Lecture Designing Organic Syntheses 2 - 08.10.14
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Dyker, Gerald
Retron of oxy-Cope rearrangement
 Welcome to the second part of the lecture, Designing Organic Synthesis. Let us start today with some general schemes of Zygmatropic rearrangements. You certainly remember the cope rearrangement and its general scheme. So we have a hexadeine system, and it rearranges under elevated temperatures to give again hexadeine. Well, essentially this is the same. But if you have some additional substituents here, then for instance here, a muffle group, and there another muffle group, then you would have a muffle group here, a muffle group there. These olefins are then twice substituted. This is generally our favorite thermodynamically. And then you have a reaction. It is an equilibrium, but in that case the equilibrium would be on that side. Here it is a reaction, it goes there and backwards, and you would notice it of course. Nevertheless, this is the general scheme of the cope rearrangement. Let us exchange one methylene group by an oxygen, then we are dealing with an oxacope rearrangement, better known as Claisen rearrangement. So a vinyl allyl ether then will give this structure. So that is, well, an unsaturated carbonyl compound, alpha beta, so gamma delta unsaturated carbonyl compound. While this is generally not an equilibrium reaction, it is on that side, it will stay there. Reaction temperatures well above 180 degrees in the typical cope rearrangement, somewhat lower, maybe 20 degrees or 30 degrees lower if you have an oxacope rearrangement, and there is another one. We have a hydroxy group here. We would have this enol as an intermediate, and this will of course, tautomerize. Finally, again leading to an unsaturated carbonyl compound, but this time it is delta epsilon unsaturated. This reaction works best if it is really an electron rich system as you should already know, if you deprotonate here, and best is, for instance, with potassium hydride. So kH in THF secondly reflux. So in that case you have the potassium alcoholate, which is reacting, and this indeed works very well in THF reflux temperature, all about 60-65 degrees or so. So at much lower temperature than the other examples. So this type of reaction is called the oxicope rearrangement. So why are we discussing these general schemes? Yesterday we discussed the general scheme of the Diels Alder reaction, and well, Diels Alder reaction, general scheme is that, and the cyclohexene we called the retron of the Diels Alder reaction. Therefore now these are retrons, retrons of the cope rearrangement, retron of the clasen rearrangement, and the retron of the oxicope rearrangement. So let us do an exercise. This is our target molecule, and as an exercise, you should find out which sigmatropic rearrangement we could use in our retrosynthetic plan for, well, later on, synthesizing this molecule, this as a target. So it's clear what you have to do is find out which retron can we find in this target molecule, this target structure. And well, this is the first task. And after you have found it out, we will discuss some hints how to solve the problem then. OK, please start with that. Find out what kind of retron do you find here. So your task was looking for retrons within this structure, so structural subunits which imply that it could be a good idea to make use of a certain rather powerful transformation, like with the other reaction or a cope rearrangement. Well, OK, which retrons do we find? Well, if you have a close look, then we have, over all, three of them. Therefore, I write it down three times. But we can discuss then each. So first of all, at least one of you found, OK, here, we have a cyclohexene moiety. So we could think about making use of a Duzalda reaction, retron of a Duzalda reaction. So and that means in our retro synthetic considerations, we should think about this substrate. And then while heating it up and hopefully it works. Well, at least it has a chance since this is an intramolecular. Unfortunately, while from electronics, it's certainly not the optimum electron rich and electron rich. Next, maybe a small problem is that you have that olefin out of conjugation to the carbonyl. Or you can synthesize that, of course. So but there are some problems involved. It is an idea. One could think about that further on. Nevertheless, let's have a look at the other possibilities. And some of you noticed that you can find delta epsilon unsaturated carbonyls. So delta epsilon, the saturated carbonyls, we can all count the positions. Let's take this as this carbon as position one, two, three, four, five, six. And here, let's also put some numbers at these positions. This will help when we go on with our considerations. So carbonyl position one, two, three, four, five, six. Indeed, delta epsilon unsaturated carbonyl group. On the other hand, let's try this way. One, two, three, four, five, six. Once again, another delta epsilon unsaturated carbonyl. So we have the retron of an oxycopere arrangement, either this way or that way. And for both examples, we should figure out what is the substrate in each case leading by oxycopere arrangement to this structure or here the same structure. Well, that means let us go back to this general scheme. We have to transform our drawing from this to that. And this is best done following these numbers. In German language, we would call it malen nach Zahlen. OK, just please try it. So again, the explanation. You figured out that in your target structure, there is this type of moiety, a delta epsilon unsaturated carbonyl compound. That means you have found the retron of an oxycopere arrangement. So then you have to transform that drawing from this unsaturated carbonyl compound essentially with additional rings or substituents attached to that type of structures, also with additional rings and substituents attached. For instance, you have to transform the drawing from something with a carbonyl group to an alcohol functionality. And then, for instance, you notice here in your target structure, you have a single bond between the positions 2 and 3. You have to transform the drawing into something with a double bond between 2 and 3. You start with a structure with a single bond between 3 and 4. And you have to transform the drawing in something with no bond between the positions 3 and 4. And we have to do that for both cases. So best is to draw that once again, but not starting with a carbonyl group. Let's put on an OH group here. We have to do the same procedure later on a second time. So we have an OH group in position 1. Between 1 and 2, there is a single bond and there should be a single bond remaining. Between 2 and 3, there has to be, finally, after that retrosynthetic step, a double bond. Here, between positions 3 and 4, we have to change from a single bond to no bond at all. Between the positions 4 and 5, we have a double bond and the positions 6 and 1 are connected. Let's do the same here for that case. OH group, between 1 and 2, the single bond remains. Between 2 and 3, we have a double bond. Between 3 and 4, the bond has vanished. Between 4 and 5, there should be a double bond. And we have again to connect positions 6 and 1 like that. So these somewhat strange drawings we have to translate. Now imagine you would put in such a structure in a molecular modeling program and press for optimization. Then you will notice that those positions get more close, of course. So essentially, in this case, we have two annihilated six-membered rings with an OH group at the bridgehead position, tertiary alcohol. And now we just claim that trans-olefins doesn't fit into the drawings. Let's just draw the cis-olefins. And we have then olefinic functionality here and there. And let's put on the hydrogen here. Or like that. It's better like that. This type of structure looks strange on the first glimpse. But wow, this is a bicyclic structure. That's it. Let's translate that. OK. The structure is the same like that. So let's think about this here one structure that is really a simplification. It depends on how easy we can transform those structures in even more simple structures. And on the other hand, does that transformation real reaction really work? So here is some problems with, well, but electronically it doesn't fit together very well. But it might work. Here, well, for the oxycoprearrangement, this and that position has to come together into close proximity. The good question is that possible at all in this structure. Well, I've put that into a molecular modeling program, and it didn't look very well in that case. Here, this position and that should get into close proximity. And it looked better. At least it looked better. So let us go on with this structure and figure out how we could simplify that within our retrosynthetic analysis. Well, this looks quite straightforward, disconnecting here, because as you know, a tertiary alcohol, we can easily synthesize by letting an organometallic reagent reacting with a corresponding ketone. So we know that, for instance, with vinyl magnesium bromide or vinyl lithium, we could get this transformation done. How about the regioselectivity of the 1,2 addition process? Does that nucleophilic vinyl reagent really approach that substrate from this side preferentially compared to the other side? The answer is, of course, yes. We have these hydrogens somewhat blocking the trajectory for the nucleophilic attack. I hope you remember the trajectory of a nortgometallic nucleophile is not perpendicular, but a little bit like that. Here you have an angle of about 106 degrees. This is called the Burgi-Dunitz angle. The nucleophile would like to approach perpendicular. However, we have the electron density of the free electron pairs of the oxygen and therefore some repulsion. And that Burgi-Dunitz angle is compromised. Well, this is sterically blocking that trajectory on the other hand. Here, this hydrogen is far less blocking that corresponding trajectory. So therefore, indeed, that vinyl group will be introduced from this side. Therefore, we get to this structure. And it is necessary that that vinyl group is on the same side as that olefin. If it would be on the other side, well, it can't reach to that position, of course. So OK, now we have this structure. And again, we have to think about how can we simplify that? I hope everyone already has noticed that we have the retron of Adil's aldo reaction within this molecule, the cyclohexene moiety. And therefore, we should think about Adil's aldo reaction like this. Well, the question is, is a ketene a good reagent, or in this case, well, ketene more reactive intermediate for Adil's aldo reaction? Good question. So what do you know about ketenes? Actually, this ketene you can generate, but it dimerizes. Forming this product, to my knowledge, you can buy that. But upon hydrolyzation, you would form this carboxylic acid, the carboxylic acid functionality, or other way around, with a carbon ni-group in better position to the carboxylic acid. And as you know, this will readily lose one equivalent of CO2. You have acetone. So this is just chemistry. You have discussed in organic chemistry too. On the other hand, it already shows that these ketenes have a tendency for 2 plus 2 cyclo addition reactions. No need for photochemical irradiation there. And well, more often than this parent ketene, this dichloroketene is used. Well, I think with a base. Maybe triethylamine, I have it in my mind, but maybe better using Hünix's base, a sterically more hindered base than triethylamine. Nevertheless, I think triethylamine will work. Elimination of HCl, and then you generate that. But if you do that in the presence, for instance, of cyclohexene, then again, a 2 plus 2 cycloaddition process takes place, giving rise to that product. OK, reductively, you can get rid of those chlorides. So the chances that you will get these older reaction with these two components, all with the corresponding dichloroketene, no chance. Therefore, we can say this doesn't work. Nevertheless, the deal's older reaction as an idea is certainly a nice idea here. And we should stay with that, but we should choose a dienophile that has a functionality which can be easily transformed to such a ketone after the deal's older reaction, and this reagent should give the deal's older reaction. So in literature, we can find examples of that, for instance, having a certain type of acetyl functionality here. TMS enol e. von Badt's side, and here, no CH3 group. Well, it's an acetyl, a mixed acetyl of the ketene. This has been indeed done, this type of reaction, and it works. And there is an alternative, this chloroacrylonitrile. So that chloroacrylonitrile would give this dissolved product how can we now transform that functionality into a ketone? Well, with KOH in DMSO and heating it up, we would get to the cyanohydrin. Finally, to the ketone, HC-AN is eliminated, and of course, the HC-AN is, well, deprotonate the reaction with the base. So let's check this drawing. Correct, no double bond is missing here. So now we have finished our retrosynthetic plan, and indeed it will work in a four-step process. I think it's enough for today. Thanks for listening. See you next week on Tuesday.
 Welcome to the second part of the lecture,  Designing Organic Synthesis.  Let us start today with some general schemes  of Zygmatropic rearrangements.  You certainly remember the cope rearrangement  and its general scheme.  So we have a hexadeine system,  and it rearranges under elevated temperatures  to give again hexadeine.  Well, essentially this is the same.  But if you have some additional substituents here,  then for instance here, a muffle group,  and there another muffle group,  then you would have a muffle group here, a muffle group there.  These olefins are then twice substituted.  This is generally our favorite thermodynamically.  And then you have a reaction.  It is an equilibrium,  but in that case the equilibrium would be on that side.  Here it is a reaction, it goes there and backwards,  and you would notice it of course.  Nevertheless, this is the general scheme of the cope rearrangement.  Let us exchange one methylene group by an oxygen,  then we are dealing with an oxacope rearrangement,  better known as Claisen rearrangement.  So a vinyl allyl ether then will give this structure.  So that is, well, an unsaturated carbonyl compound,  alpha beta, so gamma delta unsaturated carbonyl compound.  While this is generally not an equilibrium reaction,  it is on that side, it will stay there.  Reaction temperatures well above 180 degrees  in the typical cope rearrangement,  somewhat lower, maybe 20 degrees or 30 degrees lower  if you have an oxacope rearrangement,  and there is another one.  We have a hydroxy group here.  We would have this enol as an intermediate,  and this will of course, tautomerize.  Finally, again leading to an unsaturated carbonyl compound,  but this time it is delta epsilon unsaturated.  This reaction works best if it is really an electron rich system  as you should already know, if you deprotonate here,  and best is, for instance, with potassium hydride.  So kH in THF secondly reflux.  So in that case you have the potassium alcoholate,  which is reacting, and this indeed works very well  in THF reflux temperature, all about 60-65 degrees or so.  So at much lower temperature than the other examples.  So this type of reaction is called the oxicope rearrangement.  So why are we discussing these general schemes?  Yesterday we discussed the general scheme of the Diels Alder reaction,  and well, Diels Alder reaction, general scheme is that,  and the cyclohexene we called the retron of the Diels Alder reaction.  Therefore now these are retrons,  retrons of the cope rearrangement,  retron of the clasen rearrangement,  and the retron of the oxicope rearrangement.  So let us do an exercise.  This is our target molecule, and as an exercise,  you should find out which sigmatropic rearrangement we could use  in our retrosynthetic plan for, well, later on,  synthesizing this molecule, this as a target.  So it's clear what you have to do is find out which retron  can we find in this target molecule, this target structure.  And well, this is the first task.  And after you have found it out,  we will discuss some hints how to solve the problem then.  OK, please start with that.  Find out what kind of retron do you find here.  So your task was looking for retrons within this structure,  so structural subunits which imply that it could be a good idea  to make use of a certain rather powerful transformation,  like with the other reaction or a cope rearrangement.  Well, OK, which retrons do we find?  Well, if you have a close look, then we have, over all, three  of them.  Therefore, I write it down three times.  But we can discuss then each.  So first of all, at least one of you found, OK, here,  we have a cyclohexene moiety.  So we could think about making use of a Duzalda reaction,  retron of a Duzalda reaction.  So and that means in our retro synthetic considerations,  we should think about this substrate.  And then while heating it up and hopefully it works.  Well, at least it has a chance since this is  an intramolecular.  Unfortunately, while from electronics,  it's certainly not the optimum electron rich and electron  rich.  Next, maybe a small problem is that you  have that olefin out of conjugation to the carbonyl.  Or you can synthesize that, of course.  So but there are some problems involved.  It is an idea.  One could think about that further on.  Nevertheless, let's have a look at the other possibilities.  And some of you noticed that you can find delta epsilon  unsaturated carbonyls.  So delta epsilon, the saturated carbonyls,  we can all count the positions.  Let's take this as this carbon as position one, two,  three, four, five, six.  And here, let's also put some numbers at these positions.  This will help when we go on with our considerations.  So carbonyl position one, two, three, four, five, six.  Indeed, delta epsilon unsaturated carbonyl group.  On the other hand, let's try this way.  One, two, three, four, five, six.  Once again, another delta epsilon unsaturated carbonyl.  So we have the retron of an oxycopere arrangement,  either this way or that way.  And for both examples, we should figure out  what is the substrate in each case leading  by oxycopere arrangement to this structure  or here the same structure.  Well, that means let us go back to this general scheme.  We have to transform our drawing from this to that.  And this is best done following these numbers.  In German language, we would call it malen nach Zahlen.  OK, just please try it.  So again, the explanation.  You figured out that in your target structure,  there is this type of moiety, a delta epsilon unsaturated  carbonyl compound.  That means you have found the retron of an oxycopere arrangement.  So then you have to transform that drawing from this unsaturated  carbonyl compound essentially with additional rings  or substituents attached to that type of structures,  also with additional rings and substituents attached.  For instance, you have to transform the drawing from something  with a carbonyl group to an alcohol functionality.  And then, for instance, you notice here in your target  structure, you have a single bond between the positions 2 and 3.  You have to transform the drawing into something  with a double bond between 2 and 3.  You start with a structure with a single bond between 3 and 4.  And you have to transform the drawing in something with no bond  between the positions 3 and 4.  And we have to do that for both cases.  So best is to draw that once again, but not starting  with a carbonyl group.  Let's put on an OH group here.  We have to do the same procedure later on a second time.  So we have an OH group in position 1.  Between 1 and 2, there is a single bond  and there should be a single bond remaining.  Between 2 and 3, there has to be, finally, after that  retrosynthetic step, a double bond.  Here, between positions 3 and 4, we  have to change from a single bond to no bond at all.  Between the positions 4 and 5, we have a double bond  and the positions 6 and 1 are connected.  Let's do the same here for that case.  OH group, between 1 and 2, the single bond remains.  Between 2 and 3, we have a double bond.  Between 3 and 4, the bond has vanished.  Between 4 and 5, there should be a double bond.  And we have again to connect positions 6 and 1 like that.  So these somewhat strange drawings we have to translate.  Now imagine you would put in such a structure  in a molecular modeling program  and press for optimization.  Then you will notice that those positions get more close,  of course.  So essentially, in this case, we have  two annihilated six-membered rings  with an OH group at the bridgehead position, tertiary alcohol.  And now we just claim that trans-olefins  doesn't fit into the drawings.  Let's just draw the cis-olefins.  And we have then olefinic functionality here and there.  And let's put on the hydrogen here.  Or like that.  It's better like that.  This type of structure looks strange on the first glimpse.  But wow, this is a bicyclic structure.  That's it.  Let's translate that.  OK.  The structure is the same like that.  So let's think about this here one structure that  is really a simplification.  It depends on how easy we can transform those structures  in even more simple structures.  And on the other hand, does that transformation  real reaction really work?  So here is some problems with, well,  but electronically it doesn't fit together very well.  But it might work.  Here, well, for the oxycoprearrangement,  this and that position has to come together  into close proximity.  The good question is that possible at all in this structure.  Well, I've put that into a molecular modeling program,  and it didn't look very well in that case.  Here, this position and that should get into close proximity.  And it looked better.  At least it looked better.  So let us go on with this structure and figure out  how we could simplify that within our retrosynthetic  analysis.  Well, this looks quite straightforward, disconnecting  here, because as you know, a tertiary alcohol,  we can easily synthesize by letting  an organometallic reagent reacting  with a corresponding ketone.  So we know that, for instance, with vinyl magnesium bromide  or vinyl lithium, we could get this transformation done.  How about the regioselectivity of the 1,2 addition process?  Does that nucleophilic vinyl reagent really  approach that substrate from this side  preferentially compared to the other side?  The answer is, of course, yes.  We have these hydrogens somewhat blocking  the trajectory for the nucleophilic attack.  I hope you remember the trajectory of a nortgometallic  nucleophile is not perpendicular, but a little bit  like that.  Here you have an angle of about 106 degrees.  This is called the Burgi-Dunitz angle.  The nucleophile would like to approach perpendicular.  However, we have the electron density  of the free electron pairs of the oxygen  and therefore some repulsion.  And that Burgi-Dunitz angle is compromised.  Well, this is sterically blocking  that trajectory on the other hand.  Here, this hydrogen is far less blocking  that corresponding trajectory.  So therefore, indeed, that vinyl group  will be introduced from this side.  Therefore, we get to this structure.  And it is necessary that that vinyl group is on the same side  as that olefin.  If it would be on the other side,  well, it can't reach to that position, of course.  So OK, now we have this structure.  And again, we have to think about how can we simplify that?  I hope everyone already has noticed  that we have the retron of Adil's aldo reaction  within this molecule, the cyclohexene moiety.  And therefore, we should think about Adil's aldo reaction  like this.  Well, the question is, is a ketene  a good reagent, or in this case, well,  ketene more reactive intermediate for Adil's aldo  reaction?  Good question.  So what do you know about ketenes?  Actually, this ketene you can generate, but it dimerizes.  Forming this product, to my knowledge, you can buy that.  But upon hydrolyzation, you would  form this carboxylic acid, the carboxylic acid functionality,  or other way around, with a carbon ni-group in better  position to the carboxylic acid.  And as you know, this will readily lose  one equivalent of CO2.  You have acetone.  So this is just chemistry.  You have discussed in organic chemistry too.  On the other hand, it already shows  that these ketenes have a tendency for 2 plus 2  cyclo addition reactions.  No need for photochemical irradiation there.  And well, more often than this parent ketene,  this dichloroketene is used.  Well, I think with a base.  Maybe triethylamine, I have it in my mind,  but maybe better using Hünix's base, a sterically more  hindered base than triethylamine.  Nevertheless, I think triethylamine will work.  Elimination of HCl, and then you generate that.  But if you do that in the presence, for instance,  of cyclohexene, then again, a 2 plus 2 cycloaddition process  takes place, giving rise to that product.  OK, reductively, you can get rid of those chlorides.  So the chances that you will get these older reaction  with these two components, all with the corresponding dichloroketene,  no chance.  Therefore, we can say this doesn't work.  Nevertheless, the deal's older reaction as an idea  is certainly a nice idea here.  And we should stay with that, but we  should choose a dienophile that has a functionality which  can be easily transformed to such a ketone after the deal's  older reaction, and this reagent should give the deal's  older reaction.  So in literature, we can find examples of that, for instance,  having a certain type of acetyl functionality here.  TMS enol e.  von Badt's side, and here, no CH3 group.  Well, it's an acetyl, a mixed acetyl  of the ketene.  This has been indeed done, this type of reaction,  and it works.  And there is an alternative, this chloroacrylonitrile.  So that chloroacrylonitrile would give this dissolved product  how can we now transform that functionality into a ketone?  Well, with KOH in DMSO and heating it up,  we would get to the cyanohydrin.  Finally, to the ketone, HC-AN is eliminated,  and of course, the HC-AN is, well,  deprotonate the reaction with the base.  So let's check this drawing.  Correct, no double bond is missing here.  So now we have finished our retrosynthetic plan,  and indeed it will work in a four-step process.  I think it's enough for today.  Thanks for listening.  See you next week on Tuesday.
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