 Next, you write down muta rotation. Write down the spontaneous change. It is the spontaneous change in the specific rotation, specific rotation value of, value of an optically active, an optically active compound is known as, as muta rotation, okay? So what happens here, in this, the alpha D plus glucose, like for example, if you take alpha D plus glucose, it is an optically active compound, will have some specific rotation value, okay? And it's a specific rotation value, if I write down here, a specific rotation value. It is 111 degree, approximately. When you dissolve this in water, across medium, then it converts into D plus glucose, D plus glucose, okay? Now, what is the major difference in these two structures? If I say alpha D plus glucose and D plus glucose, what is the difference there? Can you tell me? The left-hand side refers to the cyclic structure, whereas the right-hand side refers to the, it's an open chain structure. Yeah. Yeah. So that's the basic difference, okay? Difference is what, this one, this one is a cyclic structure, right? Cyclic chemical structure, it is cyclic structure and this one is what? This one is open chain structure, open chain molecule. So what happens here, when you put this into water, you look at the structure of alpha D plus glucose, the hemiaxial structure, what happens in that? Look at this structure. When you put this into water, alpha D plus glucose, then H plus comes over here, right? This bond breaks and H plus goes out, we'll get a double bond here and it converts into this structure. This converts back into this structure, right? So right here, reversible, okay? So in aqueous medium, this has the tendency to convert back into this. And now, again, when this OH attacks, again, we have two possibilities. We can get alpha also, we can get beta also, right? So what happens here, what is actually happening that in aqueous medium, alpha converts into beta, sorry, alpha D plus converts into D plus or beta D plus converts into D plus and further when it attacks, this converts back into beta D plus or alpha D plus. So basically, if I write down here, you see, cyclic to open chain and then open chain to cyclic. That is what the conversion we have. And if this alpha converts into open chain and further attack takes place and this converts, if it converts into beta D plus glucose, which is possible because that C double bond over the carbonyl carbon is planar. So it is possible that from this open compound, we'll get beta and we have started from alpha. The point is this has some value of specific rotation and the specific rotation value of this is plus 19 degree, okay? Should we remember all these? What? Should we remember all these? Should we remember the general value? No, no, no, value is not required. Value is not required. You don't have to memorize the value. The point is this converts into this beta D plus and beta D plus again converts into alpha D plus. So eventually we'll get an equilibrium mixture of it. And we'll get an equilibrium, a specific rotation value. And this value is found to be around 52.5 degree. Conversion of alpha D plus into beta and beta into alpha, this is known as muta rotation. So this is the muta rotation value is this and this we can write, it is a spontaneous change, spontaneous change in the value of specific rotation a specific rotation before achieving equilibrium, okay? Got it? Yes, sir. We'll start with fructose. The similar thing we have here also so we can now go a bit faster actually. Next write down the other molecule which is fructose. Fructose, that structure is this CH2 OH and this is OH, OH, this is the structure of fructose. If you write down this, you see the number of Carrol Cartman here, it has one, two and three. And since OH on the right, we call it as what? D and it is minus actually, T minus fructose. Again, this minus sign will get from polarimeter, the experimental thing, okay? This information we are getting from the polarimeter, okay? So this is the information I'm giving you, D minus fructose, okay? Again, if you put this OH on the left-hand side, it becomes L, okay? Same thing, D and L, how to be assigned, okay? Cyclic hemiacetyl is the structure of fructose. Next, cyclic hemiacetyl structure. So in this, what happens? We have C double bond O, CH2 OH, OH, OH, OH. This is the lone pair we have here. This attacks onto this carbon and this electron pair goes onto this oxygen. So five member drink. What? Five member drink. Five member drink we'll get. So why not a six member drink? Why doesn't the CH2 OH participate? This one? Yes. This oxygen, no? If this attacks, just a second, if this lone pair attacks here, then we get what? We get C double bond H here. Why do we get five member drink here? Because... I don't know what you're talking about. I don't know what's every other carbon is... I don't know what you're talking about. D plus, D minus, okay? Fructose. So this will attack, we get a five member drink. If this carbon attacks, we get this, which is six member drink. Sir, does it have something to do with the fact that the last one's not chiral? It is, I think that carbon won't be, won't have four different atoms then. Which is... Sir, is it? Okay, just give me two minutes. Okay. We'll draw the habit structure of this and then we'll think about it. Okay. Probably they will understand because we need chiral carbon. All the five carbon should be chiral carbon over there. So just give me two minutes and I'll draw the habit structure and then we'll understand. We'll think about it. So just two minutes to go. This comes over here and we get what? OH this side and CS2 OH this side. And it's other way also it's possible. OH on the left, CS2 OH on the right, correct? And here we have H, OH, and then we have OH, H. We get hydrogen here and this carbon attached with, this carbon attached with an oxygen. This side. Is it clear? Like this? Okay. Now again, when this OH on the right, we call it as alpha D minus factors because D minus factors. So this one is alpha D minus factors. And the number of carbon atom, one, two, three, four, five and six. The beta D minus factors is what? When this OH on the left, that I'll draw here quickly. C, OH, CS2 OH, OH, OH, H. H and CS2. This is beta D minus, okay? And these two are what? These two are again in anomers of each other. Yes or no? Can we call it as anomers? Because the configuration differs at this carbon, C2 carbon. This carbon is the anomeric carbon, OH and OR, okay? This carbon is anomeric carbon, anomeric carbon configuration differs. So these two are anomers of each other. Yes, sir. What happens if this, I'm not getting it. This thing. Did you finish this? Yes, sir. Okay. Now, can you draw the Havertz structure of this? The Havertz structure of this, again given by the same scientist and the reference molecule here to draw this structure is q-ran, q-ran. The structure of q-ran is what? It is a five-membered ring with oxygen present in the ring. Double bond here. Two double bonds. Okay. So the structure of fructose if I draw, that would be like this. Oxygen, okay? This is the, see in this, the structure you see here, this structure, both first and sixth carbon is not in the ring, right? First and sixth carbon is not in the ring. Oxygen attached with second and fifth carbon, correct? So with reference to that, this is, suppose the second carbon we have, this is second, this is third, fourth and fifth. And the second carbon, if you see the structure, it has CS2 OH and OH, right? So at the third carbon you see, we can write where is this OH in the third carbon? You see that structure, the cyclic structure, the OH on the third carbon is on the left. So that will write on the top and then alternate, okay? OH on the top, here on the bottom. And then in the fifth carbon, we have CS2 OH and H. OH on the bottom and then CS2 OH on the top and H here. This, this. At the second carbon, if you write OH opposite to this OH, you see here, this is structure, if OH is opposite to this OH, then it is alpha, okay? If OH and OH on the third carbon on the same side, then it is beta. So what do we write here in this way? OH on the second carbon, if it is opposite to this OH, if I write down this OH here and CS2 OH here, this structure represents the alpha D minus fructose. And the other structure, if I draw oxygen here, OH on the top, CS2 OH on the bottom, then OH on the top, H here. And then OH here, H here, CS2 OH. This becomes beta D minus fructose. And the both the structure, since it is drawn with respect to this, we call it as fructofuranose, fructofuranose. Or is that double R or just one half? F U R, not double R. So first carbon, we have here. One, two, three, four, five and six. Number this, one, two, three, four, five and six. First and sixth carbon is not in the ring. See the actual structure you see, three-dimensional structure of this, this bond actually, this fourth and third carbon, is towards the observer. The three-dimensional structure is this. Did you understand the structure? The third and fourth carbon is towards you and this oxygen is away from you. Yes, sir. Right, so actual structure is this, this oxygen. This is the actual structure we have. Okay, one thing I wanted to understand here from you, that, and on this, they have asked question in J exam. OH here and CS2 OH here. This is alpha or beta. Sir, beta. It is beta, right, CS2 OH here and H here. Now, generally we have this structure in mind, alpha and beta is CS2 OH here and H here. What happens if I revert this molecule, means upside down. If I put this molecule exactly opposite way, then what is the structure we get? Sir, wait, how are we flipping it? See, we have this structure, like suppose you have a palm, right? And this, the finger is oxygen and this is the, look at your palm, right? The finger is oxygen and this, the thumb side, we have the two carbon here, right? Third and fourth carbon. Okay. Okay, when you flip it, right? Oh, just seeing the other side of the palm. Yeah, the other side. When you flip it, then what structure you get? Can you draw that structure? For your reference, I'll just give this number to this carbon atom. We have one, two, three, four, fifth and sixth. When you flip it upside down, right? When you flip it, then what is the structure you get? Can you draw this structure? Sir, so you'll get beta. You'll get, you'll get, you'll get alpha. You'll get alpha, how? I'm just thinking about it. Will you get alpha? No, I don't know. Yeah, you might. Yeah, you get alpha because you're flipping it through the plane, so it will change. No, but there is OH and CH2 OH are on the same side. That's on the left side, right? Did I get this? Six will become one. Six will become like one. So when you flip it, do we get the second carbon here? Yes or no? Sir. Yes, sir. Right? So when you flip it, second carbon here, CH2 OH is going down on the second carbon. So here it will be on the up. Yes, sir. The second carbon comes over here, fifth carbon comes over here. So this is the fifth carbon, correct? Third carbon will be here and fourth will be here, okay? Now, at second carbon, CH2 OH is coming on the bottom of this line we have. So here it will be on the top and OH will be on the bottom. Third carbon, OH on the top. So here, OH on the bottom, OH and H. And on the fifth carbon, CH2 OH on the top, so here it will be on the bottom. Now you look at this structure. What is the difference we have? Do you have any difference in this? It's, it looks like alpha. It looks like- No, it's not alpha. Is it alpha? How? No, it's not. The nodes are on the same side. It is actually the same thing. It is beta D minus glucose fructose only. When they ask this question related to this, they have given this structure in the exam, not this one. Usually we have this in mind that on the right side, we have second carbon, OH on the top and CH2 OH on the bottom, right? So that they have changed, they flip the molecule. So when you flip the molecule, the molecule is same only, right? So this structure you can draw this way also. So it doesn't matter if you flip it? No, they have given, yeah, actually the molecule is same. No, you have just look at the other side of the molecule. Oh, wait, there is a, oh my God, I thought that carbon number two and carbon number five are the same. The molecule won't change. No, alpha one becomes beta. Beta one becomes alpha. It is beta only. Yes, sir. Correct. So they have given this structure in the exam. Keep that in mind. Now why it does not form six-member ring? That's the question in the mind. I think only carol carbon only takes part in the reaction. Why, sir? Why? Sir, carol carbon doesn't do with reactivity, right? Optical activity you have to maintain, but... So is it because of some steric repulsion in six-membered because it's like, oh, in the same... One thing I remember that carol carbon takes part in the reaction that I remember, but the reason I'm not getting, maybe carol carbon may be one of the reason. I'll think about its reason. I'm not getting any... Yes, sir, no problem. Okay, so I'll let you know regarding this. We'll come, we'll see this later. Next class. Okay, so fine. So this one you must keep in mind. Okay, now the thing is, which, you know, you said that the question they have asked in your practical exam in... Why were they have asked this question? Why tollins this up? Proctol shows tollins test, correct? Because it can automatize into glucose or any other thing you've done.