 Good morning and welcome back to the NPTEL course on Classics in Total Synthesis Part 1. So we have been talking about total synthesis of alkaloids, we will continue our discussion on one more alkaloid today that is dendrobin and this is a very interesting tetracyclic compound. You can see that there are 3 5-membered rings 1, 2, 3, the third one is a lactone and second one is substituted pyroldine ring and the fourth one is a 6-membered ring. So this molecule was isolated way back in 1930 to almost 90 years ago as the major alkaloid constituent of Chinese ornament orchid called dendrobium nobile ok, it is almost 90 years ago we, this molecule was isolated though the first total synthesis of dendrobene was reported in 1970s. The better synthesis in terms of asymmetric as well as the shortest synthesis reported after 90s. So what we will do today, so we will talk about 3 total synthesis of dendrobene, one supposed to be the shortest synthesis less than 10 steps, the other 2 are asymmetric synthesis ok, I should say they are chiral synthesis, chiral approach, they start with a chiral compound ok, so they are called chiral synthesis and dendrobene exhibits some interesting biological activity, so they show endipiratic and hypotensive activities. So if you look at this molecule from synthetic point of view as I said, so this heterocyclic compound that itself is a big challenge and more importantly if you look at the number of chiral centres, you can count 1, 2, 3, 4, 5, 6, 7, there are 7 chiral centres of which 3, 1, 2, 3, 4, 5, 6, 6 contiguous chiral centres that means the 6 membered ring, the 6 membered ring has all the carbons chiral ok, all the 6 carbon atoms of the 6 membered ring are chiral ok. First let us start with simplest and then the best synthesis so far for dendrobene, so this is the synthesis reported by Thomas Livinghouse, he reported in 8 steps starting from 2 methyl cyclopentenone ok and the key reaction, there are couple of key reactions and the best one is the silver ion mediated cyclocondensation of an isocyanide with an acid chloride ok, cyclocondensation of an isocyanide and acyl chloride to generate the dihydro pyrrolidine ring, you have the 5 membered pyrrolidine ring to generate that ring he has used this key reaction. So from a retrosynthetic point of view if you look at this compound his first retrosynthesis was this cleavage of this lactone ok, so one side he has ester, the other side he has ketone, if you ready moment you reduce this double bond and reduce this ketone automatically the hydroxyl cyclase and then you get the corresponding 5 membered lactone. So that was the first major disconnection, then the second major disconnection was the double bond, the double bond which he has introduced. Here what he wanted to do was he wanted to use a ketone radical cyclization. So if you put a metal which can donate an electron then it will form ketone radical that radical can undergo a 1, 4 addition ok, the radical can undergo 1, 4 addition followed by dehydration one can generate not only the 6 membered ring but also the hydroxyl group which can be dehydrated. So that was his plan. Then as you can see the imine ok, imine can be methylated and hydrogenated ok to get this 5 membered ring and this is the key transformation which he thought will be really very nice if it can be exploited in total synthesis. The cyclo-contentation reaction where you have an isocyanide which on treatment with acid chloride forms this cyclic imine ok. Let us see how we made it and then before we move further on the total synthesis of dendroben from living house group, first let us see how he prepared the two starting materials one based on the cyclopentenone, other one acid chloride ok. So he started with two methyl cyclopentenone and then he did a 1, 4 addition with lithium LiCH2Nc ok. So he got a mixture of 1, 4 addition and 1, 2 addition product though this 1, 2 addition product was minor he thought it is better to isolate. So he quenched this with TBDMS chloride. So when he quenched with TBDMS chloride he could get exclusively only one product and the other starting material that is acid chloride he started from ethylacetoacetate. So where first he did the alkylation with isopropyl bromide after treating with sodium ethoxide. So you generate an anion and quench with isopropyl bromide to get this isopropyl. Then he did a quasi-Fewersky rearrangement so what he did he first he did bromination so a dibromination gave this dibromo compound this upon treatment with base potassium hydroxide. The first step is Fewersky rearrangement so which gives the cyclo propenone ok. Now the second step is the base which attacks the carbonyl and this bond migrates and the bromide goes out and in the process the ester also gets hydrolyzed to get this dicharboxylic acid ok. This dicharboxylic acid if you treat with thionyl chloride you get the corresponding acid diacid chloride. Now when you add one equivalent of methanol the least hindered one least terrically hindered one is esterified leaving the other acid chloride as such which can undergo cyclo condensation with the isocyanate which we discussed in the last slide. So the key reaction as I said it is a cyclo condensation how does it work? So if you have an isocyanate like this ok then this C minus can attack the carbonyl group of acid chloride and then you get this acylinium ion it is like you know iminium ion instead of iminium ion you have a triple bond. Now if you have a double bond at appropriate place the double bond can neutralize the positive charge on the nitrogen ok. Then if for example if Z is oxygen if Z is oxygen then this is like the higher order of managed reaction is not it? Managed reaction is you have enol and then iminium ion. So here you have a triple bond ok. So it is like managed reaction so then it can cyclize to give the corresponding imin. So that was the key reaction which leaving us used in the totals in the sub-dentropy. Let us see how this was successful. So first when you mix these two in the presence of molecular sieves at slightly elevated temperature than a room temperature the first step as usual was the addition of attack of the nitrile and the C L minus goes out. Now the C L minus which came out can attack this to neutralize the positive charge on the nitrogen. So you get this compound ok. Now what it can do the lone pair on the nitrogen can push the chloride out. So once that happens in the presence of silver tetrafluoroborate you generate this positive charge on the nitrogen. And already you have enol TBDMS so that can come and neutralize the positive charge and in the process that will become keto ok. Now you can see you started with cyclopentenol one ring. Now the second ring is constructed ok the second ring is constructed. So what is left is you have to introduce a methyl group in nitrogen and also you have to reduce the imin. So both are done in one step actually I should say one part reaction first methylation of nitrogen was done with methyl triflate you know methyl triflate is known to methylate and N methylation and followed by in situ reduction of this iminium ion was done with potassium tritersibutaxiborohydrate ok potassium tritersibutaxiborohydrate. The bulky one which is known to reduce iminium ions in the presence of ketones and esters ok. So now 2 5 membered rings are done so what he needs to do is to connect this ok connect this. So he thought he can use summerium iodine summerium iodine is well known one electron donor. So a ketone radical can form that can attack that can attack this double bond in a Michael fashion. So a ketone membered ring can be formed that was his idea but what he got was a very interesting tetracyclic compound ok. So here it is very easy to visualize how he would have got this tetracyclic compound. Now if you look at this particular compound ok you can see this there are 2 Michael acceptors one you have alpha beta unsaturated ester ok alpha beta unsaturated ester other one alpha beta unsaturated ketone ok. So the Michael addition the expected one was to undergo the alpha beta unsaturated ketone ok. It was as per his plan the Michael addition was supposed to happen at alpha beta unsaturated ketone but here if you look at this product the Michael addition took place at alpha beta unsaturated ester ok. And once that happened the hydroxyl group attacked the ester and then it formed the lactone ok. So he thought maybe he has to work around and then increase the temperature change the conditions. So simply by rising the temperature simply by rising the temperature to ambient temperature he got the required product as the major product as you can see. So this is the major product 53 percent and the simply reduced compound that alpha beta unsaturated ester was reduced so that he got 5 percent. So both are easily separable so he took the required compound and then treated with thionyl chloride. So thionyl chloride is well known for dehydration so it introduced a double bond. So now you treat with bases like DBU so that the double bond can be migrated here. So he wanted to migrate the double bond here but what happened the double bond migrated all the way to here the tetrasubstitutile bond ok no problem. Next he reduced the Adam's catalyst so the tetrasubstitutile bond was reduced to give the cis substituted compound and you need this isopropyl group in the natural product alpha ok what you got is beta. So you need alpha and you have a ketone adjacent to that so one should be easily isomerized or epimerized this stereo center. So he treatment with acetic acid could epimerize that center to get the isopropyl at the required stereo position. Now reduction with sodium borohydride came from and the hydride came from alpha so you got the beta alcohol and then the beta alcohol spontaneously cyclized to give the natural product depropyl. So overall including the starting material preparation Thomas Levinghouse took about 10 longest linear steps to complete the totals synthesis of depropyl. So this is a resemic synthesis but the shortest synthesis involve a clever cyclo condensation reaction. So now we will move to the next synthesis it is a Kairon approach synthesis reported by Shah and here he started with Kairon starting material called Carvo ok and the key reaction in this totals synthesis was generating a radical generating a radical next to carbonyl group because usually when you generate a radical next to carbonyl group that is not very reactive ok. So he could successfully generate a radical next to the carbonyl group and then he carried out a phi exo dig cyclization to get this phymombard ring ok this phymombard ring he could achieve the synthesis of this phymombard ring using phi exo dig radical cyclization. So let us see how he has done. So the first retrosynthesis was to cleave this ok he wanted to introduce the nitrogen at a later state ok and he thought this will be the precursor for that and why this precursor that is where his key reaction that is the radical cyclization. So once you generate a radical here if you replace an iodide then this can undergo radical cyclization to generate the phymombard ring. So that was the idea behind this precursor ok. Now this can be easily obtained by a 1, 4 addition. So you can have the whole unit you can add in a 1, 4 fashion followed by quenching the enolate to get the iodide at alpha position alpha to carbonyl and this can be made from carbon which is commercially available it is one of the monotrapines available in plenty and not expensive. So let us see how he did the total synthesis. He started from as I said carbon now the selectively this disubstituted double bond and electron rich double bond can be reduced with either Wilkinson catalyst or with Adams catalyst to get the reduced carbon. So now you have the enome and this enome it went through a very interesting reaction wherein the first step when you treat with methyl magnesium chloride and catalytic amount of ferric chloride it undergoes it forms a dienolate ok the dienolate that is a thermodynamic dienolate and quench with TMS chloride he got this carbon this is a known reaction ok ok. So this is the first step when you take methyl magnesium chloride and catalytic amount of ferric chloride and quench with TMS chloride you get this. This upon treatment with Lewis acid and trimethyl orthoformate to give this carbon ok. So when this happens the whole group comes opposite to this isopropyl ok. So this is how we introduce now if you look at this carefully this is the equivalent of the ester carbonyl group in dendrobene you need carbonyl group. So this is the equivalent of ester carbonyl group ok then you need to introduce a hydroxyl group to form lactone is not it you need to introduce a hydroxyl group to form lactone. So this upon treatment with LDA you can generate enolate here and then quench with TMS chloride followed by MCBBA you can introduce a hydroxyl group only thing is it is in the protected form O TMS. Now if you treat with PTSC there is paratolium sulfonic acid it forms this lactol methyl so this gets hydrolyzed and then it cyclizes to form this lactol methyl ether which is quite stable ok. This can be this molecule can be written like this ok this molecule can be written like this. Now if you look at this compound particularly the enone ok particularly the enone you can see the top face is blocked the top face is blocked by the isopropyl group. So any attack on this enone has to come from the alpha face any attack on the enone should come from the alpha face. So because of that when you add this the 4 carbon unit in your 1 4 passion followed by quenching with TMS chloride you can see this 4 carbon unit comes from the less hindered alpha side ok. Now once you have the enol TMS in one step you can convert that into iodine. So that is the precursor required for the 5-exo radical dig cyclization ok. So did it work yes when it was treated with tributyltrinhydrate with radical initiator AABN this underwent a 5-exo dig radical cyclization to get the key tricyclic carbon the key tricyclic compound required for the totals in the certain group. So now what he has to do is finally combine these two to get the 5-o pattern. So the vinyl TMS the TMS group was removed with acetic acid to get the exocyclic double bond. Now hydroboration ok so before doing hydroboration before doing hydroboration this methoxy group that is lactal methyl ether should be converted into lactate ok. So what he did he treated with MCPVA MCPVA in the presence of BF33. So BF33 you know what happens is for example if you take this compound treat with BF33 the lone pair come like this and then it will go. So basically what you will get is like this oxonium ion ok like this oxonium ion you will get ok. Now when you are adding MCPVA what will happen the MCPVA will attack MCPVA will attack and neutralize the positive charge. So that is what happens as you can see here the oxonium ion is formed which is formed in C2 was attacked by MCPVA and MCPVA also comes from the same beta cell ok. Then if you treat with DBU DBU what it will do it will pick up this hydrogen ok it will pick up this hydrogen and as you know the metocolor benzychic acid is a good leaving group. So it will pick up this hydrogen and you get the corresponding lacto. So in 2 steps the lactal methyl ether was converted into the lacto. So then as I said the next step is to introduce the 5 umber ring here. So first step was adding borane dimethyl sulphide complex. So when you do that the double bond will be hydro borate ok the double bond will be hydro borate. Now when the hydroboration takes place you can see the oxygen electron rich oxygen carbonyl oxygen can immediately attack the boron isn't it. When it attacks the boron what you get is this corresponding negative charge on borane and positive charge on oxygen. Now intramolecular transfer of hydride will take place from alpha intramolecular hydride transfer will take place from the alpha to the carbonyl so that you will get beta alcohol ok. This will give the beta alcohol after work up with H2O2 sodium hydroxide ok you get a diol. Now what you need is one of the hydroxyl group you should convert that into nitrogen ok. So how this happens you treat with mesyl chloride. So when you have primary hydroxyl and secondary hydroxyl group obviously you can selectively mesylate the primary one. So you do convert the primary alcohol to corresponding mesylate meanwhile oxidize the secondary alcohol with chromium trioxide to get the ketone ok in 2 steps you get this. Now once you have the mesylate one can easily convert that into corresponding acide. See acide is a precursor for nitrogen ok NH. So the sodium acide displaced mesyl group to get the corresponding acide. Now when you treat this with triphenyl phosphine you know this will undergo intramolecular starting a reaction that means what will happen it forms the immunopasporine and the with ketone what will happen it will undergo immunobetic like ok. So that will give you the corresponding cyclic imine ok that will give you corresponding cyclic imine. Then once you have that then sodium cyanoborahedride under acidic acid condition you can reduce the imine that means you protonate the nitrogen then you reduce with sodium cyanoborahedride. What is left now in the total sense of dendrobene is to methylate the NH. So that is simple if you treat with formaldehyde and formic acid you can easily methylate the NH to get the corresponding N methyl group. So this is the second total synthesis which I discussed about dendrobene so which took about 19 steps and it started with commercially available monotirpin called CARO. The third total synthesis very interesting total synthesis reported by Samir Zahir involved two key reactions one Poisson-Kahn reaction to construct two five number rings the both the five number rings were constructed using Poisson-Kahn reaction and second one is using a radical reaction to open the cyclobutane radical reaction to open the cyclobutane of another commercially available monotirpin called verbenone to get this isopropyl go ok. So he also took about 19 steps to complete the total synthesis but these two key reactions were very very important in the total synthesis of dendrobene ok. So his idea is as I said this is the key reaction that where you use Poisson-Kahn reaction to get the two five number rings and this can be obtained from the corresponding you know cyclic carbamate ok and the cyclic carbamate he got it from this verbenol ok verbenol by a radical cyclic composition ok. So let us see how he has done and you know what is Poisson-Kahn reaction if you have a triple bond and double bond and if you treat with dichobalt octacarb like in the presence of carbon monoxide you can get cyclopentenone. So it is a very well known reaction for making highly substituted cyclopentenones ok and not only cyclopentenone depending on the nature of the substituent the cyclopentenone can be fused with other other things ok. So now he started with verbenol which was obtained from verbenone in one step once you have the verbenol or you can also get verbenol directly from alpha pine in one step. So then you protect this verbenol with carbonyl diimidazone ok. So you get this intermediate. Now if you add methyl hydroxylamine so methyl hydroxylamine is a good nucleophile it attacks and then imidazole is a good living group so you get this. Now the NOH the OH should be converted into a decent living group so that you can generate N radical. So what was done was convert the OH into O-benzoate by treating with benzoic chloride and you get the radical precursor ok. Once you have this radical precursor you treat with radical initiator that is AIBN as well as isobutronitrile and tributyltin hydride you generate the radical and that radical is like this ok. First you generate the radical and this radical adds in a 5 XO trig fashion 5 XO trig fashion to generate the tertiary radical. Once the tertiary radical is generated now the tertiary radical can open up the strained form of the ring. You have strained form of the ring the strained form of the ring can be opened by the formation of this tertiary radical and when it does that what you get is a more stable tertiary radical. So this again it will pick up hydrogen from tributyltin hydride and it will form isopropyl and that is how in one step cleverly ok Zamizard converted this radical precursor to a bicyclic couple ok. Then you can hydrolyse this ok so bicyclic carbonate you hydrolyse to get the corresponding amino alcohol and then NH can be propogylated by treatment with potassium carbonate and propogyl bromide and the free hydroxyl should be protected so that was protected as acetate then he carried out the key reaction that is the Poisson-Corn reaction. So the Poisson-Corn reaction worked very well and as you can see here it gave the key tricyclic compound very nicely and 2 phymomer rings are formed using Poisson-Corn reaction ok. Then the double bond of alpha beta unsaturated ketone the phymomer ring can be reduced under hydrogenation condition then what you have to do if you look at is you need to introduce an ester group you see you need to introduce an ester group. Now if you look at that carbon that carbon unless you introduce a functional group you cannot introduce a carbonyl group. So if you have to introduce a functional group it is important you introduce a double bond here ok then you can do the one for addition for introducing the double bond you have to pick up this hydrogen ok this hydrogen that means you have to generate more substituted enolate is not it the more substituted enolites are generally formed by treatment with trimethylsilyl iodide and exomethyl dislycine ok. Once you make the enol TMS then treat with phenylsynyl chloride you can introduce the selenol group at that carbon. Then simple oxidation with the MCPVA first it will form phenylsynol oxide followed by elimination you introduce the double. Once you have the double bond next is very simple you have to introduce a carbonyl group one carbon carbonyl group and the best reagent one can think of is cyanide. So treatment with diethylaluminium cyanide smoothly underwent one for addition to introduce the cyanide. Now you can see the cyanide and then acetate both are in alpha position both are in the same side. So it should be possible to cyclize to get the corresponding lacto. Nevertheless before you do it you should remove the unwanted keto group you have a keto group that you do not want. So reduce with sodium borohydrate to get the alcohol and once you have alcohol one can do deoxynation using again radical reaction convert this hydroxyl into zandate ok one can convert this into zandate and or you know here you treat with this corresponding pyro acid chloride then treat with tributaryl tin hydrate and AABN you remove the hydroxyl group ok. The next step as I said you have to connect these two to form the lactone and that is done by treatment with sodium methoxide and methanol. So sodium methoxide methanol first it hydrolyze the acetate first it hydrolyze the acetate to get the alcohol but unfortunately it also epimerizes the carbon wearing cyanide ok no problem now if you treat with para toluene sulfonic acid para toluene sulfonic acid the OH will attack the cyanide to form the corresponding amelite. Now since you are adding water and then doing the reaction deoxyne this can get hydrolyzed to give the corresponding lactone ok that is nothing but the natural product tetropion. Again some is that took about 19 steps but remember this involve chiral starting material and also involve two key reactions one is the Poisson-Corn reaction to get two 5 ombre rings and the second one is opening of the cyclobutane of where we known by radical reaction to introduce the isopropyl group required in the with correct stereochemistry ok. So with this we have completed total sentence of two alkaloids and we will discuss few more alkaloids in the next few classes thank you.