 Hello everyone, I would like to welcome you all to today's lecture. I hope you had the opportunity to go through what I discussed last time. We will see very briefly what we did last time and what are the aspects of reducing agents that we discussed. So, we looked at the chemo and stereoselectivity using the Luchy reductions and sodium borohydride, cerium chloride reducing agent system was utilized to specifically say see if we can reduce say in the presence of an aldehyde a ketone or a ketone in the presence of say which is an alpha beta unsaturated system. In the presence of these two the alpha beta unsaturated system is reduced between these two this ketone is reduced. And therefore, various aspects of mechanistic considerations we did it and found out that how methanol basically gets activated by cerium chloride and releases the H plus to be attached to more basic ketone or more basic enone. And therefore, there is a selectivity in terms of reduction of ketone in comparison to aldehyde or an enone in comparison to ketone. Then we also of course discussed the stereoselectivity part of it where if one particular side of the ketone is sterically bulky. Therefore, the reduction occurs from the other side as one is normally expecting in all these cases. Then we also looked at the stereochemistry of reductions of different kinds of reductions say with lithium aluminum hydride etc. in a locked ketone where we discussed norbornane type of systems. So we take a norbornane system like this and compared with another system where we had the methyl groups on the top. And we saw that the reduction of such locked systems occurred specifically in one particular direction. In this case, for example, the reduction gives the alcohol which is on the top because the top part of this particular molecule is sterically hindered. Whereas in this case, we got the corresponding alcohol which is endo alcohol because the top part is relatively free and therefore we get the reduction from the exocyte. Then towards the end, we used lithium borohydride as a reducing agent. We tried to see how the lithium borohydride is comparable with sodium borohydride and lithium aluminum hydride. So we found out that lithium aluminum hydride is a stronger reducing agent compared to lithium borohydride which is stronger than sodium borohydride. And one of the aspects of this lithium borohydride we discussed was that it reduces esters in the presence of, so in the presence of corresponding carboxylic acid or when we compare with borane, the boranes reduce carboxylic acids to the alcohol and the esters are not touched. So basically these are different reducing agents which have different power and selectivity in reductions. Now we look at another reducing agent which is zinc borohydride and it is readily prepared from zinc chloride and sodium borohydride as you can see that the sodium chloride would be off from the mixture as a byproduct and we get the corresponding zinc borohydride. It is specifically useful in the reduction of alpha beta unsaturated aldehydes and ketones. For example, this to the corresponding allylic alcohol and there is no very little reduction which leads to the corresponding saturated alcohol. Now this is very interesting because it has been reported in the literature that it allows a highly diastereoselective reduction of carbonyl compounds if they have say if there is a hydroxy group present or an alkoxy group present in the vicinity of the particular carbonyl group that is to be reduced. It allows the reduction to take place due to the coordination of zinc ions with the carbonyl oxygen and also to the proximity based OH or OR group and thus they form a cyclic transition state and that is the reason why there is a very high stereoselectivity or diastereoselectivity depending on molecule. Say for example, if we have a case like this here as you can see that there is a carbonyl group which is to be reduced. Of course, there is another carbonyl group but this is not going to be touched because it is not a normal simple carbonyl but it is basically an amide and therefore this is less electrophilic compared to the other carbonyl group and therefore that is reduced. And now this hydro the carbonyl group is very close to the hydroxy group here which is beta oriented. So if one can see that we can expect that there is a chelation of a 6 member. So you have 1, 2, 3, 4, 5, 6, 6 member transition state and therefore the chelation occurs from the beta side and therefore the reduction occurs from the alpha side. So this is an alpha side and this is a beta side and therefore the chelation allows the availability of alpha side for the reduction to occur and therefore we get here hydrogen which is coming from the alpha side and therefore this is a major product and this is the minor product. So there is a high amount of diastereoselectivity that you can see. If we take the lower example here then we have a ketone which is immediately next to the OPMB group that is paramethoxybenzyl group and that is alpha oriented but as you can see that is pointing downward and therefore when the chelation has to occur this has to go up in a beta fashion O here and PMB. So it would look like this, somewhat like this and then what you have is a carbonyl group which is now orienting itself in a fashion that will allow the, say for example we have something like this and then you have carbonyl group coming here and then of course something like this. So what we have done it is that we had done a rotation around here so that the PMB group goes on the top and now zinc width would coordinate like this with BH4 twice here and now the reduction also occurs from the alpha side which is free but then again after re-autotating it would come to the original OPMB side. So that means this would give a product like this OPMB and of course we have here reduction taking place that means you have an OH here and alpha hydrogen like this but then we re-rotate it here and then we go to the formation of product like this where now we will have OPMB. So this is the product that is formed. This is what is the product that is formed. So it is via 5-member transition state. So you could have either a 6-member transition state or a 5-member transition state but the main idea here is that zinc easily coordinates with 2 oxygens around in the vicinity and if they are properly oriented then accordingly the transfer of the hydrogen occurs to give highly diastereoselective reductions. Now we will move on to super hydrides that are very strong and very sterically bulky reducing agents. For example, lithium triethyl borohydride is one of the super hydrides where there are alcohol groups attached to the reducing agent and they are electron releasing and therefore the reactivity of such hydrides is very high. Now one can see that if we compare it with lithium borohydride obviously this lithium triethyl borohydride is much more powerful than lithium borohydride. And thus it is called as a super hydride which also acts like a nucleophile in an SN2 fashion. So we had discussed earlier for example, if we have a carbon halogen bond like this here and if we react with say lithium aluminum hydride there will be a reduction of the carbon bromine bond to the carbon hydrogen bond. But if we take sodium borohydride and lithium aluminum hydride and an aluminum chloride the mixed chlorohydrides would not reduce this carbon bromine bond. But if we now take this super hydride lithium triethyl borohydride for example it reduces the carbon bromine bond easily and say in this particular case you have lithium triethyl borohydride and deuterium has come in the fashion that is an SN2 type of reaction and the stereochemistry of this center has inverted here. Not only bromine but mesylate, tosylate, triflate are also reduced with lithium triethyl borohydride. And as you can see here there is a reduction of the mesylate and then it leads to the corresponding methyl group formation. It also allows reduction of epoxides and of course as you can see very easily that the reduction would occur from the less ended side. So basically lithium triethyl borohydride, super hydride is comparable with lithium say aluminum hydride in many sense. But then there is a mixed reducing agent nature it has a lithium and boron. So therefore we can compare it with lithium borohydride but then in comparison to the lithium borohydride because of the electron releasing nature of the ethyl groups which are attached steric hindrance as well as the reducing power both of them contribute and therefore it acts like a super hydride and that is how it is used it is called as a super hydride. There is a very interesting example in which comparison is made with the lithium aluminum hydride and lithium triethyl borohydride that means lithium aluminum hydride and the super hydride. If we see the reduction of this epoxide here and if lithium aluminum hydride is used under the fluxing conditions in ether it is actually a symmetrical molecule say does not really matter where the reduction takes place whether it takes place here or it takes place here. But we have shown that the reduction takes place there to form the exo alcohol that means this is an exo path which is pointing up therefore the reduction is taking place from the lower side and we are getting an exo alcohol. But at the same time as you can see that there is a somewhat strange molecule which is formed which cannot be expected either if the reduction takes place here or the reduction takes place here we cannot expect this we can only expect this to form. So how is it formed we will see later. But then if we use lithium triethyl borohydride as a reducing agent which is more nucleophilic in the nature and of course because of the electron releasing nature it is electronically more dense there and therefore reduction is somewhat different. So if we see this the reduction in a normal fashion gives this particular molecule as in 93% yield whereas that was formed only in 15% and the strange molecule which we notice above in 85% yield in using while using lithium aluminum hydride gives only 0.1% of that molecule. How does this strange molecule form that where the hydroxy group is far away from the expected position. What is happening is that in both the cases the lithium plus is available but if we can see that in the case of lithium aluminum hydride because there is no electron releasing group so we have a normal ionic nature of the lithium plus. But if you look at lithium triethyl borohydride in this case because of the electron releasing nature of the ethyl groups the this particular part here is not as ionic as in the case of lithium borohydride because of the electron releasing nature of the ethyl groups. Therefore the in the case of lithium aluminum hydride the lithium plus very readily coordinates to the oxygen here and this particular bond which is present here this bond migrates and breaks the carbon oxygen bond here this carbon oxygen bond is broken. Why does that happen is because this carbon carbon sigma bond and this carbon oxygen sigma bond basically they are anti-periplanar to each other essentially and it is very easy for this kind of overlap of the orbitals to take place when there is a developing positive charge here which is possible because of the coordination or the chelation of the lithium plus to the oxygen. And therefore the moment there is a possibility of a delta plus here this bond breaks and attaches to this and generates the positive charge at this center. So if you look at it if you number these you have 1, 2, 3, 4, 5 and 6. So we have here 1 then 2 then 3 then 4 here then 5, 6 and 7. Now if you look at the same position here so you have 1 here then we have 2 here then we have 3 here then you have 4 here then we have 7 here and then this is 5 and this is 6. So that is how the bridge that has formed is basically this part here this and this is the bridge which is 1, 2, 3 this is how it has come on the top and then the hydroxy group is coming from here. So when such a migration occurs you generate a positive charge here which then gets reduced and then immediately you form this of course it can go back to the same molecule and we get a part of the reduction in at like 15%. Now this particular migration of the carbon-carbon bond to this carbon here where the epoxide opens should remind you of homo-inolate ions where basically we have a carbonyl group here the negative charge attaches in this fashion. So basically it is nothing but a kind of proper orientation of the orbitals which allow the overlap and then migration in such a way that there is a movement of the bonds leading to the formation of a cation in a slightly different way. Now the super hydride is not routinely used for the reduction of simple carbonyl groups unless there is a need to observe a high stereoselectivity because it is sterically bulky and also it is very nucleophilic. For example here as you can take an example of this kind here as the substituents next to the carbonyl groups are basically in the form of tetrahedra furan but then we can imagine that this part of the molecule which is beta oriented here due to this the hydrogens are on the top. But if you look at this part you have oxygen and which will have only a pair of electrons. So it is obvious that the bigger CH2 group blocks the beta side and therefore the reduction occurs from the alpha side. On the other hand if you look at this here the epoxide is beta oriented and therefore the reduction occurs from the alpha side and you have the reduction leading to the formation of the hydroxy group in a beta oriented fashion even though this is beta obviously. And as you can see the reducing nature of such super hydride is so much that the reaction has to be done at minus 78 degrees or at minus 20 degrees. That means that we need to control the reaction in a proper fashion so that the reaction can be done at lower temperature. There are also reduction reducing agents which are stronger than these and more bulky compared to the super hydrides they are called selectrides. So larger the alkyl groups in alkyl borohydrides there is a greater diastereoselectivity in the reduction of say prokyl ketones. Obviously it is obvious that if we have very strongly bulky reducing agents then we can expect that there is a high diastereoselectivity. Now there are two in fact there are three but more popular are two is one is called as K-selectride here TM means trademark and the other one is L-selectride. So the structures are here this is L-selectride is because of the lithium it is lithium tris second rebutyl borohydride. So we had tried thiol borohydride now we have second rebutyl borohydride and of course here we have potassium tri second rebutyl hydride and therefore because of the K we have K-selectride this is one it can easily be made from the corresponding borane. So you take tri second rebutyl borohydride and reduce it by lithium aluminum trimethoxy hydride to lead to this particular L-selectride and if we take potassium hydride then of course this K-selectride is formed. So these are two new reducing agents which have been introduced by H. C. Brown who got the Nobel Prize in 1979 for his reductions using this borane type of molecules or various kinds of borohydrides. If we take N-selectride so instead of potassium and lithium we will have sodium in the reaction in the as a counter ion. Both aldehydes and ketones can be reduced to the corresponding of course even at minus 78 degrees obviously but then both are reducing agents which are very effective both L and K and of course N but more easily L-selectride is normally utilized it because of the chelation nature of the lithium plus which is comparable with say sodium borohydride versus lithium borohydride. In the case of lithium borohydride we have seen that because of the lithium plus coordination that is a stronger reducing agent than the sodium borohydride in a similar fashion K-selectrides are less effective as reducing agents compared to L-selectride. Now we have two examples we take reduction of this which goes to the corresponding acetate after reduction and acetylation as you can see that the reduction occurs from the alpha side that means the this particular group being beta oriented the alpha side is free and therefore hydrogen comes from alpha side. As we can see the same molecule in its conformation then we can say that this group and this group are beta oriented and therefore reduction occurs from the alpha side. So you have the reduction in a highly diacetyl selective fashion and then of course you take acetic anhydride and pyridine to convert this to the corresponding O acetate. So this is an example in which we can expect the reduction to take place in a similar fashion there is a reduction which is added influence by added metal salts while using these kind of selectrides these are called selectrides and because of the selective nature of the reducing agents and if we take this example here we can see that the group here which is benzaloxy group is very close to the carbonyl group here and this is alpha oriented and if we add magnesium bromide to this we can expect that there is a relation between this benzaloxy group and the oxygen group with the magnesium bromide and therefore the magnesium bromide allows the OBN group here and so now you have magnesium and then bromide here. So you can expect such type of coordination I will not write the other part of the molecule and therefore one can see that the alpha side is blocked and therefore the reduction occurs at this carbon atom or at this carbon atom from the beta side. So you have the hydrogen coming from the beta side. So now we get 8.5 is to 1 ratio of these 2 molecules in which this is minor this is the major as you can see without magnesium bromide the selectivity was only 2.5 is to 1 therefore one can expect that such chelation by added extra Lewis acid type of molecule such as magnesium bromide makes a kind of chelated complex and which then directs the reactivity of the carbonyl group in such a way that the less hindered beta phase is then open for the reduction to take place. We will stop today at this stage and take up the remaining part of the reducing agents and reactivity of various compounds in the next class. You can go through these aspects of whatever we have discussed today and be ready for the next class. Thank you. See you next time.