Welcome to part 26 of a lecture on designing organic synthesis. Today, again, we will talk about Mannich reaction, special types of Mannich reactions. Well, let's call it magic Mannich part 2.

And we will start with the interesting Daphnifilum alkaloids. Daphnifilum, that's a plant from which more than 250 alkaloids have been already isolated and identified.

All of them with rather interesting, very complicated structures. This is one of those structures. Here we have a side chain which could be somewhat longer with different substituents.

Here we can have a double bond or it is just an isopropyl group here. And there are also some structures known deriving from some kind of skeletal rearrangement of a structure like that. So, a rather complicated structure and in his lecture on magic Mannich,

Clayton Halfcock was claiming that first of all he needed about two years of learning how to draw this type of molecule that he can analyze.

Of course, this is a joke. Nevertheless, interesting structure and with the retrosynthetic analysis, you will see that from a retrosynthetical point of view, this is not that complicated.

So, let us start. So, we have a double bond here and in addition we have a C-N bond there.

Well, we have a relationship between those functionalities and from a retrosynthetic point of view, we can claim that this bond is a strategic bond.

So, let us break this bond. Best now is let us draw the structure again. So, there is already something wrong.

Here we have two five-membered rings. The drawing is not optimal, but it is okay I think.

Well, as we said, let us assume that this is a strategic bond. Then we should say, well, here we should have the acceptor reactivity.

A cation here is of course mesomerically stabilized. So, acceptor reactivity here represented by this iminium cation.

And on the other hand, donator reactivity there, well, let us draw an olefin there.

So, and here we have the electron-rich system. Here we have the electron-poor iminium cation, electrophile, nucleophile. Of course, the C-C bond formation is also a special type of Manisch reaction.

Well, normally, the Manisch reaction, we have an enol system and this as an electrophile. Here we have an especially rich olefinic system. And, well, C-C bond formation here will leave the cation there.

Just deprotonation at the terminal position would generate this olefin.

So, next step in retrosynthesis. I hope you notice we have a six-membered ring here, here the iminium cation.

Well, a six-membered ring with an unsaturated partial structure. Well, this reminds us of the retron of the Diels-Alder reaction.

Well, the retron of the Diels-Alder reaction is the cyclohexene moiety. Here we have a cationic azer cyclohexene moiety. Nevertheless, also hetero Diels-Alder reactions are known.

So, next step in retrosynthesis is let us break this and that bond simultaneously. So, let's see where this will lead us to.

Now, our structure within the retrosynthetic analysis becomes much more simple.

That's it. So, the diene electron-poor diene, here the electron-rich dienophile,

for a Diels-Alder reaction with inverse electron demand.

So, Diels-Alder reaction would lead to this structure exactly with this side chain. CH2, CH2, then olefin here, as you see, CH2, CH2 and with this olefin.

Well, and what is missing here is this methyl group. Okay, here we have a methyl group. Now those drawings are correct again.

So, retrosynthetic analysis of this part is fairly easy. Just let us hydrolyze this iminium moiety and the enamine hydrolysis will lead us to a carbonyl compound.

So, then we have this bis-caboxaldehyde and a 1,5 relationship, 1,2,3,4,5 relationship between those two carbonyl groups.

And therefore, the next strategic bond could be this one.

Could be this one, this as a nucleophile, this as the electrophile, alpha-beta unsaturated carbonyl. Or, we could assume to choose this one, adding that to this partial structure.

Well, a Michael addition reaction, in essence. Well, I think this one should normally be the method of choice forming also the five-membered ring.

So, we will see two examples that have been realized. First, synthesis developed in Clayton-Havcock's group.

So, we have just a benzyl ether in here, which later on could be transformed to that ester.

You don't want to have that ester functionality throughout that synthesis, having to deal with various nucleophiles, which could affect the ester.

So, just an ether functionality here, you know benzyl protecting group, you can hydrolyze that off, have the alcohol functionality, which you can then oxidize, of course, until the oxidation state of the acid. So, first step, LDA.

LDA, low temperature, will deprotonate here, forming the enolate of that amide.

This five-membered heterocycle was just chosen as the nitrogen part of that amide, will be hydrolyzed off later on. So, don't care about that nitrogen heterocycle. Don't find that as a partial structure in the final molecule.

So, secondly, E is an ester, methyl or ethyl ester. So, cyclopentene carboxylic acid, ester.

What will happen? Michael addition process. This is the nucleophile, this is the electrophile. And if you add those components at low temperature, it will remain as an enolate, here an ester enolate,

until another reaction component, another electrophile is added, this iodide.

So, and, well, actually, you can get this iodide from a natural product, granule.

Just transfer the alcohol functionality into an iodide. So, the result of this transformation is, here the ester functionality, there that sidechain introduced with that iodide,

then this amide and the sidechain with the benzyl ether.

Overall yield of this one pot reaction, remarkable 87%.

Next step, reduction. So, without going into details, one has to be careful in reducing this amide,

because with various reducing agents you get to an amine, in that case. So, must be a combination of hydrolysis and reduction, because next stage within this synthesis is the diol.

I think no problem in achieving this transformation. Now, let's summarize. First, a swirl oxidation, if you have the regions for the swirl oxidation in excess,

you know, triethyl amine and oxalyl chloride, everything has to be thoroughly dried, then the swirl oxidation very often works in the range of 95 to 100% yield,

and swirl oxidation will transfer the primary alcohols into aldehydes. That's exactly what we need. So, first swirl oxidation, then they added methyl amine.

Well, for this we would need just NH3 under slightly acidic conditions,

therefore ammonium acetate would be reagent of choice for that transformation. So, here they added methyl amine. I will later on comment on that.

Third, well, acidifying the acidic acid, some ammonium acetate, they added 85 degrees 10 hours.

So, now again we have to draw the structure which resulted, and we need still the methyl group there.

So, this result was achieved with an astonishing 75% yield,

and well, that missing olefin here is not a mistake.

This was indeed a finding by the serendipity since actually they planned the synthesis with that olefin here, but simply by a mistake they added methyl amine instead of ammonia.

So, and we're surprised in the analysis it worked out just fine. 75% yield, there's no methyl group, and here is no olefin anymore.

So, what happened in that case? Well, on the stage of the reaction where you had that CH group here and the CH3 group here,

and the ammonium cation there, then an hydride is transferred. This is reduced. Here the than have an iminium cation which is simply hydrolyzed later on,

getting rise to the secondary amine here. So, it was a surprise but not a problem because also daphnophilium alkaloids are known with no olefin there.

Okay, it's part of a natural product. Of course, this benzyl ether is not naturally occurring for these structures.

So, 75%. Next, the Hepacock group started a slightly modified synthesis.

They thought, well, what about constructing daphnophilium alkaloids similar to the one we have discussed now, taking into account considerations about the biosynthesis.

Maybe you know squalene. Squalene, a polyolefinic system. Also known as starting material for various biological active compounds.

Well, also the biosynthesis depends on squalene. So, actually they planned to synthesize a functionalized polyolefinic system with no ring at all in there

and in a single preparative step forming such a daphnophilium alkaloid. They did the synthesis and they succeeded in doing so. So, again, starting from guaraniol forming this iodide.

While you do so with iodine, triphenylphosphine combination.

Then, letting this react with an ester enolate will lead to this ester.

Next step, first, protonating with LDA, again forming the lithium ester enolate.

Then, adding an electrophile, this acetal, the protecting group for the aldehyde.

This is the electrophilic center. So, tertiary butyl ester. Here we have the side chain which has been introduced by this step.

So, after hydrolysis, we will have that aldehyde here and, well, the yield of this acetal was 85%

and the hydrolysis gave a realistic 96% of the aldehyde.

The aldehyde was condensed with another ester condensation reaction.

Again, the tertiary butyl ester but functionalized with this silo group there.

In fact, that C-C bond formation process which is from the result similar to a Wittig reaction or Wittig-Horner reaction

the trans-olefin. As you know, in Wittig-type reactions you have often high selectivity for cis-olefins.

So, the result of the Peterson olefination is then this functionalized polyolefin.

Yield 73% with, well, very few steps reduction until the stage of the alcohol,

hydroxy-muffle group here, hydroxy-muffle group there. Then, swirn oxidation. Then, again, we will have this decarbox aldehyde.

So, and now treating with the base, here we have the nucleophilic center.

There, the electrophilic center with an intramolecular Michael addition reaction that five-membered ring is formed.

Well, actually this is another side chain. So, and, well, what they found out was just with ammonia, well, in the presence of ammonium acetate,

so a buffered solution, and secondly some more ingredients and dichloromethane as a solvents and details.

Secondly, acidic more, acidic conditions, acidic acid and heating up 80 degrees.

Then, 17% of the daphnophilim, alcohol, daphnophilim alkaloid was obtained actually this structure with that olefinic, with two olefins in the side chain here.

Okay, so later on, 17% yield for that process I think is quite remarkable. They could optimize it with isolating the five-membered ring.

Then they had 56% yield for this step, having that five-membered ring with the two aldehyde functionalities, they will form some kind of hemiacetal.

However, next step then with the ammonia was then another 90% yield, 90% of 56, well, that is around 50% yield if they do it as a two-step process.

And later on, then they made and they already noticed their initial mistake of applying methyl amine, which then led to this success.

They also tried methyl amine within this reaction and got then a daphnophilim alkaloid, of course missing that olefin at the isopropyl unit, well, with around 65% as a one pot process.

I think this is, well, maybe the most impressive domino process for natural product synthesis I'm aware of. Well, okay, let's leave the daphnophilim alkaloids and let's have a look at the retrosynthesis of pomeliotoxin B from Larry Overman's group.

Actually, I think we have discussed this synthesis before in

the lecture on stoichiometric organometallics within the context of organosilicon chemistry. So, retrosynthesis for this molecule, well, of course, this olefin here represents a strategic bond

and normally you can synthesize this type of olefins, for instance, a Wittig-type reaction.

However, within that Wittig reaction, you don't want to have those alcohol functionalities here. You want to have an aldehyde there, but you can't have a phosphine olet here with the alcohol functionality in the neighboring position.

An elimination reaction, of course, would occur. So, you have to change the oxidation state here within the retrosynthetic analysis and, therefore, you simply can assume that this aldehyde should be an intermediate within your synthesis.

And, on the other hand, it's certainly not a bad idea to target for a phosphine olet of this type having the alcohol functionality here protected.

So, this would give a nice Wittig-type olefin formation.

Generally, such an aldehyde, you usually synthesize by oxidation of an alkylsworn oxidation.

So, you could assume having that alcohol as one intermediate.

And, well, in this case, they decided that for certain steps, the alcohol has to be protected preferentially with as an benzyl ether,

as we have seen before with that daphnophilium alkaloid synthesis.

Well, so, secondary alcohol here, methyl group there.

Now, again, having a relationship between an olefin here and the nitrogen there.

Well, somewhat different to the example we have discussed before because the olefin was one position further away. Nevertheless, we could also apply such an iminium cation combination with an electron-rich olefin in this case.

So, iminium cation here tells us that this, aiming for iminium cation here, tells us that this CC bond is the strategic bond.

So, here the olefin. So, and now, we want that after the CC bond formation here and cation intermediate at this position,

we don't get an olefin again by deprotonation here. That would be similar to the example we have discussed before.

So, we want that selectively the olefin is formed again between these two carbons. And this becomes possible by making use of silicon chemistry.

Having a TMS group here. So, an ipsosubstitution, it is a net ipsosubstitution where with an intermediate the

TMS group stabilizes a cation in better position as we have discussed that in the lecture on stoichiometric organometallics.

Okay. Now, it's very interesting that we have a hydroxyl functionality here and an iminium cation there. And of course, in equilibrium also that oxygen will react as a nucleophile with that iminium cation.

So, let us draw the structure which is in equilibrium in this case.

It's an anoacetal. This one.

So, simply under acidic, slightly acidic conditions in equilibrium this will be protonated also sometimes at oxygen.

Making that a leaving group forming that iminium cation which then can be trapped by the vinylsilane resulting in the formation of that six membered heterocycle here.

Okay. So, next, retrosynthetic steps.

While Overmann decided that such a BOC protected amine would be just fine for his intermediates.

And having an epoxide here, you need just a nucleophile to attack. And what could be the nucleophile? Now, some vinyl metal species.

And this vinyl metal species you can get by a hydrometallation of a corresponding alkyne.

On the other hand, you might guess what will be an appropriate starting material for this kind of substrate.

Well, we would go for a bidirectional analysis and of course starting with proline as a substrate.

So, very nice and rather straight forward retrosynthetic plan that should show that even such other complicated structures can be solved reasonably.

However, it of course takes a lot of time to work out the details.

So, to test details with model compounds, of course. But finding the general outline of such retrosynthetic synthesis is not that difficult.

So, let's try an exercise also with an annihilated and functionalized heterocycle.

We again do not concentrate on considerations concerning stereoselectivity.

Of course, in all those synthesis and retrosynthetic analysis we have discussed today.

Also, considerations of stereoselectivity are important, have been important. But we skipped that. Also, in this case, just concentrate on how to form this type of skeleton.

So, again starting with just looking for functional groups that will help us to find strategic bonds.

So, that olefin here is functionalized with the nitrogen. It's an enamine. So, and if you remember how do we synthesize normally enamines?

Well, starting from an amine and a carbonyl compound and making a condensation reaction. So, and in the sense of retrosynthetic analysis means condensation for the synthesis

and hydrolyzation in the retrosynthetic analysis. This will lead us to an aldehyde here.

And here is the pyrrolidine moiety. So, with a bit of acidic catalyst at a Dean's dark trap with, well, maybe toluene as the solvent, we would get this type of condensation.

Well, with respect of the hetero deals aldo reaction we discussed before, also today, one also could get the idea.

So, this was R1 and let's disconnect these two bonds as R2.

Well, why not? Maybe this works. This amine is a nucleophile, could react delta minus with the electrophilic center here.

Then the stabilized carbon ion could attack here since after nitrogen carbon bond formation we have a cationic moiety there.

So, and then the anionic nucleophile there. Could work as a two-step, mechanistic two-step process. Must not necessarily be a concerted dissolved reaction.

The result would be similar. Could be. So, and if we would then disconnect further on, well, we would disconnect here and maybe disconnect there.

Condensation process, also condensation process.

Let's have a look the building blocks which will result.

So, here we have the building block, there, nitromethane and this

building block would result from an aldol condensation of benzaldehyde and propanol.

Okay, we could disconnect here, having again the same building block.

We can disconnect here, condensation process. The benzaldehyde, the same as here.

And we could then also disconnect here, nucleophilic addition.

Well, some, actually this is a nitroaldol reaction with an amine. So, again, nitromethane as before and this amine as one substrate.

So, compare those two structures. This is just a condensation product of that. Okay, this is now the only, not that simple starting material we have to care of, how to synthesize that.

Well, actually, I looked that up in SciFinder, how they made that. So, a Gabriel synthesis of a primary amine. I hope everyone still knows what that is.

You need a leaving group X here, and you attack that with a phthalimid. Okay, so, how to get this one? It's a bit tricky,

but very nice. Just treat that with hydrochloric or hydrobromic acid,

protonate it here, nucleophilic substitution there, and as an intermediate you will then have this one,

which simply will decarboxylate as a better keto

carboxylic acid. How to get this one? Well,

this is synthesized from this strange, but rather cheap molecule you should already be aware of as

the dimerization product of ketene. Well, it is an industrial product. You can form ketene, for instance, by flash vacuum, pyrolysis of

acetic acid, for instance. So, this plus simply this bromoethanol in the presence of

sodium ethanolate will lead us to this functionalized lactone. Now we know how to get all those

ingredients we are in need of. Well, let's talk about the synthesis that has been worked out.

So, protonate this dihydropyrrole,

that nitroolefin, then the base, this one,

diisopropyl ethyl amine called Hunig's base. Then, in addition,

a chiral catalyst deriving again from proline. Actually, this is an example for

organocatalysis, small chiral organic compounds catalyzing, interesting enantioselective reactions. This transformation

directly gives a 91% yield and 98%

enantiomeric excess of our target molecule with this well-defined absolute configuration. So, I think either way we went through

the retrosynthetic analysis, we would end up always with the same starting materials and in this case it's just a domino process and which parts are connected

first are the question of details within the mechanism. Before we stop today, or finish today, let's make a correction. CH2-benzyl group

is a phenyl-2 group which you can't simply hydrolyze off with hydrogen and palladium. We have to change that to a phenyl group also here and in the final product

we have that already. So, finish for today again we skip next week Tuesday and we'll meet again next week Wednesday. Thank you for listening, see you next week.