 Hi, how are you? Good, good. Okay, so what do you have discussed last class? Okay. Can you see the screen now? Okay, so who are not there? Let him come. I'll give you some questions on organic. Okay. Simple question. What is the name of this compound? IUPAC name. Tell me the name. What about others? One chloro first. When you write down the name, it should be alphabetically. One, two and three. Right. Three bromo, one chloro cyclohex. First we'll write bromo then chloro. Okay. Right. What is the name of this compound? Now, two or three. Why three? Why not two? That's what I'm asking. Two or three. Okay. Three chloro is correct. Why you see? Whenever we have double bond in the ring. Okay. See, we have to follow rules. We cannot go with our own emotion. Right. So rules we have to follow in nomination. So whenever we have double bond in the ring, we always give first and second position to the double bonded carbon atom. That is the rule. Right. So this carbon atom will get one or two or this carbon atom will get one or two. So numbering will be what? We are bound to give first and second position to the double bonded carbon atom. That we cannot change. Now the two options are there. We can write this as one, two and like this four, five, six, or we can write this as one, two, three, four, five, six. So in this, we'll follow this pattern, this numbering, because you're getting here at third position. From this side, you're getting here at sixth position. So this is the numbering we have. It should be three chloro cyclohexane. Understood. So that's the rule. Like whenever we have double bond in the ring, we always give first and second position to the double bonded carbon atom. One question you see. This is crown either. And we write it as X crown Y. All these questions have been asked. X crown Y. You have to tell me what is the value of X and what is the value of Y. Any idea? Number of oxygens. See, Y is the number of oxygen. Okay. And X is the number of carbon plus oxygen. Okay. So X represents the number of carbon plus oxygen atom present and Y represents the number of oxygen atom only. So X value will be, how many carbon atoms are there? Six oxygen. So Y will be six. And this will be 80. Name. IUPAC name. Tri-sino. What is the name of this one? Dynite trial. Okay. You see, first of all, the name of the first one. 1, 2, 3. 1, 2, 3. Tri-sino. Numbering will be this. In this case, the numbering will be this. 1, 2, 3, 4, and 5. So it will be 1, 5. Pantene Dynite trial. The rule says what? Whenever you have carbon containing functional group attached to the unbranched alkene. Okay. Carbon containing functional group attached to the, sorry, more than 2, right, more than 2 carbon containing functional group attached to the unbranched alkene, then we don't consider this into the parent chain. You have, you must have remembered that we always start numbering from the carbon atom of the cyano group, even for carboxylic acid also. We start numbering from the carbon atom of this functional group, right? So, but in this case, when we have more than 2 carbon containing functional group attached to the unbranched alkene, then we don't consider this into the parent chain. It will be 1, 2, and 3 tri-sino propane, right? But here we have 2 carbon containing functional group. So we have to start numbering from this carbon or this carbon. Both you can do. 1, 2, 3, 4, 5. So we have 5 carbon, so Pantene Dynite trial. Understood this? No, no, you cannot write. This is not, this is not a functional group now. We are not considering this as a functional group. So we cannot write nitrile. It is propane simply. And with propane, few hydrogen atoms are been replaced by the cyano group. Fine, but when we have more than 2 functional group present, then we don't consider that into the function of this parent chain. Then, then you can write down. Try all. Try all. The rule, the rule is, listen to me, the rule is what? When we have 2 more than 2 carbon containing functional group attached to the unbranched alkene, that is the rule we have. Okay, so we cannot write down nitrile here. Simply if you write down this, CS3, CH2, CS3, what is this? This is propane. And how this compound differs from this? You replace one hydrogen by CN group here, another hydrogen by CN group here and another hydrogen by CN group here. So in this propane, 1-1 hydrogen has been replaced by this cyano group. That's what we write. So this is the rule basically we are following. One more question we will see. This one? Others? First and second position we will give here. 1, 2, 3, 4. We cannot give 1, 2, 3, 4, 5, 6. Glow is set to look and so on. What is the name of this compound? CHO. This was single bond. CN, COOH. What is the name of all these compounds? Have you seen this kind of question? This one? First you tell me this one. What is the name of this compound? This one? These three you tell me. Ethyl cyclopropane. This one? Cyclopropane propane. Okay. And what about this one? Cyclopropane butane. Okay. First of all you see this one is correct. See whenever we have ring and alkene group attached with the ring. So the parent chain will be either ring or this parent chain depending on the number of carbon atom present here. So if the ring contains more number of carbon atom, more or equal number of carbon atom present in the branch or the substituents, then we always consider ring as the parent chain. You get this. If the number of carbon atom present in the ring is greater or equal to the number of carbon atom present in the substituents, then ring will be the parent chain. Clear? This is the rule. So parent chain is this ring here. 1, 2, 3. So ethyl cyclopropane. Correct? This will be again the parent chain will be ring only. So it is propyl cyclopropane. Correct? This is propyl cyclopropane. Here the parent chain will be the substituent because it contains more number of carbon atom. So what we will write into this? 1 cyclopropyl butane. Right? This is the substituents now. Parent chain is butane. So cyclopropyl butane, 1 cyclopropyl butane. Correct? Now if I give you this one, cyclopropane, and here we have CH2CHO ethyl, right? Because when the substituents contains functional group, then this will be the primary chain 1 and 2. This becomes the substituents now. When functional group is there, then we don't see the number of carbon atom. This is the parent chain now. This is substituents. So let 2 cyclopropyl ethyl. Correct? Understood all of you. Now the difference in this question and this question is what? That here, here the functional group attached directly to the ring. Right? Even in all these examples. But here it is not directly attached. Correct? So now for this kind of question, we have another rule here. And the rule is what? Whenever I write down this, if you don't know, write down this. Whenever carbon containing functional group is directly attached to the ring, then we use following suffix. If the functional group is CHO, then we use carbidehyde. Okay? If the functional group is COOH, then it is carboxylic acid. Carboxylic acid. If it is nitrile, then it will be carbonitrile. Right? If it is C double bond OOR, then it is carboxylate. Alkyl carboxylate basically we will write. So when this one you have to write down. So this will write cyclobutane carbidehyde. Get this. It is cyclobutane carbidehyde. Cyclobutane carbonitrile. Cyclobutane carboxylic acid. Clear? When you have this compound, one more example I will write down on the same rule. C double bond O OET ethyl. So in this we will write first ethyl here. We will write first ethyl. Then we have cyclobutane. And last what we will write? Cyclobutane carboxylate. Understood? Okay? So these are the few specific rules we have, which is important for example in the view. Others you know already basic thing that you can do. Okay? A few specific rules I have revised already. Okay? So we will start then. Miltar rotation we have done last class. So glucose, everything we have done have our structure of glucose and all those structures. Okay? So now the thing that we have already done in the last class, similar discussion we will do for fractals also. So we can go a bit faster today since we have discussed those concepts in quite detail in the last class. The same kind of thing we have for fructose here. We write down fructose C6H12O6. Now in this you see the structure of fructose is this. CH2OH C double bond O H OH OH H OH H and the last again we have CH2OH. Okay? So if you see the structure of glucose and fructose you compare, you see this part is common. Excuse me. If you leave the first two carbon atom, the last four carbon atom has the same configuration for glucose and fructose. Yes? This is one thing this is important. We will use this in few minutes later. We will use this particular concept once. Okay? So you just keep in mind the last four carbon atom has the same configuration as we have in glucose. Right? And if it is what? D or L fructose. What structure it is? D or L? How do we configure that? How do we understand? D or L? Yeah. So we have to see the carol carbon which is at the maximum distance from the primary functional group. Primary functional group is ketone into this. In glucose it was aldehyde. So one, two, three, four. At fourth position is this, the carol carbon and O H is present on the right-hand side. So we call it as D fructose. If you write O H on the left-hand side, it does L, right? And this information I'm giving you, it is D minus fructose. Like there we have D plus glucose and D minus glucose. Similarly, this structure is D minus fructose. L structure, if you draw an O H on the left-hand side, then that will be L plus fructose. Okay? Now in this, the same thing we'll discuss that is cyclic hemiacetyl, cyclic hemiacetyl structure. Cyclic hemiacetyl structure. Okay? So now you see in this also, if I write down this structure again, that will be C H 2 O H, C double bond O H O H, O H H, C H 2 O H. Okay? Now the same thing what happens into this, this lone pair, the carbon atom which is the maximum distance from the primary functional group, the carol carbon, this lone pair attacks onto this carbonyl carbon, right? And this pi bond goes here. Condonia is there? Okay, Condonia, you asked me one question, no? That benzoic acid and benzene, right? See, you cannot, okay, I'll discuss this later. I'll discuss this. I have discussed this also in the class, but I'll tell you. Okay, let it be discussed. So this will attack over here, pi bond goes here, okay? And again, we'll get the structure here like this. This is exactly same that we have discussed in case of glucose. I'll write it down this quickly, okay? So you just copy it down. O H can be this side. C H 2 O H can be on the left-hand side. And then everything will be same. H here, O H here, O H, H, H. And this is connected with oxygen here. And here we have C H 2 O H, okay? Plus another possibility is what? That this O H on the left-hand side. And that structure will be this, C H 2 O H, O H, H, O H, O H, H, C H 2 O H. And this is connected with oxygen here at this side. Okay, so when O H on the right-hand side, it is alpha, right? And since this one is what? Alpha, sorry, this one is D minus fructose. Just now we have discussed. So this one is alpha D minus fructose, correct? Because O H on the right-hand side, and this is beta, sorry, it is beta D minus fructose. O H on the left-hand side. Understood? And these two are what? These two are anomers. What are anomers? If you remember, what are anomers tell me? Anomeric carbon. And what is anomeric carbon? O H and O R group. O H and ether linkage will be there, right? So this carbon is the anomeric carbon here. This carbon, right? This carbon is the anomeric carbon. And here the configuration is different. That's why these two are anomers. Okay. Here it is, this carbon was planet. The reason I have discussed already in glucose also, this was planet. So this attack can be from the top or bottom, two different configuration possible. That's why it is alpha and beta, and they are anomers of each other. Okay, finish this. Can I go to the next slide? Yes? Okay, now for this you see, we have the another one that is Havarth structure. Havarth structure of fructose. What is the reference compound we have for Havarth structure of glucose? Turan. Now for glucose, I'm asking. Pyrene, right? So here the reference structure is, since you see here, I'll go to the previous slide. You see here, the ring that you get here, this ring contains five member ring, right? One, two, three, four, and fifth is this oxygen. This is a five member ring, right? So we required what? Five membered compound, right? And that's why the reference compound in case of this is furan. Okay. And the structure of furan is this. This is furan. And this is the reference compound we have for Havarth structure of fructose. Okay. So now when you draw the Havarth structure of fructose, that will be like this. You can draw oxygen. Okay, just I'll draw the next one. You see, this carbon is the second carbon, right? This carbon is the second carbon we have. This one is third, fourth, and fifth carbon, right? Now if you see the structure, just you refer the structure that I have given you just now, alpha d minus fructose or beta d minus fructose, okay? The second carbon you see, we have OH on the right-hand side for alpha and CS2 OH on the left-hand side. Okay? Alpha. Suppose you have to draw the alpha position here, right? So OH on the right-hand side, we have CS2 OH on the left-hand side and the second carbon. So whatever on the right-hand side will write down on the bottom. OH here and CS2 OH on the top, right? And then we'll go alternate. OH will be alternate. If you see the structure, OH will be here. Then OH will be on the bottom. And the last carbon you see, the fifth carbon contains CS2 OH and H, right? So OH on the bottom. So here we have CS2 OH on the top. And all these position, we have hydrogen. Is it correct? Now many of you see, this is the first carbon, second carbon, third, fourth, fifth and sixth. So whatever group atoms are there on the right-hand side will write down on the bottom of this vertical line. And left-hand side will be on the top. So this structure represents alpha d minus fructose. For beta d minus, what we'll do? This OH will be on the top and CS2 OH on the bottom. Okay, so I'll draw that one also. Here we have OH, CS2 OH and all other things are same, right? OH will be on the top here, OH here on the bottom and then CS2 OH on the top. And all these position, we have hydrogen present. Okay, now you see, this structure is beta d minus fructose. Okay, and the name of these compounds, we call it as fructofuranose. Okay, like there we have glucopyrinose. It is fructofuranose. Both compounds are fructofuranose, alpha 1 and this is the beta. Now the three-dimensional view, if you try to understand, this is again that thick line we have, this carbon. Means if you see these molecules, these two carbon is coming towards you and this oxygen is going away from you into the plane. Understood, can you imagine this? See these two, this molecule, this bond is coming towards you. So if you're looking from the front, you see from your whatever laptop that you have, if you're looking at this molecule from the towards the screen, it means these two carbon is coming out of the screen towards you and in the same plane is parallel to yourself, right? Like when you're watching this, when you're looking at this molecule, you are at the same plane, parallel plane and these two carbon, these two carbon is coming towards you in this plane and this oxygen is going backside of the laptop. Can you imagine this? Three-dimensional structure. So when you see this structure, this OH on the top, right? And CS2 OH on the bottom. So if you look at this molecule from the top, from the top view, if you see, then what you see in the top view? OH, OH, H, CS2H, fine? Understood. So if you're looking at this molecule from the top, then you will see OH, OH, H and CS2H. And from the bottom, if you see, you'll see CS2OH, H, OH and H, fine? So this is a structure of beta D minus structure. What? Okay, fine, can we move ahead? Now, one thing we can also, we should also observe here, when we turn this molecule, right? Turn means upside down, we are rotating this molecule. Just flip, right? 180 degree, flip. So when you flip this molecule 180 degree, so obviously, oh, just let me write down, this is first carbon, this is second carbon, third, fourth, fifth and sixth, right? So when you flip it 180 degree, then second carbon will be here, yes or no? Fifth carbon will be here, correct? And whatever atoms or group which is there in the bottom, that will come on the top and the atoms or group which is there on the top will go in the bottom, fine? Okay, so now when you draw this structure, you see what difference we get here. When you flip it upside down, so first of all, here what should we write? Here, tell me. First you tell me, which of like, this carbon is, what is the number of this carbon according to this molecule, according to this molecule? What is the number of this carbon on the right hand side? This one is fifth, right? This one is fourth, this one is, this one is, right? Now, on the fifth carbon, what should be on the top when you flip it H, because here we have hydrogen? You see, on the fifth carbon you see, hydrogen is on the bottom, right? So when you flip it on the fifth carbon, hydrogen on the top and CS2H on the bottom, fine? This is CS2H and this carbon is what? This carbon is the sixth carbon, right? Fourth carbon you see, fourth carbon, hydrogen on the top, OH on the bottom. So here we have OH on the top and hydrogen on the bottom. Third carbon, OH on the top, so here we have OH on the bottom and H on the top. Second carbon, OH on the top, so here we have OH on the bottom and CS2H on the top. This is sixth carbon and this is first carbon. So when they'll give you this structure in the exam, so obviously they will not numbering this carbon atom, they will not do the numbering like this. They will simply give you this structure, okay? So what is the difference between these two structures you observe? Look at the molecule and you see what is the difference we have? Switched, so the difference is only that, that here we have in, suppose now you imagine that there is no numbering here, only you have this molecule given, but what happens? We generally see when you draw the haworth structure of fructose, you will get this structure, yes? You'll get this structure and when you flip it, you'll get this structure. The only difference is what? Since numbering is not there, so when you give you this structure, you will get confused. Okay, the structure where we have on this carbon, we have OH here on the top, according to this, correct? But since we have hydrogen here, right? We have hydrogen here and OH here. So the point is you don't have to get confused with this because both structure are nothing but same. In one structure you are looking from the top and the other one you are looking from the bottom, right? So whether they give you OH here or H here, the structure is nothing but it is a structure of beta D minus fructose. Is it clear? So once in Je, they have given this structure, right? So don't get confused with this that here we should have OH, here we should have H and here we should have OH. If this is also given, the structure is nothing but same, only the view angle is different. Okay, here we are looking from the top and the same molecule if you look from the bottom, you'll see this. Is it clear now? Okay, so this is one important thing we have. Now another thing you see which is again important and that you write down, differentiation of glucose and fructose. Okay, how do we differentiate this glucose and fructose? Now you see glucose, glucose primary function group is what? Aldehyde, now you see the basic properties of aldehyde and ketone basically we have to compare, okay? Since it has one hydrogen present, this hydrogen that's why glucose is easily what? Oxidize or oxidizable, right? Even by mild oxidizing agent, we have discussed this already, right? Mild oxidizing agent since hydrogen is present over here, right? So all these glucose will get oxidized by the reagent called tolin's reagent or filling solution. And you can also recall that when we are doing aldehyde, we have discussed that aldehyde gives tolin's reagent and filling solution test, right? So in this also, this aldehyde, if you see, I just give you a quick recall of this C double bond O and H. When you put tolin's reagent into this, then it gets oxidized into what? Acid, aldehyde oxidized into, sorry, it is not double bond O H, right? And this is nothing but the silver mirror test we have, right? Silver mirror test, so test you must go through, it's important, okay? Silver mirror test. Filling solution, if you put this into filling solution FS, you'll get again acid that is carboxylic acid, and the color that you get here that is brick red color because of Cu plus. All these things we have already discussed, right? Brick red color due to the formation of Cu plus, right? But if you talk about fructose here, right? So if you remember, ketone does not give this reaction with tolin's reagent and filling solution test, right? Since fructose, the primary functional group is what? Primary functional group is ketone. And since ketone, if you put tolin's reagent or if you put filling solution, both of this, there is no reaction at all because there is no hydrogen present onto this, right? These are weak oxidizing agent. Since hydrogen is not present for ketone, oxidation of ketone requires comparatively stronger oxidizing agent. That's why with tolin's reagent filling solution could not oxidize this, right? So ideally if you say since fructose contains ketone as a functional group, so it should not give tolin's reagent filling solution test. Yes or no? Ideal situation should be this only, right? But fructose gives this test. This question is very important. They have asked many times in NEAT exam also, right? A few questions are in this chapter that is important. One of the question is this one. And if you remember, I have discussed this in LD Hyde and ketone. That ketone does not gives, but fructose having ketone group as primary functional group. But then also it gives tolin's reagent refilling solution test. The reason we'll discuss now, right? So first of all you write down here. It should not give this, but write down into this. Fructose contains ketone group. Fructose contains ketone group as primary functional group. Then also it gives positive test to tolin's reagent and failing solution. Okay, this is one thing. Now the point is why it gives positive test, right? Okay, so the reason you write down next and the reason you can understand by a rearrangement and that we call it as name is not important, but I'll write down. That is lobri divan brian echinstein k-e-n-s-t-e-i-n rearrangement. Okay, the name is not at all important. You can skip this also, but I have just given it once. This rearrangement because of this rearrangement fructose gives positive test, right? Now what is this arrangement? Okay, so you see, I'll just write down one reaction here. This one is glucose C double bond O, H. Here we have OH, H, OH, OH, H, OH, H and CH2H. This is glucose. Okay, now in presence of dilute alkali solution, dilute NaOH. We can use RKOH, NH4OH, right? This reaction is reversible first of all, right? And it converts into two different forms and that is this. The first one is mannose and what is the relation of glucose and mannose? Glucose and mannose. Tell me, I've given you the relation of glucose and mannose last class and I told you also if you know this relation, you can draw the structure of mannose also. No, they are not anomers. They are epimers and they call it, they call what? They call C4 epimers and what are epimers? The configuration will be different only at one carbon other than anomeric carbon. I've given you this definition last class, you can go through. You see, since glucose and mannose are C4 epimers, so configuration will be different at fourth carbon, right? You see at fourth carbon, 1, 2, 3 and 4, OH on the right-hand side, so in mannose, OH will be on the left-hand side and all of the structures, all of the configuration will be same. So here we have OH at second carbon, H, third carbon, H on the right-hand side, OH on the left-hand side. Fourth carbon configuration will be different, so OH on the right, so here it will be on the left, H will be this side. Fifth carbon will be as it is, OH, H and sixth carbon is CH2 OH. So this is what, this is mannose. Plus, it also exists, this converts into mannose and then fructose also. CH2 OH, C double bond O, H, OH, OH, H, OH, CH2 OH, right? So this is fructose. So in alkaline solution, these three molecules exist in equilibrium, right? So glucose, mannose and fructose are in equilibrium in alkaline solution, right? So if you take a molecule of fructose, then eventually there will be what? There will be, in the mixture, there will be glucose also and there will be mannose also in alkaline OH-. So you have taken fructose but because of this arrangement, fructose may convert, some part of fructose may convert into mannose and may convert into glucose. And these two contains aldehyde group as a primary function group, that's why the mixture shows positive test. Did you understand this? Yes. Okay, so that's the reason we have, that the reason if they ask, you can write in alkaline medium, fructose, glucose and mannose are in equilibrium, means when you take only fructose, it will convert into glucose and mannose also and since glucose and mannose contains aldehyde as a primary functional group, that's why the mixture shows positive test, right? Mechanism also you can understand of this reaction, okay? I will not write down the mechanism but I'll explain from here only. You see from glucose, you can draw the tautomers of this and how do you draw, you see this double bond comes over here, right? And this hydrogen will leave this carbon and this will attach onto this oxygen, correct? So you'll get here what C double bond OH and OH here, correct? Now suppose this is one, two and three and you imagine double bond here and single bond OH here. Now in this what happens, again if this double bond comes over here and this hydrogen attach onto this carbon, then what we get here? Here we get CS2OH and C double bond OH. Or if I write down only this top part here, you see? We have COH double bond COH and here it is same, right? This part is same, correct? Here we have one hydrogen already for this carbon. This is the tautomers. Now here one, two, three, if you again draw the tautomers structure of this, you will get what? This pi bond comes over here and this hydrogen will come out and attach onto this carbon atom. So this carbon already has one hydrogen, one hydrogen from here. So CS2OH will get and here we'll get what? C double bond OH, fine? So that is how it converts into fructose, okay? So this is the mechanism of this. Basically tautomers will draw, right? Here also you write down this glucose and mannose are C4EPmers. So this is the reason of fructose to show positive filling solution and pulse range and test. Generally these tests takes place in alkaline medium only, okay? Now you write down into this. So another thing is what? Since I forgot to tell you one thing in the last thing you can write down here also. Since you see this aldehyde may oxidize by weak oxidizing agent, right? We oxidizing agent and for this ketone, we require strong oxidizing agent comparatively, right? So if you take a strong oxidizing agent, then whether you have aldehyde or ketone that will get oxidized for sure, right? Understood. See, if aldehyde will get oxidized by weak oxidizing agent. So if you take a strong one, then whether it is aldehyde or ketone it will get oxidized into acid, right? Then we are not able to differentiate whether the molecule was aldehyde or ketone, correct? To oxidize this or to differentiate this glucose and fructose we usually use weak oxidizing agent like bromine water. Okay, that's the one point to write down. To differentiate glucose and fructose, we require mild oxidizing agent, mild oxidizing agent whose medium is not alkaline, whose medium is not alkaline because what happens once when we have alkaline medium? What happens if we have alkaline medium? Aldehyde, right? And then again, oxidation takes place. The point is to differentiate aldehyde and ketone, glucose and fructose we require a weak oxidizing agent whose medium is not alkaline. Example, write down bromine water. We add two with H2O. This is the oxidizing agent we use to differentiate these two compounds. Okay, glucose and fructose. Yeah, same thing. Okay, now we see a few reactions of glucose and fructose. Okay, and you also write down the way I'm writing down because every time otherwise you have to draw the structure of glucose and then the reaction, okay, so it takes a lot of time. So I'll just write down this, yeah, write down here first, reaction of glucose, reaction of glucose and fructose. Only one reaction we'll see for fructose. Getting you write down reaction of glucose and fructose. So in the middle of the page, you write down the structure of glucose first of all. So that will be C double bond O, H, and then we have OH, H, H, OH, OH, H, OH, H and CH2OH. Okay, this is glucose. Now when this glucose reacts with, I'll just write down like this, it reacts with NABH4, reducing agent. NABH4 and then water also we use. Reduction takes place and this aldehyde reduces into alcohol, correct? So here we get what? In place of this aldehyde we'll get alcohol, CH2OH and all other things will be same, OH, H, H, OH, CH2OH. Okay, now if this reaction takes place with a strong oxidizing agent like HNO3, strong oxidizing agent, then what happens? Oxidation takes place and alcohol, aldehyde, both will get oxidized. Okay, so when this aldehyde oxidizes it gets acid, all other things are same and the last carbon, here also we have COOH and here write down as it is whatever the compound it is, right? OH, H, OH, OH, H, OH, H and H. What is the name of this compound? It is gluconic acid. Okay, both will get oxidized, aldehyde and alcohol. Okay, now when you do the controlled oxidation of this, like weak oxidizing agent if you're taking, suppose you're taking weak oxidizing agent, tolerance reagent, okay, or if even bromine water also, BR2H2. Okay, then this aldehyde gets oxidized, right? And when this aldehyde gets oxidized, you will get here COOH on the top and all other things will be same, OH, H, H, OH, CH2H. This compound, the name of this compound is glucaric acid. It is gluconic, it is glucaric. I have given you this structure last class also I guess, let me check the name I think, just to see. COOH, COOH, CH2OH, if you have it is gluconic acid. The first one, this one is gluconic, sorry. This one is gluconic acid and this one is glucaric acid. We also call it as saccharic acid. Saccharic acid, glucaric acid, both are same thing. This is gluconic acid, okay? This is the three reaction we have. Now when this reacts with NH2OH, what happens when aldehyde reacts with NH2OH? H2O goes out and will get oxime, formation of oxime, right? And that will be this, that is here H2O goes out and will get C double bond NOH. This hydrogen will be as it is and all other compounds, structure will be same, CH2OH here. This we call it as oxime formation. This H2 and this O oxygen goes out as H2O and NOH joins over here, oxime formation, okay? Now when this reacts with the last one, this is important, this reacts with phenylhydrazine and the formula of phenylhydrazine is what? H2NNHPH, one hydrogen is replaced by phenyl. So it is phenylhydrazine, okay? And we take three moles of phenylhydrazine here and when you take three moles of phenylhydrazine, the product that you get here is this C double bond NNHPH hydrogen. Here we have C double bond O, C double bond N, sorry. There is a C double bond NNHPH, NHPH and then all other things are same, HOH, OHH, OHH, OHH and CH2OH. Plus, in this reaction will get pH NH2 out phenylamine. Ammonia also goes out and water forms. This molecule, we call it as osajone, okay? This reaction is actually osajone formation, osajone formation. And this product, we call it as, since it is obtained from glucose, so we call it as glucosazone, glucosazone, okay? We can write down the mechanism of this also, but that is not required, so I am not doing that, okay? So you see, three moles of this we are using, two moles we use here, one mole goes out and then this three goes out, sorry. This is the three moles we are using here. Now, the important thing here is what? You see that I have already told you that the configuration of the last four carbon atom in glucose and fructose is same, this one, right? You see the structure of fructose also, we have this configuration only for the last four carbons, right? So whether you take here glucose or you take here fructose, the compound in osajone formation will be this only. Yes or no, okay? So if you take fructose, then we call it as fructose zone, glucose and glucosazone, but actually both are same only. Glucosazone and fructose zone both are same only. Why? Because the configuration of the last four carbon atom, it is same in glucose and fructose, correct? So that note you write down here, osajone formation in glucose, osajone formation in glucose and fructose is same. In case of fructose, we call it as fructose zone, fructose zone, okay? So this is the few reactions we have which is important. Understood? Next we'll see disaccharides. These are monosecarides. Disaccharides are what? On hydrolysis which gives two monosecarides, right? Or which contains two unit of monosecarides, disaccharides. Write down the definition here first. These are the compounds, these are the compounds which on hydrolysis gives two monosecarides, which on hydrolysis gives two monosecarides. Example you see, if you do the hydrolysis of sucrose, okay, acidic hydrolysis, that's 3O plus. So you'll get glucose here and fructose. Glucose and fructose. If you take lactose, acidic hydrolysis, you'll get glucose again and galactose. If you take maltose, again acidic hydrolysis, you'll get glucose and glucose, right? Two units of glucose. Now how to memorize this you see? If you write down, see glucose is common in all these, right? If you write down sucrose, then other one will be fructose. If you write down lactose, then other one will be galactose. If you write down maltose, then we have only glucose present. Understood? Like this you can memorize. One note you write down into this. In disaccharides, monosecarides are joined by monosaccharides are joined by glycosidic linkage or glycosidic linkage, whatever you say. Glycosidic linkage. Now the first thing here you write down, first example we'll see that is sucrose right now on the heading. It is formed by the condensation of condensation of alpha D glucoparanose formed by the condensation of alpha D glucoparanose and beta D fructofuranose, alpha D glucoparanose and beta D fructofuranose. Now you see how it forms. First of all we have alpha D glucoparanose. So we can draw the structure of alpha D glucoparanose, right? Alpha D glucoparanose, what should we write here on the bottom? Here you tell me, what should we write here on the bottom? OH. Alpha means OH on the right hand side on the first carbon and whatever on the right hand side that should be on the bottom, right? And the second carbon OH again on the right hand side we have. So here we write OH here, then alternate OH here, OH here and the last carbon we have CH2 OH. You must remember at second carbon OH is always down, right? And whenever you have to write down alpha then this two carbon will be at cis position means OH will be at the cis position, OH will be down over here, right? So in all other position we have what? Hydrogen present. So this is alpha D glucoparanose, okay? Now when you write down beta D fructofuranose, so beta D fructofuranose is this. Here we have OH on the top, CH2 OH on the bottom right? So first carbon, second carbon, third carbon OH will have on the top, H on the bottom, OH on the bottom, H on the top and then on the last carbon we have CH2 OH on the top and H on the bottom. Is it clear? Now when condensation takes place then what happens? This OH and H combines together and H2O molecule goes out minus H2O, okay? Minus H2O. Now here you see the structure that you get here will be this oxygen and here we have and these two are connected by the oxygen like this. It should be like this, disconnected by this oxygen and this hydrogen will be as it is and here CH2 OH will be as it is and you can draw the other atoms, molecules, whatever it is on this carbon. Everything will be saved into this, right? So this linkage that you have, this linkage is glucosidic linkage. This bond is glycosidic bond or linkage, whatever you say, okay? Now here you see, you write down some properties of this. The first and which is important which has been asked in the exam also. It is a non-reducing sugar, non-reducing sugar. One test you must have written came on test. In that they have asked one of this question. They have given the structure like this. They have given the structure, okay, this structure and they asked whether it is reducing or non-reducing sugar, okay? So this one is non-reducing, okay? And then you write down whatever non-reducing sugar we have does not show muta rotation, does not show muta rotation. Just a second. Now you tell me one thing in this condensation which all carbons are involved. Like for this one, C1 is involved, right? For glucose, C1 is involved, C1. And for this, what is involved? For fructose, 1 and 2, C2, right? So this you have to memorize, C1 and C2 are now coming back to this. This C1 carbon, you see, it is the actually hemiacetyl structure we have, right? OH, OR and OH. And if the anomeric carbon possibly, this carbon could be the anomeric carbon. This is actually the anomeric carbon we have, you see? The structure, cyclic is this anomeric carbon, right? Here the anomeric carbon is this. Means in fructose, C2 carbon is the anomeric carbon and in glucose, C1 is the anomeric carbon, right? And sucrose forms by the condensation of glucopyrinose and fructopyrinose in which C1 and C2 carbons are involved respectively, correct? This is the point again you have to memorize. And C1 and C2 carbon is the anomeric carbon in both the compounds. So whenever anomeric carbon is involved in condensation, then the product will always be non-reducing sugar, right? Point here is what? That if anomeric carbon does not involve in reaction, then reduction takes place at anomeric carbon only, right? So that reaction is possible whenever anomeric carbon is free, whenever anomeric carbon is not involved in reaction, okay? So here you see anomeric carbon of both compounds, see? If one of these anomeric carbon is free or does not involve in reaction, then also the product will be reducing sugar. If one anomeric carbon is free, does not involve in reaction, then the product is reducing sugar, understood? And I'll discuss some more example into this, then you will see how it is reducing and non-reducing, okay? So out of this 3-1, 3 that I've told you so gross lactose and maltose, only so gross is a non-reducing sugar, understood? And if you see the previous test papers last, I don't remember actually, but second, last or previously, like the test you have written from today, if you go through the third or fourth Sunday back, okay? There you have one question and in that question, they have given this structure like this, okay? And they were asking that whether this sugar is reducing or non-reducing, very simple question. So only thing you have to see what, whether this anomeric carbon is involved in condensation or not. If it is involved, non-reducing, if one of these is present or does not involve in condensation, then the product will be reducing sugar, clear now? Okay, so this is the, like the glucosidic linkage we have. So the point you have to like the, what kind of question they ask? First of all, they ask whether they are reducing or non-reducing, shows meta rotation or not? Non-reducing never shows meta rotation, okay? And this glucosidic linkage is present between which carbon? That is also one type of question they ask. So C1 of glucopyrinose and C2 of, C2 of what? Fractofyrinose, okay? So this is what you have to, these three things you have to memorize, okay? In this, you write down, just a second, just a second, somebody. Yeah, okay. So what are we discussing? Okay, next write down inversion of cane sugar. Okay, see actually, I'll just go to the next slide, previous slide. You see in this one, glucose and fructose, this is the anomeric carbon, right? And this is the anomeric carbon, yes or no? Why it is anomeric? Because this is OH and OR attached, right? This is anomeric carbon, correct? So whenever the anomeric carbon is involved in condensation, right? Then it is non-reducing, first thing is this. So here you see in this one, this carbon and this carbon is involved in the reaction, right? So that's why the product that you get here is non-reducing. Now, if any of this molecule, in any of this molecule, if this or this carbon is not involved in reaction, we'll see the other example also, you'll understand. Means if any of this anomeric carbon is free or does not involve in reaction, then the product will be reducing sugar. Because when the reduction takes place, reducing sugar as those sugar which can reduce tollants and failing solution, right? Reducing sugar as those sugars which can reduce tollants and failing solution, tollants and failing solution. So when this molecule or this carbon is free, this anomeric carbon, then only it can reduce the tollants, reagent and failing solution test or itself get oxidized. If this is not free, then that reaction is not possible. So now just to keep one thing in mind, whenever this anomeric carbon is involved, it is non-reducing. Another example, you will understand why it is reducing, understood? So now you write down next inversion of cane sugar. Inversion of cane sugar, write down sucrose solution in this only. Sucrose means cane sugar, cane sugar is nothing but sucrose. So write down, sucrose solution is dextrorotatory in nature. Sucrose solution is dextrorotatory in nature. And on hydrolysis, see this is something like muta rotation only, okay? The specific rotation changes on hydrolysis, okay? The sucrose solution is dextrorotatory in nature. But on hydrolysis, it becomes liverotatory and it converts into D plus glucose and D minus fructose. And it converts into D plus glucose and D minus fructose. Next line, the specific rotation changes from plus 66.5 degree to minus 19.85 degree, right? So you see the change in specific rotation from positive value to negative value, right? So this change in sign here, write down after this. Change in sign indicates that inversion takes place. Inversion takes place, okay? That's the meaning. Next line, write down. That's the process of hydrolysis of sucrose. That's the process of hydrolysis of sucrose is termed as inversion of sugar, is termed as inversion of sugar. And the hydrolysis mixture having equimolar quantities of D plus glucose and D minus fructose is called invert sugar. This hydrolysis takes place in presence of any mineral acid or enzyme. In presence of any mineral acid or enzyme, which we call as invertase. Invertase, I-N-V-E-R-T-A-S-E, right? So the product we get here, that is D plus glucose and D minus fructose are invert sugar. And the inversion takes place in presence of mineral acid and enzymes. These mineral acid and enzymes are known as invertase. So this is completely theoretical. You have to just memorize this, okay? Now, next slide down, that is maltose. Second one is maltose. Write down to this. It is formed by the condensation of, it is formed by the condensation of alpha D-glucoparanose. Formed by the condensation of, I don't, two molecules of alpha D-glucoparanose. Formed by the condensation of two molecules of alpha D-glucoparanose. In which C1 and C4 carbon are involved. C1 and C4 carbon are involved. Now you see and you will understand now why it is invert reducing sugar. So two unit of glucoparanose. So draw the structure of glucoparanose here, which will be like this. And alpha D-glucoparanose is nothing but this H OH on the bottom. And then at second position, OH is on the bottom. OH on the top here. OH on the bottom and CH2 OH here on the top. You must memorize this structure because in board you have to draw this structure. Okay, so you must take care of one thing in glucoparanose at second carbon OH on the bottom and then alternate. Alpha you have to write down first carbon OH on the bottom, beta first carbon OH on the top. So this is alpha D-glucoparanose plus two molecules of alpha D-glucoparanose. So in the same structure we have to draw OH, H, OH on the bottom, H on the top and then alternate. Now here you see which all carbon are involved into this. C1 and C4 right. So first carbon is this and here the first carbon is this, fourth carbon is this. Correct? Two, three, four, five and six. One, two, three, four, five and six. You see C1 and C4 are involved. So obviously the glycosidic linkage forms between what? C1 and C4 carbon only. So this position you also have to memorize because they ask sometimes this question, glycosidic linkage present between which carbon atom, theoretical question. Options you will have C1, C4, C1, C3 like that. They'll print the option. So this also you have to keep in mind. Now this H and this OH will condense, right, H2O molecules goes out and we get the product here which is nothing but this OH on the bottom. What is this bond? What is this bond? This is glycosidic linkage. And glycosidic linkage in maltose. It is present between C1 and C4 carbon atom. This is one point you have to memorize. Correct? Now is this reducing or non-reducing sugar? Reducing. Why it is reducing? Because you see this anomeric carbon does not involve in reaction. So when you put a tolerance reagent and failing solution, this carbon will react. Okay, we'll take part in the reaction. You see in the first example, the anomeric carbon was involved in reaction. Now this anomeric carbon, one of these anomeric carbon is not involved. So the sugar is what? Sugar is reducing sugar. Understood? And when it is reducing sugar, it shows major rotation. Fine? Okay, so this is what you have to memorize. Now you understood Ramchiran how it is reducing sugar. Now the second, third one you write down that is lactose. Third one is lactose. How this? What is the mono saccharides of lactose? Glucose and galactose. Glucose is common in all. When you write on lactose, lact means galactose there will be. So it is obtained by the condensation of one molecule of... obtained by the condensation of one molecule of... beta D galactoparanose, beta D galactoparanose and one molecule of... beta D glucoparanose, beta D galactoparanose and one molecule of... beta D glucoparanose in which C1 and C4 carbon atom are involved respectively. C1 and C4 carbon atoms are involved respectively. Understood? Now with this information, C1 and C4, is this reducing or non-reducing? Reducing, right? So reducing sugar. So only... Now you see the question was so easy. They have given you the structure of glucose also, like whatever molecule it is. They have given the structure with glycosidic linkage. And then they were asking that whether it is reducing or non-reducing. Like this question was. Exact thing I don't remember, but yes, this was the question. Anyways, so in this you see the two monomer we have. Sorry, mono saccharides we have is beta D galactoparanose. Can you draw the structure of beta D galactoparanose? Okay. You help me out because I cannot see whether you can... we are drawing or not. So better you help me out. Galactoparanose it is, right? So galactoparanose, we know galactose and glucose are what? What epimers? What position? Second carbon or fourth carbon? Second, yeah, yeah, correct. Okay. So I am drawing here beta D galactoparanose. Since it is beta, so OH will be on the top, right? Beta, first carbon, OH on the top. Alpha, first carbon, OH on the bottom. So this is the first carbon. At second carbon, OH on the bottom. Okay, OH on the bottom. And third carbon, just a second, whether it is C2 or I think it is C4 epimers. Glucose and galactose. Yeah, yeah, yeah, sorry. I said mannose, no? Just a second. Mannose, I have given you C4, I have given you. It is not C4, it is C2 epimers. I made a mistake. I just corrected. That is what I was thinking. Glucose and mannose are C2 epimers. Last class you check. Glucose and mannose are C2 epimers, right? And what about glucose and galactose? You cross check once. C4, that is what I am saying. No. Last class you check. What is the relation of glucose and galactose? You tell me that. C4, that is what I am saying. And this mannose are C2 epimers. So what I have given you today? You just check once. No, no, no, just to correct it. By mistake I have given you that. Mannose are C4 epimers, C2 epimers. So just make correction over there. Here is the thing. It is not C4, it is C2 epimers. So just make some correction on this structure also. Here we have OH and here we have H. Here we have H and here we have OH. This correction you make. So glucose and mannose are C2 epimers. Now. So I am drawing beta D galactose. So OH on the top, beta. Second carbon, OH on the bottom. Third carbon, OH on the top, H on the bottom. So ideally this OH should be on the bottom if it is glucose. This is the fourth carbon configuration is different. So OH will be on the top, H will be on the bottom. And here it will be same CS2 OH on the top and H on the bottom. So it is beta D galactoparanose. Correct? Galactoparanose. Now another molecule is what? It is alpha D glucoparanose. Alpha D glucoparanose. That structure will be like this. Oxygen alpha D glucose. OH on the bottom, H on the top. On the bottom, H on the top. OH on the top, H on the bottom. OH on the bottom, H on the top. And CS2 OH here. Right? Now what all carbons are involved into this? C1 and C4. So this is C1 carbon. It is first and this is fourth carbon. The structure that you get here when H2O molecule goes out is this. And here it is attached with oxygen. So this linkage is what? It is glycocytic linkage. Again you see this carbon is available here. So at this carbon reaction may take place. So this is reducing sugar. Shows mutarotation. Done? So in all these three examples you have to keep remember that at what position glycocytic linkage we have whether it is reducing or non-reducing. Right? Now next one is polysaccharides. Polysaccharides. These are the polymers of monosecarides. These are the polymers of monosecarides. Example you write down starch, cellulose. Starch, cellulose, etc. In this there is nothing to understand. You have to memorize certain points. And this is not important also for exam point of view. Starch you write down few information about starch you should know. Starch you write down. It is a white, just in short you write down. White amorphous substance. White amorphous substance. Present in wheat, rice, potatoes, etc. Starch when you heat this at 200 to 250 degrees Celsius you will get another compound called dextrin. Dextrin and write down next on hydrolysis by enzyme on hydrolysis by enzyme. One kind of enzyme is amylase which is present in saliva. Have you heard this amylase? Amylase present in saliva? This is the kind of enzyme we have. On hydrolysis by enzyme it eventually gives glucose. And that's why you see these potatoes and all. When you eat you will feel a little bit of sweet test is there. That is why because the starch present in potatoes when it reacts with amylase which is present in the saliva converts into glucose and that's why we get sweet test. This is one information. Another one you write down. Starch solution gives blue color with a drop of what? Iodine. This you must have done in titration also. Iodometric titration. Starch solution gives blue color with a drop of iodine. Next line it has two components. It has two components. First we write down amylose. First component of starch amylose. It is a linear polymer. In this there is no branch. It is only linear polymer soluble in H2O. Few properties I am just giving you. You have to memorize this. But for J point of view it is not important. It is composed of alpha D glucose unit. Composed of alpha D glucose unit by glycosidic linkage present between C1 and C4 carbon atom glycosidic linkage. That's why you see in C1 and C4 if glycosidic linkage forms you will get a linear polymer like this. Like this and then this and then this. C1, C4, C1, C4 like this it goes. Understood? That's why it is linear polymer. Fine. So last point you write down. This question they asked in NEET exam. Number of D glucose unit present in this. Number of D glucose unit present in this can be anything between 60 to 300. 60 to 300 glucose unit are present in amylase. Second component of this is amylopectin. Amylopectin. Write down highly branched polymer properties of this. It is a highly branched polymer polymer. In this also alpha D glucose units are present and C1, C4, carbon forms glycosidic linkage in soluble in water and number of units present D glucose units present anything between 300 to 600 number of D glucose unit. Next write down protein amino acids and polypeptides. See in this one heading you write down proteins amino acids polypeptides write down this proteins are made up of polypeptides proteins are made up of polypeptides and polypeptides are made up of amino acids. So first of all we'll discuss about amino acids few structures into this. Then we'll see what is polypeptides and then we'll see how protein forms. I'll just give you a brief explanation of protein formation. It is not that important but NCRT it is given in so detail. So there are four different types of structure of this proteins we have four different stage basically. So all these states are given primary structure and secondary like that. It's given in NCRT very clearly so I will discuss those things but you have to read throughout in NCRT today only so that you will connect you will be able to connect those things. Okay but again it is not that much important for Je point of view. Only one thing is important in this some structures of amino acid and then juterium that we'll discuss now. Okay so amino acids are what these are actually bifunctional organic molecule. Bifunctional means what what do you understand by bifunctional? Two functional group right bifunctional organic molecule as the name suggests amino acids we have amine also and acid also correct. So bifunctional means we have amine present amino group NH2 and we have acids also that is COOH and this is amino. Now if you see the property of these two group this is acidic and this is basic correct. So if you write down some examples here which is CH3, CH2, CH2, CH2, COOH just write down CH here and here we have NH2. Okay so you see this is alpha position and this is beta position so we call it as what beta amino acid general name alpha and beta so it is beta amino acid. Generally the protein that we have is made up of alpha amino acids generally so we'll discuss mainly alpha amino acids only right so write down here proteins are made up of are made up of alpha amino acid and when I say alpha amino acid it means the NH2 group present at what alpha position right. Now you see the general structure of protein general structure of protein in NCRT there are around 20 structures of proteins are given those structure you must go through I will also discuss some structure now okay but you must go through those structures. Now the general structure of amino acid is what we have one chiral carbon with one group attached over here G general structure I am writing down and then one acidic group we have one amino group with this hydrogen present here hydrogen present here right so this is all these naturally occurring amino acids are generally alpha amino acids so this is also alpha amino acid you see this alpha position and this two group is attached clear now the name of this compound if this G you substitute as hydrogen right then the name of the compound will be what it is glycine it is given in NCRT also you must go through if G is CH3 alanine the name is alanine so these names you have to memorize you never know what they'll ask if G is this group CH2 with one phenyl group attached with it then this is phenyl alanine one hydrogen is replaced by what phenyl group so this phenyl alanine we'll discuss some more structure but before that you see all these amino acids are optically active except this glycine that's why this glycine is one important example we have you see this glycine when you substitute hydrogen instead of that G over there right so this carbon is not chiral anymore when you substitute hydrogen here fine yes so if this carbon is not chiral so there is obviously a plane of symmetry we have the compound will be what optically inactive yes so in NCRT those examples are given out of those examples only glycine is only optically inactive so you must keep this in mind fine when CS3 you substitute here then this carbon will be what carol carbon there is no plane of symmetry compound will be optically active right now the classification of amino acid you see I will go to the next one all of you have written this classification you see of amino acid amino acid are classified into three categories one is acidic other one is basic and the last one is neutral see this acidic and basic nature we define according to the number of acidic and basic group present if number of acidic group is more than then it is said to be acidic amino acid correct so in this acidic amino acid number of acidic group is more acidic group means what COOH more means more than the basic group group is more than the basic group basic group is what NH2 so for example you see see COOH here we have CH2 COOH here you write down NH2 and then one hydrogen the name of this compound is aspartic acid aspartic acid you see there is there are two COOH group one NH2 right again you see another structure carbon if it is attached with CH2 CH2 COOH here and then all other things are same COOH NH2 and H here this structure is glutamic acid again you see two COOH group and one NH2 group number of basic group is what number of NH3 so in this basic thing is what number of basic group is more than acidic group example you see CCOOH NH2 this is the general formula we have H here and when you substitute this CH2 for COOH NH2 CH2 for NH2 this compound is lysine L Y S I N E lysine so this is what this is the basic amino acid we have neutral one is what neutral one when we have equal number of equal number of acidic and basic group basic group see one thing you must be very clear we have done this classification on the basis of number I'm not telling you that this compound is neutral in nature or acidic in nature or basic in nature right however it is true when you say for acidic and basic thing but for neutral if the number of COOH and NH2 group are equal then also that it is not necessary that the molecule will be neutral in nature okay and that is how we get the concept of zwitterion we'll discuss that okay some more structure you see if I write down this some more structure you just write down I'm just giving you some general example okay so carbon COOH NH2 and when you write here CH2 OH instead of G this compound is serine S E R I N E serine when you write down SH here another one you write down instead of this OH you write down SH that compound is sistine C COOH NH2 H and here we have CH2 SH this is C Y S T I N E Sistine another structure is this which is which has the different pattern actually from the other one CH2 CH2 then we have CH2 N CH2 H this is proline different structure it has you must remember it actually does not does not follow the general formula we have one note you write down into this each of these amino acids each of these amino acids can exist into stereo isomeric form can exist into stereo isomeric form that is D and L capital D capital L however however all naturally occurring amino acids however all naturally occurring amino acids belong to L configuration okay even protein also consists L amino acid only okay but from all these we just go through the structures given in NCRT few common structure I have done but there are some different also maybe in NCRT just you go through okay now next you see is jitter and this is important for J point of view see the general formula we have here which is for amino acid is this only COOH G NH2 see today we'll finish this I'll take there are a few things left so I think I have to I will finish this by 730 only so I will not give you any other question from any that will not do the revision part today okay next class will do complete revision did you solve those any questions solving right last day I give you I gave you a chemical equilibrium question also okay so that also you solve okay so when you have this amino acid you put into water H2O then what happens that this compound in water exist into this form which is nothing but because this acid group that you have COOH that will lose its H plus iron CO minus H N plus H plus right now since this is the basic nature we have it has tendency to select what H plus right so there are possibilities that this H plus will taken up by NH2 and it forms NH3 plus H will be as it is I see this iron is neutral in nature yes or no neutral and dipolar also to charge point we have so dipolar and neutral in nature this iron is nothing but Jviteran so Jviteran is what it is a neutral iron right it is a neutral basically molecule dipolar in nature okay so structure of amino acid whether it exist in anionic nature or cationic nature or anionic form or cationic form it depends on the nature of the solution in which it is kept or the pH of the solution in which it is kept right so first meaning of this is what first of all you write on this point structure of amino acid depends upon the structure of the amino acid depends upon the nature of solution in bracket you write down acidic or basic the nature of the solution in which it is kept and pH of the solution also nature of pH both are related only right now what happens you see if this iron that is Jviteran is kept into basic solution so basic solution means what will have OH minus concentration there and pH will be what high correct basic solution will have high pH comparatively so now this OH minus will take this H plus ion correct and forms H2O and the molecule we get here get here is H2N C COO minus G will be as it is and H here now you see this neutral nature is not now possible in the solution correct and this is not a Jviteran because Jviteran is always neutral OH minus takes this H plus and will have negative charge over here this we call it as anionic form of Jviteran because we have one negative charge into this fine now when you put this into acidic solution acidic solution means what we have H plus concentration pH will be less low so that H plus will attach over here and will get what H3N plus C that group COOH and here we have hydrogen this we call it as what cationic form right so depending on the nature of the solution in which this Jviteran is kept it may exist into its cationic form or anionic form clear now on the basis of this there is a concept of isoelectric point so next heading you put on isoelectric point right or into this right down when we connect the electrode I'll just draw diagram first write down when we connect the electrode with an external voltage source then the cationic ion move towards the negative rod cationic ion move towards the negative rod and anionic ion moves towards the positive rod next line but at certain value of pH but at certain value of pH when the amino acid exist in but at certain value of pH amino acid exist in Jviteran exist in the form of Jviteran in the form of Jviteran then there is no movement towards any electrode right then what did you write last line does not move towards any electrode this point is known as isoelectric point right the point is what suppose we have a solution in this container right and this solution will have certain certain pH value and we have two electrodes into it and we have solution here in this solution suppose you keep amino acid into it now depending on the nature of the solution or pH value this amino acid amino acid either amino acid exist in cationic form or neutral or anionic form correct depending on the pH of the solution if it is in positive form this is connected towards suppose the positive plate and this connected towards the negative plate so when you apply electricity towards it then depending on the pH of the solution if it is if it is if it exist in positive form cationic form then it starts moving towards what negative plate and if it is in an anionic form then start moving towards the positive plate and will have some pH value here where it exists in neutral form and neutral form is nothing but what? So when it exists in the form of zwitterion then there will be no movement towards any plate or electrode correct that pH value is known as the isoelectric point clear and these values isoelectric point we don't have to calculate that is not required now suppose if I take some example if I write down C-C-O-O-H-N-H2-C-H2-C-H2-C-O-O-H what is the name of this compound this is glutamic acid acidic groups are more isoelectric point for this compound is found to be 3.22 right acidic in nature okay if you write down for lysine lysine structure I have already given you there we have C-H-2-H-O-H-N-H2 here right C-H-2-H-O-H-N-H2 here for lysine isoelectric point is it is basic in nature this one is acidic in nature isoelectric point for lysine is 9.87 which is actually true acidic so pH value should be less than 7 should be greater than 7 here right but in case of L-N-9 L-N-9 is what acidic or basic L-N-9 you tell me the structure of L-N-9 C-H-3 G is C-H-3 so we have one N-H-2 and one C-O-O-H Ramchana are you there Ramchana and Kondanya okay so when we have L-N-9 present over there so we have one N-H-2 and one C-O-O-H-G group so it should be actually what neutral so its isoelectric point should be what 7 but this isoelectric point for L-N-9 is found to be it is slightly acidic 6.02 because C-O-O-H is comparatively stronger base stronger acid than N-H-2 not acid means the basic nature of N-H-2 is less than the acidic nature of C-O-O-H means what it has tendency to lose H-plus correct and if this N-H-2 has equal tendency of the acceptance of H-plus then only it will be neutral understood my point C-O-O-H will lose H-plus very frequently but with the same frequency N-H-2 cannot accept that H-plus because it is not that much basic that's why the solution will be little bit acidic but why we are calling it as neutral that classification we have done according to the number of what number of acidic and basic group it has nothing to do with the acidic and basic nature of the compound however in case of acidic and basic amino acid this is found to be correct like see like I seen it is basic 9.87 acidic it is 3.22 but for neutral also it will be slightly acidic only so right down one point here because of a strong acidic group because of a strong acidic group neutral amino acid will not have pH value 7 so this is the discussion about amino acids now we will see the two different amino acid we have which is essential and non-essential just definition we have to go through just to write down next essential and non-essential amino acid write down essential amino acid or those acid which cannot be prepared by body ok see there are few amino acid which is prepared by within the body itself ok inside the body actually when those amino acid which can be prepared by the body itself within the body that we call it as non-essential so essential you write down first it cannot be prepared example you write down lysine phenylalanine anine phenylalanine non-essential prepared by the body itself by the body itself example you write down glycine what is G in glycine then alanine what is G in glycine what hydrogen yes next write down formation of protein or protein formation write down into this it is found by the joining is found by joining the carboxyl group it is formed by joining the carboxyl group of one amino acid one amino acid to the alpha amino group to the alpha amino group of another amino acid to the alpha amino group of another amino acid ok now you see here suppose the one amino acid we have which is this H2N see here we have some group suppose G1 and here we have carboxylic acid C double bond OOH hydrogen this when joins with the another amino acid NH2 C G2 C double bond OOH hydrogen right so in this what happens when these two combines ok again condensation takes place so hydrogen here and this OH combines H2O comes out H2O comes out and the product we get here is this H2 N CG1 H C double bond OOH NH CG2 C double bond OOH OH OH H right this is the product we get now this bond C double bond O and N this bond this bond is the peptide bond it is peptide bond or we also call it as peptide linkage when there are two peptide bonds now suppose we have another molecule here right NH2 CG3 C double bond OOH H so here also OOH and H comes out and we will have another peptide linkage here right then we have another peptide linkage here like this we will have a large linear molecule we will get means the bonding will be linear understood this is what this is one amino acid and this is the another amino acid this part this part is the another amino acid similarly we can have A3 A4 A5 A100 like this we can have a large chain right with poly with this peptide linkage or peptide bond right so when we have more than or equal to two peptide bond present then we call it as polypeptide okay and specifically if you write if you have two peptide bond then we call it as dipeptide three peptide bond then tripeptide four tetrapeptide pentapeptide like this but all these comes under what polypeptide linkage right so protein forms by this polypeptide linkage only okay now this peptides you see it has one classification also that I will write down in the next slide you just keep this in mind like this we have N number of amino acid linked by the peptide bond right we have a long chain understood this okay now I will come back to this again but first you see this classification of peptides classification of peptides the first one we have is oligopeptides and the second one we have polypeptides polypeptides and the last one we have proteins this classification is depending on the number of amino acid present this classification depends upon number of amino acid present and so in oligopeptides we have maximum nine amino acid possible two to nine amino acids right polypeptides from 10 to 100 amino acids 10 to 100 amino acids are present then it is polypeptides if we have protein then more than 100 amino acids should be present so proteins are very complex compound branched and complex compound the bonding is so complex over there okay now like I said in the like the structure of protein that we were discussing previously structure of protein you will get a long chain compound here by peptide linkage through peptide linkage and that will be like this H2N CG1 H C double bond O NH CG2 C double bond O NH CG3 CG3 C double bond O H and like this it goes so like this we will get a long chain compound right so this first thing here this is what this is first amino acid A1 this is what A2 this is A3 this is A3 right and all these amino acids are bonded together by what what is this by peptide linkage right peptide bond or peptide linkage now actually what happens since you have we are talking about proteins right we are talking about proteins so we have more than 100 amino acid present like this and since it is a long chain where we have 100 amino acid so it cannot be completely straight right since it is a long chain if long rope also if you take it is very difficult to keep that rope straight right but here it is what you see this oxygen carbon some group we have here nitrogen so obviously we have some electronegativity difference also here right so first point is what the structure of proteins is can be explained in four different stages right this long chain compound when you have only a linear chain there is no branch into it right A1 A2 A3 A4 A5 A100 110 200 like that this linear chain that you have we call it as primary structure of protein correct when there is no branch primary structure now since it is a long chain so it won't be completely straight secondly there is some electronegativity difference also here so we will have some attraction in these molecules these atoms present over here right so because of this attraction what happens this structure will be bent in the form in spiral form or helical form that you have a spring like structure you get so that is structure let me draw like I said that it won't be like the structure of the primary structure won't be in the straight chain but it will be in the form of spring right and that structure will be like this is not that good but it will be bent at some point because of the attraction due to the electronegativity difference also understood here we have if you see there we have NCC then again NCCN then again CC NCCN like this we have you see the previous one we have NCC NCCNCC like this it is going you see the previous structure A1, A2, A3 that I have written okay NCC like this we have so due to some electronegativity difference some molecules some atoms will be close enough there will be an interaction between them and then they won't be available or there won't be a straight chain compound but there will be some attraction because of that attraction it exists in helical structure or spring like structure okay and this we also call it as alpha helix okay I am not giving you the theory part into this I am just explaining you it is written in NCIT very clearly so today only you go through okay so when you talk about this helical structure this comes under the secondary structure of protein primary one what the straight chain and it is the secondary structure clear right now since again the secondary structure we have helix is also very long right it is not small very long chain more than 100 100 we are just taking a reference but it will be in lax or thousands of thousands of amino acid present in protein right so because of this you see this structure will not be also very straight it is also bent at some point of time so that structure you see that the tertiary structure that we say is something like this this helical structure will be like this but at some point it will be bent like this like this it is there some random structure will be there like this so this structure we call it as bent structure again you see we will have some electronegativity difference here also so there will have some interaction suppose this point at this point some interaction we have so this will be bent like this okay and anything it is a very long chain so again it is very difficult to be straight completely straight so it will be bent at some point of time some point like this randomly right so this bent structure when you explain this structure it comes under tertiary structure tertiary structure it comes under tertiary structure right various kind of bonds are involved into this forces are involved into this that can be intramolecular hydrogen bonding okay salt ionic interaction maybe van der Waals interaction maybe that is there are many structures possible many kind of intermolecular attraction possible here right so completely straight one is primary helical is secondary and this bent structure is tertiary right now this kind of bent structure you see first of all we have I will just draw this here suppose this is the bent structure we have and this gives you one protein like this okay sorry one protein like this so in this you see this is a one protein we have like this again because of electronegative difference we will have some charge separation on to this point that carbon, oxygen and nitrogen that we have here so we have suppose this is P1 first type of protein like this we have another structure also another bent structure also so those negative part of that bent structure will attach with positive part of this and vice-versa also so this is surrounded by another bent structure of another protein like this or another bent structure of another protein like this or another bent structure of another protein like this like this we can have any random structure possible right know it is not that good okay but it is random only right like this anything I can even draw so it's a random structure so you see this is p1 this is suppose p2 this is suppose p3 this is p4 like this one protein structure is surrounded by many other different proteins right because of some electronegative distance and positive negative charge center we'll have here and then we'll have some interaction so this is structure this is structure we'd explain under quaternary structure of protein quaternary structure of protein right so this is nothing but the protein the actual protein is this this quaternary is a very large complex random structure we have that is protein right I've given you the brief discussion of this okay in NCRT you don't forget to read it today itself okay in NCRT you see there is a structure of protein is given primary secondary tertiary and quaternary so if you go through it you will understand how this protein forms okay but again for J point of view not important but also I haven't seen the vast structure of protein but yes just go through once you can write down in your own words okay see the body temperature generally it is 37 degree Celsius right human body temperature is 37 degree Celsius when what happens when the temperature changes if suppose 37th 39 case of fever the body temperature increases right and when temperature increases the structure of protein changes because and that is why we'll have some problem then right structure of protein changes and then we'll get some problem and for that to cure that we need some medicine okay so when the temperature increases the structure of protein also changes okay this we call it as denaturation of protein okay with temperature when the structure changes a structure of protein changes we call it as denaturation of protein so this term you must keep in mind denaturation of protein right so this is it in this chapter lipids and all that is not important okay you don't need to study all those okay so for today you just go through the at least the structure of proteins and the various amino acid given in NCRT that is more than enough okay one in one note you write down here one note you write down insulin contains disulfide bond insulin contains disulfide bond consists of two consists of two polypeptide chain only thing you have to keep in mind because once in neat exam they have asked that which of these molecule contains disulfide bond so insulin contains disulfide bond that is only one thing you must remember so this is it for this chapter we are done with it okay you revise those through NCRT just go through and previous year question you see don't put much time into this chapter it is not that much important okay try to go through the previous year questions and then those things you must keep in mind because there are many theory into this anyway so one last thing we'll discuss continue ask me one question that how to convert benzoic acid into benzyldehyde right okay just a second I'll show you one thing see this one can you see this see benzoic acid to benzyldehyde conversion if you use this li alh4 you'll get CS2 OH then PCC converts into this your question was we have to convert benzoic acid into benzyldehyde without converting directly is not possible because if you remember I have discussed this that this carbonyl group you have to preserve when you directly use some reducing agent here it will convert into benzene directly because this CO carbonyl group is not preserved to preserve this first of all you can convert this into chloride also like you see this example PCL 5 you can also use SOCl2 here I have done this you just go through your notes when you convert this into chloride then this CO is preserved now after that you can use this Rosenman reaction PDBSO4 right then you'll get aldehyde here if you ask me direct conversion not possible if you use reducing agent here this will gives you benzene right so that is that we cannot do first of all we have to preserve this carbonyl carbon and for that we can convert this into chloride with the use of with the use of that SOCl2 PCL5 and all we can convert this to COCl and then we can reduce this with Rosenman reaction then you'll get benzyldehyde into this understood okay so you have to form that intermediate chloride compound SOCl2 with SOCl2 and PCL5 fine anything else what are you revising nowadays inorganic how do you revising your how are you revising inorganic so one in one day how what what one chapter do you finish fine that's fine NCRT you're doing right you just go through with NCRT for inorganic NCRT is enough little bit more information if you want you can go through some other book but just you roam around that previous year question only don't look at all those theory okay that is not helpful okay and what about 12th portion do you have any doubt in the physical portion of 12th and do you solve previous year question all of you are asking so 12th portion physical physical chemistry 12th portion do you have any doubt or do you have solved previous year question okay organic you're doing okay fine so next class I'll do some concepts of GeoC okay general organic chemistry I'll discuss with some questions basically we'll see mainly questions into that how to you know compare the stability of RS and various of the compounds right so there are seven rules we have for comparison of acidity basicity comparison we'll see okay so GeoC little bit concept we'll discuss okay 11th portion do you need any chapter revision for 11th portion what okay see do one thing more concept okay more concept and more concept to which contains the question related to equivalent normality and all equivalent mass number of equivalence okay and the electrochemistry chapter that you have there we have questions on Faraday's law so that Faraday's law also involves the concept of equivalence equivalent mass and all if you remember q q is equals to m is equals to z it will write okay z is electrochemical equivalent all those okay so when you solve those Faraday's law so that will have required the concept of equivalent mass and more concept also so once you do this more concept or equivalent mass question you try to finish that that portion also of electrochemistry Faraday's law of electrolysis okay so generally in Jay they ask questions they mix the concept of one or two chapters like in ionic equilibrium you see solubility product and then in electrochemistry also we have some question related to electrode potential solubility product and all I have done few questions in the class also okay so when you solve this solubility product you also do that electrochemistry problem also into that okay chemical equilibrium is not that tough okay only some basic concepts and leach at layer principle application you just go through okay states of matter a and b compressibility factor okay those things are important some graphical relation velocity relation that various molecular velocity vrms v average and all graph related thing those you go when go through and then we are left with thermodynamics thermodynamics thermochemistry is important again okay then we have done this enthalpy of formation and all in ionic in this chemical okay so next class we'll do that also okay GeoC concept and we'll do discuss our discussion of GeoC give screen what what yeah yeah no problem like GeoC value do we'll discuss questions only not in the form of question like I'll give you few compounds and what is the stability what is the acidic order of that like they will discuss and in physical we can discuss questions and then concept related to the questions no problem in that okay Naman what happened did you send what happened just a second what is the doubt in this h2o s3po4 gives this pk1p k2 is given pH of this of this okay this is the question of polyprotic acid okay I'll discuss this concept okay Naman because last class I haven't done with this because generally we don't get this kind of question in the exam but actually you see in this what happens this h3po4 in the first reaction okay you don't have all this question no anyways let it be I'll discuss this question next class okay I'll give you the concept of how to do the polyprotic acid this is actually the simultaneous equilibrium you have to use into this okay okay let me discuss this next class okay I'll tell you okay okay next class will start GeoC and some concepts of ironing and that will see okay okay thank you thank you