 Yeah, good morning and welcome back to NPTEL lecture series on Classics in Total Synthesis Part 1. So we have been discussing few complex natural products total synthesis and today we will continue to discuss the total synthesis of taxol. We already discussed total synthesis of taxol by Nicolau, Halton and Dany Shapsky. Today we will talk about the total synthesis of taxol reported by Paul Wender which was shortest synthesis at that time. He did it in 37 steps and let us see how he did it. And his starting material was the air oxidized product of a monitor pin called verbenone. So verbenone was oxidized from alpha pining and that was the key starting material and his total synthesis actually involved two key reactions, one photochemical 1-3 alkyl shift and second one is fragmentation reaction of an epoxy alcohol. The first one as I said, so this is the 1-3 alkyl shift. You can see here in the left hand side where the cyclobutyl group is and after photochemical reaction where the cyclo propyl is. This is the 1-3 alkyl shift. So this is a key step and it moves like this. So when we talk about the total synthesis we know how important this 1-3 alkyl shift in the whole synthetic program. The second key reaction is the fragmentation of an epoxy alcohol. The core structure only I have written there are substituents which I will discuss when I talk about the real total synthesis. Here if you look at the hydroxyl group, this hydroxyl group upon treatment with base it can come like this and open the epoxy. If you look at carefully you have already a 6-membered ring, you have already a 6-membered ring. So now this 4-membered ring, this 4-membered ring if it opens up then that will become 8-membered ring. So the left hand side will be 6-membered ring and the right hand side will be 8-membered ring. That was the key thing is almost like you know Hultens synthesis of Taxol. So the lone pair on the oxygen pushes the breaking of the middle C-C bond and opens the epoxide to get this combo. Just I leave it for a minute just to check whether everything is alright. This hydroxyl group now becomes ketone and the epoxide now it became alcohol as you know in Taxol you need a hydroxyl group and carefully you see the whole A ring, whole A ring that is cyclohexene with a methyl group here which is attached to the double bond and then two methyl group, gem dimethyl groups and you have a hydroxyl group alpha that is what you need in Taxol A ring. What is missing in this is the bridgehead hydroxyl group. In Taxol you have a hydroxyl group at bridgehead position. So these are the two key reactions which Wenders group thought about and they wanted to use these two key reactions in the total synthesis of Taxol. Let us see the hydro synthesis. The third key reaction which is different from others is the aldol reaction. So they wanted to use an intramolecular aldol reaction intramolecular aldol reaction to construct the C ring. You can see here you generate an anion and then add to this aldehyde and that will form the six ombre ring which is the C ring. So intramolecular aldol reaction is another key reaction in their approach and this compound so this is the one which I said after the rearrangement after the fragmentation or rearrangement then if you alkylate here you get this compound. When you alkylate you get this compound we will come to that little later and this can be obtained from this monocyclic compound. So here what you wanted to do is if you do your 1, 4 addition with lithium dimethyl cuprate Gilman reagent then that anion can intramolecularly attack the carbonyl group. That is how you generate this hydroxyl group which is required for the fragmentation. Now this can be obtained from where we know first you should do the alkylation here you should do the alkylation here then you do the key reaction that is the 1, 3 alkyl shift. So 1, 3 alkyl shift is very important. So then only you will get the 8 pump burrowing okay. Let us see how we did the total synthesis. First we started with where we know which is commercially available and not very expensive. So first job is to alkylate here and if you look at this enone you can generate anion only here isn't it, it is a gamma position okay. So when you generate anion and now it will be in the form of dienolate okay it will be in the form of dienolate it will come all the way here and then you will get dienolate. However when you quench that with any electrophile the electrophile will go to alpha carbon electrophile will go to alpha carbon and followed by the migration of the beta gamma double bond to alpha beta double bond. So when you treat this with potassium tertiary butoxide okay as I said dienolate will be formed. Now when you quench this with perinyl bromide so this is called perinyl bromide then what you get is the alpha alkylated product here alpha perinylated product okay is it clear okay alpha perinylated product. Next you can cleave this double bond selectively because the internal double bond is electron deficient whereas the side chain the perinyl group double bond is electron rich. Electron rich double bond can be easily cleaved in preference to the internal electron deficient double bond. So was analysis of this gives the corresponding aldehyde okay. Then comes the key reaction where when you do the photochemical treatment the 1-3 alkyl shift takes place and when 1-3 alkyl shift takes place the double bond also migrates to the other side. So this is one of the key reaction. Then you have to add a nucleophile to this aldehyde. So you have two carbonyl groups one is ketone other one is aldehyde as you know ketone is less reactive compared to aldehyde moreover here in this particular case it is little bit sterically crowded as well. So you take ethyl propylate okay this is called ethyl propylate and when you treat with LDA you can generate anion here the triple bond as a acyclic hydrogen you can generate anion. That anion can add to this aldehyde to give the corresponding alcohol that alcohol was in situ protected as TMS ether by treating with TMS chloride okay. Now you have to form the 6-membered ring okay and as I mentioned during the retro synthesis the formation of 6-membered ring starts with addition of lithium dimethylcuprate okay. So now methyl group will add here okay then followed by intramolecular aldehyde reaction okay and when you use acetic acid work up the TMS group also goes. The TMS is a labile protecting group okay. So what you should get is a 6-membered ring with a double bond and an alcohol okay. Next you have got that you do not need alcohol here later I will come back how you can selectively introduce hydroxyl group. So for time being you oxidize this alcohol to ketone using Desmartine periodinine reagent you get the ketone okay. Next you have to hydroxylate. If you look at taxol this is carbon number 9 and this is carbon number 10 okay. In carbon number 10 you have acetate okay carbon number 10 you have acetate. That means you should have a good method to introduce acetate or you should have good method to introduce at least hydroxyl group okay. So when you generate enolate with potassium hexamethyl disolocyte okay then enolate will be formed here okay enolate will formed here okay. Then this can be quenched with this oxaziridine this is called Davis oxaziridine okay this is called Davis oxaziridine and this is derived from camphor sulfonic acid okay derived from camphor sulfonic acid in 3 steps okay. Now this oxaziridine can give the extra oxygen okay this oxygen can be given here so that means once you generate enolate and then add this oxaziridine that will stereo selectively give the hydroxyl group. So you could selectively introduce alpha hydroxyl at this carbon but we will come to the stereochemistry later because we need beta okay that can be done at the later stage okay. The idea is to introduce a hydroxyl group there first okay that is done. So now you have alpha beta and such a ketone and you also have ester if you reduce with LIH both ester and carbonyl group that is the keto group will be reduced to corresponding secondary alcohol and then primary alcohol. When you treat with LIH this hydroxyl group which is alpha so that means hydride will be delivered using this handle from the same side that means the resultant alcohol here will be beta because this hydride will come from alpha so you will get beta alcohol and the ester also will be reduced to corresponding primary alcohol okay LIH reduces the ketone as well as the ester to get this compound. Next you have to protect the primary alcohol selectively because you have 2 secondary alcohols 1 tertiary alcohol and a primary alcohol as you know it is easy to protect selectively the primary alcohol in the presence of secondary and tertiary alcohol and for that what you should do is you should use your bulky protecting group okay. The bulky protecting group if you are thinking of you can use TBS chloride, TBDPS chloride, Pytal chloride so on. So here they chose TBS chloride as the protecting group so the primary alcohol was protected. Now you have 2 secondary alcohols okay that can be protected as astronaut by treating with PPTS and dimethoxy propyl okay this is a very simple reaction at room temperature one can do so these secondary alcohols were protected now. So that sets the stage for the key rearrangement or key fragmentation so already we have discussed one key reaction that is a photochemical reaction where 1-3 alcohol shift took place along with the migration of the double bond. Now for the second key reaction we have come to the key precursor what you need to do is you need to make the epoxide you need to make the epoxide here followed by the fragmentation. So treatment with one equivalent of MCPBA one could get epoxide there there are 2 double bonds but still serious selectively one can do the epoxidation at carbon 12 and 13 and this is the stage where he tried the fragmentation reaction. So for that he used a base DAPCO non-nucleophilic base and heat it and as expected the rearrangement or fragmentation took place smooth 3 to give the AB ring of taxol okay if you look at this this is the AB ring of taxol and it has almost all the functionalities okay almost all the functionalities present in A and B ring okay. So now what one has to do you have to construct the C ring followed by construction of the D ring that is oxytane then you should attach the side chain before attaching the side chain some minor functional group modifications should be done okay. So next the secondary alcohol which is formed that allylic alcohol that should be protected so that was protected as tip see-ther by treating with tip strip plate and dichloromethane once you have that now you need to introduce a hydroxyl group at Bridget position you need to introduce a hydroxyl group at Bridget position and this is the best time why because you can generate enolate here and not here okay once you reduce the double bond then enolate can be generated at that position so it is better to do it at this position okay. So it was easy and it can be done by treating with potassium tertibute oxide and oxygen and the desiccant hydroperoxide was converted into hydroxyl group by treatment with triethyl phosphate okay. So very easy and the hydroxyl group at Bridget position also being introduced now if you look at this particular intermediate as all the functional groups required for taxol in A and B rings okay what is next you should reduce the ketone to hydroxyl group okay. So this molecule if you look at it will be like this okay this molecule it will be like this so the hydride will come from the beta side okay. So when you reduce this you get alpha alcohol in taxol also if you look at carefully the structure of taxol this hydroxyl is alpha it is in taxol as benzoate okay so now the hydroxyl should be benzoated that can be done little later. Next major task is to introduce the C ring okay for that first you have to reduce the double bond okay and when you reduce the double bond it is very important you should do it stereo selectively and when you do hydrogenation okay it is a tetrasubstituted double bond and you also have another tetrasubstituted double bond and A ring you need to selectively reduce this okay and when you are doing hydrogenation and hydrogenation obviously as you know it is a cis addition so both this methyl group and the CH2OH after reduction will come in the same side. Now what you did you used cleverly craptris catalyst craptris catalyst is well known for directed hydrogenation if you have a polar functional group. So if you have hydroxyl group here see you can see the alpha hydroxyl group. So now craptris catalyst will coordinate with this alpha hydroxyl and deliver the hydrogen from the alpha side when the hydrogen is delivered from the alpha side then automatically this methyl and hydroxyl methyl these two will become beta. This is what you get hydrogen comes from the same side as that of secondary alcohol so that the methyl and CH2OH comes from the beta side. So this is a structure of craptris catalyst which is widely used for directed hydrogenation okay. So now you have methyl group your CH2OH but you should remember your primary alcohol your secondary alcohol your tertiary alcohol okay you need to protect some of them. So you can protect the primary alcohol in the presence of secondary and tertiary. So in CH2 and temporary protection with TMS chloride you protect the primary alcohol then the secondary and tertiary alcohol can be protected as cyclic carbonates. If you treat with triposgene and pyridine you can protect that as cyclic carbonate okay. So now that is protected now what you need is you need to homologate this you need to homologate the CH2O TMS then only you can carry out the intramolecular aldol reaction. So for that you can straight away oxidize you know CH2O TMS okay with PCC to get the corresponding aldehyde okay when you want to homologate the best way is you have to do wittig reaction or enol ether wittig. So what he did was he did enol ether wittig reaction. So when you do enol ether wittig reaction you get the corresponding enol ether 1 carbon extra. Normally you get a mixture of cis and trans isomer but does not matter because once you hydrolyze this you are going to get aldehyde. So one normal HCl treatment and dilute 1 it hydrolyzes enol ether you get the corresponding aldehyde. So what you have done is now you can see you have introduced one extra carbon. The hydrolysis of enol ether as well as the removal of acetonide took place when you treat with one normal HCl okay. So you got CH2CHO and the diol okay. Now if you look at this diol this particular hydroxyl group is more exposed this particular hydroxyl group is more exposed than this. So selectively one can protect this hydroxyl group with TS chloride and then you get corresponding OTS okay. Then what you need to do is you have to oxidize the other hydroxyl group, other hydroxyl group that is easily done with Desmartine periodinane. So you get the ketone and if you look at this particular structure you have 2 carbonyl groups one ketone and one aldehyde okay. Then as you know between aldehyde and ketone aldehyde is more reactive and you want to introduce a double bond here CH2CH2. So that can be easily achieved by treating with aation moisture salt okay this is aation moisture salt and that will give directly the exocyclic double bond okay. Now if you treat with allyl magnesium bromide with zinc okay. Then the allyl group adds here to get the corresponding homoallelic alcohol okay. What is left now you have to protect the secondary alcohol then you have to cleave this then you have to do the aldol reaction but for doing aldol reaction these 2 functional groups should be exchanged. One should come here and then OTS should come here okay. One the double bond should be cleaved to get aldehyde and these 2 functional groups should be exchanged followed by intramolecular aldol reaction. So before that the secondary alcohol that allylic alcohol should be protected. So that was protected as bomb chloride that is benzyloxy methyl chloride so you get the protected compound. Now if you treat with ammonium fluoride okay say ammonium fluoride removes that TS group which is more exposed as I said it is more exposed it is on the beta side so it is easy to remove TS in the presence of TIPS. Then you treat with phenyl lithium. So the phenyl lithium and the cyclic carbonates are very level protecting group. The phenyl lithium as we have already discussed it adds to this and opens and you get the corresponding benzoate okay corresponding benzoate. Now if you treat with acetic and hereditary protein you protect the secondary alcohol at C9 as acetate okay but again as I said acetate should be here and ketone should be here. So how do you do that? So you take this highly substituted guanidine okay and treat in the presence of dichloramethane. So that generally it goes through enol form enolate then intramolecular acyl transfer takes place that gives the completely rearranged product okay. So the ketone and acetate are exchanged and that sets the stage for the intramolecular aldol reaction okay. So now what is required you need to cleave this double bond selectively in the presence of disubstituted double bond okay and that can be easily done was analysis followed by treatment with triethyl phosphate so one can use dimethyl sulphide you can use zinc so you get corresponding aldehyde okay. So it is easy once you have ketone and aldehyde and treat with dimethyl aminopyridine it undergoes intramolecular aldol reaction to construct the shearing okay one you can see the shearing is constructed and these two chiral centers also fixed okay. Then the secondary alcohol was protected as trocheether okay that is quite easy to prepare the proteatic group then you treat with sodium iodide and aqueous HCl sodium iodide and aqueous HCl that is to remove the bomb group if you have to remove the bomb group selectively you can use this okay now you got allylic alcohol so what we have done so far we have made the A ring we have made the B ring we have made the C ring now we have to go for making the D ring so for the D ring you have to functionalize this exocyclic double bond what will you do you can use dihydroxylation and before that you should make sure the beta alcohol as a good leaving group but the leaving group should be in alpha position okay see if you look at the oxytane oxytane is beta that means the leaving group here should be alpha so mesyl chloride first it converted into mesylate then the mesylate was converted into corresponding bromide via SN2 reaction to get the corresponding alpha bromide okay now you have the alpha bromide next you should do the dihydroxylation so the dihydroxylation comes from the beta so you get the diol so now you have the primary alcohol and secondary alcohol okay while doing this reaction there was some migration of this benzoate to this primary alcohol so that means the primary alcohol here intramolecularly attacked benzoate here so some migration took place so you got mixture of both so they thought it is better to do complete migration so it was treated with imidazole when you treated with imidazole the complete migration took place to the primary alcohol complete migration took place to the primary alcohol now you can again you have to re-protect this diol so that was done with tri-posgene to get the cyclic carbonate okay now cyclic carbonate is protected now what you need is to convert these functional groups into corresponding oxidative so potassium cyanide methanol selectively potassium cyanide methanol will selectively hydrolyze the benzoate okay and does not stop there immediately you can take this compound and treat with unix base and reflects that will give the corresponding oxidant okay you take this compound and treat with unix base okay unix base when you reflect it automatically intramolecular centriaction takes place to give the corresponding oxidant which is required so now if you look at this all are present okay ABC everything is there with functional groups you have to acetylate the tertiary hydroxyl group and you have to open this with phenyl lithium and you have to remove this protecting group as well as this protecting group and attach the side side so these are the job left so first you acetylate the tertiary hydroxyl group so you get the acetate then you can treat with TASF so the TASF will remove the tips group that is the fluoride so once you have that then treat with phenyl lithium okay that phenyl lithium will add here and then opens up so you get back at in 3 and if you use excess there is a chance of acetate also getting hydrolyzed so you gets a mixture of back at in 3 and 10 D acetate back at in 3 as you know already we have discussed how back at in 3 has been converted into taxol in few steps and addition of phenyl lithium also please drop okay addition of phenyl lithium also please drop okay so once you have back at in 3 it is already converted into taxol okay that is how he completed the totals in this taxol and this was the shortest totals in this is that time only 37 steps starting from where we know and it was one now consider as one of the classical total synthesis of taxol okay so thank you.