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Module 1: Synthesis of Nucleobases, Nucleosides and Nucleotides

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Hello everyone and welcome back. So, we have seen how the nucleobases can be synthesized in the laboratory now comes the nucleosides. So, nucleosides basically mean the sugar plus the nucleobase so this has been done. Now we have to see how you can attach the nucleobase with the sugar. So, your target molecule is something like this here is the nucleobase it can be adenine can we guanine can be thymine can be cytosine for DNA or can be uracil for RNA. And when we are talking about DNA then this is the sugar chemistry of sugar conformation. This is for the DNA, DNA nucleoside. When if we talk about RNA then your target would be here nucleobase there will be hydroxyl here this would be your target. So first we will start with the RNA one because this is more general this as the hydroxyl group present here also then we will go specifically to the DNA. So, here what is the target is you already have synthesized nucleobase. So, a target is how to connect the nucleus with the sugar moiety at this position. If you do look at the numbering this is called the C1 prime over or 1 or simply if we can write one prime position. So, nucleobase; connection of nucleobase at the one prime position of the sugar that is your prime target now; So, there can be a lot of alternative methods I will show you that in this slide you can see so there can be other ways also to synthesize the nucleus with the sugar attaching at the C1 prime position. The first one is of course as I have talked about this is your nucleobase that you can attach the whole thing you can attach to the C1 prime of the sugar and of course it will be kind of a nucleophilic substitution reaction. So, therefore in the sugar moiety, you need to have a good leaving group so that the nitrogen on the nucleus can attack as a nucleophile and eliminate that so you can have this compound your target molecule that is pathway band this is what we will mostly follow in this. There can be other ways also one is so B and C pathway B and C both involve that you are synthesizing the nucleobase on top of the sugar. So in this case nucleus was synthesized separately sugar was synthesized separately and then they were coupled together. But here you can see that using this you are synthesizing the whole of the nucleus on top of the sugar. So, with amine-free again 2 ways one is keeping the 6 membered rings that and forming the 5-member ring here or alternatively you start with the 5 membered rings on the sugar and synthesize the 6 membered rings. So, both of these ways areas we have seen before but only thing different is that you are synthesizing the whole molecules on the sugar. We do not use these methods much nowadays most people use this method where you synthesize the nucleobase separately sugar separately then couple them together. So, the first kind of synthesis was done by is name reaction Fischer Helfrich synthesis this is one of the first kind of successful synthesis of the nucleobases nucleosides and for which Fisher is very famous actually for his synthesis of the purine especially the purine nucleosides and nucleotides. And Fischer received Nobel Prize in 19 long back 1957 for his famous purine synthesis protocols. So, in this case, what they have done is you have the sugar with a living group as I have said and then you start with this with a living group here. And for example, you start with adenine is amine here and you have this is what is your adenine and you try to couple them together. Now, look at the problems here so what is your target is basically here this is the N9 position of the adenine which should react so this nitrogen basically should react this carbon eliminating the bromide. So, this is your target. Now, what are the problems associated with this reaction? The first one is of course that you have the other hydroxyl groups in free form which can also react which can participate in the reactions giving you all the unwanted products unwanted side materials. So, the first need is to protect these hydroxyl groups so that they do not react or they do not hamper into your desired reactions. So, the first thing that they had to do is the protection of the sugar this sugar has been converted to protection means if you make the alcohol so these are basically this is secondary alcohol or secondary alcohol this is a primary alcohol. If you convert the alcohols into the acetates which are an ester basically an ester is very less reactive because esters have internal resonance within themselves. So, if their reactivity is pretty much less compared to the free alcohols. So, if you convert the alcohol to the ester that will protect the alcohol from doing the reaction. Similarly, I will not write the whole so this is AC CH 3 CO is acyl group which is usually termed as AC. So this would be OAC and you can convert this also with OAC. So, then you can protect the other functional groups which could participate in the reactions. Now your target is only here so that the bromine can be eliminated fine. Now, on the other hand, you have the nucleobase here what problem could happen to see your target is that N9 should react with the bromine. So, this is the nucleophilic center which means in this case if you use this form exactly then the lone pair of the nitrogen should act as the nucleophilic center and attack the C1 prime. Now here if you see you have an amine group also which also has a lone pair of electrons and this would be more reactive because this is the second amine and here the lone pair is very much it is inside the ring. So, the lone pair is very much involved in the aromaticity, on the other hand, this is a free amine form that is more reactive. So, if you cleave this amine-free then you will get the most unwanted reaction that this amine will react to the C1 Prime. Hence you need to protect the amine also first and this can be done using NH, the amine can be also protected by isolation. In this case, a very good protecting group is this you can use the simple acyl also as we have seen for the other one. Here this is one protecting group that is being used for the protection of the amine kind of functional groups. Here also the same thing once you do the protection the lone pair of electrons on the amine is very much involved in its internal resonating structures. So in their internal resonance so that the lone pair of electrons is not available to do the reactions with other molecules. So, that is how the amine can be protected. So, you have to start with this sugar with the protected ones and with the protected version of the nucleobase. Now if you mix them together you will try to do the reactions. You can pretty much expect to have nucleophilic substitution between this and this. But still, it does not work fine the reason is here you have OAC here you have OAC here you have OAC plus NH I am not drawing the whole production group I am writing P. So, in this case, this I am terming as P for protection so, NHP. Now if you see the sugar chemistry is actually pretty hard to do because the carbons are not very much reactive they are not many electron-deficient carbons. So, doing nucleophilic substitution is not very facile here. You have these are in this carbons these are prime or secondary alcohols here you have primary alcohol in this carbon which is our target hetero atom is attached there but that does not make the carbon many electrons deficient. If you have other electron-withdrawing groups such as carbonyl and all these things that would have made the carbon more electron-deficient number one, so, sugar, in general, are not very reactive. On the other hand, nucleobases are not reactive either much nucleobases are not very strong bases so there nucleophilicity is less and the lone pair of electrons as I have told you earlier they are involved in the internal resonating structure so it is a part of the aromatic system so they are not available much. Reactivity of the nucleobases is also less therefore if you do if you mix this and this together then you are not going to get much of your product. So, the need is that you need to activate the reactivity on the nitrogen. So, that you do by one of the old methods is that if you treat this with mercury chloride then that takes away a proton here leaves you with HCl and HCL will be eliminated NHP nitrogennitrogen. So, you have in HCl bond this is the new bond that has been formed. Now why mercury chloride we have used I will tell you. So, mercury has high electron electro positive it is an electropositive element. So, it will try to remove the electron from there it wants to be in the positive charge form. So, therefore this bond is not really a covalent bond it exists very much as a nitrogen minus and HgCL plus because my mercury is stable as a cation Hg+ form you have started with Hg 2 plus mercury chloride so this very much exists kind of an ionic bond less a covalent bond more of an ionic bond or you can call it a kind of salt. So, therefore positive charge mostly restores on the mercury and that leaves you with the negative charge distinctly on the nitrogen. So, now you have on a good nucleophile with almost a free negative charge and this is the reason that it wants to release away from the electrons. So, now if you treat this molecule with the sugar and typically they do it in organic solvent xylene 120 degree Celsius temperature that gives you the nucleophilic substitution here and you have your target molecule OAC OAC OAC here it mostly goes In here you have the NHP double bond that is it this compound but you have to get back your the free forms we protect it this protection you can do by treating with ammonia in methanol it will detect both of these both acetylate and here. This will leave you with free hydroxyl free OH here and here you will have the free adenine form so you have your adenine synthesized. Now similarly if you want to synthesize the guanine, for example, you start with this OAC protected OAC this is also protected plus you can start with the guanine which is this amine H and here instead of H, of course, you have to use the HCL same treatment xylene 120 degree Celsius and you can have your guanine. Of course, this is after so it is not the only guanine this is the 4 step second step is you have to de-protect everything else that can be done by using the same way ammonia in ethanol or methanol will give you the OH all the way it should be coming here. There is another way actually to synthesize the guanine instead of using this if you use that gives a better result actually that is why I am showing. Here instead of using the free guanine if you use this form this derivative of one in I am sorry here, the NH2 requires protection the same protection, of course, NHP so NHP and you instead of using the keto form you start with the chloride there and HgCL then what you get is this molecule. Here you have the CL N N NHP here you have OAC OAC in here in double bond here now you have to get rid of the chloride if you do the hydrolysis in sodium hydroxide a little bit water then that will replace; you will have the nucleophilic substitution kind of thing here you will have OH there and once you have H, of course, this is the isolate form so it will immediately go back to the keto form. So, this will go to the keto form in here you have NH NHP in here that will double bond here this so you have your guanine and then you can go for the deprotection. So this is your guanine. Now I will just briefly explain that say you have started here if you look at the stereochemistry of the sugar C1 carbon we have started with the arbitrary stereochemistry it could have been the after plane it could be the down the plane. So, both cases don't matter what is the stereochemistry here whether this is up the plane whether the bromine is down the plane. For ribose sugars, you always get the nucleobase that comes up with the plane. So independent of the stereochemistry of this you always get the nucleobase of the plane which is your desired stereochemistry of the nucleoside. In the nucleoside always your nucleobase comes after plane of the sugar this is the up the plane this is down the plane and that is the reason I will just explain in a brief while why this happens. This has to do with the rule that is known as Baker's rule. So, this is for the pyrimidine adenine or guanine that you can go this way.  Now if you think of the pyrimidines thymine or cytosine I will give an example of the thymine here OAC bromine now the similar way you can use the thymine with the HCL this is your metal double bond O NH double bond O so this will react will give you the correct compound up the plane it gives you the thymine. I am keeping these protection groups you can use the same methods ammonia methanol treatment to get rid of the protection. When you say you with cytosine, cytosine has one amine group here so in winter project amino group first and then do this chemistry. Now it is what we do we use an alternative method which is more facile or which is which will give you better result bit and the yield of the reactions are much higher where instead of the HCL if you treat if you have the thymine if we use the in enolate form so your idea is that to increase the nucleophilicity on this nitrogen NH. Now if you treat this with a Cylon compound TMS chloride trimethylsilyl chloride which is basically this methyl methyl methyl and this is silicon chloride, trimethylsilyl chloride silicon compounds are very, very, very useful for organics and synthesis and they have nowadays in modern organic reactions they are their use is in plenty in numbers. So, what this will do is silicon reacts very easily with oxygen with the double wound maybe single bond oxygen or double bond oxygen does not matter. Gives you eliminated reaction same thing will happen here and that will give you since if this comes here this will have to go to make the saturation. So, it basically allows you O Si Me3 similarly this will also happen here and this methyl, so this is basically the enolate form. The enolate form of thymine is quite a stable compound. Now then how can you do this chemistry? The chemistry is yes I am just writing N Me3 note the structure again you already know the structure and this, so you have the sugar with the bromine protected once. Now if you use it is very easy to break this kind of bond also salle compounds are very easy to break also. Just a little bit of treatment with water or even better is if you have fluoride f- in your solution if you use a tiny bit off- then that reacts with salle it breaks this bond back and it comes back here which creates a negative charge that will immediately react. So, this becomes a much better nucleophile when you treat this reaction mixture with a little bit of fluoride or even with water. The same thing will happen here it will be dehydrogenated and it will be hydrolyzed giving back the thymine. So, you will get your timing back here OAC. So, the enolate form of the pyramidion nucleobases you can do it with purine also with purine nucleus is, for example, thymine or cytosine since you can treat that with trimethylsilyl chloride and get this enolate form and then react it with the sugar protected sugar that will give you a very easy or the more amount of much facile synthesis of the nucleoside. So, now the question is the same thing you can do with the cytosine. When you look at the sugar I have started with this bromide and of course all the protected ones. And this has of course come the protected ones, of course, come from its original one with OH bromine and then OH here OH there. So, this can be converted into this as we have seen, now how to get this C here.  So, this you can actually get very easily if you start with all protected for example if you have the free form here OH and you randomly protect all of them treat it with acetic anhydride. Acetic anhydride is now banned in India. So similarly you can treat it with acetic acid also a little bit of high heat reflux then excess of this material excess of the acetic acid or acetic anhydride then everything will be protected. This will be OAC this will be OAC everything would become acetate because you have used not Excel excess. Now, if you use to treat it with tin bromide or tin chloride if used in chloride you get you are going to get chlorine here if you are treating with the tin bromide it is in Br4 then you are going to get bromine here. so, if you treat it with Br4 then you are going to get a reaction which will happen only here and it will give you Br. The Br will mostly come at the beta position the up plane Br, here OAC OAC OAC the reaction goes this way if you have Br4, Br4 is basically BR 3 + BR tin is a kind of a Lewis acid. So, the lone pair of electrons will go there eliminating Br- in this Br- reacts here. So, ideally, if the acetate is up the plane in the beta position then as a nucleophilic substitution rule your Brshould have come from the opposite side but it does not happen. Independent no matter what if we were there the stereochemistry here be it Beta Beta Alpha you always get a beta isomer of this compound and this comes from the rule which is known as the Baker rule.  This is what the Baker rule is it basically states that if you use sugar compounds like this have started it can have different the geometry if it is Arabinose D arabinose I am not now talking in general sugar chemistry D arabinose will have the acetate if you have an acetate this should be the 2 prime positions would be up the plane beta position. If this is a weak source then 2 and 3 both are up the plane. If you have ribose then both are down the plane and we are using ribose so both are down the plane for our case. If you have xylose then 2 prime is down the plane 3 prime is up the plane and this is the 5 prime or 4 prime is also down the plane but in all cases whatever stereochemistry you use of the C 1 prime be it a beta be it alpha when you do the reaction when you do a nucleophilic substitution reaction their stereochemistry does not matter. What matters is what is the stereochemistry at the C2 prime position if the 2 prime position is up the plane then your nucleophilic substitution can be a base this can be any nucleophile. The nucleophile will come down the plane in the final product. So, if 2 prime position is up the plane then the nucleophile at the C 1 prime position will come opposite side of it down the plane. So in this case, 2 prime position is I am just writing OAC up the plane then the nucleophile will come down the plane alpha anomer is going to be your major product. Similarly here if your 2 prime position is down the flame 2 prime positions down the plane here does not matter what is the stereochemistry of the other positions not even the C1 prime then you will have the beta anomer your nucleophile is going to come from the up the plane. So, that is what the bakers' rule is. That the stereochemistry when you do when you are trying to do a reaction at the anomeric position the C1 prime is known as the anomeric position when you are trying to do the nucleophilic reaction or a substitution reaction at the anomeric position then the stereochemistry of the final product will depend not on the anomeric atom or anomeric living group but will depend on what is the stereochemistry of the 2 prime positions and it is always the opposite of that geometry. If 2 prime position is up the plane or in beta position then you are going to have the alpha anomer. If they are down the plane you are going to have the beta anomer. So, there is a reason for this also why so same thing is happening here if you look if you come back that it does not matter what is the stereochemistry of OAC here since this is ribose sugar we are using so as 2 prime position is down the plane since this is down the plane your bromine or the chlorine or the nucleobase if you do it is going to come from up the plane and the beta anomer is going to be your preferred confirmations. Now there is a reason for it also why is that so if you look OAC here since this is involved I am going to draw this structure so always your 2 prime positions are dictating what should be the stereochemistry at the 1 prime. Now if you treat it with Sn Br4 Br3 and Br then this is going out which means you are going to have a transient intermediate O AC and this is going to be your Sn Br3 this is going to be your plus and Br- there OAC OAC. Now since you have a positive charge here it is going to leave and immediately this is going to attack. The oxygen or the double-bonded oxygen of the acetate is going to attack immediately to the C1 prime position because you are having a positive charge here which is leaving. So, that is giving you an intermediate which is this positive charge OAC OAC this kind of intermediate is going to come. So, it is an intramolecular nucleophilic substitution reaction because the acetate group of the same molecule is attacking as a nucleophile. So, if you see here the acetate group is attacking as a nucleophile and this is your living room so internal nucleophilic reactions or we call it NGP neighboring group participation. Because of neighboring group participation, it forms this kind of intermediate. So the whole of the intermediate that would be forming is down the plane. Now the Br minus that you had from which side it will come you are down the plane is blocking so only available position is up the plane. So, Br minus has to attack from up the plane because down the plane is blocked by the intermediate formation intramolecular intermediate formation. So, therefore this is going to attack and it is good it is going to come back because it is an intermediate it is not very stable you are going to get back your OAC where OAC here and then your bromine has to come from up the plane so, beta anomer. So, that is what Baker’s rule basically states. So, similarly when you are treating this with thymine or any other nucleobase if I use the OTMS the sale-related OTMS then this is going to attack again the same thing will happen Br minus will be eliminated OAC will attack from backside it will create an intermediate in the down the plane so your thymine has to come from up the plane will give you always thymine as the beta anomer. This is going to be your product. So, now that is for the sugars so far we have taken is the ribose sugar. So, this is D and ribose sugar right for RNA molecules. Now if you want to synthesize that DNA what you have to start with the deoxyribose, for example, bromine here nothing here OH OH and of course so this is for the D deoxyribose sugar we can start with the bromine version bromide added and of course you can convert it into an OAC and do the rest of the reactions like thymine and adenine addition the same way. But here there is a problem now if you start with any geometry here you are not going to get the same beta anomer that we have seen for the ribose cases. The reason is for ribose cases you had substitutions in this position which was down the plane and that was dictating the stereochemistry of the incoming nucleophile. In this case for the DNA case, you have nothing here. So, there is nothing to dictate the stereochemistry of the nucleophilic reaction over this position so if you want to get the desired beta anomer of the nucleoside there is no other alternative but to start with a specific stereochemistry at the beginning. So if you want to have the desired beta anomer then you are going to start with the specific alpha anomer alpha substituted bromine of this position and then you treat it with the thymine just giving the example OTMS OTMS and these same conditions then you have a little bit of fluoride or water either of them you can use then it is going to be pure Sn2 kind of reactions where your nucleophile will attack.