 Okay, guys, so just we'll start with the session. Let me share the screen first. Give me just a minute. Okay, so good morning guys. Today we are going to start like we discussed yesterday reaction mechanism. Okay, so like I said, the entire let me first share the screen. Now, could you share the screen now? Yes, sir. Okay, fine. So, see, like I said earlier also, reaction mechanism is like, you know, it's very important to understand the whole organic chemistry. We are not going to take reactions of one particular chapter in this particular session or for this reaction mechanism. The whole session will take will take reactions of almost all the chapters of organic chemistry. So, like I always suggest, we'll be able to cover the 60 to 70% portions of organic chemistry after this. Okay, so, like the reactions which are left after this particular chapter reaction mechanism will be like the preparation of various compounds like alcohol, alkyne, light, carbonyl compound, acid, amines, all those mainly preparation method. Few reactions are also there that we'll see in the respective chapters. Okay, so let's start with this chapter. So, like, again, I'm telling you, prepare your notes properly for this chapter. It's very, very important. Okay, so reaction mechanism. So, what is the mechanism of a reaction? For any given reaction, what do we understand with this term mechanism? Okay. So, mechanism is what mechanism is the step by step description, step by step description of any reaction. Step by step description of any reaction is the mechanism of that reaction. Step by step means what we can say here that any reaction we can have one or more steps involved. So, based on the number of steps involved, number of steps involved, the reactions are classified into two categories. Okay, based on the number of steps involved, the reactions are classified into two categories. One is simple reaction. And simple reaction, we also call it as elementary reaction. Simple reaction or elementary reaction. The other one is complex reaction. And we also call it as non elementary reaction. Just few terms that we are trying to understand. Okay, simple reactions, elementary reactions are those reactions, which takes place in one step, one single step. One step reaction is elementary reaction, complex reaction, which takes place in more than one step, two or more step. Okay, we'll see about this complex reaction in single one step reaction that is elementary reaction. In chemical kinetics chapter, okay, there also we have a section of mechanism. So, since we have more than one step involved. So, one of the step out of the given number of steps will be the slowest step, and hence that one will be the rate determining step. Okay, so in detail, we discussed this in chemical kinetics. Okay, now, suppose we have a reaction, so reaction basically what happens in the reaction, we have, suppose a molecule, suppose a molecule we have A. And on this molecule, there are certain reagents which attack, right, and then it gives us the product. This is the, like, the, like, the step, the reactions of any, any kind of reaction text is like this only. Okay, so the molecule which get attacked by the reagent, we call it as substrate. Attack takes place at substrate. Okay, and the reagent can be, are the attacking particles, these are the attacking particles, it attacks on the substrate, attacking particle. So, reagent, just a second, so reagent are the attacking particles, it attacks on the substrate during the process of the reaction. Okay, and if all conditions are fine, then this gives you the product. Okay, this is how the reaction takes place. Okay, now, like I said, there are two types of reactions, single step reaction and multi step reaction. So a little bit we'll discuss about single step and multi step reaction, a little bit we'll discuss, okay, how these reactions are different from each other. Can I go to the next page? Yes, sir. Okay, so heading you right down, single step reaction. Suppose the reaction is this A2 plus B2 gives to AB, it is a single step reaction we are assuming. Okay, so what happens in this reaction? What happens first of all, the AA bond breaks, BB bond breaks and AB bond forms. Right, so we can say this is bond breaking, bond making process. Some old bond breaks and some new bond forms, forms, right. But during the process of this reaction, we'll have a transition state here. And what is that transition state before this AA bond breaks, BB bond breaks and AB bond forms. The transition state beyond, like between these two things is the tendency of AB bond, the tendency of the formation of AB bond, this AB bond is about to form, AA bond is about to break like this and BB bond is about to break. So this is a transition state we get in between. Okay, and this pi bond, this sigma bond here, it has tendency to break, it has tendency to break and these two has tendency to form. Right, so when, so overall if you see the nature of the reaction or the property of the reaction, the first thing is what, if I ask you whether it is exothermic or endothermic, what is your answer? Exothermic. Why exothermic? Exothermic. Because it will go towards lower energy, so it will release energy. Lower energy means more stability. Anything else? So two molecules are like forming to form a single compound, so it releases energy. So it's not like that, see energy we can compare based on the bond formation. See what happens here, two sigma bond breaks and two sigma bond forms also, two AB molecule is forming and one AA, one BB molecule breaks. So it can be endothermic or exothermic, we cannot assure this thing, it depends upon the bond strength of AB and A and BB. But if it is, one thing we can say, if it is exothermic, so I'll write down here first exothermic or endothermic based on the bond strength of AA, BB and AB. Two possible graphs we have here, this is energy axis and this is progress of reaction. This is also energy and progress of reaction. So obviously when it is exothermic, then energy releases. So energy of product is lesser than the energy of reactant. So in that case the graph will be like this, it goes up and then it comes down. For endothermic the graph will be like this. So this is reactant, this is product. This is reactant, this is product. This graph is for endothermic reaction and this graph is for exothermic reaction. So why does it go up and then go down? Actually whenever the reaction takes place, the reactant molecule collides with each other. It's basically a collision. Because of this collision, the energy of both reactant molecule increases and goes to a maximum value. At this point, we call it as transition state. The highest peak of this curve represents the transition state, which is nothing but this molecule. So that's why you see transition state, whatever molecule forms has maximum energy. And hence it is highly unstable. Hence the transition states are highly unstable since they have maximum energy. And that is why it cannot be the product of any reaction. It will either convert into the reactant back or it will convert into the product based on the condition criteria that we have. Here also you see this is reactant, this is product and this is the transition state. The point I am trying to make here is if you have single step reaction, then we have only one transition state. Three step reaction, we will have three transition states, it will go like this. We will see the graph after this. But number of peak that you have in this graph, that will equals to the number of steps. It is actually the portion of chemical kinetics, we will discuss it there also. But yes, little bit of idea I am giving you. So this is it. Exothermic reaction if you have, so you see the energy of product in this also we can conclude is lesser than the energy of reactant. Here it is reverse, energy of product is greater than the energy of reactant. Delta H is greater than zero endothermic. Delta H is less than zero exothermic. So in a way can you say that the products formed in an endothermic reaction are slightly more unstable than the ones formed in exothermic? Yes, we can say that as far as energy is concerned based on that energy we can say that. But to define the stability of a product, we will see the condition of that particular reaction. But obviously comparing this product and this product, we can say this one is slightly more stable. Okay, yes. Okay, the another type of reaction we have is a complex reaction. Complex reaction, like I said is a multi step reaction. The reaction takes place in more than one step. Hence we call it as multi step reaction. Okay, so the reaction is A2 plus B2 gives to AB. It is multi step reaction. So what are the steps involved into this? Okay, the first step is the molecule A2 dissociates into atomic A. Okay, and this atomic A reacts with B, B2 and this AB bond is forms and atom B also forms here. The third step is both atom A and B again combines and it forms A. Okay, so you see the transition state of the first reaction is what? Transition state is the dissociation of AA bond when AA bond is about to dissociate. That is a transition state. It is TS1. Here the transition state is what? When AB bond is about to form and BB bond is about to dissociate. This is TS2, transition state 2. And this, the transition state is when AB bond is about to form. That is TS3. So three step, three transition state. Okay, so here you see. Here you see the Sigma bond, A bond is breaking, right? Dissociating. This step is endothermic because the bond is dissociating endothermic bond. This one we cannot say it is again based on the energy of this bond and the energy of this bond. Can be endo or exo. It's a general expression, general reaction. So all the possibilities I'm writing it down. Can be endo or exo based on the strength of these two bonds. AB bond, Sigma bond is forming. So it is exothermic step. Endowed exo. Whatever the net effect here of these energy, according to that, the reaction will be either endothermic or exothermic. Got it? Okay, so when you draw the graph here of this reaction. Since it is a three step reaction. So there are three transition history. So first step is endothermic, right? So energy of the. So what I'm assuming here to make you understand this, what I'm assuming, and it's like I said, it's an assumption, right? So overall, suppose a reaction I am assuming as exothermic. This is what I am assuming. Like I said, based on the three energies here, excuse me, the reaction can be endo or exo. But you make, you understand this or to draw a graph, I am assuming this to be exothermic reaction. So if it's since overall it is exothermic, so energy of the final product should be lesser than the energy of the reactant. Is it right? Yes. Yes, sir. Now since the first step is endothermic, so energy of the first step or the transition state we have or the reaction intermediate we have is more than the reactant molecule. Second step is endo or exos. If it is exo, it goes like this. And the third step is again exo. So it's from transition state and goes like this. Since overall it is exothermic, so this product will come down to the energy of the reactant. This is the final product. So this is AB and this is A2 plus B2. This is the TS1, this is TS2 and this is TS3. Here this one is the reaction intermediate, RI, reaction intermediate. This is also the reaction intermediate. What is the reaction intermediate in this reaction? Transition state is this. This is reaction intermediate, A atom that is forming. This is the product of the first reaction, the first step. This is also reaction intermediate. So we have two reaction intermediate here. The two things we can easily conclude here. The first thing is number of TS. What is TS? Transition state. It is equals to the number of steps involved. See this question they have asked in JEE exam. They will give you this kind of graph and they can ask some theoretical question. How many transition states are there? How many reaction intermediates are there? So reaction intermediates are what? Number of steps minus one, three steps, two intermediate we have. So number of steps minus one. Done? Yes. Now out of these three steps, the slowest step is the rate determining step. Okay. And the slowest step is the one which has minimum energy of activation. Sorry, highest energy of activation. Minimum one will be fast. This is it. So this energy should be it will go till here. Again, I'm telling you all these things will again discuss in chemical kinetics. Okay. Sir, is it possible for like TS two to have a higher energy than TS one? TS two. Yeah, it's possible. Depends upon what energy is getting exchanged in the second step. Okay, sir. This rate is very slow, slower than the first one. Then this activation energy will be high. It will go like this and come possible. Okay, sir. Okay. All possibilities are there. Can be high, can be less, can be equal also. Sir, could you repeat that activation energy part? The activation energy is the energy difference of reactant and the transition is this difference of energy. Is the activation energy. This is the activation energy of second step. So then the reaction with the higher activation energy is the fastest one? Is the slowest one, sorry. Yes, slower one. Actually, you see what happens logically you can understand. Here you have the reactant molecule. So this is actually the energy barrier. What is energy barrier? We'll discuss that in chemical kinetics. Okay. This is energy barrier. So to convert this into the product, it has to cross this energy barrier. Or this energy barrier is very high. So it's difficult to cross this energy barrier. Yeah, sir. The activation energy is high. For that reaction to process, we require slightly higher temperature. Because when you increase the temperature, the energy of this reactant also increases. So that they can cross over this energy barrier and convert into product. Okay. So high energy barrier, slow will be the rate of the reaction. So in this question if they ask you, which one is the slowest step? The answer will be step. One. Because the activation energy is maximum first step. Is it clear? Yes, sir. Right. So this is one type of question they ask. Okay. One definition here you write down reaction intermediate. Definition of reaction intermediate. These are the particles which forms. These are the particles which forms during the reaction, during the reaction and get destroyed during the reaction. And get destroyed before the completion of reaction. Okay. Get destroyed before the completion of reaction. Okay. So one thing you always understand reaction intermediate. Can not be the product of any reaction. It's not possible because those are very unstable particles. So these reaction intermediates, if the condition is fine, condition is satisfactory, then the intermediate will convert into product or it will convert back into the reactive. Right. So reaction intermediate can not be the product of any reaction. Okay. That's the one thing. Okay. Now, like this is the general thing that we have discussed about any reaction. Okay. Classification of reaction. Heading you right down. Now, on the basis of process, process, different, different types of reaction we have and we'll discuss that the first type of reaction is addition reaction, addition reaction. Okay. Why process because classification of reactions are many based on the number of steps, like I said, two types, elementary, non-elementary, based on energy, exothermic, endothermic, right? Are there many different classification of reactions? We are trying to understand the different types of reaction. Okay. So here we have addition reaction, elimination reaction and substitution reaction. So first we are going to discuss is the addition reaction. Now you'll see the basic, try to understand the basic properties of an addition reaction. Addition reaction generally takes place on alkene. Why on alkene? Unsaturated. Because it is unsaturated, right? Weak pi bond, you can dissociate this pi bond and attach any other particles, right? That's why it is. So the product of this reaction is this. Don't think about A from this side or B from that side. Okay. It's just an addition, A and B. Specifically, we are not going on to this. What kind of addition is this? This is one possibility, alkene, addition reaction possible. It is also possible on alkene. What all molecules which has pi bonds, this kind of reaction possible because pi bonds are weak bonds. C double bond, C, A and B. Further again, the addition reaction possible if the condition is there. And then it is also possible with the carbonyl group C double bond O. So it forms C, O, A, B and one bond. So this kind of reaction is addition reaction. Now what is happening in this reaction? In this type of reaction, one pi bond is converting into what? Sigma bond. How many sigma bonds? Two sigma bonds. So in this addition reaction, in all this reaction you see, once pi is converting into two sigma bonds, right? So could you tell me whether it is exor endo? Exor endo. Exor. It should be exor endo. But you're forming two bonds. Yeah, exactly. Sigma bond is higher. Sigma bond is stronger. Second thing is what two sigma bonds are forming. Even one pi converts into one sigma. In that case also it is exothermic because sigma is stronger than pi. So one weak bond breaks and two strong bond forms. That's why the property, the first property of this reaction, it is mainly exothermic, right? Delta H less than zero. What about the entropy change of this reaction? Delta S less than zero or greater than zero? So it's greater than zero. Why? Yeah, because the number of particles are decreasing. No, no, it should increase because you're going from something which is ordered and then you put it into the same molecule and also the electrons delocalize from the... But like before the reaction starts, you already have that C, C and then you have A, B, that's two different molecules emerge to form one molecule. Yeah, right. So we have two molecules initially present and that convert into one. So number of molecules decreasing and hence randomness decreases. Delta S less than zero. See randomness and property I'll tell you one thing. There are many aspects you can think about here. And I know you'll get confused if you talk about all those stability and other factors. So what I suggest you, just you see the number of reactant and number of product. Okay, that gives you an idea of whether delta S less than zero or greater than zero. If the number of product is less than the number of reactant, then we'll say that we have less randomness and dropping decreases. Okay. Now, third thing, could you tell me the condition of spontaneity? At which, what is the condition of temperature so that the reaction will be spontaneous in it? Condition of temperature means low temperature or high temperature? Low. Low temperature. Okay. So we can say that spontaneous at low temperature. I'll explain this also. Spontaneous at low temperature. See for spontaneity, what is the condition we have? Delta G should be less than zero, negative. And we know the relation of thermodynamics. We have studied delta G is equals to delta S is what delta S is negative delta H is also negative. So this negative negative positive. So condition for delta G to be negative is what the modulus of delta H should be greater than the modulus of T delta S. Right. Means this part should be less and which can, which we can achieve by lowering temperature. So low temperature is the ideal condition for this kind of reaction. So whenever you see additional reaction, it takes place at low temperature. Done. Finish. So the next type of reaction we have is elimination reaction. The second type elimination reaction. You see this example is just reverse of addition. Tell me the properties of this reaction quickly. Probably endothermic. Or this kind of reaction also possible. Delta S is greater than zero. Delta S is greater than zero. Delta G probably less than zero at high temperature. This is also elimination reaction. Okay. Let me tell you one thing about the second reaction. This is beta keto acid. You see this acid is there. Alpha carbon is this beta carbon is this. It's very important reaction. We'll discuss this later also, but try to keep this in mind. You have to keep this in mind. They've asked question on this many times in the exam that whenever you heat beta keto acid, beta keto means what? Acid we have COOH and at beta carbon, there's a keto group attached double bond O. So whenever you heat beta keto acid, this CO2 molecule comes out. Elimination of carbon dioxide and this hydrogen attached with this carbon product will do this. This you must remember. Elimination of carbon dioxide. Elimination of carbon dioxide takes place when we heat beta keto acid. Okay. Sir, I had a doubt. In addition reaction, the first one, we said we break one pi bond and get two sigma bonds. But don't we break one pi bond and one sigma bond, like AB sigma bond also? This one in the first one. The first addition reaction. Yeah, that also you can see. I was talking about substrate because substrate is the main molecule of the reaction. If you consider that, sorry, reagent also, then we can say one pi, one sigma breaks and two sigma bonds. In that case also what happens? Suppose sigma, sigma cancels out. One pi converts into one sigma, right? Sigma is stronger than pi. Then also the reaction will be able to come. Sir, but in general, should we consider the bonds in the entire reaction or only the substrate? See, generally, substrates convert into this thing, product with the help of reagents that attacks onto the substrate. So I feel personally like, you should consider reagents also into that. So it's better to say, if you say one sigma converts into, sorry, one pi converts into two sigma, that won't be wrong. But on the cipher side or precisely if you want to say, you can say one pi, one sigma converts into two sigma. That is also correct. You can say that. Okay, sir. Yeah, so better to take that. Anyways, so this is the thing. So first property of this kind of reaction is what? Could you tell me the first property? So they endotomic. Two sigma converts into one pi or one sigma also you can consider. So pi is weaker than sigma. So this reaction is what? Mainly it is endothermic. Just reverse of the first one. Endothermic means delta H greater than zero. Could you tell me the entropy change? Positive. Positive. What is the condition? High temperature. Because again, the same thing can see delta G is equals to delta H minus T delta S. Right. This is positive. This is positive to make this delta G negative. The condition is what T delta S modulus should be greater than modulus of delta H. And this condition we can achieve at high temperature. So elimination reaction you see it favors at high temperature. So whenever in entire organic chemistry, some elimination reaction is there. So generally it takes place at higher temperature, higher than the additional reaction. Okay. Right. Now the third type of reaction we have, which is substitution reaction. Substitution reaction. Substitution reaction. What happens in this? What happens in this? One atom or group replaces other atom or group in the reaction. So for example, you see we have a molecule RL and a reagent say nucleophile. Nucleophile, electrophile are the reagents. We'll discuss that also. So this one is more electronegative. I suppose RCL reaction we have more electronegative. So what happens? Here we have delta negative and here we have delta positive charge. So what happens in this? Since it is partial positive charge. So this nucleophile has tendency to attack onto this alkyl group. And when it attacks, then what happens? This leaving group goes out. Okay. So it forms RNU plus leaving group goes out as L minus. This is substitution. Nucleophile substitutes this group leaving group. Done. So what have you written before L? It is alkyl group. Any methyl ethyl group. R. So on the right. This side. This is more electronegative. This side we have RNU plus L1 lone pair negative charge. Some examples you see. Suppose we have a reaction. CH3Cl, methyl chloride OH minus. This OH minus will attack onto this carbon. Forms CH3Cl OH and Cl minus goes out. Another one you see if you have benzene ring plus an electrophile we have NO2 plus. So this will also displace hydrogen and get attached with the benzene ring. So all these are substitution reaction. Okay. Properties are what? Since the number of particles are same both side. So delta S, change in delta S is almost equals to zero. Okay. And for this kind of reaction. Enthalpy is the only factor. Enthalpy is the private factor for this kind of reaction. Okay. How do you determine the enthalpy of such reactions? We cannot say we can't determine. But see exact value for exact value you should know the bond strength of all those bonds which are forming and which are dissociating. Correct. So exact value it's not possible to determine because that value you must have that carbon hydrogen. See carbon chlorine the bond strength you must have then CO at bond strength you must have then you can find out in healthy. It's not a big deal you can find out in healthy. But whether it is delta S greater than or less then that also depends upon the bond strength of those bonds. Right. That's why like you cannot memorize the bond strength of carbon protein or carbon. Right. That's why I'm not writing down anything here by looking at the reaction you cannot determine. Okay. But yes if you have the value given you can say whether it is exorbitant. Yes sir. Yeah. For the benzene reaction when first of all NO2 is a lot bigger than hydrogen right and also when NO2 joins the ring then it will cause the mesomeric effect. So because of that the like there'll be even more delocalization. So shouldn't that increase delta S? No no no. See delta S like I said you don't have to consider delocalization. It's randomness is because of the particles more particles more randomness. See like I said you have different different factor for the assessment of delta S. And this is just a method we are using. It's not the only method. Okay. Point is when you look at the number of particles and then you can judge about this delta S. Okay this is not the only method. This is one of the method you can you get the right you know answer regarding the entropy change if you look at the number of particles on both sides. So what I'm telling you however this is not the only method but this is like you can say it is a hunch or you know a hidden trial method by which we get the right answer all the time. That's why I'm suggesting you don't think about delocalization and anything. See here also we have delocalization here also we have delocalization right. So we'll see we compare the reactant and the product what happens in between that we are not considering because eventually this converts into this and in this process what is the change in delta S. So the intermediate thing we are not considering here. So whenever you have reactions I will suggest what see the number of particles both side if it is lesser more we can think about the delta S. Okay since it is same both side so change is not that great it is almost equals to zero. Okay that's not the thing. Okay. Right one more type of reaction we have here. This is a fourth type and we call it as rearrangement reaction rearrangement reactions you see. Suppose we have this one C double bond N OH ET in presence of H plus acid it converts into this reaction like in the classification of reaction this one is not given. But this is also one type of reaction we get in organic chemistry. Sir automatism count as rearrangement reaction. Yes. This nitrogen has don't pair on it. This nitrogen has don't pair on it. And acid this H plus right down here only. This H plus comes over this oxygen and this will get the positive charge and then what happens this goes out as H2O plus H2O will also get here and nitrogen will have the positive charge. Right now to you know it supplies is positive charge this H plus comes out sorry where is double bond oh I missed something here this should be just a second mechanism is. We have H plus no so this comes over here right okay this is you see this is an intermediate we get here that we call it as nitrene. It's a rearrangement see what happens here I'll tell you this H2O goes out and then this alkyl group and then this alkyl group rearrange itself onto this nitrogen atom it comes over here. Right so this carbon will have the positive take this bond pair of electron and rearrange itself onto this nitrogen. We'll discuss this later also in a minus chapter but we'll have the positive charge here. Right this rearrange itself onto this nitrogen atom this carbon will have the positive charge and then this new water which comes out here act as a nucleophile attacks onto this. Carbon so it goes like this you see I'll tell you see double bond and when H2O goes out it has one lone pair and one positive charge here. Since H2O takes this electron pair goes out we have ET and CS3. Now this methyl group takes this bond pair of electron and rearrange itself onto this nitrogen atom and it is always the rearrangement of alkyl group here so it forms carbon with ET positive charge on it double bond and CS3 and lone pair. Okay now in the next thing what happens the H2O attacks onto this carbon as a nucleophile so it forms CS3 and we have OH2 plus ET. So how did the CH3 move the nitrogen? That's what I said it's a rearrangement reaction when you heat this this kind of rearrangement takes place when nitrene forms so in details we'll discuss this in a minus chapter. Okay this won't come now just you keep this in mind how it is happening even mechanism also we'll discuss there also. Now this water attacks onto this you'll get this and one by one these two hydrogen goes out one attached with this nitrogen and one will go in the reaction because this is behaving as a catalyst. Right so you'll get a double bond over here see what happens I'll tell you this comes over here this comes over here right H plus comes out right and when finally this to stabilize this again this H plus comes out and this will attach with this nitrogen atom. So you get NHCS3 double bond O ET so this kind of reaction we call it as rearrangement reaction just you write it down I know that you have little bit of difficulty to understand this kind of reaction but we'll discuss this later okay we'll see this later how to do this kind of questions. See what happens here we have this comes over here H2O molecules attacks onto this so we'll have O attached with carbon and two hydrogen atom on this with positive charge onto this oxygen. So next what happens this H plus comes out to stabilize the positive charge on oxygen atom and CS3 like this it comes in the last step this loan this bond pair conversion to loan pair and this pi electron shift over here this H plus which comes out this attached with this nitrogen. So you'll get NHCS3 like this it goes and we have a double bond open. So all these happens when you heat this there's no reagent here there's no attacking particle it's just when you heat this so this kind of compound which we call it as oxene. It has the tendency to convert into keto group like this double bond and OH is oxene this one. Okay now these are the types of reactions we have discussed. So I have one doubt yeah tell me so is there any enthalpy change in this whole reaction the rearrangement reaction. Just a second let me go to that yeah enthalpy change again we don't discuss here because we are heating this right we are providing energy right it takes place at that slightly higher temperature there's no bond dissociation and bond formation. Okay so it's like it's theoretically you cannot say but he has enthalpy it's not an enthalpy driven reaction like the previous one. Okay so it happens spontaneously not spontaneously it happens on its own when you heat this okay. Okay, yeah sir. Yeah, yeah, yes, yes, it's high group can also rearrange but that depends upon the position of this. Okay, this to understand in that detail I'm not going okay so you don't you know think about that because all those things will discuss later. It depends upon the position of this what is the condition there will discuss that later that you let it be known now you don't understand that there are many things before we have to and we have to discuss. Okay, so that part I am not you know discussing. Okay, now you see like I said attacking particles or reagents so what are the attacking particles we have and then we'll go to the reactions. What are the attacking particles or reagents there are three types of attacking particles. So these are the attacking particles are the positive charged or negative charged or it can be neutral also all three possibilities we have positive negative or neutral particles which initiates the reaction which initiates the reaction. Now what are the different types the first type we have electrophile electrophile then we have a nucleophile and the third type is free radical. First thing is the electrophile what do you mean by electrophiles. They like a electron basically so they're positively charged. Yeah, electrophiles means electron lovers, electron lovers or affinity towards electron that also we can see. So the first thing is what it should be electron deficient then only it can attract electrons. So it is electron deficient in nature or other possibilities what if the molecule has incomplete octet then also it can accept or attract electron vacant d orbital if it is there then also it behaves as an electrophile. And these electrophiles attacks on attacks on the electron rich side of the molecule electron rich side of the molecule. Okay, examples examples are important okay so I'm discussing it here see H plus is an electrophile. We are plus write down property it's important NO plus CL plus NO 2 plus one reaction we have seen on this CH 3 plus. So what is CL plus? But isn't that not here? Not stable but possible it's not stable but can be can be as an electrophile possible. So the reaction that it forms we are minus and we are plus forms now. Yes, that's right it's possible. So these all are positively charged so obviously let it attract electrons correct? Sir is this in some order? No, no, no I'll tell you let me finish this I'll tell you these are all positive charge no order all positive charge species obviously these are electrophiles it cannot be nucleophiles. Another example you see AL CL 3 BH 3 BCL 3, third one we have ZN CL 2 FE CL 3 SB CL 5. Now these are the three set of electrophiles can could you tell me which one is the strongest electrophile CL plus? The first set. The first set right because they have they have postage charge on it okay so these are the very strong electrophile we have this comparison you should know very strong electrophiles okay. So they have positive charge positive charge means vacant orbital they have generally P orbital or incomplete octet you see the octet is also not complete for all the ions if you compare what about this? This also has incomplete octet right, incomplete octet neutral but still it is electrophile because they have incomplete octet aluminium has only 6 electron, boron has only 6 6 so it can accept 2 electron right that's why this behaves in electrophile we cannot say electrophiles are only positive charge species okay again it has vacant orbital so this one is less electron deficient in comparison to the first set because first set has positive charge. So the positive charged one are very strong these are weak electrophiles weaker than the previous one the first one because they do not have positive charge only they are only electron deficient okay. Now these are electrophiles which has vacant D orbital the difference between the second set and the third set is what the second set has vacant P orbital and here we have vacant D or all these information guys you write down. Okay the second set has vacant P orbital this has vacant D orbital both has incomplete like both are neutral right this is also weak this is even weaker than the previous one so we can say these are the weakest electrophile we have okay. So these examples you must remember that's important apart from all these three sets the mall is few molecules we have in which the resonance is possible okay so resonance you know resonance we have discussed like. Rahul right you are the one who joined the class now right yes Rahul have you done have you gone through resonance you have discussed yes okay resonance you know. Suhas is there it's not okay fine so molecule if you know they are shows resonance then also they behave as an electrophile like you see here we can draw the. Diagram like this electron okay are you getting it Suhas any doubt no sir fine okay so you see this sulfur right. This sulfur has also tendency to accept electron because obviously oxygen is more electronegative so drag electron pair towards its side okay and this kind of structure is possible okay. So this also not very important but this also behaves as an electrophile if they ask you this question in the exam that which one behaves as an electrophile for those kind of questions you should keep that in mind. In resonance structure also this kind of thing is possible this you see CO2 O minus C plus double bond oh this also behaves as an electrophile this carbon and this okay. But aren't these like these are imaginary structures right so why would they like. That's why that's why these are the out of the all among the all thing that we have discussed these are these are the weakest structure we have weaker than the vacant D. But these are imaginary structures for a for a particular instant this kind of separation is possible agree. Okay right for that instant only it behaves an electrophile that's why these are very weak or the weakest one among the right. But why I am discussing this sometimes in the option they'll give you this SO3 which one of this is an electrophile or behaves an electrophile then you should have this information that under this circumstances it may be may behave as an electrophile. Yeah the oxygen part can act like a nucleophile right can act because but the tendency will be see it depends I'm talking about electrophile now right so if the reaction in the reaction. If the condition is there so that it behaves an electrophile then it will behave an electrophile but since oxygen also has tendency to donate electron. So for the given condition this part can also attack on to the on to the substrate okay but what kind of substrate is there according to that the behavior of this molecule. Okay but second thing what we can discuss oxygen being more electronegative element correct so it has less tendency to lose electron. Right but carbon has positive charge on it is not that stable octet is not complete so it can accept electron easily from the substrate. Yes okay but like you said we cannot you know neglect the possibility of oxygen to donate electron but yes it is difficult. Yes okay so the second type of reagent nucleophile right down nucleophile means what nucleus lover right nucleophile nucleus lover and nucleus is positively charged. So we can say it these are the electron rich particles right down these are the electron rich particles has affinity towards the nucleus positive charge species as affinity towards the positive charge species. For example you see CH3 you see this one Richard negative charge this behaves a nucleophile correct yes but here we do not have positive charge if it is positive charge then this behaves as an electrophile predominantly. Okay similarly pH O minus nucleophile or another example the similar one as we have the first one this negative charge these are the negative charge species right we can have some neutral also neutral one which are this neutral ones are what electron. So what's the last compound in the first frame it is trimethyl okay carbon this is the carbon trimethyl attack means what happens this hydrogen. This is the hydrogen. NS3 lone pair electron rich behaves as nucleophile even acid also you see the lone pair on oxygen atom behaves as a nucleophile. Amines are NH2 lone pair on nitrogen atom behaves as a nucleophile tertiary secondary amines nucleophile. Tertiary amines also behaves as a nucleophile but because of bulkiness tendency will be less. Yes H2O and anything H2O point is it molecule must have electron rich means electron density will be there some pair of electrons available for donation. So there was once a question which the which are the following conditions are for a nucleophile and there was an option a negative charge and lone pair and the answer was lone pair not negative charge. Can could you show me that question. Yes I'll send it to you. Okay there's one more term here that is nucleophilicity we'll discuss that later the definition only you write down nucleophilicity is the electron donating capacity electron donating capacity of a nucleophile electron donating capacity of a nucleophile this also depends upon the nature of the solvent. It's also depends upon the nature of the solvent okay so if I ask you NH3 and H2O which one is a better nucleophile NH3 and H2O. So does it depends on the solvent sorry. Depends on nature of solvent. No like solvent party you let it be we'll discuss that. H2O. H2O. H2O. H2O. Why H2O? Because you have two lone pairs. You have two lone pairs. But anyway it can only donate one lone pair and in H2O the O would also pull the lone pair but in NH3 that's just hydrogen so it would be easier to give the lone pair I think. See whenever you have H2O and NH3 both however H2O has two lone pairs but both lone pair water molecule cannot donate because if it donates one electron pair it becomes positive. See what happens here. If you donate H plus lone pair then it forms suppose it donates lone pair to H plus it becomes S3O plus one lone pair is present. Further donation is not possible because it is electron deficient now highly it becomes unstable. Here also the same thing it forms NS3 plus so however water molecule is 4 plus lone pair but we will compare the donation tendency of only one lone pair and then we say what that the nitrogen is less electronegative than oxygen hence electron donating tendency of NH3 is more than to that of the H2O molecule. So we can say nucleophilicity of NH3 is more to that of H2O. We will see this thing again. Nucleophilicity will discuss later the solvent effect. Third type of. Sir is there any symbol for nucleophilicity? What symbol? Like to be represented by any particular symbol. No, it's not. Free radicals are what? These are also electron deficient. Correct. We know free radicals. What is the free radical? Fine. We have discussed this in our GOC guys. Okay. So just go through it. Free radical is this. It is neutral. Electron deficient. You remember I have told you that to compare the stability of free radical. You can consider this as carbocation. Correct. So both are electron deficient. That's why I said this. However, both are different intermediates. Okay. So neutral but electron deficient. So since it is electron deficient. So we can. So we can say it is a type of electrophile. Type of electrophile. Another type of electrophile. So nuclear free radical cannot be the charged species. Unlike unlike the electrophiles. Okay. Electrophiles. There is the first case we have discussed that these are charged or neutral species. Here it cannot be charged. Right. Unpaired electron present on it. Two examples we have. For example, you see chlorine radical. One unpaired electron. Bromine radical iodine radical. You can write our own radical. Yes. Three radical. R C double bond. Oh radical. These are the examples. So how does the last one come from. This one. Yeah. It is a sale radical. Oh, this also has tendency. Let me tell you one thing. Free radical has tendency to eliminate. A neutral molecule. A neutral molecule and forms a new radical. So how this CO2 eliminates and forms alkyl radical. Here we have homolysis possible. This one electron is taken up by this carbon. One electron is taken up by this alkyl group. And this radical, this radical forms a double bond here. And this bond breaks. So it forms alkyl radical and CO2 molecules goes on. Remember if this, this property we use in that reaction. Cold bay electrolysis. Remember that. So this is the property of the radical. That depending upon the condition, it can eliminate any neutral molecule and forms a new radical. Okay. Now these are like, see it's almost like the past from the past. Almost one and one or 15 minutes we are discussing this, but these are the boundary of the reaction intermediate. This, these informations you should have. Right. Understanding plus information you should have. Now we are going into the concept like how carbocation or intermediate forms, what all reactions are there, which, you know, in which those kind of intermediate forms. And when we talk about reactions, we'll take reactions of all the chapters of organic chemistry. That's why I said it is not a specific chapter. Okay. We'll see the reactions of alkyl, alkyl light, everything. Okay. So just a second. Sir, I have a doubt. Yeah. So in the start of these types of particles. Yeah. So you've given us that CL plus can also be a nucleophile. So can this chlorine free radical also act as a nucleophile at times? And you see, we haven't a specific, you know, definition of this free radical since we have defined. So it's better to say this is a radical, not a nucleophile. Okay. So it's a choice. Basically, suppose you have options like free radical and nucleophile, then we'll choose free radical. If, if only one correct is there. Yes. Right. So that's the thing. So guys, so first we are going to start now. This, this is, these are the very important, important one. You have to understand this very carefully. So right on the topic here on the, on the heading you write on reaction intermediates. We have seen the basic definitions of various reaction intermediates, examples, everything we have discussed. Right. So first thing, the purpose here is we have to discuss what all reactions in which this intermediate forms. So carbocation, the first one, we have already discussed this in GOC, the properties of carbocation, et cetera, but quickly I'll write down here. If you have any doubt, you can ask me. If not, then we'll go ahead. Carbocation is this carbon has three bond with one vacant orbital. Yes or no. Vacant orbital means one positive charge on carbon atom. Now what are the properties? What is the hybridization of carbocation? SP2. SP2. SP2 hybridization means trigonal? Plano. Yes. Trigonal planar. It has one vacant P orbital. This is vacant P orbital. And this P orbital is perpendicular to the plane of this triangle. Right. All electrons are paired. You see it is diamagnetic in nature. There's no unpaired electron present. It has three sigma bond. It has zero lone pair, no lone pair present on to this. It behaves as a Lewis acid. Why this behaves as a Lewis acid? Electron. Electron. Sorry. Electron deficient. Electron deficient. It behaves as an electrophile also because it has only six electron in the outer motion. Six electron in complete octet. There are many properties you can write. Okay. So they ask this kind of questions at which of this statement is not true regarding a carbocation. Okay. So once you know this structure, you can conclude all these things one by one. Okay. So it's not a big deal. Now. Can all Lewis acids act as electrophiles? Sorry. So can all Lewis acids act as electrophiles? Yes. You can say. Lewis acid means what? Electron pair acceptor. Correct. Yes. Electron pair acceptor means electrophiles. Correct. Okay. These are the carbocation. And we know we have seen the reaction that wherever carbocation forms, rearrangement of carbocation is possible. Correct. Sir. So what is the last point you wrote? Incomplete. I've written incomplete. Incomplete octet. Okay, sir. Okay. Now the first thing we'll see after this. That is rearrangement of carbocation. And we have discussed it already in hydrocarbon chapter, but here we'll discuss in details. Okay. So what are the possibilities of rearrangement of carbocation? What are the possibilities we have? How rearrangement possible? Right. Shift. One example on this. I'll shift. I'll shift. There are many possibilities. Okay. So this example you see. CS2 plus. Write down the most possible. Stable carbocation in this. Done already. Let me know once you're done. I'll discuss it. Okay. Fine. So just I'll write down the answer here. Just tell me whether you have got the same thing or not. This one. Is this correct? No, sir. Do we have to move it again? Again. Three degree. Three degree. Okay. Three degree. Yeah. Here what happens? Again. Again possible. Yeah. Two times. Two times, right? Yeah. Two more times. So no, sir. So it's the same level. So it doesn't need to have the same number of half. Coming. This has one less. Coming. This is the possibility, right? Okay. Now, what happens here? Like this hydrogen. Take this bond pair of little shift on this. And that's why we have positive charges. Same thing happens here. This hydrogen takes this bond pair and shift on this carbon. We'll get here. And further again, this hydrogen comes over here. We'll get this. Right. See, a shifting of carbocation is possible. If you are getting the more stable carbocation in the next step. Means this carbocation, if it is more stable than this, then only this shifting is possible. Right. So here you can see this carbon has positive charge and it is primary. And this one is tertiary. So obviously this one is more stable. So this shifting is correct. Okay. But I would suggest what don't see primary tertiary secondary carbocation. You check alpha hydrogen. That is the best way to do this hyper conjugation you check. So for this carbon, the number of alpha hydrogen is one, but for this carbon, the number of alpha hydrogen, can you tell me how many alpha hydrogen we have here? Seven. So obviously more number of alpha hydrogen, more hyper conjugative structure, and then this one is more stable. Those who are new Rahul and Subhas, if you are having any doubt, I have the video of this, if you want, I can share that with you. Okay. So I am using all these concepts. I have already been discussed this thing in GeoC chapter. Okay. So if you have any doubt, reach out to me personally anytime. Okay. Next one you see. Next one you see here. How many alpha hydrogen here? Two. Number of alpha hydrogen here it is two. Number of alpha hydrogen here. Four. Four. Obviously this carbocation is more stable than this. Hence this shifting is correct. Further if you shift, you'll get this number of alpha hydrogen. Eight. Eight. This is even the more stable than this one. This gives you this carbocation. Correct one. Number of alpha hydrogen here. Two. Number of alpha hydrogen here. Five. Five. Number of alpha hydrogen here. Three. Three. Number of alpha hydrogen here. Eight. Eight. But you see one thing. This carbocation is more stable than this. Why I am not considering IE effect you see here. With this example you'll understand this. This carbocation is two degree or one degree. Two degree. This carbocation. Two degree. Two degree. Secondary. Secondary. Correct. Consider IE effect you won't be able to judge that stability of two carbocation here. Correct. Because number of alpha hydrogen here. That's why I'm suggesting you to check alpha hydrogen. Right. Okay. Understood. Yes sir. The first thing this has five. This has three. So this shifting itself is not possible. And hence this carbocation won't form in this particular molecule. This won't form. You can do this. Right. You can shift this. But the reaction won't do this. Are you getting it? Yes sir. The reaction will do the same. Stable plus is continuous. Right. You cannot. You can shift the hydrogen here. But reaction itself won't go through this kind of shifting. Because if this happens. This carbocation is less stable than the previous one. That's why this won't form. And hence this also won't form in the reaction. Did you understand this? So what's the maximum number of carbons that it can jump? Maximum number of? So like why can't it go directly from the five to eight? It won't go directly. One to hydride ship possible. See actually let me discuss this thing. All of you guys try to understand. Good question. I'll explain why. See actually what happens. I am talking about only the shifting of hydrogen. Which is present at the adjacent carbon. Right. Yes. No. I'm not talking about. It's like one by one shifting we are talking about. Correct. I'm not talking about the shifting of this hydrogen to this carbon. The reason of this is. That because of this positive charge. This electron pair is slightly attracted to watch this positive carbon. But that kind this kind of attraction is not possible with this hydrogen. Are you getting it now? So this Sigma electron is weakly attracted to watch this. That's why this hydrogen shift over here. Further it positive charge comes over here. Then this hydrogen shift over here provided it. The stability is more. So if you talk about the shifting of this hydrogen to this carbon or this hydrogen to this carbon that is not possible because this Sigma electron or this Sigma electrons are not attracted towards the positive charge because of the distance. Clear from. Yes. Okay. So that's the reason. Did you understand this? Correct. After shifting a carbon. If they're lesser alpha hydrogen but it undergoes resonance. Can it shift? Then it's possible. Yeah. Then it's possible. Because the point is when you draw the carbocation through shift, right? Any kind of shift that carbocation should be more stable than the previous one. Then only it's possible. That's a bottom line of all these things. Okay. Yeah. Now I'll give you some examples of resonance also. Since you have discussed this will talk some examples of resonance also. You see this one. The ring with oxygen here and positive charge. Do we have shifting possible in this? Try this one. All of you try this. One very good example. I'll show you here. Then guys. So one minute sir. So what do we have to do? You have to write down the structure with more stable carbocation if possible by hydride shift. Got it. Got it sir. Yeah. Right. Done. Done. I'll do this. You just let me know whether it is right or wrong. Okay. So I'll shift this hydrogen here or this hydrogen here. Which hydrogen tell me. So the bottom one. The bottom one. Okay. If I shift the bottom one. So we'll get a positive charge here. The oxygen might. It has lone pairs so it can put it in. Wait. No. That's good. Okay. So the thing is here you see now guys you try to understand. Could you tell me the number of alpha hydrogen in the first structure? Four. Four. How many alpha hydrogen here? Two. Two. Two. Is it possible? Yes sir but there's resonance. Yeah you see here however the hyper conjugation is less but this carbocation is resonance is stabilized by this lone pair of oxygen. Lone pair sigma vacant P the conjugation is there hence resonance is possible so this carbocation is resonance stabilized. Resonance is stabilized. The first one the carbocation is stabilized through what? through what? Hyperconjugation. Right? Clear? Gahul Subhas, this one? Yes, sir. So resonance, no resonance dominates hyperconjugation, hence this carbocation is more stable, right? Possible this conversion. Okay. Other thing you see here CS3O CH CS3 positive charge, the same logic true, right? Same logic true it is, because again this positive charge is involved in resonance with the pi electron of. So here how many alpha hydrogens are four again in the first one? First one the alpha hydrogens is two, here the alpha hydrogens you see it is three, but it is resonance stabilized also. Okay. Here the possibility is CS3O. Now you explain me this thing. I'm telling you this shifting is possible. This is correct. This one is more stable. The reason you tell me. One by one. Yes. It is extended conjugation with the benzene ring and the lone pair on oxygen. See here what happens this carbocation is in resonance with the ring, right? Here if you shift this here, we have resonance with this ring, but this carbocation is more stable because of the plus M effect of this OCS3 group, electron releasing. So extended conjugation you can say plus M effect also you can say this kind of shifting is possible, right? All of you understood quickly. Okay. One more thing one time we'll use here and that is see the shifting takes place between first and second carbon. This first and second carbon, this number is not at all related with that numbering in nomenclature. Just what we do, we give first number to the postage carbon and the azazine one the second carbon. So this kind of shifting we call it a 1, 2 hydride shift. We also call it as this carbon is alpha carbon and this is beta. So it is alpha beta shift also we call it as all our same thing. Just the name will use not more than that. One more thing you must keep this in mind that in the question wherever the carbocation forms, right? So they won't ask you this. They won't ask you to rearrange the carbocation. You have this awareness that in the reaction wherever the carbocation is forming as an intermediate, you have to have the most stable carbocation as an intermediate and then you write down the final product. Okay. It is understood what first shift I didn't get to us. One to hydride shift is this only. You see, I'll tell you what is shifting here. Hydrogen is shifting between first and second carbon, right? So since hydrogen is shifting, so we call it 1, 2 hydride shift. So in short, we write it as H shift. H is hydride. Got it? Okay. Similarly, guys, we have this thing that hydride shift is possible. Similarly, we have phenyl shift and alkyl shift. I'll give you one one example and then we'll move ahead. Okay. The second kind of shifting you write down phenyl shift, shifting of phenyl group benzene ring. For example, you see CCH2, the basic reason is guys that only that the carbocation that you get in the next step should be more stable than the previous one. In this, it is possible or not? Yes. Possible. Reason? Alkyl shift. So in both the molecules, the shifting is possible. This takes this electron and shift onto this carbon atom. So the carbocation that we get here is pH, CH2 pH, and here we have the positive charge. Why this carbocation is more stable? Because it is in resonance with the two ring. Okay. More conjugation we have here. Okay. This kind of shift, this comes over here and we get CH3, CCH3, CH2, CH3. Here we have the positive charge. You see the number of alpha hydrogen here it is 8 and here it is stabilized through I effect plus I effect. Okay. So this one is more stable. Hence the shifting is possible. But isn't phenyl like a bulky group? So how does it just move? No, here, bulkiness we do not consider. You can say that here we have the hindrance, right? When it's shipped over here, the hindrance is decreased in this structure. Stability also increased. So this is possible. Okay, sir. Okay. Two more examples I'll show you. So why isn't it possible for the hydride to shift from one of the phenyls into the CH2? Hydride from phenyl to CH2. Yes, sir. No, it's not possible. Why? Why not? I'll tell you. Distance is high. Okay. This is phenyl, right? Yes, sir. First of all, here we do not have hydrogen, right? And the bond is this. This hydrogen shift over here will have the positive charge on this and this is not at all stable. Even if you see phenyl cation, this is not at all stable. It won't form. That's why we'll discuss this later. Aryl halide may halogen atom like this bond is difficult to break. This bond is difficult to break because if this breaks, we'll get a positive charge like this. And this phenyl cation is not at all stable. Okay, sir. Okay. It is not in resonance with the ring actually because the p orbital is in different plane. It's not parallel with other p orbital. I thought it would be resonance stabilized. No, it's not. It's not. Okay. We'll discuss that in Aryl halide halogen containing compound. Okay. Now, just I want to understand two more things with you. Suppose that we have this CH2 processor right down the stable carbocation if possible. This comes under which Alkyl shifter is just general. What? So which like category does this come under? Like are you giving like general questions? It's a general shifting off. You have to get more stable carbocation. So what shift is possible? Alkyl hydride or phenyl? And the first one will be the edge move to the CH2 plus and form CH3. Yes, sir. Yeah. Okay. You see, I'll write down two possibilities here. Your answer is correct. But I would like you to show one different thing. If you shift the hydrogen hydride shift, you'll get this CH. If you shift Alkyl group from here, not from here, from here, if you shift Alkyl, I'll write down here, right? If you shift methyl, then you'll get this HC positive charge CH2 CH3 and pH. Okay. If you shift phenyl, then you'll get this. This is the only three possibilities you can think about. Yes or no? Yes. Now we have to check the stability of these three carbocation. The one which is the most stable, that shift favors here, right? So here you see the carbocation is stabilized through resonance as well as hyper conjugation is also there. Okay. So R plus H, both effect is there. Here we have only R and here we have only H. Am I right? The second thing is 2 alpha hydrogen. What? So there are two alpha hydrogens. Which one? The second one where we shift CH2. We have R plus H also here, correct? Very good. But in the first one, the H is greater than 2. Yes. So obviously this one is the least stable, yes or no? Yes sir. In this two, resonance is fine, but number of alpha hydrogen here it is what? Five. Five. Number of alpha hydrogen here? So six. Six, yeah. Six. Number of alpha hydrogen here? Two. Right? So that's why this carbocation is the most stable one. And this gives you the major product in that reaction. So how would the second one form? Because you told us that thing about the hydrogen right next to it being weakly effected, but this CH3 is two carbons away. No, it's not. You see, this bond is dissociating actually. You see, this bond we are taking. You know, this sigma bond is also attracted toward this. This sigma bond, this sigma bond, this sigma bond. Got it? Yes sir. Okay. What about this? So one of the methyl group. So that tri-methyl group will go? No, the other one will go. Tri-methyl group. If tri-methyl goes, then you'll have more alpha hydrogen. How many? How many? Last eight. Eight. Sir, but if you don't move the tri-methyl, it will get 15 or 14. Oh, wow. If tri-methyl goes, we'll get this. CH3, CH2. Yeah, move that. Positive charge. This is the positive charge. Sir, we don't move the tri-methyl. We move the other one. I'll draw both, right? Yeah, just. Tell me which one is more stable? Second one, sir. The more alpha hydrogen. Number of alpha hydrogen? Fourteen. How it is 14, guys? You don't count the one and two. This carbon, this carbon has no alpha hydrogen. This is the carbon Oh, yeah, yeah, yeah, yeah. Five here and this here. Yeah, yeah, correct, correct. Oh, yeah, I was counting it wrong. Five and this one is eight. Eight. So this one is the two. So what happens in case of more than one alkyl group is present, then bigger alkyl group will shift. That is what you can conclude from this. One note you write down here. In reversible reaction, in reversible reaction, we always write down the most stable carbocation. Reaction proceeds through the stability of carbocation in reversible reaction. Just one note, in the first structure, for the phenyl group, are there any alpha hydrogens? Which group? Penyl group. No, no, no. Penyl group won't have any alpha hydrogen. Because this carbon, which is attached with this carbon, there's no hydrogen attached with that. When you draw the ring, you will understand. Okay? So this point is very important. Reversible reaction will form the stable carbocation. Second point to write down. If the reaction is irreversible, if the reaction is irreversible, then it is controlled by, then it is controlled by, then it is controlled by, just a second, then it's controlled by the kinetics, controlled by the kinetics of the reaction, the kinetics of the reaction, comma, not by the stability of intermediate. Not by the stability of intermediate. Not by the stability of intermediate. Sir, can you explain second? Wait, wait. Just complete this line. Then I'll discuss this. Not by the stability of intermediate. Hence, migratory aptitude. Hence, migratory aptitude is applicable. Right? Migratory aptitude is applicable. Now, what is the doubt? Statement, can you explain that? I am explaining. I'll explain this. See, irreversible reaction if you have, what happens? In irreversible reaction, suppose A gives B. Okay? So, in this reaction, the rate is more important. Rate of reaction is more important than the stability of intermediate. That is carbocation. Okay? So, these are kinetically controlled reactions. Okay? These are kinetically controlled reaction. Irreversible. So, in kinetically controlled reaction, what we do? We do not use the stability of intermediate concept. Okay? But we use the migratory aptitude. What is migratory aptitude? I'll tell you. Just let me take some examples and then let me explain this. First of all, you have to understand or keep this in mind. Then migratory aptitude is applicable only in irreversible reaction. Right? Only in irreversible reaction. Correct? What all irreversible reaction we have? We'll discuss that. You have to keep that in mind. Okay? And the migratory aptitude, only three things you have to keep in mind. For hydrogen, this is a maximum. Then we have phenyl group and then we have alkyl group. This is the experimental order we have of migratory aptitude. Experimentally determined order based on the major product of the various irreversible reaction. We have got this order. So, this order you have to memorize. First thing. Second thing. Could you just how likely it is for that thing to migrate? No, no, no. That you can see. But then phenyl and alkyl, you have the contradiction into this. Okay. So, that's why I suggest you to keep this in mind. Yes, light one can migrate to what there are many things actually. It's basically the position of bond pair of electron, which side it is. Okay. But that will contradict. Actually, in case of hydrogen, the size of this is dominating. In case of this, the electron withdrawing nature is dominated. And hence, the order is this. So, two different things we use for explanation of disorder and disorder. So, better to keep this in mind. Only three things are there. It's more than enough. You're not required anything else. Okay. Whatever is there, some effects are there. Then we'll discuss that. Okay. Electron releasing tendency and all. So, for you see what I'm assuming here, now we are taking only irreversible reactions, of course. So, what exactly does migratory ability mean? Wait, wait. I'll just explain this a second. Suppose you have this molecule and the reaction is irreversible. So, we know in irreversible, the stability of carbocation is not the factor. Correct? That is what the statement I've given you. Okay. Now, just a second. Here we have hydrogen. Okay. So, here we do not use the concept of stability of carbocation. Since it is irreversible. But we'll see the migratory aptitude of the group attached to the adjacent carbons. So, we have CS3 and H attached. So, which one has more migratory aptitude? Hydrogen. So, hydrogen will take this electron pair and migrate onto this carbon atom. So, this gives you CS3 CH positive charge and CS3. However, from this example, it is not clear that we are using migratory aptitude or stability of carbocation. Correct? Because this carbocation is still more stable than this. But since you have, suppose you have an example like this. Suppose you have an example like this. CS3, here we have phenyl group and CS2 plus. Irreversible. So, what we use? We use migratory aptitude. Correct? So, according to the migratory aptitude, phenyl group should migrate. Because phenyl has more migratory aptitude than alkyl group. This takes this bond pair of electron and migrate onto this positive charge carbon atom. And here we have the positive charge. Now you see this is not the stable carbocation. Right? This is not the stable carbocation. Because if this alkyl shift over here. So, this positive charge is resonance as well as hyper conjugation is stabilized. Correct? Or the another possibility will give you what? You see this? If you have any doubt, I can have this carbocation. Okay? This carbocation you get with this. Now this is possible. The second one is possible. If the reaction is reversible, this is with irreversible migratory aptitude. Not the stability of carbocation. Obviously you see this carbocation is more stable than this. Yes or no? Yes, sir. Right? So, this is the stable carbocation. This is lesser stable carbocation. But since the reaction is irreversible, so this will form, not this one. This is what you have to keep in mind. Okay? So, sometimes what happens? Second? Sir, how is this possible? No, no, no. No, here. No, here. First of all, we are not talking. See, we are not talking about the stability of carbocation here. The stability of carbocation we are not talking since the reaction is irreversible. Right? Right? Reaction is irreversible. So, stability is not a factor. This reaction cannot go into the backward direction. But rate of the reaction is important. Okay? Rate of the reaction is important. And hence we will use migratory aptitude. So, migratory aptitude of this and this. If it is reversible, the first one will still be more right. I don't understand either. This one, this one. Sir, the first one will be more stable. This one? Second? Resonance. Resonance. Okay. Guys, don't do like this. It is resonance stability. Okay. Understood. This is resonance stability. So, that's the thing, right? Irreversible or migratory aptitude, reversible, more stable carbocation. Second thing, it is difficult to memorize which reaction is reversible, which one is irreversible. That's the challenge, right? That's what I said, Varun. So, reversible reaction has tendency to go into backward direction also. Both side reaction is possible, right? So, that's why stability of intermediate is an important factor. But irreversible reaction, backward reaction is not possible. So, at what rate the reaction is proceeding, that is the more important thing. We have other conditions also, temperature and other things that we'll see in coming lecture. KCP and TCP will discuss that. Okay. But till now, you just keep this in mind, irreversible will deal with rate of the reaction, reversible stability of intermediate, that is carbocation. Is it clear? Few things about the migratory aptitude is important for you to keep in mind. Suppose you have only alkyl group attached to the adjacent carbon, right? So, the bigger alkyl group will shift or migrate onto the carbon atom. For example, you see, if you have this CS3C, CS3, and we have this three group present CS3. So, all are alkyl group. So, here what happens? The bigger group migrate because it has maximum plus i, electron density is maximum here. So, it will be more attracted towards the positive charge carbon atom. The product here is not according to the stability of carbocation, but it is the migratory aptitude. However, it is same for the stability of carbocation also, correct? So, if you have alkyl group, the larger group migrates, right? So, right down here, due to maximum electron density, due to maximum electron density, the larger group migrates, the larger group migrates. Hence, better electron donors, hence better electron donors are better migrators. More electron donating tendency, more migratory aptitude. That does not go with phenyl and alkyl group if you see, correct? That's why I said that is the experimental thing. Hence better electron donors are better migrators. So, here if we had a phenyl instead of one of the methyl groups, will the phenyl still go? Yes, phenyl will go. phenyl is more than alkyl, phenyl will go. Because if phenyl does not shift, then the carbocation will be more stable, resonance stabilized. Okay. That's why we'll shift phenyl. Yes. Okay. Few order you must remember of, I don't, migratory aptitude on the top. Hydrogen among their isotopes has most migratory aptitude, maximum migratory aptitude. If you have molecules like this, benzene being attached with the molecule. This is not methyl, this is an open bond guys. It is attached to the adjacent carbon from this, from this side. Here we have OCS3. You need to write down the order of migratory aptitude, migratory nature of this. Sir, is that electron density applicable even for phenyl groups? Yeah, I said no. Better electron donors are better migrators. So, for the fourth one, do we have to consider static? Static crowding, yes. So, first one is lesser than the second one. Okay. Static crowding decreases migratory aptitude. So, for the fourth one, the order will be this. Sir, but won't I effect be stronger for the first thing in the fourth one? No. Actually, when you shift this, then the methyl group will provide steric hindrance. So, that won't allow the formation of bond between the phenyl and the carbon atom. Difficult to get that bond. Okay, sir. This one, which one is the maximum, has the maximum migratory aptitude? OCS3, sir. OCS3. OCS3. And why OCS3? Because here we have plus H and there's no effect here. So, it has more electron releasing tendency, one, two and three migratory aptitude. So, what's plus H? Plus H is hyperconjugation, electron releasing. Oh, but plus I also, right? Yeah, plus I is, yeah, electron releasing. All plus effects are electron releasing. So, plus H is positive hyperconjugation. Okay. What about NO2? CH3, then the plane one, then NO2. Plus H and there's no effect here. So, this is one, this is three, and this is two. And here we have steric crowding. The reason is steric crowding. Okay. Fine. A ring expansion we have discussed, right? Yes, sir. We are talking about the stability of carbocation here. So, cyclic ring rearrangement is also a factor through which the carbocation gets stable, right? So, since few of you were not present in the last session, so a little bit, we'll discuss it quickly and then we'll move on. So, rearrangement means what? Ring expansion and ring contraction. Through both way, carbocation gets stability. So, this is the natural tendency for a ring to get, convert into the higher member ring. Because the stability order is this three member, four member, five member, six member, and six member is even more stable than the seven member ring. It's a seven member ring, right? Okay. So, that's why the lower member ring has natural tendency to convert into higher member ring. Right? Because of angle strain. Okay. So, condition for ring expansion is the carbon atom which is attached to the ring must have positive charge on it. For example, suppose we have a four member ring like this and this carbon atom which is attached to the four member ring must have positive charge on it. Then only ring expansion is possible. Okay. Yes, sir. Okay. Now, how do we write down the ring expansion, the product here? What we'll do first of all, quickly I'll explain this. First of all, we'll number the carbon atom. One, two, three, four, five. So, we'll start the numbering from the positive carbon atom then to the ring. Then what we do to draw this structure, since we have total carbon here is five, we'll draw five member ring. Directly, we'll draw five member ring. Okay. So, what happens here? Actually, the bond pair either between two or five carbon or two or third carbon, any one of this bond pair you can take. Okay. This bond pair attached with this carbon atom. So, intermediate state is this. You will have this thing. This bond is about to break. This bond is about to break. And this bond is about to fall. Like this. So, this, you can say this electron pair comes over here and then this attached with this carbon atom. Like this it goes. To understand this, you can say this. This bond pair comes over five and then five attached with one. So, five attached with one. So, we'll write one here, five here, two this side, three and four. Since the second carbon atom loses its bond pair of electron. So, we'll have a positive charge on the second carbon atom. This is the more stable carbon atom. What is this? Param, by mistake you sent this to me, right? You won't break. I understood that. I'll give you some time. I'll just give you a break. Did you understand this? Subhash and Rahul. So, four member ring. You see, why this is more stable? I'll tell you. Four member ring convert into fifth member. Five member ring. That's why here we have less angular strain, more stable. A few examples I'll see. Could you write down the product in this? If ring expansion is possible? Positive charge. Possible in this ring expansion? So, isn't it already a sixth member ring? So, three will become a four and they'll share a carbon, I think. What is the condition of ring expansion? Positive charge carbon attached to the ring, right? So, for this ring, the positive present on the ring. For this ring, ring expansion is not possible. But for this one it is possible, right? Yeah. Right. Will it be a spiral compound? Yes. Got it? Yes. Okay, now again you think about it. This carbocation that you get, is it more stable than the previous one? Yes, sir. It's a four degree carbocation. Four degree you are getting? Yeah. Okay, what about this? In this expansion possible? Paramp? Yes. Did you say yes? Yes. That's a dancing resonance. Yeah, dance resonance. Yeah, sir, it's dancing resonance. So, in this the expansion is not possible? Oh, yeah. Does it look like same thing? Yeah. This part? Oh. Got it? Yeah. And this also expansion is not possible. Okay, sir. So, three member ring can't expand? No. If the postage are present like this, then it is already more stable. So, dancing resonance. Could you explain dancing resonance in short? Dancing resonance is nothing. See, actually this carbon atom is sp3 hybridized, right? Yeah. So, it has very high angle strain because the angle is 60. It should be what? Should be 109 degree approximately? Yeah. But it is actually 60. So, the angle strain is 109 minus 60, around 49 degree of angle strain is there. That's why it is unstable. This ring is unstable. But what happens? Why expansion is not possible? Because this ring is in shows dancing resonance. Dancing resonance is a special kind of resonance in which sigma electrons are involved. You see, in resonance we always define only lone pair and pi bonds are involved, right? But here what happens? This sigma bond jumps over here and forms a double bond here. So, it goes like this, CH2 and positive charge, right? Now, because of this you see, this carbon atom becomes sp2 hybridized, right? What should be the bond angle? Sp2? 120. 120, right? But since it is an open bond here, so it can rearrange itself to get the bond angle. Okay. That's why the angle strain decreases, becomes zero here and becomes stable. Okay. Rahul and Suhas, I'll share the, you have the login credential, right? Yes, sir. There is video of Geosites there. You can go through that. Okay. All these things have been discussed there properly. Yes, sir. Sir, what is the previous one did undergo ring expansion? How did it look? Previous one. Which one? This one, the one. This one? Yeah, if it did like. If it is not possible, so how do you draw the ring expansion? No, like, I mean, like, can you give us another question? Yeah, I'll give that another one. Just a second. After giving the other question, can you give a break? Okay, fine. After one question, we'll have a break. Sir, if I had two rings that were connected to one carbon that was positively charged, which ring would expand? Like two different sizes? Sir, you told us there are positive charges on the ring and there won't be any. Yeah, are you there? Yeah. Are you getting it? Where is Abhiram? Abhiram, alive. Sir, isn't the positive charge on the ring? Sir, in this case, there's no expansion that can happen, right? Yeah, I don't think there should be. My mistake at this, it should be here. No, it's fine. Presidents still happen when it's like that, when it's on a ring. On the ring, contraction is possible. But here we don't have contraction because there's six-member ring. We'll discuss that after this. Just do this one first. Sir, will it be like a six-member ring and then like a square which shares one side with the ring? So you'll get this. See, what happens? This carbon and this carbon will get attached like this, okay? So it will be like this. And where we have the positive charge? The second one here. Which is here. Got it? Yes, sir. Let's say the positive charge comes there. See, I'll tell you. You do one thing. This bond pair, suppose, this bond pair, it comes over here, okay? And then this electron pair attacks onto this carbon atom. So this bond pair, this carbon atom loses this bond pair of electron, right? Okay. That's why this carbon atom has positive charge. Okay. So it is one. It is two, three and four. Understood? Yes. You will always have positive charge on the second carbon. So I'll suggest you to number the carbon atom like this. It helps you to determine the structure. Okay, one last thing. One last thing before the break. Just two minutes. We'll finish this one. Suppose you have a ring like this. Three-membered ring. Three-membered ring. Varun, is it clear? Akshit, are you getting it? Akshit, right? Yes, sir. Advik, are you there? Advik is not there, I guess. Akansha. Sir, here there will be two possible structures, right? Two possible structures. Just a second. Akansha, you're not there. Anush. Okay, fine. Anush. Okay, Advik is there. Akshit is there. Dhruv is there. Dhruv. Sartik. Kirtan. Yes, sir. Kritika. Yes, sir. Mahit Mehul. Mahit Mehul. Param Pradyut. Pranav is there. Yes, sir. Advik. Simon, are you there? Right, right, right here. Right, try this. Correct. Sir, will it be like, a lot of, everything will start becoming into bi-cycle, bi-cycle? It forms a chain reaction. Yeah, chain reaction, that's what. Everything will start again. See, first of all, you'll get this. Here, you'll get... Sartik keeps going on. Positive charge. And then, this will be, as it is. It'll all take turns becoming a square. Yes, again, we'll have ring expansion here. Right. So, once you do like this, continuously till here, you'll get a structure which is similar to this, which is this actually. Four-membered ring. Four-membered ring. Is it clear? One by one, you'll draw. Like, I have done this the first part. Then, again, bring expansion for this. Again, bring expansion for this. And this, you'll end up getting this. Got it? Yes. Fine. If I do a structure, how would that expand? I sent it to you on WhatsApp. I'll see that. Okay, guys, take a break. We'll start at 11.45, okay? Pradeep, we'll discuss this after break. Correct? Okay, sir. Yeah, take a break, guys. So, when should we come back? 11.45. Okay. Thank you, sir. Okay. Hello. Hello, sir. Yeah. Yes, sir. Shall we start? Yes, sir. Yeah. Carbo-cut and ring expansion, we have discussed. In this, there are few cases in which the condition of ring expansion is there, but we do not use that. We do not go for ring expansion under those conditions. What is that we'll see? See, suppose you have a molecule like this, OH. And we have a positive charge, positive charge on this ring. So, we can have, first of all, hydride shift from this hydrogen so that the carbocation is resonance stabilized. Just a second. So, if you have this five-member ring, after hydride shift, you'll get this, correct? 1,2 H shift. And why this shift is possible? Because this carbocation is resonance stabilized with the lone pair present on the oxygen atom. Now, after this, again, ring expansion is possible, right? And we get a six-member ring, shooting expansion. But in this case, when you draw the resonating structure of this molecule, you will get this. Carbon double bond OH, positive charge. And here we have one H. This is a resonance structure. Resonance structure. This is the resonance structure when this lone pair forms a bond pair here, like this. Okay? Now, further, to stabilize this positive charge on oxygen, this H plus comes out. This hydrogen comes out. And we get this molecule. So, this is the product here. Point is, whenever we have possibility to get neutral molecule like this, then we do not consider ring expansion. So, where did the H go? It just leaves. Yeah. Okay. It just comes out, H plus. This is possible, this kind of thing is possible when the carbocation, carbon having positive charge is attached with the group like OH, NH2. Basically, one lone pair should be there. Right? So, here, the point I'm trying to make, here ring expansion is possible. Five can convert into six. But since by drawing the resonating structure, you are getting a neutral molecule. So, in this case, we always draw the neutral molecule. Okay. Ring expansion we do not consider. Right down. So, when carbocation is converting into a neutral molecule, when carbocation is converting into a neutral molecule, neutral molecule, then we do not consider ring expansion. But in this case, ring expansion will take place anywhere, right? Anywhere you won't take place. If it is converting into neutral molecule. Even if it didn't convert into a neutral molecule, would there be a way to have ring expansion here? Sorry, what? So, if the final product wasn't a neutral molecule, then how would the ring expansion take place? Then we'll have the ring expansion. I can give you some other example. Like, suppose if you have this thing, suppose positive charge here. Then in this, what happens? First of all, we have hydride shift. You'll get this. And then here we have ring expansion. So, in ring expansion, what happens? It is one, two, three, four, five, six. This comes over here as six attached with one. So, we have a six-membered ring where one attached with six and two have the positive charge of it. Correct? So, you see the first carbon has one methyl group also. So, this is the methyl group on the first carbon. And positive charge on the second carbon. Okay. So, after that, then the hydride shift back to the first one. Hydride shift? So, in the last step, after that, will the hydride shift due to the second carbon so that the first one has the positive charge? It will be tertiary carbocation. Just a second. So, in the last step, the hydride shift, again, two-one hydride shift will take place. Yeah. Stable carbocation. So, you'll get this, you're saying? No, sir. In the lead one. Oh, this one. In this, it's possible. Further, you can have this. Hydride shift you can take. You'll get the most stable carbocation. How can ring expansion happen in unsaturated rings? Unsaturation ring also possible. If you have a double bond here, this does not make any difference. You'll have a double bond between 4 and 5. And would a compound prefer to have resonance or ring expansion? If condition of resonance is there, then fine. Right. If it is not there, then we'll see resonance. If the carbocation is stabilized through resonance, then we do not consider ring expansion. Okay, sir. Okay. It depends upon the condition, what condition you have. Based on that, you have to check the carbocation by various possibilities, whether the carbocation is getting stability or not. Okay, sir. Okay. One more thing I would like to add here. Sir, this one. The first step itself was more stable than the second one. Here? Yes. Two, three, four. Okay, then I'll change this question. I was trying to make you understand this. Then we'll have this one. Two, two, four. Right. And here we have three plus one, four. So we'll have this. Correct? No, it's fine. Yes, sir. Okay. So we'll have a third and sixth carbon has two methyl group. Correct. So here we have two methyl group and here we have two methyl group. Here also we have two methyl group. Here also we have two methyl group. No, it's fine. No, sir. Yeah, just I was trying to make you understand how this reaction takes place. Okay. However, if you remove this methyl group, then this carbocation is itself more stable than this one. So this conversion is not possible. Okay. And to make this conversion possible, I have placed two methyl group here. Then it's possible. Anyways, so one more thing you see here, this is one possibility that whenever you have this thing, some neutral molecule possible, then we do not consider reading expansion. Even in case of neutral molecule, the six-member ring has tendency to convert into seven-member also. However, like I said, seven-member ring is not at all stable. But when neutral molecule is forming, six-member could convert into seven-member ring. Okay. Like you see, here we have the condition of resonance. Sorry, ring expansion. Okay. So ring expansion takes place here. Why ring expansion takes place? I'll tell you. Because after ring expansion, it is forming a neutral molecule. Suppose this is one, this is two, this is three, four, five, six and seven. Seven-member ring, you see this. Right. One attached with three. One attached with three. Right. So two, four, five, six, seven. Okay. And second one will have one positive charge on it and one OH. So you see again, the positive charge carbocation, carbon here, is attached with the OH. Okay. So in this case, what happens? Again, you can draw the resonating structure of this, you see. This lone pair, this resonant, this, this electron, this positive charge resonance is stabilized. So this resonating structure is this five-member ring, the double bond OH positive charge. Now H plus comes out from this and it gives a neutral molecule and that is why the six-member ring is converting into seven. This is the structure we get here. Oh, but if in the first step only there was a hydride shift, we could have gotten a neutral molecule, right? And then resonance. Here. Yeah. That is also possible. We can do that. Let's discuss this first. So we'll get a six-member ring. So we'll get this positive charge in OH. So then it will be a six-member ring. It converts into this one. So it's a five-member ring, no. So you'll get this one. Correct. Is it possible? RS, you cannot draw here, you see. Because the carbon becomes five, no. Yes, sir. So RS, you cannot draw. This is not possible. Okay. But one thing we can discuss here. Suppose here we have chlorine. Take any living room. Then if this comes over here, this chlorine goes out. Then it's possible. Living group, if it is present here, then it is possible. And in that case, we'll go with six-member ring. We won't convert into seven. But in this case, it is not possible. So this is possible seven-member ring. Right? So one note you write down. Generally seven-member ring. Generally six-member. Generally six-member to seven-member ring. Generally six-member to seven-member ring. Expansion is not possible. Expansion is not possible. But if the possibility of molecule formation is there, but if the possibility of molecule formation is there, then we can do that. Sir, which groups on the ring can give us a neutral molecule? Sorry? So which groups attached on the ring can give us a neutral molecule? Groups like OH, which has one lone pair present, at least one lone pair present. NH2? OH, yes. NH2 you can have? OCH3. No, OCH3 you cannot take. Why? OCH3 if you take, you have an url bond here then, positive charge, then this is not a good living group. Okay. Okay. That's OCH3 you cannot take. Generally OH. So you need something with a lone pair and then hydrogen attached to it? Yeah. Hydrogen attached to it. Okay. Those groups you can take. So this also note you write down, this is possible when carbocation is attached with OH, NH2, etc. Sir, but didn't we say that chlorine is also a good living group? But OCL bond is, if it is there then it's possible. OCL you can take, but OCL there is not such a nucleophile we have or molecule we have. Any living group is there. Like suppose I'll give you some other examples also. And HCL if you have, then it's possible. Hydrogen and chlorine both we have, so anyone can go out in this. Okay. Just a second. Yeah. Okay. So if the carbocation attached with groups like OH and NH2, this kind of neutral molecules may form. Okay. Okay. The another thing we are going to discuss is ring contraction. Ring contraction. Condition is what? Charge, positive charge. Must present on the ring. Expansion in expansion, the positive charge was present outside the ring. Okay. This kind of contraction is rare. Not very common. Four member convert into three. Seven member convert into six, because six is more stable than seven. Why four converts into three? You see this example. Dancing resonance. Yeah. Positive charge present onto this ring. So this will convert into a three member ring like this. CH2 positive charge, which is more stable because of dancing resonance. Okay. Can you draw the carbocation you get after ring contraction in this? So could you draw it? Just a second. One second. Yeah. So seven member thing is there. So see what happens here? Four, two, three conversion. This lone pair actually attached with this. It comes over here on this carbon and this carbon attached with this. We'll get a three member ring and CH2 outside. Okay. Similarly here, what happens? This is suppose first carbon. This is second, third, fourth, fifth, sixth and seventh. Okay. How this contraction takes place. This carbon atom, this lone pair, you can say this comes over here under a third and then this third get attached with the first. So obviously this seven converts into six members. So we'll draw a six member ring simply. First of all, in which first carbon is attached with third carbon, fourth, fifth, sixth, and this is the seventh carbon. And this first carbon is also attached with CH2, which is outside the ring, this one. This is the second carbon we have. You see here, since the second carbon loses its electron pair, so we'll have a positive charge on this carbon. Sir, can we have a methyl shift in the seven member ring? Sorry? So can we have a methyl shift in the seven member ring? I want to methyl shift. No, six member ring is more stable, right? So always if it is possible to convert seven into six, we'll do that conversion only. So but even after the methyl shift will still be a seven member ring, right? So can we do the shift and then do the contraction? No, why you'll do that shift first? See, carbocation is stability. You can do by, if ring contraction is possible, we'll do that first. After that, if it is methyl or hydride shift is possible, then we'll do that. Okay, because it is less stable than this. First, the ring has tendency to convert into the lesser member ring. After this, if you have hydride and methyl shift possible, we'll do that. Yes, here, the same thing I discussed, understood? First, we'll do ring contraction. After that, if it is shifting, is there hydride or methyl, we'll do that. Okay, so we have discussed many things about, you know, carbocation. Now, we are going to see the reaction in which formation of carbocation takes place. Formation of carbocation. And in this, we'll see the reactions of all the chapters, wherever the carbocation is possible in entire organic chemistry. Okay, so the first reaction you write down of alcohol, alcohol. Remember, when we do alcohol, I'll just give you a brief idea of this. I won't go into detail of all this, since we are discussing it here, right? That's why the notes are very important, keep it properly. Okay, so you see, alcohol when heated with concentrated acid, concentrated acid forms carbocation and intermediate. So, for example, tertiary alcohol with H+, concentrated, gives you tertiary carbocation. This is the carbocation forms. If you see the reaction of alkene with electrophiles, alkenes with some electrophiles, electrophiles like H+, Cl+, it forms carbocation. This reaction we have discussed in hydrocarbon. This is the alkene. So, two possible carbocation forms here with H+, suppose. So, it will be CH3, CH positive charge, CH3. And the next one is CH3, CH2, CH2 positive charge. The major product will be given by the carbocation, which is more stable. So, first of all, we'll form this carbocation after this, if some rearrangement is there, we'll do that. In all those reactions where carbocation is forming, rearrangement always possible. So, here the carbocation, this one is more stable, gives the major product. I'm just telling you where the carbocation is forming. This is not the final product of any reaction. It forms during the process of any reaction. Alkyl halide, another reaction you see of alkyl halide, RCl. When you heat this, heterolytic fission takes place, R+, and Cl-forms. You can heat this or this is also possible with a reagent like, we have Cl here, CS3 here, reagent like AGNO3. So, AG+, takes this Cl- and it forms a carbocation, plus AGCl precipitate forms. Okay, done here. Next page, can I go? So, one second. What happens to the NO3 in this reaction? Sorry? So, the NO3 part, AG forms AGCl, right? AG performs AG+, NO3-, so because AG+, forms, it can easily take Cl- from the molecule and forms a precipitate of AGCl. So, NO3 just like... NO3 will be there, it is present in excess, so it will present in the form of iron there and that's all. Okay, sir. Okay, so three chapters, alcohol, alkene, alkyl halide, and one more we have here, that is aldehyde and ketone. Aldehyde ketone. In this also, carbocation forms. How? You see this. Suppose we have ketone, see in ketone, the reaction mainly takes place at this carbonyl group. Because of the presence of oxygen, the carbon-oxygen bond is polar in nature. Oxygen being more electronegative, carbon-oxygen bond is polar in nature. And in presence of an acid, H+, this lone pair has tendency to take this H+, and it forms this. Further, to stabilize this carbon, this pi-electron converts, goes onto this oxygen atom, we'll get OH here and a positive charge. This is how the carbocation forms in carbon growth. Next you see the reaction, reaction which involves carbocation. The first and very important reaction we have is dehydration of alcohol. So first thing, there are four chapters in which you'll get carbocation as an intermediate, alkyl halide, alkene, alcohol, and aldehyde ketone. Now what all reactions are there in these chapters? Okay, so dehydration of alcohol. First of all, dehydration takes place with concentrated acid. Again in alcohol I won't discuss this. Okay, so keep that in mind. You can revise those reactions from this chapter, concentrated acid. Now for dehydration, we use generally, temperature generally we use more than 170 degrees Celsius. Okay, it's a temperature dependent reaction. Temperature dependent, it's a temperature dependent reaction. Dehydration gives you alkene as a product here. But I'll tell you one thing. Dehydration of alcohol can give alkene as well as ether, as we, alkene as well as ether. Ether generally forms with, when temperature is less than around 140 or 150 degrees Celsius. And alkene forms when temperature is more than 170 degrees Celsius. It is written in the book, see what happens. In some book it is written 170 and some book it is 140. So in between this, again 150 plus if you have, then alkene formation takes place. 150 minus if you have then ether forms. Actually alkene forms by elimination reaction. That's why I said it is a temperature dependent process. Elimination at higher temperature, yes. And substitution takes place at lower temperature basically. You see that one question in the group they have asked, X letter group. Trippan has asked one question just now, I have replied the same for him. You can go through that also in the group. The same reason is there. Temperature is not given in that question. If you're there in the group, you see the question was this. I'll just write down here. I'll show you the reaction right now with H plus. What is the possible product here? Options are given alkene. B is ether. C is both and D is one of these. Like this some option was there. Temperature is not mentioned here. That's why both product is possible. Now what is this reaction and what the mechanism everything will discuss. But you must keep this in mind depending upon the temperature. Two possible product. Here we have, right? So since we're talking about dehydration of alcohol, that's why for this particular reaction, I have written the temperature. Right? Okay. Understood. Now for ether also you have to use concentrated acid, right? Right. Temperature difference is only there. Generally elimination reaction. We have discussed it in the beginning. For elimination reaction, the higher temperature favors. The spontaneity we have discussed on delta G, delta H, delta S in the beginning. Right? So that is the general thing you can keep in mind. For elimination, higher temperature is required. Now, I'll take one example to make you understand this reaction. Suppose the molecule is this. CS3, CS3. Overwatch, it is an tertiary alcohol. CS2, CS3. With H plus concentrated. H2SO4 you can take, HCl you can take. The product of the reaction is CS3C, double bond CH, single bond CH3, CS3. Obviously dehydration we are considering. So temperature is greater than 170 degrees Celsius. Okay, it is understood in this. But in general, like I said, if the question is given in the book, you have to check what is the temperature there. Clear? Plus what forms? Water. The molecule goes out. And this is the major product actually. What is the minor product you get here? Is it this? This major product, you must heard the name. Major product, we call it set Jeff product. Have you heard this? Yes, sir. And this one is Hoffman. Set Jeff and Hoffman. Okay, so this is the overall reaction. You will get alkene. Set Jeff alkene is the more substituted alkene, more stable. How the reaction proceeds? We'll discuss the mechanism now. The first step of the reaction is, and it is a reversible reaction, reversible step OH CH2 CH3. What happens with this H plus? Could you tell me? It attaches with the OH and forms H2O plus. This lone pair. We take this H plus. Right. And it forms. Does this reaction involve formation of carbocation? Yes, sir. How carbocation forms since we are heating it? This goes out. There goes to oxygen and it leaves. Right. So the product here is CH3. Carbocation plus H2O. Yeah. Okay. Now this step is the slow step. And this is the RDS of the reaction. Now since you see RDS is the formation of carbocation. Always you keep this in mind. All these things are not given in any book. Okay. That's why when you go through the book, it's difficult to understand organic chemistry. Okay. RDS is the formation of carbocation. Okay. So here we'll try to get the most stable carbocation. Right. So we'll get this. And if rearrangement is possible, we'll do that. We'll get the most stable carbocation. And then in the last, we'll remove H plus. So what happens here to get the stable product here, it's a GIF product. Either this hydrogen will lose this bond pair of electron. And H plus comes out. This is one possibility. Or H plus can come out from these two carbon also. Yes or no guys? So if these two carbon loses it's H plus ion, you'll get this Hoffman. Any doubt? Or should I write down this? Write it down. Okay. So two possibility. One with this, I'm writing it down. CH3CCH3 double bond CHCH3. Correct. The another possibility is what? If these two carbons are equivalent carbon, any one of the carbon atom can lose H plus. If this goes here, then the product will be CH2 double bond, CCH3, CH2, CH3. Anyone has doubt in this? Anyone has doubt? Guys, reply please. No sir. No sir. No sir. So this product, which is the more stable one, more stable alkene, more substituted, more stable alkene is sedgef. Okay. Sedgef product you get when the carbon loses, when the H plus comes out from the carbon, which has lesser number of hydrogen, H plus comes out from the adjacent carbon, which has lesser number of hydrogen. Correct. In that case, sedgef unit. This sedgef, this is Hoffman. Now, the thing is, in this case, there are many things which you should know the properties of reaction, everything. First of all, it is an elimination reaction. Right? It is an elimination reaction. As you see, this step is the RDS, rate-determining step. So the rate of this reaction depends upon only this molecule. Right? Concentration of this molecule. So order of the reaction is one. It is E1 elimination reaction. Right? Order. What is order? We'll again discuss that in chemical kinetics. Okay? So this is E1 elimination reaction. E1, E2, E1, CB are the different, different types of elimination reaction, which we also discuss in this, you know, chapter only later. Correct? E1, when you write, I'll write down all the properties. E1, when you write, it is elimination. E stands for elimination. One stands for order of the reaction. Order is what? Order is something, it is the exponent of the concentration term in the rate-determining species. Rate-determining step. Okay? So I'll write down the properties of this reaction one by one. You just copy it down. The first property of this reaction is, it is elimination reaction. But this is for E1. Yeah, it is E1. Don't, you know, think about it. We'll discuss this later. What is the RDS? RDS is the generation. Formation of carbocation. Formation of carbocation. Of carbocation. Third one. Since RDS is carbocation formation, so rate of the reaction, ROR, ROR stands for rate of the reaction, is directly proportional to the stability of carbocation. This question they asked many times in the exam. Stability of carbocation. These are all for E1 rate. Yes, yes. Stability of carbocation. But this carbocation, which forms in the first step, means we are not considering the rearrangement, which forms in the first step. Are you getting it? Once you remove H2O, then the carbocation that you get, that stability of that carbocation will compare, not after the rearrangement. Okay. Yeah, right, correct. Okay, so the next property, in this you see the reaction to sigma bond, converts into what? One pi and one sigma, is it right? Two sigma bond converts into one pi and one sigma? Yes or no? Carbon, oxygen, bond breaks, right? Yes, sir. Yes, sir. Okay. Overall, the reaction is exothermic or endothermic? Endothermic. Huh? Endothermic. It is endo. Exothermic. Exothermic. Two sigma converts into one pi, right? So pi is weaker than sigma. Yes, sir. Right, so some amount of energy. Endo, right? I think basically breaking two bonds to form, or breaking... Oh, I got it, I got it, never mind. Just a second. It is bond dissociation. Endothermic, correct. Yeah, that's right. Correct, it's endothermic. Because you're breaking two strong bonds, but you're just forming one weaker bond. Yes, it's endothermic. Bond dissociation is energy consuming process. Two strong bonds are formed. So hence, it's endothermic. So which are the two sigma bonds? Carbon oxygen, carbon hydrogen, I think. Should we not consider the sigma bond which you formed between the hydrogen and the oxygen? Yeah, guys, what is the doubt? So like, when we move one molecule of water, then some energy will have to be released, right? No, no, no, no. Not like that, not like that. See, sigma bond is stronger than pi and bond dissociation is energy consuming process, correct? So it consumes more energy and releases less energy because pi bond is formed. Why don't we consider the energy released when you give out the H2O? Because energy releases how? When carbon oxygen bond breaks. Okay. No, energy in that process, energy does not get released. Energy consumes in that process. Bond dissociation is energy consumption. Bond formation is energy releasing process. Okay. When bond forms, energy releases. Because something, if you want to dissociate, you have to provide energy for that, no? Okay. So here, more energy you are giving in, less energy is coming out. So it's an endothermic process. So this sigma bond has been made? Sir, is the sigma bond made the oxygen hydrogen sigma bond? Yes, yes, yes, in H2O liquid. Okay. Okay, it's an endothermic. What about the entropy change? Delta S greater than or less than? Agree because you're forming oxygen. So what temperature favoured this reaction? High or low temperature? This is positive. High temperature. Yeah, T delta H should be more than delta H. So we require high temperature. That's why you see, initially only I have discussed this, that this kind of elimination reaction favours by high temperature. Right? So in this two molecules are forming. So it is also entropy driven process. And the last one is the amount of H plus is not changing. Concentration of H plus is constant throughout the process because in the last step H plus get removed, right? So H plus concentration is not changing. Means the acid behaves as what? Catalyst. Yeah, these are the properties of these elimination reactions. Am I clear with this? Okay. So basically whenever you get the question, you can write down the dehydrated product. Okay, if the temperature is given accordingly. Sometimes they ask question, like they ask you to compare the rate of dehydration. Okay, compare the rate of dehydration. So how do we compare? More stable carbocation forms. We'll compare the stability of carbocation forms in the first step. So I'll give you some questions on this. Sir, you told the concentration of H plus doesn't change. But since H2O is being given out, won't the concentration change? No, I'll explain this. See, what happens here in the last step, what happens? H plus comes out, no? That I understood, but since you're getting elimination of water, so your volume will increase, right? Water elimination takes place by the consumption of H plus from acid. Correct? Yes. But the same amount of H plus has been released in the last step. So he's saying that you're getting more solution so then the concentration should decrease because you have the same amount. The number of moles of H plus remains the same, only the concentration is changing. Yeah, because volume of water increases. No, H plus, H2O also if you have, it will present in the form of ions only, H plus and OH man's a polar solvent. Right? So whatever the conjugate base, suppose you are taking H2SO4, correct, for the reaction. So it gives H plus and HSO4 and HSO4 minus or SO4 minus 2. Right? So when the H plus comes out, that will combine with, see, in the solution, it's an electrolyte. So it is not like H2SO4 will be present there, but there will be H plus and SO4 minus 2 minus present. Correct? 2H plus and this ion will be present. Correct? So when water forms, so water also gives H plus and OH minus. So in the solution, we have same concentration of H plus. It is not getting consumed effectively. You don't think water is a molecule and it is forming and then water is present over there. Right? I am telling you what? That you have H plus OH minus SO4 2 minus present after the reaction. Right? So this H plus and SO4 2 minus will be equal to the same concentration of acid that we are using. Yes, sir. Okay. Water also gives H plus. So basically what, how you can think this, that first step H plus is getting consumed and the last step H plus is getting released. So this will maintain the concentration. Okay. Right? Okay. So yeah, next is, so we were discussing about this thing. Order of, just a second Karthik. So which phase you want? This one. Properties. Write it down. Clear? Done. Okay. Compare the rate of dehydration. Write down the heading. Compare the rate of dehydration. Molecules are OH attached on cyclohexane. Rate of dehydration. Done. First one. What is the answer? It will keep increasing. Yeah. So 4, 3, 2, 1 is there. So 4, 3, 2, 1. Because of the resonance and. Yeah. So the second one will be more. I think. Yeah. Third one. So first one is more. First one is more. Yeah. I didn't understand the second one. Bridgehead position thing. Last one. So the last one will be first. I do it. The third one. Yeah. Okay. So it is 3 degree alcohol, right? So tertiary carbocation more stable than this. This is right. 3 degree, 2 degree, 1 degree. This is also right. Okay. Here the carbocation forms over here, right? Which is a bridgehead position. You see, I have also discussed this in GOC. This bridgehead position, the carbocation is not at all stable. Because this carbon is sp2 has tendency to get 120 degree bond angle. Correct. So this bond has a high strain over here. It has tendency to go this way. Right? But which is not possible because of this bond present with this carbon atom. That's why this thing is a bridgehead position and carbocation as bridgehead position is not at all stable. This is more stable than this. Here what happens? The, this one, the order is also right. For the last one, this is not correct. And the correct answer for this one is 1. This is 2. This is 3. And this is 4. Why? Because we told the final group can't be. Yeah. This is phenyl cation, not at all stable. This is in resonance, more resonance here than here. Here we have hyper conjugation stability and here we have not such a picture. The order is this. Why doesn't resonance take place on the phenyl? See, this, this pi electron delocalized because of the p orbital here, here, here, here are parallel, right? Yes, sir. But the p orbital which forms this carbon and OH bond, right? This p orbital is not at all, not at all is this thing parallel with the another orbital like this. So this is the orbital of this thing, carbon, and this is the orbital of oxygen. This forms a sigma bond here, right? And this orbital is not parallel with the orbital with p orbital, with this carbon atom. You see, this p orbital is parallel with this p orbital, right? So for this carbon, one p orbital is already in conjugation with this p orbital, you see? In this p orbital, like this, no. This kind of conjugation is there. Okay. Right, this kind of conjugation. But this carbon has another p orbital in different plane actually, which forms this OH bond. So when you, when you take this out, the p orbital here is not parallel to the p orbitals which is present here and conjugation with each other. Are you getting it? Yes, sir. So let me just draw the, this structure. See, if you have a benzene ring, so this p orbitals are parallel. Delocalization takes place in this and will get this. But apart from this p orbital, we have one more p orbital on this carbon atom, right? Which forms the bond with OH. So do you think this p orbital is parallel to the other p orbital? No, see it's very clear, no. Because this p orbital is parallel. That's why when you remove this OH, the positive charge here is not in conjugation with this. Got it? Yes, sir. Got it. So phenylketon is not a stable throw resonance. You must keep this in mind. Okay. Now one more question in this we'll discuss. The rate of dehydration, if you have this CS3CH CS3 OH and suppose if you have CS3CH CS3OD, in which one the rate of dehydration is more or less, just a second. One change is there, just a second, just a second. Here we have D3. D3. The middle is also D or H1? No, that's fine. So I have a set clip out. Is it possible to form HOD? Why not? Is the deuterium one more stable? Yeah, second one. Second one will have a faster rate. Why? Deuterium has a stronger plus. It will stabilize the carbocation. See here what happens with H plus? You'll get this CS3CH positive CS3 and here with H plus, you'll get CD3CH positive CD3. Plus you'll get HOD. H2. Even if you write D2O, that also won't make any difference. So here it's D, sir. In the second one, you'll have CH3. Where? Oh, this one. So now here we do not have any change into the two reactions, right? Because carbon-oxidant bond you have to break, no? Both of them. Next step, what happens? One of the hydrogen comes out. So here you have to break the CH bond and you'll get this, correct? Here what happens? You have to break the CD bond and you'll get this. That is the only difference. So what is the difference in two reactions? Can you tell me? Sorry. Here we have H plus and here we have D plus. What is the difference in the two reactions? Where we have the difference? Reign of CH and CD bond. CH and CD bond. Because till here, till here, we have the same thing, correct? Formation of dissociation of CO bond, CO bond here. So which bond is more strong? CH or CD? CD rate. CH. So we have more, this thing, hyperconjugation in which we have more hyperconjugation, CT3 or CD3 or bond length of CH and CT, which one is more? CD rate. Bond length of CH is maximum. Sir, but can't we use plus I? Why do we use plus I? To stabilize the carbocation. So what? Let me explain this. First of all, to understand this, you should know the bond length of CH is more than to that of CD and then CT. So dissociation of CH bond is easier than CT bond. Correct? This is bond length. Bond dissociation of CH is easier than CT. Then you can say sir, this dissociation is easier, so this will be the higher rate, correct? Yes sir. Yes or no? Guys, reply please. Yes sir. But actually this won't affect the rate of the reaction. Both reactions will have equal rate, almost equal rate. Why I'll tell you? Because this is the difference we have in this step, second step. But this is not the RDS of the reaction. RDS is the formation of carbocation, no? Yes. Rate, data mining step is what? Formation of carbocation. And this step is same for both the molecule, correct? That's why they have almost equal rate in this reaction. So when you say almost equal, they're not exactly equal process. No, no, no, no, like this is what like I am telling you. Okay, because precisely you can say the rate will be equal. OD bond may affect this bond strength little bit. Okay, so precisely if you say the rate will be slightly different, but that we do not consider actually. In books if you say they have written they have equal rates. Okay, but if you have to choose between equal, unequal, greater or less, we'll go with equal and only. Yes sir. More precisely I am telling you that, not more than anything else is not there. Okay, so point here is, point here what you have to remember that you probably say that this one is higher, this one has based on the bond length, but this factor is not at all applied here because this step is not rate determining step. Got it? So rate if you have to compare we have to focus on, we have to focus on what? RDS. RDS of the reaction, correct? Sir, but I have a doubt sir. Yeah. Sir, but in the, in the deuterium one shouldn't the carbocation be more stable because D gives more plus I effect. So then shouldn't this, shouldn't, isn't that affecting the RDS? No RDS, see actually this is the stable one, right? CD3 has more plus I. Yeah. And CS3 has more plus I, but hyper conjugation is there. So we'll, we'll talk about hyper conjugation, no? Isn't that equal? What? Hyper conjugation. No, hyper conjugation. Oh, how can you say that? Hyper conjugation is, is more in CS3. Oh. Hyper conjugation is more in CS3, but you can say this carbocation is more stable than this, revise GOC ones. Okay. I have done this. Okay. This carbocation is more stable than this. Then you can say, but RDS reaction is what? RDS is because of dissociation with carbon oxygen bond. Carbon oxygen bond, okay? And that is unaffected by the presence of isotopes. Okay. So this D won't make any difference. Okay. That's why the rate will be same. Yes, sir. Okay. So one thing you write down here, removal of H plus or D plus, removal of H plus or D plus in the last step does not affect the rate of the reaction. Okay. That does not affect the rate of the reaction. Hence, kinetic isotopic effect is not applied. I'll tell you what this effect is. Okay. Hence, kinetic isotopic effect is not applied. What is kinetic isotopic effect? Kinetic isotopic effect is, we call it as due to the presence of isotope, the rate of the reaction gets affected. Then the kinetic isotopic effect is there. Right? So basically kinetic isotopic effect is what presence of isotope change the rate of the reaction. Since the rate of the reaction is not changing here, then the kinetic isotopic effect is not applied. So it's very, there's a very thin line between the two, let me explain this again. Rate of the reaction, previously we were discussing about the stability of carbocation, correct? This carbocation is slightly more stable than this because of hyperconjugation. Hyperconjugation we have here also. Here also. Here we have more hyperconjugation than this. But that won't affect the rate of the reaction to that extent. Mainly it is because of the dissociation of this bond. And since, and we have to discuss this with respect to the isotope, that is deuterium and hydrogen here. So presence of this isotope does not affect the dissociation of this bond. That is why we are saying that the kinetic isotopic effect is not applied. And hence the rate of the reaction will be same. Right? You can say, so this carbocation, the first one is more stable. So the rate should be more for this. Okay? But we are comparing with respect to the isotopes. Okay? That won't affect the rate of the reaction, presence of isotope. Clear? Yes, sir. Okay. So guys, we'll wind up here. Just you write down the heading, we'll start from the next class. I don't want to, you know, we have two, three minutes, but I don't want to leave things in between. Just you write down the heading, we'll start this from the next class. The name of the reaction is pinacol, pinacolone. Have you heard this name? No, sir. Pinacol, pinacolone rearrangement. We'll discuss this in the next class. Okay, so heading you write down, we'll start from this. Sir, can you show the previous slide one? Yes, just a second. Done? Yeah. Okay, so go through this, whatever we have discussed, just go through this one. And GOC, you must revise. Okay. We'll have a session next week and then I'll share some assignment on this. Okay, because we have to cover a few more things and I can share some assignment with you on this. Okay. Okay, sir. What? So how many classes in this chapter today? It is a big chapter, Ruchid. So we'll try to finish as quickly as possible. So like, I think two or three more classes we'll have. Focus. We'll finish two classes to PAKKA. We'll see three also possible. Will we start 12 in the summer or when school starts? We'll continue. We won't wait for school because we have to discuss many things in chemistry. Your school stuff is not that much. For a competitive exam, there are many things to discuss. So I won't wait for anything, but what a school is doing another thing. Okay, so I focus on the exams like JEE, another competitive exam. Okay. Okay, guys, thank you. Thank you, sir.