 Hello everyone, I would like to welcome you all for today's lecture. And before we proceed for the remaining part of the topics, we would like to recap of the last lecture in brief. As I have shown it here that what we learnt in the last class was the use of polymethyl hydrosilane. This is the structure and we studied its reactivity in the reactions of carbonyl compounds in the presence of tetrabityl ammonium fluoride. Then we also looked at the reactivity of strikers reagent which is what it is and it permits as you will remember 1, 4 reduction of enones because that is a very important reaction. And in this connection, we also saw the effect of silicon based reducing agents both this triethylsilane and also polymethyl hydrosilane. Then we also looked at the silicon based radical reactions. In that context we saw both reductions using the silane based reagents and also C-C bond formations using radical based chemistry. In that context, we introduced this tritstrimethylsilane which is a sterically hindered hydrogen containing silane and also tetrafinyl disilane which is also a very nice reducing agent and also it allows C-C bond formation to take place. Now we looked at the reduction of halides like Rx where x can be especially bromine and iodine and then in the presence of benzoyl peroxide it gives the corresponding Rh basically replacing X R bond by H R bond. What happens is that the benzoyl peroxide as I discussed last time upon heating or phytolytic conditions decomposes eventually to phenyl radical by the loss of carbon dioxide. That phenyl radical then picks up the hydrogen from here generating this TTM radical. This TTM radical then reacts with the Rx to form R dot which is what is important and then TTMx which looks like this that means the silicon X bond is formed and generate this R dot. This R dot then reacts with TTMS that is tri, tritstrimethylsilane which provides hydrogen and it forms the Rh which is the reduced product and of course we recover the TTM radical. So this is how the chain continues and we have the reduction of Rx to Rh. Then we also looked at Barton McCombie reaction or it is also known as Barton deoxygenation reaction in which we can convert a hydroxy compound to the corresponding hydrocarbon. Particularly it is very useful for the conversion of a tertiary hydroxy compound to the corresponding hydrocarbon that means the carbon hydroxy bond is replaced by the carbon hydrogen bond going via the corresponding thio carbonyl compound of this kind here where R1 is S methyl O phenyl etc. We made use of that in the conversion of a sugar derived thio carbonyl compound like this to the corresponding deoxygenated product of this type here but here we use tetra phenyl disylane as a reagent in place of the tributyltin hydride. However wherever we use tributyltin hydride as a reagent one of the main disadvantages is that we have to use excess of tributyltin hydride which is toxic, expensive and very difficult to remove from the reaction mixture. It is in that context I also mentioned another alternative usage of catalytic amount of tributyltin oxide that is 7.5% as a radical source in the presence of poly methyl hydroxyl oxane as the hydrogen source. What I mentioned last time was that we can take a hydroxy compound like this which can be converted to the corresponding thio carbonyl derivative by using this reagent phenyl chlorothione format in the presence of pyridine and then of course we use catalytic amount of tributyltin oxide in the presence of PIMHS and AIBN of course and N butanol as one of the very important solvents in toluene and the refluxing condition that allows the deoxygenation to take place to form this particular product. Of course the side products turn out to be carbonyl sulphide which goes away and this tributyltin phenoxy compound basically what happens in these reactions is the tributyltin oxide reacts with PIMHS to form the tributyltin hydride which then of course in the presence of AIBN forms tributyltin radical and that leads to the formation of the deoxygenated product from this thio no carbonyl derivative and the side product which has come out from the tin based reagent that is phenoxy tributyltin derivative then that reacts with PIMHS to regenerate the tributyltin hydride so and of course there will be some silent derivative so this is how the catalytic amount of tributyltin oxide is utilized to kind of carry out the reaction for the deoxygenation of alcohol so the corresponding hydrocarbons under these conditions. Now likewise halides also react with the tributyltin hydride and AIBN to generate radicals and get involved either in reductions or C-C bond formations. Now this chain process could lead to the polymer formation since we are discussing about the radical based reactions and reductions at the same time it is possible that we can look at the C-C bond formation using the radical based chemistry. So what is happening is that in these cases if you have an Rx here and you react it with tributyltin hydride in the presence of radical initiator like AIBN as we discussed last time it would form a this particular radical which is what we call it as initiator dot that interacts with the tributyltin hydride and it forms INH which is what is this particular molecule or radical interacts with the hydrogen and forms this and releases the tributyltin radical which interacts with the Rx to be reduced and then R dot is formed and tributyltin X is released and then this radical interacts with the tributyltin hydride to go to the reduced product. Now this is something that we had also seen similar to silicon based reactions. Now what can happen is this particular R dot can react with an olefin of this type where there is a double bond between A and B and when this radical attacks onto this you can generate a another radical here like after the C-C bond formed which then can abstract hydrogen radical and release this particular molecule which is what is a C-C bond formed molecule. On the other hand if this radical interacts with the double bond and there is no tributyltin hydride to stop the reaction then we can expect a polymer formation of this kind because this radical which is formed RAB dot that again reacts with the next molecule to form RAB AB dot that continues further to form a polymer. So such polymerizations are known in radical based chemistry for example polyethylene or many such kind of molecules are formed basically by such reactions. But then from the organic synthesis point of view we also need to have both the alternatives that is either we reduce the Rx to Rh or we have the stopping of the reaction at this stage when one C-C bond is formed. If we have excess of hydrogen source then the polymer formation could be minimized as you can understand that if we have say R, A and B dot form in order for this reaction to get arrested we need to have tributyltin hydride in a sufficient quantity so that that reacts to form the corresponding terminated product after one C-C bond is formed. What has been done is that a large number of studies have been carried for these kinds of reactions and it is found that the reaction depends on steric factors and also the nature of the substituent Z on the olefin. What is found that if one takes an olefin of this kind where there is one Z substituent and if Z happens to be an electron withdrawing group then the formation of the first C-C bond could be arrested. See what is found is that if Z is edge the rate is if it is considered as one then if you take ester as Z an electron withdrawing group the rate turns out to be 450 and if some more electron withdrawing group like aldehyde then the rate is 2300. So basically what it means that the reaction can be arrested at the first C-C bond formation and reaction could be quenched by tributyltin hydride if we have electron withdrawing group attached to the double bond. So if we take a catalytic cycle then we have this r dot which is formed from this rx and if this r dot reacts with the olefin then we generate another radical here which in is situated alpha to the Z group next to the Z group and if that has a rate k1 and then the k2 is the one in which the tributyltin hydride interacts and it releases the end product like this where hydrogen is attached to the carbon where there was an radical alpha to the Z group and of course then you release tributyltin radical and then that reaction continues. So in this respect what has been found that if k1 which is shown again here if k1 is not fast enough then if this particular rate is not fast enough then r dot obviously will abstract the hydrogen from tributyltin hydride and the reaction will end in the reduction and this is the product that is going to form but if k2 is not fast relative to successive addition so if this particular rate here is not fast enough then what will happen is this radical will not get quenched but it will then react further and then oligomer or the polymer will form. So what is important is that this radical should be less reactive towards the original double bond than r dot is towards this that means this particular radical should not react with the next olefin but then that particular rate should be less than the rate in which the r dot is reacting with the double bond and then this particular one should be more reactive towards the tributyltin hydride than r dot is towards it. Very clear that this should react faster with tributyltin hydride than r dot reacts with tributyltin hydride because if r dot reacts with tributyltin hydride you will get the Rh but if this reacts with tributyltin hydride then you would get the the reduced product that is this here the reduced after the C-C bond formation. So this is how it is done by Z group being an electron withdrawing group. Now exactly how why and how the electron withdrawing group allows such a reaction to occur that means to stop the reaction after the first C-C bond has formed and then the reaction is arrested. What is found is that alkyl radicals with electron releasing groups behave like nucleophiles and react fast with electron deficient olefins and vice versa. So suppose if you have say r dot which is a carbon radical and it behaves more like a nucleophile and that reacts with say you have an electron withdrawing group such as CHO attached to the olefin then this reacts faster and or you have an radical which is having electron withdrawing group here or electron deficient radical then it reacts faster with say you have an electron rich olefin that is how it is vice versa. Alkyl radicals with electron releasing groups behave like nucleophiles and react fast with electron deficient olefins and electron withdrawing groups which they are attached to the radicals they react with electron rich olefins. Now in order to enhance the possibility of C-C bond formation and stopping after the first C-C bond has formed what needs to be done is that and also at the same time to make sure that the reduction that is Rx does not get reduced to Rh this could be one competing reaction but what we want is a C-C bond formation so we need to have this particular bond to be formed or as we wrote earlier then you have A and B here. So basically we want to stop the reaction after this radical has formed which then quenched with the tributyltin hydride to form this. So the competition is between this and this formation for that you have to decrease the concentration of tributyltin hydride that means the concentration decreases means the in the solution of the reaction you increase the dilution. If we increase the dilution that means the concentration of tributyltin hydride has decreased. Now for this there are two pathways one is to add very slowly a dilute solution of tributyltin hydride that means at a particular time the amount of the tributyltin hydride which is present in the reaction should be as less as possible and this is done using syringe pump it is a kind of device it is a pump in which you take a very dilute solution of the tributyltin hydride and time it out to add a very tiny drop in say over a period of 3 minutes or 5 minutes you can work it out and add very slowly drop wise over a period of say 2 hours or 3 hours so that in the concentration of the tributyltin hydride at any time is very less of course you have to use more than one equivalent of the tributyltin hydride but then it is put it in a solution which is very dilute solution so that the possibility of reduction is reduced. One can also generate tributyltin hydride in situ by using sodium cyanoborohydride or even sodium borohydride but sodium cyanoborohydride is a better alternative than sodium borohydride because sodium borohydride does react to the carbonyl group and along with the sodium cyanoborohydride or sodium borohydride one can use tributyltin chloride in catalytic amount so what exactly happens is the catalytic amount of tributyltin chloride reacts with say sodium borohydride or sodium cyanoborohydride to make the corresponding tributyltin hydride and this tributyltin hydride then under the conditions where we use AIBN as a catalytic amount and of course we either heat or we fertilize the reaction medium and then of course we generate the tributyltin radical. Now this tributyltin radical will then abstract the iodine from here as I have shown in here to generate this cyclohexyl radical and then cyclohexyl radical will add on to the acrylonitrile to form the corresponding CC bond. So basically what is happening is that during the process when we regenerate the tributyltin iodide from the starting material then of course this tributyltin iodide is then reduced with sodium borohydride which is used in more than one equivalent as I have shown here is 1.3 equivalents of sodium borohydride. So this is how by using catalytic amount of tin material tin based starting reagent and of course using excess of reducing agent like sodium borohydride or sodium cyanoborohydride can be made use of particularly to avoid the generation of a large amount of tributyltin hydride or a large amount of tin impurities in the reaction mixture which are difficult to remove. Now in addition if we have then 1 and 2 having opposite polarity then it is possible that the reaction will be facile because they complement each other in terms of electronics. Now another example of opposite polarity can be seen that we have a halide in which there are two electron withdrawing groups like esters are attached and we have this electron rich double bond. So it is clear that when tributyltin hydride reacts with this particular chloride you generate this radical here and this radical is having two electron withdrawing groups therefore it is much easier for it to react with electron rich double bond where the radical is generated at this center. So basically what is formed is this type of molecule in which you have a radical which is now next to the oxygen and therefore it is a stabilized radical because of the lone pair of electrons on the oxygen here. So this is how the example of opposite polarity can be taken up and as you can see the yield is very good 80%. It is also found that the carbon bromine bond of the original starting material RBr and say acrylonitrile the react equally fast with tributyltin radical. But then what is found is that Ri reacts 10 to 100 times more than the RBr and therefore many times the possibility of using Ri is explored. However there is a little problem sometimes some of these iodic compounds are not as stable as RBr at particularly high temperature during the reaction. So one has to look at the compromise between the two and then carry out the reaction but the use of dilution avoids the reduction of the radical at the first stage and therefore C-C bond formation occurs and then subsequently the reaction stops. One can look at the radicals reactivity if you have a nucleophilic radical and if you have an electrophilic radical then this one can look at the singly occupied molecular orbital obviously would be of higher energy because the radical which is generated if it is having an electron donating group then its orbital having a single electron would be of a higher energy. And complementarily if the olefin has electron with drawing group say here, electron with drawing group is attached to the olefin then of course its lumo would be of a lower energy and therefore there is a possibility of interaction of the somo that is singly occupied molecular orbital with the lumo that is lowest unoccupied molecular orbital because energy gap would be less and therefore the reaction occurs and then the product forms of this time. On the other hand if we have an electrophilic radical then we expect that we use say you have an electron donating group attached to the olefin and if the R dot is having an electron with drawing group so the singly occupied molecular orbital will be of lower energy and therefore it will go down. On the other hand the lowest unoccupied molecular orbital will be of higher energy. On the other hand the homo the highest occupied molecular orbital would be of also higher energy but then the somo of the electrophilic radical and the homo of the olefin would be easy to interact with the energy will be less and therefore one generates the corresponding radical with this kind of arrangement like this. And therefore these are the important frontier orbital interactions of two different types of radicals with two different types of olefins one with electron with drawing group and the other with electron rich olefin. At the same time we can also look at the couple of more aspects of it. One is that as we discussed the electron with drawing groups lowers the energy of lumo of the olefin and somo of the nucleophilic radicals is of higher energy. That is how for example if you take simply the cyclohexyl radical it reacts 8500 times faster than the same radical reacting with an ordinary olefin having no electron with drawing group. Tertiary radical which is electron rich reacts faster with acrylonitrile than the corresponding say for example you have a secondary radical or a primary radical in comparison to that this tertiary radical reacts faster obviously because of electron releasing nature of the substituents on the tertiary radical. On the similarly electrophilic radicals will have low energy of the singly-operative molecular orbital and therefore somo-homo interactions are dominant. Finally in this regard we have one example in which if we take a molecule like this where there are two hydrogens which are likely to be abstracted by different radicals. So we compare chlorine radical versus say methyl radical. Methyl radical has high energy somo and so it attacks the hydrogen bond with low energy lomo that is lomo which is alpha to the carbonyl group. So you have this is the alpha position and this is the beta position. So this particular hydrogen CH bond would be of low energy lomo and therefore the methyl radical reacts there. On the other hand Cl dot will have low energy somo and so it attacks the CH bond with highest energy of Homo which is what is this particular CH bond which is away from the carboxylic acid. So even such kind of abstraction of hydrogens are also affected by the similar type of reactivity as we discussed with the nucleophilic and electrophilic radicals reacting with double bonds having either electron withdrawing or electron donating group. So we will stop it at this stage and then we will carry on the discussion in the next class on some other aspects of reactivity of the molecules using radical chemistry or other reactions. So we stop it here and you try to study whatever I have told today and till then goodbye and thank you.