 Hello everyone, welcome to today's class. We will briefly go to matter that we discussed last time. If you recall we discussed initially the reductions using lithium aluminum hydride and aluminum chloride and of course we saw how the selective reductions can be carried out. Then we proceeded to look at the reactions of redol and the redol comparison with lithium aluminum hydride and we saw how important is this reducing agent because it is better than lithium aluminum hydride in many sense for especially from handling point of view. And it reduces the acid to the corresponding alcohol but interestingly when you take epoxy alcohol both redol and aluminum, lithium aluminum hydride they lead to selectivity which is different from that type of dye ball. We then of course discussed the mechanism involved in those reactions and saw how in the lithium aluminum hydride and redol case intramolecular reduction of the epoxide gives 1, 3 diol as the major product and in the case of dye ball it is an intermolecular reduction that allows the formation of 1, 2 diol as the major product. And then towards the end we discussed luchy reduction and as I mentioned after a lot of combinations the luchy and gemal found out that cerium chloride 7H2O that means hydrated cerium chloride along with sodium borohydride very comfortably gives the reducing system which allows reduction of say particularly carbonyl which groups which are alpha beta unsaturated can be aldehyde or can be ketone and that leads to specifically reduction with the reducing agent to the corresponding allylic alcohol. This is something very interesting because it is very easy to perform this reaction and this selectivity is very high. Then of course we also discussed how the ketones oxygen and also alpha beta unsaturated ketones or aldehydes oxygen here is more Lewis basic than the corresponding unsaturated ketones or say aldehydes. So this difference in the Lewis basic character of ketones and alpha beta unsaturated systems was also exploited by the luchy and gemal in their reduction. So we will look at some aspects of it now in this lecture today. Initially it was felt during the reaction that more electrophilic aldehyde gets protected as aldehyde rate and then gets regenerated during the workup. That means what is happening is that if suppose you have a molecule that has say aldehyde here and also it is a part of a corresponding ketone. So you have an R group here. So this particular compound allows the reduction of the ketone over say for example aldehyde. This is the selectivity that is particularly found in the case of reaction of a molecule that contains both an aldehyde and the ketone and the normal conditions aldehyde being more electrophilic in nature compared to the ketone should get reduced with sodium borohydride or with any reducing agent. But the combination of reagents such as cerium chloride heptahydrate and sodium borohydride allows a kind of aqueous medium because the reaction is also carried out in methanol as a solvent. So in methanol as a solvent everything is miscible and at that time initially it was felt that the aldehyde gets basically formed as hydrate and gets blocked and does not react something like this and the ketone remains as it is. So this is the intermediate that was proposed that under the conditions here when we have a cerium chloride 7H2O at that time when molecule containing both aldehyde and the ketone is exposed to these conditions the aldehyde basically gets blocked as a hydrate and now the reduction takes place on to this because this particular oxygen atom is more Lewis basic and therefore the cerium plus 3 coordinates with this which is readily available. And then during the workup the aldehyde is regenerated that means once the reduction has taken place this is what originally it was felt. Later on more detailed studies of these mechanisms of involved in this particular reaction were carried out and it was found that methanolesis of sodium borohydride occurs to form methoxy borohydrides that means as I have already mentioned earlier that we take sodium borohydride normally sodium borohydride is easily used in alcoholic medium but I also mentioned the earlier time that sodium borohydride does react with methanol or ethanol but slowly. So under these conditions it is expected that what may happen is that the methanol reacts with sodium borohydride and forms different species that means you have NaBH say for example it can be 4-n that means it can be 1, n can be 1, 2 or 3 or something of that kind forming this species and of course the corresponding hydrogen goes off. So this is the methoxy borohydrides which are expected to form this particular junction when sodium borohydride is used in methanol along with cerium salt. This sodium methoxy borohydrides are obviously harder reducing agents because now you have methoxy groups attached to it and therefore the electron withdrawing nature of the methoxy group due to inductive effect makes the corresponding hydride here of the sodium borohydride as harder based on acid based principle, hard soft acid based principle. And thus it allows 1, 2 reduction to take place. It is very clear that a species of this kind should be involved because it is expected that even methanol has coordination with the cerium plus and of course now the modified methoxy borohydrides then interact and act as a reducing agent to this where the carbonyl oxygen has a kind of interaction with the H. Now because you can expect that now this can become delta plus and therefore there is a coordination of the oxygen with the H plus and this is the path which is coming from the methanol. So basically what is happening is cerium chloride activates methanol in this particular fashion and thereby makes the proton very labile and a harder acid. Now on the basis of HSOB principle what we expect that the reaction of an alpha beta unsaturated ketone may proceed in this particular fashion. So what is the initiation that happens that is when an alpha beta unsaturated ketone comes in contact with the proton that is being released from methanol which in turn complexes with the cerium plus 3 salt and therefore this more basic ketone of the alpha beta unsaturated system the more basic oxygen interacts with the harder H plus and thereby increasing the electrophilicity of this particular carbon atom making it a harder Lewis acid and to which hydride then attacks. This hydride is basically made up of try methoxy borohydride in which there could be one methoxy or two methoxy or three methoxies it could be a mixture of all all the three types of methoxy borohydride and therefore they form a harder Lewis base in terms of H minus which then interacts with the harder Lewis acid at this particular carbon atom which is due to the coordination of the oxygen which is a harder base with the harder acid here H plus released by the complexation or as a result of the complexation of methanol with the cerium 3 plus salt resulting into an allylic alcohol. This mechanism has been well studied and reported in these two papers and we can therefore assume that based on the HSAV principle the reaction proceeds in this particular fashion. So this is what is usually expected that the reaction allows the reduction of an alpha beta unsaturated ketone to specifically form the corresponding allylic alcohol. If we take various examples of the Luci reduction we find that as I mentioned earlier one of the examples is that if we take a ketone and an aldehyde only the ketone gets reduced and the aldehyde remains unreacted. If we take an alpha beta unsaturated ketone having an aldehydic group then the ketone is reduced to the corresponding allylic alcohol and the aldehyde remains unaffected. So these are the two examples which I discussed where initially what was felt that the aldehyde group gets blocked under the reaction conditions as a hydrate. But then the mechanisms which we discussed where mixed methoxy borohydrates are formed where it is expected that the ketonic oxygen is more Lewis basic and similarly alpha beta unsaturated ketones oxygen is more basic than the corresponding aldehyde here. And that is the reason why the aldehyde does not react with the reducing agent and specifically the ketone or the ketone of an alpha beta unsaturated ketone gets reduced which is exactly opposite to what normally one would expect because aldehyde is more electrophilic than ketone or aldehyde is more electrophilic than the corresponding enone. In another example in which an enone was taken and a ketone was taken in the same molecule as you can see that the enone gets reduced to the corresponding alcohol allylic alcohol and the saturated ketone remains as it is. Again we can think in the same fashion that the ketone here is more electrophilic in the nature whereas enone is less electrophilic because of the double bond. And exactly on the basis of the larger Lewis basic character of an enone oxygen versus saturated ketone the reduction of the enone occurs. And if one takes for example an example of this type where an enone is there and an ester is there we have already discussed that the ester carbonyl here of course is much less electrophilic here and enone because of the double bond being there reacts specifically to form this. And here also as one can see that there is a stereoselectivity involved. Since this particular carbon-carbon bond is alpha oriented the hydrogen which is coming is from the beta side. So not only that there is a chemo selectivity these are the examples of chemoselective reduction. So because you have one type of aldehyde or ketone and the there are two different types of carbonyl compounds in each case and therefore it is a chemoselective reduction. At the same time this also stereoselective reduction chemoselectivity and stereoselective reductions occur. So both these are very important aspects and a number of examples have been reported in the literature where the utility of the sodium borohydride cerium chloride combination which is known as Luci reduction is used in a number of synthetic transformations. Now we see the stereochemistry in reductions of ketones. If we take locked ketones we find that reduction occurs from less ender side. So for example if we take a case of this bicyclic ketone which is two north bourna known or it is also called as bicyclo 2 2 1 heptane 2 1. Now if we see the positions of the hydrogen around the carbonyl group then we will see that the endoside of the molecule is sterically more hindered than the exoside. For example there are three endohydrogens which are at the lower side of the molecule whereas there is only one hydrogen which is blocking the face of the carbonyl group from the exoside. Therefore when a reducing agent approaches this particular molecule it sees more steric hindrance from the endoside and therefore it reduces the carbonyl group from the exoside or the hydrogen is transferred from the exoside resulting into the formation of the endo alcohol. On the other hand if we take a somewhat similar example but with different substitution like for example camphor. In the camphor case the carbonyl group now has a slightly different endo and exo selectivity. Here as you can see that the exo phase has two methyl here on the top which are blocking the face of the exoside. On the other hand the endoside has only the three hydrogens which were present in the case of two north bourna known. But because of the closeness of the methyl groups here to the carbonyl group the exoside is sterically more hindered and therefore the reducing agent approaches the carbonyl group from the less hindered endoside resulting into an exo alcohol. So basically what is happening is that in long cases where there is no possibility of any confirmation change the reduction always occurs from the less hindered side. If we take less rigid cyclohexanones then we obviously get mixtures where we can see how the distribution of the products occur. For example if we take a case like this where of course we have put tertiary butyl. So it is a kind of locked because the tertiary butyl group would always remain in an equatorial orientation of the molecule and therefore when the lithium aluminum handed attacks as you can see what is found is that the reduction occurs mainly from the axial side. That means the hydrogen approaches from the axial side leading to 92% of the equatorial alcohol. That means that since the OH group is bigger than the hydrogen obviously and the aluminum part would react with alcohol of course when the reduction occurs there is a chelation with the lithium plus and then eventually the O- which is going to form would interact with the AlH3 which is released and that will give O AlH3 and upon workup we get the corresponding OH. But then with the lithium aluminum hydride is a small reducing agent it prefers to attack in such a way that the larger OH or the O AlH3- goes to the equatorial side. But then if we say take a hindered ketone something like this where now deliberately we have put a methyl group here. This is the methyl group whereas here it was only hydrogen. So this was not that much sterically hindered. So if you put a methyl group here and it is not a locked particularly molecule as such but it has a methyl group which is axial. So if this is reduced as you can see that the major product is the one that is between the two it is 45, 55 but even then it is slightly more than the equatorial alcohol. So you can see that the preference of reduction occurs from the equatorial side so that we get axial alcohol. So the axial alcohol formation occurs because the equatorial side is somewhat. So this side is somewhat less hindered compared to this side and therefore we get 55 percent of the alcohol which is axial alcohol via the reduction from the equatorial side whereas the axial side reduction gives 45 percent. Now if we increase the bulk of the reducing agent because lithium aluminum hydride is relatively small in terms of steric size and therefore if we increase say lithium aluminum triad dorsary bitoxy hydride or lithium borohydride or any of these kind where there are big bulky groups present on the reducing agent and there as you can see that reduction occurs preferably from the equatorial side and then 99 percent more than 99 percent of the axial alcohol is formed because of the this particular portion being sterically hindered and therefore reduction preferentially occurs from the equatorial side this is the equatorial hydrogen. So this is how one can expect that the reduction of cyclohexanones can be predicted. Now we have the lithium borohydride as a reducing agent which can be easily prepared from sodium borohydride and lithium bromide. We take the combination of the two and then as you can expect that the sodium bromide will precipitate off from here if sodium and then lithium borohydride will form and there will be sodium borohydride bromide as a byproduct. So we get the lithium bromide borohydride very readily because of the lithium plus nature which is harder compared to sodium plus the reduction occurs not only of aldehyde and ketone but also as it chloride epoxides this should be ester and the lactone. So we can expect the reduction of these types of molecules to take place but it is interesting that we are basically trying to compare sodium borohydride with lithium borohydride here. So because the there is a difference in terms of the sodium versus lithium plus and lithium plus is harder therefore it is able to reduce even like epoxide esters and lactones. But it is not sufficiently since even then there is a borohydride part which is relatively mild reducing part and therefore the nitro, cyanide or carboxylic acids are not reduced. And lithium borohydride is found to be soluble like 4 grams in 100 grams of ether or 21 grams in 100 grams in tetrahedro furan. It is mostly used in the reduction of esters and lactones to diols as we have already shown it here that can be easily done. But if we have a molecule in which there is an acid to be reduced then of course we use borane, diborane or BH3 say THF complex or THF complex or any such thing which is the source of BH3 reduces the acid to the corresponding alcohols. So we see it very clearly that if we take lithium borohydride it reduces esters in presence of acid, in presence of carboxylic acid. But borane reduces acids in presence of ester. So this is something that it is important for all of you to remember that lithium borohydride and diborane they complement each other in terms of reactivity. And as I mentioned esters are easily reduced, similarly lactones are reduced to the corresponding diols. One of the best uses of lithium borohydride is to deprotect chiral auxiliary like we can see that we can take this particular molecule in which this part is a chiral auxiliary this part here. And because of this part we can make an anion here with say LDA and say add methyl iodide here as an electrophile we can introduce here the methyl group. So after the methyl group has been introduced here which is based on the side chain which is present and this is an optical reactive and therefore this is a highly diastereoselective alkylation. And when we need to remove the auxiliary from here what is done is simply use lithium borohydride which is soluble in ether or THF and we can completely reduce this particular part which is an easily reducible part with lithium borohydride equal into ester type of thing is and then we can get the corresponding CH2OH. So this is one of the best examples where lithium borohydride has been used in terms of deprotecting the chiral auxiliary. So if we look at the reactivity profile sodium borohydride is much less reactive which is comparable with lithium borohydride which is more reactive than sodium borohydride as we see in this in these examples and which is more now again comparable with the lithium aluminum hydride and lithium aluminum hydride is much more reactive than lithium borohydride. So we have the lithium aluminum hydride versus lithium borohydride which is then compared to sodium borohydride the reducing power keeps decreasing and that has to do with the lithium plus here and of course comparison with sodium plus and boron versus aluminum. So this is how the reactivity profiles have been found. So we will stop it at this stage today and you can go through these reactivity profiles of sodium borohydride, lithium borohydride and lithium aluminum hydride and try to understand what I discussed and we will continue with the reductions of various kinds in organic chemistry using other reducing agents and see how do they compare with the reducing agents that we have discussed so far. So we stop it here and thank you see you next time.