 Ok. Sir, but did we say that even in one carbon there is no complication? What? The intensity, this is in conjugation over here. It is not complete conjugation. No, there is no resonance structures for this compound. We can draw resonance. Sir, but if it says that the algorithm is not supposed to draw resonance. No, this part is not involvement. But we have resonance till here. Sir, there will be like three counting that. Yeah. This, you can draw this side or you can draw this side also. But this carbon is not involving resonance. Ok. So when you draw the trotomers of this, one of the H plus from SP3 hybridized carbon atom comes out and it joins onto this oxygen atom. Ok. So and how this pi electron shift you see, this pi electron comes over here and this pi electron comes over here. Sir, there will be two structures. Yeah. Sir, it is conjugation, not hyperbolic. Sir, then what is conjugation? Conjugation means this is the condition for this. Pi sigma pi. Pi sigma pi. Pi sigma lone pair. Pi sigma positive just gets over. All these things have discussed conjugated system. Right? So this is the conjugated system. This H plus comes over here, you see here. This H plus will be here. This pi electron comes over here and this pi electron comes over here. The structure we get here is phenol. So all there will be like two things. You can have it shift this side also. Alternate, like this, this shifts come over here, this comes over here. You can do the other way. That also you can do. This comes over here. This comes over here, this comes over here, this goes over here. Same thing. Ok. But in this compound you see, this is aromatic compound, right? Yeah. Phenol is aromatic. So converging of this will be 100% here. Means after some time this won't be present in the mixture. We'll have pure phenol we'll get. Because aromatic compound is most stable compound. And this convert into this concrete. Ok? Phenol. So is thought awareness that the hydrogen always leads from the sp3 carbon? Always from sp3. All? Yeah. If you have conjugation then we'll shift the carbon at least. Till we get sp3 hybridized carbon. Ok. One more example I'll write down for you. We have double bond O and this. Can you write down the product into this one? Yes. So here there will be two things. Yes. Draw up both the structure. Yes. Do we have two structure possible in this? Yes sir. Two structure possible? So one of the structure is this. Light. This one is structure. Another structure is what? Yeah the other one. This is less unstable. This is more unstable. I mean the heat. Stability behind. Otherwise you'll get confused. Why wouldn't that be less stable? Stability we'll discuss. We'll wait for those. This is a creation of the electron. I'll show that. These two structure possible. How do you draw the structure? You see this hydrogen either this hydrogen takes part in trotomerism or this hydrogen takes part in trotomerism. Both possible? Both possible. Because here it is conjugation. So this carbon, this carbon, this carbon, this carbon. SP2 hybridized. No conjugation is there. Conjugation is there but no trotomeric possibilities. So what happens here? This edge comes out. Right? This edge will attach onto this oxygen. Yeah. And this sigma bond comes over here. Makes a pi bond. This pi bond sits over here. We'll get another pi bond here. Okay. This pi bond sits over here. And this pi bond comes onto this oxygen. Which takes O minus, which takes this H plus under structure. Okay. Alternates the pi electron, pi bond, you have to stick by one carbon. Okay. Now what is? Other way, sir. Other way means this one. This one, you see, this H plus comes out. Okay. And this sigma bond makes a pi bond here. And this pi electron sits over here. Which takes this H plus over here. There's a carbon hydrogen bond you have to break. That's sigma bond. You have to convert that into pi. And you have to shift the pi bond by one carbon. Okay. This stability. The stability will be up there. Just wait. Unless, first of all, you should know how to write down the product. Okay. If you talk about stability, this lone pair. Right. Here we have, see, this lone pair is in resonance with this carbon. Okay. Right. Here we have, see, this lone pair is in resonance with this pi electron. And here we have, you see, this one, we have extended conjugation. Okay. This lone pair, pi, pi, pi, we have extended conjugation. But here we have conjugation till this part only. Sir, don't you have conjugation there also? We have here, you see, that's what I'm telling you. We can draw conjugating or resonating structure in this manner we can draw. Right. With pi, sigma, lone pair. Or you can also draw with pi, sigma, pi in this direction, right? Two different directions, the resonance is there. This is cross conjugation. Cross conjugation. This is cross conjugation. But here we have extended conjugation. Pi, sigma, lone pair. Sorry. Lone pair, sigma, pi, sigma, pi. This is extended conjugation. Resonance, resonance, which one? Extended. That is more stable. Extended conjugation is always more stable than cross conjugation, right? This is more stable one. It is extended and this is cross conjugation. Right? So extended is more stable than cross. Major product will be the second one. Can you draw the structure in this one? Chartermeric structure? They both are in the same. Which one? Only one product is possible. Both sides will be the same. Is it more than something there? No. You will get an aromatic structure. There is only one kind of environmental involvement. That is the only one. Shifting a pilot or is it different? You get only one. What is the structure in India? Autonomous structures and resonance in the region. The only one that can do it is the auto-online. When this hydrogen is involved in total momentum, right? So this hydrogen comes out, right? And this sigma bond makes pi bond here. And this pi electron shift over here. This pi electron shift over here. This pi electron shift over here. And this pi electron shift over here. And this edge glass comes over here. So product is what? This is the whole thing. We go this 4th and I have the same. 4th and I have the same. We buy 4th and I have the same. Sir, I would get that. We buy 4th, sir. We buy the 4th question. We buy the 4th question. 1, 2, 3, 4, 3, 4. OH is the number of the time. Sir, it should be 5 hours. Sir, if we take the hydrogen, it will be the other way around. This size. Yeah. Only the position of pi bond will be different. But the structure will be different. Got it? Clear this one? Sir, benzene attached to anything will go through there. When the other one has function from there, it is going through there. That's the whole thing I look in the structure. This one can I look in the structure. This one is water. It's very complex. Sir, it's okay, sir. It's okay, sir. Sir, it's okay, sir. Sir, it's okay, sir. Sir, it's okay, sir. Sir, it's okay, sir. Good. Sir, it's okay. Sir, it's okay. Sir, it's okay. Sir, it's okay, sir. The same number. X, Y. X, Y. X, Y. I don't know what you mean by that. Sir, is there something like benzene or what? No, no, that we have. Abalbond, Abalbond, benzene, benzene. Benzene, phenone, phenone. It's a phenol, phenol, phenol. If I just wanted to use this. Hydroxybenzene. Hydroxybenzene. Don't you use the prefix. There's another thing that's more important. We call it benzene 1-O. I thought the prefix is only if there's another more important. Or is it benzene? Very, very important. Except the prefix of this is hydroxybenzene. So if we have CHO, we use the prefix. I thought we only use the prefix if there's another thing that's more important. If there's another thing that's more important. That is 1-O-3-hydroxybenzene-hydroxybenzene. Yes, sir. If there's no CHO, then we don't use hydroxy. We use all. That's right. We have only one functional group. So we use some fix of that. So it's just benzene 1-O. You can say that. Benzene-O we can say. It's obviously the position of this. Anyway, so this is, in this one, what is the part we get? Tell me. Oh no. No, no. So phenol, phenol. Also here again then. Yeah, phenol, phenol, phenol. Benzene. This one here. So there's two things more important. Carbon acid. Sir, won't that turn into carbon acid? Oh, it's too much. If we try carbon acid. In the text here. Texting the last carbon acid. Oh, yeah. See, in this one, there's no total monosin. No, that's it. No, you're right. Because this is not hydrogen. It's just Cs3 here. H plus can come out. Cs3 plus can come out. Because, see, the effect of this slope, at this carbon you can consider. But at this carbon it's not possible because the sense is too much. So electronegativity. But here we went from there. Yeah, because we have fast-pacation. Yeah, we have conjugation. No, that's why. Sir, can we try for tautomerism? That carbon ends up coming from a high point. Stop trying. It's not possible here. It's not possible here. We say not tautomeric is correct. Tautomerism is not possible in this because this carbon does not have any hydrogen. Okay, hydrogen is there. That's fine. Understood. Now, we have already discussed how to write down the product. We see the mechanism here. In acetic medium, how the reaction goes and what is the major product. Tautomerism is possible in both acetic and basic medium. Okay? So acetic mediums, mechanism will see first. Acetic medium means what? We have H plus. Suppose the molecule is this. The simplest one we are taking. Cs3 C triple bond O. Cs3. And the tautomeric structure we have already done. H plus. The structure is Cs2 double bond. COH. And Cs3. This is the structure. Now, how do we get this structure? What are the different steps we have of the reaction, how this reaction proceeds? So first step, what happens in first step? This Cs3 C triple bond O. Cs3. And the H plus from acid. Acetic medium, it is acid gives H plus. Right? So H plus get attached onto this oxygen atom. Since this has two lone pairs, it attacks onto this electron-deficient hydrogen. And this H plus attached onto this oxygen. Okay? This forms Cs3. C double bond OH. Cs3. And since oxygen loses its electron pair, this oxygen will have one positive reaction. Okay? Right? Now, to stabilize this oxygen, what happens? This pair of electron lone pairs shift onto this oxygen because oxygen is an electronegative element. Right? And positive charge of oxygen is highly unstable. So this pi electron will stabilize this oxygen atom and this will shift onto this oxygen. So this we get here, Cs3. C-O-H. Cs3. So isn't this a reaction as opposed to automerism? Because none of the maintained like Automerism is a reaction. Yeah, but like this use an external acid. Yeah, but like this use an external acid. As in like usually the hydrogen comes from within the dude. So but in the first one, why do you need the H-plus? Because see for this reaction to take place, you must have some medium. Like that medium you have to provide a reaction medium. But the H-plus is not actually being used. It is not being used. It will come out eventually it will come out from the reaction mixture. So it is not getting consumed. It is behaving as a reaction in this reaction. Okay? So with this mechanism you understood. See, is this correct? Is this correct? Yes. Double bond is correct. But there is one extra Yeah. So if there is an extra acid. That's correct. Yeah. But it's a different problem. Yeah. I did not finish it. This converts into this. The path of reaction that's what we are discussing. The mechanism of reaction. Okay? So this is a carbocation. You see here it is a carbocation and carbon will have positive charge because this carbon here we have one electron of the carbon atom and one electron of oxygen. And this oxygen takes both the electron of carbon atom also here. So carbon uses one electron. That's why carbon has positive charge. So we get a carbon atom. So why again carbon atom? Because positive charge on carbon atom. No. This one. Oxygen has positive charge. It is highly unstable. So to to to to. Okay. Now in the next step in the next step what happens the molecule we have CH2C OH CS3 and one oxygen is here. This H plus comes out because H plus is behaving acid is behaving as a catalyst here. So it won't get consumed in the reaction. The concentration of acid must be constant right throughout the reaction. So this H plus comes out and leaving this bond there of electron behind this carbon has positive charge. So H plus comes out into this and we get CH2 double bond CH3 plus H plus So concentration of acid is not changing here. In the first step H plus get consumed in the last step H plus releases. Right. This is step the second step is the RDS of reaction. You know what is RDS. RDS is rate determining step. Okay. Slowest step is always the RDS of the reaction. Right. If the reaction is based by two three different steps the one which is the slowest step is the RDS of the reaction rate determining step. Okay. Now in this step we always define the major products according to the RDS of the reaction. You should know this. This is not RDS. Right. RDS is this. Now in RDS we get alkene. Correct. So we try to get most stable alkene. If it is more stable acidic medium what we will try to pump. If the medium is acidic most stable alkene will pump. Most stable alkene is the major product. Alkenes is the major product. So if the cotomer cotomer is more than two alkene then acidic medium will try to get the more stable alkene. In this only one possible. Okay. So like I have this. I know that. See this is the mechanically understand this. I got this. I got that. This is the RDS. But then if this is the RDS then like this has to be formed over here. Like this first step takes very little time. Right. And then after that it has to do that. Do that this one. It has to form that. Because that is far. Right. And then how is the like how can you say that this will be the like less stable product? Because it's I didn't say that. I didn't say that. It's less stable. See first of all all these are intermediate of the reaction. Yes. These are not the product. So intermediate has to convert. Obviously intermediate can never be the product of any reaction. Carbocation you never get carbocation, carbonyl and free radical as the product of any reaction. It forms during the process of the reaction and eventually it converts into the final product. Okay. Because these are highly unstable species. Intermediates are highly unstable. You cannot get as an intermediate as a product of any reaction. Okay. So this step I'm writing it down to make you understand. It is not like obviously this takes very little time. This what I said this is the slowest step. It's time taking for this conversion is obviously more than the time taken over here. Because of common iron effect until this happens. Where is common iron effect? Because of see like if I count that as an iron which one? H plus? No sir that this one. Yeah. This one. Because there will be too much of that so then it won't go forward with the action. After this you are saying after this this is the slowest process. No. Actually I have written this this way. Actually after this only H plus counts out and you get this product. Yes sir. Okay. That's why this is one step another step is this. So one step is formation of an intermediate. Second step is conversion of intermediate to product. That's how it goes. Okay. So over here the only alkyne is this one. Is this in this case? In this case if you have more than one alkyne possible then the major will be the one which is more stable. Why I am emphasizing on this thing because if suppose in case this step is the slowest step. Try to understand this. If this step is the slowest step then in this step we are getting a carbocation. So we will try to form most stable carbocation and the most stable carbocation gives you the major product of the reaction. So now after the RDS is the most stable product. Yes. This is RDS and this gives an alkyne so we will try to get most stable as the major. So why is it I didn't have this. No, no, no. It's not since this step is slowest that's why we will try to figure out the most stable alkyne. It has to exist for the longest time. It takes maximum time if we come by the time here and here the time required here it is more than this. That's why it is slowest step and since in the slowest step alkyne is forming that's why the stability of alkyne gives you the major product. More stable alkyne the major product of the reaction. It will then go faster because it will be much more eager to become more stable. That's why the reaction goes in forward direction. But it goes very slow, right? It is going slow but this has maximum tendency to convert and see we are saying that it is slow. You cannot even imagine and the reaction gets over why it is slow because this takes a little longer time than this and what is it? The product of the RDS This product see this product you will get this product you will get the most stable one also and the least stable one also if it is possible. But which one is major? The one which is most stable that is the major product. Because in different products there is a step one or step two in both the RDS can either be RDS No for acidic medium this is the RDS One is the RDS why will it go to step two because step one is already the major step one is the major product. I am saying if it was the RDS Yeah if it is see it is the RDS we are getting carbocation here so we will try to get most stable carbocation first and that stable carbocation gives to the major product of the reaction because intermediate these are intermediates so it cannot be the final product of any reaction it has to convert into some other product but if it is most stable that gives you the major product we will consider the stability of carbocation if the first step is the RDS of the reaction which is not the case here So even if the first step was the RDS once the reaction reaches the continue right Yes obviously because it is an intermediate it has to convert into product but if we will discuss some examples you will understand So also if that step is the lowest like that that thing exists for the longer time as in the thing that is taking the most the kind that is taking the most time to convert doesn't that exist for longer than the product if the product forms it will be there in the mixture right it will be there right but this one obviously this step is slowest so this takes longer time than this comes we are comparing this it is a relative thing also you need the most stable carbocation and the most stable carbocation No no no that is what you are not taking we are not considering this because this does not you know affect the rate of the reaction rate of the reaction is affected by this step removal of H plus okay formation of carbocation does not affect the rate of the reaction we will discuss a chapter in 12th grader's chemical kinetics it is based on the rate of the reaction only so rate of the reaction is defined by the slowest step of the reaction so if I ask you at home you have to consider this step not this one okay because this defines the rate of the reaction so also this stability like increase uniformly like that being the least stable and just the most stable which one like this is like no no we cannot compare the stability of this right the medium is acidic we will have some stability but that that won't affect the formation of the product because as in like in every step but you don't see this is neutral right and this is carbocation okay so if you compare the stability of this obviously neutral is more stable but in the medium in the acidic medium the reaction goes this way lone pair of oxygen take this H plus and forms a carbocation if we have more than one carbocation possible then we say okay this carbocation is the more stable once we'll form this carbocation and we'll get the product according to the more stable carbocation but you cannot compare an intermediate and reactant you cannot compare you can compare two different reactant which one is more stable two different product you can compare two different intermediates you can compare yes what is the carbocation problem we'll do that we'll do that these are the examples we have discussed enough okay that is not required