Elimination Reaction Mechanism Introduction

All right. So let's quickly summarize what we did last uh couple of sessions. Just two sessions we had one and a half hour or 1 hour and 45 minutes. So what we learned was that uh when we talk about mechanism in organic chemistry okay there are different ways reaction can happen. All right. When you look at the reaction u you know you need to identify which of these four or five ways reaction might be happening and once you guess it right whether it is SN1, SN2, E1, E2 free radical mechanism if you guess it then there is extremely high chance that you'll get the answer correct. All right. So unfortunately there are at least seven or eight ways the reactions can happen in the organic chemistry. But good thing is that you know these seven or eight ways if you understand really well it can cater to infinitely many reactions. So that is why we are doing this. So these type of reactions broadly you can classify into three or four types. Substitution, elimination, addition right and substitution are typically of whatever we have learned last two classes that is related to substitution right. So first we discuss about SN1 reaction right? SN1. Can anybody talk about SN1? Whatever you may have understood, feel free to speak about it. Anyone? >> There's formation of kion and SN1. >> Okay. There is a formation of carocation. What else? Sir, >> it's a two-step reaction. >> Twostep mechanism. What else? >> So they can go uh they can undergo a rearrangement to gain stability? >> Rearrangement of the carocatan? All right. The rearrangement can happen. Then out of 3°, 2° 1°ree which has a higher chance of SN1? >> Uh three. degree >> three degree out of R I R C L RF which one is highest chance of SN1 >> R >> right I >> because leaving group should be a good leaving group otherwise carocatine now how does carocatine gets created is the nucleophile creates a carocatan or it creates on it itself >> by itself. >> By itself just like when you dissolve NaCCl into water Na+ and Cl minus gets created by itself. Right? Same way due to polarity of the bonds when you dissolve this in a polar solvent then we you know in SN1 carocatine gets created on its own right. What is a typical nucleophile here? Solvent itself is a nucleophile. Typically solvent itself is the attacking reagent. Typically all right. So if you dissolve R I into water what will happen? If it is SN1 alcohol can gets created on its own. Right? Now how you will ensure that the alcohol this alcohol let us say if you have this alcohol under goes SN1 reaction how will you ensure that? So diga switch on your webcam. Huh? How will you ensure that this will undergo the uh the reaction SN1 reaction O will um it under go solosis where H attaches to O and that's a good leaving group and it leaves and then the nucleophile will attack. >> Correct? O by itself is not a good leaving group. So in case of al you know alky helides the I minus cl minus they are stable so they come out on its own. O minus oxygen to carbon bond is very strong as such. So it is not a good living group. You need to first make it a good living group by ensuring that there is an acid that attacks O. Okay, this is an example of catalyst base reaction. There's a catalysis that is happening here. Okay, so H minus is increasing sorry H+ is increasing the rate of SN1 reaction. So this is O2+ and now this O2 is a good leaving group and hence carocation gets created like that. Okay. So this is what it is right. This is the SN1 reaction. So you know how it is right now. Let's talk about SN2. before SN2. What about the stereochemistry here? >> This is a reemic mixture. >> It is near remic. Which one uh will be higher in value or percentage? Inverted product or retented product? Retention or inverted? >> Retented. Retented product sir. No, it is inverted because this I minus when it is about to leave it is very near to the R and in that process only the nucleophile attacks. So nucleophile doesn't want to be near negative charge I minus Cl minus. So it tries to attack from behind initially and once I minus leaves then it can attack both ways. So retention is lesser than inversion. Okay. SN2 we'll quickly do that. In SN2 first let's talk about uh chemistry inversion happens mainly. Right now what how will you know that uh with a particular nucleophile SN1 will happen or SN2 will happen? By looking at it, how will you know that? >> So if the nucleophile is strong, it look good too. >> Nucleophile is very strong. So it requires a strong nucleophile. The solvent is not a nucleophile. In the solvent there is KOH, Na H2. These things are the attacking reagents are dissolved in smaller quantities in the solvent to attack and create SN2 reaction. Right? Now uh this um over here also you know you can say that RA I RC this is how it is because leaving group leaves that will always accate the reaction then it is a onestep reaction on one-step reaction. So when nucleophile attacks the bond is formed and broken at the same time right. So there is a complex that gets created. Uh let us say this is how it is. This is what it is. Then a nucleophile comes it attacks here and this leaves. All of this is happening simultaneously. Now when it is if it is happening simultaneously can this nucleophile come and attack from here. Nucleophile is typically negative charge or it has a lone pair and this is an electronegative thing electronegative atom. So nucleophile want to be as far as electronegative atom if it is there at the time of reaction. So that is why in SN2 inversion happens. So nucleophile will attack from 180° behind wherever the living group is. So inversion happens. Okay. Now you can see there is lot of crowding is also going on because at the same time nucleophile is also attacking, leaving group is also there. Everything is happening together. So SN2 doesn't like crowding and that is why you know here 1°ree SN2 reaction is faster than 2° faster than 3°. Okay, I hope uh you understand very clearly difference between SN1 and SN2. There is no carocatine formed in SN2. There is no question of rearrangement, nothing. Okay. Now what what we learned next was SNI. Remember SNI? What is what was that? anyone >> sorry nucle nucleophile substitution >> that is internal nucleophilic substitution in a way but actually uh you know it's not uh that that step this is typically with SO2 or CO2 so there is a sulfur which has vacant orbital in SO2 sulfur has a vacant orbital so This lone pair attacks the vacant orbital. All right? And this Cl goes away. Right? Then you will see this is the intermediate that get formed. Uh R O S O C L there is H also. So that is why there is positive charge over here. Then this H will leave. Now till now uh there is no uh nucleophilic attack on the alcohol. In fact alcohol was a nucleophile for SO2. This lone pair was attacking sulfur. Are you getting it? The lone pair of oxygen was attacking sulfur. So alcohol was a nucleophile till now. Now you will see that although this mechanism is not very important but you should understand that such kind of thing you'll see later if you study higher order organic chemistry. This lone pair will shift here and create SO2. This Cl minus released and this Cl minus will attack R. This is intermolecular attacks. So this is RCL plus SO2 gets formed. So you can see that Cl minus comes from the same molecule same molecules it comes and attacks the R. So that is why it is SNI. Okay little weird but this is what it is. Then we discuss about uh what we discuss about you remember SN2 A R we discuss about nucleophilic attack on the benzene right to typically benzene doesn't have a nucleophilic attack in order to accelerate that what we do anybody remember in order to augment the SN2 mechanism in arm I don't pressure. >> Yes. If it is let us say you know uh if if you are talking about this this and you just directly want to introduce then you require a very high temperature and pressure. Okay. Why it is so? Because due to misomeic effect this has a double bond characteristic. This one has a double bond nature. So it will not come out. This is let's say the ali or sp2 hybridized carbon chlorine bond. So that there that is second reason. Main reason is that it has a double bond characteristic. Okay. Now if I want it to be happening during uh in the room temperature then what condition should be there? >> Electron >> electron withdrawing group should be there where in which locations? >> Or paraposition >> or paraposition. So if you have more electron withdrawing group or ortho or para positions then the nucleophile can attack easily because the intermediate that get formed it has negative charge. So electron withdrawing group will stabilize it by you know uh by not letting it localize in one location. So electron that get generated during the attack it can disperse itself. So this is the reason why uh electron withdrawing group is required for SN into a R reaction. Typically benzene will give electrofphilic reactions. A electrofile which is let's say positive charge will attack nucenzene ring because benzene ring is typically electronrich. It has entire cloud of electron at the top at the bottom. So a positive charge entity can easily attack but a negative charged entity like nucleophile if it has to attack it should have electron withdrawing group which will decrease the electron density in the benzene otherwise it will not be able to attack it. You require extremely high temperature and pressure. Okay. Then we also learn about the benzene mechanism. These are all you know you just need to know little bit about it. When we will learn about the benzene in detail you will see that later on. Then towards the end we learn about the SNTH. You remember SNTH? >> Tetra. >> Tetraal intermediate. Typically it is formed when it is typically formed >> carbonal compound >> carbonal very good for carbonal compounds. Okay. What is a carbonal compound? C double bond O. So let us say this is the thing L is a leaving group. This is a carbonal compound and this is the nucleophile. This I mean this is uh you know you can easily guess where nucleophile will attack because this oxygen is electronegative it will pull the electron density due to the minus I effect and the carbon will be delta plus this is delta minus. So nucleophile has extra pair of electrons. So it'll attack this carbon. But carbon cannot have five bonds. So which bond is weakest and to be broken? Pi bond. So this pi bond will break and negative charge can go to oxygen. Oxygen can hold the negative charge because it is electronegative. This is what will happen. And you know every time there's a carbonial group there is a C double bond O and a nucleophile attacks whether it is uh alihide whether it is ketone whether it is acid you'll see the same mechanism. So what we are learning is going to help you in all the chapters that are going to come. So you should understand how it happens. So nucleophile will attack the carbon and then what will happen? It will have suddenly four bonds. four bond as in it will have the tetrahedral nature. This is the nucleophile. This is L. Now what will happen is that this negative charge will again come and form a pi bond and this leaving groups group leaves. Okay. And instead of leaving group you will end up having the nucleophile. This is what will be found. Now of course in this mechanism carbon to leaving group bond should be weaker than carbon to nucleophile otherwise why it will happen. Okay. So this is the summary of what we have done in the last uh two sessions. Okay. Today we'll talk about the elimination reactions. Okay. This is all substitution. Of course there are maybe some finer details here and there but uh more or less we are done with the substitution. So whenever substitution happens uh one of these mechanisms we'll be utilizing then now we are going to talk about the elimination you know elimination there are two kinds of mechanisms E1 and E2 substitution broadly you know you can say that majority of the substitution is either SN1 or SN2 similarly majority of elimination is E1 and E2. Now looking at the uh looking at the uh substrate, looking at the solvent, looking at the reagent, nobody's going to tell you whether it is SN1, SN2, E1 or E2. You need to think and recognize okay elimination will happen here. No, substitution will happen here. So you need to pay attention to the reagent kind of reagents and the solvent. Okay. So let's proceed. Write down even even reaction. Okay. So uh let me show you a typical elimination reaction. So this is the typical elimination reaction. This is uh hogen. I'm most of the time I'm using uh halides because halides are very good leaving groups. So almost 50% of alky allides will be over by the time we'll finish the reaction mechanism itself. So x is a helide this. Okay. In step number one, carocatine gets created. This leaves. Okay. Right. In step number two, in step number two, the electrons of this carbon hydrogen they come here to form a pi bond and H+ is released. Okay, this is what you can see. Okay, we'll we'll discuss how and why it happens. But this is the E1 mechanism. What happens? First the leaving group leaves. Carocetime gets created and then H is extracted. H+ is extracted. H+ is gone. But the electron between carbon and hydrogen two electrons because of that bond will get paired and create carbonarbon double bond. Okay, this is the mechanism. Now you can see that the first step the step one do you recognize it is same as SN1? It is exactly same. So since it is same the rearrangement can happen. Please write down in even rearrangements are possible very well but the step two is not same. So what makes uh you know how will you know whether in step number two substitution will happen or elimination will happen that is what is the game here you need to understand that right so we'll you know we'll write all the details in fact the first step please write down first step it is a slow step in SN1 also that was a Slowest step formation of carocatine. Right. All right. In the second step there's abstraction of proton. I mean H+ is proton only. Electron is gone. So only one proton is left. So you either call it H+ or proton. It's very very small. Uh this thing size of a proton is H+. Okay. Anybody has any doubt? Anyone? >> No. >> Now u the step one is same. Step number two, what is happening is you can see that the entropy has increased. Entropy increases in case of the elimination. Why entropy increase? Because the more number of products alken is there, H+ is gone. There is more chaos over there. Many things are coming out right. So when entropy is increasing then what is that condition that favors increase in entropy? Increase entropy means E2 sorry E1 elimination is going on. Decrease in entropy means substitution is going on. Substitution is what? There are two three uh reagents that get substituted. So entropy went down. In elimination things are coming out. More numbers are getting created. So entropy is increasing. So tell me at higher temperature more entropy is favored or at a lesser temperature more entropy is favored. >> Higher temperature more entropy is favored. >> You can just take an example of evaporation. Right? When you boil the water the entropy increases lot far lot more. So here also I mean if you want to understand using thermodynamics you can write delta G you know gives free energy delta H minus T delta S. Okay. So now gives free energy you guys remember or not. So if delta G is negative means reaction will happen. If delta G is positive means reaction will not happen. This delta H accounts for the uh thermodynamic you know variable as in the product is more stable or not that is delta H. Okay. But many reactions are endothermic in nature. Venations are endothermic then also they happen they happen because this particular thing T delta S this T delta S compensates for stability so that is why in endothermic reaction if you increase the temperature the reaction will proceed in the forward direction okay the products are less stable but still it is getting formed so basically I mean I don't want to tell you unnecessary many complicated stuff because of increase in temperature the delta S is positive for the elimination. So if you increase the temperature T delta S term increases and because of that delta G becomes more and more negative and hence reaction is favored for the elimination. Right? So if you just want to remember like a thumb rule at higher temperature elimination is favored. Okay. So if I write if I draw a delta here it means I am heating it up and when you heat it up this delta symbol is there along with some nucleophile assume elimination will happen. Is it clear to all of you? Anybody has any doubts? No doubts, right? Can you draw the energy curve of this for E1? Quickly draw that that curve. Try done all of you in this uh how many intermediates are there assuming rearrangement doesn't happen how many intermediates carocatin is the only intermediate If there is one intermediate, there will be two transition states. Okay. Two intermediates, four transition states. Double the number of intermediates. Transition state transition states has the highest energy. So using that rule, you can draw it and it will look similar to your SN1. This is your uh trans uh this is your intermediate. These are your transition states. Transition state one is when the leaving group bond is getting broken. It is in the process of breaking. Leaving group is leaving. That process of leaving is a transition state. And the second transition state the second peak is when the uh this thing when carbon hydrogen bond is breaking. All right. Okay. So now over here what do you think is a nucleophile in in S? I'm sorry. Another uh thing that you will notice that in SN1 the solvent is the nucleophile. In even typically it is not okay but anyways the main differentiator is the delta sign. The delta sign in E1. In E1 the nucleophile act as a what? Anybody guess? Nucleophile act as what does it act as? Nucleophile does what? Where it attacks? Where the nucleophile attacks? Look at this. >> It's just carbon atom. >> No, >> wait. No. Nucleophile doesn't attack. >> Nucleifile attack. That is in substitution. In substitution, nucleophile attacks carbon. In elimination, nucleophile attacks where? >> Hydrogen. >> Hydrogen. Hydrogen. It attacks hydrogen to extract H+. Now who extracts H+? What do you call it as? What do you call something which takes H+? Who takes H+? >> Oxidation >> base. Base base. It is base. Okay. So uh in elimination it's it's about base. So sometime they will write uh strong base stuff like that when elimination is happening. So always remember substitution nucleophile attacks carbon elimination nucleophile attacks hydrogen. Okay. Now uh it is possible to have more than one product. It is quite possible. All right. So can you tell me over here? I'll just give you a simple example right now. This is a situation. Tell me what will happen. Substitution elimination. >> The elimination. >> Elimination. >> Elimination will happen. uh draw the two products that can come out. Two products that can come out. By the way, whatever elimination we are talking about till now is uh you know it's beta elimination. Beta elimination or one two elimination. Why we are calling it so? Because the leaving group comes from the first carbon and hydrogen is going out from the adjacent carbon. It's not that leaving or hydrogen group are attached to the same carbon. They're attached in two adjacent carbons. One carbon leaving group comes out other carbon hydrogen comes out. So that is it is called beta or 12. There are 1 one elimination, 1 13 elimination also but we will focus on 1 12 elimination only because that happens majority of the cases. Huh? Have you drawn how many adjacent carbon you can see to this? >> Four. >> Yeah. Four >> adjacent to this four. This is chlorine. >> Three, right? three. But aren't these two exactly same? Whether you go this way or that way, the path is same. So, uniquely you can say two. So it can happen uh let me uh it can happen that hydrogen can come out from here or from there. You can say sir from here also hydrogen can come out. It is same thing as hydrogen coming out from here. All right. It is the same thing. So if hydrogen comes out from let let me name it 1 2 3. If it comes out from two, what will happen? You will get this type of alken. Can you see that? This is what you'll get. But if hydrogen comes out from here, you'll get this. This one, this one is less substituted alken. This one is more substituted alken. Which one is more stable? Less substituted alken or more substituted alkan? Which one is more stable? You have learned in alken right? Which one is more stable? >> More substituated alken. >> More substituted alkenes are more stable. Okay. >> Typically. Now why so? Because of the hyper conjugation. because the uh the uh there is an overlap of sigma orbitals of this carbon hydrogen bond with the pi orbital of this carbon carbon. So there is an overlap. So you know why these overlap creates stability because the charge is always unstable if it is located at a very small place. If you let it spread, if you let it spread, charge is stabilized. Concentrated charges are unstable. So when overlap happens, the electron gets space to move around. So when it does that, it becomes stable. That that is why benzene is very stable, right? The there is a pi cloud electron can move around here and there. Anyways, so this one is more stable, more substituted. So we call this as a Hoffman product. You you remember uh have you ever heard these things? Hoffman sad Jeff. >> Yes sir. >> Sad Jeff. I mean pardon my spelling. I don't know. Okay. So you know in this even mechanism the Hoffman product is always more okay why because typically you know from carocatan it gets time and it will move towards the more stable product only. Okay fine. So let me ask uh you the some questions. The the first question is theoretical question. So all of you hear it and you answer then okay correct statement for even reaction. Option A it is a two-step process. Option B rearrangement is possible. Option C good leaving group favors even. Option D. All of these. >> Option D. >> All of these. Good. All of these. Okay. Uh let me give you another question that there can be little little tricks involved. Okay. So be uh attentive there. Which one gives three alkanes after dehydrogenation? What is dehydrohalogenation? This is dehydrohalogenation only. Even what we did hydrogen is removed. de dehydrogenation hogen is removed dehalogenation. So dehydro dehydrohalogenation hydrogen and hogen both are gone that is dehydrohalogenation. Okay. So your options are CH3, CH2, CH2, CH2, Br I mean I'm not asking what is the major product whatever may be percentage of product which of them gives three products CH3 CH3 P R CH3 CH2 CH Br3 CH3 CH2 CH Br try we can name it. This is first, second, third, fourth. So option four others. >> So option four. >> Are you sure? Let's see option four what happens over there. So CH3 CH H CHB Br and this is phenile. So which hydrogen can come out? This one can any other hydrogen can come out other than this? Any other comes out other than that? Adjacent to this carbon is what? This one or phenile. Can phenile hydrogen come out? Can hydrogen from that phenile group comes out? Huh? Who said uh three? Answer it. No, there's a phenile group like that. There there is hydrogen over there. No doubt there is a hydrogen. But this hydrogen 1 2 3 4. No, it doesn't have a hydrogen at this location. So only one hydrogen can come out. Do you all agree? only one hydrogen. I mean even if hydrogen would have been there from benzene sp2 hybridized carbon hydrogen will not come out. Okay. Now if this hydrogen comes out how many products you will see how many products would be there? 1 2 3 How many? >> One. No, there are two. One will be cis, other will be trans. In even in even both cis and trans can get created. Look at the intermediate. There's intermediate, right? So this bond this bond can rotate and then uh when it is rotated cis can become trans can become cis. So it can create both cis and trans. Are you getting it? Now let's look at option number three. CH3 CH H CH Br2 H. If I take out this hydrogen, this one, can there be cis and trans? If I take out this hydrogen, then it will look like this. Will this alken have sis entrance? Why not? Because for cis and trans on both the carbons two unique groups two different groups should be attached. Here both of them are hydrogen only. So it will not show cis and trans geometrical isomer. Okay. So that is why if this hydrogen goes only one product but if this one goes can it show cis and trans then now there will be two products CH3 C H double bond C H CH3 this is trans and you can make sis also so the answer is three not four. Okay, write it down somewhere that in even sis and trans both can get generated. Okay, next question. Tell me what product you will see. Delta is there. So elimination will happen. Even try Then anybody what will happen? So would it be a cyclutane ring with like C double bond CH and CH3? >> Cyclutane will remain as it is >> forms a double bond with the >> CH2 and then single CH2. >> Let us see that. Okay. First step is what carboine will get formed. C++ C3 CH3 Now this is a 2° carboatine can it become 3° possible or can this overall compound be more stable if the ring expansion happens? If this alkyle shift goes to this carbon, if it shift to that carbon, you'll see that this is CH3. CH3 and this it has become a five membered ring. positive charge will go there because this bond has been broken. So this carbon will attain the positive charge. So it's a five member ring. Now uh has anybody tried this ring expansion? You guessed it correct? Anybody? That's okay. This is positive charge. Now can this carocatine become more stable? Is it possible to be like 3° carocatine from here? Possible. Can I shift some hydrogen or something to make it a threederee carocatine? Anyone? Okay, this methile you can see can shift and this will generate CH3 CH3 plus. Okay. And now this hydrogen as well as this hydrogen are next to this carocatan that can be removed and double bond can be shifted here or there. If you shift this one, if you remove this hydrogen, more substituent product will be formed. So since we're talking about a major product, the product that will get formed majorly is this. Anybody has any doubts please ask anything. Shall I proceed? Yes. >> Can you repeat the whole mechanism once more? >> See this is a carocatine that get formed. This is clear I guess right. Then this compound to gain the stability ring expansion whenever there's a chance it will happen. Four member ring is very strained. So there's an alky shift. There's a ring expansion that happens. So from here this will get formed and the positive charge comes here. Is it clear till here? Till here it's clear. No. How you know positive charge will go there? Because this bond has been broken. This bond has been broken for this carbon and for that carbon both. But for this carbon one bond has broken and another bond has formed. But for this carbon it is just that one bond is broken and nothing else happened. So for that carbon positive charge it has lost one electron a positive charge. Then this is a twoderee carocatine and if you shift methile you'll get 3°ree carocatine and that is why there's a 3°ree carboatine and hence this product >> this >> okay any other doubt anybody nothing okay now this was even reaction with highlights Can anybody tell me how the even reaction will happen with alcohols? Carotine needs to be formed. How it will happen? >> So is it by adding electron withdrawing? >> Adding what? >> Electron withdrawing groups. No, that is not um don't confuse with benzene. What we used to do with uh SN1 for alcohol >> you add H+ and then >> add H+. Exactly. Exactly. Okay. So for even mechanism with alcohol it is already written above for that we will have either these you know when these things are written H3 P4/ heat or second H2SO4/ heat delta is important. You can see that whenever there's a heat or delta or written higher temperature, your mind should immediately think that there is an elimination that is going to happen. Okay? Otherwise, uh substitution can also happen, right? And here uh there is no nucleophile as such. Okay? just that H2SO4 and heat the reaction will proceed with the elimination. Okay. But in the previous case there was this uh you know in SN1 H2SO4 was there and then there was a nopile which attacks. Okay. Fine. Fine. So let's uh let's take certain examples. All of you do this. CH2O if you write concentrated H2SO4 okay then you don't need to write heat also otherwise H2SO4 heat let me write heat you know so that you don't get confused Anybody got it? Done. This carocatine will get formed. Then what will happen? Anybody? >> Shift. >> Huh? What will shift? >> The positive charge will shift. >> Hydride. >> Hydride shift. This hydide shift will happen. Hydride is H minus. Okay. H minus shift there. This becomes CH3. This is plus 3° carocatine from 1° directly it is becoming 3°. So that will be favored. Now hydrogen the adjacent carbon hydrogen can be from here and can be from there. If it is from here more substituted alken will get formed like this. If it is from here less substituted alkan will get formed like this. So this is major. Okay. Anybody has any doubts as such? Anything? Nothing. Okay. I'll give you uh maybe one more example. Do this. Then see uh these questions are not uh very straightforward. So in case you're not getting it, it's okay. But learn from it. Okay? If I were a student, I may also be not getting it. So this O will be gone and positive charge will be here. Do you all agree this carocatine will be formed? Now can this carocatine which is 2° becomes 3°? Is it possible? Why not not? >> So for there's no adjacent hydrogen atom for the tertiary carbon. >> No even alkal shift can happen. Right. >> You have these two alkyle. This one and this one. These two alkyes are there, isn't it? So, which one will shift? >> Methile. >> It is ethile that shifts. Okay. Always remember this migratory aptitude. Hydride is highest. Then phenile is the highest. Phen hydride though it is so little because of its size it can shift very fast. Then if alkaly shift has to happen then 3° then 2° and then 1°ree. Okay, let's not get into reasons every time. Just remember this. Okay. Now this alkyle will shift over there and you have positive charge there and there's a hydrogen here which whose electron can create a pi bond. and more substitute alken gets formed like this. Okay, fine. I guess next question. These questions are more important in theory. Okay, so you should be able to solve questions. So, next question is this. This is well-known famous question. Delta is there elimination will happen. Are you done? Okay. First, let me make it straight like this. So I will ask you only this is CH2+. Now what do you think should happen? Unmute yourself. >> The hydrate shift. >> Hydrate shift will happen. it will gain this will help it gain much greater stability apart from being uh 3° what else is special about that carbon what >> it's a conjugated system >> it's a conjugated system double bond single bond positive charge is a conjugated system double bond single bond double bond that is also conjugated double bond single bond negative charge that is also conjugated conjugated it's a cross conjugation from left hand side and as well as from the right hand side now what do you think will happen let's see >> yeah it have like a resonance and form a benzene ring something like that >> okay see one product is that a double bond gets formed here only right that is one product if I ask you next product is there a possibility ility of another product. You can say sir this is there is a hydrogen here at just carbon hydrogen but then you can't have a double bond here. So that is why it can't take this hydrogen and also it is allelic hydrogen sp2 hybridized carbon hydrogen that can't be taken just like that. So due to conjugation you have to see what is the next possible structure. So when you shift it over here double bond comes here. This there's a positive charge now goes there. So you can see a 3°ree carocation converts to 2° carocation but more than 2° 3° this conjugation is much better because it is continuous conjugation. Here it is what crossed they are coming in way of each other. But anyways uh that is not most important thing. The most important thing is a product is very stable because now it can take this hydrogen and form benzene ring. So you can see now this will get formed right? That is most important thing. The product is so stable that once it starts getting formed entire thing will move towards that only. Okay. Okay. One last question before we go to E2. E2 is a small uh this thing >> sir. Uh so the previous question uh it has two products formed or just one >> very little amount of uh this this product will get formed. >> Okay. >> Because the second product is so stable that almost entirely will be but then suppose there's a question what will get formed only one option is correct then you have to choose the major product only. Okay, if they ask you write down all the products, then you have to write all the possible products, right? And then you have to also see cis and trans. Sis and trans are possible only when you'll have two different groups attached to this carbon and two different group. If let's say these two groups are same. If these two are same, no cyr. These two should be different. It should be XY here also or here could be Y Z or something. It can't these two cannot be same for CIS entrance which we have done in the GOC in detail. Okay. Anyways, CH2O CH3 you have H2SO4 concentrated concentrated H2SO4 is a great dehydrating agent. I mean it will you don't need to write heat with alcohol at least. done. First carocation will be this. Then what will happen? this hydride will shift. Huh? So you'll have CH3 CH3 plus now it can take hydrogen from this carbon as well as from that carbon. From where it will take more A or B? Hy hydrogen it will take from A or B? >> From B. >> B will create this uh alken. And if you take from A, this alken will be created. So where it will take A or B? This is from A. This is from B. >> So would it be A? >> It'll be A. More substituted alkenes are always preferred. If it can get formed. So in even more substituted alkenes will get preferred. things will go towards uh more stability or thermodynamic uh thermodynamics will drive here. Kinetics will not drive or entropy will not drive situation here. Okay. So I'll move ahead to E2 now. So this is E1. How will you identify whether it is E1? there's a delta sign or concentrated H2SO4 is written concentrated H2SO4 for the uh for alcohol. Okay, for highlights it is always delta sign with a weak nucleophile E1 will happen. Okay, now we'll write down E2. E2 just like SN2 is a biomolecular. Write down biomolecular reaction elimination reaction. Okay. So again here it is beta elimination that we'll be talking about. Beta is what we discussed. Uh beta is what? The alpha carbon is what? Alpha carbon is the carbon in which leaving group is attached in which hogen or O group is attached. That is your alpha. Beta is what? Next to the alpha. Next to the alpha could be left hand side or right hand side. Okay. So that is it can have let's say two or three beta carbons. Alpha alpha can have two or three beta carbons. So this is leaving group. This is hydrogen. It'll create carbon carbon double bond. This is what we have learned in E1 also. What is so special in E2? In E2 everything happens simultaneously. It's a biomolecular elimination. Just like in SN2, uh the reaction depends on concentration of the nucleophile as well as the substrate. The red determining step will have both together in SN2 as well as in E2. In SN1 and E1, red determining step has only substrate. Nucleophile doesn't come in the red determining step. If you increase the concentration of nucleophile rate will not get affected for SN1 and E1. Okay. Now coming to E2. Let's see the typical mechanism. This is H. Now everything is happening simultaneously. And uh you know here instead of calling it a nucleophile which attacks we call it base. It is better to call nucleophile as base since it attacks H instead of carbon. If it attacks carbon then you say nucleophile. Why will you say nucleophile when it takes H+? H+ is taken typically by the acids. Sorry, base. So you'll call it base. So this base takes this H and it just takes H+. There are two electrons here which are creating the bond. These two electron will go there and start creating the pi bond. And this leaving group will take its lone pair and leaves. And all of this happens simultaneously. Okay. And then you know this uh carbonarbon double bond gets created. Okay. So if you look at the transition state it is somewhat like this. the base. This is bond formation. This is bond breaking. This is bond formation. This is bond breaking. So bond formation, bond breaking, everything is happening together. Okay. Now let's write a couple of point then I'll come back to the stereochemistry of it. Write down uh E2 reactions. Do you think it this E2 reaction can happen with uh alcohols? O can it happen with that? Can O leave like this? O cannot leave like that. So with alcohols you have to catalyze and when you catalyze it becomes even. So with alcohol E2 will not happen. You need good leaving group which can leave on its own. When base attacks it can leave. Okay. When the base attacks you can't have acid and base together in a mixture and then you'll be like okay base will attack hydrogen. Acid will catalyze the alcohol and then good living group. It doesn't happen like that. Okay. So please write down E2 reaction needs good leaving group. It need not be helide though need not be helide but typically it is helides because helides are good living groups. Hellightes are good living groups. Fine because Br minus such a big size bromine bromine iodine they are big in size and so if it is a negative charge negative charge will spread in entire volume and it becomes stable that's why it is a good living group. Okay. So what are the reagents here? What kind of base are there? Please write down uh hot reagent that will be written for you to understand that E2 is going on. Hot alcoholic hot alcoholic solution of Koh. Alcoholic is what? A polarroic or polar aproic. Aroic. >> Alcohol are aproic. They are proteic. They have a hydrogen attached to the uh electrogative atom. Okay, which is oxygen. Or you can have let us say EO minus in EO H. Okay. You you can have M EO minus in MO. MOH is what? MOH is solvent. MO minus is a salt. Meus salt dissolved in MOH. Can you dissolve MO minus in EO? You can dissolve. Dissolve. But now you have two nucleophiles or two bases. EO will also do something and MO minus will also do something. So you can have multiple products. So that is why EO minus in EO or MO minus in MO. Okay. Or simply you can take hot alcoholic solution of KOH. That also will do. You you need a very strong base. A very strong base. Na H2. NH2 minus is extremely small uh extremely strong base. NH2 minus can take uh the H+ and become ammonia. Then you have these kind of bases like tertiary buttoxide in tertiary butile alcohol. Okay. Fine. Is it clear to everybody? Clear? No. Okay. Now, how it is different from SN2? SN how will you recognize whether substitution by molecular substitution will happen or biomolecular elimination will happen? How will you know? Looking at the reaction I told you already. >> So here you need a base. >> Good leaving. >> There also everything was that way is only but only one thing that you'll see is heat. In all the elimination reaction heat is one thing which favors the higher entropy and elimination will go on. Of course these reagents are also important but most important thing is heat. If heat is not given at room temperature substitution is preferred elimination is not preferred in in SN2 just a strong nucleophile is needed needed in E2. Strong nucleophile plus heat is needed both. Okay. I hope it is clear to all of you. Now let's talk about stereochemistry as in uh what happens to how it attacks at what angle it attacks what should be the geometry so that E2 happens and stuff okay so let's take this thing there's a leaving group S1 first of all the you can see there are four atoms written here you have hydrogen carbon carbon and the leaving group. So all four atoms H C L should be in the same plane. That is the first requirement. Okay. Fine. First thing. Second is L and H. There could be many H. I'm talking about the H where the base will attack. Hydrogen where the base will attack. I'm talking about that hydrogen. That hydrogen and L they should be uh I mean I'm telling in a very crude way they should be like 180° away. If leaving group is down hydrogen should be up. The reason is pretty simple. The base has a lone pair. Leaving group also comes out with a lone pair. They will ripple each other. So base will attack however much away it can from the leaving group. Right? So H and L are 180° away from each other. Fine. How will you know it is 180° away from each other? Have you seen uh this kind of projection? This is the same compound. I can write like this. If you see this kind of projection, this is L and this is H. Then it become very simple. You know that is 180° away from each other. Here also it is evident. Here also it is evident. Here also it is evident. But when you write like this new man projection sorry this is Fer projection in Fisher projection it is not straightforward. So you need to either convert it this way or this way then E2 reaction you have to check. Okay you'll see when uh you'll have uh the reactions. Okay please write down couple of points. Point number one. Point number one, the rate of the reaction depends on rate of reaction depends on concentration of base. I'll write like this rate is equal to constant time base as well as uh the uh substrate. Same thing was with SN2 as well but instead of base we were calling it nucleophile because nucleophile attacks carbon base attacks hydrogen. Okay, second point is rearrangement cannot be possible. Rearrangement is not possible. There is no formation of carocatine itself. Forget about rearrangement. Okay, that is second point. Third point. Third point, write down a I also have to write down. No rearrangement. First, second, third. If bulky base like tertiary buttoxide is used then the hydrogen comes out. from less crowded carbon leading to less substituted alken. Bulky base is like you know tertiary betoxide. How does it look like? I'll show you. This is tertiary oxide. Okay. Now imagine this attacks something like this. CH3 Wait, C uh Br CH2 CH3. Now look at this. this hydrogen and this hydrogen. So this base will attack which hydrogen? One first one or the second one? >> First one. >> First one. So it can't go to the second. It is so bulky itself. So even though more substituted alken is more stable but here because of the steric hindrance this alken will be the major product less substitute will be major product. This happens with the bulky one. Tertiary buttoxide whenever it is there you have to be careful that most stable alken doesn't get formed. less hindered alken gets formed. Okay. Okay. Fine. I hope it is clear. Okay. Now tell me um now we'll solve numericals. Okay. We are done with both E1 and E2. Whatever time is remaining couple of numerical and then we are over done. Uh suppose this is your chlorine. Okay. Hydrogen. Your hydrogen. Okay. Now you are using the alcoholic a and heat. What will be formed? Which hydrogen it will attack first? Tell me that. The first one or the second one? Which hydrogen it will attack? Alcoholic KOH. >> First one. >> Second one. Right sir. >> Which hydrogen is 180° away from the chlorine? First or second? >> First one. >> First one. So it attack first one. Suppose now hypothetically speaking 180° opposite hydrogen is not there. All the hydrogen are towards the side of the chlorine. Then it will not be able to attack only. Nothing will happen. It must have a hydrogen. You can write it down. It must have a hydrogen which is like 180° away from the leaving group. must for E2. Okay. So the alken will get formed and this hydrogen and this chlorine oh sorry chlorine is gone. This will happen. Okay. In fact it is since it is sp2 habitized then you don't need to draw vagon dash with sp2. You can say that it comes in the same plane. This is the hydrogen. But what is gone is this. This one is gone. Okay. Uh next I have a theoretical question. Uh carefully uh listen and answer. Okay. Correct statement for E2 reaction is it is a two-step process. It is a unimolecular reaction. Strong base favors carbonion is formed during the reaction. Which one? >> Strong base favors. >> And what about unimolecular? >> That's a biomolecular reaction, right? >> It's a biomolelecular. Two molecules are involved. Okay. So only that one statement is correct. Okay. Then uh >> so I have a doubt. >> So you had mentioned that all four atoms the H both the C and L must be in the same plane >> right? Uh and uh if they're in the same plane we have to take the uh dashed wedge right because the leading group and H should be in the same plane. >> Correct. I think the solid and the dash they're not in the same plane. >> They are in the same plane. Let's say this one line is going up, one line is going down. I can have a plane like this which will have both lines going up and going down. So my plane is going like this and it covers whatever is above as well as whatever is below. uh which of these cannot but then anyways you know the most important point is the hydrogen and the leaving group they should be 180° away from each other always if that is not there reaction will not happen okay which of these cannot undergo E2 reaction I'll write down this Holy. Answer. A B CD. Which one? >> Will it be C? >> Reason. There's no hydrogen present. >> Okay. All of you agree this. >> I thought it was steel. >> Why it will it has No, it has two hydrogens's here. You need beta hydrogen here. Look at this. Is the carbon from where the bromine will come out. Adjacent to it only one carbon. Does it has does it have any hydrogen here? there's no hydrogen so no point uh talking about that will come out. So you need a beta hydrogen otherwise nothing will happen. Okay. Now tell me rate of reaction here. Uh when when you talk about halides R I, RBR, RC, RF, which one will give the highest rate of reaction in E2. Obviously the living group should leave easily. So this in fact the carbon to florine bond is very strong. So it doesn't fall under E1 or E2 we with RF it is a different thing altogether. Okay but it is slow rate is slow. So that is why we can include it. Clear? I hope it is clear. Okay. Now let's talk about rate in terms of 3° 2° 1°ree alkyle helides for elimination E2 which one will have the highest rate you remember SN2 which one had the highest rate 3° 2° 1°ree >> one >> one degree Over here what? Over here also you expect something >> degree. Yeah. >> But that is not true. Over here it is 3° 2° and 1°ree. This is little counterintuitive but if you uh understand clearly how substitution happens. This is R1 C R2. This is leaving group. This is R3. Nucleophile comes attacks the carbon and the leaving group leaves. But over here in elimination what is going on is this the base is coming it doesn't attack 3° carbon here in elimin sorry in substitution nucleophile has to attack thirdderee carbon in elimination base attacks The carbon next to the threederee carbon, it has to take hydrogen from the adjacent carbon. Getting it? Is it clear? Right. And then this bond is getting broken and uh then you know uh what else? Maybe this bond is getting formed. So hyper conjugation is getting created. So more it is a 3° carbon. So it can stabilize this intermediate that get formed in a better way. So crowding doesn't affect it so much. Even though it is a 3° halide, it doesn't have to go and attack the 3° carbon in elimination. It has to attack the carbon adjacent to the threederee carbon. Clear? So please remember this is something which is different from the SN2. Everything else is very similar but when it comes to the rate of 2° 3° this is different from the SN2. If you compare E2 and SN2 now one last thing is the energy diagram. How many transition state you have here? >> One. So >> one. Do you have any intermediate that an intermediate compound is created like a carocatine? No. No intermediate means only one transition state. One intermediate has two transition state. Two intermediate will have four transition states. So there is no intermediate. So it's a straightforward curve like this. it this reaction right now I'm showing it to be uh exothermic sorry endothermic it can be exothermic also but since we are providing heat and it is kinetically favored not thermodynamically favored typically we assume elimination to be endothermic so that is why when we show endothermic product slightly above we draw okay but it need not be endothermic all the This is the transition state. The main transition state that is there. Okay. Right. So that's it for today. I will share some assignments. We have done substantial amount. Now uh unfortunately yesterday class got cancelled otherwise we would have completed the mechanism by now. But uh was addition mechanism done in your school or only substitution and elimination was done? >> Only substitutions. >> Sub the the elimination was not done. >> Did elimination. We did substitution and elimination >> but addition was not done. Okay. So what we'll do next class? We'll quickly finish the addition and straight away jump to the alkalhalides. Now alkaly helite I mean can you able to relate to your school whatever you did in alkyle helites more than 50% is already over right so now we'll move faster and every time when we do the chapter I'll keep reminding you that this mechanism we have seen this mechanism we have discussed like that we will proceed okay so that's it for today I'll share the assignments okay bye >> thank you sir thank you Thank you, sir.