 Good morning and welcome back to the NPTEL lectures on Classics in Total Synthesis Part 1. And we have been discussing about total synthesis of natural products and in last class we talked about total synthesis of 4-membered ring which is not a natural product but very important compound called Cubane. So today we will continue along this and then we will discuss about natural product synthesis particularly dealing with 4-membered ring. Today's natural products are called endiandric acid and you can see here when you look at these two natural products B and C there is nothing common but when you have a closer look at you will find that these two have a 4-membered ring, is not it? And this 4-membered ring in the case of endiandric acid B and A you can see B and A this 4-membered ring is attached to a 5 and 6-membered ring, is not it? This is 5-membered ring and this is 6-membered ring okay and same thing you can see in endiandric acid C also it is attached to a 5 and 6-membered ring. But these 5 and 6-membered ring are attached to another 6-membered ring in a different fashion okay and endiandric acid has a diene which you can call it a cyclohexadiene and also a side chain which is another diene okay. So these interesting natural products which were isolated as resumes was reported in 1980 by Black and if you closely look at these 4 natural products okay particularly let us start with endiandric acid A and B. A and B have 2 cyclohexene moieties can you see 2 cyclohexene moieties see whenever you see a 6-membered ring with a double bond that is cyclohexene substituted cyclohexene immediately one reaction which will come to your mind is Diels-Alder reaction one reaction which will come to your mind is a Diels-Alder reaction and in endiandric acid you can see another 6-membered ring. This is a 6-membered ring with a cyclohexene okay with a double bond 6-membered ring with a double bond a cyclohexene ring. So that means all this endiandric acid endiandric acid A, endiandric acid B, endiandric acid C could in principle be thought of being made from corresponding dienes and dienophiles through a Diels-Alder reaction to support that if you look at endiandric acid D already you can see a diene okay you can see a diene. So in this diene you have 2 double bonds and if you are using this as dienophile one of the double bonds can act as a dienophile or if you are using this as diene this can act as a diene and the other one can will act as a dienophile okay. So from the biological activity point of view they show very good antibiotic activities and from stretcher point of view as you know they have 4 fused rings and overall 8 stereogenic centers though this molecules were isolated as resimines nevertheless it is very important to fix the stereochemistry. Those days whenever group isolate natural products they also try to propose a synthetic pathway a synthetic pathway in the sense how these natural products would have been made by nature. So that we call it as biosynthesis. So blacks group they propose a hypothesis which is biosynthesis what they proposed was if the nature has started with this tetraene having 2 side chains in the terminal carbons okay. Now this tetraene you can say octatetraene upon heating under thermal condition as you know when you see a conjugated tetraene which reaction will come to your mind and you see a conjugate conjugated tetraene which reaction will come to your mind a thermal electro cyclization reaction okay. Since it is a 8 pi electron then under thermal condition corn rotatory cyclization is allowed okay. So that should give you this cyclo octatetraene okay. So this is formed if you can see the corn rotatory cyclization corn rotatory as you know it is same side. Now if you rotate this combo along this axis by 180 degree if you rotate this combo along this axis by 180 degree you should get this combo is not it. So what black proposed was after this corn rotatory 8 pi electro cyclization still if you look at these 2 compounds you can see a conjugated trine okay conjugated trine. So the conjugated trine under the same thermal conditions it can further undergo an electro cyclization reaction. But it is 6 pi electron so the allowed is this rotatory electro cyclization. So that should give you this combo though this was not isolated he proposed that it is endoendric acid E and that should be the precursor for one of the endoendric acid. And the other one okay if it does the same thing that is this rotatory 6 pi electro cyclization then one should get another natural product though it was not isolated he named it as endoendric acid D and that should form the precursor for making some other endoendric acid. So the key thing was starting with a tetrion which undergoes successive 8 pi and 6 pi electro cyclization that leads to the fused 6 and 4 membrane okay that was his proposal. And if you start with a different substituent here if you start with a different substituents at the terminal carbon then again this will undergo another 8 pi con rotatory electro cyclization to give this cyclo octa trine. And again as I said you rotate it by 180 degree and then you will get this compound both as expected one can easily propose that this can undergo a 6 pi this rotatory cyclization to give endoendric acid F okay and same way the other one can give endoendric acid G. Now if you look at endoendric acid F it has a diene as side chain it has a diene as side chain and in this cyclo hexa diene this double bond perhaps can act as a diene of oil. If that happens it can undergo an intramolecular dielsal reaction to give another endoendric acid whereas in the case of endoendric acid G what you see is this will act as diene and the side chain alpha beta unsaturated carboxylic acid will act as a diene of oil. So what will happen this will act as a diene and this will act as a diene of oil and both will undergo an intramolecular dielsal reaction to give another endoendric acid okay. So as I mentioned this is what he proposed so endoendric acid F should give endoendric acid B through this intramolecular dielsal reaction and endoendric acid G which can be written this way okay just I leave it for a few seconds for you to understand okay. Now this upon intramolecular dielsal reaction should give endoendric acid C okay this is what black proposed after isolating this natural products that from synthetic point of view all this can be made by combination of pericyclic reaction that is electrocyclicization and dielsal reaction. So taking this as a challenge Nicholas group they proposed a beautiful retro synthesis and later executed a couple of years later they could complete the total synthesis of all this endoendric acid. So their idea is both endoendric acid A and B can be made from one common intermediate okay you can see X can be homologated to get endoendric acid B otherwise you know you can easily convert this into endoendric acid A. The key reaction is only the intramolecular dielsal reaction the key reaction is only the intramolecular dielsal reaction. Now this intermediate which can be written like this now how you can make this intermediate okay so he thought if we break this bond if you break this bond now you have two different substitutes which is attached to the cyclobutane okay one is CH2Y the other one is CH2X okay this idea is Y can be easily homologated. Then if you look at this carefully endoendric acid D also can be obtained from this particular intermediate okay if you homologate X you will get D if you homologate Y you will get this intermediate okay. Next the endoendric acid G that also can be obtained from the same intermediate it is just a matter of functional group manipulation okay some functional group manipulation and then transformation one can get all the endoendric acid from this particular intermediate okay and if you look at endoendric acid G already I told you black proposed that this upon heating will give endoendric acid C. So overall to make all this endoendric acid what one needs is to prepare or synthesize this key intermediate. So from one starting material it should be possible to make all the endoendric acid. So this is what Nikolov proposed and this is endoendric acid E F G D and all this he wanted to make from this intermediate okay. And this intermediate how one can make as I said the CHO should be differentiated because it is important then only one side you can homologate. So he made it as X and Y and this can be obtained from this cyclo-octotriene and this cyclo-octotriene can be made from the octotetriene with two substituents at the terminal carbon okay. Now if you look at this carefully if you break this bond if you break this bond and convert these two double bonds into triple bond if you break this bond and convert the adjacent two double bonds into triple bond you will get this. But from synthetic point of view forward direction how will you do if you do a Linder catalyst hydrogenation okay. Then these two triple bonds these two triple bonds will be reduced to cis-alkene these two triple bonds will be reduced to cis-alkene that will give you this tetroid okay. And how will you get this symmetrical compound it is very easy again you break this bond that is the bond between two triple bonds which is normally obtained by Glacier coupling okay. So that will lead to this as the starting material okay the starting material is now very simple so what he did was he started with this allylic alcohol having a triple bond at the end now Glacier coupling gave that diene okay. Once you have this diene then Linder catalyst treatment gives you the tetroene but what happens when you treat this diene okay when you treat this diene with Linder catalyst it does not stop there. What happens it undergoes first the 8 pi con rotatory cyclization as predicted and proposed by black and it does not stop there it goes further it undergoes 6 pi rotatory electro cyclization to give this intermediate and this is a product okay directly you can see in one step he gets this key intermediate which is required for synthesis of all endangering acid. So now having got this bicyclic compound with two CH2OH which are opposite to each other the next step should be to differentiate these two CH2OH is not it. So if you can differentiate these two OHs then you can functionalize on the other side. So with that idea he was trying to protect he thought this CH2OH will be more you know open okay then it will be easily protected but in reality when he was trying to protect this CH2OH selectivity was poor both the hydroxyl groups could be protected by TBDPSC. So what he has to do he has to follow an indirect method where one can protect one of the hydroxyl groups. So what he did was he carried out an Iodo ethyrification reaction Iodo ethyrification reaction. So when you have a double bond when you have a double bond and CH2OH adjacent to that then when you treat with iodine the double bond the iodine will add to the double bond it will form iodonium ion and in the presence of base this OH lone pair can attack and open the iodonium ion. So this is a well known reaction like iodolactonization it can form Iodo ethyr. The advantage is by using this you can protect the double bond as well as hydroxyl group okay it is a dual protection you are protecting the double bond as well as the hydroxyl group. Once you have done that now you can protect the other primary alcohol so that was EC TBDPSC with base you can protect the primary alcohol as TBDPSC ether once that is protected now you have to regenerate the double bond and the alcohol that is easily done with zinc dust and acetic acid okay zinc dust and acetic acid will release the double bond as well as CH2OH. So basically originally he was expecting this reaction to be very clean in one step he should be able to selectively protect one of the primary alcohols but he could not so he needed 3 steps Iodo ethyrification reaction followed by protection of the other primary alcohol then releasing of the primary alcohol as well as double bond by treating with zinc dust and acetic acid okay. So now he has differentiated the 2 CH2OH what is the next step now you have to homologate the other alcohol first you convert that alcohol into a good leaving group so first is easy to may convert into bromide then you need to homologate so simple essential displacement reaction with sodium cyanide in the presence of more polar solvent DMSO you could get the cyanide so you homologated that CH2 to CH2 cyanide now cyanide can be easily reduced with the dye ball to get aldehyde then that aldehyde upon Wordsworth, Horner, Imans, Wittig reaction you get the dye in okay so that aldehyde now is converted into the dye in as we discussed during the heterosynthesis once the dye in is formed then this can undergo DL cell reaction with one of the double bonds of the cyclohexane okay. So what you need to do is just heat it so take this compound take with toluene and it undergoes in the molecular DL cell reaction IMDA type 1 to give tetracyclic compound okay this tetracyclic compound now what is required you need to remove the protecting group and whatever functional group you have to do you have to attach removal is very easy you use Tbuff any fluoride source will remove the TBDPS group now you got the CH2OH again like how you have done on the left hand side convert the hydroxyl group into a good leaving group convert the hydroxyl group into a good leaving group that is bromide and then followed by displacement reaction with cyanide you get CH2 CN now you hydrolyze you get indiandric acid it okay simple hydrolysis of cyanide will give you carboxylic acid and that compound is nothing but indiandric acid. So from this common intermediate the first natural product which was isolated from that is indiandric acid has been successfully synthesized okay. Then to go for indiandric acid B so the cyanide if you reduce with di-ball you get the aldehyde then followed by your witty stabilized witty you get the alpha beta unsaturated ester then again hydrolysis of the ester will give you indiandric acid B okay. So from the same intermediate so now Nicholas group would synthesize two natural products one is indiandric acid A the other one is indiandric acid B. Then you have to go for indiandric acid D so the indiandric acid D is slightly different than indiandric acid A and B so what you did the same cyanide which was the original intermediate for synthesis of indiandric acid A and B. Now you keep it as such the cyanide is not attacked but remove the TBDPS group with Tbaf you get the alcohol then you hydrolyze the cyanide hydrolyze the cyanide to ester with methanol HCl then you convert the CH2OH into cyanide a two step standard protocol converting into CBR4 converting into the corresponding bromide followed by SN2 displacement you get the cyanide now treat with di-ball okay. So when you treat with di-ball there are two possibilities you know one the cyanide can be reduced to aldehyde at the same time ester also can be reduced but between cyanide and ester cyanide is more reactive towards di-ball so you get only the cyanide being reduced to get the aldehyde then again you do the vitic so you get the corresponding di-ing. So what you should do now hydrolyze the ester that will give you indiandric acid D okay if you hydrolyze the ester you will get indiandric acid D and for indiandric acid C what you should do you have to reduce the ester with di-ball to get aldehyde then you do the stabilized vitic reaction to get alpha beta unsaturated ester now if you do the intramolecular DL cell reaction okay this compound can be written like this now if you heat this okay now if you heat this with toluene you will get the intramolecular DL cell reaction product alpha beta and such ester now the ester upon hydrolysis you will get corresponding natural product indiandric acid C. So if you look at the whole sequence what Nikola has cleverly used was Lindler-Catrice hydrogenation of triple bond two triple bonds into cis double bond and during that period a corn rotatory you know 8 pi cyclization followed by this rotatory 6 pi cyclization takes place to give a bicyclic compound where the cyclobutane is already there then from this common intermediate from this common intermediate he could successfully convert the diol into the corresponding OT V DPS and the other one as a living group from that intermediate he could go all the way to complete the total synthesis of indiandric acid A, indiandric acid B, indiandric acid C. So this is one of the classical examples wherein the proposed biogenetic pathway has been successfully you know established and executed in synthetic laboratory by clever design of starting material okay. So I will stop here regarding the synthesis of 4 membered natural products and then we will move to synthesis of 5 membered natural products from tomorrow okay. Thank you.