 So, good morning everyone and welcome back to this NPTEL lecture series on classics in total synthesis part 1. So, we have been discussing about total synthesis of natural products having 6 home birds of units. In the last lecture we talked about total synthesis of carbonone wherein oxidative phenolic coupling and intramolecular Dielsall reaction as the key reaction to you know construct three rings in one part. And today we will move to another very interesting natural product called Maybelline and why synthesis of Maybelline is important, why this natural product was important. If you look at this natural product Maybelline, it is a fungal metabolite and it was isolated from the cultures of aspergillus terres. This is the starting point for the development of many cholesterol-lowering drugs like Atovastatin, Rosatatin and many such drugs were made or inspired by this particular natural products. And the key pharmacophore of Maybelline is this hydroxy 6 home bird lacto. They use this key pharmacophore and changed only this whole unit. Now, let us see the total synthesis of Maybelline reported by Derek Clive from University of Alberta, Edmonddon and when you look at this molecule, you can see it is a bicyclic compound and also having a 6 home bird hydroxy lactone. Overall, it has 8 chiral centers and also a conjugated diene. And this hydroxy 6 home bird lactone, if you have a close look at it, this is nothing but an aldol, is not it? So, nothing but an aldol. So, there is a possibility of dehydration, it is a labile beta hydroxy lactone and these are some of the challenges one would face when you have to plan for total synthesis of such molecule. So, the synthesis which I am going to talk about today was reported by, as I mentioned, Derek Clive from Edmonddon though the synthesis is little longer, but one would learn lot of new reactions in this total synthesis and his total synthesis involved few key reactions. One, the use of Yvonne's chiral auxiliary for making a 6 home bird ring and it also involved use of Yvonne's chiral auxiliary to get introduced this chiral center and this chiral center. So, he used Yvonne's chiral auxiliary exclusively to introduce few chiral centers and he used the McMurray reaction, we saw McMurray coupling recently. So, he used a McMurray coupling to introduce this particular double bond. So, the McMurray coupling is nothing but if you have a ketone and if you treat with titanium zero, so then it gives the dimeric product where you get a double bond. Basically, it involves 2 steps. First, it goes through the corresponding diol. So, it is almost like pinnacle coupling, you have a ketone and then treat with titanium zero and you get the diol and the diol is still connected to the titanium and it undergoes elimination of titanium dioxide to give the double bond. This is a well known reaction, one can sometimes stop at diol and sometimes it will go all the way to the corresponding alkene. So, from his retrosynthetic point of view, the first disconnection was this C double bond O. His idea is this precursor, if you remove the protecting group here as well as this astronyne, the primary alcohol can be selectively oxidized. Primary alcohol can be selectively oxidized to aldehyde. Once that aldehyde is formed, this alcohol can intramolecularly attack that aldehyde to form lactol which can be further oxidized to give the corresponding lacto. So, he thought this should be the precursor for making maybe no lead. Now, this can be obtained from this intermediate. So, what he is planning is to use this aldehyde. This aldehyde, if you generate anion and quench with this aldehyde, you can introduce this side chi. So, the aldol reaction and this can be obtained from this lacto. So, now, one can generate a carb anion and make enolate and then quench with this iodide that should give this particular intermediate. So, there are two key reactions, one is aldol, another one is alkylation. Alkylation followed by aldol, one get this precursor and here as I said the McMurray coupling will give the corresponding bicyclic compound. And how these two fragments can be made? The fragment iodide can be made from this homolylic alcohol and this can be made from malic acid. I will come to that how this is made from malic acid. This lactone, whenever you see a lactone and a double bond in one of the rings, then one can think of using iodolactonization. So, if you have this double bond and carboxylic acid and this can undergo iodolactonization followed by elimination of H i, one should get the double bond. And this in principle can be obtained by intermolecular dielsol reaction between butadiene and the dienophile which is attached to Yvonne-Skyroloxalate. So, that way the first step itself is the asymmetric dielsol reaction using Yvonne-Skyroloxalate. Now, let us see how he made each precursor and you combine A, B and B, C and so on. First he started with this known compound which is made easily from Yvonne-Skyroloxalate by attaching propionic anhydride. So, then you alkylate by treating with LDA, you generate anion and congeal with allyl bromide, you get this intermediate and this upon most analysis, you get the corresponding aldehyde and that aldehyde you protect it as acetal using ethylene glycol and para-tolybe sulfonic acid. Now, you can remove the chiral auxiliary. Once this is done, you remove the chiral auxiliary to get the corresponding alcohol. Now, from this alcohol to this aldehyde, so you could do in few steps. First, swarn oxidation, you get the ketone, then do the vitic, you get the corresponding double bond, then remove the acetal, you get the corresponding aldehyde. So, that is the fragment A. So, you have the fragment A, now we will see how the fragment B was synthesized. Again, he started from Yvonne-Skyroloxalate and here you generate anion by treating with butylythium and congeal with clotyl chloride. So, now you have successfully attached the dienophile to Yvonne-Skyroloxalate. The next key step is the intermolecular Dielsall reaction with butadiene, so that was done using diethylaluminium chloride as a Lewis acid. At minus 10 degrees, you get this trans isomer or isomer. So, next step, you have to cleave this, so that is normally done with lithium-benzoate. So, now you have the ester, that ester can be exchanged with OME by treating with LI OME. So, you get corresponding methyl ester. So, this methyl ester, if you look at this structure, these two substitutions are trans. But in mevinolene, you need methyl and ester, they are cis to each other. Not only that, the ester should be having one more extra carbon atom, here this is directly ester, what you need is CH2 COH and also cis to each other, methyl and the CH2 COTME should be cis to each other. So, the first step is, you know, you have to isomerize. So, the isomerization is done using base, strong base, then you treat with lithium-aluminium hydride. So, ester is hydrolyzed to corresponding alcohol and the tosylation, you convert this primary alcohol into a good living group and treat with sodium cyanide in DMSO and SN2 reaction this way you homologate. Okay. The cyanide upon hydrolysis, you give the carboxylic acid and that is set for now the key iodolactinization. So, treatment with sodium iodide and catalytic amount of MCBVA, so you get the corresponding iodolactone. This iodolactone, now upon treatment with DBU, it undergoes elimination to give the fragment B. So, we have synthesized fragment A and we also have synthesized fragment B. Then what you should do, you have to prepare this fragment C, then you have to combine fragment B and C followed by combining with ABC. So, for the fragment C, he started with commercially available malic acid. So, this is malic acid. Okay. So, he took this malic acid and reduced with borane. So, borane is known to reduce carboxylic acid to corresponding primary alcohol. So, now upon complete reduction, both the carboxylic acids are reduced to primary alcohol. So, you see two primary alcohols and one secondary alcohol. Okay. So, one can easily protect these two primary alcohols or one can easily protect one primary alcohol and this one secondary alcohol. If you treat with ketones or protected ketones, then one primary alcohol and the secondary alcohol will be protected and it should be 5-ambered ring. Okay. That is faster. The formation of 5-ambered ring is faster. At the same time, if you treat with benzaldehyde, then the formation of 6-ambered acetyl is faster. So, that you will get. So, you can choose whichever you want accordingly. You can use either ketone or you can use benzaldehyde. Okay. Since he needs the 5-ambered ketol, he exchanged with this 5-ambered ketol to get the 5-ambered ketol, then the primary alcohol was protected as TBDPS ether. Okay. Then you remove the ketol using 80% acetic acid. You get back your diol and the primary alcohol can be selectively methylated in the presence of secondary alcohol. So, you do that and then treatment with base. Here, benzal-trimethylammonium hydroxide. So, it forms the corresponding epoxide. Okay. Then, open this with vinyl lithium. Okay. Open this with vinyl lithium and with copper cyanide. So, it is opening of the epoxide from this side. So, you will get the corresponding vinyl group. So, now what you got is homo-allelic alcohol. Okay. From homo-allelic alcohol to this iodine, stereo selectively. It was done again using another key reaction. So, we already discussed iodolactonization. Okay. This is analogous to iodolactonization where instead of COH, what they have used is OCOO minus. OCOC double bond OOO minus. They have used this and opened the iodonium ion. How they have done? It they took this homo-allelic alcohol and treated with butyl lithium. Okay. So, the butyl lithium forms O minus. Then, when you add carbon dioxide, it forms this OCOO minus. Then, when you add iodine, the iodine will form here. The corresponding iodonium ion, isn't it? And here you have OCOO minus. This OCOO minus will open this and you will get the corresponding iodocyclic carbonate. Okay. Iodocyclic carbonate. And this Iodocyclic carbonate. Now, if you treat with dry acetone and para-tolivine-sulfonic acid, two things happen. One, this carbonate is hydrolyzed and you get back the diol. Okay. Your carbonate is hydrolyzed to get back your diol and the diol is protected as acetone. The diol is protected as acetone. Then, you protect the free primary alcohol as TB, DPS, ether, tercibutyl, diphenyl, silyl ether. So, what he has done, what we have discussed so far is how to make the three fragments namely A, B and C. So, he has made successfully these three fragments. The next step is to combine B and C fragments. So, this fragment B, if you treat with LDA, one can generate anion. So, that can form the corresponding enolate and then enolate can quench. When you quench the enolate with this iodide, so you get this fragment. Okay. Next step is treatment with diol. So, diol, if you look at this, you have a lactone and one equivalent of diol at low temperature, this lactone can be reduced to corresponding hydroxy aldehyde. Okay. You have a hydroxyl group and you have aldehyde. Okay. Basically, lactone to lactol, lactol is nothing but hydroxy aldehyde. Then this aldehyde upon further treatment with manganese dioxide. So, the aldehyde remains same whereas the allylic alcohol, the allylic alcohol will be oxidized with manganese dioxide to form the corresponding alpha, beta unsaturated keto. Now, from here to here, which reagent one can use? So, what is happening here? If you look at carefully between the left hand side compound and the right hand side compound, what is missing? I will give you 20 seconds. Just check. From the left hand side compound to right hand side compound, what is missing? Which functional group is missing? And how that can be achieved in a single step through a well-known reaction? Okay. If you look at carefully, you do not see this aldehyde. Isn't it? You do not see this aldehyde in the product. Normally, such decarbonylation of aldehydes are done using Wilkinson catalyst. Wilkinson catalyst is known to decarbonylate. So, that is what he has done. So, once you have this, next step is again, you generate anion, make the unolate and quench with the fragment A. So, that will give you this alcohol. Okay. So, what you have to do? So, you have the precursor for the lactone and now you have to combine this ketone with this double bond. Original plan is you have keto and aldehyde and then do a McMurray coupling. But if you have to do, you have to protect this hydroxyl. So, the hydroxyl was protected as TES ether that is triethyl silane ether. And then for the preparation or for the synthesis of the ketone aldehyde which is required for the key McMurray coupling, the double bond was cleaved using vasoanalysis to get the ketone aldehyde. The ketone aldehyde then upon treatment with titanium trichloride and potassium naphthalenide, it generates titanium zero and the titanium zero facilitates the intramolecular McMurray coupling. So, that facilitates the intramolecular McMurray coupling to form the bicyclic ring. So, now if you look at this compound, the trans diene is formed using the key McMurray coupling. Now this particular intermediate as all the chiral centers is this particular intermediate as all the chiral centers required for the synthesis of plus may be no lead. So, what needs to be done is as I said this TBDPS group should be cleaved, the TES also should be cleaved and the lactone should be formed, is not it? That will complete the total synthesis of may be no lead. So, you treat with tetrabutyl ammonium fluoride is known to remove silal protecting group. So, you can remove both TBDPS and TES. So, you have the primary alcohol and the secondary alcohol and now you protect the primary alcohol again. Now you protect the primary alcohol again as TBDPS ether then you have to attach the side chain here, the secondary alcohol you have to attach the side chain. So, that side chain was attached by using mixed anhydride, mixed anhydride of this carboxylic acid. So, now that will give you the corresponding the last but one recursive. So, now what is left is only the formation of six-membered lactone. So, it is quite easy again remove the TBDPS using tetrabutyl ammonium fluoride, you get the alcohol and then alcohol was oxidized under sworn condition to get corresponding aldehyde then removal of this asteroid under acidic condition will give this alcohol will add to the aldehyde to form the corresponding lactone. So, now if you look at this intermediate and compare with the structure of nebulonin what is missing is this lactol should be oxidized to corresponding lactone. So, that was done using feticin reagent. So, feticin reagent is nothing but silver carbonate on silite. So, which is a well-known reagent for selective oxidation of lactol to corresponding lactone without touching other hydroxyl group. So, the feticin reagent gave the final natural product that is femenolin. So, that is how Clive's group could successfully complete the total synthesis of nebulonin and if you look at this molecule as I mentioned. So, this is a key starting point for the synthesis of several cholesterol lowering drugs like autostatin, lipid and so on. And to summarize Clive et al reported very highly stereocultural synthesis of natural product namely nebulonin and the key reactions involved in this synthesis are McMury coupling to form the cyclohexene and Yvonne-Skyroloxylery was used for Diels-Alder reaction and alkylation and also Iodolactonization was used to get a lactone and a double bond. And another interesting reaction which was used is Iodocarbonate formation Iodocyclic carbonate formation. So, that was also very very interesting reaction to create one more chiral center. The whole sequence to accomplish the total synthesis of meminolin involved the longest linear sequence of 27 steps which is understandable considering the complexity of this molecule and yield was about 1.27 percent starting from Yvonne-Skyroloxylery. Nevertheless, this was one of the clever synthesis and may not be very efficient, but it involved many key reactions to accomplish the total synthesis of meminolin. So, I will stop here and then we will discuss more about natural products having six-form word subunit in the next lecture. Thank you.