 Hello everyone, I would like to welcome you to today's class. What we discussed last time we would recap in a brief way today. We started with zinc borohydride based reductions last time and we saw how the chelation with zinc allows highly diastereoselective reductions of ketones and we tried to look at some examples in which the chelation is seen both as a 6 member and a 5 member transition state. Then we looked at the super hydrides and we saw how the like lithium-ry ethyl bromoborohydride is introduced in and compared with lithium borohydride which is more sort of ionic than lithium triethyl borohydride. Since the electron releasing nature of ethyl group makes this as a stronger reducing agent this is called as a super hydride because it is much stronger than lithium borohydride. And then we also saw how it can be compared with lithium aluminum hydride where we saw the reactivity was quite different from the lithium triethyl borohydride. And then we also looked at towards the end the selectrides and in the case of selectrides we saw 3 different types of L-selectride, K-selectride and N-selectride. Of course, these two are not really so much popular as the L-selectride is because of the lithium plus being the stronger collating ion than potassium and sodium. And therefore L-selectride is a reducing agent of a choice in which we have the secondary butyl group as a bulky substituent on the lithium borohydride paste reducing agent. And therefore L-selectride was obviously a choice for the reductions in which high diastereoselectivity is expected. And we saw some examples we also discussed and looked at that these are used only in cases where high diastereoselectivity is required at low temperature. And we then looked at the temperatures where the some cases minus 78 degrees can also be used because of the high reactivity of such molecules. They are called selectrides because there is a selective reduction and therefore they are called selectrides. So these names like superhydrides, selectrides etc have been given based on the reacting nature of such molecules. Now we look at even more strong reducing agent where even more steric hindrance has been introduced. For example, this is called as LS-selectride and this is called as KS-selectride. L stands for lithium and S stands from the trasiamyl part that is tri iso-amyl part in which the S letter has been taken up and this is how it is called as LS-selectride as you can see here. So this is KS-selectride and this is LS-selectride in which the groups which are attached are bulkier than secondary butyl basically it is iso-amyl or tri iso-amyl group which is present. Now we take a case something of this kind which is reported in literature in 2003. Now if we take this bicyclic molecule which is now reduced and as you can see the reduction leads to this particular hydroxy group being alpha. If we see the conformation of these molecules, this molecule we see that if we make it as a trans type of molecule in which now we have a nitrogen here and hydrogen here then we can put the oxygen at this stage methyl group is alpha therefore we can put it in this way. And now if we put the carbonyl group which is here and then the substituent C5H11 which is beta oriented is put it here C5H11. Now very clear that this particular conformation is not a preferred conformation because the bulky C5H11 is axial. Therefore, we would imagine that this particular conformation is preferable as you can see that the C5H11 now is in equatorial orientation and the methyl here this is axial hydrogen this is methyl this is also equatorial this is equatorial hydrogen and we can remove this part here and look at the methyl being equatorial. So both the groups are equatorial and now in this case what is happening is that this particular CH2 group is axial and therefore the reduction does not come the reducing agent does not approach the carbonyl group from the lower side because lower side is basically offering a steric hindrance and therefore the reduction takes place from the top side that is the beta side and therefore the reduction leads to the hydroxy group being alpha oriented and the hydrogen is coming from the beta side. This is very easily seen by this particular conformation and so there are two reasons why this conformation is preferred one is of course you have a substituent which is expected to be in such a way that the bulky substituent prefers equatorial orientation and then between the two of them that we consider the trans as well as the cis type of di-calone type of molecules we prefer the cis one and obviously in all these cases wherever we have discussed the carbonyl group reduction with any reducing agent we have to keep in mind that the reduction takes place at 107 degrees angle which is the rule according to or the descriptions made according to the Burgi Donitz kind of hypothesis or the observations. So in any case the reduction has occurred in a very highly diastereoselective fashion mainly because of the very large bulky groups which are present on the this LS-selectride as we discussed earlier it is whenever there is a preference of L-selectride versus K-selectride we prefer L-selectride in a similar fashion when we have LS-selectride versus K-selectride people prefer LS-selectride where there is a solubility problem. Now we go to another reducing agent which is sodium cyanoborohydride not only alkyl groups which we saw in the case of lithium triethyl borohydride or selectride or LS-selectride we saw different alkyl groups which are being put in order to increase the electron releasing nature of the borohydrides and thus making them stronger and stronger reducing agents. But not only alkyl groups that can be put as substituents but even electron drawing group such as a cyano group has been introduced and attached to the boron part of the sodium borohydride which is easily made by reacting sodium borohydride with hydrogen cyanide and that leads to a sodium cyanoborohydride. So what is the purpose of such a reagent that is something that we need to understand it. Sodium borohydride is obviously the simplest reducing agent and that reacts with carbonyl compounds readily because it is ionic and in ethanolic or ethanolic solutions it reacts. But when we put cyano group so cyano group is an electron withdrawing group this particular cyano group is an electron withdrawing group and obviously it reduces the nucleophilic nature of the borohydride part that is now in this particular case BH3 is much less than the sodium borohydride case. It is obviously not easy to reduce a particular carbonyl group with a compound that is having an electron withdrawing group as a nature cyano group as an electron withdrawing group and therefore in order to increase since we have decreased the nucleophilicity of sodium borohydride by putting cyano group. So we need to increase the electrophilicity of the substrates that is the carbonyl group and what is interesting is that because the nucleophilicity nucleophilic nature of the sodium borohydride has been reduced by putting cyano group. So the sodium cyano borohydride now is not nucleophilic enough. So it is found that it does not decompose in a solution even up to say BH3. That means it is stable under acidic conditions and that is the advantage and it is also soluble in THF methanol water HMPA that is hexamethyl phosphoramide DMF that is dimethyl formamide and these they do not react with these particular solvents. Now just for comparison this is a phosphoramide and this is a phosphorus triamide but this is more used as a solvent and this is used as a kind of phosphine in comparison to triphenylphosphine or thialkylphosphine. What do the sodium cyano borohydride allow the reactions of that is if we take a halide such as RI or RBR or R-tosylate where there is a fairly good leaving group. You have an iodide as a leaving group, bromide as a leaving group or a tosylate as a leaving group or even mesylate as a leaving group. So we have paradolene sulfonyl or this methyl methyl sulfonyl. So sodium cyano borohydride reduces these molecules and the leaving groups go and then hydrogen is introduced at the R position. So I can see here at in HMPA at 70 degrees this kind of molecule which is somewhat sensitive molecule we can reduce this particular carbon bromine bond and introduce here carbon hydrogen bond. In this case is one thing which is important to remember that the reducing agent sodium cyano borohydride is stable under acidic conditions up to pH 3. So what are the reactions that are done? As I mentioned if we have decreased the reducing ability of sodium borohydride by introducing a cyano group and thus the sodium cyano borohydride is less nucleophilic therefore we need to increase the electrophilicity of the molecules which we need to reduce. And for that purpose since sodium cyano borohydride is stable up to pH 3 to 4 we can add acid into the molecule where carbonyl group gets now protonated to form the corresponding oxonium ion which then now is fairly good electrophilic in nature to which sodium cyano borohydride then reacts and then your hydrogen is transferred as a hydride and the corresponding alcohol is formed. So we can also do in this fashion that we put H plus and then we have sodium borohydride and obviously we can introduce the deuterium here and form the corresponding deuterated alcohol. So this utility of sodium cyano borohydride to the corresponding deuterated alcohol is also utilized it. Now in all these cases what we have seen is that we have introduced the acid to the reaction medium in order to increase the electrophilicity of the carbonyl group. We can also utilize the nature of this sodium cyano borohydride which is stable under acidic conditions in such a way that for example if we take an alcohol of this kind and see that the alcohol which is a tertiary alcohol can easily be reacted with a Lewis acid and could be ready to form a sort of carbocation which is a tertiary carbocation. In this particular case it is a rigid molecule and therefore the alcohol when it coordinates with a zinc bromide it releases the OH path and then sodium cyano borohydride attacks from the same side and the hydrogen is introduced where the OH group was present. Basically only because the reagent sodium cyano borohydride is stable under acidic conditions. So it is basically what we are talking is that you have a sort of carbocation that is formed. So carbocation allows the reduction to take place. Even here in this case as we can see that this carbonyl group is reduced to the corresponding alcohol and the at pH 4 the reduction allows the approach of the hydrogen coming from the beta side here because of the ester group which is alpha oriented. So you have a beta hydrogen to come at the carbonyl carbon. So not only it is reducing it at lower pH but it is also of course following the same principle of stereoselectivity. Now in this case in the last example here the aldehyde is also reduced again at pH 4 in methanol as a solvent. So when can go all the way. So you have a possibility of going via a carbocation. We have a possibility of protonating the carbonyl groups and therefore under these conditions the sodium cyano borohydride being stable can allow the reaction to take place. Now because carbocation can be formed as we can imagine that we have an enamine and this enamine can be reduced to the corresponding saturated molecule basically because you have the enamine which is kind of nucleophilic in terms of the fact that we can move the electron density to the proton. So if we have under acidic conditions the reaction to take place in the presence of sodium cyano borohydride then we can expect an intermediate to form something of this sort. So this is the intermediate that will form which then gets reduced under the conditions to basically have the reduction taking place at the center and then there is such molecule is coming. So this particular hydrogen is coming from the acid and this particular hydrogen is coming from the sodium cyano borohydride and that is how we get this saturation of the double bond. In a similar fashion this is an emine which is very easy to understand that it gets protonated under the conditions to form the corresponding positive charge on this center and therefore the nucleophile sodium cyano borohydride would attach it here and then your hydrogen will come there. So this hydrogen comes from sodium cyano borohydride and this hydrogen comes from the corresponding acid. Therefore such reductions are quite useful in order to which cannot be easily done by means of sodium borohydride because such possibility of formation of an ammonium ion or a carbocation is not there. If we now look at this aspect of it here of examples in which we have done the alkylation that is very interesting. Now I would like to discuss it in detail about it that how does this particular alkylation of amines takes place. So let me remove this particular part of it which we discussed just now and look at it here. If we take this particular molecule and react with formaldehyde and of course sodium cyano borohydride and under these conditions what happens is first the formation of an something of this sort occurs after the nitrogen of the molecule methyl interacts with the formaldehyde. That means there is a condensation with the formaldehyde and an emine is formed and this ammonium ion then gets reduced at this stage here with sodium cyano borohydride to form here NHCH3 and of course the corresponding carbonyl group is present and the corresponding ester is also present. Now here there are two aspects one of course we have deliberately added formaldehyde therefore the reaction occurs by condensation of amine with the formaldehyde and we form this secondary amine. Now if we continue the reaction secondary amine also will undergo condensation and form the this type of CH3 and here you have a double bond CH2 positive charge. So the second condensation occurs of the secondary amine here and again the reduction takes place at this center with sodium cyano borohydride here leading to the formation of dimethylamino group. So what we have done is we have introduced say you have an R and then you have NH2 and we have gone stepwise so to form RNH CH3 and then you have RNCH3 and CH3. So that is because of the formaldehyde and sodium cyano borohydride and that is a very straightforward and an easy way of introducing a dimethyl group on an primary amine using the properties of sodium cyano borohydride reducing a cationic species. So the interesting way of converting a carbonyl group to the saturated sort of molecule like this hydrocarbon here. What are the ways by which such a reduction can be done? That means if we have a carbonyl group of course we can carry out Wolf-Kishner reduction and then we have another possibility that we protect the carbonyl group here as a diethioketal and once we have the diethioketal the reduction can be done by Renny-Nickel, a special Nickel catalyst which on the surface of which hydrogen is absorbed and that allows the cleavage of this carbon sulfur bonds by the hydrogen and then you get the corresponding molecule in which the oxygen is replaced by two hydrogens. Now there is another way by which we can do the same reaction by converting carbonyl group to the corresponding n-tosyl hydrazone that is what the reduction is that we are discussing and if we use a proteic source in the in the DMF and heat it we can get the same reduction but then it is via tosyl hydrazone. Now what is the mechanism and the acidic condition the nitrogen the first nitrogen here of the tosyl hydrazone gets protonated here and we generate a ammonium ion here to which now sodium cyanoborohydride donates hydride here and the double bond moves with the loss of tosyl group here it goes off leading to the formation of this particular intermediate which then is losing a proton and goes to another molecule like this which can now be expected to be somewhat like this that we have loss of nitrogen in this fashion and then under the proteic condition the anion which is formed here is grabbed by the proton which is present here. So the blue hydrogen is coming from the sodium cyanoborohydride and once that allows the formation of this proton proteic species here this can rearrange by the loss of nitrogen and the proton is coming at this stage this is how the reaction occurs. Now if we take alpha beta unsaturated system we have seen that the normal tosyl hydrazones are reduced to the corresponding molecule which in which the carbonyl group has now through tosyl hydrazone be replaced to the corresponding two hydrogens. But if we take an alpha beta unsaturated system which is from molecule like this if we take a molecule like this we can convert into the corresponding tosyl hydrazone which then under the similar condition forms the corresponding hydrazine type of molecule like this which then undergoes rearrangement in this fashion to move the double bond to this position which is this was a conjugated system. Now what we have found that under these conditions it undergoes deconjugation and the hydrogen is coming at this stage here the hydrogen is coming here. So the original hydrogen had come at this stage here this should be blue in color and then we have a rearrangement that means now what we have started we started with this molecule in which we had the carbonyl group like this and eventually what we have got is this. So it is not only the reduction of the carbonyl group to the corresponding hydrogens which we would have expected to give something like this if it was a normal reduction then we would have expected the two hydrogens to come here but that is not the case what we have got is the two hydrogens which have come are basically at this position here eventually. So instead of getting this molecule which we did not get it we got this molecule via the tosyl hydrazone. So this is how the reactions of sodium cyanoborohydride take place. So we will stop it today at this stage we have seen various aspects of these reducing agents and now we will take up the other reducing agents next time you can go through the notes of today's class and get ready for the next class till then bye and thank you.