 So, welcome back in the last class we were looking at acid-base equilibria and if you consider this as a generic reaction for acid-base equilibria, you have 2 acids here HA and BH plus which is the conjugate acid of this base. So, I told you understanding the relative acidities of both of these is very important. So, to predict the relative acidities there are 2 guiding principles you can use. For neutral acids more stable conjugate base implies stronger acid, for cationic acids more stable acid would be the weaker acid. And then we had looked at all of these factors which affect acid strength, so we had looked at solvent, a solvent which solvates the ion more. So, in the case of a neutral acid if the ion is more solvated that means you have a better acid whereas for a charged acid if the acid is more solvated it would be a weaker acid. Then we had looked at electronegativity where if you have the acid Hx, so greater the electronegativity of X better would be the acidity. Then we had looked at examples as to where resonance plays an important role in increasing the acidity. If you have a conjugate base which can be stabilized by resonance then you see an improvement in acidity. We had also looked at bond strength so within a group, so within a group what you see is as the bond strength decreases the acidity increases. So, if you compare Hx where X are helides what you would see is Hi would be greater than Hf. And I told you that you can use this to compare only when you have atoms within a particular group. Then we had looked at electrostatic effect where if you have your conjugate base say you have, so if you have a positively charged species here let us say R+, that would stabilize this conjugate base. Then we had looked at hybridization where sp would be more acidic than sp2 would be more acidic than sp3. And the last thing we had seen is stabilization due to aromaticity which leads to improvement of acidity. So, if the conjugate base is stabilized by aromaticity you see an increase in acidity or lowering of pKa. So we had looked at all of these factors in the previous class. So now what we look at is the catalysis by acids and bases. We would be looking mainly at Bronsted acid base catalysis. So Bronsted acid base catalysis means you have a proton or a hydroxide. So proton generally in the form of H3O+, or the protonated solvent that you are working with that is what would be important and the hydroxide. And here what happens is the proton or hydroxide due to interaction with the reactant lowers the energy of the transition state for the reaction and thereby you have an accelerated reaction. So it acts as a catalyst and of course at the end of the reaction your proton or hydroxide is regenerated. What you observe is so these are called Bronsted acid catalyzed reactions. What you would see is there will also be examples where you will have the acid or base being consumed. So if you have your reactant plus say H+, when you get back your product you do not get back your H+, it is in some form incorporated in the product. So in these cases these reactions are called acid or base promoted reactions. They cannot be called catalyzed because your acid or base is not regenerated at the end. So for classification they are classified as specific catalysis and general catalysis. So what do each of these terms mean? Let us look at each of them in a little detail. So the first is specific catalysis. Now many people get confused with the terms specific and general. What you need to remember is in specific catalysis proton transfer occurs before the rate determining step. So proton transfer is not involved in the rate determining step. Proton transfer occurs before the rate determining step. So in other words the proton transfer is very, very fast. Now what exactly is a specific acid? Specific acid the term itself means protonated form of reaction solvent. So it does not matter what acid you are adding in the reaction, you are only talking about the protonated solvent as the factor which improves or which shows up in the reaction kinetics equation. So what do you mean by protonated form of reaction solvent? If you are looking at water as your reaction solvent the protonated form would be of course H3O+. That is the specific acid. If you are looking at doing a reaction in DMSO, this is the structure of DMSO. This protonated form is your specific acid. If you are talking about doing a reaction in acetonitrile this is your specific acid. So again I want to emphasize this specific acid is the protonated form of the reaction solvent. You can add any other acid in the reaction, you can add some small amount of acetic acid, it does not matter. What matters is only the protonated form of the reaction solvent. It is very, very specific to only the protonated form of the solvent. Again specific base is the conjugate base of the reaction solvent not any other external base that you are adding. So if you consider water the conjugate base would be OH-. If you are looking at DMSO the conjugate base would be this carbanion here. If you are looking at acetonitrile this is the base that you are talking about. So these are all the specific bases that you are talking about when you do the reaction in these particular solvents. Now let us try to get a deeper understanding of what it means that no other concentration of acid or base has an impact on the reaction rate and it is only the protonated form of the reaction solvent. So to keep it simple let us look at a reaction in water. I have just chosen water for simplicity. You can choose any other solvent and it would boil down to the same kinetic expression. So let us give this reaction so you have your reactant being interacting with the protonated form of water which is your specific acid. So you have H3O plus giving you the protonated reactant. The next step conversion of the reactant to the product is your rate determining step. So this is the slow step and the protonation takes place fast and all these steps are also fast. So this is a typical expression that you would see for specific acid catalysis. I repeat for specific acid catalysis proton transfer is fast. So it occurs before the rate determining step alright and the reaction rate only depends on the protonated solvent. In this case it would be H3O plus. Now let us derive the kinetic expression for this. So if you remember when we were studying kinetics we were looking at rate loss. So when we were looking at rate loss we had done derivation of rate loss for complex reactions. So let us try to derive the rate law for this. So in this case to simplify it we will just start from the rate determining step. So the rate for the reaction would be given by K which is the rate constant for this process into concentration of Rh plus right. So based on what we had studied for rate loss this is the rate expression. Now what we usually do is we move back and try to express Rh plus in terms of Rn H3O plus. So I will give you a hint I will give you the final expression. So this is what the rate dp by dt the rate expression is given by this where in the numerator you have the rate constant K which is this concentration of R concentration of H3O plus and in the denominator the term that you have is the acid dissociation constant of Rh plus alright. So with this information I want you to look at your screen carefully and try to derive this from this form that is given here. So I will ask you to press the pause button and do this derivation yourself before you start working it out with me. As I have told you again and again the best way to learn is to practice yourself. Once you practice you can check your answers by comparing with me. By practicing you will also come to know what the mistakes it is that you are making. So go ahead and try to work out this derivation. So you can check your answers now. So by definition the dissociation constant Ka of Rh plus deals with this equation. So you are looking at Rh plus plus water giving you R plus H3O plus right and this could be the K equilibrium for this process. So now the K equilibrium for this process is given by R concentration of H3O plus divided by Rh plus we had done this and I had told you if you take water on the left side what you get would be the Ka or dissociation constant would be given by Rh 3O plus divided by Rh plus. So in other words if I were to write Rh plus in terms of these it would be concentration of Rh plus will be equal to R into H3O plus divided by Ka of Rh plus right. So now if I plug in this value of Rh plus in this equation let us call this equation 1 you would get rate is equal to dp by dt is equal to Kr into H3O plus divided by Kar H plus. So now we have derived the rate law for this reaction which is catalyzed by an acid in which the proton transfer step is fast and occurs before the rate determining step. So what do you see here now when you look at the rate law the acid dissociation constant is a constant for the particular acid Rh plus concentration of R is your reactant. So what you would see is you do not have any term corresponding to the acid. So what you have is essentially you can say this as K observed it only depends on concentration of the protonated solvent which is H3O plus here. Now one might argue like what if I add a small amount of acetic acid to this if I add acetic acid you would say that instead of H3O plus maybe I will have the acetic acid there and it would be something different. So based on the definition of specific acid catalysis in this case you have specific acid catalysis because the rate only depends on concentration of H plus which in turn depends on the pH of the reaction medium. So if you maintain the pH of the reaction medium you essentially are maintaining the reaction rate. So now if you argue with me that I am adding now a little acetic acid. So now we have added an additional term which is HK right. So if you are adding acetic acid the question is now will this still follow specific acid catalysis. So now you need to remember that this step is again fast all of these are fast. So what we have is based on the original definition you have proton transfer occurring before the rate determining step. So by definition this should follow specific acid catalysis. But now let us mathematically derive to see is it true. So like before you would write the rate would be given by K into concentration of Rh plus because that is what is involved in the rate determining step right. A- is just a byproduct of the first reaction so it does not show up. So you have the rate determined by concentration of Rh plus. Now let us see if we can write Rh plus in terms of R and HA. Remember you also have water in the medium. So it is not just the HA disintegrating to H plus and A minus it is slightly more involved than that. So you need to introduce water here. So this is the final expression where you can talk about the rate in terms of the first equilibrium which is this. So this is K equilibrium the dissociation constant of the acid HA. So again let us try to derive this. I will ask you to derive this first yourself. So what I want you to do is I want you to press the pause button on the screen and try to derive this expression yourself. So let us see if you have the answer. So here again let us first try to write it in terms of the equilibrium constant. So the equilibrium constant for the reaction K equilibrium is given by Rh plus times A minus divided by R into HA correct. So in terms of this what you would have is you will have Rh plus equal to K equilibrium into R into HA divided by A minus. So that takes care of these two terms and this we have already. Now we would like to write HA by A minus. This ratio is determined by the acid dissociation of HA. So KAHA so the equation that you are looking at is HA plus water giving you A minus plus H3O plus right. This is what gives you the acid dissociation of the acid HA. So now the KAH value is given by A minus into H3O plus divided by HA remember the concentration of water is included in the term KAH. So now that you have this term what you get is you get the ratio of HA by A minus is equal to H3O plus divided by KAH. So if I plug in the value of Rh plus and this HA by A minus into this equation what I will get is the final equation that you have here. So here you can substitute for K equilibrium R and HA over A minus is given by H3O plus by KAHA. Now what do you observe here? What you observe here again is despite the fact that you have added an external base here again it reduces to the earlier form where you have K observed into R. So it essentially depends again on the pH of the reaction and what was your reaction solvent water and protonated form of water is H3O plus. So we have mathematically derived here that the specific acid catalysis is seen whenever you have a reaction where the proton transfer occurs quickly before the rate determining step. So now that you have an understanding of specific acid catalysis what I want you to do is I want you to do the same derivation for specific base catalysis and here again what we would be looking at is we would be looking at the rate would be given as this is K here again the base would deprotonate much faster as compared to the rate determining step. So the rate is given by K into concentration of R minus alright and now you can write R minus in terms of the base and just like we had done for the previous slide where we had added acetic acid in water here again we have added a base in water. So this is what your final expression will come like. So we are again writing it in terms of acid dissociation constant of Rh into concentration of Rh and H3O plus. So you can go ahead and derive this quickly now that we have seen the earlier example this derivation must be quite quick for you to do. So we will quickly go ahead and do this derivation. So here what you have is the rate of formation of the product is given by this expression. So when you look at the dissociation of Rh you are looking at Rh plus water giving you R minus plus H3O plus correct. So now Ka of Rh would be given by R minus concentration into H3O plus concentration in the denominator you have Rh water as I said is included in this. So now you can write R minus so R minus is equal to Ka Rh into Rh divided by H3O plus. So you plug it into the equation above so if you plug this value into this equation you will get the final equation for the rate. So here again you would see that the rate does not depend at all on the concentration of the base. So we have seen how by deriving the rate law again we are connecting to a concept we had learned earlier which is deriving rate laws for reactions. So when we derive these rate laws what we see is that when you have specific acid or specific base catalysis where proton transfer occurs quickly before the rate determining step you do not see a dependence on concentration of the acid or base. The kinetics depends only on specific acid or specific base which is the protonated solvent okay. So now if you look at so if you have understood this concept I will ask you to just make a very simple plot. So what will the plot of K observed versus concentration of HA where HA is the acid that you are adding say in water which is the solvent. So what will this plot look for when you are doing specific acid catalysis alright. I will repeat the question is there on your screen so you can press the pause button and write what the plot will look like in your notebooks. So let us see if you have this answer correct. So if you have let us say a K observed value what you would see is that the plot would be a straight line as I told you it does not depend on concentration of HA. So this would be a straight line. So let us say this is the plot at pH is equal to 2. Now if I do the same reaction at pH is equal to 6 or 7 okay let us say I am doing the same reaction at a pH equal to 7. What will happen to this plot I want you to think about it. So I want you to draw the plot when I do the same reaction but now I am varying the pH I have the pH at 7 what will the plot look like. And it will be a straight line because it does not vary on concentration of HA. But now will the K observed be higher or lower at pH 7. pH 7 means right if you are doing say the reaction in water the concentration of protonated solvent which is H3O plus in pH 7 would be greater or less than pH 2. pH 7 is less acidic right compared to pH 2 if you I mean has a lower concentration of H3O plus right compared to pH 2. pH 2 is more acidic. So what will happen is you will have a lower K observed value but again it will still be a straight line. So this would be at pH 7 alright. Now I want you to think what will the plot of K observed versus B look for a specific base catalysis. So we have already looked at specific acid catalysis here now we are plotting K observed versus B what will this look like. So again let us start with pH 2 and pH 7 now that we have this example on the left you can go ahead and draw it for two pHs pH 2 and pH 7. So what you would see is at pH 2 in this case you will have a lower rate. If you remember the expression in this case the H3O plus concentration was in your denominator and we are looking at a base catalysis reaction. So having an excess concentration of H plus will not help you. So the K observed would be lower at pH 2 again just by the definition specific base catalysis what that tells you is that it would be a straight line it will not vary with the concentration of B and what you would see is it will be higher at a higher pH alright. So you see a complete switch here where when you are doing specific base catalysis the K observed value is higher at a higher pH. The last thing that I want you to think of before we end this class is in the case of specific acid catalysis how do you think the log of K observed will vary with pH alright. So remember now on your X axis you have pH and not concentration of the acid HA. So I want you to think how it will change with the pH. For specific acid catalysis we saw in the previous slide that for lower pH you have a higher K observed and for higher pH you have a lower K observed. So what you would see is as the pH increases this value will come down alright and if you look at the specific base catalysis you would see the exact opposite trend. What you would see is that as the pH increases the K observed value will increase and what you would see is this will have a slope of minus 1 and this will have a slope of 1 alright. So we will stop here and in the next class we will look at the general acid and base catalysis and we look at reactions where you can actually see the specific acid catalysis and specific base catalysis. Thank you and see you in the next class.