 Hello, good morning. Just give me a minute, I'll start then. Hi, Rithvi Ramcharan. How are you? How was your exam? Good morning, Saimeer. Okay, so how was the experience? How was the exam? Just a second. Yeah, so tell me how was your experience? How was the exam? Jamie, chemistry was easy, right? Moderate. So what is the marks you are expecting? Anyone above 200? Anyone above 200 you are expecting? Okay, tell me one more thing. When is your board exam? Have they declared the date? March 2nd, okay. And how many of you are, you know, I think some of you are obviously planning to write JEMN in March also, right? Okay, so you must have seen the paper, right? Yeah, so all of you who are planning to write JEMN in March also, right? We have enough time, more than a month we have to prepare, okay? I know in this time you have board exam also, but you can, you know, you can do that because if you have seen the paper, all the paper if you have seen, they did not ask any tough questions, right? The questions were not tough. In chemistry what I can see that there are some factual questions are there, okay? Like, you know, the pH of rainwater or like the isotopes of hydrogen, there are very factual questions are there. And if you see those questions from inorganic especially, those are mainly from NCRT, right? So that's why I told you already that for inorganic chemistry NCRT is more than enough. If you have studied line by line NCRT, you can do all those questions, right? So JEMN has been changed a lot over the year now, right? Since you can also understand like this earlier last year what happened, they have to prepare 30 questions in, you know, 30 questions in chemistry, 13 physics and 13 maths, right? In one year they have to prepare only 30 questions in each subject, right? So they can work on these 30 questions, so obviously you'll get better questions over there. But now what happens you see from 8 to I think 13, almost a week you have exam, right? 7 days, 6 or 7 days, correct? And each day you have to shift. Now if I'm talking about chemistry itself, there are 60 questions total in one day. In 7 days it will be 420, roughly I'll take 400 questions only, right? 420, so around 400 questions only. Similarly you have one more exam in March, there also you'll have 400 questions more. So in one subject they have to prepare now 800 questions, right? So if you think like this earlier where they have to prepare only 30 questions, now this 30 becomes 800. So where they'll collect these 800 questions? They cannot give you some new questions into this, okay? So they have to pick up the questions from here and there, from the books or from any other exams, okay? So that's why the questions, you may feel that a few questions were repeating. Is it there? Is it like that? A few questions, have you felt that the questions were repeating? You have solved all those questions like from any other book or in problem solving session. You must have seen those questions. Yes? So that's why, that's why I, from this one month, yeah exactly. And in chemistry also there were questions like which of these molecular molecules according to MOTR is stable or not. So question was HE2 plus then we have HE plus something like that. The question was there, the based on bond order, okay? So I can see there are many questions were there which you have seen somewhere here and there, okay? Yes, many were similar, many were similar. So that's what I think. So the point I'm trying to make is what? Again, they have to prepare this 800 questions. So you are not going to get any new or difficult questions, okay? You will get easy questions only and those questions you must have solved already. The only thing is that you have to keep on revising those things again and again, okay? From this exam what we can say, right? So that's what the thing is, okay? So what I suggest to you in this one complete month that again you have to concentrate on board exam also. But parallely you have to do inorganic chemistry, again NCRT you must solve, okay? And wherever you get some factual question, especially in your organic question, factual question, you just write it down in a diary or notes, okay? So that in one month you have, you know, a good number of factual questions so that you can revise this question in the last one week, okay? So like this you have to prepare. Numericals I think I just wanted to know from your side. As far as numericals were concerned which have been asked in, which has been asked in JMN exam, do you get any difficulty in solving those numericals? Like I'm talking about chemistry. Do you get any difficulty in solving those numericals? I'm talking about this only in general terms. I know all questions obviously you cannot solve. Yes, so the point is numericals were very direct and formula based if you see. All direct formula, yeah. So that's what, that's what. So you don't have to worry with numericals, okay? If they ask very tough numericals that would be only probably one or two questions. Don't worry with that, okay? So what do you do in this coming one month? Just you revise basic formula, right? And theoretical questions also. For theoretical questions what I'll suggest to you that you must go through the previous year, NEAT exam questions, NEAT and AIPM, sorry, this AIMS question. NEAT AIMS, whatever the medical examination chemistry sports are there, that you must solve. You'll get a fair bit of idea over there, okay? Because I have seen many questions that have been asked in Jamie in this year. They have already asked in NEAT exams, right? So that's what the point I'm trying to make that in this one month, try to focus more on, you know, very basic question, formula based question, theoretical questions and NCRT for an organic chemistry. How do you solve this theoretical and formula based question? The best way is to solve all NEAT exams or AIMS paper, okay? You will have fair bit of, because in NEAT and AIMS exam, if you see they ask the question like this only, formula based question, a little bit of trick was there, but not that much stuff, okay? So that's what I think you should do in this one month. Understood this, okay? Stop thinking about what has already done, okay? Now, what experience you had from this Jamie exams? With that experience, what changes you should do in your strategy? Think on that for a day, okay? Take your time and then start working on that, the new strategy, okay? So that is very important. Okay, anyways, now we'll start the class here. You see, whatever question I have seen, there were no any questions, which is based on tautomerism. Did you see any question based on this tautomerism? I haven't seen any question based on tautomerism this one week, okay? So there are high possibilities that next time you will get question from tautomerism, okay? And you also work on this, whatever portion are left now, see, just you talk all of you with each other, right? And try to figure it out that in chemistry or in physics or in maths, what portion they haven't touched, right? Last year also what happens, offline exam was there, let me tell you this, offline exam was there. In offline exam, they haven't asked anything about, you know, polymers. Hardly they have asked anything about polymers, okay? Twitter and all. This thing they haven't asked. So one of the students here in Centremer Academy, that time I was newly joined over here. So she asked me, sir, what should be the important topic for online exam? So I have gone through the paper of offline exam, and then I figured out that polymers, they haven't asked any question, much question from polymers. So I suggested her that you should revise polymers once, and then polymers, Twitter and concept is important. So you must revise at least that much, okay? And then some formulas of that. And in online exam, they have exactly got one question, which is based on Twitter and concept, okay? So that is what I'm trying to, the point I'm trying to make, that you all, you know, talk with each other and try to figure it out, okay, like we are also working on it, and try to figure it out that which portion in math, physics and chemistry they haven't touched in this exam, okay? Those portions you try to cover first, okay? Because there are high possibilities that you will get questions from them, okay? That you must do. Anyways, so we'll discuss a bit of concept of tautomerism. We have already discussed this thing, but since this is a revision class, problem solving class, we'll solve some problems also, right? But since tautomerism, again, it is a factual question, because if they ask you some questions like, what is the percentage in all content, and which product is stable and all, where the in all is stable or keto is stable. So for that, that is based on the experimental thing, right? And from few, you know, a few logic you can put into that, okay? So that's why I'm discussing tautomerism first, and then we'll see some questions of any of the 11th grade topic, because I am assuming this since you have board exams, you are already in touch with 12th grade topic, right? So I just wanted you to be in touch with 11th topic also, okay? So that's why we'll solve some questions of 12th grade also, okay? But today we'll see some 11th grade topic only. So here we are starting with tautomerism today. See, tautomerism is what? Tautomerism is a special type of functional isomerism. I am not going to dictate you anything into this because we have already done this, okay? So whatever the important points are there, I'll write it down. If you want to write it, you can note it down, okay? So this is a special type of functional isomerism. Since we have the chain-in functional group, right? Keto converts into Enol, right? In this, what happens? The isomers, they exist in dynamic equilibrium. Isomers exist in dynamic equilibrium. So the isomers are in this type, isomers are interconvertible, interconvertible, okay? So they can convert from one form to another form, right? But their structures are different, okay? These kind of isomers are this phenomenon is tautomerism, right? Now, how many different types of tautomerism are possible here? So I'll give you some examples into that. They may ask you this question also, okay? So you see, first of all, the very basic and, you know, we all know this type of tautomerism that is keto-enol tautomerism. Keto-enol. So keto-enol is nothing but this CH3 C11O CH3 and this converts into CH3. This is keto-enol. One more type we have that is nitroso-oxyne, okay? In this nitroso-oxyne also, tautomerism is possible. Like you see the example here, single bond N double bond O. This is nitroso. Oxyne will be what? One H will come over here and we'll get double bond here, right? So it will be H2 C double bond NOH. So this is nitroso-oxyne. Have you heard about imine inamine? This is important, imine inamine. Immine is nothing but this. We have CH3 NH. This is imine. And inamine is H2C double bond NH. NH2, sorry. Why it is inamine? Because Ene stands for this double bond and this is the amine group. That's why it is inamine. Next one is amido-imidol. Amido is this. H2N C double bond O. And imidol is azohydrazone. Azohydrazone. Azo, what we have? We have N double bond N, right? CH3. This converts into H2N single bond N double bond CH2. Double bond CH2. And the last one we have here, that is diso-nitrosoamine. That will be aryl group N double bond N OH. Conversion of this into aryl NH single bond N double bond OH. All these type of compounds, tautomerism, possible, right? The mechanism of this, we have already discussed. I'll just discuss it again quickly. Here you see all these adjacent carbon, sp3 hybridized we have here. This hydrogen, any hydrogen or alpha hydrogen, right? We can say these hydrogens are alpha hydrogen. Alpha hydrogen are involved in tautomerism. First of all, this thing you have to keep in mind. So the sp3 hybridized alpha hydrogen we have here. These hydrogens are acidic. Alpha hydrogen are acidic. Acidic means what? These molecules can lose this hydrogen easily as H+. So what happens here? If this hydrogen comes out as H+, right? Oxygen already has lone pair on it. Similarly, in all these cases, oxygen has lone pair here. Here nitrogen has lone pair, sorry, one lone pair only. Oxygen has lone pair like this it is. Here this nitrogen has lone pair, right? So because of this lone pair of electron, this H+, has tendency because once this molecule lose this H+, ion, so the ion that is left over here is H2C, single bond C double bond O, CH3 with negative charge over here, right? Now this H+, from this alpha hydrogen has tendency to come out and then again attach to this molecule, right? So when H+, suppose when this H+, wants to attach with this ion here, so there are again two possibilities. One is that this H+, may attract it towards the lone pair of this oxygen or it may attach to its initial position which is nothing but the carbon atom, okay? And this process is spontaneous, we are not doing anything into it. All these things happens on its own depending on the stability of these two products, okay? Which one is, suppose this one is more stable and has more tendency to attach on this oxygen atom, not on this carbon atom, right? This is acetone, the meaning of this is what? Suppose in a bottle if you take acetone, right? 100% acetone suppose if you have in this bottle, right? Initially the composition of this is 100%. So after some time if you try to find out the composition of acetone, it would be less than 100%, right? On its own, we are not doing anything into it so what happens, some of this part of acetone has been converted into enone by this reaction, okay? So when this H+, attach over here, you will get this product, when H+, attach over here, you will get the initial product, correct? See all these molecules, the tautomerism is possible. Mechanism if you see, because this mechanism is not required but one thing you just keep in mind, because tautomerism is possible, tautomerism is possible in both acidic, basic medium, acidic and basic medium, right? I am not discussing the mechanism in all this, that is not required, okay? So I am leaving that part, okay? If you want to know the mechanism to discuss the mechanism, I will discuss that also. Let me know if you want me to discuss the mechanism quickly, okay? So you see the base catalyzed mechanism, first of all we will see in presence of base, how this reaction proceeds. Suppose the molecule is this, RCH2. So the base that you have here, which is OH- this OH- will take this alpha hydrogen, right? This OH- because the lone pair here, this will take this alpha hydrogen, right? And this is a keto form. So when it takes this alpha hydrogen, so what we get here? And this bond, it takes H+, and this sigma bond shift over here. And in this case, this pi electron will shift onto the oxygen atom, correct? So the product we get here is this RCH double bond C, single bond O-, single bond O-, and R. Next what happens? This ion, we call it as enolate ion. This is enolate ion, which further converts into, because we have this H and OH- forms what? H2O, right? So this H2O only has tendency to lose H+, ion and that H+, will attach on this oxygen or this carbon depending on the stability, correct? So suppose this negative charge we have here, we have three lone pair here. This will attack onto this hydrogen and this bond comes over here. So we'll get what? RCH double bond COH, R plus OH-, minus the base will be as it is when the reaction is, right? This is the enol form. So this is how the reaction proceeds, the mechanism we have. One thing you must keep in mind in this kind of reaction, that the carbon across C double bond O-, I'll write it on here, this you must copy, the carbon across carbonyl group, across carbonyl group, which contains more number of hydrogen, which contains more number of hydrogen, take part in the reaction, will take part in the reaction. Meaning is what? Suppose if I write down this molecule, RCH2, C double bond O, CH3, then the product in this is what? In base catalyzed reaction, we have RCH2, single bond COH, double bond CH2, carbon which contains more number of hydrogen atom, from there only the hydrogen take part in the reaction. Understood this? Yeah, I'm doing that. This is the base catalyzed mechanism, okay? So did you understand this? Can I move on? So this is very important point we have in base catalyzed mechanism. Must you keep this in mind? Because in acid it is reverse, okay? That's why I have given you this in note, okay? So in base catalyzed, the hydrogen comes out from the carbon, which has more number of hydrogen, alpha carbon, right? But in acid it is reverse. See this? Acid catalyzed mechanism you see. Suppose reaction is this, RCH2, C double bond O, C and R will take, generally only will take, but I'll take CS3, that would be the better option way I think. Let me write this CS3 only over here. Since it is an acid catalyzed reaction, what are we also going to use? So acid will give H+, correct? So whenever if you have acid catalyzed reaction and carbonyl carbon is there, so the lone pair of that oxygen takes part in the reaction. So this will attack onto this H+, and this hydrogen will attach over here, okay? So what we get here, you see, RCH, COH, double bond CH3. This will have the positive charge, right? Now in this what happens, the H2O molecule, because we are taking here H3O+, acid with this water. So the H2O molecule that you have, oxygen has lone pair. This lone pair has tendency to attack in any one of this carbon atom, right? And in this case, what happens, what you have to keep in mind, that this water will take hydrogen, which from the carbon, from the adjacent carbon, which has lesser number of hydrogen, right? So this oxygen or lone pair will attack onto this hydrogen, because it has two hydrogen, here it is three, right? Then this bond pair goes over here, and this pi electron will shift onto this oxygen, right? So the product we get here is RCH double bond COH and CH3. This is the enol form, this is keto, right? So in this mechanism, acid catalyzed, what you have to keep in mind, that the hydrogen comes out, hydrogen comes out from the adjacent carbon, adjacent carbon, which has lesser number of hydrogen. This is the difference we have in acid and base catalyzed mechanism, right? Few examples you see on this, questions. Suppose the molecule is this, this is in presence of acid, and this reaction is in presence of base, first one. Second one, acid, and this is base. Tell me the product. I mean, suppose it is, I will not write down this. Suppose this is first and second carbon is this, and this is first and second carbon is this. In the first question, hydrogen comes out from which carbon? For acidic reaction, this product. Certainly for this reaction, acidic medium, hydrogen comes out from which carbon? From the second carbon, right? And basic medium, hydrogen comes out from the first carbon. Similarly, you can write down the product here also. Fine, can we move on now? Okay, next you see the very important point we have here, that is stability of enol form. And in this section, whatever I am going to discuss, that you write down properly, okay? Because these are based on some logic, but again, percentages in all content and all, you have to memorize because I have seen few questions in other different books, that sometimes they have asked the percentages in all content also, okay? So whatever example I am going to discuss here, these examples are very important, okay? You will get this question direct, okay? Or if you are not getting the same question, you will definitely have a fair bit of idea if you have this data in your mind, okay? That's why these are important. So stability of enol form, this is very, very important, okay? First case in this, in case of monocarbonyl, carbonyl compound, this is where we have only one C double bond O group. For example, you see CS3 C double bond OR, right? The product of this will be what? This hydrogen comes over here, directly what we can write. This hydrogen will jump over here and we have a double bond over here. So the product will be CH2 double bond COH and R. Now if you see the percentage in all content over here, this is very less, 10 to the power minus 4% of enol. Rest is keto form only. Means here, the keto form is more stable, keto form is more stable than the enol form, right? So the reason behind this, if you see, here you have carbon-carbon double bond, right? The reason of this you see, the first thing, that in this keto form, we have C double bond O and the enol form, we have C double bond C, okay? So the bond energy of C double bond O is, suppose I'll write down this bond energy one and this bond energy two. So the bond energy of C double bond O is quite higher than the bond energy of C double bond C. Okay, so that's why this bond is difficult to break and this bond is comparatively easy to break, right? Hence this one is more stable than this. So keto form is generally more stable than the enol form, right? But we should have some condition over there. Also that we'll discuss one by one. Another reason here it is what? That in this carbonyl group C double bond O, this is resonance stabilized also. Means there are possibility of resonance over here. So C plus and O minus, right? This particular type of separation of charge is comparatively more stable than carbon-carbon. If you try to separate the charge over here, we have C plus and C minus. So if we have positive and negative charges on the adjacent carbon atom, then this is highly unstable compound, right? C plus and C minus is highly unstable compound we have, okay? So that's why this keto form is more stable than the enol form. Understood this? Okay, these are the two condition you have to keep in mind. Just a second. So keto form is generally more stable. Yeah, it's in basic medium. Yes, Shweta. It is in basic medium. Generally, we do this tautomerism in basic medium only, okay? If nothing is mentioned, it means it is in basic medium. And if they ask you, you see, actually you try to understand one thing, right? Actually, you see, they don't ask this question. Like, you know, why I am not, why I said that the mechanism is not important, the reason I'll tell you now. They never ask this question. Then what is the product you get in this tautomerism of this? CH2, 3, C double pond O, CH2, CS3. They never ask the product here, right? What they'll give you? Sometimes they'll give you A, B, C, D, four different, you know, product here, and they'll ask you the percentage in all content of all these product, like 10 to the power minus four, right? 10 to the power two, like this they'll give you the option. So they talk about the stability of the tautomeric product. Understood my point. So they usually ask about, because it is no point, like if they ask you this question, whether it is basic medium or acidic medium, if nothing is mentioned, then you can take this as basic medium. But if they have mentioned already the medium, you know what should be the product, correct? So that you can do easily. I am trying to focus on that if they ask you the stability of a tautomeric product, then what all logic you can put in, understood? So that's the point. But whatever the medium it is, you can apply those logic for all these products. Only the product will be different, but stability if you talk about that logic is applicable for all the product, whether the medium is acidic or basic, right? Now similarly you see this question. So I'm giving you all these questions in a series, right? So that you can memorize this. All these examples you must memorize. In this, when I write down the tautomeric product, it would be this. We have a double bond here and OH, correct? Right? Now, if you see the percentage in all content of this, that will be 1.2% here. Okay. So when you compare the stability of these two, obviously this one is more stable, right? Because 1.2 is this means 98.8 is this molecule, right? So, but when you compare these molecules, like the first example that I have taken, which is nothing but this one, which is the chain in all form, the first example which I have written that is CH2, double bond COH and R. This molecule was the product in the first example. So if you compare the stability of these two, then this cyclic one is more stable than the chain one here. The reason of this is what? So first of all, you write down here, this compound is more stable as compared to more stable as compared to aliphatic in all form, right? It means if you're comparing these two. So whatever comparison you are doing just to keep this in mind because I have seen these questions exactly same question they have asked directly many times in the exam. Okay, test paper exam or in the books, okay? That's why I am picking those these examples, okay? So if you compare the stability of these two, this one is more stable. The reason behind this is what? Because of the, this is because of, because of free rotation, free rotation of OH bond. So what happens here? This OH bond that you have, this entropy, we call it as free rotation of OH bond that will increase the entropy of the system entropy of the molecule. This entropy, we call it as zeroes entropy, G E R O S. The name is not important, right? But the point is what this molecule, if you see, we have a double bond here and then OH. This bond has a free rotation, right? We'll have a free rotation over here because of this only its entropy increase. That rotation is here also. But since this is a linear compound we have, so because of these molecules present here, it will be a bit hindered but the increase in entropy here, it is lesser than the increase in entropy here, right? That's why this aromatic enol form is more, sorry, the cyclic enol form is more stable than the aliphatic one, okay? So just, this logic is, you know, it is not that important, okay? It is not actually this entropy is not in our flavors also. Okay, so you can, you know, let it be, it's not required. But the thing you have to keep in mind, if you have cyclic enol form, it is actually more stable than the aliphatic enol form. Now, how do you compare the stability of the other products you see? In this only, if I give you this molecule you see, the tautomeric product will be this, this double bond will be as this, here we have a double bond and then we have OH. So now we compare the stability of these two, obviously this one is more stable. It is almost, you know, more than 90% stability of this we have composition. It is not lesser in concentration. Again, you have mixed two things. The concentration of this, with respect to this I'm telling you, so it is 98.8%. Okay, so if you compare the stability of these two, this one is obviously more stable. Content is also more, right? What is the composition of this one? 10 to the power minus 4%. So this is obviously more than this. So I'm comparing the stability of the product of two different tautomeric reaction. Okay, that's what I told you. They'll give you four different product here, reactant here. And what is the in all content? They'll give you the option over here. So this is what we have to compare the product. That's why I'm comparing the product here. The first product is this, 10 to the power minus 4, it is 1.2%. When you compare this, this one is even more stable than these two, right? Because this is aromatic, right? This benzene ring is aromatic. So its stability is more than 90%. Okay, that is also one thing we have. So this is possible when we have mono-carbonyl group present. Fine, now we'll see the second case where we have di-carbonyl group, means two carbonyl groups are present, right? So case two you write down, in case of di-carbonyl compound, in this also we'll consider open chain first and then we'll see the closed chain, okay, cyclic compounds. You see if the reaction, if the molecule is this C double bond O and C double bond O, carbon. So when you have carbonyl group present at adjacent carbon atom, so we have lone pair here and here also we have lone pair, right? So we have got lone pair, lone pair repulsion. To reduce this lone pair, lone pair repulsion, what happens? One of this side, one of this group or this group will rotate, right? Across this carbon-carbon single bond. So you see here, to reduce the lone pair, lone pair repulsion, the group, the half of this molecule flips in a different direction so that the repulsion reduces. So what happens here? Once lone pair, lone pair repulsion, the molecule rotates, half of the molecule actually rotates. So when this rotation takes place, then the molecule is this C double bond O and C. This is the molecule we have. Now in this when you write down the in all form of this, that would be possibly this. Only one C double bond O group will take part, both will not take part in the reaction, that is not possible, right? So we will have this product, correct? This is one case we have discussed. Now another case is what when we have cyclic compounds? Why I am discussing both together that you see, then we will see the percentage content of this. Cyclic compound like this, of course, OH double bond O. This is the product we get. Now here you see what happens. This molecule, here the rotation is hindered, right? Here the rotation of this like this, the rotation here is not possible in case of cyclic compounds, because the rotation is hindered because of this ring, correct? So the C double bond O cannot go to the opposite side here. When this rotation is hindered, this molecule will get stability through this hydrogen bonding. So that's why this compound is highly stable than this and the content of this is almost around 100%. This will be 0%. But here since there is no bonding, so stability of this is quite lesser than this, so it will be 0% and this will be 100%. Remember again one thing in mind and we'll discuss that truth you just give me a few more minutes, okay? When methyl group is in between then we'll discuss, okay? We are coming to that only. Few more minutes, right? So you don't have to again compare these two. What you have to keep in mind that in cyclic ketone, right? The enol form is more stable than it is in open chain compound, right? So this comparison you must have. You must keep in mind, right? So because of hydrogen bonding, so one factor is what? See the first factor is what? The bond energy of C double bond O and C double bond C we have discussed, right? We have discussed about entropy also which is not in our syllabus, but we have discussed this. We have discussed now hydrogen bonding, the third factor. All these factor affects the flexibility of the enolic form, right? So that's why these three factors we have discussed so far, okay? In the ring what we can write here we have restricted rotation, right? We have restricted rotation here and that is why the molecule or the product is hydrogen stabilized through hydrogen bonding, okay? Now the second case in this we have in this the case B when we have 1,3 di ketone di ketone. For example, you see the molecule is CH3 C double bond O CH2 C double bond O, CH3 This is what you were talking about Shruti, right? Now you see this, see again because this carbonyl group this hydrogen is also acidic this hydrogen is also acidic and this hydrogen is also acidic because of all this carbonyl group. Out of these three 1,2 and 3 which one is the most acidic acidic hydrogen? Can you tell me? The previous one see actually what happens first of all we are comparing this and this, okay? So since hydrogen bonding through hydrogen bonding the stability will be there so all these molecules will convert into enol form like this, okay? And here what happens, see again it is why it is 0% or 100% this is the factual thing we have it's a fact, okay? Like I said it is a spontaneous process we are not doing anything into it you take this molecule in a sample you take this molecule in a vessel okay and after some time you try to measure the content of this in the same vessel for this molecule if it is changes, right? then some of this has been converted into enol form but in case of this 1,2 diketone that we have and if the molecule is a chain molecule we have aliphatic compound then it won't convert into the enolic form, right? Because there is no you know since the flip is there because of this repulsion and through this flipping the product that we get the enol form we get the diketone stable because of see here we have pi, sigma, pi conjugation is there right here and here also we have conjugation like this lone pair, pi and sigma right? so this lone pair is coming into bond pair and if you see here the stability here number of covalent bond is more over here right? one thing is what if you consider you see this first of all why it is 0 and 100% that is of experimental fact okay that's what I have given you this but if you try to understand this two things you can see over here one thing is what this is resonance is stabilized, correct? first of all we have two C double bond O group here here we have only one and we know C double bond O the bond energy is more than to that of C double bond C so we have two C double bond O and this one is more stable second thing is what? this is resonance is stabilized right? when I say resonance you can ask me sir here also we have resonance here also we have resonance correct? so this resonance then this lone pair is involving resonance this is one type and this is another type so this gives you the cross resonance so cross resonance is highly less stable than the normal one so with all these fact you see also this one is more stable through bond energy also this one is more stable since this compound is more stable here so this compound has very less tendency to convert into its in all form that's why it is zero percent correct? now the next one that we were discussing is one three diketone right? so you see here like I said which of these hydrogen is most acidic actually this which is the middle one because this hydrogen is affected by both carbonyl group equally right? but the effect of both carbonyl group on this and this hydrogen is not that much okay? so this is the most acidic hydrogen we have and this is the methylene group and we call it as active methylene group active methylene group active methylene group so whenever the methylene group is present between two carbonyl group it is always active methylene group and hydrogen always comes out from this carbon not from these two to get the stable one okay? so the product that we get here is this one of this hydrogen comes out from here and that will attach any one of this oxygen atom so the product will be this CH2 C single bond OH double bond CH single bond C double bond OH CH3 okay? so that percent is contained of this is 76 percent here rest is this one so this one is more stable here the enol form here it is more stable so you cannot say that always keto form is more stable it depends on what compounds what group you have now why it is more stable so there are two three logic you understand the first one here it is intramolecular hydrogen modeling okay? so intramolecular hydrogen modeling how it takes place you see this molecule I am going to draw it over here CH double bond C here we have CH3 and here we have CH3 so this is first carbon second carbon is this sorry we have CH3 here then we have CH3 here this CH3 here and we have this so we have intramolecular hydrogen bonding present here like this and because of this only it is more stable because of intramolecular hydrogen bonding yes yes yes always hydrogen comes out from the active methylene group in this case one three diketone always so like I said in this there is no we don't have any thumb rule into this depends on what molecule you have because of the minus R group of this minus I group of this these two has electron with drawing group now so the effect of for this carbon atom to lose H plus is dependent on the nature of I group here so both I group is you know it is strongly affects the acidity of this carbon atom this hydrogen atom because of both this carbonyl group this hydrogen becomes most acidic so obviously this will come out as H plus always so first reason is what intramolecular hydrogen bonding possible like this there is one more reason we have and that is the large size conjugation or large conjugation also you can say large conjugation which means I will show it here only this lone pair is in conjugation with this pi bond and this will come over here this will come over here and then this will come over here large size conjugation we have this kind of conjugation is not here can you see any conjugation here because this resonance also this is more stable third part is what due to hyper conjugation stability due to hyper conjugation it depends on the number of alpha hydrogen so there are three alpha hydrogen present so you can draw three hyper conjugative structure right so that's why this is more stable so these are the three things we have intramolecular hydrogen bonding conjugation which is nothing but resonance and hyper conjugation all these affects the stability of the enol group right so this is very very important this active methylene group that we have in this there are many different you know possibility like we can have different different group present like this you see the second example I'll take into this one instead of methyl suppose we have phenyl group pH C double bond O CH2 C double bond O pH its enolic form is what pH C single bond O H double bond CH C double bond O pH right now in this what we can say because of this phenyl group we have what we have extended conjugation with this also we have conjugation with this because of the extended conjugation here this compound is highly stable than the previous one the previous product and its stability is almost around 96% its composition is almost around 96% its very stable this one so it depends on what groups are attached with it suppose if I replace this by hydrogen then the product will be H so you compare the stability of this this and the previous one where we have CS3 here right what happens here because of this hydrogen which is smaller group here in comparison to CS3 the hindrance is less over here the previous the first example that I have taken when we have CS3 present here correct so we have less hindrance but there is no extended conjugation into this one okay so obviously this one is more stable than the first one but less stable than this second one so first one is 76% this one is 96% so this one will be roughly around 84% conjugation okay one more example you see here in the same right CH3 C double bond O CH C double bond O CH3 here also we have CS3 so what is the difference between the first and fourth the first one we have CS2 here and here we have CS3 correct so when you draw the in all form of this the product will be CS3 COH double bond C CH3 C double bond O CS3 right so when you draw the structure of this here also the intermolecular hydrogen bonding is there so this carbon is here this carbon is here C double bond COH single bond C double bond O here we have CS3 here we have CS3 and here we have CS3 so here also we have intramolecular hydrogen bonding right but at the same time what happens we have repulsion also in these three methyl group I am comparing the first and this one the first product and this product right I am continuously telling you we are comparing the stability of product correct so this one is highly stable because here we have extended conjugation in the first one we don't have extended conjugation the first one is 84 this one is 96 here we have here we have less hindrance then we have CS3 present over there so its composition should be more than the first one so first one was 76 so this one is 84 right here if I replace one of this hydrogen atom by CS3 we will get this product but since this methyl group is what we have the steric hindrance over here then this composition is lesser than or it is very less than very like it is the composition of this is very small right and that would be somewhere around 7% ok so very less than the first one ok because of this hindrance of this methyl group right however hydrogen bonding is there right so again you can ask me sir hydrogen bonding is there then why not this should be more than this you can ask me but there are two factors for hydrogen bonding is providing stability to this but the steric hindrance is providing instability to this right because of this hindrance this molecule is becoming unstable so which two factor is dominating depending on that the composition of the product will be right so here this methyl group is the hindrance among this methyl group is dominating over the hydrogen bonding factor so that's why it is only 7% right again if you replace this methyl group by any other larger group then the composition of enol will be 0% for example I'll give you one more here if this methyl group is replaced by any larger group then the composition will be of enol will be 0% so it is probably the fifth one suppose we have C double bond C and here we have isobutyl group isobutyl isobutyl so this group is obviously it is quite large right so for this case the percentage enol content will be 0% enol because of the hindrance in these three right so hindrance is also one of the factor we have steric hindrance one more example you see we have CH3 C double bond O CH2 C double bond O O ethyl okay this is what this has minus M effect weak but yes it has minus M effect just a second just make a small change here this is not 7% this is 44% actually make this change this is 44% lesser than the first one so it is 44 not 7% by mistake I have written that okay make this correction not 7% this is 44% 7% is this one I guess because of this minus M effect of this group the percentage enol content is also lesser and that product will be CH3 COH double bond CH single bond C double bond O O ethyl group right this percentage of this group will be 7% here minus M effect of this so what happens here see you can also think like this you may get confused I'll just explain that part also you can think sir we have this is the minus M effect we have so it will you know withdraw the electron pair toward this side so the removal of H plus will be easier from this carbon is it understand this tell me right what is the doubt you may have that this is the minus M group so this can withdraw the electron towards its side so the removal of H plus will be easy over here right so if it is easy then this would be more acidic and percentage in all content should be more correct but the point here is what because of this minus M group that you have it will reduce the electron withdrawing tendency of this carbonyl group because the carbonyl group is directly attached to this okay you see this is also pulling electron towards its side this is pulling electron towards its side this also has pulling electron withdrawing tendency so both of these group are competing with each other right if this group is able to withdraw electron if it is then this will withdraw electron from here from this group the effect of this group will be more over here right so what we can say the minus M group that is present over here that will reduce the electron withdrawing tendency of C double bond O it's not minus M it's plus M plus M or minus I we can say because of the lone pair on this oxygen correct yes tell me quickly is it clear because it has plus M and minus I not minus M correct so which one is dominant over here plus M right plus M is dominating so when plus M is dominating so it has electron withdrawing tendency so when it has electron withdrawing tendency so electron withdrawing group of this electron withdrawing nature of this group will get reduced now means electron withdrawing nature of this group from this carbon atom that will be less because it is getting electron from this side already correct so because of this electron releasing group the electron withdrawing nature of this group from this carbon atom that is methylene group will be less and hence the tendency to lose X plus will also be less and hence the percentage in all content will be less is it clear tell me is it clear the point here it is what whatever factor we have discussed till now that hydrogen bonding, steric hindrance and all this is the another factor that is electronic effects depending on plus M and minus M you can understand whether the tendency to lose hydrogen H plus ion is more or less and then you can say okay this is again another case we have now the next example you see which is this one this molecule it's double formula is what double bond here this bond will be as it is and we have OH here means this hydrogen will jump over here you'll get this now you tell me which one is more stable here in these two in all form is more stable why so see this molecule has 4 pi electrons and hence it is anti aromatic so it is unstable the percentage content is 0% so this is the another factor we have aromaticity is it clear now another factor we got here is aromaticity by in all form in all form is the compound is gaining aromaticity it means it is the most stable case which we have discussed already in case of phenol the first case correct so this is the another factor aromaticity also you have to check if possible like this example you see the 8th one I have very strong feeling that you will get one question from this in the coming exam in March because they haven't asked any question on this so you may probably get one question on to this so must remember all these examples okay in this the in all content in all formula will be this this compound is aromatic compound and hence it is 100% one more example you see in all content will be this this is also anti aromatic 0% understood this few questions you got this time with aromaticity I think these kind of questions was there which of these molecules are aromatic like this few questions were there I guess minus this kind of questions were there in the exam I saw few papers I think this was the one of the questions which one Swetha the first one this is actually two all these are conjugation right all these are conjugated system two four six eight pi electron we have directly so four and pi rule it is following correct but the total since all these are in conjugation so total number of pi electron is what two four six eight will count all these bond because it is not more than two ring that we have here we have discussed aromaticity in the class no there we have discussed if we have more than two ring present and will not count the you know will count only peripheral pi bond this one they have asked this question no in the exam another example you see in this the possible product will be this or depending on the probability of these two product which one is major which one is minor first one and the second one okay why the second one is minor major sorry second one is second one is more stable see actually in this molecule this is the major product and this is the minor product more substituted which one amok which one second one you are saying major no second one is minor I will tell you why see no matter what answer you are giving now whether you are able to give the right answer or not but from today onwards after when we finish this class you must keep all these factors and examples in mind okay don't miss it you see here when you have double bond here right this carbon becomes sp2 hybridized right means the bond angle should be what bond angle should be 120 degree right and because of this cyclo propane that you have here the bond angle here is 60 actual bond angle is 60 because of hybridization it should be 120 so bond angle should be 120 because of hybridization but it is 60 over there we obviously have little bit of angle strain over there angle strain you understood this bond has tendency to form 120 degree of angle but it is bonded to 60 degree only there will be a strain in this bond so that it will convert into 120 right so that is the angle strain we have and when angle strain increases stability decreases more angle strain less will be the stability right but here you see what happens here also you can say sir there is angle strain correct but this carbon atom is sp3 hybridized so for sp3 hybridization bond angle is 109 degree 28 minute so obviously angle strain here it will be lesser than the angle strain here correct so because of this less angle strain here this one is comparatively more stable than the second one is it clear all of you understood this so what all factors we have discussed so far we have discussed hindrance we have discussed edge bonding we have discussed aromaticity then we have discussed angle strain we have also discussed effect electronic effects M effect and all electronic effects so all these factors affects the stability of in all form correct so you must keep all these factors in mind and we have discussed around 10-12 examples if you remember this 10-12 examples you can solve any question based on this they will not ask you anything more than this whatever I have said so far okay you must remember this 10-12 examples and all these factors that will be good enough okay on the same thing angle strain I will give you one more question one more example we will see this one also I will write down this is a possible product one product is what when we have double bond here and here we will get OH means what when this hydrogen will take part into tautomerism this double bond O will be as it is oxygen and this bond will be other another possibility is what when this hydrogen takes part in tautomerism then we get what OH tell me the stable one or one and two which one is more stable first one is more stable the reason do we have hydrogen bonding possible in one and two angle strain hydrogen bonding is possible yes hydrogen bonding is possible here you see here also hydrogen bonding but hydrogen bonding is here also in both the molecule hydrogen bonding is possible okay but first one is the major product we have the reason being what because of less angle strain right so the reason for the first compound to be more stable is less angle strain not hydrogen bonding because hydrogen bonding is present in both the compounds correct right so all these factor you must remember okay you can have any molecules but when you remember these factors you can understand and with this in all content also you remember whatever examples I have given you percent is in all content you must remember because factual question you don't you never know what they are going to ask next okay so percent is in all content you must remember I have seen few question papers of you know career point they have asked this question directly in matching J advance type in matching matching match the column they have asked this question they have given four compounds here and the percent is in all content you have to match those right I have seen those questions in a question paper of test paper of career point okay so you must remember that percentage in all content and then whatever the factors which affects the stability of in all content correct so this is it for tautomerism okay if you remember these all these examples one more thing which I will not discuss now but just I'll tell you we have discussed in biomolecules fructose right reaction of fructose we have discussed and it contains the major functional group here it is ketone but then also but then also it gives tolling's test then also it gives tolling's test the reason I have discussed already because because of tautomerism only the fructose form aldehydes and that aldehydes gives tolling's test converts into glucose and mannose if you remember right remember this fructose due to tautomerism only actually converts into glucose and mannose right and hence it forms tolling's and it shows tolling's test correct so they may ask you in tautomerism they may ask you this question also fructose related question also so if they ask this they'll cover both the chapters biomolecules and tautomerism both cover they'll cover over here okay so this fructose part tolling's test you must remember it is important okay understood this so we'll take a break now okay after this we'll start in 11 after 15 minutes we'll start okay 11 15 11 20 we'll start okay we'll solve some questions we'll solve some questions now okay tautomerism we have finished we'll solve some questions on some chapters I'll give you some questions to solve correct can we take a break now okay so we'll start at 11 20 have you whatever whatever you want okay we'll start at 11 20 okay hello can we start now okay j e previous year paper you have solved no at least five years already yes j e previous year paper you have solved right correct so in this one month you try to solve past at least past five years neat questions do you have that question paper like previous year paper for need exam chemistry do you have that okay so if you have that then you start solving that okay if you don't have let me know I'll just send you the question papers tell me the answer I think this one we have solved already let it be this question we have solved B is correct I'll give you another question I think this set of questions we have done okay you see this question equilibrium we have done I think B is correct you see this question the density of density in gram per ml of 3.6 molar H2SO4 solution solution having 29% H2SO4 by mass 1.64 1.88 1.22 1.45 I have given you one direct formula of this in solution chapter itself is it C 1.22 Simon got C Amog what is the answer Amog, Vishisht, Vaishnavi, Shruti Kushal what is the answer Ramchandra what is the answer tell me Sanjana Aditya what is the answer C is correct 1.22 check your calculation Aditya see we have this question suppose we have 100 gram of solution right in this 100 gram we have 29 gram of H2SO4 and rest is what 71 gram is solvent right so molarity is given so with molarity you can find out the volume of solution right density will be what mass by volume so mass is 71 divided by volume that you have here this is what you have done 1.22 is the answer it is 29 by 98 into 100 1000 so volume is equals to molarity is 3.6 into 98 divided by 29 into 1000 so density will be 71 into 3.6 into 98 29 into 1000 is it 21 or 20 it is 29 percent not 21 Shweta is it correct correct 1 by this that is what I was also thinking so density will be mass by volume so 71 divided by 29 into 1000 into 3.6 98 is this 71 is correct we cannot take 71 here because you see density of solution we have to find out 71 is the mass of solvent so we should take here 100 so you see this becomes what 100 into 3.6 into 98 divided by 29 into 1000 so you see what is the here what is the formula we can write this 29 is what it is a percentage weight of solute so this becomes density is equals to 3.6 into 98 divided by 29 into 10 okay and molarity we can write from this is molarity right we can write what molarity is equals to density into percentage weight of solute into 10 this whole divided by 98 is the molecular mass of solute this is the formula we have density percentage weight of solute 10 divided by molecular mass so when you solve this you will get 2g per mn correct now next question did you understand this can we move on you have to plot this plot of I have done this in the class tell me the answer plot of 1 by xa 1 by xa and vapor phase we have 1 by ya is linear is linear whose and intercept are you have to find out the slope and intercept I will give you the option p0b by p0a p0b minus p0a divided by p0b option b p0a minus p0b this is the vapor pressure of actual of the pure one all this p0 term that we have p0a minus p0b p0b p0a by p0b p0b minus p0a divided by p0b p0b by p0a p0a minus p0b divided by p0b tell me the answer if you do not remember you can derive this also in one single step what is the answer ok I will give you one hint the partial pressure of a is equals to we can write a mole fraction in liquid phase into total pressure and that will be equals to mole fraction into vapor phase into the vapor pressure p0a you are getting a ok see this so this we can write the mole fraction of a in liquid phase is equals to the mole fraction of a in vapor phase vapor pressure divided by total pressure similarly mole fraction of b is what xb is equals to yb p0b by pt right this is one and this is two if you divide one by two because what happens here this total pressure we have to eliminate right or you can also put p0a xa plus p0b xb that also you can put ok so xa by xb is equals to what ya p0a by yb p0b we do not need b right xb yb we do not need so xb we can write 1 minus xa and yb we can write 1 minus ya into p0a by p0b now you can solve this and you can form the equation of 1 by xa and 1 by ya ok you will get option c as the right one yes correct Saivet option c is correct understood all of you next question right you see this how many ml 0.1 molar at cl are required to react are required to react completely with completely with 1 gram mixture of na2co3 and naHco3 containing equimolar amount of these two options are 157 0.8 ml 0.15 0.78 ml 210.4 ml 105 0.2 ml solve this got it ok what you do I will just give you some hint 1 gram mixture of na2co3 and naHco3 you have assume one of these is x so other one will be 1 minus x since they have equal number of moles so equate their number of moles you will get x from there two different reactions both na2co3 and naHco3 will react at cl will react with hcl separately like this you see see what happens here and na2co3 combines with hcl gives what we will get over here naCl plus h2o plus co2 right if you balance this we will get two moles of hcl two moles of naCl similarly naHco3 also combines with hcl forms naCl plus h2o plus co2 so both in the both reactions the volume of hcl we are using here since the volume of hcl we have to find out right and two volume of this we are using so when you compare this and this the volume of na2co3 will be used here that will be twice to that of naHco3 because the ratio is 1 is to 2 here 1 is to 1 here for 1 mole we use 1 for 2 mole we use 1 so like that so what you have to apply here the m1v1 for hcl is equals to m2v2 for na2co3 plus m3v3 for naHco3 right this volume you have to find out okay nothing see the volume of this is not given naHco3 and na2co3 so we will assume the volume as 1 and 1 over here but the point here is what since the reaction reacts completely what is given completely reacts right so two volume of this reacts with two volume of this okay that's what the point we have okay so two complete reaction to complete this reaction we require two volume of na2co3 and one volume of naHco3 correct so here the molarity of hcl is given which is 0.1 into v1 we have to find out we will write volume of na2co3 is 2 molarity we do not know plus 1 into molarity of this we do not know v1 we have to find it out so what we can say since we have 1 gram mixture of this na2co3 and naHco3 we have out of this suppose 1 gram of na2co3 we have then 1 minus x gram of naHco3 will be there so equal number of mols we have so number of mols will equate x divided by the molecular mass of this will be 106 is equals to 1 minus x molecular mass of this will be 84 you solve this for x will be 0. 5 578 approximately this is the mass you will get the number of mols of na2co3 will find out because molarity we have to find out and that will be equals to the number of mols of naHco3 which is equals to 0. 5578 divided by 106 and this will get 0. 005 26 mols right molarity will assume volume will assume as 1 over here so concentration molarity will be this only number of mols is nothing but the concentration that will substitute over here so 2 into 0. 005 26 plus 0. 005 26 you solve this equation you will get v1 as 157 157.8 ml so this is the answer we have here yes right option A is correct next question you see the aqua solution next one an aqua solution of 2 percent weight by weight weight by weight non-volatile solute solute exerts a pressure exerts a pressure of 1.004 bar boiling point of solvent find out the molecular mass of solute molecular mass of solute is equals to what options 0.3 655 36.55 41.34 40.16 what is the answer 41.5 41.5 yes it is close so c is correct what happened others I think you did not revise this chapter properly these are direct formula waste actually should I explain it we do not have time tell me the last question what happened what happened you see this first of all I have already told you that in this type of questions just write down the data first okay so you see exerts a pressure so the solution pressure is given that is p is equals to 1.004 bar the solution pressure is given right now the mass of solute so the weight of solute is what I assume 2 gram so weight of solvent will be 98 gram right and since it is aqua solution it is so the molecular weight of solvent is 18 which is nothing but water molecular weight of solute we have to find out what formula we can apply p0 minus p divided by p0 is equals to number of moles of solute nb by na number of moles of solvent this formula we can apply right this is the formula we have p0 minus p p0a minus pb is minus p divided by p0 is equals number of moles of solute by solvent which is nb by na so this pressure is given already 1.004 number of moles of solute you have weight by molecular weight so molecular if you know this number of moles you can find out molecular weight also that is what we have to find out and number of moles of solvent is already there 98 by 18 correct only thing is not given here which is p0 it is not given directly but indirectly it is not given because it is saying that exerts a pressure of 1.004 bar at the boiling point of solvent so what is the boiling point of solvent solvent is H2O boiling point is 100 degree celsius and we know water at 100 degree celsius will be at equilibrium right equilibrium which is vapor form right so this is the equilibrium condition and that is only possible when the vapor pressure of solution is equals to the atmospheric pressure correct vapor pressure of solution is equals to atmospheric pressure so the vapor pressure of solution will be what is equals to the atmospheric pressure which is nothing but 180m and in bar if I write down it will be 1.013 p0 is this I have done this kind of question that when equilibrium is given or boiling point of water when solvent boiling point is given pressure p0 we always take is equal to the atmospheric pressure which can be 760 mm or 1 bar 1.013 bar till here is it clear tell me now you have to substitute this value here p0 is 1.013 minus p is 1.004 divided by 1.013 is equals to number of moles of solute divided by number of moles of solvent you substitute this value you will get number of moles of solute from here which is nothing but the weight of solute is 2 gram molecular weight you have to find out ok and then you will get the answer so in this kind of question solution question you just write down the data and you try to apply the formula ok you will understand what you have to do what is given and what you have to find out ok but first I always told you this that first you write down the data like this you will understand what formula you have to apply and one more thing you must keep in mind that when the pressure of pure solution is not given ok then you have to take pressure is equal to the atmospheric pressure there is no other way to do this ok if it is not given then you must think on this that the pressure must be equal to atmospheric pressure then according to the pressure given unit of the pressure you can take the value 760 mm or in bar ok this vapor pressure of pure solution it is the experimental thing you cannot tell that the pressure of solution will be this ok it should be given in the question if it is not given then by any means they will make you understand this that the solution pressure is equal to the atmospheric pressure ok and that is how they have mentioned over here that is what they have mentioned over here boiling point they have mentioned boiling point is only defined when the pressure exerts by the liquid is equal to the atmospheric pressure ok the vapor pressure of the liquid is equal to the atmospheric pressure then only the liquid starts boiling it means the pressure of solution here is equal to the atmospheric pressure and that is nothing but 1.013 bar is it clear Is it clear? Did you understand this? Right, so we'll wind up the class here only, okay. Next class, we'll discuss few things in hydrocarbon. Hydrocarbon, we haven't discussed, right? Did we discuss hydrocarbon? So next class, we'll discuss few reactions of hydrocarbon and see some questions onto that, okay? Because hydrocarbon reactions are very important for all other organic chapters, okay? And I'll suggest to you again to solve last five-year question, if you have solved, just go through the NEET or any AEMS paper, last five years, papers of NEET and AEMS exam, okay? Chemistry, at least you must see that, okay? Because there you'll get some factual question. And again, one thing, whatever the factual question you get, you just note it down in your notes, okay? Don't forget to do that, okay? So that you can revise those factual questions in last one or two days when the exam was there. Correct, so we'll stop the class here only. Thank you for joining, see you soon. Bye and happy bundle, all of you, enjoy.