 So, good morning everyone, welcome back to the NPTEL lecture series on classics in total synthesis part 1. So, today we will continue our discussion on total synthesis of terpenoids and we will focus on two total synthesis of one complex natural product called gibberlich acid. So, this gibberlich acid as you can see here it is quite complex, it has about 5 rings and several chiral centers and this plant hormone belonging to called gibberlings and it was produced by plants and from synthetic point of view when you look at this molecule it poses several challenges. It has 8 stereocenters you can see here 1, 2, 3, 4, 5 that means all the 5 carbons of the middle 5 membered ring are chiral centers okay then you have one here 6, 7, 8 okay so there are 8 chiral centers in this molecule and more importantly it is not only the chiral centers 8 chiral centers which can create trouble as such this molecule is very sensitive towards acids and bases okay. So, that is another reason why synthesis of this molecule created lot of challenges to the chemist. Nevertheless the first total synthesis of gibberlich acid was reported by another than Nobel laureate E.J. Coray and his synthesis involved deal salt reaction, intramolecular deal salt reaction, iodolactonization, epoxylactonization and radical circulation. So, these are the key reactions which Coray has successfully used in the synthesis of gibberlich acid. So, let us see this retrosynthesis this idea was this particular lactone okay this particular lactone can be made from this intermediate using iodolactonization as the key reaction okay using iodolactonization as the key reaction. You have a carboxylic acid here okay and you have a double bond so one can easily plan iodolactonization I will come to that what is iodolactonization in a couple of minutes. Then he thought these two hydroxyl groups okay can be introduced with the help of double bond. If you have a double bond here then one should be able to introduce two hydroxyl groups which are anti to each other okay and also you can hydrolyze this lactone and oxidize you will get this dicharboxylic acid okay. So, he thought this could be made from this particular intermediate okay now if you look at this intermediate his idea was to use an intramolecular Diels-Alder reaction okay. So, you can see a diene here and a diene of 5 the intramolecular Diels-Alder reaction 4 plus 2 and you also can see the chloride here is not it that can undergo elimination okay to generate one more double bond this double bond okay one more double bond this double bond can be generated after the Diels-Alder reaction followed by elimination okay. Now this can be obtained from the corresponding alcohol okay. So, this is the key intermediate this is the key intermediate in the synthesis of Zeppel like acid proposed by E.J. Kori how this key intermediate can be made when you have a double bond here one can think of using a Wittig reaction is not it. So, by methyl Wittig it should be possible to homologate this Aldehyde to the vinyl group now you have an alpha beta unsaturated Aldehyde okay wherever you see an alpha beta unsaturated Aldehyde one reaction we should come to our mind is Aldehyde reaction okay. So, that means if you have di Aldehyde basically if you have a di Aldehyde like this then this can undergo an Aldehyde followed by dehydration you should get this alpha beta unsaturated Aldehyde. But how do you get this 1, 5 di Aldehyde you can do a cleavage of the double bond you can do a dihyde deoxylation followed by cleavage with sodium paravidate one can get the corresponding 1, 4 di Aldehyde. Next when you look at this molecule carefully you have this bicyclic 3, 2, 1 system bicyclic 3, 2, 1 system and this we wanted to cyclize okay this is through the keto Aldehyde okay one can cyclize to get the corresponding alpha hydroxyl ketone okay then closer look at this molecule one can see cyclohexene one can see cyclohexene when you see cyclohexene in any compound obviously 4 plus 2 cyclohexene reaction should come to your mind is not it. So you thought you can use an intermolecular 4 plus 2 cyclohexene to get that intermediate okay this is a known compound and this you thought you can prepare from this O Allyl that is 2 ortho O Allyl anisol which can be prepared from Coacol which is commercially available okay. So this is the first retrosynthesis and you also propose another route for the key synthesis of the key intermediate A and that you thought you can make it from intermediate B by 2 alkylation okay one alkylation will lead to the CH2OH the other alkylation will give an aldehyde which can be homologated. So this ketone can be obtained by a radical like reaction here and hydroboration and oxidation regioselective hydroboration and oxidation should generate the keto and this double bond you can see here this could be obtained by a 3-3 sigmatropic rearrangement okay I will come to that when I talk about the real synthesis and this intermediate can be obtained by an intermolecular DL sol reaction okay by an intermolecular DL sol reaction you can see if you cleave this this is the diene of fire and this is the diene okay that diene can be obtained from cycloventa diene. So these are the two pathways Coray had proposed to synthesize jibbellic acid and let us see how he executed the total synthesis of jibbellic acid. Before that as I said this synthesis as well as the next synthesis of jibbellic acid by MRO both involve iodolactinization as the key step. So what is iodolactinization? So if you have a carboxylic acid if you have a carboxylic acid and a double bond in the same molecule at appropriate place okay then when you treat with iodine the double bond first it will form the corresponding iodonium ion okay. Now if you can deprotonate this hydrogen of carboxylic acid using mild bases like sodium bicarbonate okay then the COO- which form immediately can open this to give iodolactin. Obviously the 5-ambered lactone is preferred okay and this goes through what we call it as 5-exotec cyclization okay. This is another example you can see there are two double bonds one here and one here and if it cyclizes here it will form a 5-ambered ring and if it cyclizes this side it will form 4-ambered ring okay. So obviously 5-ambered is more favored so you get 5-ambered iodolactone likewise one can also do if you have epoxy see for example the same thing if you use MCPBA and base like sodium bicarbonate. What will MCPBA do? MCPBA will epoxidize the double bond then sodium bicarbonate will generate carboxylate that can open up okay you can get intermediate like this okay that will open up and you will get a 5-ambered lactone with CH2OH okay. So iodolactinization you can do epoxy lactonization you can do in fact one can also use phenyl selenyl chloride if you use phenyl selenyl chloride you can call this as phenyl selenolactonization all are possible okay. What you need is a double bond and carboxylic acid which are separated by two carbons okay so that you can get a 5-ambered ring. Now let us see how he synthesized the tricyclic intermediate. So he started with glyacol so that is this compound okay OIME and OH and one can use sodium hydroxide and allyl bromide you allylate the free phenolic hydroxyl then you heat it so you heat it at 230 degree Celsius it undergoes glycentry arrangement okay as you know allyl phenyl ethers undergo glycentry arrangement to give the corresponding allyl migrated product okay. You protect this hydroxyl as mem ether now you can do a dihydroxylation okay and followed by ferrovalid cleavage you get diol and cleave it you get the di aldehyde that you that you reduce it you get the corresponding CH2 CH2 oxides if you look at the dienophile required for the dielsol reaction you need a two carbon unit here so that is done. Now you can protect that hydroxyl as benzyl so you deprotonate that hydroxyl group with sodium hydride and quench with benzyl bromide you get the corresponding benzyl ether then comes the removal of mem group so that you can do with trifluoresetic acid and followed by oxidation of this phenol to the benzoquinol okay. So that can be done with this catalyst so oxygen as the oxidizing agent and with this cobalt catalyst one can oxidize this phenol to corresponding benzoquinol okay. Once you have this then you do the dielsol reaction so this is a known compound that is a diene and this substituted benzoquinol is a dienophile and then do the dielsol reaction you get the corresponding bicyclic compound okay. So now how many chiral centers are fixed of course it is resemic so 1, 2, 3, 3 stereo centers are fixed then protect the free hydroxyl group as THB ether. So dihydropyrin and treat with para toluene sulphonic acid you protect the primary hydroxyl as THB ether. Now before we move further I will just briefly explain how the dielsol reaction gave these 3 stereo centers okay, stereo selectively okay. So this is the transformation and as you know dielsol reaction gives mainly the kinetic endo product as a major product. So you can draw like this transition states you can see in this case you keep the diene like this and the dienophile comes from the bottom and this double bond is below the diene. If it is away then it is XO this is below so that is why you get the endo product because of the secondary orbital interaction. Now after the dielsol reaction can you draw this structure just to connect these 2 bonds okay and here also between these 2 double bonds okay that is double bond A and double bond B only double bond A acts as dienophile and double bond B does not because the double bond A is more electron deficient than double bond B okay the double bond A is more electron deficient than double bond B because of the presence of electron donating methoxy group okay. So this can be rewritten like this okay I will keep it for 30 seconds so that you know you should be able to draw the conformation properly so that you will arrive at this particular structure okay. So it is like this now if you do it then you will know this will be beta and this hydrogen also will be beta so both are similar and here this hydrogen you can see that also will be beta that means this CH2OH will be alpha okay done. Now you have made the bicyclic count so what is next you have to reduce these 2 and before that you have to protect the primary hydroxyl so as I said the primary hydroxyl group was protected as THP ether by treating with dihydropyrin and PPTS. Then you can reduce this ketone selectively over the other ketone because of the presence of lone pair of methoxy group this carbonyl is more electrophilic so that it can attract hydride faster than the other carbonyl so one can easily reduce and also the structure is like this isn't it so the hydride will come from the top that will lead to alpha alcohol okay. Then protect the alcohol as methoxy methyl chloride that is CH3 OH CH2Cl then reduce the enome at low temperature that is alpha beta and ketone so reduce that at low temperature to get the corresponding allylic alcohol. Now the allylic alcohol if you missile it okay so allylic alcohol if you missile it you get the corresponding mass elated products. Then sodium bicarbonate you know it is a good living group so you can use this lone pair on the methoxy group to eliminate that and in the process what you get is corresponding enome okay so this could have been easily done on the alcohol with acid but as you know this THP and MOM both protecting groups are sensitive to acid that is why he has to convert that allylic alcohol to missile it and then use sodium bicarbonate that is base mediated hydrolysis to get the corresponding cyclohexene. Then you have isolated double bond then conjugated double bond conjugated double bond is selectively reduced under rhodium catalyzed condition. Now if you use lithium and ammonia okay so lithium ammonia is known to remove benzyl group okay so the benzyl group is removed but at the same time you have a ketone also isn't it when you have ketone lithium ammonia also will donate electron to the carbonyl group so as a result the carbonyl group will become ketyl radical okay then that will eliminate this OM group and you will form enolate you will form an enolate that will become ketone then that ketone also will be further reduced so what lithium ammonia does here is 3 things one it removes the benzyl group you get the corresponding OH then it adds one electron to the ketone first so that removes the MOM group and in the process it forms enolate and enolate becomes ketone and then ketone is further reduced to corresponding hydroxyl group and oxidation of the diol with chromium trioxide pyridine you get the corresponding keto aldehyde okay. Now you carry out this McMory coupling so the McMory coupling on this keto aldehyde gives the third phymometry which is phymometry okay now this is tertiary alcohol this is secondary alcohol okay so secondary alcohol is oxidized under modified swan condition to get the alpha hydroxy ketone and again this bridge head hydroxyl was protected as mem ether okay so now this whole side is taken care so what we need is modification on the left hand side okay. So you have a double bond if you treat with asthmatic peroxide in the presence of stoichiometric amount of NmO you get a diol and that diol if you cleave with let it acetate you get the diol degree okay so as we have noticed in that transition this is so this is one of the key precursor for making the key intermediate A so once you have this then one can carry out an intramolecular aldol reaction with bases like dibenzyl ammonium trifluoro acetate so that gives the corresponding alpha beta unsaturated aldehyde. So once you have that alpha beta unsaturated aldehyde simple vitic reaction will give the corresponding double bond so now you have the diene which is ready for undergoing intramolecular diol salt reaction okay. So what we need is you need to remove this protecting group and attach the dienophile for the intramolecular diol salt reaction okay once you have the THP ether treat with acetic acid you remove the THP and you get the key intermediate A. The same intermediate is prepared by two more methods the second method is started with four benzyloxy cyclohexanone four benzyloxy cyclohexanone then that is worth Iman's modification so he did this vitic modified reaction he got the alpha beta unsaturated ketone and then vinyl 1, 4 addition to this enone you get gamma delta unsaturated ketone the same thing one can also think about using a Klaisen rearrangement to get gamma delta unsaturated ketone or aldehydes we will come to that when we talk about Klaisen rearrangement then the double bond is cleave double bond is cleave with osmium tetroxide and sodium perovideate so this upon aldol reaction this upon aldol reaction you can see 1, 2, 3, 4, 5 so it will form a 5 umber ring okay so now you have a spiral system okay 6 umbered 5 umbered both are fused through spiral fusion then one can think about adding a 2 carbon unit so that is what he did a vinyl magnesium bromide along with cupra psiodate it underwent a 1, 4 addition and then with triethyl orthoformate and ethylene glycol he protected the ketone as the ketol then he did a selective hydroboration okay on the double bond to get the primary alcohol that alcohol was mesylated to get the primary mesylate so now what he needs is you have to remove the benzyl group oxidize the alcohol to ketone then connect this okay so the benzyl group was hydrogenalysif to get the alcohol okay then PCC oxidation gave the ketone then treat with potassium terributoxide to give the corresponding tricyclic compound if you look at this tricyclic compound you will see there is 1 carbon extra okay what you need is what you need is this isn't it so this particular ring instead of 2 carbons you have 3 carbons am I correct instead of 2 carbons we have 3 carbons so the 2 carbons how you can get it from 3 carbons so you do treat with methyl lithium then dehydrate to get the corresponding alkene okay then what you will do you can do a O's analysis to get the corresponding ketone okay now if you treat with base there are 2 places it can generate anion one here another one here if this adds here you will get the 5 ombre ring whereas if B adds it will give 7 ombre ring okay but with sodium hydroxide and ethanol one could get predominantly the 5 ombre ring and once you have this aldol okay then you can carry out a Bayer-Villegar oxidation okay so the Bayer-Villegar oxidation can see the acetyl group COCA3 will become OAC okay then hydrolysis of OAC will give the bridged hydroxyl and oxidation under one condition to get the ketone and remove the ketol okay before that of course you have to protect the bridged hydroxyl as memmeter then do the vitic on the ketone and remove the ketol you get tricyclic intermediate D which has already been converted into intermediate A okay so how did you do that now if you treat with sodium hydroxide and quench with ethyl formates you can introduce an aldehyde on this side okay now this on treatment with potassium tertiary butoxide and methyl iodide so this is 1 3 carbonyl isn't it this is 1 3 carbonyl so it will exist in the corresponding enol so that enol become the enol methyl ether when you treat with potassium tertiary butoxide and methyl iodide okay then you introduce another aldehyde on the other side of the carbonate okay again you follow the same route sodium iodide ethyl formate introduce aldehyde on the other side now you have to reduce this ketone and as well as this aldehyde so this is done with redol so what is redol redol is sodium sodium bis okay there are two sodium bis metoxy etoxy aluminum hydrate metoxy etoxy aluminum hydrate so that reduces both ketone and aldehyde to get the diol now if you treat with acid so what will happen this will become protonation will take place at this voltage it will become H2O plus then this lone pair will push this double bond here and in the end what you get is alpha beta unsaturated aldehyde which upon vitic you will get the corresponding triene okay then the third route which is supposed to be the shortest route and here you use Diels-All reaction and 3 3 sigmatropic rearrangement as the key reaction for the Diels-All reaction the diene was prepared in a single step so butyl lithium treatment gives cyclopenta dienyl anion and quench with 2 bromo allyl bromide followed by DBU treatment gives a mixture of these two isomers gives a mixture of these two isomers this upon treatment with this dienophile okay between these two electron withdrawing group one is acetyl and other one is ester which is more electron withdrawing acetyl or ester acetyl group so that means the acetyl group will go to end opposition okay so this upon Diels-All reaction with this particular diene gives the Diels-All product 53% then there is another product very interesting product in 20% so what is this product how it was formed this again was formed by Diels-All reaction only difference is here this is formed by normal Diels-All reaction and this is formed by hetero Diels-All reaction so what is hetero Diels-All reaction now the dienophile okay the dienophile acted like a diene okay the dienophile acted like a diene since one of the alkene is a carbonyl group you can call this as hetero diene okay hetero diene and your original diene now acted as dienophile okay so that intermolecular hetero Diels-All reaction gave B in 20% yield so the major product which is obtained by normal 4 plus 2 cycloaddition this upon treatment with TMSO tripleton base formed enol TMSC3 this molecule when you look at it you can see I have given already the number 1 2 3 1 2 3 so that means this can undergo a 3 3 sigmatropic rearrangement okay this will undergo a 3 3 sigmatropic rearrangement to give this interval I leave this for few seconds so that you can visualize okay the same intermediate can be redrawn like this the same intermediate can be redrawn like this okay this is the 5 1 but ring and you can see the 5 1 but ring and this is a 6 1 but ring and you see the 6 1 but ring okay and you have the 2 bromalyl group so that we have and you have CH2 next CH2 then enol TMS then ester ester okay so this undergoes a 3 3 sigmatropic rearrangement to give this material okay now if you do sodium chloride DMSO what happens this is nothing but a protected form of beta keto ester protected form of beta keto ester okay so the beta keto ester can undergo elimination under sodium chloride and dimethyl sulfoxide reflex incantation okay so that gave corresponding ketone then one can use 9 BBN so that regioselective hydromeration can take place to get alcohol and that can be oxidized with PDC to get a di ketone then you carry out a dibutyl copper reaction so it exchanges with this and undergoes intramolecular nucleophilic addition to the carbonyl to form the 5 1 but ring now if you protect the hydroxyl group as memmether so that gives the intermediate P okay so what I will do I will stop here now then tomorrow I will continue our discussion on the total synthesis of gibralic acid reported by Coray starting from the intermediate A how he accomplished the total synthesis of gibralic acid and I also will discuss one more total synthesis of gibralic acid reported by amada okay so thank you