 So, in the last class we had wrapped up acid-base catalysis, so we had looked at both specific acid catalysis as well as general acid catalysis. So, specific acid catalysis is where proton transfer occurs before the rate determining step whereas general acid catalysis is where proton transfer occurs in the rate determining step. Now remember specific acid means the acid is the protonated solvent, so it is water it is H3O plus, so that is why it is called specific acid and specific base whereas general acid can be any acid that you add in the reaction medium that is why it is general. And we had looked at the example of acetal hydrolysis and we had seen that in which cases you see specific acid catalysis and we had looked at an example where you see general acid catalysis. Then we had looked at how you can correlate the rates of these reactions following general catalysis with the pKa, so this is called as the Bronsted catalysis law, so for a general acid the law K is given by minus alpha pKa plus C whereas for general base the law K is given by beta pKa plus C, so this is similar to a linear free energy relationship, so on the left you have the rate constant for the reaction which is being protonated by the general acid and on the right you are looking at the ability of the acid to protonate water because that is what your pKa is. So you are correlating this thermodynamic parameter with the rate constant K. Now the negative value of alpha is because the rate increases for a general acid as the pKa decreases because the general acid becomes a better acid, so we had also looked at what alpha and beta signify, alpha and beta are called sensitivity constants and they indicate the extent of protonation in the transition state of the rate determining step. So typically the values lie between 0 and 1 and we had seen that in some cases you see a negative value, the negative value indicates that your rate constants for protonation does not follow the same strength as the pKa, so the acid which is strongest would actually undergo a slower reaction, so that indicates a negative alpha value and we had seen an example where this is observed, so typically but the values lie between 0 and 1. So now we will move on to other types of catalysis. The first we will do in today's class is catalysis due to binding. Now catalysis due to binding we had looked at reaction coordinates before and you can express catalysis due to binding using this generic equation, so you can have your reactant in this case your reactant is A and let us say you have the reactant binding to the catalyst, so you will have the complex of reactant with catalyst and then you will have this being converted to the product B which is still bound to catalyst and then ultimately you will get B plus catalyst. So this is how you can write this entire process, so this here is the uncatalyzed reaction and we had done this reaction coordinate before, so this is just like revision for you, so can you then go ahead and draw the reaction coordinate diagram for the catalyzed reaction, you can press the pause button and draw it in your notebook. So let us see if you were able to draw this properly, so the first thing you have is the complex between the reactant and the catalyst and as I said since it is binding to the catalyst it indicates a lower energy than A, so you have this then you have conversion of A cat to B cat, so that if it is actually catalyzing the reaction what you would see is you would see a decrease in activation energy and then finally you will have formation of your product. So here what is important if you remember we had told you, so I had told you that for the reaction to be catalyzed efficiently by the catalyst what you need to have is you need to have a tighter binding with the transition state compared to the reactant, so you have to have and we had looked at different scenarios if you remember as to how the reaction coordinates will look if it does not bind tightly to the transition state. Now in this case what happens is because it binds tightly to the transition state you have a lowering in energy as compared to the uncatalyzed reaction, so it is very important that your catalyst binds tighter to the transition state as compared to the reactant. Now if you were to think about it similar to a concept that you had studied earlier remember initially when we were looking at various factors affecting reactions I had told you it is always the relative difference remember we had spoken about climbing a mountain and if you have to reach, so if you need to reach the peak faster you would need to either lower the height of the peak or increase your starting point, so you either increase your starting point so that you have a shorter distance to cover or you lower the height of the peak so that you have a shorter distance to cover, so you can think of this similar to solvation which you had studied earlier, so when you had studied solvation earlier if you remember I had told you what is important is the relative stabilization, so say I have a reaction, so this is my reactant and this is the transition state and suppose the transition state is quite polar and now I do the reaction in a more polar solvent and let us assume the reactant is not as polar as transition state, so what you would see is that for your reactant the lowering in energy is less whereas for the transition state it is more, so this is a more polar solvent, so now what you see is you have a lowering in activation energy, so always remember it is these relative differences which are very very important when you study kinetics, it cannot be absolute just a very tight binding to reactant will not help your reaction, there needs to be a tighter binding to the transition state, so now with this understanding let us look at specific examples and one of the classic examples is seen in biology when you look at enzyme catalysis, so in enzyme catalysis, so here we are representing the reactant as S which is a substrate, this is a common terminology used which is why I am using that, E is your catalyst which is the enzyme, ES is your enzyme substrate complex, so your reactant catalyst complex like before, EP is your enzyme product complex, so this is similar to the catalyst B complex you had seen before and finally you have P plus E, so if you are getting confused I will just write below this so that you are not confused what each of these corresponds to, so enzyme is your catalyst, substrate is your reactant, P is your product alright, so then the rest of it will make sense to you, so the rate law for this is given by the Michaelis-Menten equation shown on the right, now you might think that you need to memorize this Michaelis-Menten equation but then you do not need to do that, remember earlier we had studied how to derive the rate loss, so you can do that very systematically using whatever you had studied earlier for chemical reactions, so what I want you to do now is given that this is the rate determining step and applying the steady state approximation for ES what I want you to do is derive the rate law for this enzyme catalysis, so the transformation given on your screen, so you can press the pause button and work it out, so let us see if you were able to work it out, so the first thing that you have is the rate is given by Kcat concentration of ES, now assuming that the enzyme substrate complex once it is formed it is consumed to give you the enzyme product complex, you can apply the steady state approximation, so applying steady state approximation what are the things which lead to formation of ES, so you have K on concentration of enzyme into concentration of substrate is equal to K of, so these are the things which lead to consumption of ES plus Kcat concentration of ES, so now when you think of a reaction essentially you will have all the enzymes which are unbound then when you add the substrate you will have the substrate binding to the enzyme, so now what this would lead to is that you will have some enzymes which are free and some which are bound to the substrate, so at any time the enzyme concentration would be given by E total minus enzyme substrate concentration, so then you can substitute here, so what you will get is K on to concentration of total enzyme which is E total minus enzyme substrate complex into substrate complex would be given by again K of enzyme substrate plus Kcat enzyme substrate, so from this you can get the concentration of enzyme substrate, so enzyme substrate will be given by, so I am not going to show you all the steps but it would be given by K on enzyme total concentration into substrate concentration divided by K of plus Kcat plus K on substrate concentration, alright. So now if you divide this whole thing by K on which is from the numerator if you bring that down what this would simplify to would be enzyme total into substrate divided by K of plus Kcat by K on which is given by Km plus substrate, so now if you substitute this value of Es into the rate equation given here you will get an expression similar to what is shown here, the only difference being what you would see is here you have Vmax when you substitute this Vmax would be given by enzyme total into Kcat and this makes sense correct because when will you have Vmax you will have Vmax when you have a very very very very high substrate concentration, so essentially what will happen is if you look at this expression for Es what you would see is this expression for Es because the substrate concentration will be very very high the denominator becomes equal to almost substrate concentration because substrate concentration would be much greater than Km, so what will happen is you will have both of these getting cancelled, so you will have concentration of Es equal to total concentration, so then if you put that in this equation the rate will become Kcat into E total concentration which is the maximum rate or Vmax, so here what I have shown you is you can very very systematically derive the Michaelis-Mentem equation using whatever you had studied earlier when we were studying kinetics, so it is not that complicated. Now a lot of times when people compare enzyme activity, so if I were to see that which is a better enzyme when I am trying to figure out the two catalysts or the two enzyme efficiency is compared using this term Kcat over Km, so people look at the value of Kcat over Km and say that okay this enzyme has a greater Kcat over Km that means it is a better catalyst. Now why do you think that is the case I want you to think about it, alright? So when you had remember I told you the case of Vmax is when you have a very high substrate concentration, so you have maximum catalysis then. In general in our body what you see is the amount of substrates that you have are in the nanomolar concentration, so it is very very very dilute. So now let us see what happens to this rate expression when you have a very very very dilute solution. So when you have a dilute solution essentially what you would have is, so in our body concentration of S is low. So what you will have is you will have Km becoming now greater than concentration of S. So now Km is greater than concentration of S, so now what you would see is in your expression you would have the rate determination, the rate dependence directly related to Kcat over Km. So you can see that by substituting in the expression, so now you have the Km value becoming much higher than substrates, so you can ignore the substrate concentration and then what you would see is in your reaction, so your reaction rate is given by Kcat times Es, so that would be given by Kcat into Et times substrate divided by Km. Now enzyme total is how much enzyme you are taking that would be similar for two enzymes, the substrate concentration is very low, so what would be important would be the ratio of Kcat over Km. So hopefully this gives you an idea of how binding is seen in biology for catalysis and it is very very important in several several biological functions. Now let us look at the other factor which contributes to catalysis which is proximity. So when you think of a chemical reaction, you think of an electrophile, you have a nucleophile and you have a solution. So you have all these molecules in solution and you have your electrophile and nucleophile which have to meet for your reaction to take place. So you need a collision to take place for the reaction to occur and because of the collision the transition state that you get has a greater order. So entropically formation of this transition state is not favored because essentially you have greater disorder when the molecules are alone and then they join together and now you have greater order in the transition state. So what is seen is that if you bind the two reactions, the two reactants, so you bind the two reactants with some sort of a linker, this loss of entropy is lesser, so the reaction is thermodynamically more favorable because what is seen is since now the reactants are bound, so you do not have as much disorder as you had with the free reactant. The second thing is having them close to each other improves the probability of collision and that is why the reaction rate enhances. So both these factors help in catalyzing reaction by keeping the two reactants close to each other. So now let us look at examples where you can see this. So here what you see is the reaction that is taking places formation of the anhydride. So what you have is, you have this coming in and then it forms the tetrahedral intermediate which then comes back and pushes out this phenoxide. So here what you see is both of these are tied together by a linker. So this reaction would be quicker than a reaction where you have two different acids coming together to form an anhydride. So the cyclic anhydride would be easier to form. Now let us compare three of these cases. In the second case what you see is the length of the linker is slightly shortened by so slightly means by one CH2 unit. So now you are forming a five membered ring instead of a six membered ring and in the third case you have put a double bond. So the number of carbons is the same but you have now put the double bond and the geometry of the double bond is cis. All right? So these are the three scenarios here and I want you to think of how the kinetics will change. So if you put, if you were to compare the reaction rates for these which do you think would be faster? So think about this proximity effect. You can press the pause button and spend more time thinking about the answer. So all right? I will give you the answer and let us see if you can understand why. So the relative rates are given by one for the first example where it was forming a six membered ring and then for the two five membered rings. So what you see is when you do not have a double bond the rate is slower as compared to the case where you have a double bond. So why is this so? So what you would see is when you have a shorter linker you are improving the proximity of this O minus to the electrophilic carbonyl center which is why you have an enhanced rate of 230. All right? Also the approach is very important when you think of these reactions. So probably you have a better approach in this case. So what you see in the third case is that now you are really restricting the proximity by putting an sp2 center. So having an sp2 center makes these two highly proximal to each other. So now there is a greater, greater enhancement of the reaction rate. So it is almost 10,300 times the first case. So in this case what you see is even orientation is improved for the reaction to take place. Now here the double bond geometry is cis. What do you think will happen if I take a trans double bond geometry? Obviously the rate will decrease because now I am keeping them far away from each other and unlike the previous case where the molecule had flexibility with the trans geometry it does not have that flexibility. So it will be more difficult for it to reach this electrophilic carbonyl center. Another example is shown on your screen. So here you are forming the lactone and not the anhydride. So here again you will have the O attacking the carbonyl carbon. So here you have an aromatic ring which again similar to the sp2 case that we had seen earlier. It helps in keeping the two reactants proximal to each other. Now what is seen is when you add this gem dimethyl group here to form now the same six-membered lactone you see an enhancement in the rate. And this is a very standard observation that is seen quite often. And this effect is called the Thorpe ingold effect. So what the Thorpe ingold effect does is that when you put this gem dimethyl group so gem is for geminole. So having this two groups so you have the CH3 CH3 which are large what it does is it reduces the bond angle. So reduces it relative to the case where you just have two hydrogens. So compared to the two hydrogens when you have the dimethyl group because it is larger and needs more space this bond angle becomes reduced. So having this bond angle reduced improves the proximity of the oxygen to the carbonyl group. So here you see how catalysis due to proximity can also be thought of like a binding effect where you have the two reactants bound to each other. So we will stop here and in the next class we will look at the other types of catalysis which I had shown you on the first slide of catalysis. So thank you and see you in the next class.