 Hello guys good morning. So guys today in this session we are going to start the second part of the substitution reaction and the second part of substitution reaction and that is SN2 reaction. Okay, so if you remember last class we have discussed about SN1 reaction. Right, SN1 reaction we have discussed that there is this reaction is of two steps. First steps involves the formation of carbocation right and in the second step the nucleophile, nucleophile is anything that is negatively charged or neutral molecule which has lone pair present on it. Okay, like water, energy, etc. Okay, so in the second step the nucleophile attacks on the carbocation and we get the product. Okay, this is what we have discussed in SN2 reaction. We have also discussed the effect of the effect of solvent, no substrate, the effect of solvent we have discussed we have discussed the effect of substrate. We have also discussed the nucleophilicity order right, we have also discussed the nucleophiles. Okay, all these factors affects the substitution reaction. Okay, these factors affect substitution reaction SN1 as well as SN2. If you look at the solvent factor, in the solvent factor we have discussed that SN2 reaction is favoured by polyaprotic solvent. Okay, polyaprotic solvents are those solvents in which there is no active hydrogen present. Means what? Hydrogen is not attached with, hydrogen is not attached with the electronegative element that is oxygen, nitrogen, etc. Okay, these things we already discussed. So let us understand what is SN2 reaction right and what is the difference between SN1 and SN2. Okay, so write down the heading SN2 reaction like SN1 that is nucleophilic substitution first order reaction. Similarly, this one is nucleophilic substitution second order reaction. Nucleophilic substitution and two stands for the order of the reaction that is second order reaction. You don't have to think about order over here. Okay, that is a part of chemical kinetics. So that we will discuss in class 12, they won't ask any questions based on this. Okay, in KVVOS. You don't have to think about this. What is order now? Okay, so just let it be. Now, look at this reaction. If I write down one general reaction here of SN2 type, we have a nucleophile, right, nucleophile and a substrate. Substitute is an alkyl group and a leaving group attached over here, right. So what happens in this, this is suppose we have delta negative and delta positive charge delta negative and delta positive. So this nucleophile is attracted towards the positive charges species like this. So this nucleophile wants to attack onto this carbon atom, which has a partial positive charge. Okay. And what this nucleophile want this one that this leaving to goes out and and this will make a bond with this carbon atom over here. Right, so we'll get a transition state over here. Transition state is this where this nucleophile is trying to attach with this alkyl group and leaving group is trying to go out. This is a transition state we get. Okay, this is a transition state. Transition state is the intermediate state of any reaction. Transition state is the intermediate stage of any reaction. Okay, means these two reactants react and before forming the product we get a complex compound, right, which is highly unstable, right complex compound that we call it as transition state. If there is condition sufficient condition that this transition state can convert into the product and this will convert into product which is this finally this bond forms and leaving group goes out. This is a product. Now I have written this in two steps this comes over here this comes over here but actually the reaction is single step reaction. Means this we won't show in any reaction overall the reaction what we'll write and you minus less RL this converts into any you are an L minus single step reaction. I have written this transition state to make you understand but it is not a two step reaction. Okay, got it. So this is one thing. Now one more example in this you see suppose we have a substrate which is this we have carbon carbon attached with a leaving group with this is a molecule substrate right and one nucleophile we have here. This nucleophile attacks on this carbon right and we get a transition state here. Right, the transition state is this these three atoms of molecules becomes into the same plane. And this side we have any you. And this side we have leaving you. Okay, so when you look at this, the carbon is is no tends to form five bonds here 12345 bonds here. Which is not possible. Right, that's why this transition state forms for a fraction you cannot even imagine this forms and this breaks down into the product or the reactants. Okay, this is not possible. Hence it is highly unstable structure. It is highly unstable structure. Okay, so this converts into finally this converts into this bond goes out and the nucleophile attacks from the backside. Right, carbon X, why this is this is the problem we get plus L minus goes out. So this is SN2 reaction. Okay, now what is the properties the characteristics of this reaction you see characteristics of this reaction. For example, it is the single step single step reaction. Okay, order of the reaction is true. Okay, there's no carbocation formation, no carbocation formation. Okay, rate of the reaction directly proportional to the leaving group ability, leaving group ability. We can also say it is inversely proportional to the bulkiness of X, Y and Z. Okay, bulkiness of X, Y and Z. What is the meaning of this bulkiness of X, Y and Z. XYZ if it is bulky, this XYZ if it is bulky group, then this will not let the nucleophile attack on the carbon because of hindrance. It provides hindrance to this nucleophile. Okay, and hence this attack of nucleophile is difficult and the rate of the reaction increases rate of the reaction is also directly proportional to the strength concentration, millenium strength and concentration of nucleophile. A strong nucleophile can attack on the carbon atom easily. That's why more tendency to attack mode will be the rate of reaction. This one you see the last one that I've written. Let's see one example on this. Suppose we have two reaction that is CH3O- and this reacts with CS3Cl, SN2 reaction we are considering. Okay, and the second one is BHO- plus CS3Cl, SN2 reaction we are considering. Okay, if you consider the product here, the product will be this attacks on to this carbon and this chlorine goes out. This attacks on to this carbon and this chlorine goes out. The product is CS3O- CS3 plus Cl minus goes out in this. Here we get BHO- CS3 plus Cl minus goes out in this also. Now the point is which one A and B which one has higher rate for this reaction. See both the reaction you see the difference is here CS3O- and BHO- so because of this nucleophile only we can compare the rate of the reaction in this case. Because this is same, this is same, this only affect the rate of the reaction. So we consider the nucleophiles. So look at the two nucleophile here. The nucleophile are CS3O- and BHO-. This BHO minus is this. We have a benzene ring. You know CS3 has plus I effect, releases electron and this has minus M effect this phenyl which shows it withdraws electron towards inside. So here what happens in this case, the electron density on oxygen atom is less. Electron density on oxygen atom is less. Less in comparison to this. Here the electron density is more. Oxygen has more electron density. More electron density means it can easily attack onto this carbon atom. Okay, on this carbon atom. So what happens here? This has more density to attack. Hence the rate of first reaction A is more than that of that is how we compare the rate of the reaction. Okay, the next property of this reaction is, it is favours in polar-aparotic solvent, polar-aparotic solvent. Okay, it leads to Walden inversion. Walden inversion. Walden inversion means the configuration of an optically active compound R becomes S or S becomes R. The change in configuration is inversion. And this we call it as Walden inversion or umbrella inversion. This last point, this Walden inversion, it is valid for optically active compound, for optically active substrate. Otherwise there is no sense of talking about this Walden inversion. So this is all about SN2. Okay, SN1, SN2 we have discussed. One general comparison you should know. One general comparison you should know that SN1 and SN2, this both are competitive reactions. Means, in any reaction both competes with each other in order to form the product. It's not like in one reaction only SN1 is possible. We can say SN1 is, you know, we get product, major product according to SN1. And we also say, we can also say that the tendency for the reaction to go under SN2 is less. It is not zero. But it is less. Similarly, in SN2 reaction, the tendency for the reaction to go under SN1 is less. So SN1, what happens if you try to recall, last class we have discussed, SN1 reaction is, it is what we can say, it is a formation of carbocation takes place into this. So it leads to the formation of carbocation first of all. And in this one, there is no carbocation. That is the basic difference. So it depends upon the substrate actually. So if the substrate can give a stable carbocation, right, the substrate can give a stable carbocation. Then the reaction goes under SN1 reaction. If other things are same, right, if other things are same, the reaction goes under SN1 reaction. Okay. If the carbocation is not stable, then it goes under SN2. I'll discuss this with some example. Okay. So for example, you see if you have alkyne halide, if you have alkyne halide, alkyne light, for example, you see, we can take CH3, CH3, CH3, CL. Another one we can take CH3, CH, CL, CH3, CH3, CH2, these three compounds. If you look at these three compounds, okay. So this carbon attached to three carbon atom, so it is tertiary alkyne halide, 3-degree. This one is secondary alkyne halide, 2-degree. This one is primary alkyne halide, okay, 3-degree, 2-degree, 1-degree. Now we know the 3-degree carbocation is more stable than 2-degree and 1-degree because of the hyperconjugation effect of this alkyne halide. So the carbocation forms over here, if you see, the carbocation that forms in the first case is if it forms, then it will form this one. This is the carbocation it forms and in the second case, the carbocation is CH3 whole twice, CH positive charge and it is CH3, CH2 positive charge in the third case. So number of alpha hydrogen, if you look at here, that would be 9, number of alpha hydrogen here, that would be 6, number of alpha hydrogen here, that would be 3. Okay, so due to hyperconjugation, the stability order is this, without any doubt, more alpha hydrogen, more stability. Right, so since the stability order for carbocation is what? We have already discussed it in GOC, that is 3-degree, 2-degree and 1-degree. This is the stability order. So if you have 3-degree alkyne halide, this has major tendency to go under SN1 reaction because it gives stable carbocation. 1-degree carbocation is not that stable, so primary alkyne halide goes under SN2 reaction, mostly, if other conditions are same. Okay, so this thing you can keep in mind. 3-degree goes under SN1 and 1-degree goes under SN2. Like I said, these are competitive reactions. Okay, so it's in 3-degree also, there will be a little bit tendency of SN2, but that won't give you the major product. Hence we say that it goes under SN1 reaction. Okay, 1-degree goes under SN2, still they have little tendency to go under SN1, but major product is according to SN2, hence SN1 we are neglecting. Okay, but what about this 2-degree and 2-degree carbocation? Sorry, 2-degree alkyne halide. 2-degree alkyne halide can go under SN1 or SN2 both, but this depends upon the solvent present into this. SN1 possible when the solvent is polar protic. Polar protic then SN1. SN2 when the solvent is polar aprotic. Example of polar aprotic solvent, we have DMSO. It will be written in the question. DMSO. DMSO stands for dimethyl sulfoxide. So dimethyl sulfoxide is DMSO. It is polar aprotic because there is no hydrogen present to this oxygen atom. That's why it is polar aprotic. Polar protic is any solvent you can take. Water is polar aprotic solvent, oxygen-hydrogen bond. Alcohol is like CS3 over H methanol is polar aprotic solvent, CS3 over H bond. There are many examples like this. Okay, this one we can have CCL4 also, polar aprotic, benzene also, polar aprotic solvent. Okay, so there are many examples of this too. So in case of secondary alkyne halide, you have to be a little bit careful. Okay, depending upon the condition you will get the answer according to the mechanism, whether it is SN1 or SN2. But always keep that in mind that in the option you will get, you'll have one of the options according to SN1 and one of the option according to SN2. So you have to be very careful that what is the solvent given over here. Okay, let's see this example. Suppose we have a compound say CCL, CS3, CS3, CS3 and this is, you know, it is reacting with a nucleophile. Suppose we have over H minus and the solvent we are taking is H2O. In this one, in this one, the solvent I am taking here is CCL4 over H minus with CCL4, strong nucleophile over H minus. Okay, and CCL4. So this reaction is also SN1 because it is 3 degree. This is also SN1, 3 degree. The product of the reaction would be COH. The product is same here. The product is same, but the path is different. Now the path is also same here. The point I am trying to make is what? We are taking different solvent here. But then also the mechanism is SN1 since it is 3 degree. Okay, but here the SN1 tendency rate if you compare here. Rate if you compare of the formation of product A and B. Rate of formation of A and rate of formation of B if you compare. A will have more rate because the solvent is polar protic here and polar protic solvent favors SN1 reaction. This is polar protic and this is polar of protic. So difference of solvent we have then we write down the order with respect to solvent. Polar protic more rate towards SN1 reaction. Hence this is, no, A has more rate over here. Suppose if you have primary alkyl halide, CS3Cl with OH minus and H2O. Okay, then CS3Cl with OH minus with, suppose we have again DMSO. Then the product is CS3OH and this is also CS3OH. Mechanism is again here we have SN2. Why? Because this is 1 degree. 1 degree. So this is also SN2. This is solvent is polar protic, but still it is SN2 because it is 1 degree. Product is same, rate if you have to compare. So rate of A here is less than to the rate of B because B is what? B is polar or protic favors SN2 reaction. If nucleophile is different, like in this case you see, suppose we have CS3Cl plus CS3CH2OH. Okay, CS3Cl CS3CH2OH minus. So here you see some solvent we are taking, same solvent we are taking here. Solvent is not the constraint here, same solvent. Obviously if you change the solvent also here, then it depends. Then it will be factored actually. You cannot compare in that case. Because nucleophile also you are changing, solvent also you are changing. Okay, so one thing must be common. Then only we can compare based on the uncommon part given in the reaction. Okay, so now when you look at this nucleophile, so this one is a better nucleophile, better nucleophile. Why? Because it is charged, negative charge is present. So it has more tendency to attack here. Right, here also we have lone pair but there is no charge over here. Because of lone pair it behaves as a nucleophile. But this has lone pair and negative charge also, three lone pairs this oxygen has over here. We can also compare more lone pair here, more tendency to attack. Hence it is a better nucleophile. So better nucleophile you will get some product here, that product would be an ether, that is CS3, O, CH2, CS3, O attacks onto this and we will get the product. And here also we will get the same thing, CS3, O, CH2, CH3, H plus goes out and here CL minus goes out. Yeah, CL minus goes out. And H plus combines with CL minus forms HCL here, forms HCL here. But this is not the important one, the important thing is that which one has a better rate over here. Okay, so rate if you compare rate of the reaction A and the rate of the product B, product A and product B. B gives produced by a better nucleophile, so rate of B is more than product A. Some more questions you see to compare the rate of SN2. We have different L-type line. IOD is a better living group, better living group. All these things you can take as an information. However we have discussed it before also, the order of the living group. But for you, you are doing it first time. You should take it as the information. Okay, this happens like this. This is a better living group. This is a poor living group. These things you can understand this one. Okay, at least twice and or twice you need to do this, then only you will be able to get it. In the first time you will get only 40%, 50% of the things, not more than that. Okay, so twice twice before the exam you have to do. So this one is a better living group. So order of the rate would be this, better living group can go out easily and hence more will be the rate of reaction. This one if you compare, when alkyl halide attached directly with the ring, okay benzene ring, we call it as aryl halide, general name, aryl halide. This molecule particularly, this one is aryl chloride. Okay, general name is, general name is aryl halide. Okay, now aryl halide, the carbon and halogen bond in this, the carbon and halogen bond is difficult to break because of partial double bond characteristics. Because of the partial double bond characteristics. Okay, this pi electron, the lone pair that we have here, it is actually in resonance with the ring if you see. This comes over here, this goes over here. Pi sigma lone pair, the conjugation, hence we develop here a partial double bond characteristics. Because of this partial double bond characteristics, the carbon halogen bond is strong. Right, if you look at the strength of the bond, then carbon single bond is, you know, the strength is least, then we have double bond, and then we have triple bond. This is the strength order we have of the bond, single bond, double bond and triple bond. So if we have a partial double bond characteristics here, so it is stronger than the carbon halogen single bond which is present over here. You see in this case here, there's no resonance, we don't have partial double bond characteristics here. Right, so aryl halide, the bond is stronger, here the bond is weak. And since the bond is stronger, it is difficult to break this bond and when it is difficult to break this bond, this chlorine has less tendency to go out as a leaving group. Right, so in this case we can say what, aryl halide, the halogens are a poor leaving group in case of aryl halide. Right, not in general, particularly in this case. Right, hence a poor leaving group. Okay, you must be thinking few minutes back we were talking about that the halogens are a good leaving group. Now I'm saying it's a poor leaving group. That's what the organic chemistry is. Okay, you cannot generalize things over here. Everything will have some exceptions here and there. Okay, like in this case it happens this way. In this case it happens this way. Right, generally in most of the cases, halogens are a good leaving group, but in case of aryl halide it is not. Okay, so here it is a poor leaving group. This one is a good leaving group. Better leaving group means can go under SN2 reaction easily. Hence the order of SN2 reaction is this. This one you see, in this molecule you see what is the difference we have. We have to compare SN2 order. Right, the difference in this two molecule is what? The difference is this methyl group, the position of this methyl group. It is on this carbon which is attached to the chlorine. Hence here what happens, suppose you have a nucleophile and here the nucleophile has tendency to attack on this carbon and here the nucleophile has tendency to attack on this carbon which is attached with the halogen. But since we have one methyl group present over here, here you see what happens, two hydrogen. Here we have one methyl group instead of one hydrogen. So this methyl group produces more repulsion to this nucleophile. It is relative basically. Here you see the repulsion of the two hydrogen atoms to this nucleophile is lesser than the repulsion of this methyl group and hydrogen present over here. So we can say here, here we have more steric crowding, more steric crowding and hence the repulsion is more over here, less rate towards SN2 reaction. So the rate of SN2 reaction in this case would be this. This one you see, this alkyl halide, this one is one degree alkyl halide and this one is two degree alkyl halide. And we know one degree alkyl halide has more tendency to go under SN2 reaction. Hence the order of, this question you see, in this we need to write down the product. We have BR here, one BR present here and we have one methyl group. We are using OH- as the nucleophile, only one equivalent of it. There is only one OH- we have, not more than one, only one equivalent of it. Then what would be the product in this one? I hope you have done it. So since they have only one equivalent of this nucleophile OH- obviously this is a nucleophile we have. So this nucleophile has tendency to attack onto the carbon here or here. Since only one equivalent we have, it will either attack on this carbon or this will attack on this carbon. Depends upon which carbon, which alkyl group is more reactive towards SN2 reaction. So how do we compare that? Easily if you see this carbon atom. This carbon atom is secondary because attached with two carbon. This carbon atom is tertiary, 3 degree. So if you are considering SN2, this is what we are considering. You are not considering SN1 here. I am asking you to write down the product towards SN2 reaction here. So if you are considering SN2, then this OH- attacks on the less hindered carbon atom which is 2 degree. And then this VR- goes out and the product here is, this VR will be as it is and here we get over. Since we have only one equivalent. But suppose the same reaction we are considering, nothing is mentioned. Nothing is mentioned. And we have one equivalent of OH- one equivalent and solvent we are taking H2O. Water is the solvent we are taking. So since this water is a polar-protic solvent, this is what polar-protic solvent and hence this favors SN1 reaction. SN1 reaction, again you have to compare secondary or tertiary. Where the SN1 reaction is more, obviously tertiary. So what happens, the product here, first we get a carbocation, this VR will be as it is. First we get a carbocation and this carbocation in the next step, the nucleophile OH- attacks onto the carbocation and we get the product here which is this. So this reaction is, since it is H2O, it is SN1 reaction we are considering. So this is how we will go for this kind of question. If solvent is not mentioned, SN2 is given. If solvent is mentioned, then the mechanism will be SN2. We will get this product carbocation and then if this carbocation is forming, so we will always try to get the most stable carbocation. For the given molecule, the most stable carbocation will form. This one you see OH-1 equivalent of SN2. I am giving you SN2 product. So what happens, obviously it is aryl halide, this part you see, this bond has partial double bond characteristics, double bond character, difficult to break. This one has single bond can break easily. So this OH- this will attack onto this carbon atom and this goes out as a leaving group, SN2, no carbocation formation. So the product here is, so overall what have we discussed, we have discussed this that in any reaction there are chances of both SN1 and SN2. Hence both SN1 and SN2 reactions are competitive reactions, depending upon the substrate, whether it is primary or tertiary, depending upon the nucleophile, whether it is weak nucleophile or strong nucleophile, depending upon the solvent, whether it is polar protein or polar protein and hindrance and all. Based on all these factors we can analyze and conclude whether the reaction goes under SN1 or SN2 reaction. SN2 reaction, again I am telling you, it is a single step reaction, no carbocation forms into this one. SN1 is a multi-step two step reaction, first step carbocation forms and the second step a nucleophile attacks on the most stable carbocation. So whenever the carbocation forms in any reaction, okay, so we always try to get the more stable carbocation, okay. By hydride or methyl shift, last class I have discussed on this, what is hydride and methyl shift, we haven't gone into so depth of it. We only now consider the facts of the portions of KVPY, however again I am telling you that the syllabus is not defined, but all these things that we are discussing it is based on the previous year paper of KVPY, okay. So always keep that in mind. Now in general what we can write, suppose we have the last thing we will see, suppose we have a nucleophile will have a solvent and product, product by means what would be the possible mechanism. If the nucleophile is neutral, if the nucleophile is neutral that is weak, right, or the nucleophile is negatively charged. So I have already discussed the negatively charged nucleophile are strong nucleophile, OH- stronger than H2O, CS3O- stronger than CS3OH, okay. So neutral is weak, negatively charged are the strong one. So neutral nucleophile plus we have polar-protic solvent, polar-protic solvent. If the nucleophile is neutral, we have polar-protic solvent, this definitely leads to the carbocation formation and hence we get the product by SN1 mechanism. If we have negative charged nucleophile, strong one, and the solvent is polar-protic, then this won't wait for the carbocation to form. We can add directly on the partially charged, partially positive charged carbon atom and the product will gain by SN2 mechanism. So these kind of, you know, a comparison you must know. One last example you see here, suppose we have a substrate, this one. This entire concept is very dynamic actually, we cannot generalize anything, okay. It depends upon what we are taking. Now we see this, okay. If this form, this molecule forms a carbocation. What it forms? This Cl- goes out and we get a CH2 plus here, plus Cl-, right, leaving group Cl-. This is the carbocation and if you look at this carbocation, this carbocation is stable carbocation through resonance. Of this ring, right, resonance, it is stable carbocation. But at the same time, this carbon is what, this halide is what? It is one-degree halogen, primary halide, primary halide. Because it has with only one carbon. So you see, we have discussed that the primary halide goes under SN2. If you look at this particular information, then the mechanism should be SN2 because it is primary halide. But if you look at the stability of carbocation, the mechanism should be SN1. And we know SN1 is possible when there is tendency of forming stable carbocation and this carbocation is quite stable. So which, you know, mechanism is possible over here, right. So in this case, if you take, in this case, it depends upon what nucleophile you are taking, right. If the nucleophile is weak, weak nucleophile, then the formation of carbocation is possible and the mechanism is SN1 here. Because weak nucleophile, less tendency to attack, right. So if the nucleophile is weak, you are taking a nu minus weak. Since it is weak, so this will wait for the charge to develop, then it will get attracted towards this. It won't get attracted towards the partial positive charge since it is weak. But if it is a strong nucleophile, suppose the nucleophile is strong, then it is strong enough to attack on the partial positive charge of this carbon. Then this won't wait for this bond to break and form the carbocation. And hence the mechanism is what? Mechanism is SN2 in this case. But if the nucleophile is weak, the mechanism is SN1 for the reaction. Okay. So like I said, this two things are very dynamic, depending upon the molecule substrate nucleophile solvent, we can decide whether it is SN1 or SN2. This requires a lot of practice for the understanding of these two reactions to get the right answer. It requires a lot of practice with practice only you will understand obviously certain things you have to keep in mind that this is in this case this kind of things happens. The reaction would be this and product according to this reaction is. Okay. So this is it for SN1 and SN2. Next class, we'll see some examples, some problems on this, how to solve problems based on SN1 and SN2. And some more, no, good examples with this. Okay. So thanks for today. Take care.