



S1: good morning. today we're going to be talking about polyprotic acids, which are acids that have more than one easily ionizable hydrogen, and among the most important of these are the amino acids. which exist in proteins and peptides, and so we're going move to those as soon as we, cover some of the basics, with, some simpler, polyprotic acids. i have on the board acetic acid, as the, archetypal carboxylic acid, and then i have two... diprotic acids, succinic acid and adipic acid. these are the common names for these, dicarboxylic acids. their scientific names, are, one-two-three-four butane, dioic acid, and one-two-three-four-five-six, hexane, dioic acid. however, they're usually called succinic acid, and adipic acid so we'll, stick with their common names. now notice that we have two P-K-A values once we have, two carboxylic acid groups, and i've given the P-K-A values here, and we want to compare them then to the P-K-A value of the single carboxylic group we have in acetic acid. and you can see, that, of course these two protons are equivalent and so the first ionization can be either one of those, and so, either one for the first ionization, has the P-K-A value of four-point-two-one. but once i have removed one of the protons, then the other proton becomes significantly less acidic. and so we want to talk about, first why the first proton, is more acidic than the one in acetic acid, and then secondly, why does the, it becomes so much harder to remove the second proton once we've removed the first one. and i think you know the answers to both of those questions, why is succinic acid more acidic, in its first ionization than acetic acid.
<P :04> 
S2: resonance?
S1: hmh? [S2: (isn't it resonance?) ] i just didn't hear.
S2: is it resonance?
S1: well the resonance if i draw the, structure of the anion in each case so this is the anion for acetic acid the acetate ion... and lemme draw the corresponding, anion, for succinic acid. and it wouldn't have mattered which of these protons i removed, i have similar resonance in both of 'em. i can draw a resonance contributor in which i delocalize the charge to the other oxygen in either case. so resonance is not the, reason for the difference. it's a, good reason for why it's as acidic as it is, but not for the difference. <P :04> any other ideas? yep?
S3: you have (inductive and that's supposed to) take away the negative charge a little bit (xx) 
S1: and what is the thing that's the indu- doing the inductive effect?
S3: the oxygen atoms.
S1: the other carboxylic acid group here. okay? so i have the partially positive carbon these oxygens more electronegative than carbon and so i have, an electron withdrawing effect, through the bonds, of one carboxylic acid group stabilizing the carboxylate anion in the other case. so this, base, is weaker, than the base, in, acetic- acetate ion. and where the base is weaker, the corresponding conjugate acid is stronger. well why is it so hard to remove the second, proton? why does the P-K-A go up, with the second one? <P :08> yep?
S4: because there's two negative charges after that?
S1: because i already have a negative charge in the molecule so let's just write a little, equation for, the removal of the second proton and i'm going to put the, hydronium ion... that has resulted sort of up there to remind us, that we're trying to pull this positively charged species away from an ion that now is doubly, negatively charged. and so we have to exert more energy it is, harder to do, to remove that proton from something that is already negatively charged. so the first proton is easier to remove, than it was for acetate ion, but the second proton is harder to remove. now notice what happens with adipic acid. the first proton, is, more like the proton in acetic acid, in other words the P-K-A has gone up relative, to the P-K-A of succinic acid, for the first ionization, but the P-K-A for the second ionization has gone down. what's going on here?
<P :13> 
S5: the dipole is, weaker when more carbon atoms (xx)
S1: i have a l- greater distance between the two carboxylic acid groups. so my inductive effect, if i ionize this one first, the inductive effect of the second carboxylic acid group is weaker, than it is in the case of succinic acid. because inductive effect falls off with distance, so with the greater distance between the two carboxylic acid groups, i have less effect, on sta- the stabilization of the base resulting from the loss of the first proton. what is also true then about the loss of the second proton?
S6: (negative charges) (xx) 
S1: again with negative charges are further apart from each other, and therefore it's not as hard to remove a proton from this end of the molecule, when this end is negative, as it was with succinic acid where the charges were closer together. okay? any questions about what we've done so far? <P :05> okay i'd like to move on then to talk about amino acids, as po- polyprotic acids and the amino acids that we're gonna talk about are the alpha amino acids the amino acids that are found in peptides and proteins. and they're interesting to us of course because they are the components, of, the very important, substances such as enzymes muscle tissue cellular tissues, which are made up of, proteins... there's alanine, representative of a simple... amino acid... and i'm drawing it this way first, to emphasize the fact that it is an alpha amino acid here, it is the carboxylic acid group, and the carbon adjacent to it, (will) you remember we talked about, is the alpha carbon, and i've got an amino group, on that alpha carbon so this is an alpha amino acid. and it has, two sites of aci- possible acidity-basicity, um, this, is the, carboxylic acid group and the P-K-A there if we compare it to acetic acid, we would expect it to have a P-K-A of approximately four-point-eight. on the other hand, we have the basic site here which can be protonated to give us an acid and that is sort of like ammonia, ammoniamine, methylammoniamine, and that has a P-K-A, we would expect of about nine-point-four. so if we compare this these two structures, we will see that the structure to the right, is the one that we should be, considering... because, if we consider this as an equilibrium, between two acids, this is the stronger acid this is the weaker acid and the proton resides on, the weaker acid. and so we have an equilibrium, where the proton is mostly on the nitrogen, and only a little bit on the oxygen atom of the amino acid. so the, true structure of the amino acid, is in this form, where it is an internal salt, the proton has been transferred from the oxygen to the nitrogen, and we call that form the zwitterion. the internal, salt where we have the negative charge and the, positive charge within the same, species not two separate ions that can, migrate apart from each other. <P :04> well let's write, a series of equilibria for alanine and i'm going to start with the form that will exist at low P-H. where i have both of the basic sites in the molecule protonated so at low P-H, i will be protonated at the nitrogen and protonated at the oxygen. but as i add base to the system... i will deprotonate and if we look up the actual P-K-A values of the, two acidic protons here, it turns out that this one is about two-point-three-four, and this one, is nine-point-six-nine. (okay,) so when i add base to the system, which proton is going to leave first the one on oxygen or the one on nitrogen?
SS: 
S1: the one on oxygen is gonna leave first. so the first deprotonation of this polyprotic acid and this is why i'm calling it a polyprotic acid i've got two protons here that i can lose well the first ionization, gives me that, species, and if i were to add acid to that, it would go back to being, this species here. okay? so at low P-H i have this at intermediate P-H i have that form, and i can go on and add more base and that will deprotonate, the next, acidic proton so i get to the form that i have at high P-H. let's take a moment to consider, these two P-K-A values. why is, the carboxylic acid group in an amino acid, more acidic, than the one in acetic acid? <P :08> what's going on there?
S3: positive charge (xx) increase (xx) stabilize the (xx) 
S1: so i have the positive charge on this nitrogen, exerting an inductive effect, stabilizing, the negative charge, on the carboxylate anion. okay? so that this is, considerably more acidic, than what we expected by just comparing it to acetic acid. the amino group, it depends on whether you compare it to ammonia, ammonium ion which has a P-K-A of about nine-point-four, or methylammonium ion which is about ten, it lies somewhere in that range. and you can argue, two ways, you can say well it's a little more basic than you expect these electrons are a little more available because they're close to a negative charge, or you can argue well... you have the carboneal group there still, therefore it makes it a little more, unavailable, it lies right at the cusp of where you expect it to be and so it's a little hard to make an argument about that as well. well we're interested, in two numbers. one of them is, that for a technique that is very often used in analyzing for amino acids or peptides proteins and in separating them. and that's called electrophoresis, where you put a substance on a gel, and you put two electrodes on it, and then you watch to see where, the compound migrates, in this electric field. and there is, a point at which no migration takes place. and that P-H is called the isoelectric point. the point in whi- at which the charges within the molecule are balanced, so that it migrates neither to the positive electrode nor to the, negative electrode. <WRITING ON BOARD DURING NEXT 1:00 OF UTTERANCE> so one, value we're going to be interested in, is, P-H for what we call the isoelectric point. <P :15> the point at which <P :08> the amino acid <P :07> does not migrate <P :05> to either electrode <P :06> in an electric field. <P :09> for example, at low P-H the amino acid has a net positive charge, and therefore, it will migrate to the negative electrode... and at high P-H it has a net, negative charge, and therefore it will migrate to the positive electrode. and i can manipulate the P-H to do to make it do either one of those things. but at this intermediate value whatever that is, i have a situation, where the charges are balanced within the same molecule, and therefore, the compound is stationary, in an electric field. this also happens to be the point at which the amino acid is least soluble. so you can sometimes get it to precipitate from solution, at this isoelectric point. so that's another reason that we're interested, in knowing what that isoelectric point is so that we can isolate, (a) compound. the other thing that we are really interested in, especially when you get into biochemistry, medicine, is what is the form of the amino acid, at physiological P-H? what is the form in which it exists, the cells and the tissues, enzymes things like that? so that is the second P-H value that we're gonna be interested in. and for the, sake of argument it varies from tissue to tissue, but let's just take, a P-H of six-point-five as physiological P-H. okay...? well last time... we derived, the equation for the acidity constant, of a weak acid, the K-A, and from that then <WRITING ON BOARD DURING NEXT :31 OF UTTERANCE> by certain manipulations we arrived at the, Henderson-Hasselbach equation, which said that the P-H was equal to the P-K-A <P :05> plus the log, of the ratio, of the concentration of the, conjugate base of the acid... to the concentration of the conjugate acid itself. <P :04> so this is a relationship that relates the P-K-A of the particular acid we're talking about to the P-H of the solution. <WRITING ON BOARD DURING NEXT :32 OF UTTERANCE> and we went further and said that when, at the special point at which we had an equal concentration of the conjugate base and the conjugate acid, so when the concentration of the conjugate base equaled that of the conjugate acid, then the log... this term the log of the concentration of the conjugate base, over that of the conjugate acid, was equal, to zero... so that allows us to cancel out this term... and we have this one special point, at which, the P-H of the solution, is equal to the P-K-A of the acid that we're talking about. okay? so now let's see what, it looks like, if we try to plot <P :06> what kinds of species we have, at different P-Hs, for ana- for alanine, (xx) for the amino acid, (alanine.) so i'm going to take, P-H as this axis and on this axis i'm going to plot the relative abundance, of the different species there whether the, concentration of the, conjugate, base or that of the acid. so at some point i have hun- a hundred percent, of whichever species i'm talking about, down here i have zero percent, and in between, that point i have equal amounts of the two species... so i start with alanine at low P-H. <P :06> and almost all, of the, species in solution are this form, of alanine. the_ form in which it has it's doubly protonated. okay, but as i add base, the concentration of the, acid falls, H-A falls and for every molecule of H-A that is deprotonated i get one molecule of A-minus formed, so i have a loss of H-A, and a gain of A-minus. okay? and our equation tells us that at a P-H value equal to the P-K-A value, which is two-point-three-four for the first proton that i'm taking off, so somewhere around here... the concentration of A-minus will equal the concentration, of H-A. and then, the concentration of H-A will continue to decline. and growing in then starting from almost zero, i will begin to see the conjugate base, forming. and again at this point, half of the species in solution will be that of the conjugate base. then i continue to add more base, i'm now at this point, that's what we said at intermediate P-H, here, and i continue to add base let me remind you that i can go this way too with acid, i continue to add base, and so at the P-K-A value corresponding to this proton which is the one i'm now going to remove, at nine-point-six-nine, i again have a crossing point. so all in this region, it is that species and then it starts falling off, comes down to half of it's value, over here, and tails off and the third species builds in. and i'm deliberately drawing these longish tails, to remind you, that this doesn't happen all at once. that all three of these things are really in equilibrium with each other we're just talking about what predominates at any one, P-H. and so somewhere around here we begin to see, some of the third species coming in and at high P-H then, it becomes the predominant species. so here we have that species this region we have predominantly that species... down here we have this species... now with this we can answer two questions, one is, what is the isoelectric point? and that is the, point at which we have a... maximum for our... balanced ionic species the zwitterion. which is actually if you wanna be more accurate the arithmetic mean of the two P-K-A values that, border it. okay? so if we took the arithmetic mean of two-point-three-four and nine-point-six-nine, you would get, the, isoelectric point for alanine. and the other question was, what form does the amino acid exist in, at physiological P-H? and physiological P-H we defined as six-point-five. okay? so again the zwitter- zwitterionic form exists at physiological P-H... do you have any questions about what i've done so far? yeah?
S7: when you say physiological P-H of the zwitterions, (ones) what about the (alpha) classes of blue (lines) doesn't, the uh basic form also exist? 
S1: yes, but the, you know the blue is just beginning to grow in at those P-Hs, there's very little of it. so what is the predominant form there? that's what we're talking about. and if i did it accurately if i calculated at every point what the concentrations were, i may have exaggerated how early the blue really begins to appear. but the point i'm really trying to make is that it doesn't suddenly become blue. that there is some blue, way back... any other questions? <P :05> okay well is this true of all amino acids? and the answer is no... alanine is what we call a neutral. but other amino acids have additional amino groups, called other basic groups, or other acidic groups on the side chain. and therefore, they tend to be, not just diprotic as alanine is but triprotic and so on. so let's take a look at lysine <WRITING ON BOARD DURING NEXT :12 OF UTTERANCE> as an important basic amino acid. it's very often found in the active site of enzymes, and so we're interested in knowing, what form it exists in, at thou- those positions. <WRITING ON BOARD DURING NEXT :32 OF UTTERANCE> <P :04> so lysine has two amino groups. one of them is at the alpha position <P :07> the other one is at the <P :04> epsilon position. <P :09> so it is known as an epsilon amino, acid as well as an alpha amino acid. and here are my three acidic protons, so i have three different P-K-A values for them, this one is <WRITING ON BOARD DURING NEXT :08 OF UTTERANCE> two-point-one-eight... this one is eight-point-nine-five <P :04> and this one is ten-point-five-three. okay? so this one is almost like methylamine, this one is more acidic than en- methylamine and this one is definitely more acidic than acetic acid. so i've drawn the form that exists at low P-H... and we can do the same thing we did, with alanine, add base and explore what happens, at each stage. so as i add base to the system i'm gonna lose one of these protons, which one's gonna go first?
SS: <UNINTELLIGIBLE ANSWERS>
S1: the one on the carboxylic acid, the one with the lowest P-K-A so i have an acid-base equilibrium here, as i add base i go to this form if i were to add acid i would go back that way this is at some intermediate P-H. but notice this is not the zwitterion. this is not what exists at the isoelectric point, because i have two positive charges and one negative charge. so this is not balanced in its charge yet. <P :09> but i can add some more base to it, (set up) another equilibrium, what's the next proton that's going to be removed? which nitrogen? the alpha nitrogen or the, epsilon nitrogen? [SU-F: alpha ] the alpha nitrogen is going to be the one to go next. notice that this is the zwitterion. this is the form in which i have a balance of charges there. and also presumably then the form that exists at the isoelectric point. well i still have another proton i can remove relatively easily i'll set up another equilibrium and this is the form that exists at high P-H. where i have removed all of the easily ionizable protons. this is a triprotic acid, and i have three ionizations, for it. well let's do the same thing we did for alanine, draw a curve for this <P :23> so here are my axes again. of relative abundancies on on- abundances on one axis, P-H on the other. okay? and once again i start with the most acidic form, the one at low P-H where everything is protonated, and i have about a hundred percent of the specie(sic) up there, but then very soon, at a P-H of two-point-one-eight as i add base, i have lost half of that species... it continues to die off as i add more and more base and a new species, comes in. so at that P-H, half of the s- species in solution is this, monodeprotonated form... which increases as i make the system more basic but then, at a P-H corresponding to the P-K-A of the next proton that's going to be removed, in other words at eight-point-nine-five, so right around here this species, starts falling off and, a third species, comes in, crosses at that point... increases in concentration... then at the P-K-A corresponding to the ionization of this, third proton in other words at ten-point-five-three, you see this happens very fast. this falls off, and my final most basic species comes out. now the structures are too big to write here so what i'm going to do is i'm going to label them, this is structure A <P :04> this is structure B, this is structure, C... and this is structure D. okay? so we have A, predominating in this region... we have B, predominating in this region, here we are at C, and we finally get, to D. so let's ask our two questions. <P :05> what is the isoelectric point, for this species? well, the species that exists, at the isoelectric point is C, so our isoelectric point is, here. so it has a high i- isoelectric point. and what is the form that exists, at physiological P-H? now remember we defined that as six-point-five. so which form is it? B, is the species we have at physiological P-H. so at physiological P-H... notice that the, amino group, at the epsilon position, of the amino acid lysine is protonated. in other words lysine is positively charged at physiological P-H. and this is significant. for example in the structure of a protein, to have the positive charge on one lysine molecule interacting in an ionic fashion with the negative charge of a carboxylic acid group somewhere else in the molecule. so you have ionic, intermolecular type of interactions, nonbonding but still, helping to shape the structure of the protein. okay? so the overall shape of the protein, is a multiple, of these various interactions through the molecules you have a question?
S8: you said that the isoelectric point was also the arithmetic mean, is that only for diprotic?
S1: no it's the arithmetic mean <INDICATING TWO POINTS ON BOARD>
S8: oh just between those two.
S1: between those two. what is the H-A and which_ what is the A-minus, for that particular range okay? any other questions about what i've done. <P :06> okay next time we'll start talking about carbon acids, and enolate ions and then move on to their reactions.
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