 Hello everyone, welcome to today's class. I hope that you had the opportunity to go through the last class. And we will briefly look at what we did last time. We started looking at the nomenclature of the sin and anti-aldol which were also called as Erythro and trio respectively. And we saw that if we draw a main carbon chain as zigzag form and then of course we can then expect that the once we have this here then whatever the substituents are here for example here methyl hydroxy or a methyl in this direction therefore we can anticipate and say that okay this is a sin-aldol. And likewise anti-aldol and of course these are called as Erythro or the other one is called as a trio that we looked at it. Then we also saw that how Z-inolate can give high selectivity in terms of sin-aldol. We took all the 4 cases and we saw how the large ratio of the sin-aldol is likely to form when we start with the Z-inolate. Then we also looked at the various features that are needed for obtaining Z-inolate. And we saw that the formation of the Z-inolate is much better, much easier and also the formation of the corresponding sin-aldol from Z-inolate is also very much easy. Then this was the work that David Evans has reported and then the work related to the formation of the E-inolate was also taken up by other people and especially H.C. Brown and as we saw that it depends on the size of the substituents say you have here R 2 groups that is 2 R groups attached to boron and or triflate if this R 2 R group which is 2 substituents has R they are large in size and then they allow the E-inolate to form in case we use small base. And the small base as we discussed forms irreversible complex with the boron and therefore it allows the formation of the E-inolate. We discussed the transition state and in the case of Z-inolate what we have also seen that if the base is very large then the base like such as this Huynix base diisopropyl ethylamine and such a base then allows the formation of the reversible complex with the boron and therefore that leads to the formation of the Z-inolate as the major product and based on these Z-inolate and E-inolates that we can anticipate to form syn and anti aldol products to form. So now we will look at what are the other auxiliaries that are required to get highly enantioselective reactions. So far we were talking about the racemic products where there is a possibility of getting different types of products in terms of their syn or anti aldol. But now we will also look at how the enantioselective reactions can be done with which kind of chiral enolates. So as I have repeatedly been telling that David Evans name and David Evans of course has introduced this oxazole denones which we have earlier also considered. So now what we have is if we take a ketone and take a compound such as this in which this particular part is attached to the auxiliary that is made from some of these amino acids. If you start with S-valanol or R-valanol then as you can see from here this particular auxiliary which is shown here is prepared from this particular S-valanol and if we convert this particular oxygen and nitrogen and make it as a carbonyl group here then of course we get auxiliary like this and from the nitrogen you can then attach with the help of say suppose if we have RCH2 COCl and if we have NH here we deprotonate then we can make a bond between the two of them and that is what would lead to if we just turn it around this is what is the product that is going to form. So this comes from S-valanol likewise if we take R-valanol then this will be beta right now this is alpha oriented it can be beta oriented if we take from R-valanol and of course we can use many of these kinds of chiral auxiliaries to make different types of oxazole denones like this where the substituents here would be fixed in terms of their configuration. So this is how the with these various kinds of chiral auxiliaries from different amino acids have been reported and of course this is non-ephedrine and the corresponding oxazole denones can form. So let us see what can we do with that. Now as you can see here that when such a compound is deprotonated we expect this type of intermediate to form the reason for that is the it forms a 6 member chelate with the metal and once that happens as you can see here the R group here the R group and the large auxiliary here are basically transoriented to each other. So we can expect that actually this and this is the same essentially there is no difference and when the group attaches to the any electrophile is reacted with this enolate as we can see that since this large isopropyl group is alpha oriented then the incoming electrophile which is going to attach on to this particular carbon atom because this has to break this particular oxygen metal brown will break and this enolate will then be having a nucleophilic center on this particular carbon atom. This nucleophilic center will then react with an electrophile where which is an electrophile here and of course you will get the product. It can also go onto the other side that is on the top basically what it means that when it when the enolate is going to react with the electrophile it would like to be oriented away from this particular large isopropyl group. Since isopropyl group is alpha oriented therefore the attachment will take place from the beta side. So for example if we take this particular compound and react with LDA in the presence of THF or sodium hexamethyl disilazide. So basically it is sodium plus hexamethyl disilazide you have N- and Me2 Me3 Si twice. So it is hexamethyl disilazide and then when that reacts with allyl bromide you see now it gives a beta orientation of the allyl group and then we can hydrolyze it with the base and therefore the hydrolysis takes place here and we get the corresponding acid in a very high anionsioselective pure product. Of course then auxiliary comes out and you can re-cycle the auxiliary. So this is how the enolate reacts with an electrophile in an alkylation fashion. If we take for example this type of auxiliary here like this in which fennel as well as the methyl both of them are alpha oriented this is even better because now you have two big larger groups oriented in alpha fashion therefore if we do the deprotonation here and react with an electrophile the major product will form here like this because these two are alpha oriented therefore the electrophile comes from the beta side and this is the alpha side. So as you can see here that under two different conditions the selectivity is 93 is to 7, 98 is to 2 and normally the lithium salts are better because when this forms this kind of enolate when we have auxiliary like this and the enolate that is going to form would be better in terms of alkylation when it is lithium than when it is potassium or sodium and therefore that reflects also because this bicyclic enolate has a better effect in terms of selectivity because then there is no rotation around here. But if this particular chelate formation is somewhat loose then there is a rotation around here and then that can lead to the enolate to form and that will lead to less selectivity that is why this particular selectivity as you can see one is 98 is to 2 other is 93 is to 7. Therefore generally as we have seen lithium and boron enolates are better in terms of achieving better selectivity. Now we look at the chiral-aldol reaction we saw so far chiral reactions but that was alkylation reaction. Now this is a little bit chiral-aldol reaction is somewhat complex then we will have to look at it very, very carefully. So the same way as we did in the earlier case if we start with this particular product which is formed from the oxazolidinone auxiliary in which isopropyl group is alpha oriented. Now you react this with now the eunich space diisopropyl ethyl amine and BU2BO3TF. We discuss how this can lead to the formation of Z-enolate. So now what we have here is this is Z-enolate and in which the methyl group and the auxiliary are of course trans to each other. Now once this Z-enolate is formed there is a better chelation between boron and oxygen just the way as we saw in the case of lithium and therefore this is a bicyclic enolate where everything is prefixed. Now here as you can see that the Z boron enolate which has a bicyclic structure involves the boron and the oxygen of the auxiliary in chelation. Therefore the boron enolate and the auxiliary itself has a chelation in it. But when the aldehyde comes into the reaction mixture when it is dropped in the reaction mixture at low temperature of minus 78 degrees then further reaction to occur in to give aldol product there has to be a chelation between boron and the aldehyde oxygen and therefore this particular chelation has to break down. So it is very important that the how does this chelation break and how does the new chelation occur that becomes a very important part. So now it is something that we already know that the geometry of the enolate determines whether syn or anti aldol is going to form. And of course as we have seen that Z enolate will give syn and aldol and E enolate will get anti aldol and stereochemistry of the auxiliary that means here this particular auxiliary the absolute configuration of the auxiliary would also determine the absolute configuration of the aldol products. Now we look at the oxazolidinone basically is an amide and this particular part is essentially a carbamate and it is a very large size and therefore the Z enolate formation occurs preferentially. Even though there is a chelate in the boron enolate like this is what is the boron enolate this is the boron where the metal is the aldol reaction will occur when the chelate forms at the breaks. Otherwise there will be no Zimmerman-Traxler transition state something that we will discuss soon. With this chelate gone with the chelate that we saw it up with the enolate and the auxiliary there is a repulsive dipole-dipole interaction like for example once this breaks then there will be repulsive dipole-dipole interaction between the oxygen atoms because this oxygen atom and this oxygen atom there will be once this is broken from here then we will see how these have dipole-dipole repulsion and that allows the confirmation to flip by 180 degrees. The result is still the aldol product but relative to the oxazolidin the phase of attack is opposite when we were doing alkylations. So, we will see that how does that happen. So, this is how it is. So, we can write something like this here as you can see we can write the oxazolidinone with alpha oriented isopropyl group we can turn around 180 degrees and make it look like this also therefore now it is beta oriented. We have not broken the bond but we simply changed 180 degrees. Now when this comes in contact with the aldehyde here now this particular oxygen boron chelation breaks and then what happens is of course there is a dipole-dipole repulsion here there is a delta negative here delta there is a delta negative and therefore the orientation has a dipole repulsion and this turns around there is a rotation around here here to form this so that the carbonyl group comes down. So, now this or in this enolate is the dipole is here whereas this dipole is here therefore there is no interaction between the two of them. In this situation now the ketone has a chelation with the boron here and at this situation the attack since this is alpha oriented and therefore the attack occurs from the beta phase from the top phase and the enolate reacts with the aldehyde with the concomitant chelation and the C-C bond formation takes place from the beta safe phase and that happens then of course the boron will be transferred to the oxygen and this oxygen boron bond is now broken by the oxidation with hydrogen peroxide and then of course the boron is relieved taken out of the system and oxygen comes out in the form of OH and is released as the corresponding aldol. Of course, we can write the same thing in a different form by just turning around by 180 degrees. So, this is how the syn diol is formed as the major product. Now, we can look at this in a slightly different way from the conformation angle. So, we can look at the Zimmerman Traxler transition state. For example, if we look at this is how we started, this is the product that this is the enolate that we started the same enolate can be written up in this particular fashion where as you can see that the methyl group here is the methyl group here double bond is here then the enolate and the methyl group are in the same direction as you can see here they are in the same direction here and then when the aldehyde comes into the contact with the boron then it has of course, the R group will remain in the equatorial form as we saw in the last time how the two possibilities of aldehyde in the transition state it can be either axial or equatorial, but the axial one is sterically more hindered therefore, it prefers to be equatorial. Here there are again two choices that we had to see which direction the attack would occur from as we say that it will attack from the top side that is seaside or the reef phase sea phase or the reef phase as you can see here the oxygen will have a chelation with the boron and therefore, this chelation will be somewhat like this if this happens the R group is equatorially oriented and therefore, the enolate will attack from the top side which is what is sea phase because you have this is the oxygen of the aldehyde this is the R and this is hydrogen therefore, we have a anticlockwise rotation of this particular phase and therefore, this is the sea phase where the enolate is going to attack. Once the attack takes place R comes into this particular form OH comes from here of course, after the hydrogen peroxide reaction C-C bond formation takes place methyl is here axial hydrogen is equatorial and of course, we get the auxiliary attach to the carbonyl group there which we can also write something like this that the hydroxy group is going behind and the R group is coming towards us and the hydrogen is in the plane of the blackboard or this particular board and there of course, we have a methyl group in the plane hydrogen is coming towards us and this particular bond is going behind and the same thing can be written up in this way you can see that like for example, if we turn the R group which is beta oriented bring it to the into the plane of the board below then the hydroxy group will go up and therefore, it is beta and similarly for example, if we bring the methyl group down this side then this particular broken bond here which is the alpha bond goes into the plane if it goes into the plane that means it is coming towards if you just rotate it around here like this then it comes into the plane towards us in that situation the methyl group will of course, come down and then that is why it is looking like a beta oriented here. So, basically it is a synaldol and we can also write the same synaldol like this. So, this is the way the synaldol forms and the synaldol forms. Now, we look at the reversal of diastereoselectivity. Now, how do we get the antialdol? If we have to get the antialdol, how do we get it? Now, what has been done that if we start with say for example, this type of enol boron enolate and react with aldehyde. Now, if we add extra Lewis acid such as diethylaluminum chloride then what happens is that this particular chelation does not break. And the aldehyde comes in to contact with this enolate in this particular fashion because the aldehyde oxygen has this extra Lewis acid interacting with its oxygen. That means the aldehyde oxygen interacts with this. So, it does not have any need to go and break the boron enolate and then interact with the boron. If that happens then of course, we have such a situation which is coming into the picture and when such a situation comes of course, in this case also as we can anticipate that the carbonyl group would be away from this particular enolate here or this particular double bond it would be away to allow the repulsion to be less the dipole-dipole repulsion. And therefore, the perfect orientation of this particular group would be somewhat like this where R, H and ketone would be oriented in this way that is a carbonyl group will be oriented in this fashion. When this happens, this allows the formation after the hydrogen peroxide and of course, hydrolysis of the auxiliary that this particular part give you the acid methyl group is here, R is here, H is here and OH is here. And this is if we trans can convert into the final orientation it would look like an anti aldol. So, this anti aldol formation is essentially taking place because we have not broken the enolate, the boron enolate and therefore, this diethyl aluminum as a Lewis acid allows the orientation of the transition state in such a way that we get the anti aldol which is the reversal of the diastereoselectivity. Now, we will try and look at the enol silyl ether that means silyl enolates with having a silyl group. Now, in this case, Ireland-Kleisen rearrangement is an important reaction which depends on the geometry of the enolate. We also saw the geometry of the enolates influencing the formation of the sin products in aldol and anti aldol. Of course, there we also saw that the Lewis acid made a difference. But in this particular case, if we start with allyl alcohol and its ester that is allyl ester where you have a possibility of deprotonation here and if the deprotonation is done and the enolate is trapped as a silyl ether or silyl enolate and if the reaction is done at low temperature in the solvent THF then what is found is that we get this type of enolate of course, but where this OX group and the R group are trans to each other and here we have in the presence of HMPA and THF as a solvent the enolate gets changed. It forms OX and the R group, they are now cis to each other. So, this is a cis enolate and this is a trans enolate. Now, that allows the formation when the rearrangement takes place, the collision rearrangement takes place here. Then what happens is that the product that is going to be formed is such where the R1 group here and the R group here are cis to each other and in this particular case, R group and R1 group are trans to each other. So, this is a basically the process in which two new asymmetric centers are being created here. There is an asymmetric center here and there is an asymmetric center here. So, these new asymmetric centers are as a consequence of the formation of the enolate because this double bond geometry is fixed. In both the cases, the only difference is between the geometry of the enolates and based on the geometry of the enolates, the configuration of the final product is of course determined. Obviously, in these cases, we are talking about only the archiral molecules and therefore, this product will be anyway archiral unless we use chiral auxiliary into it. So, we will stop it at this stage in today's class. You could go ahead and look at what I have talked in this today's class and be prepared for the next class. Till then, bye. Take care.