 So the structure elucidation of glucose is right down the next heading is structure elucidation. How do we come to know about the structure of glucose? That's what the thing we are going to understand. The first heading you write down is structure and this is just a reaction of all of the right down because there are many reactions of glucose to reduce the elucidation. The first reaction here, the first reaction that we use is this one you write down here. The first method from which we come to know about the structure that glucose when we do the reduction of glucose with red phosphorus and HIE. This gives you n-hexane, normal hexane. There's no branch present into this. So this reaction when you get normal hexane, it means there's no branch in glucose. We have a straight carbon chain on which the different hydroxyl groups are attached. So red phosphorus and HIE we use for all those molecules which contains oxygen that converts into alkane. This is one thing. The first reaction gives us the information that there is no branch present. Now for the other reaction I will draw the structure here of glucose 1, 3 and 4. Here we have CHO, primary functional group, OH this side, OH on the left, H on the right, OH on the right, H on the left, OH on the right, H on the left and CH2. So when this reaction, this molecule reacts with NH2OH, what is NH2OH? Hydroxylamine. And when this reaction takes place, the product here we get is, see this aldehyde combines with this oxene, H2O goes out and we get what? And we get CH double bond NOH that is oxene. And we have seen this reaction, that carbonyl compound with hydroxylamine gives oxene. Everything will be same, every other thing will be same, OH on the right, OH on the left, then OH on the right, OH on the right, H, CH2, OH. Yeah, did. Okay, this oxene. And when oxene forms, then this confirms the presence of carbonyl group. Okay, carbonyl group, carbonyl group we know now but we do not know whether it is aldehyde or ketone. Okay, whether it is aldehyde or ketone. Even one more reaction confirms the presence of carbonyl group, which is nothing but the reaction with HCN. What we get with HCN? Can you tell me? I have done this reaction in the class. With HCN, the carbonyl carbon. We will get here COH, CN, H. Is it clear? I have done these reactions in the carbonyl compound chapter, so I am not discussing this here. HCN on carbonyl compound gives cyanohydrin if you remember. Yes? When carbon contains hydroxy and cyanoglue, we call it as cyanohydrin. So this reaction also confirms the presence of carbonyl group, cyanohydrin that forms. Now, we know that from these two reactions that there is a carbonyl group present, but which carbonyl group it is, whether it is aldehyde or ketone, that we do not know. And to understand that, to confirm that what we do, we do the oxidation of this. Because we know what? We know aldehydes can easily oxidize because aldehyde contains hydrogen on this carbonyl carbon. So oxidation of aldehyde is easier and easily oxidized. But for ketone, we require better oxidizing agent. It is a bit reluctant towards oxidation because there is no hydrogen present towards oxidation because there is no hydrogen present on the carbonyl carbon. So the point is with mild oxidizing agent, oxidation of aldehyde is possible, but ketone won't get oxidized. Okay? So for that, we use the reagent here is bromine water, Br2H2O. Br2H2O is a, you know, weak oxidizing agent which can oxidize only aldehyde, not ketone. Okay? So when you take this Br2H2O, it converts aldehyde into acid and the product here, this CHO, converts into COOH. Every other thing will be same and CH2OH. This compound, the name of this compound is gluconic acid. Call it as gluconic acid. And this confirms this oxidation, confirms the presence of aldehyde. Since it gets oxidized, so it confirms the presence of aldehyde. Okay? Can I move to the next page? All of you have done? Yes, sir. Now to confirm how many hydroxy group are present, we'll do acetylation. To confirm, number of hydroxy group will do acetylation. What is acetylation reaction? Reaction with anhydride. So, okay. So what we get in this reaction? We'll get this. First of all, this we are using excess. Okay? So we have sufficient amount of anhydride. And if you see the reaction of anhydride, it gives you CH3COOH, you get CHO, all these one, two, three, four and five. So here we get COAC, acetyl group. Here we have hydrogen. And this will have four times. And then we have CH2OAC. So we then can, so we can then do a titration with CSCOS to find out the number of hydroxy groups. Hydroxy group. Correct. So actually, if you know the number of CS3COOH that you are getting, which is five over here, because you see one hydroxy group gives one CS3COOH, two, three, four and five. So we get here five moles of CS3COOH. That confirms that there are five hydroxy group are present in this structure. It confirms the presence of five hydroxy group. Okay. So hence there are five hydroxy group present. That gives us the information of the number of hydroxy group present. This is the structure elucidation of glucose. Now the question is why this OH on the right and why this OH on the left, that is not in our syllabus. Even it is the part of some masters and bachelor. So that is not required. For that, obviously we need to understand the optical behavior of this carbon, this carbon and all. That is how only we get this, but that is not required for this. Okay. Any competitive example. Okay. So this is how we confirms or we get to know that how many hydroxy group present. And if there is any carbonyl group, what carbonyl group we have oxidizing agent we use for that purposes BR2 bromine water BR2 with H2O. And this is the structure. One last thing I'll tell you here. If I take BR2 with H2O, we are getting gluconic acid over there. Okay. What happens if you take a strong oxidizing agent like HNO3. So if you take HNO3, it is a strong oxidizing agent. This will oxidize aldehyde and alcohol. This alcohol and this aldehyde. And the product we get here is. So we'll get like a chain of. No, we'll get aldehyde acid here and here. And here we have OH H H OH. So we do this in a control way basically. So point. This is not important, but the only thing here you need to keep in mind that this molecule, we call it as gluconic acid. This name you must remember gluconic acid. Or we also call it as saccharic acid. Gluconic is when we have COOH here, CS2 OH here. When both side we have COOH, that is gluconic acid. Sir, can HNO3 oxidize those secondary hydroxy groups? So can it be like a row of the very symmetrical row? If you take very excess, large amount of HNO3, then it is possible. Sir, it will be nice and symmetrical, sir. What? Sir, the product will be very symmetrical, right? Yes, yes. So that is actually not stable because on that next carbon will have COOH and H here. So maybe hydrolysis is also possible over there. So that you won't get a stable structure over there. Not hydrolysis dehydration, sorry. H2O molecules goes out. Okay. Dan, all of you? One second, sir. Sir, but if there are more than three carbon atoms between the carboxylic acid group, shouldn't both CO2 and H2O go out? What were three carboxylic acid where? If there are three carbon atoms between the carboxylic group, shouldn't CO2 and H2O go out? If you heat that possible, that is a hydration effect. Even if all these get oxidized, okay, then the structure is not stable. There are very high tendency of getting dehydrated. Yes, sir. Because we have H2O, we have COOH present, so that won't give you the stable structure. And yes, whatever you are saying, the heating effect, if you have two carbon atom or three, then CO2 or H2O go out, that is true. Yes, sir. Dan, can we move on? So basically the reaction thing you must keep in mind. Yeah.