 Hello everyone, I would like to welcome you all for today's class. We will briefly look at what exactly we did it last time and then proceed further. What we looked at last time was the reductions of alkynes, nitriles with di-ball H and we saw that we can stereoselectively reduce alkynes to cis or trans alkenes and we also can reduce the alkynes and convert them to corresponding alpha beta unsaturated acids which could be cis or trans depending on the conditions that we discussed. We also looked at the conversion of nitriles to the aldehyde or ketone depending on what nitrile we take. I mean in the case of nitrile it will be of course aldehyde with the di-ball and how we could stop it from further reduction to the corresponding amine. Then we also looked at the one rep amide utility especially to stop the reaction at the aldehyde stage or a ketone stage where a nucleophile can be added to this and it can allow chelation as one of the parts by which we can have the stopping of the reaction at the aldehyde or ketone stage. For example if we have here Li plus then of course we can have the reduction of it leading to the aldehyde formation. Now the over reduction or over reaction of nucleophiles is stopped because of the chelation that is here. So we discussed various types of nucleophiles that can be added to the wine reb amides and apart from the hydrogen addition like reducing agent we can also take say you have methyl magnesium bromide or any Grignard reagent that can be used and the reaction can be stopped say in this case up to the corresponding methyl ketone. Then towards the end we discussed the reductions with lithium aluminum hydride we also discussed the reductions using lithium tri tertiary alkoxy hydride and lithium aluminum hydride in presence of aluminum chloride where the use of aluminum chloride modifies the reducing agent to say for example you can have this or you can have this and because of the electronegativity of the chlorine the reducing power of lithium aluminum hydride which is now modified to having one hydrogen or two hydrogens or two chlorines or one chlorine is reduced and accordingly the selective reductions can be carried out. What selective reductions can be carried out is something that we look at it once again today is we discussed this mixed chloride hydrides which act as reducing agents but less powerful than lithium aluminum hydride. We discussed that the CX bond X is halogen can be reduced to the corresponding CH bond using this lithium aluminum hydride. But then the same reaction is not possible to be done using a combination of lithium aluminum hydride and aluminum chloride. On the other hand if we take another example of this type here say you have a bromide and an ester in the same molecule here if we do the lithium aluminum hydride base reduction then lithium aluminum hydride reduces this carbon bromine bond introduces this hydrogen here and at the same time ester is reduced to the corresponding CH2OH. But if we take a lithium aluminum hydride in combination with aluminum chloride then this modified mixed chloride hydride reducing agent only reduces the ester and does not touch the carbon bromine bond. So, there is a selectivity in terms of reducing power of this combination of lithium aluminum hydride and aluminum chloride. It is very clear that the mixed hydride chloride for example if it is something like this then the chelation of the ester takes place more easily than the direct nucleophilic addition of the hydride to the carbon bond to reduce the carbon bromine bond and replace it by hydrogen. Now it is also seen that we have an alpha beta and saturated ester something like this and if we reduce it with lithium aluminum hydride we get a mixture of allylic alcohol and the saturated alcohol. At the same time if we use the mixed chloride hydride reducing agent like lithium aluminum hydride aluminum chloride we only get this allylic alcohol as the major product. So it is very obvious that again here same thing is happening that only the chelation allows the reduction to take place. So once this happens of course you have a delta positive here and a delta negative here then the hydride reduces this and goes further for the completion of ester to the corresponding primary alcohol. Now there is another reducing agent which is comparable with lithium aluminum hydride but is in some sense better than lithium aluminum hydride which is called as redol which is reducing aluminum basically it is a sodium bis 2 methoxy ethoxy aluminum hydride. So this is the methoxy this part is methoxy ethoxy so and an aluminum hydride. So it exists as bis 2 methoxy ethoxy aluminum hydride which is known as redol and also it is called as vitride by the name of its discoverer which is wit. So the vitride or the redol is more stable in air and reductions are comparable with lithium aluminum hydride. One of the problems with lithium aluminum hydride as I mentioned last time is it is very unstable towards water and therefore not easy to handle but in comparison to that redol is more stable in air and it is also soluble in aliphatic and aromatic hydrocarbons because of the increased hydrophobicity that is not the case in lithium aluminum hydride. Lithium aluminum hydride is somewhat more ionic and more polar than redol and therefore this needs to be dissolved in ether or THF whereas this is possible to dissolve in aliphatic and aromatic hydrocarbons. Now this redol allows reduction of the acids to the corresponding alcohol similar to the reductions with lithium aluminum hydride and also allows in a similar way reduction of alpha beta unsaturated ketones to the corresponding allylic alcohols. And these vitride can also be available as a solution in the market and therefore one can easily take it and use it for reductions of various kinds. We will also see how further things can be done. For example when the reduction of an alpha beta unsaturated ketone is carried out with redol in the presence of copper 1 like cuprous bromide. This 1, 4 reduction takes place. Last time I told that this is 1, 2, 3 and 4. So 1, 4 reduction has taken place and essentially what is happening is it is more or less like a copper hydride type of species that is expected to be involved and then copper hydride being softer adds on to this in this fashion and we have 1, 4 addition where we can expect species like this to form and then this leads to the saturated ketone. So this is something similar to adding say dimethyl cupidate or such copper based reagents. So in a similar fashion we can take a somewhat more difficult example and reduce it to the corresponding saturated ketone. So there are many such examples in the literature where such copper hydride type of species is believed to be involved in the reduction of alpha beta unsaturated ketones to the corresponding saturated ketone using redol. There is another interesting reduction which is very specific and selective reduction that is of epoxy alcohols. This 2, 3 epoxy alcohols so you have 1, 2 and 3. So this 2, 3 epoxy alcohol is very easily available by a sharpness epoxidation of the corresponding allylic alcohol and therefore the utility of these 2, 3 epoxy alcohols become very important and thus their reduction to give the corresponding 1, 2 or 1, 3 diol is important and therefore if there are reagents that can selectively reduce 2, 3 epoxy alcohols to either 1, 3 diol or 1, 2 diol then of course the reaction becomes more meaningful and synthetically useful. What is found that if one takes redol or lithium aluminum hydride both of them are as I said comparable in terms of reactivity. They lead to 1, 3 diol as the major product and 1, 2 diol as the minor product. On the other hand if one uses diol for the same reduction then what is found is that the 1, 3 diol is found in minor amount and 1, 2 diol is formed in the major amount. Now how does this reaction occur? Why should there be a selectivity? Now, lithium aluminum hydride when it comes in contact with any alcohol as I mentioned earlier is something that leads to the expulsion of hydrogen gas after the OH of the alcohol or water reacts with the lithium aluminum hydride. So, you have lithium aluminum hydride as species which is having a negative charge here and a positive charge here and when it comes in contact with ROH we get here ROLI and of course AlH3 and hydrogen comes off. So, same thing happens even in this case hydrogen comes off and now as this species can interact with AlH3 which is released here and can form the corresponding negatively charged lithium plus species. So, this is exactly what is here but this is coming from the epoxy alcohol. Now, we can also write the same thing in this particular form here as we can slightly rotate the bond and now we can see that the reduction of the carbon oxygen bond of the epoxide occurs in such a fashion that the hydrogen from this particular part of the reducing end attacks at the carbon bond sorry carbon atom of the epoxide carbon bond where we have an SN2 type of reaction that means the hydrogen, carbon and oxygen of these species should be sort of collinear then you have the proper SN2 type of reaction taking place and this is exactly what happens. So, you have an intramolecular reduction after the lithium aluminum hydride has reacted with the alcoholic path of the epoxy alcohol and the reaction occurs in an SN2 fashion in an intramolecular fashion. At the same time when we take di-ball and react with these epoxy alcohols. So, in a similar fashion as lithium aluminum hydride when the di-ball comes in contact with the epoxy alcohol then also you have a hydrogen here then if di-ball comes in contact with this OH we can expect this type of species to form and this also loses this is a negative charge and this is a positive charge this also loses hydrogen and forms an intermediate of this type. Now, this particular intermediate where there is an aluminum species will have a coordination with the oxygen that is of the epoxide oxygen and it forms a kind of chelate complex which is strongly held because aluminum here is trivalent and therefore it is highly electrophilic therefore the oxygen of the epoxide coordinates with the aluminum and forms this. Now, since this aluminum path as against the aluminum part of the lithium aluminum hydride here does not have any hydrogen for allowing the reduction to take place. Therefore the another part of the similar molecule or di-ball then reacts to this part in an intermolecular fashion. So, there is no intramolecular fashion and therefore the possibility of attack of the hydrogen or the hydride to this particular carbon atom would be much better if between the two this carbon is attacked which is relatively sterically free compared to this particular carbon. And thus the reduction here leads to 1, 2 diol whereas in the case of in the case of lithium aluminum hydride it leads to the 1, 3 diol basically leads to 1, 3 diol. So, this is how the reductions occur of the 2, 3 epoxy alcohols with lithium aluminum hydride or di-ball. And since lithium aluminum hydride is similar to redol the redol also gives 1, 3 diol as the major product very similar in a fashion that lithium aluminum hydride does it. Now we have another reducing agent which is known as luchi reduction which is named after the discoverer which is luchi and it allows 1, 2 reduction to take place and particularly when there is a alpha beta unsaturated ketone. So, the reagent that is utilized is a combination of sodium borohydride and cerium chloride which is hydrated with 7 waters. This specifically allows the reduction of even alpha beta unsaturated aldehydes to the corresponding allylic alcohols. We discussed earlier in one of the lectures that the reduction of alpha beta unsaturated ketones and alpha beta unsaturated aldehydes with lithium aluminum hydride is not straightforward. It all depends upon many factors. But then we also said that alpha beta unsaturated aldehyde reduction can take place with some other reducing agents and that is what the one that we are now talking as the sodium borohydride cerium chloride based reducing agent which allows the reduction of the alpha beta unsaturated aldehydes also to the corresponding allylic alcohols. Of course, the ketones alpha beta unsaturated ketones also get reduced. Now this Gamal and luchi are the two people who initiated examining the reduction of alpha beta unsaturated carbonyl compounds with various metal salts along with sodium borohydride as a reducing agent. As we can see here that there are possibilities of all the three types that one can anticipate. So all three types of products are possible. So you have allylic alcohol saturated ketone and completely reduced saturated alcohol. So a lithium and copper was found to give the corresponding completely reduced product as the major product. And if one uses cobalt and nickel salts along with sodium borohydride, they give this saturated ketone as the major product along with saturated alcohol and of course a lot of starting material remains unreacted. It is possible that nickel 2 plus and cobalt 2 plus based reduction proceed via a different pathway but since a large amount of starting material Sm, Sm basically stands for starting materials which is nothing but alpha beta unsaturated carbonyl compound. And if that is the case then this is not a very good reaction. So they did with some other salts and particularly with lanthanide salts when they carried out the reaction like lanthanum, cerium, samarium, europium, yttribium, yttrium and they all gave the reductions in good yield from 86 to 97%. And particularly it was found that it should not be europium it should be cerium. So the cerium plus gave the corresponding allylic alcohol in about 97% yield. So what was then discussed or discussed in the papers that were published later on is how does the reaction occur. What was initially found and accepted that if this complex is formed for example one could expect that we have an enone like this and it is obvious that the metal salt which is used for the reaction with whatever the atoms are attached to this. Now we can expect some sort of chelation with this and then we have a delta positive here and of course you have a BH4 minus from the corresponding sodium borohydride. And one can expect the reduction to take place either at this carbon to give the corresponding allylic alcohol or onto this carbon to eventually reduce the double bond and make the corresponding enolate which leads to the saturated alcohols or saturated ketone. That means the double bond gets reduced in one case in the other case the carbonyl group gets reduced. So what was proposed what if this complex if this particular complex type of should be CH3 and if this type of complex is reversibly formed that means under thermodynamic condition it attaches to the carbonyl and then again detaches, again it attaches if that happens then there is a possibility of attack onto this particular carbon atom. So 1, 4 reduction is more easily feasible if it is reversibly formed that will of course depend upon the nature of the M plus that is there and of course the groups which are attached. But if it is irreversibly formed that means that immediately attaches to the oxygen and because of the electronegativity of the oxygen and electropositive character of the lanthanine salts the moment carbonyl group attaches to that and if it is an irreversible attachment then it is obvious that 1, 2 reduction would be much more facile something like this. So it is much easier to get 1, 2 reduction. So therefore the reaction depends upon thermodynamic conditions and kinetics. Because of the E plus I effect of the two flanking alcohol groups the carbonyl oxygen of a ketone is considered to be more basic than that of aldehyde. It is obvious that if a ketone has two alkyl groups the plus I effect would increase the electron density on the carbonyl oxygen in comparison to aldehyde in which there is only one alkyl group. And therefore the cases where ketones and aldehydes are compared obviously ketones will form more stable Lewis acid, Lewis base complex with electrophilic metal salts than aldehydes. This is exactly what is exploited in Luchy reduction. In Luchy reduction for example cerium 3 plus salts are utilized for the selective reduction of ketone over aldehydes and that is because of the stable Lewis acid Lewis base complex formation with ketone in comparison to aldehydes. In a similar fashion when we compare an alpha beta unsaturated carbonyl compound with saturated carbonyl compound we can see that while in the case of saturated ketones it is only the inductive effect that increases the viscosity of the oxygen. In case of alpha beta unsaturated ketone it is the electron density from the pi system that is in conjugation with the carbonyl group and we can write a resonance structure like this in indicating that more electron density is residing on the oxygen of the alpha beta unsaturated system compared to the oxygen of the ketones and therefore it is very clear that an alpha beta unsaturated ketone would form more stable Lewis acid Lewis base complex with say cerium salts in comparison to saturated ketone and therefore alpha beta unsaturated ketones are reduced more preferentially in comparison to the saturated ketone. So these are the things which have been exploited in the in the Luci reduction. We will stop it today here and look at the further reactions of the Luci reagent with different types of carbonyl compounds and what exactly is the role of solvent in modifying the Luci reagent and how do they affect the selective reductions of various types of carbonyl compounds. So you can go through these aspects of the reagent that I have discussed and we will see you next time. Thank you.