 So good morning in the last lecture we talked about total synthesis of morphine by Gates and Larry Waverman's group. So today we will continue our discussion on the total synthesis of morphine by 2 more groups. The first one is from Parker's group and the second one is from James White and in the case of Parker's total synthesis he has used a tandem cyclization where you generate a radical. So that undergoes 6 endo and when it comes back it removes the phenyl thioretical and overall he took about you know 11 steps to complete the total synthesis of morphine from commercially available isomanily. Let us see is retrosynthesis and how he has planned this synthesis using the key radical cyclization reaction. So the first disconnection is this bond where this can be introduced using a hydroamination reaction first you remove the tosyl group followed by intramolecular hydroamination one should be able to introduce this particular bond. As well as you know later you want to have to introduce this double bond for that you have a hydroxyl handle so that can be used to introduce the double bond. Then this is the key reaction the key reaction is Phi XO key radical cyclization reaction you have a bromine so it can generate radical here. So that radical can add first Phi XO then this can add a 6 endo okay 6 endo and to form this radical that radical when it comes back this thio phenyl group will go out. So that is how the tetracyclic core structure of morphine was planned. And this can be obtained by a simple Midsnub reaction on this alcohol this corresponding phenol is the nucleophile okay and further retrosynthesis the fragment A can be obtained from isovaniline in few steps where basically you have to introduce a phenyl group and then homologate the aldehyde the other fragment that is fragment B can be obtained from this epoxide where you know you have to you know if you open this epoxide you will get this allylic alcohol and this can be obtained from metamethoxy phenethylamine using birch reduction and hydrolysis as key steps okay. So what are the key reactions one is birch reduction which we have already discussed the other key reaction which is used is Midsnub reaction. So Midsnub reaction is nothing but another method for making ester you start with the carboxylic acid and then treat with alcohol in the presence of triphenylposphine and diethyl aciodicorboxylate you form the corresponding ester this is the diethyl aciodicorboxylate one can also use diisopropyl diisocorboxylate. So basically what happens in this reaction this diiso will be reduced to get the corresponding hydrogen derivative and the triphenylposphine will be oxidized to triphenylposphine oxide and the overall process of esterification from carboxylic acid and alcohol is after forming the ester you get a water molecule. From the water molecule hydrogen goes to this diethyl isodicorboxylate and the oxygen goes to triphenylposphine oxide basically it is a dehydrating reaction and one can also call it as redox reaction because the reagents which we use for this particular transformation is dead and triphenylposphine the triphenylposphine gets oxidized and dead gets reduced. So during this redox reaction your alcohol and carboxylic acid are coupled to form the corresponding ester the mechanism is first the triphenylposphine attacks the nitrogen of diethyl isodicorboxylate then the other nitrogen picks up hydrogen from carboxylic acid. So you get a carboxylate anion and the positive charge is on the triphenylposphine. So now the other substrate oxygen of the hydroxyl group attacks the triphenylposphine. So you get RO pph3 plus this is the key intermediate in the mechanism for Metsunobi reaction as well as many related reactions. So this R1 the nucleophile can attack the R1 from the backside so that the RO bond can easily break to form triphenylposphine oxide. So that is why this attack of RCO2 minus is an SN2 reaction. So the carboxylate attacks the R1 from the backside and that forms the ester and triphenylposphine oxide. Here are some examples where you can see this is a menthol. So one can do Metsunobi reaction and look at the stereo center it is exactly opposite. Similarly so this is again secondary alcohol and you do Metsunobi reaction and followed by hydrolysis. If you hydrolyse then you get completely inverted hydroxyl group. So how the fragmented A was synthesized by Parker it took isovaniline and then brominated at this carbon using bromine and iron then he did this stabilized vitic to get the corresponding thioenol ether. So that is a fragment A which is used for Metsunobi reaction. For fragment B he started with metamethoxy phenyl ethyl amine and metal ammonia reduction that is birch reduction gave this diene and if you hydrolyse this enol ether you will get the ketone and also the double bond will migrate to give the more it was substituted and conjugated enol. So you get this alpha beta unsaturated ketone and during the process if you use tosyl chloride okay you can protect this amine as NH tosyl then methylite this NH to get N methyl as you know in morphine you need this N methyl okay. Then you have this enone that enone can be reduced using Lucia condition that is sodium borohydrate, cerium chloride you get the allylic alcohol and using the allylic alcohol serious center of course this is relative it is a resemic compound you can direct the epoxidation using the alcohol so you get the epoxide delivered from the same side of the hydroxyl group then you treat with Lewis acid titanium isopropoxide which opens the epoxide to get the corresponding allylic alcohol okay. So then between these two this can be protected selectively using TBDMS triplet so that is fragment B. So once you have fragment A and fragment B then carry out the Midsnopper reaction okay so first step is the Midsnopper reaction for combining the fragment A and fragment B to get this bicyclic compound. So this set the stage for the key phi exo radical cyclization followed by 6 endo and elimination of the cation phenol. So when you treat with ABN TBDH this will happen and before that this bulky TBDMS group can be cleaved using fluoride source to get the corresponding alcohol then you do the key reaction. So that key reaction as I said first it forms the radical that radical undergoes phi exo to give this intermediate and again this radical will further undergo 6 endo to give this radical now this will come back when it comes back you eliminate the phenyl trio radical okay. So now you form 4 rings the last ring is the 6 membered people in ring so that is formed by lithium and tertiary butanol in ammonia so here what happens first the endo cell group gets cleaved and then the hydroamination takes place to introduce the fifth ring. So now you have all the phi rings in correct place so what is to be done is you have to introduce a double bond here and also demethylate. So what is done you do swan oxidation so that will give you dihydrocodinone. So the dihydrocodinone has been already converted okay in the last lecture I talked about this dihydrocodinone has been already converted into morphine so this completes the formal synthesis of morphine by Parker's group. So Parker took about 11 steps to complete the total synthesis of morphine key steps are metronome reaction and tandem radical cyclization and he also started with simple starting material isovenylene which is commercially available and metametoxyphenylamin which also can be easily prepared the overall yield of this is 11.63 percent which is significantly higher than other methods reported as thus far okay. That brings me to the fourth total synthesis of morphine which was reported by James White why I want to discuss this was the earlier synthesis if I look at first one was racemic synthesis and the second and third were asymmetric synthesis but they were they synthesize the naturally occurring morphine whereas James White's group they wanted to synthesize the enantiomer of naturally occurring morphine that is plus morphine okay. So the key step in the total synthesis of James White's group is carbenoid CH insertion okay so that is another clever use of rhodium acetate catalyzed carbenoid CH insertion okay I will come back when I talk about the total synthesis. So the retrosynthetic analysis the first step is the reduction of the ketone to allylic alcohol and that can be obtained from this particular alcohol as you know oxidation and then introduction of double bond you will get this. The second key step is the Beckman rearrangement if you have a ketone and the Beckman rearrangement you can introduce this NH then you can methylate and the second key step is the CH insertion. So the CH insertion of that CH2 bond was done from this starting material. So you can see this one gets inserted at this carbenoid okay. So let us see how it was done when we talk about the total synthesis and here this CO bond was done using intramolecular SN2 like reaction and that can be obtained from this ketoldehyde using a Robinson annulation sequence. So if you look at this cyclohexenome as you know when you have cyclohexenome one reaction which should come to your mind is Robinson annulation sequence. So the Robinson annulation sequence was used to introduce this cyclohexenome and that can be obtained from this particular compound using asymmetric hydrogenation okay as you know Nairobi has developed several methods. So you have one modified version was used to highly stereo selectively reduce this double bond and this can be obtained from manually okay. This is the simple retro synthesis put forward by James Y and let us see how he has successfully accomplished the total synthesis of plus morphine starting from manually. And this is the mechanism for carbenoid CH insertion. So if you have a diazo compound and then if you treat with either copper or dirodeum tetracetate you get this carbenoid and that inserts to any CH bond and you will get like this okay. And there are many examples in the literature one of them is shown here. So here you have a diazo compound either it can insert here or it can insert here and this is the major product and this is the minor product so both are possible okay. Now let us see how White's group started the total synthesis and how they accomplished they started with esomanoline and then you do a Stobey condensation okay the Stobey condensation was done with dimethyl succinate, fluorimethoxide and dimethyl succinate. So you get this particular alpha beta unsaturated ester which is required for asymmetric hydrogenation at the same time you also have the CH2COH which is required for cyclization either at this carbenoid okay. So the asymmetric hydrogenation was done using this catalyst that catalyst is little complex okay so this is what you know the ligand which is used with the chiral the rhodium species so that gives this isomer this is a high enantiomeric excess was obtained and then the next step is the Fertilicab cyclization okay. The expected one was so whether it can undergo cyclization here but unfortunately what happened this cyclized at this curve so not ortho to hydroxyl group but it went to para so it says that the para is more reactive so in order to force this carboxylic acid to cyclize at ortho with respect to hydroxyl group you need to block the para position. So that was simply easily achieved by bromination so when you do the bromination it goes to the para position then you do the Fertilicab's acylation drama molecular Fertilicab's acylation now you can see it underwent it can undergo only at one place okay. So that is how this substituted tetralone must prepare then once you have this you do not need the bromine is not it the bromine served its purpose of forcing the carboxylic acid to undergo Fertilicab's acylation ortho to the hydroxyl. So reductive removal of bromine gave this tetralone the next step is the Robinson annihilation sequence. For doing the Robinson annihilation sequence first you have to hydrolyze this ester to carboxylic acid then you introduce an aldehyde here okay that was done by treatment with potassium hydride and methylformate so you introduce the aldehyde then you do the Robinson annihilation sequence when you do the Robinson annihilation sequence with methyl vinyl ketone so first the 1 4 addition takes place 1 4 addition takes place with methyl vinyl ketone and at the same time the carboxylic acid adds to the aldehyde to form this lactol type okay. Then sodium hydroxide will cyclize here to get the corresponding cyclohexenone at the same time the decarbonylation also takes place the decarbonylation takes place to give this tricyclic compound. So once you have this tricyclic compound so what is to be done you methylate this so esterification with disomethane you get the corresponding methyl ester and one has to be careful because you also have a phenolic hydroxyl group. So both carboxylic acid and phenolic hydroxyl group can be methylated if you treat with disomethane but careful treatment with 1 equivalent of disomethane one can selectively methylate the carboxylic acid and not the phenol. So once that is done then you have to cyclize this isn't it so that is done by treating with bromine firsts but as you know when you treat with bromine bromine also will go here isn't it yeah that is what happened because you brominated both this untreatment with DBU okay you can imagine how untreatment with DBU it will cyclize here it is possible because DBU also can isomerize the double bond. So it goes through this deconjugation double bond was deconjugated then the cyclization can take place you know it is an intramolecular SN2 reaction then followed by again migration of the double bond okay now double bond migrates again you get the tetrasubstituted compound but it is in conjugation with carbonyl group. So that is how it was done so what is next you have to reduce the carbonyl group to corresponding alcohol okay. So allylic alcohol was done then this bromine you do not need okay so reductively remove the bromine under hydrogen analysis condition your 4 rings are ready okay then the 5th ring white as I mentioned has used an intramolecular carbinoid CH insertion as the key reaction isn't it so that means you have to convert or you have to hydrolyze the ester to carboxylic acid convert that into diaso ketone okay. So first you protect this free hydroxyl group okay as mom ether then you hydrolyze the ester to carboxylic acid you have the carboxylic acid it is easy to convert a carboxylic acid into diaso ketone in 2 steps convert this into acid chloride by treating with oxalyl chloride to form the acyl chloride now you treat with diosamethane so you get the corresponding diaso ketone. So once you have this diaso ketone next what you do is treat with dirodium tetra acetate. So the dirodium retro acetate first it will form the corresponding dirodium carbinoid and it will undergo insertion at this CH so that will give you directly this compound okay. Now you have the pentacyclic compound so what is missing is you have to insert an NHM isn't it so that is normally done using a Beckman rearrangement you have a ketone then once you have ketone you can think of using Beckman rearrangement to carry out the ring expansion. So for that what you should do you should convert the carbonyl into oxyne so that was done by treating with hydroxylamine to get the oxyne and oxyne was made as a good leaving group okay by treating with para bromobenzene sulfonyl chloride so N-brosylate was formed then you do the Beckman rearrangement by treating with acetic acid. So that will give the Beckman rearrangement product that is the corresponding 6-membered lactam. The 6-membered lactam now one can easily methylate by treating with sodium iodide and methyl iodide you get the corresponding N-methylated compound and the carbonyl group also should be removed and the mom group also should be cleaved. So the mom group was cleaved using HBr in acetonitrile and the hydroxyl was oxidized and the Desmartine paraionine condition to get the ketone and the double bond was introduced okay in one part using phenyl acetyl chloride and mystic acid and reduce the ketone that is this one as well as remove the carbonyl that is lactam to corresponding amine was done in one step with lithium aluminium hydride and now what is left is the removal of the methoxy methyl group so that will give the morphine that was easily achieved using Lewis acid VBr3 at very low temperature one can do demethylation so that gave the corresponding natural product that is plus morphine which is the enantiomer of the naturally occurring morphine. So this is the first total synthesis of unnatural morphine okay so this was reported in 1997 in JOC again the starting material like other synthesis of morphine was isovaniline and he used two key reactions one is the carbinoid CHN session initially then later he used Beckman rearrangement to expand the 5-ambered ketone to 6-ambered lactam. So overall it took about 28 steps and yield was considering that it is 28 steps 3% overall it is quite good okay so with this I will stop and we have completed the 4 total synthesis of morphine and we will discuss some more synthesis of alkylides okay thank you.