 Hello, good afternoon. Hello guys, how are you doing? Please type in your name all of you. Okay, so good afternoon everyone. Your exam is finished, right? 12th board exam or one exam is left and that is probably on 2nd of April, right? Okay, so today you see we are going to discuss, yeah, we are going to discuss few questions and it is of general organic chemistry. GOC will discuss, okay. So there are few more things which we need to understand and discuss there. So with questions only, I will try to explain those things in between, okay, wherever it requires, okay. So because theory we cannot cover, there are so much of theory in GOC also, so we do not have that much of time so that we can cover all those things. But I will give you the questions like we were, no, holding this kind of session, no, it is a similar kind of things we will do over here, problem solving session it is, right? So with that particular problem what theory is required, right, and some additional theory, right, that we will discuss today in this session, okay. So we will start today with these questions. So we got this, not this one, this is again, yeah, this one, okay. You see these two questions, tell me the answer. The first one is A, right, yeah. So the correct answer is option A and why do we have this option A? The reason behind this is what? Like I said with question only we will try to discuss those important, no, concepts. Since the electronegativity of fluorine is maximum, yes, since the electronegativity of fluorine is maximum, that is why option A is correct, fluorine and then oxygen and then nitrogen, right. So please take care of this thing that we are comparing the first atom here, F, O and N, okay. I will come to the second question but first you tell me here if I ask you the minus I effect of this NH3+, this two, the minus I effect of NH3+, and NR3+, suppose the first compound that I have written is A and B. Which one has more minus I effect, A or B? Which one of these has more minus I effect and why A, why NH3+, will have more minus I effect? I will discuss this, right, a bit later. You tell me this also OH and OR, A or B, minus I effect, minus I effect, A again, right. This two, minus I effect and these two also, minus I effect. For this one, COCl, for this one, so fourth one is A you are saying, Aditya Amogh, fourth one is A, third one you are saying B, first one you are saying B and second also you are saying B. So the A, third one is A, minus I effect, Amogh, third one is A minus I effect, okay fourth, right. So you see if third and fourth I will discuss first it is straight forward. The difference here we have chlorine and hydrogen and chlorine has electron withdrawing nature. So obviously COCl as a whole will withdraw more electron, okay. So this is the order. Similarly this R group will have electron releasing effect. So this will decrease the electron withdrawing tendency of this whole group in comparison to the COOH and the COOH will withdraw more electron, right. This two if you see R, this alkyl groups are electron releasing group. So this will also decrease the electron deficiency of nitrogen and hence this group withdraw less electron comparatively from this, right. But the order here is found to be this, right. NR3 plus is more shows more minus I than NH3 plus and here also the order is this. I think all of you have given A for these two, yes. These two are actually exceptions, okay. These two are actually exceptions. So this you must remember. Only these two groups we have exceptions. Logically if you think alkyl group has electron releasing tendency. So this group should have less minus I than this group. Similarly here also you can see. But the order is found to be reversed, right. It is not that logic that we are applying here. First of all you have to memorize this, okay. If you ask me the reason for this like I said it is an exception. But in few books if you study they have written what this are the carbon here in nitrogen, carbon nitrogen and carbon nitrogen. There are p orbital involved in bonding here. p orbital involved in bonding, right. So for this nitrogen if you talk about the percentage availability of orbital here. This nitrogen we can say since p orbital is involved. So percentage s character of this nitrogen is more, right. Since comparatively p orbital is involved in bonding, right. So for this the percentage s character is more. It is more available over there. But here hydrogen has s s s orbital. So here the percentage s character will be less if you compare with this one, right. Since this group has more percentage s character and that is why the electron with drawing tendency is more. The same logic will apply here also, right. So this is written somewhere in the book, few books it is given. But like I said reason they won't ask you you have to memorize these two exceptions here or minus i effect. And for all of the groups the logic that you are putting in for these questions similarly you have to do, okay. There is no exceptions apart from these two. Got it. Yes, let me know. We will move forward then. NR2 and NH2. For that it will be same like NR2 and sorry, R2 and NH2. For this, this will have more minus i effect, okay. So only these two are exceptions. Like if you apply the concept of percentage s here also then you will get confused. That's why I told you these are exceptions, okay. Correct. Next compound the polarization of electrons in acrylene. Acrylene molecule formula is this. It is vinyl aldehyde C double bond OH, right. So this has electron with drawing tendency. So oxygen will have delta negative, carbon will have delta positive. Option D is correct. This is what you have said all of you. Okay, next question we see. Tell me question number 8, 9, 10, 11. Eighth one is B. I won't discuss this because it is tertiary, right. Answer is B. Reason is very straight and simple. It is three degree. This is two degree, one degree and one degree. If they ask you to write down the order, order will be option B most stable than A and C and D, the stability is almost same, right. Why ninth one option A is correct? Ph group act as an electron donating. Final group are not electron donating Aditya. Final groups are electron with drawing. You say, yeah, final group is electron with drawing. You must remember this. This is a very important order. If you, you know, do you have the minus i order with you? NF3 plus, NF3 plus has maximum electron with drawing tendency. It is the most powerful minus i group NF3 plus, okay. Final, if you ask me, the order is this C triple bond CH. Triple bond is most electron with drawing. Then we have phenyl group and in the last we have alkene CH double bond CH2. This order you must remember. This is very important. Got it? So the ninth one you see option E answer is correct, but it is not because of inductive effect. What is the reason for this stability? Why you are saying C Amogh, Loretta, Kushal? Why it is C? Tell me. Tell me the reason. Why it is C? Guys don't do this kind of mistake. Okay. See, you try to understand one thing. Okay. We will discuss this question a few minutes later. Okay. First you try to understand this. Do you know this thing? Suppose this compound, okay. And here we have, suppose OH present, phenyl I am taking and at this carbon, we have NO2 attached with it. Right. Okay. NO2 attached with it. Now you tell me here in this molecule, what all electronic effects, what all electronic effects are there? The question is this. What all electronic effects are there? You know what is electronic effect? You know what is electronic effect? Electronic effects are inductive, resonance, hyper conjugation, electromeric. Correct? All these are electronic effects. Now you tell me at this position, NO2 is attached. So because of this group, what all electronic effects are actually acting here? Resonance and inductive. Correct? And what kind of resonance we have here? What kind of resonance will have here? Plus R or minus R? Plus R or minus R? Vashnavi, only resonance. Only resonance, Vashnavi, only resonance, time here. You know what all groups shows plus R or minus R? See, okay. I will just, can I discuss these questions a few minutes later because I think you need to know something over here. You are confused, some of you are saying plus R minus R. So I think some basic discussion requires over here. Okay. So can I discuss few concepts over here, then we can go to these questions? Or you want to solve questions directly? Tell me guys first. Okay. See, I will come to this page again. First you see this. I will go a bit fast. Okay. I will go a bit fast. See, first of all, this is the benzene ring. Any group is attached over here. All of you know this. This is ortho position. This is meta and this is para. Even this is also ortho and this is meta. We have two meta and ortho position, one para position. Okay. The thing that we are going to understand here, that which groups can show plus R and minus R, that we are going to understand. And because of that, what happens? Okay. So like I said, plus R or plus M are electron releasing group, ERG minus R or minus M are electron withdrawing group, EWG. Right. Now, electron withdrawing group are those groups which can satisfy these three conditions basically. What are these conditions? I will tell you condition. First condition is what the first atom with which the benzene ring is attached. I am taking random some molecule here. Between first and second atom, we should have multiple bond present. Correct. And this is the case with, like I said, multiple bond. So we can have triple bond also. Right. And both for this, the electronegativity of Y should be more than the electronegativity of X. If this kind of group is attached to the benzene ring, then they are electron withdrawing group. Correct. Like for example, C triple bond and cyanide group. Right. Nitrogen, benzene ring is attached with this carbon and between first and second atom will have multiple bond and nitrogen, obviously it is more electronegative than carbon. Okay. Second example, you see N O2, N double bond O and this O. First and second atom, we have multiple bond. Oxygen is more electronegative than nitrogen. Electron withdrawing. Correct. One more condition we have here, which is usually we don't counter this thing, but you should keep this in mind. Like suppose the atom with which the ring is attached, any element, I will write here, suppose M, this should have vacant P orbital, vacant orbital on it. For example, we have CH2 plus or vacant P orbital or we have VH2. It has vacant P orbital on it. So these three, under these three conditions only group can show electron withdrawing nature. Understood. Is it clear? Till here. Right. So if you keep this condition in mind, you don't have to memorize that group. This group shows plus R or minus R. Just you see N O2, you know, N double bond and O. So this condition is satisfying. So it is electron withdrawing. Is it clear? Tell me first. Okay. Right. Now, electron releasing groups you see, like I said, ERG, electron releasing group. For this, the condition is what? We should have the molecule or the atom with which the molecule is attached with the ring that must be electron rich. Then only it can donate electron. That's a simple logic you can understand. Those groups which are electron rich, those groups can only donate electron, which are electron deficient cannot donate electron. So for that, what are required? The atom with which the ring is attached atom must have, must have at least one lone pair on it, at least one lone pair on it. For example, you see halogen, how many lone pair, three lone pair, right? Electron releasing group. We can have OH again, electron releasing group. We can have NH2, one lone pair on nitrogen, electron releasing group. We can have OCS3, OR I can write, electron releasing group, right? We can have OCOCS3, CS3, electron releasing group, NR2, electron releasing group, NHCOCS3, double bond OCS3, electron releasing group. All these groups you see, the first atom has one lone pair of electron. All these will have electron releasing nature, understood, right? Here also you can write down few more examples. Aldehyde group, C double bond OH, electron withdrawing, acyl halide, electron withdrawing, right? C double bond OOR, electron withdrawing. So just the condition is what first and second atom we should have multiple bond, correct? Now coming back to the second thing. Now you understood which all group shows plus or which all shows minus R that you can, you know, understand, right? By looking at the group. Clear? Okay. Another point here is what, like I said, you see this OR and OH are electron withdrawing group, electron releasing group I am telling you here, right? But if you remember the first slide here, the first slide, I have given you this example and I asked you, these are electron withdrawing group I told you here, right? OH and OR. So this is because of when I talk about the electron withdrawing nature, I am talking about inductive effect, right? But here we are talking about what? Since this oxygen has lone pair of electron on it, so this can also show resonance and these are, here it is, electron releasing group, correct? So if you take this example OH, right? So when the benzene ring attached with OH, so we will have an electronegativity difference between that carbon atom of ring and oxygen of OH group. So since we have this electronegativity difference, so obviously this bond pair of electron will be slightly shifted towards this oxygen atom and that is the electron withdrawing nature, which is minus I, correct? At the same time, since oxygen has lone pair of electron also, so it can also show electron releasing nature due to, due to resonance, that is plus R effect. Now the point here I am trying to make is what? One group can show minus I also, plus R also, minus I also, minus R also, plus I also, minus R. Any combination should be there, like one group can show plus I, minus R, minus I, minus R, plus I, plus R, minus I, plus R. Any of this combination is possible at the same time. But now the point is which effect is dominating? According to that only we will decide the property of the molecule or the stability or acidity, whatever it is, right? So whenever you have resonance which is plus R or minus R, this is very important, okay? This is very important. Whenever you have plus R minus R or plus I, minus I effect in that group, then always we have what? Resonance dominates over I effect. Resonance dominates over I effect. Is it clear? This you have to memorize. Why resonance is dominating? Can you tell me the reason behind this? Why resonance is dominating? Okay, is it clear till here? Plus R, minus R, plus I, minus I. And whenever you have resonance, I effect, resonance always dominates, correct? But here also we have one exception. Why resonance is dominating? Because in resonance we have complete transfer of electron. But in inductive effect, we have partial shifting of electrons, correct? Inductive effect, we have partial shifting of electrons, shifting of electrons. But in resonance we have complete transfer. So that is why it is dominating, clear? Let me know if you understood, okay? Okay, now another thing here you see. See if you have this ring, right? OH is present here. When you draw the resonating structure for this, when you draw the resonating structure, I am not going to draw all the structure, you can draw it on your own. You will get this, right? And further if you draw the resonating structure will have negative charge at alternate position, right? So in this case which is the plus R we are considering here. So because of plus R effect, ortho and para position becomes electron rich, becomes electron rich. Similarly, if you draw for minus R when you draw the resonating structure, I am assuming that you know this, how to draw this structure. That is why I am not drawing this, okay? If you still have doubt, you can ask me, I will draw one structure. Because of minus R effect, ortho and para position becomes electron deficient. Because then we will have positive charge at ortho and para position, right? So and in these two condition we can easily see that meta position is unaffected, is unaffected, right? Now the thing is what? Meta position is unaffected, it means any group which is present at meta position, any group present at meta position, meta position can show only inductive effect. It can be plus i or minus i. Is it clear? So whenever a group present at meta position, there is only plus i or minus i possible. Resonance is not possible. Even hyper conjugation is also not possible, okay? Only inductive effect possible. One more thing I told you here that the first atom when it is attached with ring has lone pair, then it shows plus R, right? We will take an example of halogen x. It can show plus R and minus i. Which one is dominating here? Plus R or minus i? Which one is dominating here? See, like I said, plus R and minus R always dominates over inductive effect with one exception and that exception is this only. In case of, no, in case of halogen, minus i effect dominates over resonance, okay? And that is the only exception. Apart from this, in all other cases, in all other cases, plus R or minus R always dominates inductive effect, right? So here you write down this condition is true except halogens because of high electronegative of halogen atom, okay? That is the one thing you must remember, right? So what we understood now? How to identify that the group shows plus R or minus R? First thing is that, okay? Multiple bond or what? Or lone pair present in the first atom. That you have to keep in mind and in resonance, only ortho and para positions are affected. Due to plus R effect, ortho para position becomes electron rich. Minus R effect, ortho para position becomes electron deficient. Meta position is unaffected, right? Unaffected due to resonance. And any group present at meta position shows only what? Only inductive effect. Now on the basis of this few questions I'll give you so that you can understand the concept and how to apply the concept, okay? This question you tell me, that stability order you have to tell me. Here we have CH2 positive charge. CH2 positive charge. Here we have OCS3 attached. And here we have OCS3 attached. Ortho para directing it's fine. I'm not talking about ortho para directing, same here. I'm talking about which effect is, which electronic effect will be there? Because of halogens, which effect will be there? That is what I'm talking about, okay? Now you tell me these two, which one is more stable here? Suppose this is A, this is B. Which one is more stable? First you tell me here, you see here, which effect is there at this position? Which effect we should have here? Which effect we should have here? Tell me. Plus R or minus R? Plus I. Sundaria, inductive effect, plus I or minus I? Plus I or minus I? Four-week, tell me. In the first one, which effect we have? Plus I or minus I? See, this oxygen, yeah, we don't have resonance here, fine. So we'll have I effect here, right? But which I effect will have? Oxygen is more electronegative than this carbon, right? So here we'll have minus I effect, not plus I. Here we'll have minus I. See, you have got confused with this lone pair, I guess. Oxygen is electron withdrawing, oxygen is more electronegative than this carbon atom. So this has electron withdrawing nature. We cannot write plus R here. Oxygen has lone pair, but we cannot write plus R because it is at meta position and resonance is not possible there at meta position. So we'll have only minus I. Here if I ask you, here we have ortho position plus R and minus I also possible, both at this position, correct? But we know plus R dominates over I effect in case of OCS3. So this effect will be dominating. According to R effect will find the stability of this compound and this according to I effect. Now we know resonance always dominates the inductive effect. The order in fact is this, resonance hyper conjugation and then I effect, okay? So because of resonance, the stability will be more, correct? So the order will be what? B is more stable than A. Understood this, got it all of you? What about this? This too. Which one is more stable? A, B. Which one is more stable? Which effect will have here at this position? Which effect will have? Tell me. Which effect will have here? First of all, you try to identify the effects, okay? Yeah. So here we have minus I is very simple and straight. Here we have only minus I. Here we can have what? Minus R and little bit of minus I also will be there. But obviously minus R will be dominating, right? And all these negative effect, whether it is minus I or minus R, that will destabilize the carbocation, okay? So here we have minus I, it will destabilize it, minus R will destabilize it, minus R dominates minus I. So this will destabilize more, right? Hence A is more stable than B. Got it? Suppose if I write down one more compound into this only and which is nothing but this CH2 plus, right? Furthermore, one more compound if I write down. CH2 plus and CH3. Suppose this is C and this is D. Now which one is the most stable carbocation here? Finial group is minus I. Finial group is minus I. But if it is in conjugation, then minus R also possible depends what compound we have, okay? So if finial group is in conjugation, then minus R also possible because of conjugation. If conjugation is not there, minus I. I did not get you Amogh. How do we know CH2 plus resonance is more dominating over NO2 resonance? No, I am talking about the stability of this carbocation, Amogh. The question is this. The stability of this carbocation, you see, I understand what is your doubt. The thing is what is common in all these questions? Suppose you get this question and the question is which one is most stable, right? So what is common in all these? This CH2 plus on benzene ring is common. Now the stability will change decrease or increase because of this CH3, this NO2 and this NO2. Obviously it is understood that when you add these groups at different, different position, then what is the effect on the stability of this carbocation? That is understood from the question. Now you got it, Amogh. We are talking about the stability of this carbocation at different, different groups we are attaching at different, different position. Got it? Okay, now you see the actual molecule is this, okay? And in this, if you attach CH3, NO2 at different, different position, then what happens on the stability of this carbocation? So obviously this NO2 and this NO2 has electron withdrawing nature. So this will de-stabilize the carbocation. So obviously A and B are more or less stable than C, right? And since here we have plus i that will stabilize this carbocation, D will be most stable. Understood this? Yes, tell me. So whenever you get this kind of question, this kind of question is very common in this, you know, CET, Comet K, whatever you write, those kind of exams, Manipal, Bitsat, this kind of question is very common and straightforward you will get. The only thing you have to be very clear with that at what position, what effect is taking place, right? For example, one more question I'll give you, this is also an interesting comparison, right? The only thing is what, let me freeze this first. Whatever the position you have, you write down the effects possible, plus i, plus r, minus i, minus r. Always remember one thing, resonance effects are dominating, minus r will de-stabilize the carbocation, minus i will also de-stabilize. Since resonance is dominating, so this will de-stabilize more. This will be more stable. Accordingly you have to go with this. Now one more question we'll see here and then we'll move on to the questions that we were discussing. CH2 plus, right? CH2 plus, CH2 plus. And what should we do? Suppose here if I attach chlorine, right? Here if I attach chlorine and here if I attach CH3. Tell me the order A, B, C. Which one is most stable? A, B, C. C is the most stable, fine. Answer is correct. C, B, A answer is correct. You tell me first here, here what effect we have? Here what effect we have? In C, what all effects possible? We'll have plus i here and at the same time we'll have plus h also, hyper conjugation also possible. And hyper conjugation is dominating over i effect. This will not change your answer, okay? But if they ask you what all effects possible at ortho and para position hyper conjugation is also possible. No resonance is not possible because here it is not in conjugation. There is no, you see there is no lone pair, there is no double bond. Sondarya, resonance is not possible. Methyl does not show resonance. Ortho position, we can have resonance hyper conjugation i effect all. But in case of methyl resonance is not possible. We'll have hyper conjugation and i effect. Here we have only i and here we have minus i and plus r, correct? Minus i and plus r. So minus i dominates in case of halogen, right? Plus h dominates over plus i. If you have any confusion in this hyper conjugation, we can discuss this also. You let me know. Next class, we'll discuss this hyper conjugation also. But I want to solve those questions also. That's why I'm not going into it. But the point is for hyper conjugation, we require alpha hydrogen. Even alpha chlorine also shows hyper conjugation. But in that case, we'll have minus h not plus h, okay? So if you want, you let me know. We'll discuss this also. So you see this plus h because of plus as the stability increases. So c will be most stable. A and B both shows minus i. And we know inductive effect are distance dependent effect. So effect of this chlorine, since it is closer to the CH2 plus will be more. And effect is what? Electron withdrawing. So A will be least stable then. So we'll have B and then A. Anyone has doubt in this? You let me know. Understood? Okay. Now can we solve that question? Okay. Now you see, like you said, you're considering i effect. That's why it was wrong. You see, what all effects are possible? And then you tell me the stability order. A molecule, you draw this. We have CH positive charge. And there's two phenyl group attached here, right? Here we have only one phenyl group. And here we have only i effect. Here we have only plus i. Here we have plus i. And here we have resonance. Here we have what? Do we have resonance here? No, we'll have only plus i. But that would, that is also very less, very negligible plus i effect. And this is also we have what? Resonance, right? So obviously we know because of resonance, the stability increases of this carbocation, right? And since we have two phenyl groups, so you can draw resonating structure with this phenyl group also and this phenyl group also, right? That is why this one is the most stable. More resonance in a, more number of resonating structure you can draw. And that's why it is most stable. Understood this one? Let me know if you have any doubt. Or if you have done, you solve the next one. Penyl ketine does not show any inductive effect. Penyl, yeah, it shows minus i is Vandhria. Order I have given you know, you see this is the order. Alkyne and then phenyl and then alkyne minus i effect. Tropylium cation, Amogad, so it's D. What is tropylium cation? Okay, sir, fine. Tropylium cation is this. This is tropylium cation. 1234567 compounds, right? This is tropylium cation. If you see, this is an aromatic compound. This is an aromatic compound. Okay, but the answer is not this. This is also one thing. Have you heard about dancing resonance? Have you heard about dancing resonance? No bond resonance, right? So you see two things, let me tell you first. Yeah, the answer will be A for this one. The answer will be A. However, whenever you have aromaticity, the molecule is highly stable, right? But here in this case, what happens? Dancing resonance, if you talk, we have tricyclo-propyl carbocation, this one. Tricyclo-propyl methyl carbocation, this one. Okay, fine. This one what happens, this cyclo-propyl ring is highly unstable because of angle strain here, correct? Highly unstable because of angle strain here. So if you talk about this CH2 plus, right? To gain stability, what happens? This sigma bond comes over here and it forms this positive charge here, double bond CH2. That is why we call it as, see, this is a special type of resonance in which sigma electrons are involved. No bond resonance is hyper conjugation. Sigma electrons are involved here. You see, usually in resonance, only lone pair and pi electrons are involved. There is no sigma electrons involved in lone pair, except in this case, cyclo-propyl, correct? So this kind of resonance, this bond, you know, this bond may jump from here also. That is also possible. So this we call it as dancing resonance, okay? And because of this, and why it happens, because this is highly unstable because of angle strain, right? So whenever you have to, coming back to this point, whenever you have to assign that stability of carbocation, we'll first give preference to dancing resonance, dancing resonance and then we'll see aromaticity. Then we have aromatic compounds, okay? But here in this case, what happens? If you try phenyl compound, then there are so many resonating structure you can write, okay? And that is, this is actually an experimental comparison. However, it is an aromatic compounds, then also because of mode number of resonating structure, try phenyl carbocation is more stable here, right? Why I have discussed this? Because if you see this kind of molecule, then its stability will be maximum. It will overpower the aromaticity also, this tricyclo group that you have. Similarly, here also you see, if you compare the, you know, stability of these two, obviously this A and if it's this D, A will be more stable because you have three cyclo-propylene, again more resonating structure you can draw. Did you understand this? Here, tropelium ion is this, however it is aromatic, but try phenyl benzene compound, right? And even phenyl also here it is aromatic, right? And because of mode number of resonating structure, A will be more stable. Got it? Okay? So this is an important comparison, this triphenyl you must remember this. This one is important, okay? Next one, 11th, question number 11. Fourth one is what? It is 3 degree benzylic, right? This one is 2 degree, this one is 1 degree and this one is benzylic carbocation, right? So stability will be what? This, 2, this and then this. So 4, 3, 1, 2, this is the order, option A, right? All of you have done this. Next question we'll see, tell me this, least stable resonating structure. Why does A like charges close to each other? Yeah, correct answer is A. All of you have done it, option A is correct because two positive charges are at adjacent atoms nitrogen and carbon, this will be highly unstable compound, right? You remember I have taken a class for relative stability of resonating structure. There are 7, 8 rules I have discussed, mode number of pi bonds, charge separation, octet, aromaticity, all these I have discussed. Okay, yeah, you must revise that, okay? Because those kind of topic it requires some revision, okay? You must revise that. Question number 28, what is the answer? Hyper conjugation is sigma pi resonance. Why it is sigma pi resonance? 28, 1 answer will be B, not A. Hyper conjugation is sigma pi resonance. We call it as sigma pi resonance. You see, I'll take one example. Why it is sigma pi resonance? One simple example I will take here. First of all, you see, there are two terms for hyper conjugation. I'll just, yeah, I'll just explain this hyper conjugation. Name you must remember. It is sigma pi resonance and we also call it as no bond resonance. Both are same thing. One thing you have to memorize that hyper conjugation is only possible in case of alkene or alkyne with carbocation and carbonyl, sorry, pre-radical, pre-radical. In these three cases only hyper conjugation possible, okay? And for hyper conjugation, the necessary condition is what? We need, we need sp3 hybridized, condition is this, sp3 hybridized alpha hydrogen, okay? More number of alpha hydrogen, more will be the hyper conjugating structure and more will be the stability. I'll come to that question number 29. First all of you try to understand this, okay? For example, you see, suppose if I take this CH2H, single bond CH, double bond CH2. Now in this example you see here that we have alpha hydrogen, right? This is alpha position, so we have three alpha hydrogen. What happens here, how this hyper conjugating structure will draw up? This sigma bond comes over here and forms a pi bond and this pi bond jumps onto this carbon. So we'll write what CH2H plus double bond CH, single bond CH2, one lone pair on it, negative charge, okay? Now you see why it is sigma pi resonance, because this sigma bond converts into pi, you see? Sigma bond converts into pi, correct? That's why it is sigma bond resonance. Why no bond resonance? Because there's no bond present here, but H plus will be there only. It will be there only, it won't go anywhere else. It is not coming out, right? Actually what happens, this three bond that you have, carbon attached with three hydrogen, some, this sigma bond, this sigma bond or this sigma bond continuously takes part into the hyper conjugation. When this bond breaks, this bond forms, when this bond breaks, this bond forms and when this bond breaks, this bond forms, like this it goes. So this is hyper conjugating structure, which is possible only in case of carbocation, free radical carbonyl. Now the point here, I'll just discuss few more things into this, which is again important. You see, this bond in this sigma bond is breaking and it is converting into pi. So for all those molecules where sigma bond is weak, hyper conjugation will be more, right? For example, if I ask you the hyper conjugation in this CH3, CD3 and CT3, the order of hyper conjugation is this. CH3 will have most maximum hyper conjugation. Why? Because the carbon hydrogen bond length is maximum here, then carbon deuterium and then carbon tritium. All these are small, small information. You must have this. You never know on the basis of this, I can ask questions, especially all these in Manipal and all exams, any exams, even in JEMains also, right? This is the order of hyper conjugation. CS3 will have maximum, then CD3 and then CT3. But I did not write down over there. The I effect order is reverse. That also you write down if you want. I effect of CT3 is maximum, that is electron bridging effect of CT3 is more than CD3 and then CS3. This is the order of plus i and this is the order of hyper conjugation. This is one, again, information I am giving you. Why this order is there? Because carbon hydrogen bond length is maximum and we can see here, sigma bond converts into pi, it breaks down. So we should have what? Least stable bond, right? Now, since we have three alpha hydrogen here, if they ask you, number of hyper conjugative structure, that will be equals to number of alpha hydrogen. Number of hyper conjugating structure is equals to the number of alpha hydrogen. If they ask you total structure, total number of structure, then we have three hyper conjugating structure and one structure is this itself, the number of alpha hydrogen plus one. This is also you keep in mind. Got it? Got it? Did you understand this? Okay. So like you see, I said that plus i minus i plus r minus r, all this we have. Similarly, like that also will have plus h and minus h also possible. Plus h and minus h also possible. I will draw two more structure and then we will again move back to the questions. Okay, let you understand this. Plus h again means electron releasing ERG minus h electron withdrawing. Okay? So for example, suppose we have this ring and on this ring, we have ps3 group present hydrogen and here also we have hydrogen. Now, when you draw the hyper conjugating structure, we will draw like this. This pi bond comes, sigma bond comes over here, pi bond will go here and the structure will be this. I am drawing only one structure. You can draw other if you want. The h plus will be here. Single bond h, single bond h. This is the structure and furthermore structure you can draw. Okay? So this is what, since it is giving electron to this ring, so it is plus h effect, positive hyper conjugation. Okay? Similarly, minus h also possible. Minus h effect, like we have these groups CF3, sorry, just a second. So CF3, TCL3, TBR3 and CI3. Minus h, like I said, electron withdrawing. Out of this four, only CCL3 chose minus h effect and this is the only molecule we have or group we have, which can show minus h effect. Right? And that is only you have to keep in mind. And how do we draw the minus h structure here that also will discuss one, double bond, double bond, double bond and we have TCL3. So what happens here, this pi electron comes over here and this sigma bond goes on to this chlorine and we get this TCLCl. We have chlorine negative charge. Here we have positive double bond and double bond. Right? Now to sum up this, what we can say, due to hyper conjugation also, ortho and para position gets affected. Metaposition does not get affected here, whether it is plus h or minus h. That is why, that is why again, like plus r or minus r, hyper conjugation, whether it is plus h or minus h is only possible at ortho and para position. Metaposition, there is no effect and that is why I told you that time that at metaposition, only i effect is possible. Right? Why CS3 does not show because of carbon fluorine bond is very strong. Right? Both have compatible orbital 2P2P, very strong carbon fluorine bond and hence this bond does not break. Carbon fluorine bond does not break. Carbon bromine, carbon iodine bond is weak, but the electronegativity difference is not that much or you just memorize this. Reason is not required. Only CCL3 can show minus h effect and minus h only possible plus h only possible at ortho and para position. Tell me, did you understand this? Now whenever you see any questions in the exam, okay, whenever you see any questions into the exam, you write down all the effect possible and accordingly you will do this. Like I said, the overall order is what? Resonance dominates over hyper conjugation over inductive effect. Okay? So when CS3 present at ortho position, like in this example, CH2 plus, when CS3 present here, right? So what are effects possible? Since resonance is not possible, but we know hyper conjugation possible, we know i effect possible, but which one is dominating? Hyper conjugation is dominating. Accordingly we will decide. Okay? Suppose we have CCL3 present here. This question you tell me the stability. CH2 plus, right? Here we have CCL3, CH2 plus and here we have CF3. Which one is more stable? A and B. The more examples you solve onto this, the more clarity you will have. Tell me which one is more stable? A is more stable, right? Okay, A is more stable. Answer is correct. Okay? Your answer is correct. Now you tell me in molecule A what all effects possible here? What all effects possible? Only minus i. Yes, minus h and minus i. In molecule A, two effects are possible which is minus h and minus i. Which one is dominating? Minus h is dominating. So for this molecule we will decide the stability according to minus h effect. What all effects we have here? In molecule B, in molecule B, what all effects? Only minus i. Yes, because CF3 does not show hyper conjugation, right? CF3 does not show CBR3 CIT. So that's why you see this organic chemistry little bit of information you should have. Otherwise you will attend the question, but you will end up with the wrong answer, right? So this kind of information you must keep in your mind, okay? CF3 since it does not show hyper conjugation, so only minus i possible. Minus i will destabilize this carbocation and minus h what? Minus h will also destabilize, right? Order will be this. No, it is not true. Order will be this. Minus h will also destabilize. Minus i will also destabilize. But minus h since it is dominating will destabilize more. B is more stable. Got it? Understood? Hyper conjugation is there, but it is not plus h. It is minus h. So that will destabilize. Got it? Understood? Okay, so we will go back to the question again. Yes, so 28 answer B is correct. Okay, it is Sigma Pi resonance. For this one, 29, tell me. What happened, Vaishnavi? What is the confusion? Tell me. Tell me, tell me. What is the confusion? What is the confusion you have? Tell me first. In question number 29, the charges are all, all we have, everywhere we have positive charge. Everywhere we have positive charge. Yeah, right, Vaishnavi. You just go through it again and if you have any doubt, you text me. Everywhere we have positive charge in 29. This is very good question. Very interesting. If you understand this question, you will get your concept properly. Okay. Okay, now you see this. Vaishnavi, you also try to understand. With this also, you can get your concept clear. This is the first molecule, right? Now, tell me here, what all effects are there? Tell me what all effects are there? I'll write down all the molecules. What all effects are possible here? Let me ask you one by one. Do we have hyperconjugation possible? Do we have hyperconjugation possible into this? See, I will explain. Here we have 3-hydrogen, alpha-hydrogen, 3-hydrogen, alpha-hydrogen. So, yes, we'll have hyperconjugation possible and that is also plus H with total 9-alpha-hydrogen, correct? This is one thing. Sorry, not 9, 6. 6-alpha-hydrogen. Okay? Plus H effect possible and 6-alpha-hydrogen. Another thing you see? This oxygen has lone pair on it. Lone pair, sigma, positive charge. So, we'll have resonance also here. Resonance also possible and since this oxygen is more electronegative, so this will also withdraw electron. So, we'll have what? We'll have minus i effect also possible. Clear? Now, these three effects, which one is dominating? Resonance is dominating, right? Let it be here. Now, in this question, do we have hyperconjugation possible? Yes, possible. That 2 plus H with how many alpha-hydrogen? 3 here and 2 here. So, we'll have 5-alpha-hydrogen. Do we have i effect possible? No, it's not possible. Do we have resonance possible? No, it's not possible, right? So, this is only plus H effect. Now, here also you see, same thing we'll have. We have hyperconjugation, but number of alpha-hydrogen will be here 3, right? Then we have minus i and then we have plus r, sorry, resonance, r effect, right? Plus r, we can say, electron releasing. Plus r, you can say. Now, here it is what? Only hyperconjugation possible with number of alpha-hydrogen is 2. That 2 only plus H possible here. Okay. Now, we know because of resonance that 2 plus r effect here, plus r, plus r here, the stability of this carbocation will increase, right? That will be maximum. This 2 you see, this 2 will have lesser stability than first and third because in first and third we have resonance also and in this, we don't have resonance, D and D. You are considering this one for i effect. Then you also consider i effect for this. Then you also consider i effect for this or i effect for this. Here we do not have. But we are taking this as a molecule. This as a molecule. This as a molecule. The whole molecule we are taking. Another point is what? If you see, this carbocation is what? 2 degree. This is also 2 degree, right? So, that also if you consider, it will be everywhere, i effect, okay? So, along with resonance here and here, so the one which is most stable is the first one because of resonance and then resonance we have here also. But here, apart from resonance, we have hyperconjugation, which is here also. But here we have only 3 alpha hydrogen. Here we have 6 alpha hydrogen. With more number of alpha hydrogen, more resonating structure you can draw and that is why this first is most stable and then we have third, right? If you compare the stability of this 2, obviously 2 is more stable because it is 2 degree carbocation. It is 1 degree carbocation. All other effects are same. Then we have 2 and then we have 1. Now you tell me any doubt, any way you can do, if you consider i effect also, that is not the dominating effect, okay? Because we have hyperconjugation there. So, we have to, yeah, it is option 4. It is option 4, right? Option 4 is correct. What happens I mean? What are you saying? This is the right answer, option D, right? So, you have to understand this in what all conditions various, you know, various electronic effects possible and then depending on the dominating effect, I have written wrong. First we will have maximum, it is right only, you know. First is this maximum, this is first and then second and then third and then fourth, okay, okay, fine, fine. This one will be the fourth, right? Yeah, correct. First, third, second, correct, yeah. Understood. So, like this you try to understand the various effects which is dominating according to that. Suppose another one thing like why I said this question is very good because you see in this two, if you consider resonance, both will have resonance, right? Then next factor we will see. Next factor is what? Hyperconjugation. So, hyperconjugation is this, we have six hydrogen, we have only three. Obviously, more resonant to more hyperconjugative structure, more stability. Got it? All of you, tell me fast. See, this kind of questions is very important whether you are going to write especially in this exams of CET, Komet K, Manipal, Bitsat, VIT, any other exams, this kind of questions will be very direct. The only thing you should know that what all effects are possible at what place. Then according to the order RHI, you can do the questions. Okay, next question. Do this, 18, 19 and 20. What happened? Question number 18, direct question. You will get trans product into that. Preparation of alkene. Yeah, you will get trans product because it is very direct. 19th one is, Simon is getting D, Amogh is getting C, Andhraya is getting D. Okay, so most of you are getting C. Okay, see this. First of all, basicity is nothing but the tendency to donate this lone pair of electrons, which we have here also and here also. Now you see what happens if this lone pair it, it donates, then the nitrogen will have the positive charge here. Right? And because of this positive charge, you see. So many of you have been forced to discuss this. With respect by the military forces, I am in conversation about a number of the demands that have been raised and committed to them that we will, we will give them the status of chenille, which they do not have and if you are talking about corruption. Also, sir, you didn't mention Robert Rodgers. You told the other thing. What, what about this? What, what about this? It should be investigated as a single person. And so the answer applies to everybody. No problem. Which is saying that it is possible for the grocery industry to have a company or graphite, the CPI director, 130 at night, is sad. The Supreme Court says, bring the man back. Sacks him again. The president with France says the prime minister of India has called me personally instead of in a meeting with Mr. Anil Ambani should get the graphite contract. The prime minister doesn't comment about it. Everybody knows that the prime minister stir up 30,000 crores and give them to Mr. Ambani. No country knows it. But why is there, why is there an investigation? Investigate everybody. Investigate Mr. Bhadra. And the first person to say, investigate the prime minister. And be absolutely crystal clear, Mr. Narendra Modi for one minute hasn't opened his mouth to graphite. And the proof is, HL has been making a graph for 70 years. He has never made an aircraft in his good life. HL gave 3,000 crores of its own money to the government of India. Anil Ambani has 35,000 crores of debt that he has not paid off. Yeah.