 Good morning and welcome back to this course on Classics in Total Synthesis Part 1. The last week we talked about you know brief introduction to organic synthesis, total synthesis, semi synthesis and also we discussed few natural products having 3-mumbled ring as one of the course structure. So now we will move to 4-mumbled ring. So when you talk about 4-mumbled ring there are quite a few natural products having 4-mumbled ring and of course when you talk about non-natural product in the first and then foremost non-natural product should come to your mind is Cubane. So Cubane is a very interesting compound and how many 4-mumbled rings are there in Cubane? Look closely, you can see there are 6, 6 4-mumbled rings fused together and when you talk about one 4-mumbled ring itself is quite difficult because of its ring strain and here you have 6 4-mumbled rings and they are fused together that means the real challenge to synthetic chemists to make this Cubane is very, very high. You know when you talk about total synthesis the molecule need not be very complex even small molecules can give its own trouble. So one such case is Cubane. It gives different types of challenges compared to other big molecules. The first and foremost challenges as I said there are 6 4-mumbled rings which are fused together and second challenge which is not a real challenge but from retrosynthetic point of view these are important challenge. Cubane does not have any functional group. See always when you talk about retrosynthesis you look at a functional group, you look at a strategic bond then only go for retrosynthesis. When the target molecule does not have a functional group then it is a problem. So you need to introduce a functional group and so on and obviously it is very, very strained. We have 6 4-mumbled rings fused together and how much strain it will have you know. So highly strained compound and molecular weight is 104 as you all know low molecular weight compounds are always difficult to synthesize. So these are the 4 major challenges when you talk about Cubane total synthesis and how this 4-mumbled rings can be synthesized. So when you again talk about 4-mumbled ring and the first reaction which should come to your mind immediately is a photochemical reaction it is 2 plus 2 cycloaddition reaction. Reactions are widely applied for making 4-mumbled ring. So now as you can see this is the home of alkene and the loom of other alkene then you can see nice overlap to form a 4-mumbled ring. So there are several examples. So this is the simplest example 2 ethylene molecule giving simple cyclobutane. There are many complex reactions carried out using this 2 plus 2 cycloaddition before we actually go into the total synthesis of Cubane. I just thought you know show you one very interesting example which was reported much earlier is the preparation or synthesis of Cookson-Dion. So this Cookson-Dion involves a sequential thermal and photochemical reaction. It is done in 2 steps okay it is done in 2 steps starting from cyclopentadiene and benzoquinone you can see this is the diene okay and this is the dienophile this double bond is the dienophile and this undergoes a 4 plus 2 cycloaddition reaction to give this bicyclic adduct. In principle it is a tricyclic adduct but we consider this. Now this can be redrawn like this okay now when you redraw this molecule then you can see this double bond and this double bond they are very close to each other okay these two double bonds are very close to each other then on shining light this undergoes a 2 plus 2 intramolecular cycloaddition to give this very interesting strained diode okay this is called Cookson-Dion because it was prepared by Cookson first time and in 2 steps remember cyclopentadiene, benzoquinone and you get 100% atom economy reaction okay whatever you started with in the starting material and you can see all that here okay 100% atom economy okay you do not lose anything. So this is the beauty of all these cycloaddition reactions when you do cycloaddition reactions you get 100% atom economy okay. So then if you want to do cyclobutane synthesis you start with first cubane and look at how many form umber rings are there as which we have already seen there are 6 form umber rings are there. Now we know form umber rings can be made by photochemical reaction the next question is how many form umber rings you want to make using photochemical reaction? How many form umber rings you want to make by photochemical reaction? So for example you do a simple retro synthesis and then see whether you can make 3 form umber rings okay 3 form umber rings basically you are making only one form umber ring but while making automatically you are making one form umber ring you are making 2 more form umber rings see here the form umber ring will be formed between these carbon atoms but while doing that you get this ring as well as this ring. So overall you are making 3 form umber rings but by carrying out only 1 2 plus 2 cycloaddition reaction okay so logically this sounds good okay so the cyclobutane see this cubane can be disconnected and this could be the precursor for making cubane but in reality as you know in synthesis one can write retro synthesis based on the knowledge of wide range of organic reactions but when you go to lab many of them would not work here you do not have to even to even to go to lab you can say straight away this scheme is not good why? One this reaction is reversible okay because of you know strain this will be reversible and it can come back second the stability of the starting material starting material may not be very stable and third and foremost is this can undergo instead of intramolecular 2 plus 2 cycloaddition why it should undergo only intramolecular 2 plus 2 cycloaddition it can undergo intermolecular 2 plus 2 cycloaddition 1 molecule can undergo 2 plus 2 with another molecule and that can undergo 2 plus 2 with another molecule if that happens as you know finally you will end up in getting a polymer okay so though this idea is very good to construct 3 4 membered rings in one step by doing only 1 2 plus 2 cycloaddition reaction there are problems okay though it is logical but in practical this is not a good scheme. So how to avoid this what are the various solution possible first of all as I mentioned when you look at Cubane, Cubane does not have any functional okay so you need either a functional group or a strategic bond because that only will act as a tool handle for further retro synthesis so whenever you do not see a functional group or whenever you do not see a strategic bond in your target molecule please remember first thing you should do is introduce a functional group okay once you introduce a functional group that functional group will be your handle okay using that functional group you can further disconnect so that should be your first priority yes we have to introduce a functional group then which functional group okay there are so many functional groups okay so here again you have to break your head which functional group you have to introduce well you can introduce a functional group but at the same time you should also know in the end when you go forward that functional group should also be removed easily it should be easily introduced and easily removed okay these are two things we should know when you talk about choosing a functional group okay yes it should be easily made that is very important from simple, simple reaction and from another easy precursor so one way to think about is when you introduce a functional group can you think of another reaction that means you are going to make a four-membered okay you are going to make a four-membered ring so can you make this four-membered ring either via ring contraction or ring expansion okay either via ring contraction or ring expansion and during the process you should also introduce a functional group it is clear you are thinking about ring contraction or ring expansion when you do this reaction you also introduce a functional group okay it is 2 in 1 so then next question is whether you want to do ring expansion or ring contraction which one is better ring expansion means you should start with a three-membered ring and then go to four-membered ring and for ring contraction you should start with a five-membered ring and end up in four-membered ring so these are two options and which one will choose and as I said not only you should ring expand or rig contract but also introduce your functional group in this same reaction okay so let us see what are the reactions possible okay what one can try for example if you are talking about ring expansion take cyclopropanol okay take cyclopropanol if you treat with dyesamethane okay it can undergo a ring expansion reaction to form cyclopropanol but the problem is in the end product in the end product where you see cubane cubane all carbon atoms are sp3 all carbon atoms are sp3 okay here you have one sp2 carbon atom so then again you have to you know remove this ketone so it is little bit tedious however when you think about converting a five-membered ring into four-membered ring okay five-membered ring into four-membered ring that is ring contraction one reaction you should come to your mind is Fevarski's theorem okay alpha halo ketones on treatment with base can undergo Fevarski rearrangement the ring contraction so if you take a five-membered ring with a halogen at alpha position with respect to the ketone now the Fevarski rearrangement can give a cyclobutane here all the carbons are sp3 all the carbons are sp3 now if you remove the carboxylic acid if you decarboxylate you will get cyclobutane so it is very obvious between these two that is from ring contraction and ring expansion method from cyclopropanol to cyclobutanol and cyclopentanol to cyclobutane the second method that is ring contraction by Fevarski rearrangement is much much better in terms of getting cyclobutane with all sp3 carbon atom okay so that is what we choose so it is better to go with ring contraction strategy for the synthesis of cubic okay. Now next question so you know you identified to use Fevarski rearrangement as the key reaction to make cyclobutane so that it will form cyclobutane with one carboxylic acid so can you use only one carboxylic acid can you think of two carboxylic acid so suppose if you make the precursor as this so then you can also think about Cubane having two carboxylic acid okay so then you have to do two Fevarski rearrangement in the same part okay two Fevarski rearrangement in the same part so which one will simplify if you take one carboxylic acid if you take one carboxylic acid then the precursor will have one 5-ambuteric okay precursor will have one 5-am but in fact ideally speaking there will be two 5-ambuteric okay whereas if you take two carboxylic acid you will have more 5-ambuterics okay you will have more 5-ambuterics that means by doing this if you convert more 4-ambuterics into 5-ambuterics then your job is becoming much simpler okay it is better to construct 4-ambuteric in the end rather than in the beginning so obviously between these two choices one should prefer to have two carboxylic acid or in other words introduce two carboxylic acids to Cubane then start the Petro synthesis okay so then again when you want to introduce two carboxylic acid where will you introduce like aromatic system you can have 1-2 position 1-3 position and 1-4 position isn't it? So this is 1-2 carboxylic acid then you have 1-3 and also you have 1-4 okay so all are possible because theoretically theoretically all are possible but in from the experimental point of view from the synthetic point of view which one will be easier? Which one if you take and then move forward that will give you simpler starting material okay so that is what the major point about Petro synthesis isn't it? Between these three which one will be the best precursor okay so when you look at this we know what is Fever's Kirchner you also should know what is Quasi Fever's Kirchner See Quasi Fever's Kirchner is the one where carbonion is not formed or enolate is not formed here if you take this alpha heliketon what happens when you treat with sodium methoxide okay the sodium methoxide attacks the carbonate okay then this bond migrates and then your bromide goes and that will give you directly this 5 ombre ring so in Cubane we will talk about Quasi Fever's Kirchner arrangement and not Fever's Kirchner arrangement is a broad one Quasi Fever's Kirchner arrangement is subset of Fever's Kirchner arrangement okay so this you should know and if you are using one carboxylic acid this should be the precursor okay now you can see OAT or OH okay OH will attack here and this bond will come and it will remove the bromide and that will give you the carboxylic acid one carboxylic acid and ideally if you look at carefully from left to right okay left to right when you see where the bromine was earlier where the bromine was earlier the same place carboxylic acid comes okay this is for simplicity you should remember where the carboxylic bromine is there the same place carboxylic acid comes okay. Now look at this compound there are 2 bromines and their relationship is 1 2 is not it there are 2 bromines their relationship is 1 2 so that means at the end of this Quasi Fever's Kirchner arrangement you should get the corresponding 1 2 carboxylic acid we are just extending what we have discussed here okay wherever bromine is there carboxylic acid will come so if the bromines are 1 2 related and in the end you also will get 1 2 carboxylic acid. So now look at this example this is 1 3 okay then that will give you 1 3 carboxylic acid there is another possibility you know this 5 membered ring this 5 membered ring is here in this example 5 membered ring is on the right hand side but both will give 1 3 carboxylic acid because look at the relationship between the 2 bromines they are 1 3 related okay. Now for 1 4 so this is 1 possibility that will give 1 4 carboxylic acid and this is another possibility where the cyclopentadone is on the left hand side. So both will give you 1 4 carboxylic acid. So now among these 3 options which one will choose among these 3 options which one will choose and then see how that will give you a proper retrosynthetic pathway to simpler starting material. So what we can do is let us take this precursor okay so this is a compound which upon decarboxylation will give you cubane and this in principle can be made from this using Coase Fever Schere arrangement. So now let us take this as the target molecule okay then go further okay. So this is what we have taken and how many 4 membered rings are there now in this? How many 4 membered rings? 2 4 membered rings in cubane we have 6 by doing 2 Coase Fever Schere arrangement in 1 step in retrosynthesis. We removed 4 4 membered rings we have only 2 4 membered rings because that is a significant reduction in the complexity as well as removing the strain okay. So the Coase Fever Schere arrangement retrosynthesis has helped. Now what we should do? We should do a photochemical disconnection. So you should form you should look for 4 membered ring and try to do a retrosynthesis. So first let us take the front 4 membered ring okay. If you make this front 4 membered ring using 2 plus 2 cycloaddition will you consider this as the precursor? Now you see there are 2 double bonds this can undergo 2 plus 2 cycloaddition and interestingly if you look at this when you do this 2 4 membered rings are formed okay. So 1 2 2 4 membered rings are formed basically so far what we have discussed is in 2 steps you remove completely 4 membered ring. Coase Fever Schere arrangement you removed 4 4 membered ring and using photochemical reaction you removed 2 4 membered rings okay. So now if you look at this you can look at this compound you can see only 5 membered ring no 4 membered ring okay. Now same compound instead of making the front 4 membered ring suppose if you make the side 4 membered ring using 2 plus 2 okay then see what is the precursor? Obviously this is the precursor again there are no 4 membered rings in the precursor both are 5 membered ring. Now my question is just look at this assume that this is A and this is B my question to you is whether A and B are same A and B are same are they same I will give 30 seconds just to see whether these 2 are same and how many of you agree? Yes they are same okay if you redraw this structure if you redraw this structure this is what you get and same B if you redraw this structure you get this compound. Now you see both are same both are same so it does not matter whether you make the right hand side 4 membered ring or the front 4 membered ring using 2 plus 2 cycloaddition both are starting from the same compound okay. So now what you have to do you have to do a retrosynthesis of this compound as such you can see this is obviously a deal solder product. So that means what should be the precursor? The precursor is the same diene dienophile both are same one will act as a diene other one will act as a dienophile okay as you know cyclopentadiene is not stable. So as soon as it is formed it should undergo 4 plus 2 cycloaddition reaction. So how we do it is simple we start from cyclopentenone okay you start with cyclopentenone and carry out allylic bromination allylic bromination with NBS under photochemical condition. Now you add bromine so bromine will add to this double bond so you get a tribromocom okay. So in 2 steps you introduce 3 bromines. Now if you treat with triethylamine what will happen you know these 2 protons are acidic is not it? So this can undergo elimination this can undergo elimination. So minus 2 HBr will give you your starting material that is 2 bromocyclopentadiene and as I said that once you prepare it is unstable immediately it will undergo 4 plus 2 cycloaddition and you get the next step very easy very easy starting from commercially available cyclopentenone okay in 3 steps you get this bicyclic compound. So now one can think of doing 2 quasi-favorsky arrangement in one step but they face lot of problem so they wanted to go sequential. So what they did they protected both ketones okay then selectively they removed one of them that is this side so they could keep the top carbonyl group being protected okay. Then next step is the 2 plus 2 cycloaddition reaction. So this can be written already we discussed is not it this can be written like this okay then if you do 2 plus 2 cycloaddition reaction you get this compound. So what we have done so far is we have introduced 2 cyclobutane ring 2 cyclobutane ring then we have to introduce 4 cyclobutane rings. So what you can do carry out the quasi-favorsky arrangement. So at this position you will get carboxylic acid and the 5 ombre ring also will become 4 ombre ring. So now what we have done how many 4 ombre rings we have done 4 4 ombre rings we have done first we did using photochemical condition you did 2 and using quasi-favorsky arrangement you did 2 more okay. Next what you should do obviously you have to remove the handle the handle here is carboxylic acid. The carboxylic acid first you convert into acid chloride and then treat with terributal height of peroxide so you get the peroxycopene. Then heat it with cumene so this was reported by Philip Eaton from University of Chicago I will come to that little later. So then this upon heating with cumene it undergoes decarboxylation and you get the corresponding cubane with ketol and bromine intact. Next you have to remove the ketol then do the quasi-favorsky ring. So why when you do this now you introduce all the 6 4 ombre ring. If you look at this synthesis carefully each step that is when you are introducing a cyclobutane ring each step you introduce 2 cyclobutanes each step you introduce 2 cyclobutane. So once you have the carboxylic acid treat with thyroid chloride and the terributal height of peroxide and followed by heating you get the cubane. So this is one of the classical synthesis of a non-natural product called cubane and this was reported by Philip Eaton and Thomas Cole from University of Chicago in 1964 and afterwards several derivatives of this cubane was synthesized and used particularly army people were interested in several derivatives of this cubane. So with this I think to summarize what we have done is today we briefly started with 2 plus 2 cyclo addition reaction under photocomical condition then we did a proper retro synthesis for cubane then using quasi-favorsky rearrangement and 2 plus 2 cyclo addition we could successfully construct 6 4 ombre rings of cubane and then completed the total synthesis of cubane. See you. Thank you.