear that these normally derive from, that yields aldo-reaction with singlet oxygen, was an endo peroxide, and then reducing it.

And then with some methods, the enantiomers are separated.

And here, from the corresponding alcohol, only one alcohol functionality is protected. The other is transferred to give an acetal like that.

And photochemically, a radical center is generated at this position. Photochemically, the carbon iodide bond is broken.

So, what will happen? Radical center here, C-C bond formation.

It is clear that a cis-anilated system will be formed. So, we have a defined stereo selectivity at this position.

Then, a radical center is here. And the reaction was performed in the presence of an excess of this reagent, a reagent which can trap the radical.

So, first step, intramolecular C-C bond formation as a radical process. Then, this radical reacts in an intermolecular trapping process forming the C-C bond from here to there.

Having a radical center here which is also stabilized by mesomerically with that carbonyl group.

And this center then is also trapped with a radical hydrogen making use of that tributyltin hydride.

By the way, for producing materials that are producing pharmaceuticals, you have to avoid strictly organotin compounds.

Because organotin compounds are rather toxic and even just amounts that you can just

detect with AAS spectroscopy are already enough to provide, to prohibit selling those pharmaceuticals.

Therefore, for a synthesis where you will indeed use these pharmaceuticals, you have to avoid tin. And there are some reagents with silicon hydrides which also works.

But this is the initial approach and the most easy one. So, next step. We want to get rid of that TMS functionality.

No problem. Just heating it up at 140 degrees. Rearrangement takes place.

This has a name. The TMS enolate is formed. This is the so-called Ruck rearrangement.

And for the target molecule we are already approaching, you need the transformation to an alpha-beta unsaturated system.

And this has been done with palladium acetate, not as a catalyst but as a reagent.

Okay. So, an expansive test. However, prostaglandins in pure form are so expansive that one even could afford using palladium acetate stoichiometrically in such a synthesis.

So, stoichiometrically, what happens? This is the nucleophilic center. This is the electrophilic center. A C palladium bond will be formed there at this position and then simply a beta hydride

elimination as we know from the usual palladium chemistry, also in catalytic processes that will occur then. I'm pretty sure if you use palladium acetate in catalytic amounts and add some appropriate

oxidizing agent, for instance quinone or something like that, you can also get that done catalytically. Okay. And this oxidation is called the Segusa oxidation. So, let's draw that intermediate completely.

So, the next steps. Just three steps.

First, with S, B, Na, H. So, what is that? I haven't seen that before either but the name is clear.

This derives from this binathtyl moiety. I assume that is simply binol with an aluminum hydride. So, oxygen, oxygen and aluminum hydride.

Overall, because it's an atropical compound, this is then a chiral hydride donator.

So, this will react with the carbonyl group and will reduce that enunciate selectively. So, this transforms the alpha beta unsaturated ketone enunciate selectively to this chiral allylic alcohol.

So, in this case a diastereoselective fuel. So, what else? Next step.

HCl, water, THF. All these reaction conditions will hydrolyze that TMS, not TMS, silyl ether.

On the other hand, it will also hydrolyze here the acetal. So, that means we will have another OH group there.

So, what remains after hydrolyzation of that acetal? We will have an aldehyde functionality here.

And this aldehyde functionality is used for a vittic reaction with this ullite

since this type of ullite is known to give preferentially a cis olefin.

Then, after hydrolysis, we get to this result. So, and this is a prostaglandin with the abbreviation plus PgF2 alpha.

Over these three steps, 54% overall yield is obtained and starting from here to get to that oxidized product, overall yield 58%.

So, that means almost 30% yield starting from that. Okay, you need a couple of steps to get to this intermediate in the complete synthesis.

But, I think, rather impressive. So, next very famous synthesis. It's the hirsutein synthesis of Dennis Curran from 1986.

The other stuff that we were talking about is chemistry of 1770s of the last century.

For the hirsutein synthesis, we only have a look at the key step. So, it was performed as a racemic synthesis.

So, and this is the substrate for the radical chain process. Again, the radical chain process starter, catalytic amounts.

We need a tin hydride. Heat it up. And the first step, as usual, will be a five axial trick reaction. Radical is formed here. I'm pretty sure that also with just photochemically one should be able to start that reaction.

So, C-C bond formation from here to there. Stereochemistry at this position then is clear.

Because, well, this chain is above the blackboard. So, the radical will also approach from above.

Radical center there. And now, the next step, five axial dig cyclization at the diagonal center.

Having the radical here, which is, of course, then trapped by the tin hydride.

So, and the overall yield of this product, and this is already the natural product, although racemic is 80%.

Very nice reaction, I think. From the same group, the current group, a related reaction making use of

the fact that with samarium diiodide and a carbonide group, for instance, naldide.

Such a key to radical is formed, which can participate in a radical chain reaction, of course.

So, Dennis Caron made use of that in a synthesis of a related natural product related to his routine. And I think the other one who is renowned for samarium chemistry, that is Gary Moolander at Boulder, Colorado.

And I think we discussed that type of chemistry in the lecture about stoichiometric organometallics. Well, so now, here the aldehyde functionality.

So, very similar set up difference is that acetal here and the aldehyde there.

So, with samarium diiodide and afterwards hydrolysis, you need at least two equivalents of that.

And after hydrolysis, this product was obtained by far more functionalized than his routine, or the higher functionalization makes it more sensitive.

Nevertheless, about 60% yield was isolated of that, also quite impressive, I think. For the next synthesis we have to discuss, we have to introduce another highly interesting reaction, radical reaction, the Barton reaction.

You, of course, know Sir Derek Barton, who has won a Nobel Prize, not for this radical reaction, but for analysis of conformers.

So, the Barton reaction, you start with an alcohol, treat that with nitrosyl chloride in basic medium, pyridine.

And as a result, you obtain a nitrous acid, ester.

Photochemically, the oxygen nitrogen bond is cleaved. You get an oxy radical and the nitrosyl radical. The oxy radical will abstract radical hydrogen at this position.

You have this terminal at the terminal position, that radical, and in the solvent cage, the nitrosyl, this will recombine to give this nitrosyl compound.

This will rearrange the tautomerization to finally give the oxime, and you can make use of the oxime by some further transformations.

For instance, hydrolysis, giving a carbonate compound, or reducing it to an amino group. So, in the next retrosynthetic analysis, one made use of this Barton reaction.

Well, a complicated name of a target compound, perhydrohistrionycotoxin.

We don't need to keep that name in mind. Well, it's once again a synthesis from the Cori group.

So, the target molecule is this one, a spirocyclic compound.

So, and retrosynthetic analysis.

Well, let's simplify this structure. So, this amine, we could envision that this derives from an amine like that, with some pentyl organometallic, pentyl grignier, for instance.

That would work. Well, in that case, with grignier, you should protect your H-group, or on the other hand, pentyl grignard is not that expensive.

Just use two equivalents, the first one will get hydrolyzed by your H-group, the second one would react here. So, this is of course interesting. You can envision that this amine functionality could derive from a lactame.

And of course, you should know that lactames like that can be formed by a Beckmann rearrangement. You remember, caprolactane?

So, and therefore, retro Beckmann rearrangement within our retrosynthetic analysis.

So, that means from the six-membered ring derives from a five-membered carbocycle as an oxime.

Okay, and now, imagine where do we get this oxime from? Well, from a Barton reaction, from this.

What about the selectivity? Why is this activated and not that?

So, let's have a look. Another drawing, chair conformer.

With this conformer, it would be clear that the CH2 group is in close proximity of this nitrozole astro.

Well, unfortunately, that conformer is not the preferred one since these two groups are both in axial position. And that is the reason why this transformation unfortunately only worked with a 20% yield.

Main product should be the radical reaction, radical oxidation at that side chain.

Nevertheless, it is elegant, unfortunately, with not a very good yield. But, okay, reasonable. So, I think this is enough for today. Have a nice weekend.

See you next Tuesday.