 So, now we will move to the synthesis of another important vitamin called a biotin. This biotin plays a very, very important role in physiological process like gluconeogenesis and fatty acid biosynthesis. So, obviously a lot of interest was there in early 60s, 70s, 80s to synthesize this biotin. So, today we will talk about one total synthesis of biotin and when you look at this molecule immediately you can see that there are three contiguous stereocenters. So, 1, 2, 3. These three are chiral centers and they are contiguous and there is one tetrahedral thiophilin ring. You can see completely reduced thiophilin ring and there is a 5-carbon side chain having carboxylic group at the terminal end. Then you also have a cyclic urea, you have a cyclic urea. Then if you look at the whole biotin, the difficult part is the carboxylic acid and these three chiral centers. So, you are going to introduce these three chiral centers stereoselectively so that biotin can be made in naturally occurring form. So, the first synthesis was reported by Roche group and here they used a very interesting intramolecular 3 plus 2 cycloaddition of a nitrone and an alkene. They started with naturally occurring amino acid, cysteine, actually they used a dimer of cysteine, cysteine and they tried this intramolecular 3 plus 2 cycloaddition reaction. So, what is this 3 plus 2 cycloaddition between nitrone and alkene? This nitrone, when you look at nitrone, nitrone is nothing but if you have an emine and if the nitrogen is oxidized, okay, the oxide of nitrogen, okay normally you know we talk about n-oxide, n-methylmorphylene, n-oxide, tri-methyl amine, n-oxide. So, here nitrogen of emine if it is oxidized then that is called nitrone and this nitrone if you look at carefully it has one electron deficient end and one electron rich end. So, it is a distributed over 3 atoms, okay. So, when you do a 1, 3 dipolar cycloaddition with an alkene and if the alkene is electron rich then this is what the stereochemistry of the product which you get, okay. So, obviously this is electron rich and this is electron deficient. So, you can see the bond forming between these 2 carbon atoms. Likewise, when you use electron deficient dipolar of ions, okay then what will happen? This is electron rich and this is electron deficient. So, this is where the bond will form and the stereochemistry will be reversed, okay. So, this nitrone alkene cycloaddition has been successfully used to make 5-hombard ring which you can call it as exoxazolidine, okay. But contrary to 4 plus 2 cycloaddition that is our Diels-Ald reaction. Here the secondary orbital interaction is not there or it is very, very minimal. As you know when you talk about 4 plus 2 cycloaddition or Diels-Ald reaction the secondary orbital interaction is important for getting endoisomer as the major product. Since this nitrone alkene cycloaddition that is 3 plus 2 cycloaddition does not or involve very little amount of secondary orbital interaction you do not have to get only the endo product as the major product. In fact, you get exo product as the major product and the formation of exo or endo is mainly controlled by your substrate or if you are using a catalyst. Only it gives the exo product and if you take this cyclic nitrone and treat with dipolar file like acrylonitrile then you get this as the major product this is nothing but exo product. And here this is a bottom phase approach of dipolar file that means you have the nitrone like this and your dipolar file attacks from the bottom, okay to give this as the major product. And there are several intra-molecular 1-3 dipolar cycloaddition between a nitrone and alkene present in the same substrate is reported for example you can see this molecule this can be redrawn like this. Now this O- will attack here and this double bond will attack and the positive charge on the nitrogen will be neutralized to get this tricyclic ring, okay. And this is a isoxazolidine and this endo bond can be easily cleaved with zinc and you can also cleave it with hydrogenation condition you can also cleave it with LAH. So that will give you amino alcohol and this is another example this is intermolecular and still you can see there are three chiral centers, okay. Three chiral centers one can fix during this 1-3 dipolar cycloaddition depending on the nature of your substrate, okay. Now again if you cleave this eno bond, okay so that will give you this amino alcohol and when you use LAH not only the eno bond gets cleaved but also the ester so you get two hydroxyl group one primary and the other secondary in addition you get a secondary amino group, okay. And this also can be rewritten like this. So I also will show you another example of intramolecular nitron-alkene cycloaddition because this is important so that you will understand the 1-3 dipolar cycloaddition which is involved in the synthesis of biotene as the key step, okay. So this starting material is called citronellol commercially available this all treatment with methyl hydroxyl amine it forms a nitrone and the nitrone undergoes 1-3 dipolar cycloaddition and it goes via this nitro you can see this can be drawn like this a chair like transient state you can write and followed by formation of this 5-ohm buttery. Again you can cleave this you will get corresponding amino alcohol. Now let us see how this biotene was synthesized what was the retro synthesis by Roche group. So what they did was they introduced an extra hydroxyl group in biotene that was a first retro synthetic you should not call it as disconnection but as introduction of additional functional group required for better disconnection. So why they introduced a hydroxyl group because they wanted to use a 1-3 dipolar cycloaddition as I said when you do a 1-3 dipolar cycloaddition between nitrone and alkene you get isoxazole and if you cleave the NO bond you get amine and alcohol. So that is why they deliberately put this hydroxyl group so that this hydroxyl group along with this nitrogen you know it can form during the dipolar cycloaddition. So then they said if you can look at this complex structure okay so this is formed if you cleave this NO bond if you cleave this NO bond you will get this okay. But this isoxazole can be obtained from the nitrone this nitrone and the double bond okay. Now this O- will attack here and the double bond will attack so it will form a 5M buttery and this nitrone I leave it for sometime so that you can visualize any of you can discuss more when we talk about the total synthesis. Now you just see can be obtained from this particular compound and which can be made from commercially available amino acid called cysteine okay and from cysteine you can acylate with this carboxylic acid. So basically what you are doing is you are going to start with L-cysteine methyl ester and this acid chloride which you can make it in 3 to 4 steps and it is a known compound okay. So now let us see how the Roche group synthesized biotein so they started with L-cysteine methyl ester then acylated to get this compound okay. Then the this is a dimer you should know that this is a dimer S, S and the dimer. So you can cleave the SS bond with metal. So for example if you use zinc and acetic acid that SS bond will cleave and then you will get corresponding SH. So the SH spontaneously will add to the triple bond SH will add to the triple bond and you will get in thioether okay. Then what you do you have to reduce the ester to aldehyde because aldehyde is required for making nitro okay. So you reduce the ester to aldehyde in one step with di-ball then you treat with benzyl hydroxyl amine okay. So when you treat with benzyl hydroxyl amine this aldehyde will form a nitrone okay. You can see the nitrone now. Now once you have this nitrone this undergoes an intramolecular 1, 3 dipolar cycloaddition that will give you the corresponding high succes already okay. So I leave it for a few minutes so that you know you can understand how this 5 membered ring is formed. Already I put the arrows properly so it will be easy to understand. Nevertheless from the stereochemical point of view I leave it for 30 seconds so that you can understand. Once you get this isoxalidine obviously the next step is the cleavage of the ennobar okay. So you cleave it with zinc and acetic acid you get the corresponding amino alcohol and the amine is already protected as n-benzyl and you get the free hydroxyl group. So you have to protect the amine once again. So that was done with chloromethyl formate in the presence of base like sodium carbonate. So now the amine is fully protected. Then you can hydrolyze this amide okay. The hydrolysis of this amide was done with barium hydroxide in reflecting dioxin water to give the corresponding carboxylic acid and amine okay the amide was cleaved. Once this amino acid is formed then what happens the lone pair on the amine attacks this carbomate then OME comes out that leads to the formation of this urea derivative cyclic urea derivative okay. This NH2 attacks the carbomate carbonyl OME comes out and you get the corresponding the urea derivative cyclic urea derivative. Can you redraw this as this bicyclic compound? You can see first the cyclic urea yes you have done the cyclic urea. Then attach this tetrahedrothiophen okay and you can see both ring junction hydrogen or beta so you have written beta and next you have the 5 carbon side chain that 5 carbon side chain with a hydroxyl and carboxylic acid okay. Same thing if you rotate it by 180 degree rotate it by 180 degree you will get this rotate it exactly by 180 degree. So the thiophene will come down and the urea the tetrahedrothiophene will come down the cyclic urea will go away. So if you look at this structure now what is missing or what you do not want. So there are two things you do not want one you do not want this hydroxyl group two you do not want this benzyl group. So if you can remove the hydroxyl and benzyl group that will lead to the formation of biotech okay. So how it was done? You first try to remove the hydroxyl group when you treat with thionyl chloride first the carboxylic acid will become acid chloride okay. Then that hydroxyl also will attack the thionyl chloride and it forms the half thionoyster okay. So this you have a lone pair on the sulphur of tetrahedrothiophene that will intramolecularly attack and your OSOCl will go out. So that will form a three-membered ring with sulphur having a positive charge. Now what will happen? The chloride which comes out okay. So now if you see here the SO2 will come out which is a neutral molecule the chloride which comes out again it will attack. So it is a double SN2 reaction on that carbon so you get again the beta chloride okay. Now the chloride can be removed with sodium borohydride and HPR water not only hydrolyses the ester but also removes the benzyl group to give biotech okay. So to summarize this in 1982 Roche group discussed the first amylocelective totals in this is a biotech and they use the amino acid L-cysteine which is a dimer of L-cysteine and they used an intramolecular nitrone olefin cyclo addition to fix the stereochemistry of nitrogen and the hydroxyl group. The hydroxyl group as you know it has to be removed but nevertheless to fix the stereochemistry of CN bond they use the intramolecular nitrone olefin cyclo addition reaction overall this synthesis was done in 11 longest linear steps and the yield was impressive of about 8%. Even though the molecule looks small overall yield of 8% is considered decent one for such molecules okay. So now we will move to the synthesis of other natural products having 5-unbuttering. So we have been discussing about total synthesis of penicillin type antibiotics and we will continue our discussion on one more natural product close to this. So that natural product is called Lacta-cysteine this was isolated in 1991 by Omura and his group and the structure was elicited by the various spectroscopy techniques particularly the X-ray was very helpful and if you look at this molecule closely it has 2 amino acids so 1, 1 here and then second one is here. So 2 amino acids and then N-acetyl-cysteine is also part of this amic acid derivative. The first total synthesis of Lacta-cysteine was reported by E.J. Coray a year after it was isolated and his synthesis involved many key reactions which I will discuss when I talk about the retrosynthesis as well as his synthesis. The first obvious disconnection of Coray's retrosynthesis of Lacta-cysteine was to cleave this part that actually simplifies the natural product so into 2 fragments. So now this carboxylic acid if you look at carefully so you have a carboxylic acid here and then your Lactam and of course you have 2 hydroxyl groups here there are 4 functional groups a carboxylic acid a 5 membered Lactam and 2 hydroxyl groups both are secondary. So if you look at this how he has made this compound from this particular compound you can see you have the hydroxyl group is still intact the second hydroxyl group is still intact and this one that is CH2O TBS. So that will form carboxylic acid that will become carboxylic acid is a latent functional group transformation then if you cleave this if you remove this CH2 then this NH can cyclize with this this N can cyclize with this that will form the 5 membered Lactam. Now let us see how he made this precursor. So this precursor he made it from here so just you have to generate anion and then quench with an electrophile and this is nothing but you know if you have to make this you can make it from another commercially available amino acid called serine. Serine is nothing but so this is serine so from serine one can make this in 2 steps so that was a commercially available starting material also. You take this compound and treat with LDA so LDA generates anion here and quench with isobuteroid. So when you quench with isobuteroid now you can see he has introduced one more chiral center okay already there was one chiral center here second chiral center here so he introduced the third chiral center which is very very important. Now you treat with trifluoroacetic acid so the trifluoroacetic acid removes this protecting group so that gives you the corresponding N benzylated amine then the CH2 OH so you remove that and then you get the amino alcohol so that was redrawn like this okay leave it for few seconds so that you know you should be able to understand the CH2 OH is beta so it is beta and then NH benzyl is there. Take this compound and then treat with TBS chloride so TBDMS chloride you can protect the primary alcohol okay you protect the primary alcohol as TBS ether then treat with toluene sulfonic acid and formaldehyde this is NH and this is OH okay you can see NH OH and if these two are protected with formaldehyde again you will get a five umber ring is not it again you will get a five umber ring so that is what happened okay you can see you protect this amino alcohol then the ester you could reduce with lithium borohydrate to get the primary alcohol then sworn oxidation will oxidize the primary alcohol to corresponding aldehyde. So this is the key fragment which you could make in few steps starting from commercially available amino acid called serine then you take this 2-dimethylphenol 2-6 dimethylphenol and then treat with propionic anhydride you get the corresponding ester that compound upon treatment with LDA you generate anion you generate anion then quench with this aldehyde okay it is an aldol reaction basically an aldol reaction and you get this aldol okay. Now from here how he goes to this one carefully observe okay you have the aldol next you have to remove this CH2 you have to remove this CH2 at the same time you also you have to remove the n-benzyle so if first if you do hydrogenalysis what will happen the benzyl group will be cleaved the benzyl group will be cleaved so that automatically once it cleaves it will attack this carbonyl group and you have a good living group 2-6-dimethylphenol phenol is a good living group okay and that will give this interpretation okay. So it is easy to visualize because I have to I have rotated 180 degree so that is why I am just leaving it for sometime from the natural product side you know you have to write like this so that is why I have rotated 180 degree I will leave it for sometime okay this portion this portion has come to the right side okay that is why everything is exactly opposite TBS was removed with HF then swan oxidation gives the corresponding aldehyde okay swan oxidation gives the corresponding aldehyde next step is oxidation of the aldehyde to corresponding carboxylic acid following Pinnick's protocol then treat with 1-3 propane diethyl what will happen when you treat with 1-3 propane diethyl 1-3 propane diethyl in the presence of HCl this CH2 that is formaldehyde is not it that CH2 when it is cleaved under acidic condition it is formaldehyde that formaldehyde will be protected with this 1-3 diethyl and leaving out your amino alcohol basically you are cleaving that protecting group with 1-3 propane diethyl with HCl okay. So now as I said there are 4 chiral centers in lactose cysteine lactose cysteine 1, 2, 4 all 4 are done okay now we have to attach the thiol side chain so that you can do from cysteine okay the cysteine N is protected as acetate and then the carboxylic acid is protected as allyl ester okay. So now you couple this with carboxylic acid so you get the corresponding thiol ester okay. So what is left you have to remove the allyl group okay so how do you remove allyl group palladium catalyst so if you treat with tetraxyl palladium in the presence of armic acid so you get lactose cysteine so basically this was a very elegant total synthesis starting from commercially available amino acid called serine. See there are 2 amino acids he used one serine another one is cysteine okay other chiral centers he used based on these 2 amino acids. So this was the first total synthesis reported by E.J. Coray and the key reactions are aldol reaction mainly aldol reaction and then 2 types of protections he carried out so that was important while carrying out the next aldol reaction. Overall this synthesis was accomplished in 14 steps and he has to use few protecting groups say when you use once protection you are adding 2 reactions if you use more protecting groups accordingly you have to multiply by 2 because 1 step is required for introduction and 1 step is required for removal. So that is how the number of steps has increased to 14 and with the overall yield of 6 percent then the second total synthesis was reported by Baldwin in 1994 2 years after E.J. Coray reported the total synthesis and here again he started with another commercially available amino acid glutamate and let us see how he made this. His retrosynthesis again as you know the cleavage of CS bond is the first thing then he wanted to introduce the hydroxyl group here using a dihydroxylation method if you have a double bond then you can do dihydroxylation and then selectively cleave one of the hydroxyl group okay. So he wanted to do that and then this chiral center he wanted to introduce using aldol reaction okay. So that aldol reaction will give this compound and this can be obtained from amino acid for glutamic acid okay. Now let us see how he started and then how he got the synthesis of lactocysteine. So glutamic acid pyroglutamic acid and treatment with thionyl chloride ethanol you convert that into ester ethyl ester and ethyl ester can be selectively reduced in the presence of lactam using lithium borohydride to get the corresponding primary alcohol. This upon treatment with the para toluene salphonic acid and benzole guide you get the bicyclic ring and one more chiral center is fixed okay. With this one can generate anion here and quench with methyl iodide to get the methyl group then you introduce a phenyl salenate group. Introduce a phenyl salenate group here by treating with LDA and quenching with phenyl salenate bromide followed by treatment with ozanalysis you introduce or you oxidize the phenyl salenate group to phenyl salenate oxide followed by elimination to get that double bond. And as I mentioned once you have the double bond the next step is the dihydroxylation. So the dihydroxylation before doing the dihydroxylation he wanted to do an aldol at this carbon. So at this carbon means you know alpha beta gamma carbon because if he does dihydroxylation at the double bond then the gamma position it will be difficult to oscillate or do aldol reaction. So what he did he wanted to do the aldol reaction first. So he treated with the TBS triflate and then 2,6-lutidine. So that form the dienolate the dienolate was quenched to form the corresponding TBS ether. So this is almost now we can see it is like pyrrole ring is not it that pyrrole. Now if you treat with isoputroldehyde in the presence of Luisa I said like tin tetrachloride the aldol reaction takes place at gamma carbon. So that is how we can see this stereocenter was fixed. So now you need one more hydroxyl group at this beta carbon. Again the free hydroxyl was protected followed by aspen tetroxide treatment you got the diol. Now you have a tertiary alcohol and secondary alcohol. The secondary alcohol should be intact but tertiary alcohol should not be. So what he did cleverly he treated with CDI that is carbonyl diumetazole. So that reacted with tertiary alcohol to form the corresponding thio derivative. So now this type of derivatives are known to undergo deoxygenation. If you take this compound and treat with tributyltrinhydrate in the presence of AABN. So that will undergo deoxygenation. So only problem is this methyl group that chiral center is not one isomer. Nevertheless you treat with sodium hydroxide. So sodium hydroxide will hydrolyze the acetate as well as it will epimerize this. Sodium hydroxide will hydrolyze the acetate as well as it will epimerize the carbon adjacent to the carbonyl group. Done then this protecting group can be easily cleaved by hydrogenalysis. So you get a free amino alcohol. You have come up to very, very key intermediate now. How many chiral centers are fixed? Four chiral centers are fixed starting with one chiral center of pyroclutamide. Then protect the primary alcohol as T acetyl and secondary alcohol as acetate. Then remove the TS with 40% HF to get the primary alcohol. The primary alcohol what you want in lactocysteine is carboxylic acid. So in one part you can do with the Jones oxidation without touching the acetate, without touching the chiral center the primary alcohol can be oxidized to carboxylic acid in 15 to 30 minutes. Once you have the carboxylic acid then the acetate group can be cleaved with sodium hydroxide solution. Now you can see all the four chiral centers. All the four chiral centers are fixed. So what E has to do? E has to couple this carboxylic acid with a thiol and that thiol should be derived from cysteamine. So it took cysteamine then NH was protected as acetate and then carboxylic acid was protected as allene ester. Then this thiol can undergo coupling with carboxylic acid. Yes it went and it formed a corresponding thiol ester and what is required for the total synthesis of lactocysteine is to remove the allene group. So the allene group can be easily removed with tetracase triphenolposterene palladium and the formic acid and triethylamine lead to lactocysteine. So this is one of the simple and straightforward total synthesis of lactocysteine starting from pyroglutamate and it was also reported 3 years after the isolation of lactocysteine and the key reactions involved in the synthesis of lactocysteine by Balvin are stereoselective aldol reaction. If you look at the siloxipyrrol the aldol reaction took place at the gamma position. Gamma position that is the first key reaction where the quaternary center was incorporated with the chiral center. Then the this side aldol reaction and finally attachment of the cysteine. Overall this sequence took about 20 steps slightly longer than what Coray has reported. Nevertheless the starting material is commercially available that makes huge difference and it is not expensive and the overall yield compared to Coray's. Coray's was 6% and this is about 4.27%. Okay. So with this we complete the total synthesis of 5-umbered you know antibiotics. So we will move to the next natural products having 6-umbered as the core structure. Thank you.