10.04.18 Competing Reactions
CC Attribution 3.0 Germany:
You are free to use, adapt and copy, distribute and transmit the work or content in adapted or unchanged form for any legal purpose as long as the work is attributed to the author in the manner specified by the author or licensor.
Dyker, Gerald
Click Chemistry, Convergent Strategy, Retron
 So welcome to the lecture on competing reactions. When we are synthesizing a target compound, you of course are always aiming for a high yield as well as a high purity. Both could be a problem, but most often the high yield is the problem. So for achieving a high yield, we can optimize, we can try to optimize reaction conditions to optimize the working up procedure. Usually you will lose some of your yield during working up, it's clear. Even a simple column chromatography, usually you lose about 10% of your material. And well, a really good idea is to avoid side reactions, the competing reactions. And those competing reactions, having a closer look on what is going on within your reaction flask, this is the main subject of this lecture, this semester lecture. It is important of course in academic labs, but even more important in labs in industry. Just of course to avoid waste and to optimize your industrial process. So this lecture will be accompanied by the book of an industrial medicinal chemist, Sarah Gosser-Durwald, and the title is Side Reactions in Organic Synthesis. And of course I will put in into this lecture even more information from recent original papers connected with this subject. So let us first have a look on lessons we have learned in the preceding semester. That was the lecture in designing organic synthesis, retrosynthesis. Who of you has listened to that kind of lecture last semester or maybe a couple of years ago? At least some of you. So a good plan for the synthesis of a target compound tries to minimize the steps. With every single step you usually lose material. So minimizing the steps because 100% selectivity is very seldom, maybe never reached at all. So actually there are only a very few really reliable good reactions in organic synthesis. Well at least the famous chemist Barry Sharpless was claiming this information. And he wrote a rather important article which is interesting to read and you should look that up. So from Sharpless the article Click Chemistry. Actually the name Click Chemistry was introduced with this article in Angerwante Chemistry. Click Chemistry Diverse Chemical Function from a few good reactions. So what are these few reactions Sharpless would classify as good? So the examples. First of all, cycloaddition reactions. For instance, wall A deals alder. Of course you all know the alder reaction. For instance, cyclopentadiene and such an acrylic ester at lower temperature usually ando selective would give this by cyclic product. Or B13 dipolar cycloaddition. Sometimes called the Hüskern reaction named after Professor Hüskern from Munich University. And one prominent example is of course the first step within the ozone nozzles. Adding ozone to an olefin. But nowadays the Hüskern reaction that has been most made use of is the following. An alkyne plus an azide. Very often with a copper mediated, mediated cycloaddition. Giving rise to such a triazole. And nowadays some people forget about one, three dipolar cycloaddition and Hüskern reaction and call this process the click reaction. This article about click chemistry made this reaction famous again and more and more people noticed that if they have an alkyne with something on there and an azide with something else on there. Very reliable you could put those two components together and having a covalent tether in between. And it almost doesn't matter what you have on one side and on the other side. You can connect that by that click reaction or better called on Hüskern type reaction. A lot of biologists and biochemists make use of that type of process. So cycloadditions. Then addition reactions to C-C multiple bonds, double or triple bonds. For instance, in case of olefins, electron rich olefins, the epoxidation and the dihydroxylation. I hope you remember that the azimetric cis dihydroxylation was also invented by Barry Sharpless who won the Nobel Prize for this highly useful synthetic procedure. Then nucleophilic substitution abbreviated with as an reaction, especially at epoxides is also in Sharpless list of those few good reactions. Carbonile chemistry, especially non-aldo carbonyl chemistry. For instance, condensation reaction of an amine and an alderheid. Urea formation. Lots of examples but an easy one would be this. An isocyanate plus an amine. Also a reaction that proceeds nicely usually with a very, very high yield. Let's add more examples here. Lactonization. While an example, heating up such a hydroxy substituted carboxylic acid, heating it up and a bit of catalytic acid. Acid, well actually you already have the acid, it's a carboxylic acid and well, it will readily undergo a condensation reaction to form a lactone ring. And head arene formation. Well, maybe you should think about various heterene formations you know. Well, let's see one example. This condensation reaction for formation of a furan. So, this is a list that has been originally shown in that seminal publication. Well, we should ask why are those reactions so reliable with high yields? Obviously, they are highly selective because with a high selectivity you can achieve a high yield. And this has of course something to do with the energetics of this processes. So, important is the thermodynamic driving force of usually better, bigger than 20 kcal per mole. And this is reached if there is some extra energy profit within the reaction compared to other normal processes in the range of 50 to 60% yield. So, let us think about what is the extra energy profit in those examples. Have a look at the yield's older reaction. A concerted process where not a single bond is formed but two CC bonds are formed simultaneously. So, one conjugated double bond is broken but at the same time one double bond is formed and two CC bonds. Not only one bond is formed, two. Gains more energy. Here in this one three dipolar cycloaddition process, well, in principle the same is the case, two covalent bonds from this nitrogen to that nitrogen to here and there is formed. But at the same time this profits from aromatization. Okay, this is a het arene. The same as here. Het arene formation profiting from aromatization. Next one. So, epoxidation and dihydroxylation, two co bonds formed simultaneously. Well, actually the dihydroxylation proceeds usually also via a cycloaddition reaction in the first step if you have a closer look to the mechanism. Then carbonyl chemistry, okay. Well, at least this profits also from entropy. Oh no, no. Okay, the entropy profit we have here. Within the lactonization. So, kinetically it's easy because it's an intramolecular process and well from one molecule to molecules are formed and entropy goes up. So, and the higher the temperature and the higher the entropy gain is also profit in the net energy. Okay. So, explanation why that imine formation is so good. Wow. Must admit I don't have the precise answer yet, but at least this is an equilibrium reaction. And within the equilibrium reaction you will be able to trap the water for instance within a dean's dark trap and therefore then driving the reaction almost to completion. Okay. So, the urea formation. This reaction clearly profits energetically by the formation of a donator acceptor system conjugation is also energetically rather low. So, within that seminal paper. Sharpless brought a nice example where through a couple of steps all working very well because they are from his list of few good reactions. You can follow the building up of a high complexity. But step by step with very nice yield always around 90% at least more than 80%. So, let's have a look at this example. From this cyclohexadiene and you should all know this is made by a birch reduction of benzene. A double epoxidation. Forming this double epoxid. Well, here, Sharpless didn't give a yield because you get mixtures of stereo isomers. Okay. So, the overall yield of this this epoxid, regardless of the stereo chemistry is also above 80%. But you have that cis one and you have the trans one. While the trans one is the major isomer, but under certain reaction conditions, you can get this up to close to 50%. Okay, this bis epoxid is then treated with sodium azide ammonium chloride in water under reflux temperature. And then highly stereo selective. This double azide is produced with a 97% yield. Next step, offering acetylene dicarboxylic acid diethyl ester for the Hüskern cycloedition. And two triazoles will be formed. Well, the apple ester I will abbreviate with just E. This step also works with 97% yield. Then, catalytic amounts, in this case of a sterically hindered amine as a base, diisopropyl amine. If I remember it correctly, this is called the Hünig base, not Hüskern. Hünig was a famous professor at Würzburg University. So, 85% of the lektronization product. So, where could the lektronization take place in tremolecular ester formation? This ester is too far away from that one clear, but this one is rather close. So, there is the place where the ester formation will take place. So, this here remains unchanged. But here is the newly formed lektron functionality. Then, last step of the synthesis reported. This is offering this secondary amine as nucleophile and again in 85% yield. So, one equivalent. It will react at the most reactive ester functionality and it is this one, the lektron is opened again. And you notice it already takes some time. It takes some time to draw that final product due to the rather high complexity. So, and again, 85%. And originally, they started with 5.0 gram of that and could isolate at the end 20.2 grams of that one. Certainly, far enough for biological testing of that highly functionalized molecule. So, in the preceding semester, we went through the lecture on designing organic synthesis, retrosynthesis. And let's have a look about the essentials, what we have learned there. Well, within retrosynthetic analysis, it is clear that a convergent strategy is usually much better than a linear strategy. So, convergent strategy usually superior to a linear one. Except you can do every single step above 95% and have a synthesis robberter like Merrifield peptid synthesis or something like that, of course. But generally, convergent strategy, let us assume we have those building blocks for our synthesis from A to H. So, in the first step, we put together A and B having then a covalent bond between C and D, E and F and G and H. Then, sorry, putting this building block together with this one and that with the last one, having then two bigger building blocks and in the last step, while it's clear this is our target molecule. And let us assume that every single step works with unrealistic 80% yield. Then, the convergent synthesis will deliver this product with 51% yield. The alternative pathway is putting together A and B, then next step plus C, next step plus D and so on until we finally reach the target molecule. The overall yield, it's easy to calculate, will be around only 21%. So, this illustrates convergent strategy generally is superior. Let's have a look at a nice and rather easy retrosynthetic analysis. Chikimic acid, it's an important intermediate in the biosynthesis of certain aromatic amino acids actually drawn to the blackboard now is the unnatural enantiomere, so ant chikimic acid. So, making use of those well-known good reactions, then a retrosynthetic analysis, this is the retrosynthesis arrow. So, we are disconnecting strategic bonds, which can be easily formed and disconnecting it within a gedanken experiment will lead to a simpler structure and then we will go on. Well, okay, we have three hydroxide groups here. Getting rid of two of those would be a good idea, simplifying the structure since the cis dihydroxylation is one of those good to excellent reactions and those two are on the same side of the ring. Clearly, should be a good idea to incorporate this retrocystic hydroxylation having in mind that with AD mix, azimetric dihydroxylation, that is such a mixture you can already buy. For instance, if all the other companies, we have this intermediate. Okay, thinking about this molecule, how could we simplify that? Of course, we could eliminate now water, then we have a much simpler structure, water and benzoic acid. But it's not a good idea within that retrosynthesis because we have no idea to do this reaction selectively. So, but imagine that this molecule could derive from this one. And let's draw that. Somewhat nicer than we have this structure and well a similar one I have already drawn to the blackboard there. Okay, retrodial solder. Well, and then we have simple starting material. We should keep in mind that during the real synthesis, we should work with some protective groups. Otherwise, we get into problems. For instance, from such a structure, water will be indeed easily be eliminated and then we are in that thermodynamic sink. Okay, so how was the synthesis on the other way around actually done? So it started indeed from furan and some acrylic acid derivative. Okay, let's put on on OR. It's a bit more complicated. And then with a chiral Lewis acid, then in non-selectively the formation of this ando deals older product was achieved. 67% yield well have to use rather large except excess of the foran with an astonishing 97% non-selective and non-selective excess under of course, nicely optimized reaction conditions with an optimized Lewis acid. So with a sterically hindered base with lithium diesel acid NTHF at low temperature minus 78 degree. What will happen? The most acidic proton, this one will be abstracted and then that low temperature, this oxygen bridge will be opened. And then we are in danger that a subsequent illumination will take place than at least than acidifying the crude product. So the newly formed alcohol, lithium alcohol late has to be trapped with a silo protecting group. And this is done that by adding TBS triflate, so a silo protecting group with an additional added base present to 6-lutidine nitrogen hydrocycle at low temperature and then that silo oxy group sits at the only remaining or yeah the only remaining carousel center. In between we had three carousel centers here we have only one. It doesn't matter because then the cis dihydroxylation will take place on the opposite side of the ring. And well since they had already that stereochemical control element, it was not necessary to take the ad-mix as a metric dehydroxylation mixture containing already the carousel information. It's not necessary. We have already carousel information in our molecule. So, osmium tetroxide, NMO as oxidizing agent, 0 degrees and then the S2 could be isolated. Hydrolyzation of the ester and deprotecting here then produced the, finally the end-chicchemic acid. One other very important term we have learned in the preceding semester was the term retron. So, you remember of those who attended that lecture, remember, certainly remember that such a cyclohexene moiety, if you notice the cyclohexene moiety within your target molecule, you should think about applying a desolder reaction for the synthesis of this molecule. Because having this as a partial structure strongly suggests that you should make this retrosynthetic transformation around actually using a desolder reaction. And even if you don't have a cyclohexene, but just a cyclohexene moiety, you could think about, within your retrosynthetic analysis, adding the olefin functionality that you then generate the retron. And this retron of a desolder reaction. You can't do so because you know a simple and reliable method to get rid of a double bond, simply catalytic hydrogenation. We have in the preceding semester talked about the retron of an oxy-cope rearrangement. So, what is the oxy-cope rearrangement? It is a cope rearrangement with an alcohol-late at this position, usually a potassium alcohol-late, and already above 25 degrees, but for sure above 55 degrees. But cope rearrangement would take place. Please remember that normal cope rearrangements take place above 160 degrees or so. A cytifying will then give rise to this final product. So, what about the retron of the oxy-cope rearrangement? If you notice in your target structure such a moiety, a delta-abselon unsaturated carbonyl, then you should take into consideration that this could be formed by an oxy-cope rearrangement starting from that. This might be much simpler to be synthesized than that. In this case it is rather complicated to have the overview, and therefore it makes sense to put on numbers to the position. This will help to solve related problems. Also in this case one could think about having for instance something like that, a key turned with a chain like that. If you don't have a double bond there within your retrosynthetic analysis, you could also think, well let's introduce this one, then we have a possibility to do that. Because you will find a way after maybe protecting that carbonyl group to hydrogenate that off, or making use within your synthesis of that. So, here we have a target molecule. So, and within the retrosynthetic analysis of this target molecule, some chemists who knew about the retron of the oxy-cope rearrangement got the idea that for doing such an, or applying an oxy-cope rearrangement, they have to introduce both an additional carbonyl functionality and an olefin. To make it easier to find the solution, I will show at which position they decided to do that. Here the olefin, there the carbonyl group. So actually this was done in the group of Leo Paquet, the most famous natural product synthesis group, made extensive use of oxy-cope rearrangements. And now as an exercise, please try to figure out how could you make use of an oxy-cope rearrangement for synthesizing this tricyclic system. I know it's easier for those who already went through the lecture on retrosynthetic analysis, but nevertheless it's a nice exercise, but fits here. So it makes sense to have those numbers here, because to solve the problem, it really helps drawing by numbers or in German Malen nach Zahlen. So let's try it here. Where do we have the position 1? So please keep in mind, we noticed we have the carbonyl group here and the olefin there. And while position 1 is the carbonyl carbon, we have now the choice. 1, 2, 3, 4, 5, 6 is one possibility, 1, 2, 3, 4, 5, 6 is the second possibility. For retrosynthetic analysis, both are right, but one will lead us to a much simpler structure and the other way, this one, to an equally complicated structure. So let's follow therefore this route. And some of you have followed that route and could then compare, and the others who did this route could also make at home this one and compare it. So 1, 2, 3, 4, 5, 6. So here we have the retron of the oxycopere arrangement. So let's draw that structure once again and just following the numbers by adding something or wiping something out. So once again 1, 2, 3, 4, 5, 6. And now we have always to compare. Here we have between position 1 and 6 no bond, but here we have a bond. So in retrosynthetic analysis we have to form a bond between 1 and 6 and form an OH group here. But let us start with the OH group here. So from carbonyl to OH. We had between 5 and 6 a bond. This bond will migrate to the position 4 and 5, between 4 and 5. Here we will have a double bond. Don't worry about that 5 bonded carbon. We have to eliminate another bond between 3 and 4. This bond is wiped out. We have to form a single bond between the position 1 and 6. And we have to form a double bond between 2 and 3. And yes, here is a problem. That muffler group. Okay. Let's wipe out the muffler group. Just wipe it out. You have to look, can we introduce that selectively? Okay, so let us translate that. If you would put that up into a force field computer model and push the optimize button. What will happen? Well, a biscyclic system would be drawn. So where do we have the muffler groups? Here and there. Okay. Right. So and with maybe potassium tibutoxide and heating it up, then this should be transformed into. This product and from here to there, we have to solve the problem how to add the muffler group. Okay, we have a selectivity problem. But hopefully we find an appropriate base to selectively deprotonate here and then just a mathelite there. Okay. So now. Next step should be easy. Disconnect here. Plus some organolithium compound like that. Okay. Thinking about how to synthesize a molecule like that. Actually in the preceding semester, we had a similar problem we discussed. One could think about making a deal's order reaction. We have a cyclohexene moiety. But within the retrosynthetic analysis, this would mean that we would make use of this diene plus that ketene, but unfortunately, ketenes just don't undergo four plus two cycloaddition reactions, but two plus two cycloaddition reactions. So we can solve the problem if one changes to that one. Okay. Making use of the reaction and then have an alpha-chloro-cyano compound and exchanging the chlorine against an OH group. And then you have a cyanohydrin and well, if you hydrolyze that, you have a ketone. Rather complicated and well, in addition, one should think about, yeah, okay, it should have the right regioselectivity for the cycloaddition. However, the packet group had the idea for another protocol or another route. So, they had the idea that, well, let's change oxidation state at various positions. Let's go back to the alcohol functionality here. Having an alcohol here, whatever, stereochemistry. And such a bond could have been formed. This is the idea by an aldol addition. But of course, we then need the carbonyl group. Well, okay, change oxidation state here. Like that. Okay. Then, retroaldol leads to this alderhut. Well, if you have that alcohol functionality here, first protected, reduce that one, eliminate here. Having there the olefin, deprotect, and then oxidize there. Okay, this would be synthesis. How could you get something like that, an alderhut here? Well, this is a classic idea having an allylic functionality at this position. And from here to there, it's an ozoneolosis. How could you introduce that alder group there? Well, should be something related to some kind of Michael addition process. And the best way here is the Zakkuri reaction. Maybe you have heard about that. Zakkuri reaction, it is an allylsilane. And then as a catalyst, titanium tetrachloride or another Lewis acid, which activates, which coordinates to the carbonyl group, activates the electrophilicity of this position. And then this is the nucleophilic center, and the reaction undergoes with an elimination here. So, last steps. How could we synthesize that? Well, an alpha-better unsaturated ketone one should always think, of course, about an aldol condensation. So, and finishing the retrosynthetic analysis, well, disconnecting here for a Michael-type process, then you have that alder height and here you would have an alpha-better unsaturated ketone. And you would notice that with an alder height, it wouldn't work very well because the alder height has the highest alpha acidity and at the same time, the highest carbonyl activity. So, you wouldn't choose the alder height as a real nucleophile. So, what you need is a symphonic like that, a donated reactivity here, and the synthesis equivalent, and most easily could be an enamine like that. So, I hope these examples from retrosynthetic analysis helped those to remember what we have learned in the preceding semester and will motivate others sooner or later to take a course in retrosynthetic analysis. Thanks for listening. See you tomorrow in the same lecture hall, but already at 3 p.m.
 So welcome to the lecture on competing reactions.  When we are synthesizing a target compound,  you of course are always aiming for a high yield as well as a high purity.  Both could be a problem, but most often the high yield is the problem.  So for achieving a high yield, we can optimize,  we can try to optimize reaction conditions to optimize the working up procedure.  Usually you will lose some of your yield during working up, it's clear.  Even a simple column chromatography, usually you lose about 10% of your material.  And well, a really good idea is to avoid side reactions, the competing reactions.  And those competing reactions, having a closer look on what is going on within your reaction flask,  this is the main subject of this lecture, this semester lecture.  It is important of course in academic labs, but even more important in labs in industry.  Just of course to avoid waste and to optimize your industrial process.  So this lecture will be accompanied by the book of an industrial medicinal chemist,  Sarah Gosser-Durwald, and the title is Side Reactions in Organic Synthesis.  And of course I will put in into this lecture even more information from recent original papers connected with this subject.  So let us first have a look on lessons we have learned in the preceding semester.  That was the lecture in designing organic synthesis, retrosynthesis.  Who of you has listened to that kind of lecture last semester or maybe a couple of years ago?  At least some of you.  So a good plan for the synthesis of a target compound tries to minimize the steps.  With every single step you usually lose material.  So minimizing the steps because 100% selectivity is very seldom, maybe never reached at all.  So actually there are only a very few really reliable good reactions in organic synthesis.  Well at least the famous chemist Barry Sharpless was claiming this information.  And he wrote a rather important article which is interesting to read and you should look that up.  So from Sharpless the article Click Chemistry.  Actually the name Click Chemistry was introduced with this article in Angerwante Chemistry.  Click Chemistry Diverse Chemical Function  from a few good reactions.  So what are these few reactions Sharpless would classify as good?  So the examples.  First of all, cycloaddition reactions.  For instance, wall A deals alder.  Of course you all know the alder reaction.  For instance, cyclopentadiene and such an acrylic ester at lower temperature usually ando selective  would give this by cyclic product.  Or B13 dipolar cycloaddition.  Sometimes called the Hüskern reaction named after Professor Hüskern from Munich University.  And one prominent example is of course the first step within the ozone nozzles.  Adding ozone to an olefin.  But nowadays the Hüskern reaction that has been most made use of is the following.  An alkyne plus an azide.  Very often with a copper mediated, mediated cycloaddition.  Giving rise to such a triazole.  And nowadays some people forget about one, three dipolar cycloaddition and Hüskern reaction and call this process the click reaction.  This article about click chemistry made this reaction famous again and more and more people noticed that if they have an alkyne with something on there  and an azide with something else on there.  Very reliable you could put those two components together and having a covalent tether in between.  And it almost doesn't matter what you have on one side and on the other side. You can connect that by that click reaction or better called on Hüskern type reaction.  A lot of biologists and biochemists make use of that type of process.  So cycloadditions.  Then addition reactions to C-C multiple bonds, double or triple bonds.  For instance, in case of olefins, electron rich olefins, the epoxidation and the dihydroxylation.  I hope you remember that the azimetric cis dihydroxylation was also invented by Barry Sharpless who won the Nobel Prize for this highly useful synthetic procedure.  Then nucleophilic substitution abbreviated with as an reaction, especially at epoxides is also in Sharpless list of those few good reactions.  Carbonile chemistry, especially non-aldo carbonyl chemistry.  For instance, condensation reaction of an amine and an alderheid.  Urea formation.  Lots of examples but an easy one would be this. An isocyanate plus an amine.  Also a reaction that proceeds nicely usually with a very, very high yield.  Let's add more examples here. Lactonization.  While an example, heating up such a hydroxy substituted carboxylic acid, heating it up and a bit of catalytic acid.  Acid, well actually you already have the acid, it's a carboxylic acid and well, it will readily undergo a condensation reaction to form a lactone ring.  And head arene formation.  Well, maybe you should think about various heterene formations you know. Well, let's see one example.  This condensation reaction for formation of a furan.  So, this is a list that has been originally shown in that seminal publication. Well, we should ask why are those reactions so reliable with high yields?  Obviously, they are highly selective because with a high selectivity you can achieve a high yield.  And this has of course something to do with the energetics of this processes. So, important is the thermodynamic driving force  of usually better, bigger than 20 kcal per mole.  And this is reached if there is some extra energy profit within the reaction compared to other normal processes in the range of 50 to 60% yield.  So, let us think about what is the extra energy profit in those examples.  Have a look at the yield's older reaction. A concerted process where not a single bond is formed but two CC bonds are formed simultaneously.  So, one conjugated double bond is broken but at the same time one double bond is formed and two CC bonds. Not only one bond is formed, two.  Gains more energy.  Here in this one three dipolar cycloaddition process, well, in principle the same is the case, two covalent bonds from this nitrogen to that nitrogen to here and there is formed.  But at the same time this profits from aromatization. Okay, this is a het arene. The same as here. Het arene formation profiting from aromatization.  Next one. So, epoxidation and dihydroxylation, two co bonds formed simultaneously.  Well, actually the dihydroxylation proceeds usually also via a cycloaddition reaction in the first step if you have a closer look to the mechanism.  Then carbonyl chemistry, okay. Well, at least this profits also from entropy.  Oh no, no. Okay, the entropy profit we have here.  Within the lactonization. So, kinetically it's easy because it's an intramolecular process and well from one molecule to molecules are formed and entropy goes up.  So, and the higher the temperature and the higher the entropy gain is also profit in the net energy. Okay.  So, explanation why that imine formation is so good.  Wow. Must admit I don't have the precise answer yet, but at least this is an equilibrium reaction.  And within the equilibrium reaction you will be able to trap the water for instance within a dean's dark trap and therefore then driving the reaction almost to completion.  Okay.  So, the urea formation. This reaction clearly profits energetically by the formation of a donator acceptor system conjugation is also energetically rather low.  So, within that seminal paper.  Sharpless brought a nice example where through a couple of steps all working very well because they are from his list of few good reactions.  You can follow the building up of a high complexity.  But step by step with very nice yield always around 90% at least more than 80%.  So, let's have a look at this example.  From this cyclohexadiene and you should all know this is made by a birch reduction of benzene.  A double epoxidation.  Forming this double epoxid. Well, here, Sharpless didn't give a yield because you get mixtures of stereo isomers.  Okay. So, the overall yield of this this epoxid, regardless of the stereo chemistry is also above 80%. But you have that cis one and you have the trans one.  While the trans one is the major isomer, but under certain reaction conditions, you can get this up to close to 50%.  Okay, this bis epoxid is then treated with sodium azide ammonium chloride in water under reflux temperature.  And then highly stereo selective.  This double azide is produced with a 97% yield.  Next step, offering acetylene dicarboxylic acid diethyl ester for the Hüskern cycloedition.  And two triazoles will be formed.  Well, the apple ester I will abbreviate with just E.  This step also works with 97% yield.  Then, catalytic amounts, in this case of a sterically hindered amine as a base,  diisopropyl amine.  If I remember it correctly, this is called the Hünig base, not Hüskern.  Hünig was a famous professor at Würzburg University.  So, 85% of the lektronization product.  So, where could the lektronization take place in tremolecular ester formation?  This ester is too far away from that one clear, but this one is rather close.  So, there is the place where the ester formation will take place.  So, this here remains unchanged.  But here is the newly formed lektron functionality.  Then, last step of the synthesis reported.  This is offering this secondary amine as nucleophile and again in 85% yield.  So, one equivalent.  It will react at the most reactive ester functionality and it is this one, the lektron is opened again.  And you notice it already takes some time.  It takes some time to draw that final product due to the rather high complexity.  So, and again, 85%.  And originally, they started with 5.0 gram of that and could isolate at the end 20.2 grams of that one.  Certainly, far enough for biological testing of that highly functionalized molecule.  So, in the preceding semester, we went through the lecture on designing organic synthesis, retrosynthesis.  And let's have a look about the essentials, what we have learned there.  Well, within retrosynthetic analysis, it is clear that a convergent strategy is usually much better than a linear strategy.  So, convergent strategy usually superior to a linear one.  Except you can do every single step above 95% and have a synthesis robberter like Merrifield peptid synthesis or something like that, of course.  But generally, convergent strategy, let us assume we have those building blocks for our synthesis from A to H.  So, in the first step, we put together A and B having then a covalent bond between C and D, E and F and G and H.  Then, sorry, putting this building block together with this one and that with the last one,  having then two bigger building blocks and in the last step, while it's clear this is our target molecule.  And let us assume that every single step works with unrealistic 80% yield.  Then, the convergent synthesis will deliver this product with 51% yield.  The alternative pathway is putting together A and B, then next step plus C,  next step plus D and so on until we finally reach the target molecule.  The overall yield, it's easy to calculate, will be around only 21%.  So, this illustrates convergent strategy generally is superior.  Let's have a look at a nice and rather easy retrosynthetic analysis.  Chikimic acid, it's an important intermediate in the biosynthesis of certain aromatic amino acids  actually drawn to the blackboard now is the unnatural enantiomere, so ant chikimic acid.  So, making use of those well-known good reactions,  then a retrosynthetic analysis, this is the retrosynthesis arrow.  So, we are disconnecting strategic bonds, which can be easily formed  and disconnecting it within a gedanken experiment will lead to a simpler structure  and then we will go on.  Well, okay, we have three hydroxide groups here.  Getting rid of two of those would be a good idea, simplifying the structure  since the cis dihydroxylation is one of those good to excellent reactions  and those two are on the same side of the ring.  Clearly, should be a good idea to incorporate this retrocystic hydroxylation  having in mind that with AD mix, azimetric dihydroxylation,  that is such a mixture you can already buy.  For instance, if all the other companies, we have this intermediate.  Okay, thinking about this molecule, how could we simplify that?  Of course, we could eliminate now water, then we have a much simpler structure,  water and benzoic acid.  But it's not a good idea within that retrosynthesis because we have no idea  to do this reaction selectively.  So, but imagine that this molecule could derive from this one.  And let's draw that.  Somewhat nicer than we have this structure and well a similar one I have already drawn to the blackboard there.  Okay, retrodial solder.  Well, and then we have simple starting material.  We should keep in mind that during the real synthesis, we should work with some protective groups.  Otherwise, we get into problems.  For instance, from such a structure, water will be indeed easily be eliminated  and then we are in that thermodynamic sink.  Okay, so how was the synthesis on the other way around actually done?  So it started indeed from furan and some acrylic acid derivative.  Okay, let's put on on OR.  It's a bit more complicated.  And then with a chiral Lewis acid, then  in non-selectively the formation of this ando deals older product was achieved.  67% yield well have to use rather large except excess of the foran with an astonishing 97%  non-selective and non-selective excess under of course, nicely optimized reaction conditions with an optimized Lewis acid.  So with a sterically hindered base with lithium diesel acid  NTHF at low temperature minus 78 degree.  What will happen?  The most acidic proton, this one will be abstracted and then that low temperature, this oxygen bridge will be opened.  And then we are in danger that a subsequent illumination will take place than at least  than acidifying the crude product.  So the newly formed alcohol, lithium alcohol late has to be trapped with a silo protecting group.  And this is done that by adding TBS triflate, so a silo protecting group with an additional added base present  to 6-lutidine nitrogen hydrocycle at low temperature and then  that silo oxy group sits at the only remaining or yeah the only remaining carousel center.  In between we had three carousel centers here we have only one.  It doesn't matter because then the cis dihydroxylation will take place on the opposite side of the ring.  And well since they had already that stereochemical control element, it was not necessary to take  the ad-mix as a metric dehydroxylation mixture containing already the carousel information.  It's not necessary.  We have already carousel information in our molecule.  So, osmium tetroxide, NMO as oxidizing agent, 0 degrees and then the S2 could be isolated.  Hydrolyzation of the ester and deprotecting here then produced the, finally the end-chicchemic acid.  One other very important term we have learned in the preceding semester was the term retron.  So, you remember of those who attended that lecture, remember, certainly remember that such a cyclohexene moiety,  if you notice the cyclohexene moiety within your target molecule, you should think about applying a desolder reaction for the synthesis of this molecule.  Because having this as a partial structure strongly suggests that you should make this retrosynthetic transformation  around actually using a desolder reaction. And even if you don't have a cyclohexene, but just a cyclohexene moiety,  you could think about, within your retrosynthetic analysis, adding the olefin functionality that you then generate the retron.  And this retron of a desolder reaction. You can't do so because you know a simple and reliable method to get rid of a double bond,  simply catalytic hydrogenation.  We have in the preceding semester talked about the retron of an oxy-cope rearrangement.  So, what is the oxy-cope rearrangement? It is a cope rearrangement with an alcohol-late at this position, usually a potassium alcohol-late,  and already above 25 degrees, but for sure above 55 degrees.  But cope rearrangement would take place. Please remember that normal cope rearrangements take place above 160 degrees or so.  A cytifying will then give rise to this final product.  So, what about the retron of the oxy-cope rearrangement? If you notice in your target structure such a moiety,  a delta-abselon unsaturated carbonyl, then you should take into consideration that this could be formed by an oxy-cope rearrangement starting from that.  This might be much simpler to be synthesized than that.  In this case it is rather complicated to have the overview, and therefore it makes sense to put on numbers to the position.  This will help to solve related problems.  Also in this case one could think about having for instance something like that, a key turned with a chain like that.  If you don't have a double bond there within your retrosynthetic analysis, you could also think,  well let's introduce this one, then we have a possibility to do that.  Because you will find a way after maybe protecting that carbonyl group to hydrogenate that off, or making use within your synthesis of that.  So, here we have a target molecule.  So, and within the retrosynthetic analysis of this target molecule,  some chemists who knew about the retron of the oxy-cope rearrangement got the idea that for doing such an, or applying an oxy-cope rearrangement,  they have to introduce both an additional carbonyl functionality and an olefin.  To make it easier to find the solution, I will show at which position they decided to do that.  Here the olefin, there the carbonyl group.  So actually this was done in the group of Leo Paquet, the most famous natural product synthesis group, made extensive use of oxy-cope rearrangements.  And now as an exercise, please try to figure out how could you make use of an oxy-cope rearrangement for synthesizing this tricyclic system.  I know it's easier for those who already went through the lecture on retrosynthetic analysis, but nevertheless it's a nice exercise, but fits here.  So it makes sense to have those numbers here, because to solve the problem, it really helps drawing by numbers or in German Malen nach Zahlen.  So let's try it here.  Where do we have the position 1?  So please keep in mind, we noticed we have the carbonyl group here and the olefin there.  And while position 1 is the carbonyl carbon, we have now the choice.  1, 2, 3, 4, 5, 6 is one possibility, 1, 2, 3, 4, 5, 6 is the second possibility.  For retrosynthetic analysis, both are right, but one will lead us to a much simpler structure and the other way, this one, to an equally complicated structure.  So let's follow therefore this route.  And some of you have followed that route and could then compare, and the others who did this route could also make at home this one and compare it.  So 1, 2, 3, 4, 5, 6.  So here we have the retron of the oxycopere arrangement.  So let's draw that structure once again and just following the numbers by adding something or wiping something out.  So once again 1, 2, 3, 4, 5, 6.  And now we have always to compare.  Here we have between position 1 and 6 no bond, but here we have a bond.  So in retrosynthetic analysis we have to form a bond between 1 and 6 and form an OH group here.  But let us start with the OH group here.  So from carbonyl to OH.  We had between 5 and 6 a bond. This bond will migrate to the position 4 and 5, between 4 and 5.  Here we will have a double bond.  Don't worry about that 5 bonded carbon.  We have to eliminate another bond between 3 and 4.  This bond is wiped out.  We have to form a single bond between the position 1 and 6.  And we have to form a double bond between 2 and 3.  And yes, here is a problem.  That muffler group.  Okay.  Let's wipe out the muffler group.  Just wipe it out.  You have to look, can we introduce that selectively?  Okay, so let us translate that. If you would put that up into a force field computer model and push the optimize button.  What will happen?  Well, a biscyclic system would be drawn.  So where do we have the muffler groups?  Here and there.  Okay.  Right.  So and with maybe potassium tibutoxide and heating it up, then this should be transformed into.  This product and from here to there, we have to solve the problem how to add the muffler group.  Okay, we have a selectivity problem.  But hopefully we find an appropriate base to selectively deprotonate here and then just a mathelite there.  Okay.  So now.  Next step should be easy.  Disconnect here.  Plus some organolithium compound like that.  Okay.  Thinking about how to synthesize a molecule like that.  Actually in the preceding semester, we had a similar problem we discussed.  One could think about making a deal's order reaction.  We have a cyclohexene moiety.  But within the retrosynthetic analysis, this would mean that  we would make use of this diene plus that ketene, but unfortunately, ketenes just don't undergo  four plus two cycloaddition reactions, but two plus two cycloaddition reactions.  So we can solve the problem if one changes to that one.  Okay.  Making use of the reaction and then have an alpha-chloro-cyano compound and exchanging the chlorine against an OH group.  And then you have a cyanohydrin and well, if you hydrolyze that, you have a ketone.  Rather complicated and well, in addition, one should think about, yeah, okay, it should have the right  regioselectivity for the cycloaddition.  However, the packet group had the idea for another protocol or another route.  So, they had the idea that, well,  let's change oxidation state at various positions.  Let's go back to the alcohol functionality here.  Having an alcohol here, whatever, stereochemistry.  And such a bond could have been formed.  This is the idea by an aldol addition.  But of course, we then need the carbonyl group.  Well, okay, change oxidation state here.  Like that.  Okay.  Then, retroaldol  leads to this alderhut.  Well, if you have that alcohol functionality here,  first protected, reduce that one, eliminate here.  Having there the olefin, deprotect, and then oxidize there.  Okay, this would be synthesis.  How could you get something like that, an alderhut here?  Well, this is a classic idea having  an allylic functionality at this position.  And from here to there, it's an ozoneolosis.  How could you introduce that alder group there?  Well, should be something related to some kind of Michael addition process.  And the best way here is the Zakkuri reaction.  Maybe you have heard about that.  Zakkuri reaction, it is an allylsilane.  And then as a catalyst, titanium tetrachloride or another Lewis acid,  which activates, which coordinates to the carbonyl group,  activates the electrophilicity of this position.  And then this is the nucleophilic center,  and the reaction undergoes with an elimination here.  So, last steps.  How could we synthesize that?  Well, an alpha-better unsaturated ketone  one should always think, of course, about an aldol condensation.  So, and finishing the retrosynthetic analysis, well, disconnecting here  for a Michael-type process, then you have that alder height  and here you would have an alpha-better unsaturated ketone.  And you would notice that with an alder height, it wouldn't work very well  because the alder height has the highest alpha acidity  and at the same time, the highest carbonyl activity.  So, you wouldn't choose the alder height as a real nucleophile.  So, what you need is a symphonic like that,  a donated reactivity here, and the synthesis equivalent,  and most easily could be an enamine like that.  So, I hope these examples from retrosynthetic analysis  helped those to remember what we have learned in the preceding semester  and will motivate others sooner or later to take a course in retrosynthetic analysis.  Thanks for listening. See you tomorrow in the same lecture hall, but already at 3 p.m.
Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_0_3071.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_3072_5675.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_5676_5999.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_6000_6755.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_6756_7643.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_7644_8795.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_8796_15263.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_15264_15599.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_15600_18563.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_18564_19835.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_19836_20003.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_20004_20135.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_20136_29747.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_29748_29795.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_29796_29843.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_29844_30671.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_30672_33131.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_33132_33299.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_33300_33443.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_33444_36995.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_36996_39287.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_39288_39407.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_39408_40331.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_40332_41783.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_41784_45839.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_45840_46967.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_46968_47675.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_47676_47747.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_47748_48503.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_48504_49199.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_49200_49547.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_49548_51035.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_51036_53363.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_53364_53699.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_53700_54503.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_54504_57083.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_57084_60635.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_60636_62807.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_62808_64931.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_64932_66203.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_66204_66335.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_66336_67151.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_67152_68099.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_68100_68471.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_68472_69203.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_69204_69539.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_69540_72035.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_72036_73715.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_73716_75611.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_75612_76511.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_76512_76955.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_76956_77063.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_77064_77243.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_77244_77795.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_77796_78791.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_78792_79367.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_79368_81347.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_81348_82403.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_82404_83531.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_83532_85415.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_85416_86327.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_86328_86807.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_86808_86867.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_86868_87071.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_87072_89795.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_89796_90467.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_90468_95735.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_95736_100115.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_100116_100643.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_100644_101255.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_101256_101363.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_101364_101627.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_101628_105479.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_105480_105923.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_105924_107531.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_107532_107579.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_107580_108167.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_108168_110303.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_110304_110555.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_110556_110939.png Lecture_Competing_Reactions_1_Prof_G_Dyker_100418_uhNQk3ejmnk_110940_112006.png