 So, good morning everyone and today we will talk about one very interesting alkaloid called starosporine as you can see. So, this is the structure of starosporine actually originally they assign the enantiomer of starosporine as the structure of starosporine. It was isolated from streptomyces starosporis by Omura and colleagues in 1977 and if you look at this molecule it is a very interesting combination of 3 important substructures one obviously you can see this is a sugar unit ok. Then you have 2 indoles ok one here one on the left hand side. So, 2 indole mites and then sugar in the bottom then you have an isoindolinone ok. So, these 2 indoles are connected to isoindolinone on the northern hemisphere and sugar portion in the southern hemisphere ok. So, this is a very very interesting and challenging alkaloid and it also showed very interesting biological activities. For example, it showed protein kinase C inhibition activity at very low concentration a nanomolar concentration it showed activity and because of that it was expected that it could be a potential anti proliferative agent. And the first total synthesis of starosporine was reported by Dynashevsky and here the major synthetic challenge when you look at this molecule was the connecting these 2 indoles ok the 2 indoles to the sugar portion ok. So, you have to connect in such a way that you have to get only the required enantiomer that has been the major challenge and this all these challenges in addition to this were overcome by Dynashevsky in his total synthesis and his retro synthesis actually started by introducing a carbonyl group here ok. So, later he thought it can be selectively reduced and when they reduce it may be possible to make both the starosporine as well as the other epimer and he also connected this oxygen and nitrogen by a carbonyl group. So, it is cyclic carbamate. So, this is a precursor while making starosporine in the total synthesis of Dynashevsky that was further disconnected as you can see clearly what he has done is he has removed this Cn bond ok. So, this can be made using iodo. So, once you make iodonium ion here ok on the double bond then the nitrogen can open the iodonium ion to form a Cn bond as well as the CH2i that i can be easily removed with tributyl tin hydride and this as you can see you have a hydroxyl group that is because why that hydroxyl group is required you start with glucose ok. So, that is why the hydroxyl group is there and also you can see here now the 2 indole portions can be clearly seen and this can be cyclized through a combination of electrocyclic reaction followed by aromatization ok and this can be easily made from these 2 fragments one is from bisindole mighty the other one is derived from the sugar mighty basically it is derived from glucose ok. Now let us see how he started his synthesis and how overall he accomplished the total synthesis of starosporine. He started with this di bromomiliumide ok and then he protected this nitrogen as n-bomb ok by treating with benzyloxymethyl chloride and base then he did 2 successive 1, 4 addition followed by elimination first with this 3 indolile magnesium bromide. So, the addition of 3 indolile magnesium bromide took place in a 1, 4 fashion while coming back the bromide will go then that NH was protected as SEM ok by treating with sodium hydride and SEM chloride. Now the second 1, 4 addition and elimination was done with another indole 3 magnesium bromide. So, now you can see the 2 indole moieties are attached to the meliumide ok. So, what needs to be done now you have to add the sugar unit in the southern side. So, before that you have to prepare the sugar unit. So, he started with L-glucose opposite to the one which is commercially available. So, the L-glucose was treated with carbonyl diamethazole to form this cyclic thiocarbonate ok. So, there is a 1, 2 diol and then when you treat with CDI that is carbonyl diamethazole it can form the cyclic thiocarbonate. The reason for making this cyclic thiocarbonate is to remove these 2 hydroxyl groups to introduce a double bond. So, that was easily done when you treat with a trimethyl phosphate. So, the trimethyl phosphate or triethyl phosphate is known to do this job. So, you could easily introduce the double bond ok. So, this is called L-glucol ok that was a starting material in fact ok. The L-glucol once you have the primary alcohol can be easily selectively protected as tip seether by treating with sodium iodide and tips chloride. Then you have the 2 secondary hydroxyl groups. This upon deprotonation with sodium iodide followed by addition of trichloroacetamide trial it forms trichloroacetamide ok. Then you treat it with BF3 etherates. So, when you treat this with BF3 etherate what happens you can see the Lewis acid can coordinate with this nitrogen then the lone pair on the oxygen of L-glucol can come and then expel this trichloroacetamide. So, that will give you this intermediate ok. Now, this nitrogen can utilize the positive charge on the oxygen. So, you get a phi Hombard right ok and it also comes from the same beta side. Now, if you hydrolyze this just if you hydrolyze this you will get hydroxyl group and this side the amide ok. Then the next step is you simply treat with base ok. If you treat with bases like sodium iodide what will happen it will form anion here that will intramolecularly attack the carbonyl group and CCl3- which is a good leaving group. So, that will lead to the formation of the carbamate ok. Once the cyclic carbamate is there then this NH is protected as bound. So, by treating with sodium hydrate and benzyloxy methylchloride ok. Now, the primary hydroxyl group which was protected as tips can be released. So, that was done easily by treating with fluoride source like Tbuff. Then you protect the same primary alcohol with a sturdy protecting group like paramethoxybenzyl group ok. Then you have to epoxidize this double bond ok. This was done with dimethyl dioxidane. So, when they did they got a mixture of these two epoxides where the required one that is A was the major isomer. So, he took both the epoxide ok and then treated with the bisyndole which he has already synthesized. So, now you take this compound and then treat with sodium hydride. So, it forms the N minus and that opens the epoxide. So, when it opens the epoxide you get this as the major product. So, I have written the sugar unit with the standard chair like conformation ok. So, once you have this the next step is to remove the hydroxyl group ok. So, now once you formed the Cn bond using this epoxide opening next the hydroxyl group is normally removed via xanthate formation. So, it was treated with thioposgene to get the half ester and this further addition of pentafluorofenol gives this intermediate ok. This is a treatment with tributylinhydride and it undergoes elimination of the whole group to give a CH2 at that carbon basically you have removed the hydroxyl group there ok. Now, what is the next step? You have to connect the second indole ring with this sugar unit ok. So, the first step is to remove the PMB ok. So, because basically you want to convert that into CH2i. So, first remove the PMB group with DDQ to get the primary alcohol. So, once you have the primary alcohol then remove the SEM using Tbuff to get the NH then you treat with photochemical condition ok. Then you subject this compound under photochemical condition in the presence of catalytic iodine. Why catalytic iodine and aerobic condition? This reaction should be done because once you do this electrocyclic ring closure of this hexatriene you get a cyclohexadiene ok. This was oxidized with catalytic amount of iodine under aerobic condition to get this aromatic ring ok. So, now you can see the top portion is done ok. You attached this indole with millimide. Now, you have to connect these two carbons. So, these two atoms nitrogen and carbon. So, how we did? So, you converted the primary alcohol to CH2i then treated with base. So, the base underwent elimination of HI to form the corresponding enol ether ok. So, once you have this enol ether, it is a treatment with iodine and bases like potassium tertributoxide. First, iodonium ion will form then this N- will attack and open the iodonium ion to give the CH2i. At the same time you can see the other indole also is attached to the sugar units. So, what is left? Now, the iodide should be removed to have the methyl group. So, that was done with tributyltin hydride. So, you can see this is the structure which you would have seen in star sporing. Now, remove the iodide using tributyltin hydride to get the methyl group ok. So, now what is to be done? You have to remove this carbonyl and also remove one of this carbonyl and clear the bomb group ok. So, these are the things he has to do to complete the total synthesis of star spore ok. First, he has tried the hydrogen aliases. So, hydrogen aliases will remove both bomb group ok. So, to get this carbomate as well as the melimide, then you selectively protect the carbomate nitrogen as N-bock. Then you do protect the melimide nitrogen as N-bock by treating with sodium hydride and benzyloxymethyl chloride ok. Now, you have to cleave or remove this carbonyl group. So, that can be done by treating with CCM carbonate and methanol. So, you get an amino alcohol and the amine is protected as bock. What is to be done? The NH should be made as N-H-Me. So, that means the bock should be removed and methylation should be done. So, this was done by treating with sodium hydride and dimethyl sulfide to introduce the methyl group. After introducing the methyl group, then hydrogen aliases removed the bomb group. And if you treat with TFA, trifluoracetic acid is known to remove the tertiary-butyl oxycarbonyl group that is the bock group. So, the bock group has removed and you have the NH-Me and OM-Me. So, only one thing is left that is selectively this carbonyl should be converted into CH2. To complete the total synthesis. So, what he did? So, he reduced with sodium borohydride. So, it was known earlier similar compounds when you reduce sodium borohydride, one of the carbonyl groups can be reduced to corresponding alcohol, alcohol in the sense CHOH which we can call it as amidol ok. But what he got was he got a mixture of both carbonyls get decreased in one is to one ratio, there is no selectivity. Nevertheless, he moved ahead and further treatment with phenyl selenine in the presence of catalytic amount of toluene sulfonic acid, para-toluene sulfonic acid, he could cleave this COH bond and replace with hydrogen. So, that means he got a mixture of the naturally occurring starosporine plus its other isomer in one is to one ratio. So, this is how he completed the total synthesis of starosporine and as I said he started with L-glucol for the glucoside fragment and the top fragment he started with dibromo N-protected melimide. Then he attached the two indole units through one-four addition followed by elimination. Then the next key reaction which he used was the electrocyclic ring closure followed by aromatization to construct the central aromatic ring. Overall, he took about 28 steps starting from L-glucol and yield was close to 0.5 percent. Nevertheless, this was one of the very good total synthesis on an alkaloid which is quite complex. However, as the number of steps were more there are many groups which were interested in the total synthesis of starosporine to improve as well as reduce the number of steps. So, one such report came from John Boott's group that was reported a year after Dainishevsky's report on the total synthesis of starosporine and here he took only 13 steps and he used few interesting domino reactions. So, his retro synthesis was basically the first retro synthesis was to remove the N-edge. So, that can be introduced by reductive ammunition and the second retro synthesis was the ring expansion reaction. That is, you start with the 5-ambered that is the sugar unit you start with the 5-ambered then you do a ring expansion to get the 6-ambered ring and this can be obtained by an acid-catalyzed cyclization of bisindole with this 5-ambered compound. And here this 5-ambered compound that 5-ambered highly substituted furonose was made from this diosat compound which is made from methyl acetoacetate. So, commercially available methyl acetoacetate which upon diosat transfer you get this compound and that is the starting material for making this 5-ambered and this was made from this glycinester. So, let us see how he has done the total synthesis. First the substituted furonose potion was prepared from this diosat compound on treatment with this chiral allylic alcohol and dirodium tetracity followed by treatment with Lewis acid like BF33. So, what happens as you know first it forms a rhodium carbonate and since you have allylic alcohol chiral allylic alcohol. So, that immediately attacks the rhodium carbonate carbon. So, it forms a negative charge. So, that undergoes you know a rearrangement to give this. So, I will leave it for few seconds to understand this mechanism. So, he could get this in 2 steps and yield is quite high 77% for such interesting rearrangement. Next he wasanalyzed the internal double bond to get aldehyde followed by treatment with para toluene sulphonic acid and methanol. So, what happens this is protonated then methanol attacks this carbon and this attacks this carbon. So, that is how he got this furonose compound. So, the furonose compound is easy then the other component that is bisyndolmite he started from the protected glycine and then treated with mono ester of malonic acid. So, he coupled this NH with this carboxylic acid using dicyclohexane carbodimide. So, once this amide was formed his next step was to treat with sodium ethoxide and ethanol. So, sodium ethoxide methanol will generate an anion here that will add to intramolecularly to the ester and this OET will come out okay. So, this is what you get after this sodium ethoxide ethanol treatment. Now, if you treat with acetonitrile and reflex it if you look at this ester. So, this is nothing but beta keto ester okay. So, that undergoes decarboxylation. So, that will give you a carbonyl group here and this enol. Then treatment with mesyl azide and base one can introduce this diosogru. So, standard method if you have a keto ester or keto amide or 1, 3 diketone you can introduce the diosogru by treating with tosyl or mesyl azide in the presence of base. So, once you have this take this compound and then treat with bisindo okay in the presence of diodium tetra acetate. So, that gives first you know from here it goes and then attacks here and then it forms an enol then that undergoes the electrocyclicization reaction and followed by elimination gives the corresponding that northern hemisphere bisindole moiety okay. So, this bisindole moiety is ready and the other side the glycoside that is the furonous portion also is ready now he has to connect this bisindole with this furonous moiety okay. So, what he did? He took both and then treated with camphor sulfonic acid. What he got was the five ombre ring once you treat with camphor sulfonic acid it protonates and then it comes out. So, that leads to the formation of auxonium ion okay. Now, since you have another lone pair on the oxygen at adjacent carbon. So, this can open so that will give you a methyl ketone and then aldehyde the nitrogen intermolecularly attacks to give this compound in the ratio 2 is to 1. Why 2 is to 1? You look at this carefully okay because both indoles can attack okay that is why you get this mixture. So, you take this compound and treat with again reflects it. So, then nitrogen can attack the carbonyl group okay nitrogen can attack the carbonyl group that OH can attack. So, you get the five ombre ring okay. So, again in the same ratio 2 is to 1. So, this is what you need if you look at this structure. So, this is what you need for the total synthesis of starosporic. So, you take this compound then treat with lithium borohydrate. So, lithium borohydrate selectively reduces ester in the presence of lactam. So, you get primary alcohol without touching the lactam. Then you take this primary alcohol and carry out morphotoxidation. So, morphotoxidation is nothing but you take an alcohol and treat with DMSO and DCC. So, you oxidize the primary alcohol to aldehyde then you treat with BF3 ethyrate. So, BF3 ethyrate it triggers a rearrangement. What happens now the lone pair comes like this and this bond migrates okay. So, that is more nucleophilic is not it that bond migrates when that bond migrates you get this 6-mumbered and this hydroxyl originally here it becomes ketone that aldehyde becomes lactam. This highly stereo selective rearrangement gives the 6-mumbered that is the pyronose ring okay. So, what is to be done now you have to reduce the ketone not simply reduce the ketone you have to treat with nitrogen and then reduce it that means you have to do a reductive amination okay. So, that was done by treating with hydroxyl amine first you get aldehyde then the aldehyde was methylated then he did hydrogenation with Adams capitalist. So, you got correctly you got directly the amine then that amine was treated with this particular reagent here what happens this NH2 attacks this carbonyl and acetate goes out. So, basically you get NHCHO the NHCHO can be in situ reduced with borane dimethyl sulfide to form NHCH3 okay. So, that leaves only the removal of this dimethoxybenzyl group once you remove the dimethoxybenzyl group that will give starosporine. So, the dimethoxybenzyl group was removed by treating with trifluoroacetic acid. So, that gave that natural product starosporine. So, if you look at the total synthesis of starosporine reported by John Wood. So, he accomplished it in 13 steps with an overall yield of close to 10 percent which is significantly higher than the first total synthesis reported by Tanisher. And second thing is if you look at this synthesis it involved two times rearrangement first when you did the dirodium tetra acetate catalyzed rearrangement okay. The second rearrangement was when you had hydroxy aldehyde this upon treatment with Lewis acid the furonose ring was converted into the pyronose ring. So, this synthesis involved these two rearrangements as key reaction to complete the total synthesis okay. So, with this we completed the total synthesis of starosporine using two different strategies one reported by Tanisher ski the other one reported by John Wood okay. So, we will see you in the next class okay. Thank you.