 Good morning and welcome back to the NPTEL lecture series on classics in total synthesis. In the last lecture we talked about total synthesis of few trichunanes and we will continue our discussion on total synthesis of more trichunanes today and a few more lectures we will focus on the same and today's lecture what we will do we will take one key reaction and how this key reaction was used successfully to synthesize a few trichunanes and that key reaction is radical cyclization. As you know radicals can be easily generated from corresponding alkyl halides or nitroalkanes. So one of the most common methods for generating radicals is take an alkyl halide and then treat with tributyltin hydride and AABN. So AABN is nothing but also this isobutero nitride you take this compound and easily one can dehalogenate if you have an alkyl halide that halide can be replaced by hydrogen and the mechanism is very simple when you take an alkyl halide or nitroalkane in toluene or benzene when you reflex it then you had this isobis isobutero nitride. So if you look at the structure of isobis isobutero nitride there is a nitrogen in to which is a good living group. So EC extrusion is possible. So this is what you observe when you do this reaction when you take this halide in toluene or benzene and reflex it then when you add this isobis isobutero nitride immediately you can see the extrusion of nitrogen bubbles you know you can see nitrogen bubbling and that will generate this radical. So this radical what will happen since you add tributyltin hydride which is 1 to 1.1 equivalent of tributyltin hydride you add to this reaction mixture. So that will immediately abstract the hydrogen of tributyltin hydride and it will form the corresponding tributyltin radical and already you can see the isobis isobutero nitride became simple butonitride. Now the tributyltin radical will react with your alkyl halide and then form tributyltin halide and alkyl radical that alkyl radical will pick up hydrogen from tributyltin hydride and then it form Rh. So now what will happen this tributyltin radical again it will pick up it will react with Rx to form alkyl radical that alkyl radical will further undergo you know hydrogen abstraction from tributyltin hydride so the cycle will continue. So what you need is you need only a catalytic amount of azobis isobutero nitride so that is a radical initiator but you need more than 1 equivalent of tributyltin hydride because that is what which replaces the halide in your system. And when you use tributyltin hydride as I said you can also remove nitro group if you have a nitro group the nitro group also can be cleaved instead in some nitro group what you get in the product is hydrogen okay. In addition what will happen when you have an alkyl halide and also an acceptor in the same system okay also an acceptor in the same system then this radical instead of abstracting hydrogen from tributyltin hydride it can add to the acceptor it can add to the double bond it can add to the triple bond so when that happens the radical will not be at this place okay so once the radical is generated here then the radical adds to this double bond and it the radical comes to now the primary carbon of the double bond okay at that position that CH2 radical will abstract hydrogen from tributyltin hydride. So in the process you can see if you start with an open chain you end up in getting a ring and this is the process and when you do this you also will see in many papers people write like this notation phi xo, 6xo and some number they will give and xo and endo and then they will give trig, dig and so on and what are they? xo you would see, endo you would see, dig, trig, ted, dig means diagonal, trig means trig represents trigonal, tet represents tetrahedral. So that means so this is sp3 carbon, this is sp2 carbon, this is sp carbon. So these are the carbon atoms which are accepting the radical, if you are acceptor has sp carbon atom then you write trig, dig, if you are acceptor has sp2 carbon atom then you write trig, if you are acceptor has sp3 carbon atom then you write ted. Then what is xo, what is endo and also you will see some number in front of xo or endo. So what that number means? That number means when you do the cyclization what is the ring size? The size of the ring is represented by n. Suppose if you are forming a 5 umbil ring then you write 5, if you are forming a 6 umbil ring then you write 6, if you are forming a 3 umbil ring then you write 3. So the n represents the size of the ring formed after the cyclization. What is xo and endo? So when the radical is formed now the radical adds to the double bond it can add in 2 ways. First way it adds like this and the final radical, the final radical that is after the cyclization if it is outside the ring, if it is outside the ring then the whole process is called xo. And same thing when the cyclization takes place and the final radical if it is part of the ring, it is inside the ring then it is called endo. N represents the size of the ring formed and xo means the final radical is outside the ring, endo means the final radical is inside the ring. Then the trig and then dig as I said since we are talking about radical cyclogen and acceptor we will talk about only sp and sp2 carbons. So this is sp2 carbons, the acceptor is sp2 carbons. So this means it is trig, xo trig. This is sp carbons, both are sp carbons, the acceptor has sp carbon atom. So then the cyclization you have to write dig. Depending on the ring size you put the number before the xo. This is what proposed long time ago and also Baldwin proposed a set of rules which are the reactions allowed which are not allowed based on the literature. So according to him all xo trig, all xo trig reactions are allowed or favored. What is not favored is 3 endo, 4 endo, 5 endo. These are not favored in the case of trig that is if you have a double bond and if you are carrying out radical cyclization then 3 endo, 4 endo, 5 endo are not favored. The first earliest ring where endo is favored is 6 endo, only 6 endo is favored. Coming to sp carbon atoms you have exactly opposite to what trig is. All endo dig reactions are favored, all endo dig reactions are favored but 3 xo and 4 xo dig are not favored. Only from 5 xo dig onwards all xo dig reactions are favored. So these are rules which you can remember or no problem when you carry out reactions automatically you will come to know whether your reaction works or not. If it does not work then go back and then see why it did not work. It may be because of these rules but though these rules are used extensively there are many exceptions as is the case with many rules. Then what about the VJ chemistry? So when xo and endo are allowed for the same substrate which one will be favored? Xo will be favored or endo will be favored. Then when your substrate has substituted after the radical circulation what will be the stereochemical relationship? So the VJ chemistry and stereochemical outcome of radical circulation can be easily explained. See for example if you take this compound now 5 xo will give a tertiary radical whereas 6 endo will give a primary radical. So from the stability point of view you know the tertiary radical is more stable so that is why this is more favored that is 5 xo is more favored than 6 endo. Coming to the stereochemistry see for example you have a double bond and you have a halogen. So it forms 5 xo trig since you do not have substituted at the end so 5 xo trig is more favored. Now you put the substituted at the same carbon as the halides. When the radical cyclization takes place this will become a methyl group is not it? This will become a methyl group and what will be the stereochemical relationship between the R and methyl? What will be the stereochemical relationship between R and methyl? So what you should do? You should draw a chair like conformation. See this is a chair like conformation and when you do that you put the radical and also put the R group in the equatorial position of the chair conformation. So you draw the chair conformation and put the R group in the equatorial position. Now when you cyclize, when you cyclize this is what you get. What you get? R and methyl are cis to each other. Now the R group can be here, here, here. What you should do? Accordingly you have to draw the chair conformation and then put the R group in the equatorial position accordingly. Accordingly you put the R group in the equatorial position then do the cyclization. Then you draw the cyclopentane and then look at the methyl group and your R group. So I will give one more example. So now what I have done? I just moved the R group to second carbon and again you draw the same chair conformation and this R you put in equatorial position. Do the radical cyclization and as you can see here in this case the methyl and R are trans to each other. So these are the major products you also will get the other product. So in this case you will get trans, in this case you will get cis also but these are the major regio and stereochemical outcome. So the regiochemical outcome is based on stability of the radical whereas the stereochemical outcome is based on putting the alkyl group or the substitute one in equatorial position and drawing the chair like conformation and then see the final outcome. So now we will see how this radical cyclization reaction has been successfully used in the total synthesis of natural products. As we are talking about trichunanes I will explain how this particular radical cyclization has been used in the synthesis of linear, angular and propellants. One example which we will see is kirsutanes and kirsutane belongs to you know linear trichunane. So this is one of the earliest kirsutane type trichunanes which are isolated. It was isolated in 1976. I will not go into the details of this because there are 4 different types of linear trichunanes and this 4 different types vary based on the position of the 4 methyl groups. The 4 methyl groups you can see here these 4 methyl groups are located in different places for these 4 different skeleton. The first natural product where a tandem radical cyclization was used as the key step was kirsutane. So this is a Cisketer pin called kirsutane and this molecule current has cleverly used a tandem phi xo radical cyclization as well as a Claisen rearrangement to prepare the starting material for the key radical cyclism. This was one of the earliest examples of Pauline radical cyclization. So once you see a double bond here then you also see 3 3 phi membered rings. So one can easily think about phi xo radical cyclization reaction. So what you thought was this could be easily made from this precursor. So the kirsutane can be easily made from this precursor. So his idea was this bromine on treatment with AABN and tributyltin hydrate should generate a radical here. That radical should undergo first phi xo cyclization to give a radical here. So the phi membered ring is formed and this phi xo trig because this is sp2, this is sp2, isn't it? So trigonal phi xo trig and the phi membered ring is formed that leads to another radical. Now you have an acceptor. This time it is a triple bond. So that means it is dig. So that will be phi xo dig. So it is a combination of phi xo trig and phi xo dig. All this happen in one part. So you start from this and one part in principle you should be able to convert this into natural product. So that was the key reaction which Karan has proposed. Now this compound can be easily obtained by simple homologation. So what you need is you need to add a triple bond to this. So some function group transformation followed by adding this triple bond you will get the radical cyclization precursor. The next key step is the nucleophilic attack. The nucleophilic attack of this whole unit nucleophilic attack of this whole unit to this lactone. So now the nucleophilic attack here and the double bond will migrate and this will open up. It is like a centiprime. So that will give you the carboxylic acid. And this lactone, if you look at this lactone, whenever you see a 5 membered lactone, one reaction which should come to your mind immediately is iodolactonization. So now you have a double bond here. So what one can do is after iodolactonization you can eliminate. And this can be easily obtained from this acetate through a glycine rearrangement which I will discuss during the synthesis. Let us see how this bicyclic lactone was made. It started from commercially available 2 methyl cyclopentenol. And the first step was to use Lucia reduction that is sodium borohydride serine chloride to reduce only the ketone of alpha beta ansatur ketone to get the cyclopentenol. So it is a methyl group at 2 positions. Then it treated with acetic anionide. So which acetylated the free hydroxyl group to form the corresponding 2 methyl cyclopentenol acetate. This on treatment with LDA and quench with TBS chloride. So you can write this compound like this. Now you see this as acetic proton. You can generate anion with LDA and if you quench with TBS chloride this is what you get. So if you look at the substrate carefully if you look at the substrate carefully. So this is having a 1,5 diene. It is having 1,5 diene. You can see 1, 2, 3, 4, 5. So when you have 1,5 diene then that is a substrate for cope or glycine rearrangement. Since you have oxygen part of this, so this is glycine rearrangement. So this is easy. You can see. This will undergo this 3,3. Sigma tropic rearrangement upon heating to give the corresponding rearranged products. Sigma delta unsaturated ester alpha, beta, gamma, gamma delta. Always you know when you do such a glycine rearrangement you will get a gamma delta unsaturated system. So this upon treatment with phenyl-cellenyl chloride. So you can think about using iodolactonization or phenyl-cellenolactonization. The phenyl-cellenolactonization is slightly better than iodolactonization just for the reason that the introduction of double bond is much easier. It can be done at 0 degree. So that will give you the corresponding selenolacton. As you know once you have a selenol group then treatment with hydrogen peroxide at 0 degrees. One can easily eliminate the phenyl-cellenic acid to introduce the double bond. It is in elimination to get the corresponding bicyclic lactone. So the bicyclic lactone was obtained in 6 steps from commercially available 2-methylcyclo pentenol. And the next step is to make the nucleophile. Make the nucleophile and that should undergo SN2 prime reaction. For that you started from again commercially available 1, 3 diol having a gem dimethyl group. And selectively one can protect one of the alcohol because this is a symmetrical alcohol. So you protect one of the alcohol as corresponding THP ether. So it is also very simple and straightforward. So you protect this as a THP ether. Then the other alcohol you mesilate and convert that into a Bromocomp. So this is the second precursor. Now you convert this into lithium. Convert this into lithium. So you can treat with tertiary-butyl lithium. I know those days he has used lithium naphthalene to convert that into lithium, lithium derivative. And now you make it as copper. So that now it can undergo SN2 prime reaction to give this carboxylic acid. And you also introduce this 3-carbon unit. So next step is you have to homologate. You have to introduce the triple bond. You need a triple bond here. So that can be done before that this THP group. So the THP group has served its purpose. So once the protecting group served its purpose it is better to remove. And the problem with THP group is it will give additional stereo center. THP is nothing but if you see, so this is THP. And then you can see there is another chiral center here. So because of this extra chiral center you will get a diastereomer. You will get a mixture. So that NMR will not be clean. So whenever you use THP, whenever you use a THP ether and if you have a chiral center in your molecule, remove the THP as clearly as possible so that you will get a good spectrum. So then remove the THP and then you get the primary alcohol. Now the free carboxylic acid can be easily reduced with LIH to get another primary alcohol. So now if you look at this molecule you have 2 primary alcohols. So both you convert into triplet and then convert that into corresponding iodide by treating with tetrabutyl ammonium iodide. The tetrabutyl ammonium iodide converts these 2 into CH2I here and CH2I there. Now what you need to do is you need to homologate here with the triple mole. Now when you treat with TMS as clean and butyl lithium. So this is a neo-pentyl system. This is a neo-pentyl system. So when you had lithium trimethyl sialyl acetylite, the neo-pentyl system is not that reactive. It is very, very difficult. So here this nucleophile can attack only at this carbon. So what you get is the corresponding triple bond and TMS and for the radical cyclization you do not need this TMS. So just to remove it with the fluoride source. So CCM fluoride will remove the TMS and that sets the stage for the key radical cyclization. So the radical cyclization as you can see here it undergoes a tandem phi exo trick followed by phi exo dig radical cyclization to give a natural product irsutine. So if you look at the overall process, the total synthesis of irsutine was made from commercially available 2-methyl cyclopentenome and it involved 3 key reactions, glyzendry arrangement, ascent to prime substitution and tandem radical cyclization. And the yield, overall yield for this whole sequence was close to 8% and considering that it is a 14 step process, 8% is a very, very good yield. After successfully synthesizing irsutine, he wanted to extend this methodology to another closely related natural product called capnally. If you look at capnally and irsutine, immediately first look at these 2 molecules you will feel that both are same but it is not. You need to have closer look at this molecule. In irsutine you have methyl group here whereas you have hydrogen. In irsutine you have hydrogen here, you have methyl group here. In irsutine the dimethyl group is here whereas dimethyl group is here at capnally. So there are subtle differences between these 2 natural products and what Karan wanted was he wanted to extend the same tandem radical cyclization to capnally also. So what he did? So he also started from cyclopentenome. This time he does not need the methyl group here and followed the same process to get this bicyclic lactone. But he needs a methyl group here. So for that first what he did? He did a 1-4 like addition with methyl, magnesium bromide and copress bromide dimethyl sulfide. So it opened to give this carboxylic acid and then followed by hydrolactanization and elimination he could get this bicyclic lactone. Then another ring opening with this berygnaud reagent and copper. So this is commercially available. The corresponding bromide is commercially available. So make the berygnaud and add and you get this carboxylic acid. Reduce the carboxylic acid to get the corresponding primary alcohol. Then you mesilate and convert into CH2I and now based on the earlier experience you do not need even TMS acetylene. You can directly take lithium acetylene dimethyl complex. So that will give you the triple bond required for the radical circulation. Now what we need is you have to remove this, convert it into gem dimethyl group and also the halide. So Jones oxidation directly oxidize the protecting group first hydrolyze the protecting group to aldehyde and then oxidize the aldehyde to carboxylic acid and then carboxylic acid was methylated using diazomethane to get the corresponding methyl ester. Now if you take excess methyl magnesium bromide and add to this ester it will give corresponding tertiary alcohol. The tertiary alcohol was converted into corresponding iodide by treatment with primethyl silyl iodide. So that gives the key precursor for the tandem radical cyclization. So you have the radical here. So you can generate the radical from this iodocompound and then that will give phi xo followed by another phi xo. First one is phi xo trig, the second one is phi xo dig. So that gave directly the natural product that is capnoline. So again using the same radical cyclization, same tandem radical cyclization currents group could successfully achieve the total synthesis of capnoline and here they started from the commercially available cyclopentenone, two cyclopentenone and not with methyl group and overall yield was almost same as in the case of heat city. The total number of steps was 14 steps and with an overall yield of 8 percent. So we will continue our discussion on the radical cyclization, how this radical cyclization has been successfully used in the synthesis of more triquinones in the next lecture. Thank you.