 So, welcome back. In the last lecture, we had looked at kinetic isotope effects. Specifically, we had looked at primary kinetic isotope effects. So, this gives an insight into the rate determining step and whether the bond that you are making the substitution at is cleave during the rate determining step. So, the principle of primary kinetic isotope effect is that the C-H bond that is substituted breaks in the rate determining step. Effect is seen due to the symmetrical vibrational stretch. Effect is maximum for a symmetric transition state and the theoretical maximum value is 7. And we had also seen how if you have a K-H over K-D value greater than 2, it is called a primary kinetic isotope effect. And we also saw that a linear transition state shows a greater primary kinetic isotope effect than a non-linear transition state. And this is because in a linear transition state, the difference between the C-H and C-D zero point energies is very less as compared to the difference between the C-H and C-D zero point energies for a non-linear transition state. And we had looked at the application of kinetic isotope effect in determining the mechanism of a reaction. So, today what we will be doing is we will be looking at what is called as a secondary kinetic isotope effect. So, in the secondary kinetic isotope effect, the isotopic substitution occurs at an alpha or beta position of the bond that is breaking in the rate determining step. So, we are not looking at the C-H bond that is actually breaking in the rate determining step, but we are looking at bonds that are at the positions which are alpha or beta to the C-H bond that is breaking. So, normally when you think of it, it usually involves hybridization changes. So, if you look at it here, you have an example where you have the substitution at this hydrogen for an sp3 reactant. Now, once you have X leaving, what you generate is the carbocation and X minus. Now, when you see it here, the C-H bond is still intact and if you remember the examples that we had done in the previous class, we had seen that in these cases what you observe is a small kinetic isotope effect. So, some of you might be wondering although this bond is not breaking, why do you see a kinetic isotope effect at all? Because one would assume that okay, this bond is not breaking. So, it does not matter whether you put C-H or C-D. So, the question is why do you see an effect at all? Another example is here, the second reaction you have hydrogens that are substituted and once, so this is a concerted process. So, in the product, you have the hydrogens with the bond still intact, but here you have a change from sp2 to sp3. So, there is a hybridization change and what is observed is that this also leads to a kinetic isotope effect. So, one thing which will be very obvious to you is that if you compare magnitude that is the value, the magnitude for primary kinetic isotope effect would be larger than a secondary kinetic isotope effect, but then you still observe a kinetic isotope effect whenever there is a change in hybridization. So, in the first example, you had sp3 going to sp2 and in the second example, you had sp2 going to sp3. So, now whatever we had studied in the previous lectures about the primary kinetic isotope effect, one thing you must understand by now is that when you talk about kinetic isotope effects, the vibrational levels are very important. So, in order to understand now secondary kinetic isotope effects also, we should look at the vibrations that are involved. So, now let us look at the origin of a secondary kinetic isotope effect. So, we will look at the vibrations that take place. Now, since the bond is not breaking, the stretching vibrations do not make as much of an impact as you had seen in the case of the primary kinetic isotope effect, but what plays a very important role is the bending vibrations. So, now to understand the bending vibration, first I would like to ask you a simple question. So, if you compare the bond length of CH versus CD, which bond would be longer? So, again I repeat the question. So, if you were to compare the bond length of CH versus CD, which bond would be longer? So, you can think about this question by pressing the pause button on your screen, alright. So, when you compare CH versus CD, the CH bond is longer than the CD bond. So, let us assume that this is your CH bond and this is your CD bond. So, the CD bond is smaller than the CH bond. Now, when you think of bending, it involves movement of the bonds. So, if you imagine a CH bond and let us imagine that the bond is moving from left to right like this. So, this is my CH bond. Now, if I compare my CD bond, so this is a smaller bond, I am going to do the same movement. Would you expect, so I have an arc that has come here, would you expect this arc to be smaller or bigger? So, now I have a smaller bond and I am again going to bend it from left to right. So, what do you think this arc will be? Will it be smaller or bigger? Let us find out. So, now I am going to move it from left to right. So, when I compare both of these, what you would observe is that in the case of the CH bond which is longer, it has greater freedom to move, right. Whereas, when I have a shorter CD bond, it has lesser freedom to move, alright. So, you can also try it in your homes by taking a small string. So, you can cut the string into different sizes and then you can try holding the string on one side and seeing how much you can move, alright. So, use two different strings and you can see how much the string moves. So, what you see is there are different types of vibrations possible both for SP3 as well as SP2. So, in the case of SP3, there are two types of bending vibrations that are shown. One is an in-plane bend and one is an out-of-plane bend. Similarly, for SP2, so if I draw arrows to show it, so an in-plane bend, you will have this CH bond moving in the plane of your screen. In the out-of-plane bend, you can imagine this bond moving out of the plane. So, you can imagine it moving perpendicular to the plane of your screen. So, I am using the dash and wedge that you usually see when you represent molecules and show their stereochemistry. So, what this means is this is an out-of-plane bend where the CH bond is coming out of the plane of your screen and into the plane and in I mean going behind the back of your screen. Similarly, you can have in-plane bends for the SP2 and you can have out-of-plane bends. So, in out-of-plane bends again, I will show it with a dash. So, I will show it with a dash and a wedge. So, now let us think of the scenario where you have CH versus CD. So, this is longer as compared to the CD bond. So, now we already saw just with the flexible bond which I had shown you earlier that when you compare CH versus CD, CH can move for covers a larger area in doing so. So, let us try to understand the bending vibrations looking at the SP3 carbons. So, in the SP3 for those of you who have not seen, this is what the SP3 looks like, it is a tetrahedral. So, what I have here is I have these 3 atoms in a plane and then I have the wedge coming closer to me and the dash going away from me. So, this is what the representation looks like when you draw the stereochemistry of the molecule. So, now let us see what happens when you have CH versus CD bond. So, let us first consider the case of in-plane vibrations. So, when I consider the case of in-plane vibrations, so as I said I have these in-plane. So, I move the bond like this. Now, because the CH bond is longer, it covers a larger area. Remember the arc that we had shown here. Now that it covers a larger area, what happens is the steric interaction that it faces with this atom here is more. Now, if I compare it with a CD bond, so if I compare it with a CD bond, again I am doing the in-plane bend, it covers a smaller area. Here the smaller arc, now because it covers a smaller area, the steric interaction with this atom here is minimized. So, if we were to look at the first case shown here which is the in-plane bend, I have, this is my CH bond and I am moving it in-plane. So, you have interactions with this atom here, alright. So, this interaction is minimized in a CD bond which is smaller. So, when you compare a CD bond, this bond length remains the same because we have not done any substitution here. What happens is, this movement is smaller, so this interaction, steric interaction is minimized alright. So, now let us look at the out-of-plane bend. So, in the case of an out-of-plane bend, now we will start with CD. So, in the case of an out-of-plane bend, what I will be doing here is I will be moving this up and down, okay. Remember, this one atom is closer to me and another one is behind me. Let us try to see it again. So, I am moving this up and down. So, let us see it from a different angle. Let us see it from this angle so that it is easier for you to see. So, this is how it is moving. So, now when I have CD, as you can see it moves less. Now, if I compare that with CH, it can actually move more alright. And this case what is happening, it is interacting with these two atoms here alright. So, when you compare the out-of-plane bend, so when you have the case of CH or CD, there are these two atoms that they interact with. And again in the case of CD, since the bond is smaller, again this bond length does not change. So, this interaction is better because the sterics again here in both these cases, this is minimized as compared to the case of the CH alright. So, again to revise in the case of SP3, you can have an in-plane bend and out-of-plane bend. And in both these cases, when the CH bond moves, it meets some atoms. And what is seen is that, so in every case, what is seen is that when you move the CH bond, it meets some atom and which is why, so you have a wave number of 1350 centimeter inverse alright. So, this is directly proportional to the energy that you are talking of the vibration. Now, let us look at the case of SP2. Now, SP2 is very interesting as compared to SP3 because here what you have is, you have all three atoms in one plane. So, you have all these atoms in one plane unlike the SP3 where you also had atoms coming close to you and away from you. Here you have everything in one plane. So, here let us see what happens when you have CH versus CD. So, let us make the substitution. So, this is our CH bond and now we are looking at the in-plane bend. So, we are looking at the plane of the molecule. So, the in-plane bend would be moving in the plane of the molecule. So, in the in-plane bend, what you see is you have interaction with both of these atoms. So, as it moves in the plane, it has interactions with both of these atoms. Again, we can draw it here. So, in SP2, you have the atoms in one plane. So, you have the in-plane interactions with both the atoms. So, here again in the case of CD, this would be smaller. But remember in this case, since the atoms are further apart, the effect is not as pronounced as your CH bond. So, in this case, you have the CD bonds again interacting with both these atoms. So, here again, let us look at the out-of-plane bend. So, in the out-of-plane bend, if you have, so in the out-of-plane bend, what you have is this is the plane of the molecule. So, you have the bond move up and down. Now, what is interesting in the case of your SP2 is that there is nothing above or below the plane of the molecule. So, if you were to look at it with a different view, so you have all of these in one plane and I am moving this like this, what you see is as this moves, you have nothing in this region and you have nothing in this region. So, there is no steric problem that you are talking about. So, in general, the energy for this bending vibration is lower. So, here again, if you replace a CH bond with a CD bond, what you would see is it would move lesser. So, now let us look at what happens here. So, here in the out-of-plane bend, there is no significant steric effect. So, that is why if you compare both of these energies, you would see that the energy for the out-of-plane bend is actually lower than the in-plane bend for the SP2 case. So, now that we have an idea about how the in-plane and out-of-plane bend works for SP3 and SP2, let us see what influence this has on the kinetic isotope effect because that is what we are trying to understand. We are trying to understand the secondary kinetic isotope effect. So, here let us look at the first case where we are looking at a reaction step involving a change from SP3 to SP2. So, this is similar to the example that we had seen on the first slide where you have and then this is an SN1 like reaction. So, you have a carbocation that is generated and let us say that we are making the substitution at this hydrogen. So, what happens in the case of SP3 and what happens in the case of SP2? So, in the case of SP3, if you remember the in-plane and out-of-plane bend, in both of these cases, you have significant steric interaction. That is why both of them were 1350. So, in the case of SP3, in all these cases whether you have your out-of-plane bend or you have your in-plane bend, what you see is you have steric interactions with all these three atoms. So, here if I compare CD versus CH, the vibrational levels, which one would be lower in energy? So, again for SP3, both in-plane, so the net effect you see is a combination of in-plane and out-of-plane vibration. So, for SP3, when you replace CH versus CD, which would be lower in energy? You can think about it and press the pause button on your video. So, alright, let us see if you are able to get this answer. So, in the case of SP3, what you would notice is that the CD and CH energy difference is large because CD has less destabilizing interactions as compared to CH. Now, your transition state would have a mix of SP3 and SP2 character. Now, what happens in the case of SP2? In the case of SP2, if you remember what we had seen earlier, in your in-plane bend, you still have some steric effect. So, CD would be lower in energy than CH. But in the out-of-plane bend, since it is not meeting any atoms above and below, it does not matter whether you put a big bond, a small bond, a very small bond. So, it does not matter. So, CH over CD will not have a very large impact if you are talking about SP2. So, in the case of SP2, the energy difference would be less. I just extend this a little. So, now your transition state would be somewhere in between. So, when you draw your transition state, we are looking at the vibrational energy levels here. So, this is what your transition state will look like. So, now when you compare the activation energies, so if I compare the energy from CH versus CD, so this is E1, this is E2. Again, the same example that we had seen earlier. So, we are talking about climbing a mountain. If you are already at a higher point and your destination is almost the same, then you have a smaller distance to travel. So, what you would see is, CH would be greater than KD. So, typically you would see what is called as a normal kinetic isotope effect while going from SP3 to SP2 and that means the value would be CH over KD greater than 1 because CH is greater than KD and this value would be somewhere between 1 and 2. As I told you, a much larger effect is seen in a primary kinetic isotope effect. So, secondary kinetic isotope effect, magnitude is not large, but then you do see an effect. So, in this case, you see KH is greater than KD. It is called a normal kinetic isotope effect. Now, what happens when you go from SP2 to SP3? So, the example that we had looked at earlier was the Diels Alder reaction. So, if you have, let us see, these are the hydrogens that we are substituting, reacting with some sort of a dienophile. So, the product you get is cyclohexene and then you have these hydrogens here and the R group here and if you are making substitutions here, so here again the bond is not breaking, we are going from SP2 to SP3. So, now let us look at the vibrational levels again. So, in this case, now that you have understood the case of SP3 to SP2, this would be easier, the transition state energy level. So, let us write all the vibrational energy levels. So, I would like you to do this exercise yourself. Now, that I have shown you the potential energy diagram, I want you to sketch that out in your notebook and draw the different vibrational levels for the reactant, the product and the transition state which would be somewhere between the reactant and the product. So, go ahead, press the pause button on the video and work this out yourself. So, let us welcome back, let us see if you are able to do this, if you were not you can also try to follow what I am doing. So, here again this is SP2, this is SP3. So, in the case of SP2, the difference in energy between CD and CH is not much. So, if you did not understand this, you can go back, you can rewind back and look at the explanation for SP3 to SP2 and whereas in your product you have a greater difference in energy. So, your transition state now would be somewhere in between these two. So, you would have an energy difference here. So, now let us try to figure out the activation energies. So, this is the case of CH, E1, case for CD, this is E2. Now, in this case what is happening is you have a reverse scenario. So, imagine again the mountain where you are starting from almost the same point, but then one person has to climb a much higher peak as compared to the second person. So, who will climb quicker? The person who has to climb the lower peak. So, in this case what you have is you have E2 less than E1 because CD has to climb a lower peak as compared to CH. So, E2 is less than E1 which implies Kd is greater than Kh. So, essentially what that means is Kh over Kd is less than 1. Alright? So, this is what is called as an inverse kinetic isotope effect. Why inverse? Because usually you would expect the CH to be faster than CD, but in this case what you observe is it is the reverse. So, it is called as an inverse kinetic isotope effect. So, now we have understood the basis for the secondary kinetic isotope effects involving hybridization changes. So, sp3 to sp2 gives you a normal kinetic isotope effect that means Kh over Kd is greater than 1 and sp2 to sp3 gives you what is called as an inverse kinetic isotope effect where Kh over Kd is actually less than 1. So, now to understand this further. So, now let us see if you have understood this concept. So, shown are some reactions on the left and some values on the right. So, these are Kh over Kd values. So, what I want you to do is look at the reactions, write their mechanisms and match the value on the right. So, see if you can go ahead and do this exercise. We will come back in the next lecture where we will check if the answers that you have are correct. So, thank you and see you in the next lecture.