Hello, guys. Good evening. Can you hear me? Good evening, sir. Yes, sir. We can hear you. Yes. Give me a few minutes. We'll start. Yes, sir. Hi, sir. Good afternoon, sir. Okay. Hello. So tell me, where were we in the last class? Havod structure of Huran, sir. What? Havod structure of Huran. Huran. Right. Alpha D fractures. Alpha D fractures. So we have done Havod of Glucose, right? Yes, sir. Okay. Huran also, I think we have done that flipped Vala structure, correct? Yes, sir. The next term we have to see in this chapter is, write down all of you. It is muta rotation. Muta rotation. Write down. It is the spontaneous change. It is the spontaneous change in the specific rotation value of, in the specific rotation value of an optically active compound. An optically active compound. Okay. So you see, suppose I am taking an example of alpha D plus Glucose. Alpha D plus Glucose. Its specific rotation value is 111 degree. Experimentally this is, right? And you don't have to memorize this. 111 degree. Now alpha D plus Glucose means what? What is this structure? Alpha D plus Glucose. Over it is down. Sorry? Over it is on the right. Yes. Alpha D plus Glucose, correct? Cyclic one. Yes, sir. Glucose means that straight change in the structure, correct? Yeah. Alpha D plus Glucose, it means it is the semi-acetyl structure. Cyclic one, correct? Yes, sir. Keep that in mind. Now you see what happens when you dissolve this in water. S2O. Then what happens? Hydrolysis takes place, right? And the ethyl linkage we have in alpha D plus Glucose, the hemi-acetyl structure, the ethyl linkage we have there, the linkage of O like this. It is there, no? And then we have this structure like this, right? So this ethyl linkage we have here, it gets hydrolyzed, means H plus comes over here and this bond breaks, right? Means what is the point here that when you hydrolyze this, this bond breaks and it converts into D plus Glucose. Okay, backward reaction over here. D plus Glucose, okay? And this structure is what? This structure is open chain structure. End out in this? No, sir. This is an open chain structure, right? Now, when you have this structure of Glucose, let me, you know, draw this here. This open chain has the structure, that is a cyclic structure. For example, this is Glucose, OH on the right, OH on the left, OH on the right, OH on the right, CH2, OH in the bottom and CHO. Here we have hydrogen. So, first we have cyclic hemiacetyl then it converts into this open chain structure. When you have this open chain structure, then this also has tendency to convert into closed chain, that is cyclic hemiacetyl structure, correct? So, it is like two and four. This is converting into this and this again converting into cyclic structure. That is possible, right? So, when this converts into, see here, when this converts into, again, the cyclic hemiacetyl structure, then we have two possibilities. One is what it can either form, alpha-vala form or it can either form beta. Yeah. Yes or no? Yes, sir. And all these are reversible reactions both way it can be. Right, now you see for this compound, the specific rotation is found to be this. It is plus 19 degree. Now you see, so what we can say in aqueous solution, this molecule, alpha-d plus glucose continuously converts into beta and beta continuously converts into alpha. This is happening because of their difference in a specific rotation, correct? So, in aqueous solution, alpha-d plus glucose is in equilibrium with beta-d plus glucose. Right, in aqueous solution. Because it has different specific rotation, this has different specific rotation. Right, so we must get an equilibrium value. Right, so the equilibrium value of equilibrium value of specific rotation is 52.5 degree, right? So, this continuous change of alpha into beta and beta into alpha due to difference in their specific rotation is called muta rotation. Is it clear? So, muta rotation is what? It is a spontaneous change in the specific rotation before establishing equilibrium. Yes, I got it. So, basically D plus glucose is an intermediate it's like facilitating the conversion. Ah, right, right. Okay, sir. Means, this has no value, actually. This has no value. This will either exist in this form or this form. This wants to converts into this. This just via this, it will go into this, that is it. Okay. So, it forms for a fraction of second and then it converts into this. Again, it converts between this and then into this. It's like the road between two destination. No one cares about the road. The destination and the road that we have. Okay, sir. Okay, so this is what we have. In aqueous solution, that's why we say in aqueous solution, we have this equilibrium. So, this is spontaneous and since D plus glucose into water, correct. So, it isn't spontaneous change. This is spontaneous change because of difference in their specific rotation. This phenomenon, we call it as mutarotation. Sir, but equilibrium value is not the exact half between 111 and 19, no sir. So, it's alpha D glucose mode. We already know this beta D plus glucose is more stable than alpha, right? Yeah. So, the equilibrium value should be closer to the value of beta. Okay, that to it. What is the average we are getting? Average is we are getting 60, correct? 65, yeah. It's more stable, so it will dominate. Okay. That's why this value is closer to beta. Oh, yeah, okay. So, this kind of question that the question that you asked this, they can frame question on this till now they haven't asked. But on the basis of stability we can say that the equilibrium value should be lesser than or falls in the range of what? This value and the average of this two. Are you getting it? Yeah, that will be convenient, yes sir. So, that kind of question they may ask in the exam till now they haven't asked. Right, they cannot ask you the specific rotation of alpha D plus glucose. You cannot memorize the specific rotation for any compound. But in general, logical questions like the question is why it is not exactly half or average of it. These two. The reason is what? The stability of this. We know beta form is more stable. Based on this stability thing, they can ask you this question. Range. Yeah, yes sir. If it's like a dam-derm stable, then it will be 119 itself, otherwise it will be somewhere in the earth. Not 19, exactly 19 though it's not possible. But even more closer. Yes sir. 20 something Yeah. Sometimes they ask you this question, not with this. Other question that they ask, that's like suppose fructose you have, glucose we have. So, how many isomers possible for this? Okay, number of isomers. Sir, but so many possibilities are there. No optical we can consider. So many other things we can consider. Stereo isomers. Okay. Stereo isomers with optical activity. We'll see. So we have three parallel centered in fructose. Correct. The answer will be but all those three we can have RN as possible. So two to the power. Sir, one second. Two to the power N no. N is the number of carols. Suppose we have this. I'm taking an example of fructose. I'm taking an example of fructose. Suppose we have C H2 OH C double bond O and your head. Basically, this is the carols. Okay. Middle three. Yeah. So what is the possibility? Yeah. Yeah. SS. Yeah. SS. SSR. Okay. RSS. And then what we can write SSR then SSR like this you can count the number of combination possibilities. Yeah. Then we can have SRR then RSR then SSR SRS Yes, sir. Yes. It is the possibility. So you don't have to do like this three centers, I don't know more than this. It was founded like 90 years ago. Other than that, I don't know more than that. It's a Mohanlal or someone. Mohan Bhagwat is the chief. Mohan Bhagwat is the chief. Anyway, that's not our discussion. Let it be. Otherwise, things will go away. Fine. So, till now guys, we have discussed about glucose, fructose structure. All those have the structure of glucose and fructose. We have discussed the structure elicitation of glucose. What all structures we have? Reaction of glucose also we have seen. Glucose to glutamic acid and all that one you mean, sir? Reactions we will see. One reaction we haven't seen, we will see that reaction. So, before that one very important reaction we have, that is differentiation of glucose and fructose. Okay, and it is not d-y-y-dx, okay? So, it is how do we differentiate these two molecules? Tell me, what is the basic difference between the glucose and fructose? C double bond O, ketone linkage. Sorry? Ketone linkage, C double bond O. What is ketone linkage? I think you can say the major functional group is stone in fructose and aldehyde in glucose. Yeah, that's what I wanted to say. Correct? Yes, sir. When you want to differentiate glucose and fructose, you have to have the relation or reactions which can differentiate an aldehyde or an ketone, correct? Right, because the major functional group is the primary functional group is aldehyde and ketone. The reaction of aldehyde and ketone with the help of that we can differentiate glucose and fructose, correct? Yes, sir. So, we have glucose and fructose. The primary functional group is an aldehyde here and primary functional group is ketone, okay? Now, you see one thing, it is the property of aldehyde that these aldehydes are easily oxidized. Right now, aldehyde are easily oxidized. Easily oxidized means what? In comparison to fructose. Sorry, in comparison to ketone. So, it can be easily oxidized. It means it is a good reducing agent. Yes, sir. Aldehyde itself gets oxidized. It means it is a good reducing agent, correct? But the oxidation of ketone is very difficult. Right? It can be done in drastic condition. Right, we can take acetic KMNO4 for this purpose or K2CR207, acetic KMNO4 or K2CR207. Let me just put one thing over here that after finishing this chapter, we are going to do a chapter called Grignard reagent, its reaction, oxidation and reduction. So, in that particular chapter, you won't see any chapter by this name, okay? In any book, right? So, we are going to see the reaction of Grignard reagent, how Grignard reagent reacts in various conditions, right? Plus, the various oxidation reduction reaction in the entire organic chemistry, right? So, there in that chapter, we are going to see some reactions of aldehyde, how it gets oxidized, what all reagents we have, okay? And same thing with ketone also, okay? Various oxidation and reduction reaction. Reaction we will see for aldehyde and ketone. They are in details, okay? Plus, we also see the oxidation of alkenes, diols, all those things, correct? So, there we will discuss these things in details, okay? That is one thing. Sir, will we also do alcohols, phenols and those kind of chapters? Yes, we will do. See, alcohols, phenols, phenols we will do. It's not like those chapters we do not do. Point is reaction of Grignard reagent and all, you will get these questions into the exam, right? So, it's always better to understand those things separately, right? Yes, sir. You won't have a chapter called Grignard reagent reactions and all, right? So, that's why I have planned this. You will have this chapter separately. And what all reactions we have covered into this, we won't continue with that in the same chapter. Like, suppose aldehyde's oxidation is to finish in the cure, right? So, when we do aldehyde and ketone, we will skip those reactions. We will see that chapter and you can see over there. Okay, because it's the same thing, we cannot repeat. No, because phenols also we have to finish by October. Yes, sir. Right, so that is the reason, okay. Okay. So, we will definitely do an alcohol phenols reaction. Yes, sir. Okay. E. So, what I was talking about, that fructose has primary function with ketone and its oxidation is quite difficult. Oxidation of ketone, usually they won't ask question on this. It is not that important as we have aldehyde co-oxidation. Okay. But oxidation of ketone is done in very drastic conditions. Like, we have to have acidified KMNO4, acidified K2CR2O7. Very strong oxidizing agent requires. Okay. So, that's why oxidation of this is not that easy. Okay. What we can say, it is not oxidized by, not oxidized by mild oxidizing agent. So, mild oxidizing agent can oxidize this aldehyde but it cannot oxidize this ketone. Right. So, suppose you have a mixture. Okay. You have a mixture and you need to find out whether it is aldehyde or ketone. Okay. What we have to do? We will take a mild oxidizing agent. And we will try to do the reaction. Correct. If the reaction is taking place, it means it is glucose not fructose. It means it is aldehyde not ketone. Clear? Yes, sir. Okay. So, there is a change in color. That change in color, you know, it confirms the presence of aldehyde there. Okay. And hence, we can differentiate aldehyde or ketone in any mixture. Okay. So, what we'll do here, the various reagent we use, suppose you have a carbonyl compound. Okay. Suppose I'll write down here. Carbonyl compound, we do not know whether it is aldehyde or ketone. This carbonyl compound, if you allow this to react with a reagent called tolin's reagent. Okay. Or we have another one that is failing solution. What is tolin's reagent and failing solution? We'll discuss this in next chapter. Okay. Just now you write it down. Okay. Tolin's reagent and failing solution. This carbonyl compound, if it converts into acid. Okay. That is carboxylic acid. This also converts into carboxylic acid. If it is happening, it means this carbonyl compound is, this carbonyl compound is what? Aldehyde. Aldehyde, not ketone. Oh, glucose to glutamic acid that CHL converted. Right. Aldehyde, not ketone. And this is how we can differentiate any aldehyde or ketone. So there is a color change here. We also call it as silver mirror test. Important. Silver mirror test. Right. Detail me. We'll see later. Right. So point I'm trying to make, like with these solutions, we can differentiate aldehyde or ketone. Okay. That is one thing. Now one note you write down. So what we can say. So fructose contains ketone as a primary functional group and glucose contains aldehyde as a primary functional group. So fructose won't give this reaction. Yes or no. It won't give us a, because, because ketone does not give this reaction. So fructose will also not give this reaction. Right. Right. Over here. So fructose contains ketone as fructose contains ketone as the primary functional group. Then also it gives positive tolerance test. Right. I'm coming to this. Then also it gives positive tolerance test and failing solution test. Now the natural question is why. Correct. See, usually this thing is correct that ketone does not give this test. That is fine. Okay. But in case of fructose, it is not, you can say it is an exception. Why exception? We'll understand that. Okay. But this thing you must remember. Okay. Fructose this question they have asked many times in need and J exam. Okay. Which of these compounds shows positive tolerance test. One of the options is fructose. Right. So fructose. It contains this thing. A ketonic group. Then also it shows positive tolerance test and failing solution test. This you must remember. Okay. Important. Now why this is happening. Okay. So heading you right down next. Heading you right down. Why fructose gives positive tolerance, positive tolerance, P O W L E N S. Why fructose gives positive tolerance and positive failing solution test. Okay. So first of all, you have to keep this in mind. These both reactions. These reactions takes place in alkaline medium. Okay. Both one. Alkaline medium. This one also. And this one also. Okay. It takes place in alkaline medium. Why sir? Why can you have to see the reaction in this? Okay. Let it be for now. Okay. Just let it be. Yes. This is the next chapter. You'll understand. But not a big deal. Yes. Okay. So suppose you have taken fructose. Right. For this test. I'll show you one. Reaction first of all. All of you draw this compound and then we'll see this. This compound is. Copy this. Then. Yes. So once again sir. Okay. Done. Yes. Yes. Okay. So what is happening here? We have taken fructose. For the test. Suppose you have taken the compound which you do not know whether it is glucose or fructose. Suppose you have this. Okay. Or if you try to understand from here what happens. This part you see glucose. I'll just write down the top part of it, which is. C double bond OH. And here we have. Here we have. OH. And these are. In. Basic medium. What happens because the medium for tolerance and. This is the solution. Collins test and filling solution test is. Basic medium. Right. Alkaline medium. Alkaline medium. What happens. This molecule goes under. Okay. And how it goes you see. One. Two. And three. Third say hydrogen comes out and this will attach on to one. Or you have pi electron. Yeah. Okay. This forms this one. C at C OH. Double bond C. OH. And below the things will be as this. No problem. Right. And this carbon becomes sp2 hybridized. This carbon becomes sp2 hybridized. Now when it since it has tendency to go in backward direction. Also. Right. So when this goes into backward direction means when hydrogen. From this oxygen comes out and attached onto this carbon. This hydrogen may attach from this side or from this side left. Right. So if it has had from the right side, then the OH will be left side, which is this structure. Yes or no. Okay. What I'm telling you suppose this reaction since it is reversible. Right. So it has tendency to go in backward direction also. Tendency to go in backward direction also. Correct. Now when it goes in backward direction, how it is possible when. Hydrogen from this oxygen comes out and attached onto this carbon. Tell me yes or no. Yes. Yes. Okay. So now when this hydrogen will get attached onto this carbon, this may get attached from this side or from this side because this carbon atom is what it is sp2 hybridized. No doubt. Yes. So when this attached from this side, this gives you glucose, glucose, but when this attached from right side, it gives you mannose. Correct. So from here, what on the compound that you get is mannose D mannose. Is it clear? Copy this. Okay. Understood this. No doubt. Now another thing you see here. That further from this molecule again, tautomerism is possible. Right. This is tautomerism. And when tautomerism takes place here between, suppose this is one, this is two, and this is three. So what we get here, you see this hydrogen from here, it comes onto the third carbon and the pilot prawn jumps over here. So what did you get? You see on the top, we have CH2 OH, because H OH we already had this edge goes onto this department. So it becomes CH2 OH and this becomes what? C double bond OH rest is same. Any doubt on this? No. So what happens here? What happens here? You see this is nothing but fructose. Now the point I'm trying to make over here from all these diagram is what? That when you take fructose, right? For or any, suppose glucose or fructose, if you are taking for this test to differentiate the two, right, college region test. Since the medium is alkaline, then this fructose has tendency to convert into, to convert into glucose. And glucose also has tendency to convert into mannose. Or we can also say these three compounds are in equilibrium when the medium is basic. OH minus. Right. One second sir. So once I can write down the structures, then we will repeat it, whatever you want. Just a second sir. Yes, I could repeat the last sentences that you are telling sir. Yeah, what I am saying is suppose you have taken glucose or fructose for that college test. Yes sir. Ideally glucose will get this test because it has aldehyde. Yeah. Fructose for this test. Fructose also shows positive college test because the medium is alkaline and in alkaline medium, fructose converts into mannose and glucose by tautomerism. Right. Mannose and glucose by tautomerism. Hence, since mannose and glucose has aldehyde as the major functional group, primary functional group, hence this solution shows positive tautomerism. Overall, the meaning of this is what? Suppose what happens? You have taken fructose in a beaker. Right. And you are adding TR. What is TR? Tolenzia gene. And you have taken fructose here. So you have taken fructose, but in the solution, what it does, it converts into glucose and mannose. And this we do not know because we have the information, what information we have that in solution, we have fructose only present. But since the medium is alkaline here, so fructose converts into glucose and mannose. In fact, these three compounds are in equilibrium in the solution. And because these two compounds has aldehyde as a major functional group, that's why the solution, the solution shows positive tolerance reagent test. So here also it's effectively the ROH group, that's like aldehyde group that's showing positive, what did ketone do? No, ketone does not show. Ketone does not show this test. But why the question is fructose having ketone as a primary functional group. Then also it shows the test. The reason is what? That in alkaline medium, it converts into glucose and mannose, which has aldehyde as a primary functional group. That's why this shows positive test. Yeah, yes sir. Very important. They have asked this question many times, the reason behind this. Done? Yeah, done, sir. So this rearrangement that we have because of conversion of fructose into glucose and mannose, this rearrangement we call it as, we have a very small name here. And the name is, right, it's a very small name, Lowbury D. Van Brien Kinstein rearrangement. So that's even bigger than my name. That's why I said it's a small name. Yes sir. This entire rearrangement is called that sir. Sorry? Glucose to fructose mannose is called the Lowbury rearrangement. This equilibrium of the three is Lowbury D. Van Brien Kinstein rearrangement. So one note you write down. Done all of you? One second. Copy this. Done sir. Done? Yes sir. One note you write down. Note right down after this, the reaction of toluene reagent and filling solution, the reaction of toluene reagent and filling solution takes place in alkaline medium, takes place in alkaline medium, where fructose convert into mannose and glucose, fructose converts into mannose and glucose, which has aldehyde as the primary functional group, which has aldehyde as the primary functional group. Hence they shows positive tolerance and filling solution test. Hence they shows positive tolerance and filling solution test. Done? Yes sir. Yes sir. Now from this rearrangement one thing we conclude is what, that we cannot use any reaction to differentiate glucose and fructose. We cannot use any reaction which medium, whose medium is this alkaline. Because whenever the medium is alkaline, fructose will convert into glucose and mannose and shows the test. So we should use a reaction or an oxidizing agent whose medium is not alkaline. Correct? Hence to differentiate glucose and fructose, we use bromine water as the oxidizing agent. It's a mild oxidizing agent and bromine water we use for this process. Write down. Sir, does fructose taste sweet? Sorry? Does fructose taste sweet? I think no. Means it's not as glucose is. Then we can eat it and see, why do we have to do so many tests and all that? What? We can eat those samples and say, this is more sweet, this is glucose, this is less sweet, this is fructose. What should we do? Law, P, D, Van, try and make constant rearrangement. See, we can test and we can say. But when we have like the proof, for the proof we must have some, see there are many different ways. Okay? So whether it is a salt or cocaine, right? You can test and say, correct. Both looks white powder. But you can test and say it is a salt or cocaine. So that is also one of the way, but technically or you know, logically what is the reaction by which we can differentiate the reaction is this. But I think the salt, the fructose is not that sweet as sweet as we have glucose. Yes, sir. Right? But the chemical reaction, which differentiates the two is this. Okay? We use bromine water for that. Okay, okay, sir. Then you can also say, so why do we need to differentiate these two? Let it be. It's all hard discovery. They have done so many research, since they have, you know, given their time. So they want us also to know all these things. And that's why these things are in our slaves and we have to study all these. Okay, sir. What happens? Suppose if you do not differentiate these two, then what is the, what went wrong? Nothing. Right? Yeah. The sun in that case also will rise in the east. Yes, sir. Nothing goes wrong with that. But the thing is, all the scientists, they have given a lot of effort into this. So these things are there in our slaves. We have to study all this. But yes, obviously when the test is different, we can test and we can say these things. This particular thing is this. But since they asked question based on the reaction. So we should know what reaction we have, which differentiate the two that is. Yes, sir. Right. One more thing is very important here. Since this glucose and this thing, uh, fructose, you see they, we can draw this through totomerism. Correct. So we can also say D glucose and D fructose are totomer of each other. These questions they will ask you directly. D glucose and D fructose are totomer of each other. Correct. So what is the oxidizing agent we use to differentiate glucose and fructose? Romin water. Romin water test. Romin water. Second point you must have to keep in mind. We use those oxidizing agent whose medium is not. Alkaline. Alkaline. Why? What happens in alkaline medium? It goes into conversion and converts into mannose. Okay. Yes, sir. Now next you write down the reaction of glucose. Okay. The reaction of glucose. We have one very important reaction in this. And we call it as OSA zone formation. Don't write this. Now I'll show you.  Complete phase. The blank page. We'll draw the glucose structure here and we'll see the, all the reactions. Okay. So you take some space, like need some space for this. Otherwise we have to draw the structure of glucose again and again. So heading right down the reaction of glucose. When this reaction takes place with NABH4. In presence of H2O. NABH4. It is a mild reducing agent. This is a reaction of this also in the next chapter. So it reduces aldehyde into alcohol. And the reaction gives you this CH2 OH on the top. This is the only change we have dressed. All the things are same. Okay. So I'm not drawing the other part, which are common. Okay. When this is allowed to react with a strong oxidizing agent, say HNO3, strong oxidizing agent. It oxidizes this and the below one. This compound, we call it as saccharic acid. Okay. When you take here, tollens reagent, failing solution or bromine water, BR2 H2O. Then this get reduced and it oxidizes aldehyde into acid. So saccharic acid is used in that sugar-free thingy, you know. Yes, correct. Yellow color dabba. Right, right, right. Grandfather uses. Okay. So you get this and this compound, we call it as gluconic acid. Gluconic acid. Okay. Next, when this allowed to react with hydroxyl amine, NH2 OH. So NH2 OH, it forms oxene. We have seen this reaction and OH here. Hydrogen and all other things are same. No change in this. Okay. Oxene. One very most important reaction we have is, is. Osa-zone formation. This is important. You must remember and Jay, they have asked question on this. Osa-zone formation. Osa-zone formation. We take three molecules of phenyl hydrazine for this process. NH2, NH, pH. You see, NH2, NH2 is hydrazine. Yes, sir. When you displace one hydrogen by phenyl group. So it becomes phenyl hydrazine. Yeah. So in this one, the product that we get is this C double bond. N, NH, pH. Here also we get C double bond and NH, pH. And then here it is same, HOH, CH2 OH. This is the compound we get. Remember, we are using here three moles of it. It is important. This compound we call it as glucosa-zone. Yes. So what is that orange compounds name? This one, gluconic. Okay. So one second. This group, Osa-zone formation, this involves both condensation and tautomerism, both reaction. H2O evolves, NH3 evolves. Okay. Aniline also forms. So if I write down all the other products in this reaction, we get pH, NH2, aniline. NH3 forms. And H2O also forms in this reaction. Yes, sir. Okay. Yes, sir. Now you see the last reaction, how it proceeds. Okay. I won't draw the entire structure, but just the important one I'll draw so that you can understand. C double bond OH we have here. And on this carbon, we have OH on the right, H on the left. And these things are let it be. It is same. So when it reacts with NH2, NHPH, phenyl hydrazine. So first mole we are using over here. We are using three moles of this. So first mole is this. This H2 takes this oxygen and goes out as H2O. And it forms what? C double bond N, NHPH, COH. Osa-zone. Now after this we have tautomerism here. One, two and three. This comes over here. And this hydrogen will go on to this nitrogen. So the tautomeric structure here is this. The double bond here. Okay. Now what happens? This bond pair you see. This OH bond pair. This comes over here. This pi bond goes on to this. And this goes out as a living group. PHNH. And this PHNH takes this hydrogen. It forms PHNH2. So in the next step what happens you see? C double bond NH, H. Here we have single bond C, double bond O. And bottom everything you see. Plus we get PHNH2 here. Here we get this plus water. Now after this we use. Two moles of phenyl hydrazine. NH2, NHPH. See out of two moles. One mole. Just a second. So this you see two hydrogen takes up this oxygen like the reaction we have here. And we get C double bond N. NHPH on this one. This hydrogen and this NH2 combines and goes out as NH3. Okay. So we'll get here the product is C double bond N. NHPH. One mole here we use. And C double bond N. NHPH. Second. Mold we use here. And then this thing is same. Plus H2O. Plus. This is what the structure topic is done. So why do we need two moles then? Sorry. Okay. Two places that reaction happens. Yeah. Here on this one and this one. Okay. Done. Finished. Yes, sir. Okay. So this one, the last reaction you see the OSA zone formation is the most important reaction. Okay. One thing also you see, since we are using glucose, so the compound that we get here, the final compound is Glucosa zone, the name of the compound. Even if this reaction takes place with fructose, right, then also we'll get the same kind of compound, but the name is fructosa zone. Because the bottom part here, it is same for fructose also. This bottom part is same. Only the change we have here. Right. So we'll get the same compound, same molecular formula, but the name is fructosa zone. Basically Glucosa zone and fructosa zone are the same thing, but fructosa zone forms with fructose. Glucosa zone forms with glucose. That is it. Right down this line. OSA zone formation in glucose and fructose is same. When we use fructose, then the compound we get is known as fructosa zone. So can you repeat? OSA zone formation in glucose and fructose is same, but when we use fructose, the compound that we get is known as fructosa zone. Yes, sir. Next, right down the heading, inversion of cane sugar. That is sucrose. Right down. Sucrose solution is, sucrose solution is dextrorotatory, is dextrorotatory. But on hydrolysis, but on hydrolysis, it becomes hydrolysis. It becomes levorotatory and it converts into, it converts into D plus glucose. It converts into D plus glucose and D minus fructose. So what happens on hydrolysis, you see? We have sucrose and we know sucrose is made up of one unit of glucose and one unit of fructose. So when it goes under hydrolysis, under hydrolysis, it forms D plus glucose, one unit of D plus glucose. That is D plus glucose, one unit of D plus glucose and one unit of D minus fructose. That is what it forms. So you see we have two opposite configurations, plus and minus. So the net change we have, that confirms the inversion of cane sugars. Inversion is what? Inversion is the change in configuration. It changes from plus to minus or minus to plus. Okay. So write down the next point into this. If you see this specific rotation, this point to write down, if you see this specific rotation, it changes from, changes from plus 66.5 to minus 19.85 degree. So this conversion we have from plus to minus, this change confirms the inversion. This change in sign confirms that the inversion takes place. Copy till here. Done. Write down next line. One second, sir. Yes, sir. Done. Next line. That's the process of hydrolysis of sucrose. The process of hydrolysis of sucrose. Process of hydrolysis of sucrose is termed as inversion of cane sugar is termed as inversion of cane sugar and the hydrolyzed mixture and the hydrolyzed mixture having equimolar quantities of D plus glucose and D minus fructose having equimolar quantities of D plus glucose and D minus fructose is called invert sugar. It's called invert sugar. Inverse, sir. Not inverse. Invert sugar. Okay, sir. Invert sugar. Next line. The hydrolysis takes place in presence of the hydrolysis takes place in presence of any mineral acid or enzyme takes place in presence of many mineral acids. Enzyme takes place in presence of many mineral acids and enzymes which are known as invertees. I N B E R T A S E. Sir, can you repeat the spelling? Yeah. I N B E R T A S E. So this question also they asked what invert, what is invertees? Okay. And what invert is we use? So we can use enzyme. We can use any mineral acid for this hydrolysis. Okay. So this we are done for. We're done for this thing. Mono saccharides. However, this comes under disaccharides to close on a part, but this is it. Now we'll discuss about disaccharides. What are disaccharides? These are two monoseccharide units, which on hydrolysis gives two units of monoseccharides. All of you write down the heading. These are the carbohydrates. These are the carbohydrates, which gives, which gives two units of monoseccharides, which gives two units of monoseccharides on hydrolysis. Okay. So the example for disaccharides we have sucrose, maltose, lactose, examples of disaccharides. General formula C 12 X 22 O 11. This is the formula for disaccharides. We already have seen this reaction. So, sucrose goes under acidic hydrolysis and forms glucose plus lactose lactose goes under acidic hydrolysis and forms glucose and galactose other than glucose plus glucose. Yes. Maltose on acidic hydrolysis gives two units of glucose. One note you write down. In disaccharides, monoseccharides are joined by disaccharides. Monoseccharides are joined by glycosidic linkage. All monoseccharides are joined by glycosidic linkage. So two, three different informations you should know in this particular section, right? What kind of question they ask based on that? Okay. So first thing you see, I said what, that it is attached by the glycosidic linkage, means between this glucose and fructose will have a glycosidic linkage present, which forms sucrose. Right. Similarly, here also we have glucose and fructose, a glycosidic linkage forms lactose and here also we have glycosidic linkage forms maltose. Right. What exactly is glycosidic linkage? Wait, wait. So first of all, you should know that, you know, the glycosidic linkage present between which carbon atom because the glucose we have four, five, six carbon fructose we have six carbon atom. So which carbon atoms involves in the formation of glycosidic linkage that you should know? What is glycosidic linkage will tell you? What is glycosidic linkage will tell you? Okay, sir. Okay. So glycosidic linkage, the carbon atom you should know plus based on this linkage, we can say that given disacrides is a reducing sugar or non-reducing sugar. Reducing sugar are those which can reduce tolerance and failing solution. Okay, based on that we can say. Okay. So these two informations are very important that which carbon atom involved in glycosidic linkage and which sugar is reducing and non-reducing plus which one shows mutarotation. These three informations you should have. Okay. So first disacrides we are going to see how it forms. The first one you write down sucrose. Take two minutes I'm coming back just a second. So sucrose we know it is made up of one unit of glucose and one unit of fructose. So like I said we should know which carbon atom involves in this bonding between the glucose and fructose. So first of all I'll draw the structure of this. So write down sucrose. It is formed by the condensation of alpha D glucose pyranose beta D, beta D fructofuranose. Okay. So that information was very general information that it is made up of glucose and fructose. Glucose we can have alpha beta both. Fructose we can have alpha beta both. So here we have alpha D glucose and beta D fructose. Clear. So this is also important. So then what are pyranose and furanose? I discussed about the structure of glucose. It is made up of pyranose. Okay. Got it. That is structured with glucose pyranose.  That is called fructose furanose. Now I remember. So basically glucose pyranose and fructose furanose are the structure of glucose and fructose. Yes, sir. Yes, sir. Okay. So you have to draw alpha D glucose pyranose structure and the system. So let's draw the structure. Alpha D glucose pyranose means the first carbon, OH on the bottom. Second carbon, OH on the bottom. Third carbon on the right. Fourth carbon on the left. And then the fifth carbon, CH2, OH on the top. Right. Yes, sir. No, sir. So this is alpha D glucose pyranose. What is beta D fructose furanose? Based on furanose. That structure is this. Beta means at the, this carbon, OH on the top and CH2, OH on the bottom. And then third one, OH on the top. Third one, OH in the top, OH in the top, edge in the bottom, bottom and here in the top. And then here we have CH2, OH on the top and edge on the bottom. So this is the structure we have that we already know. Okay. We know the structure. Now, when you want to, if you want to draw the structure of glucose, obviously condensation takes place, means when these two reacts, H2O molecules evolve into this. Water molecules evolve. So water molecules evolve when OH and edge combines. Right. So what happens here, if you look at this carbon, this carbon is the first carbon. This is second carbon, third, fourth, fifth and sixth. This one is first carbon. This one is second carbon, third, fourth, fifth and sixth. So when the reaction takes place, the first carbon and second carbon, first carbon of C1 of alpha D glucoparenose and C2 of butadifractoheronose involved in the reaction. So this edge and this OH and this edge combines, forms H2O and this carbon attached with this carbon, with this oxygen. So the structure that we get here is this. So what's up with you? Please repeat, sir. Sorry? I didn't understand. Sir, could you please repeat? Oh, wait, let me draw this. Oh, yes, sir. Okay. So here you see, like I said, C1 and C2 carbon takes place in the reaction. Like, no, it involves in the reaction. Okay, C1 and C2 carbon. C1 of alpha D glucoparenose and C2 of butadifractoheronose. So when these two combines in the reaction, this OH and edge combines, forms H2O and this goes out. Okay, sir. Okay. This C1 and C2 combines with this oxygen. So this bond here you see, this bond we call it as glycosidic bond. Glycosidic bond. So you should know first of all that glycosidic linkage present between which carbon atom. Okay. So it is C1 of glucoparenose and C2 of butadifractoheronose. Here we have H on the top. No doubt. Draw this. No, sir. Draw this structure first. I won't ask why the oxygen glycosidic linkage is so big. Glycosidic linkage is? So big. It looks big because I have drawn it big. Yes, sir. That's what I won't ask. Okay. You won't ask. Understood. Done, sir. Finished. All of you are done? Yes, sir. Okay. Now, so first information you should have that sucrose forms by which compounds alpha D glucoparenose and beta D butadifractoheronose. First thing is that. Second thing, which carbon atom is involved? C1 of this and C2 of this. Okay. C1 carbon, you see this carbon, I have already discussed it. This carbon is the anomeric carbon. Anomeric carbon. Even this carbon is also anomeric carbon. What is the definition of anomeric carbon? Change in only that compound gives entire change. Like with different configuration only at that compound like that. That are anomers. Anomeric carbon is the carbon which has OH and OR group attachment. Yes, sir. Okay. Yes, sir. Now, the third thing that you should know is that which carbon atom is involved, correct? So when anomeric carbon of both monosecrites are involved in the reaction. Okay. When anomeric carbon of both monosecrites are involved in the reaction, then the sugar that we get, which is this sucrose here, becomes non reducing sugar. In other words, non reducing sugar. Means it cannot reduce tolling C agent or ferring solution. No, sir. Before that you told one big sentence. What? When anomeric carbon involved in the reaction, both anomeric carbon involved in the reaction of the monosecrites, then the compound that we get, the disacrite that we get is non reducing. Involves in? Involves in the reaction or condensation. Yeah, condensation. Yes, sir. Then the disacrite that we get is non reducing. Disacrite. Yes, sir. Once again, sir. It is non phase that tolerance test. Yes. It cannot reduce the tolerance. So it won't produce a silver metal. Yes. Non-sugar does not show mutarotation. And in this molecule you see, we have two OR group attach, OR and OR, same for here also. So these structures are hemiacetyl, but this one is acetyl acetyl. What is hemiacetyl? We have one OH and one OR. Here we have two OR group attached. So both are acetyl. This is acetyl acetyl. Sir, OR is O up, O down, sir. What is, what, what, what? Like the two OR linkage is shown. This one and this one. Okay. What about the over and pentagon, sir? It is also acetyl acetyl. No. The suprose is this structure, the complex one. Yes. The entire structure is acetyl because we have OR, OR group here. OR, OR group. Okay. Okay. Like pairs, two at a time basically. Sorry? Two at a time basically. Yes. Because we have disacrite, so two molecules. No. Two acetyl structure. The entire structure is acetyl. Okay, sir. Then. Yeah, yes, sir. One more point to write down. Note. Reducing sugar. Reducing sugar. Must have free aldehyde or ketone group. Reducing sugar are those sugars. Which has free aldehyde or ketone group. All monosecrites are reducing sugar. Got it? Yes, sir. Okay. Like we have havert structure of locos, havert structure of fructose. This structure is the havert structure of suprose. Oh, yeah. Yes, sir. Similarly, we have for other disacrites. Okay. Maltose, lactose. You should know all these things. Like, you know, which carbon atom is involved in the glycosidic linkage. Okay. Whether it is reducing or non reducing sugar. If any, like we see, you see in this two compound, we have two anomeric carbon, this one and this one. If one of the anomeric carbon is not involved or any anomeric carbon is free, not involved in this condensation reaction, then the, then the compound that we get, the disacrite that we get is a, is a reducing sugar. Okay. So you should know which carbon atom is involved and accordingly you can say whether the compound is reducing or non reducing. Okay. Now the second one you write down. Sir. Yeah. But since in this, both the anomeric carbons are involved. It's a non reducing sugar. Right. Okay. That's why it is non reducing. Yeah. Yes sir. Second one you write down. You can die. We have multiple write down. It is formed by two glucose. It is formed by the condensation of the condensation of the condensation of two molecules of two molecules of alpha D. Glucopyrinose two molecules of alpha D. Glucopyrinose in which Glucopyrinose. Glucopyrinose in which C one and C four carbon C one and C four carbon atoms are involved. Okay. C one and C four carbon atoms are involved. Okay. So now the first information you have that it is it forms by the combination of by the combination of glucose. Glucopyrinose. Okay. And we know in alpha D. Glucopyrinose the first carbon that we have is the anomeric carbon. Right. Yes sir. Okay. So the first molecule is involved but with the second C four is involved in the reaction. Right. It means the second molecule the C one carbon is free. Not involving glycosidic linkage. Right. I mean first one C four is free. Yes. Second one carbon is free. It means this disaccharide with maltose. Okay. It is what it is reducing sugar because the anomeric carbon is one of the anomeric carbon. Okay. So first of all we'll draw the structure for this alpha D. Glucopyrinose is this. Okay. Alpha D means O H down H up O H in the top O H on the bottom C H two O H on the top H on the bottom and H here. Okay. One, two, three, four, fifth and sixth carbon is this. One, two, three, four, fifth and sixth carbon is this. So like I said C one and C four carbons are involved into this one. So this H and this O H C four carbon here it combines and eliminates water molecule. So the structure that we draw here is this horizontal glycosidic linkage. So this linkage is what? Glycosidic. This one is the glycosidic. You see which carbon atom is involved fourth one and the first one. Anomeric carbon is free. Anomeric carbon is free. And hence this sugar is reducing sugar and reducing sugar shows mute our relation. One second sir. Done sir. Done? Yes sir. Just a second. Yeah. Okay. See anomeric carbon is the carbon atom which has O H and O R group attached to it. One hydroxy group and one ethereal linkage is the anomeric carbon.  Anomeric carbon. Similarly maltose we have done. The third one we have right now. Lactose. Lactose. Right now it is obtained. It is obtained by the condensation of it is obtained by the condensation of one molecule of one molecule of beta D, beta D galactopyrinose, one molecule of beta D galactopyrinose and one molecule of one molecule of beta D, beta D glucopyrinose, glucopyrinose. Okay. So you know the structure of this beta D galactopyrinose. We know beta D galactopyrinose and beta D glucopyrinose are C4 etymers. Okay. Means whatever the structure of beta D glucopyrinose you have at the fourth position you change the configuration. Right. Means suppose in beta D glucopyrinose if the O H at the fourth position fourth carbon is coming down then in galacto it should come up. It should go up. Right. So you can draw the structure that way. So the structure of this you see here we have H and O H on the top since it is beta O H here H on the top, O H here H on the bottom. Here we have H on the bottom and O H on the top. C4 etymers of this H CH2 O H. So this is beta D galactopyrinose and the reaction of this with beta D glucopyrinose which is this O H on the top, O H on the bottom, H on the top, O H on the bottom, O H on the top, H on the bottom, then O H here in the bottom, H on the top, CH2 O H and H. So when the reaction takes place you see this H and this O H combines and forms H2O eliminates water molecule and the structure here we have this, draw the structure. Done? Yes. So one second sir. So there's actually very different from the bio molecules we learned last year. Because the reaction part you have to understand now. Yeah and that was in bio we learned structures and not DNA, RNA, not these kind of things. Yeah that we won't do here. Yes sir. Done sir by the way. Okay. So could you tell me is it a reducing or non reducing sugar? Reducing. Reducing because carbon is free here. So it is a reducing sugar. Very common question. Reducing sugar hence shows a new rotation. New rotation. Yes sir. Next slide down, polysaccharides. Right on. These are the polymers of polymers of monoseccharides. Polymers of monoseccharides. Example. Example we have starch. Oh yeah. Cellulose etc. General formula. So what about titan pectin and all? Sorry? Examples of polysaccharides. What about titan pectin and all? That also you can write. Fine. Okay. C6H10O5. Dimmel formula. Two properties of starch you write out. It is a white amorphous substance. It is a white amorphous substance. Present in beet, rice, potatoes etc. Starch on hydrolysis. Starch on hydrolysis. By enzyme. On hydrolysis. By enzyme. Enzyme. Amylase. Enzyme amylase. This amylase is present in saliva. Yeah. So that's why when we eat these things like rice, potatoes. This starch hydrolyzed by amylase. Okay. And on hydrolysis it converts into glucose. That's why we get sweet test of it. Okay. So the starch present in beet, rice, potatoes. On hydrolysis by enzyme. Amylase. It eventually converts. It eventually converts into glucose. And provides sweet test. Next slide down. Start solution gives blue color with a drop of iodine. Gives blue color with a drop of iodine. And we use this in titration. How is that? In titration to indicate as an indicator we use. And just a doubt isn't iodine already blue? No, I don't do so. It's brown red, brown blueish, brown black blue. That color. Right, right. It's not blue. Okay. The starch solution gives blue color with iodine. Yes sir. Okay. So you see this has, this has two component. Okay. Right down. It has two component starch. The first component we have here. We call it as amylose. Amylose. And the second one is amylopectin. We'll see that. Okay. Amylose is a linear polymer. There's few properties you have to memorize. It's a linear polymer. And soluble in water. Linear polymer soluble in water. And it is composed of this point is very important right now. It is composed. Composed of. It is composed of. Alpha D glucose unit composed of alpha D glucose unit. In which in which. C1 and C4 carbon in which C1 and C4 carbon is involved in glycosidic linkage. Composed of alpha D glucose unit in which C1 and C4 carbon are involved in glycosidic linkage. Number of D glucose unit we have here. Number of D glucose unit is anything from 60 to 300. This is the number of D glucose unit we have from 60 to 300. Okay. Since alpha D glucose is the monomer we can say over here. Right. The reputating unit we have. So if you draw the structure of this, that would be this one. Okay. This is the reputating unit we have here. Like this 60 to 100 units are combined. This oxygen is within this and this often outside the reputating unit. So this is the reputating unit we have like this 60 to 100 units combined to form amylase. Done. Yes sir. Done. Okay. This is the first component of polysaccharides here. Now the second component right down and we call it as amylopectin. Amylopectin right under this. It is highly branched polymer complex structure we have here. Composed of composed of alpha D glucose unit alpha D glucose in which again C1 and C4 carbon are involved. C1 and C4 involves carbon atom involves in glycosidic linkage. Okay. It the number of units here it is more than the previous one number of units here it is anything from 300 to 600 and this is insoluble in water unlike the previous one. Sir. Yeah. Sir if the same basic unit forms both of these compounds like the same glucose pyroglucanose ring then why is one soluble the other not soluble in water. But it is branched one but the previous one is linear. Okay. But in this we have branched and complex structure number of units are also there. Yeah. Done sir. Next right down. Next we have to discuss here protein then amino acid polypeptides proteins amino acids and polypeptides. Okay. CD proteins are made up of polypeptide. Okay. And polypeptides forms by the combination of amino acids. Okay. So it's like the basic unit we have the basic unit is amino acid. And from this we get polypeptides where we have the peptide linkage and from this polypeptides we get a very complex structure and that we call it as protein. Okay. So first we'll discuss amino acids. Then we'll see what is polypeptide and then we'll see what is protein. So we'll start this amino acid after the break. Okay. Take a break now. We'll resume the session at 6.40. Hello. Hello sir. Yeah. Okay. Shall we start? Yes sir. Yeah. Okay. Amino acids. Okay. Amino acids, polypeptides and proteins.  Okay. So first we'll see what is amino acids. And then how amino acids combines by peptide linkage forms polypeptides and then protein formations. Okay. Amino acids you see as the name suggests it has two functional group. One is a mines that amino functional group NS2 and another one is acid. That is carboxylic acid COOH. These two functional groups are present in amino acids. Hence, we call it as bifunctional organic molecule bifunctional organic molecule two functional group present into this. Okay. For example, CH3 CH2 NS2. Right. So this compound we call it as beta amino acid. Beta amino acid. Why? Because the amine group present at beta position. Right. Amines are present at beta position. Correct. Protein that we have protein it forms by forms by alpha amino acid. Hence we'll discuss only, hence we discuss only the alpha amino acids means the amine group present at the alpha carbon. Okay. You see the general structure, the general structure of this we have. We have a carbon and one COOH group attached with it. So this is the alpha carbon with respect to this as you see. And this alpha carbon has NS2 present one hydrogen present here. And one group are G. Any group here is possible. So depending upon this group, this G will have different names. So when G is equals to G, if you replace by H, then the name of the compound is Glycine. Glycine. Glycine. When G is equals to CH3 if you place, when this compound we call it as LNI. Ah yes sir, LNI. When this H is also is replaced by pH, then it is phenylalanine. Alenine. Alenine sorry. Penylalanine. Correct. So like this, we have different, different structures possible. Okay. Now here you see in your book NCRT, if you see there are 20 such structures given. Around 20 such structure given. Okay. All the given structures are optically active except this one. Because this hydrogen. It has two hydrogens. This carbon is not a caril carbon. The glycine is optically inactive. This is the very important one, must remember this. And all the structure given in NCRT, you must remember. Must keep that in mind. Sir, is there in part one or part two sir? Sorry. Is there in which part? In the bio molecule. Okay. Okay. I don't have the appropriate name. Okay. So maybe first or second fact that you have to see. I don't remember. Okay, sir. Correct. So there. I don't have that book with me. Because I left it in school and logged on again before I could collect it. What? I don't have organic NCRT with me sir. I left it in school and logged on began before I could take it. Okay. So this are the general structure we have here and this is okay. We have another structures also. We'll see that there is a structure called proline. Okay. It does not follow the same pattern. The general structure that I have drawn. I'll show you that structure also. But first you see the classification of amino acid classification of amino acid. Okay. Three different types. We have here classified in three different types. The first one is acidic. Then neutral amino acid. And then basic. Acidic neutral and basic amino acids. Okay. Acidic amino acid are those assets in which the number of acidic group is more. Write down. These are the compounds or amino acids. Which has more number of acidic groups, a number of acidic group that is COOH. The number of acidic group is more than basic. Like for example, you see we have NH2, CCOOH. This thing is common in all the molecules. Right. Now when this GU plays here, CH2COOH. This compound, the name for this compound is aspartic acid. We have two acidic groups, COOH, COOH and only one NH2. Hence it is an acidic amino acid. Another example you see. Carbon with COOH, hydrogen, NH2. And here if you put CH2, CH2, COOH. Again more number of. Glutamic acid. That is glutamic acid. Yes sir. Okay. Similarly for basic amino acids, the number of basic group is more. And here we have equal number of acidic and basic. Number of COOH and NH2. The example for this one, we have seen glycine is one of the examples for this. For basic one, the example is NH2. Even allenine sir. COOH and here we have CH2 for NH2. Okay. So more number of basic group we have here. This compound is basic amino acid. The name for this compound is lycine, L-I-S-I-N-E. Copy? No sir. A little bit left. Two minutes sir. No sir. The other structures we have, CH2OH, C, carbon with COOH, NH2, H and in the top we have CH2OH. This compound, the name for this compound is syrene. Carbon with COOH. Sir. Yeah. Syrene basic or neutral. Syrene is acidic I guess. See, when you see the number of acidic group here, there is equal, right? COOH and NH2 equal. So it comes under neutral amino acid. Hydroxychrofoil. Yes sir. OH is slightly acidic because of alcohol group. But OH is slightly acidic. Right. Overall the behavior of this is acidic in nature. But like the definition of, you know, the classification of amino acid we have, it depends upon the number of acidic group and basic group. Okay. OH, if you consider into this acidic behavior, then overall the molecule will be acidic in nature. Obviously basic it is not. Okay. So that's all the acidic behavior we have. Correct? Yes sir. Okay. Now the another one we have COOH and here we place NH2 and here instead of OH, if I place CH2SH. The name of this compound is systeine. Okay. This is also acidic, slightly acidic. The other molecule is proline, which does not follow this pattern that in the pattern that we are looking at now. Systeine has, so proline the structure is this, we have a five member ring in which the nitrogen is present in the ring with HOH. This carbon has COOH. This is proline. Does not follow the same. All these amino acid, if you see, except the first one glycine. Okay. All are optically active. Write down. All these amino acids except glycine, each of these amino acids except glycine is optically active. And exist in and exist in D and L forms, D and L forms, D and L forms. However, however, all naturally occurring amino acids belong to L configuration. However, all naturally occurring amino acids belongs to L configuration. And protein also consists of L amino acids. Okay. Forms with L amino acids. Protein consists of L amino acids. Done, sir. Okay. Now. The next concept we have that is jitter. You see, you have any amino acid that we have any amino acid we have. For example, if I'm taking the general formula COOH and here we have. Okay.  Then this carboxylic acid releases H plus iron and forms. It releases S plus iron in water. And since this nitrogen is basic in nature, it has tendency to accept this H plus and becomes neutral. So it again converts into an H three plus carbon. This is neutral. Okay. This neutral iron here, we call it as jitter and this is dipolar also because we have to pull the site. So it is dipolar. So one minute, sir. So what exactly is this thing then? See, any amino acids, since we have both end acidic in and basic in. So one end has tendency to lose H plus and other has tendency to gain H plus. Yeah. In water what happens this kind of a neutral iron it forms, which we call it as jitter and. Oh, okay. Sorry. But the basic definition of ions any charge carrying species now. So we have this charge. Overall it is neutral. Overall the total negative charge equal to a positive charge. But this ions that we have here in this form. We have one more thing to discuss it. We haven't finished it. Okay. Okay, right on next point. The structure of amino acid. The structure of amino acid depends upon the structure of amino acids depends upon depends upon the type of solution in which it is kept depends upon depends upon the type of solution in which it is kept. Means the pH of the solution, whether it is acidic or basic. Okay, depends upon the pH of the solutions. So depending upon the nature of the solution, whether it is acidic or basic. Jitter ion either it exists in an ionic form or it exists in cationic form. I'll show you how Jitter ion is this carbon with COO minus and is three plus. H and here we have the group is Jitter ion. Now when you place this into a basic solution, OH minus. Or we can say high pH value, high pH value. Okay. So what happens this OH minus takes H plus from this and this three plus iron right and it forms carbon with COO minus. And it's to H and G. So this is the anionic form of Jitter ion. So Jitter ion is the neutral one. Okay, this is Jitter ion like I've already discussed Jitter ion. But depending upon the nature of the solution, it either exists in anionic form, or it exists in cationic form like you see here. This acidic or low pH means this H plus combines here with this oxygen and it forms carbon with the group COOH and is three positive and H. So this is the cationic form of Jitter ion. So Jitter ion is neutral, but it exists either in cationic or anionic form depending upon the nature of the solution acidic or basic. So two minutes. Okay. Again, you see based on the pH of the solution in which it is kept. Okay. We define one more term here, which we call it as isoelectric point right down the heading isoelectric point. Right down this is what happens.  So depending upon the pH of the solution first of all you see Jitter ion is neutral right but depending upon the pH of the solution it will either exist in cationic form or anionic form. Okay. Yes. In this solution if you place an electrode or any rod and we connect this with an external source of voltage, then if it is negatively charged, then it moves towards the positively charged electrode, positively charged and moves towards the negatively charged. So we have an a potential here a value of this potential of the battery that we are connecting. Where the or we can say one more thing a certain value of pH where this amino acid, the Jitter ion that we have it does not move towards any of the right so what point I'm trying to make this to copy this down first of all solution containing Jitter ion. And after this you write down on applying a potential difference across the electrode and applying the potential difference across the electrode across the electrode. The cationic the cationic form move towards the negative charged electrode anionic form. It moves towards the positive charged right so this will either move towards the battery part or the anionic part of them. Now, if the pH of the solution will adjust in such a way that there is no movement of these iron towards any electrode. Okay, next point to write down, but at certain value of pH at certain value of pH amino acid amino acid does not move towards any electrode does not move towards any electrode does not move towards any electrode and all the amino acid and all the amino acids convert into Jitter ion. And this pH value, we call it as iso electric okay so overall what happens in this, we know this fact that in a given solution pH value, Jitter ion presents either in acidic in cationic form or in an anionic form and in that case it will move towards the oppositely charged electrode. Okay, but if we adjust the pH in such a way that all this cationic and anionic form. Exist in the form of Jitter ion. Okay, Jitter ion I like I told you it is a neutral charged. It is a neutral molecule right total positive charge is equal to total negative charge bipolar neutral. Okay, so the if the pH adjusted in such a way that all the cationic or anionic form of the Jitter ion converts into the neutral form of it means which is the actual form of Jitter ion. Total positive equals total negative. Okay, at that pH value where all this cationic and anionic form converts to Jitter ion. Since it is neutral, right positive negative charges present neutral, it does not move towards any of the electrode. And when it does not move towards any of the electrode, the neutral part we have here that particular pH value is the isoelectric form just you need to know the definition of it. Right, you cannot memorize there are different different values we have isoelectric point of different amino acids. So we cannot memorize the value just you should have the understanding of what is isoelectric form. Understood, few examples you'll see. Okay, it's isoelectric point is 3.2. This is the acidic one right. If you take the basic one. Okay, the name of this compound is what it is. Plutamic acid. Plutamic acid. This one is. And this one is LNI. You see LNI if you look at the number of acidic and basic group it is equal, right, but its pH value is not seven not neutral, it is acidic, because the acidic behavior of COOH dominates the basic behavior of. Yes. Okay. So this is for amino acids we have discussed. Now we have to see peptides and how these peptides forms with amino acids. Okay, so write down the heading. This is peptides and proteins together on the classification of types. Three types we have of this. The first one is oligo peptides. Second one we have polypeptides and third part of this is nothing but proteins. All these differs in the number of amino acid it contains. Okay, like for example, oligopeptides contains contains two to nine amino acids. The number of amino acids will be two to nine. Polypeptides contains 10 to 100 amino acids. Protein has more than 100 more than 100 amino acids. And the structure is very complex. Okay. Next heading you write down a structure of protein structure of protein we have we have four different steps here for the structure of protein steps. We can say the four different stages here to understand the structure of protein. The first step is we call it as the primary structure primary structure right down in this right down. In this structure, the amino acids joined together in this structure amino acids joined together by peptide linkage joined together by peptide linkage. This is also like condensation reaction, where the two molecules combined together by elimination of water molecule. For example, you see, if I write down this NH2 with carbon and this carbon has C double bond OH. Here we have hydrogen and here we have the group. Okay. Suppose we have G1 here, COOH. When this combines with the another amino acid, for example, this one, it's not this. For example, we have HNHCG2HCOOH. So when these two combines, then what happens? This OH and H combines together forms water and eliminates and the bond that we get here is C single bond C double bond O and this is bonded with nitrogen HCG2COOH. Here we have hydrogen G1 and this is NH2. Okay. This C double bond O and NH, this bond that we have here, this bond we call it as this is peptide linkage. Got it? Yes, sir. Okay. This is peptide linkage. The primary structure is based on this linkage. Like this, we have third amino acids combined this side, then fourth, fifth, and so on, up to what? Up to nine amino acids. Got it? Right. So all these nine amino acids, two to nine, combines with peptide linkage and that gives the primary structure of protein. Correct? So let's see this. So we can write down the structure. We have H2NCG1HC double bond O connected with NHCG2C double bond O. H and so on. Right. So this here you see, this is the first acid we have. This is the second one. This is the third one. Similarly, fourth one, fifth one, and so on. Okay. So this is the first amino acid. A1, A2, A3, and so on. This is the first amino acid, A1, A2, A3, A4, A5, till A9. This and this structure, this peptide linkage structure that we get, we call it as primary structure of protein. Okay. Like this, the molecules combines and we get the primary structure of protein. Now, since this, we have nine different molecules, nine amino acids attached together. So it is a long chain compound and hence it cannot be in a straight line. Okay. And it exists in helical form because long chain structure, straight chain is not possible. Right. So it forms a helical structure. You see here, we have nitrogen, carbon, carbon, nitrogen, carbon, carbon, nitrogen, carbon, carbon, and so on, like this. Okay. So write down next line. This is a long chain compound and exist in, and exist in helical form, helical structure, exist in helical structure. This helical structure, we can draw like this. Like this we have helical structure. Where this, we have NCC like I told you, NCC, then again NCC, like this kind of structure we have. Roughly I am drawing this, okay. This kind of helical structure we have. This helical structure, we call it as alpha helix. It is like spring, spring-like structure. And this spring-like structure, we call it as secondary structure. Secondary structure. Next, write down this alpha helix structure. This alpha helix structure is also not a straight chain. It has, write down like this, the alpha helix structure has also not a straight chain. But it is also bent at some point due to the different forces present. Due to the different forces present. Okay. It also bends at some point due to the different forces present. Okay. So this alpha helix structure, since again it is a large compound. It presents in a bend form, bend-like structure. For example, you see this kind of structure we have. Then it bends like this. Okay. Then it goes like this kind of structure we have. Okay. So this kind of random structure we can draw, it is a bend-like structure. And this bend-like structure, we call it as, this we call it as tertiary structure. Right. So actually what happens in this compound, there are different forces involved. We have intermolecular hydrogen bonding. Okay. We have van der Waals interaction also. Right. Depending upon the molecule, suppose the amino acids has sulfur atom present into this. Then we have disulphide linkage also. Like in case of cysteine, we have disulphide linkage. So there are different types of polarity in this molecule we have because of difference in electronegativity. And hence we have hydrogen bonding possible, iron-salt interaction there, van der Waals interaction there. All these kind of interaction possible here. Because of this interaction, what happens suppose at this point, this and this ion are attracting to each other. And this will bend toward this. Similarly, this bending takes place and this gives the tertiary structure. This bending happens because of the attraction of the various ions present in this helical structure. Correct. So like this we have, there are different, different tertiary structures there. It is a one unit of this tertiary structures. Like this we have n number of units possible of tertiary structures where we have the charge separation randomly at some point. Right. So suppose here we have the positive charge. So this will attract the negative charge of the other similar units. Okay. Suppose somewhere we have negative charge. So this will attract the positive charge of the other units of this tertiary structures. Okay. So the point I'm trying to make here that when you get this tertiary structures, we have charge separations and this kind of n number of tertiary structures will have which are very close to each other because of the plus minus attraction, charge attraction. Okay. So what happens, this gives you one tertiary structure. Suppose the tertiary structure is, is suppose this one randomly I'm drawing this structure in just a second. Suppose the tertiary structure is this. Okay. This kind of tertiary structure you have. One tertiary structure is like this and we know in this one we have what and I have drawn like this, but it has actually what it has helical structure. Just a second. Okay. So it has helical structure like this. This kind of helical structure is there bent at various points because of various attraction interaction. Correct. This is the first structure P1 it is suppose this is actually protein. But it is a component of the large structure, the complex structure of protein. It is not the protein, but protein is this kind of n number of structures combined together by various interaction. So this suppose if I write this as P1 and this is, you know, this is what we can say this is surrounded by the other structures like this you see surrounded by the other structures like this. This kind of other units possible which surrounds like this. Randomly I'm drawing because there's no any fixed shape for this kind of. Okay. So suppose this is this is P1 this is P2 the another protein. This is a third one protein fourth one and fifth one. So around this also we have certain protein structures like this.  This is structure is actually it is the quaternary structure quaternary structure. Yes, sir. Okay. This is what this is this is a polypeptide because we have n number of peptide linked into this in this we have n number of peptide linkage and so this entire thing is what this this this this everything is a polypeptide right so in quaternary structure what happens due to due to the attraction due to the attraction of other polypeptide other polypeptide chain connect with connects with tertiary structure. Okay. So when you get this tertiary structure we get a bigger molecule here. The bigger molecule that you get which has very complex structure. This bigger molecule is nothing but protein. Yes, sir. Okay. So this is a this is the structure of protein we have now the structure of protein changes temperature the structure changes with temperature. Okay. Right down this point here. The body temperature you say it is 37 degree Celsius right and if the temperature changes to 39 degree Celsius by some reason the structure of protein changes right so structure changes with the temperature structure of protein this term we call it as denaturation of protein. So what is denaturation it is a change in structure of protein with respect to temperature. So but when we get fever does it happen? Yes. So when there is a change in the structure that reflects with fever means when the protein structure changes okay because you suppose you have something or suppose you know when some reaction is there or suppose if you do not have your breakfast and sometimes you feel you know not you know comfortable because the SEL that secretes to digest the food that's probably ruptures the some bond over there because the acidic means correct changes the structure of protein which affects in the you know in the body temperature or some other changes that and for which we take medicine to settle down those things. Okay so everything that happens to our body body it is basically because of the change in the structure of proteins okay because that is the basic unit or by any means if you eat something which is not you know good for the body that will change the structure of protein or you know rupture some bond over there right that will reflect into the fever or any other you know problems that occurs into the body. Okay so this is what the basic thing we have okay structure of proteins they won't ask question based on this okay you won't see any this is basically the part of biology okay for chemistry mainly we have those reactions that we discussed and the type of linkage we have and all those things the only one type they ask into this one that protein contains what peptide linkage basic thing is that when you know the primary structure that forms that forms by what the peptide linkage correct two amino acids come okay so that is one question that asked in the exam okay okay now there are two things into this one one is essential and the other one is non-essential okay I mean oh yes this was taught and just as fun fact my last one thing that you already know that peptide went to you know peptide bond forms we call it as dipeptide okay or peptide linkage when one molecule combines we'll have when two molecule combines one peptide linkage when three molecule combines we have dipeptide linkage similarly we have tripeptide tetrapenta like that okay for example you see this one see this one when two molecule combines you'll get one peptide linkage right another molecule this side so two peptides so this one two peptide linkage there we call this dipeptide similarly tripeptide tetrapeptide and so on right so proteins we have n number of such peptide linkage okay so protein forms like we have already discussed by joining the carbonyl group of one of one amino acid and alpha amino of another alpha amino group of another amino acid that structure already I have shown you that in one of the molecule we have COOH that is the OH of carbonyl group combines with the H of the amino group okay that is how the protein forms two terms we use here that is essential and non-essential amino acid right down non-essential amino acids right down essential amino acids are those only definition we have here essential amino acids are those which cannot be prepared by body okay body does not prepare these amino acids for example we have phenyl, alanine, valine, tryptophen, lysine another example we have non-essential amino acids are those acids which is which is prepared by body serine for example okay within the body inside the body for example serine, glycine so actually this is a mnemonic for this essential private tim hall sorry for essential is a mnemonic PVT private tim hall private MIT hall yes sir I think maybe okay one last thing right down one last thing right down test of proteins test we have two tests for proteins sometimes they have asked the name of this what test we use for proteins and all the first one is we have Milans test Milans test and the reagent we use for this test we call it as Milans reagent Milans reagent so what is Milans reagent right down Milans reagent is mercurus and mercuric nitrate mercurus and mercuric nitrate in nitric acid mercurus and mercuric nitrate in HNO3 nitric acid this test is given by is given due to the presence of phenolic group okay so what is Milans reagent this we must remember okay you'll get direct question from this right because there is not much to understand Milans reagent mercurus and mercuric nitrate in HNO3 the test is given due to the presence of phenolic group on addition of Milans reagent on addition of Milans reagent in protein the white precipitate turns brick red on heating the white precipitate turns brick red on heating this remember second and the last we have white precipitate turns brick red on heating the second test we have we call it as nitro preside test nitro preside test write down the protein contains SH group okay the protein contains SH group SH groups gives this test gives this test on addition of sodium nitro preside on addition of sodium nitro preside the violet color the violet color turns sorry violet color on addition of sodium nitro preside the violet color appears okay that is the test of proteins this compound is sodium nitro preside not much important sometimes they have asked formula directly what is sodium nitro preside the violet color appears that violet color confirms the presence of protein test is not that important okay just name you should know so this is it for biomolecules okay so we have done with it I think next class will start with polymers we'll finish this polymers also and then we'll start with Grignard reagent reactions we can finish this polymers and I think yes okay and then after finishing this we'll start with we'll start with what Grignard reagent reactions another okay any doubt yes sir yes sir I have a couple of organic not organic physical doubts tell me first once I go get my mom's phone to send you I don't have it with me two minutes sir yes I just came I'll send it to you now sir sir I sent it one to you sir you did not get it just like sir you might come having some problem I should have sent now no you read it also which one the second one sir 52 question 52 the value of delta H not for the cell reaction is twice that of delta G not at 300 can be the inner cell reaction is the MF the cell reaction is 0 delta S not of the cell reaction per mole of B formed at 300 can be so T can I what is the doubt in this suppose of the our MF is 0 correct yes sir you like will try it sir I don't think you'll get the answer I wasn't getting the answer see MF is 0 correct delta so you need to find out not of the cell using last equation did you find E not yes sir once I can say I I don't I don't have the solution with me there are different notebooks probably we could start again see I'll give you the idea the reaction is this a solid gives a 2 plus of course B 2 plus at plus 1 molar and B what is the reaction a gives a 2 plus plus 2 electron B 2 plus 2 electron gets B solid so the reaction is a plus B 2 plus gives a 2 plus plus B solid yes sir now we use last equation E cell is equals to E not cell minus 0.059 log of concentration of a 2 plus a 2 plus by B 2 plus these two things are given you know given in the question oh yeah yeah yes sir okay E not cell from this correct yeah and from this E not you'll get delta G not minus N of E not N is 2 right is twice of delta G not yes others other can leave now okay that's the last they're doing this last question if you want you can see this otherwise we'll leave now thank you sir delta H not is equals to 2 delta G not is given the question yeah use this form delta G not is because delta H not minus B delta S not yes sir I was doing this same thing I did I didn't do anything different what is that delta E not cell you'll get tell me wait sir I have to calculate now you have done this yeah that's pretty long back sir actually I keep trying this was when before the previous came on only so I keep trying the question again again it gets lost in my notebook once I can say nothing it's not that hard you know cell minus 0.0591 by 2 log of what is it 1 by 2 right log of 2 log of plus by B2 plus so there'll be 0.313 correct no sir 0.313 you saw there's no A2 plus is given 2 by 1 2 by 1 log 2 is 0.313 3010 see value I do not know you can find out values okay check the value then might be calculation 0.0591 into 0.3010 0.0088 0.0088 what what is this value E not total E not is 0.0088 so you find out delta G not I did this okay now I'll show you delta G not equals minus 2 into 96500 E not yes I see this is the problem I'm getting 1698.4 that's 1.6 kilojoules okay the answer is 11.6 what you're getting 1.6 1.69 11.62 you must have missing some points here and there see 1.6 11.6 it's close what you're getting 1.16 no I'm getting 1.6 only I'm not getting 1.16 and one more question I had question you get log of 2 could you tell me what is this we have A what A plus 2 only is it plus 2 it's A N plus yes that's what I can see it's A N plus no and here what we have A B 2 N plus that's what the mistake we made it is twice of this no N plus plus N electron and B 2 N plus plus 2 N electron right you have to multiply this by twice that's allowed the number of electrons hence it should be A square here okay so if it is A square then you'll get here a log of 4 when you solve this you get E naught cell from this okay sir now after this then E naught equals if I know G naught then G naught equals delta H minus delta S yes and one more question one more question I'll see that I have some work now right now I have some engagement 8.15 okay but you try this and send me the solution see one thing I will do that you can do in one more way because E cell to solve this what we can see E cell is what this expression you get the entire expression is this yeah and E cell only we can write delta G by N F right this we can further write delta G naught by N F times minus of it from this only you can directly get delta G but if you are not getting it find out E naught cell from this expression then this formula delta G naught is equals to minus N F E naught cell and we know delta H is equals to what given delta H naught is equals to 2 times 2 delta G naught if you substitute it here so what we get delta G naught is equals to T delta S final expression will be this yes if divided by T gives you the answer delta S yeah okay so so just confirming and delta G naught N is equal to N F is 96500 E naught whatever we calculate okay think the way you can solve this you will get the same answer F is 96500 E naught sir huh okay ok you send me that question I will see that ok I will send it to you sir ok sir thank you sir bye sir