00:00:00 Our first speaker is Professor Jeremy Knowles of Harvard.

00:00:16 Professor Knowles received his education at Oxford at the BA in 1959, the MA and DPhil

00:00:30 in 1961 at Christ Church College, was on the faculty at Oxford until 1974, and then moved

00:00:43 to Harvard.

00:00:44 His field is bio-organic chemistry.

00:00:49 He has studied the mechanisms of many enzyme reactions.

00:00:54 In particular, he has elegantly investigated the stereochemistry of a number of reactions,

00:01:04 including phosphoryl transfer reactions.

00:01:13 Professor Knowles.

00:01:28 It is a pleasure this morning to help to celebrate the extraordinary contributions of Bob Woodward

00:01:37 to chemistry.

00:01:40 I remember a couple of months before Bob died, there was a meeting that we attended in London

00:01:49 on penicillin 50 years, actually it was 51, but the English were a little slow, 51 years

00:01:56 after Fleming.

00:01:59 And Bob, the schedule had appropriately put him last on the day's program.

00:02:10 And after his lecture, a mere hour, he was coming down the aisle receiving, as it were,

00:02:22 the accolades from a number of people who wished to give him them about his talk.

00:02:30 We were going out to dinner, and so I was quietly waiting right at the end.

00:02:35 And as he walked down between these multitudes, it was a sort of gauntlet of eulogy, because

00:02:45 each person had to do a little better than the last.

00:02:48 So there would be Professor Woodward, that was a most exciting lecture.

00:02:55 Professor Woodward, that was formidable.

00:02:58 Bob, they would say, that was inspired, divine.

00:03:06 And Bob, with his Mona Lisa-like, enigmatic smile, received all these compliments almost

00:03:17 silently.

00:03:21 And some of us, however, felt that perhaps it was a little bit much.

00:03:27 And when he reached me, I said to him, Bob, that was quite a good lecture.

00:03:35 And there was one of those splendid Woodwardian pauses, and he puffed on his cigarette.

00:03:42 And he said, thank you, Jeremy, you might even be right.

00:03:50 We must be grateful, I think, all of us, for that unremitting quest that Bob had for

00:03:59 yet higher standards.

00:04:04 I want to talk this morning about the transfer, the nucleophilic attack on phosphorus of phosphate

00:04:11 monoesters.

00:04:14 Now there has been an immense amount of work, largely of the physical organic kind, that

00:04:23 have delineated three extremes of mechanism, as shown on this first slide.

00:04:34 There are three extremes of mechanism in which a phosphor group, here depicted without double

00:04:42 bonds or charges for simplicity, is transferred from a donor, D, to an acceptor, A, in a simplified

00:04:50 reaction, as shown at the top of the slide.

00:04:53 And there are three formal extremes, pathways, that this process might occur.

00:05:00 By analogy with a carbon chemistry, an SN2 reaction, the phosphorus chemists choose to

00:05:07 call this process in-line associative, both for obvious reasons.

00:05:12 The reaction may proceed via a pentacoordinate transition state, as shown there.

00:05:21 In analogy with an SN1 reaction, there are cases firmly believed to proceed via monomeric

00:05:28 metaphosphate, via a dissociative path, in which the bond cleavage, this bond precedes

00:05:35 the bond making, and there's an explicit intermediate of monomeric metaphosphate, recently explicitly

00:05:42 demonstrated by Westheimer and his group.

00:05:46 Thirdly, and with no analogy in carbon chemistry, is the so-called adjacent associative process,

00:05:55 in which the acceptor, the attacking nucleophile, attacks the phosphorus adjacent to what will

00:06:02 become the leaving group, D, and there is then the necessity, according to the rules

00:06:08 of Westheimer, for at least one pseudo-rotation to allow the leaving group to leave from an

00:06:16 apical position, resulting in the formation of the product.

00:06:23 Now these three pathways could be distinguished, as we shall see, from one another, if one

00:06:31 could distinguish between these formally, amongst these formally equivalent peripheral

00:06:36 oxygens, these three, thereby making the phosphorus chiral, and this can be done using the three

00:06:44 stable isotopes of oxygen, O16, O17, and O18.

00:06:50 Of course, in the late 60s, Usher and Eckstein had made this distinction using O16 sulfur

00:06:58 and O18, but for a number of reasons, not all of them relevant today, we decided to

00:07:04 attack this using the three stable isotopes.

00:07:09 These are depicted here and in all subsequent slides by oxygens of increasing density.

00:07:18 If one considers again the three mechanisms, using now a chiral phosphoryl group, you can

00:07:26 see you expect three formally different stereochemical consequences.

00:07:32 Obviously, this first one, the SN2-like process, will lead to inversion of the configuration

00:07:38 of phosphorus.

00:07:40 The dissociative pathway, if this monomeric metaphosphate were to be free and symmetrically

00:07:47 solvated, would result in the racemization of the configuration of phosphorus, since

00:07:53 the acceptor A could attack from either face of the planar metaphosphate moiety.

00:08:01 And finally, the adjacent associative, as we shall see in a moment, will result inexorably

00:08:08 in retention of the configuration.

00:08:10 And the fact that these three are different at least allows a clean distinction amongst

00:08:17 the three pathways.

00:08:19 What I want to do today, however, is to assure you that the stereochemistry of phosphorus

00:08:25 really is as diagnostic as the predictions on this slide imply.

00:08:32 If we're going to solve the stereochemical course of a reaction, of a phosphoryl transfer

00:08:37 process, whether it be enzymic or chemical, we have to be able to do two things.

00:08:44 First, we must be able, by a general route, to synthesize a molecule chiral at phosphorus

00:08:53 and of known absolute configuration at phosphorus.

00:08:57 Then occurs the reaction, here as on this slide, an enzymic reaction, but this of course

00:09:02 could be any chemical transformation.

00:09:06 And then the second problem is to analyze the configuration at phosphorus in the product.

00:09:12 So we need a general method of synthesis and a general method of determining the absolute

00:09:18 configuration of stereoanalysis, the absolute configuration at phosphorus.

00:09:26 We have devised a general method of synthesis, which is summarized on this slide, for the

00:09:31 molecule, which it becomes the key molecule in our analysis, as we shall see, the phosphate

00:09:38 ester of propane diol, in fact, of s-propane diol.

00:09:46 This synthesis involves five steps and is a general one and is amenable to the synthesis

00:09:53 of more exotic molecules, such as ATP, chiral at the gamma position, and other interesting

00:09:59 intermediate metabolites that I shall not discuss today.

00:10:03 Between the synthesis, one uses primary sources of isotopes, water, of course, H2O18, and

00:10:09 this will become important later, one can buy in almost complete isotope excess, nearly

00:10:18 100% O18, so essentially whenever you see an O18, it's O18.

00:10:24 However, the best O17 you can buy, and this is still true, even though this slide is now

00:10:30 a year and a half old, is about half of it only O17, and the other, there's a quarter

00:10:38 of it O16 and a quarter of it O18.

00:10:41 So in this synthetic scheme, in fact, when I draw such a phosphoryl group, chiral, only

00:10:48 half of them are chiral, because, of course, only half of them, in fact, have O17 in the

00:10:54 indicated position.

00:10:57 Now the power of asymmetric organic synthesis might lead one to suppose that there was no

00:11:04 problem in the assignment of stereochemistry.

00:11:07 One could be quite relaxed about the fact that indeed one had made this molecule, but

00:11:12 of course we must face the analytical problem.

00:11:16 This molecule is indeed chemically what is put on the slide there, and it indeed contains

00:11:23 the appropriate, as you can see from these numbers, contains the appropriate isotopic,

00:11:28 has the appropriate isotopic composition.

00:11:31 The question is, what is the configuration at phosphorus, and can we determine it independently?

00:11:38 I want to spend the next few minutes running rather rapidly through the simpler of the

00:11:44 two available analytical methods.

00:11:49 The problem amounts simply to determining whether the isotopes are arranged in order

00:11:56 of heaviness in an R or an S sense.

00:12:01 But of course, because of the free torsion about that PO bond, as shown here, the oxygens,

00:12:13 the peripheral oxygens, are torsionally equivalent.

00:12:16 And if we are to determine the R or S sense of the isotopes, we're trying to hit a moving

00:12:22 target, and that is obviously the first problem that has to be overcome.

00:12:28 We must create a stationary target to simplify the problem.

00:12:34 And this we do by exploiting that hydroxyl group and closing the ring to a five-membered

00:12:40 ring as shown here.

00:12:43 If we take a reagent, nondescript for the moment, and activate one of the oxygens, and

00:12:50 via a reaction of known stereochemistry, here depicted in line stereochemistry, close the

00:12:58 five-membered ring, we obtain the cyclic diester, here with the loss of O17, of this kind.

00:13:06 Now you see the targets, although we've lost one of the targets, the O17, they are now

00:13:13 immobilized with respect to the rest of the molecule.

00:13:18 However, of course, any reagent we use will not discriminate between O16, O17, and O18.

00:13:27 Isotope effects notwithstanding, those will be very small, and certainly no analytical

00:13:32 method we should wish to use dare rely on such very small effects.

00:13:39 And so, what we will in fact obtain by such a ring closure process are three molecules

00:13:50 in one-to-one-to-one ratio, and an inexorable, inseparable mixture of these three species.

00:14:00 So starting with a phosphoryl group, for the moment R at phosphorus, using a reagent, diphenylphosphoryl

00:14:08 imidazole, that very smoothly, and with adequate gentleness, ring closes with the removal of

00:14:15 either O18, O17, or O16, we change what was one formidable problem into three, a mixture,

00:14:25 it must be said, of three somewhat simpler problems.

00:14:30 We have now fixed the isotopes in space, but these, the isotopes, let's look at this top

00:14:38 one, the O17 and the O18 are chemically almost identical, but they are topologically different,

00:14:45 and it's the topological difference that of course interests us.

00:14:50 That is, that O17 is on the same side, syn, if you like, of the five-membered ring to

00:14:57 the methyl group, since this carbon center is throughout, or was S. The O17 is on the

00:15:05 same side, and the O16 there is on the opposite side.

00:15:09 What we must do is to provide, is to accentuate the same side and opposite side difference,

00:15:17 and label chemically the, the relevant oxygens, and what we do is to allow these, this mixture

00:15:23 to react with diazomethane, methylating, of course, all six positions, there, there, there

00:15:31 and there, there and there.

00:15:36 The problem you see now, we obtain six molecules, you may not think at this point the problem

00:15:41 is getting any simpler, but because we've started with one very difficult problem, we've

00:15:48 gone to three simpler ones, and now six even simpler, but the mixture is still with us.

00:15:56 Now in our original approach to the problem, which was mass spectrometric, which I shall

00:16:02 not discuss, we separated the syn set from the anti-set, this syn set of cyclic triesters

00:16:10 from the anti-cyclic triesters, in a somewhat heroic exercise, since these molecules are

00:16:17 relatively unstable, that is to say they're extremely unstable.

00:16:24 However, in the method I want to discuss this morning, we don't need to do that, because

00:16:30 I'm going to discuss the method using P31 NMR, and there are three features of P31 NMR

00:16:37 that allow the solution to this problem.

00:16:41 These three features discovered in the laboratories of Mildred Cohn in Philadelphia, Gordon Low

00:16:47 in Oxford, and Ming-Do Tsai then at Purdue.

00:16:53 And these workers had shown that, in the first instance, that any molecule containing a bond

00:17:03 between oxygen-17 and phosphorus, because largely, actually, of the electrical quadrupole

00:17:10 moment of oxygen-17 interacting, it so rapidly relaxes the dipolar P31 nucleus that, in effect,

00:17:21 you do not see the P31, the line is so broad for any species containing a P31 O-17 bond

00:17:30 that you don't see those phosphoruses.

00:17:33 So what we can see immediately is that the phosphoruses disappear from the NMR spectrum

00:17:39 any time you see an O-17 phosphorus bond, and I've crossed out the ones where there

00:17:46 is such a bond.

00:17:48 And so, in fact, the six molecules in the NMR spectrum become two.

00:17:54 So the problem begins to simplify.

00:17:57 So we should only see these two.

00:17:59 This is a syn compound, and that's an anti-compound.

00:18:03 And in the unlabeled species, those differ by about 0.1 ppm in the P31 spectrum.

00:18:11 The second feature of P31 NMR that we exploit is the fact that, when you replace O-16 by

00:18:21 O-18, there is a small but detectable upfield shift, so that these molecules, the P31 spectra

00:18:30 for both of these molecules are somewhat upfield from where they would have been if these had

00:18:37 been O-16s, unlabeled species.

00:18:41 And the third feature of P31 NMR that we exploit is the fact that the upfield shift due to

00:18:47 O-18 is proportional, almost, to the bond order, so that this one with P double bond

00:18:55 O-18 is shifted upfield about twice as much as that one.

00:19:01 This, indeed, is shifted about 0.02 ppm up, and that about 0.04 ppm upfield.

00:19:10 And so, in summary, the spectra, when you put all this lot in the P31 NMR tube, in the

00:19:18 NMR tube, and put it in an extremely high resolution P31 NMR spectrometer, you

00:19:25 should see the following.

00:19:28 If S and A, if those little arrows are the positions of the unlabeled species, when we

00:19:35 have P double bond O, a large 0.04 ppm upfield shift, and P single bond O-18, a small one,

00:19:43 0.02 ppm.

00:19:46 That is what one would see if the original phosphoryl group had had the R configuration.

00:19:55 What about the enantiomer?

00:19:56 Can the distinction really be made?

00:20:00 Well, I duplicate what you've just seen here on the left.

00:20:04 There's the large upfield shift for the P double bond O, and on the right

00:20:14 is the analogous pair for the S phospho ester.

00:20:20 And you can see, I hope, that here, in contrast, of course, the O-18s are in the opposite place,

00:20:26 in the other place, and there is a small upfield shift for syn and a large upfield shift for

00:20:31 anti.

00:20:32 And if you half close your eyes and look at these two, you can see, of course, immediately

00:20:36 that they would be both qualitatively and quantitatively distinguishable.

00:20:42 The problem, therefore, should be solved, except you will remember I told you that the

00:20:49 best O-17 you can buy is only half O-17.

00:20:53 So the problem, unfortunately, is a little more complicated, and that is illustrated

00:20:58 on the next slide.

00:21:00 If we start here on the left-hand top corner with an R phospho, and we obtain these three

00:21:07 cyclic diester species that we have seen before, whenever I write an O-17, half of the time

00:21:15 it's O-17, and I don't see the P-31 resonance.

00:21:19 However, a quarter of the time, remember, it was O-16.

00:21:24 So a quarter of the time, I will see unlabeled species, 16-16.

00:21:30 A quarter of the time, that O-17 is, in fact, O-18.

00:21:35 So I see some of this chap.

00:21:39 This compound, only containing 16 and 18, is always 16 and always 18, and therefore

00:21:45 all of it carries through to the right-hand column.

00:21:50 And finally, here again, when we have the doubly-labeled species, cyclic diester, whenever

00:21:58 that O-17 is O-17, we don't see the resonance.

00:22:03 But a quarter of the time, it's O-16, which transforms into that, and a quarter of the

00:22:09 time, it's O-18, so we see some doubly-labeled molecules.

00:22:14 The end result of all this is that instead, when we hoped, out of these three here on

00:22:19 the left, only to see the middle one, that is to say, one line, in fact, we shall see

00:22:26 always four lines.

00:22:29 The most intense of them will be the one we wanted, but the four lines will be in the

00:22:34 approximate ratio of 1 to 2 to 4 to 1.

00:22:40 Since we have one line from syn and one line from anti, we now predict an eight-line spectrum,

00:22:45 and the demands upon the P31 NMR spectrometer are rising rather sharply.

00:22:52 So this is what we would expect.

00:22:54 Here are the compounds again on the bottom, and you can see that if on the left, here

00:22:59 is the syn, the set of four signals from syn.

00:23:02 This is a predicted spectra, simply.

00:23:05 Here's the unlabeled.

00:23:06 There's an upfield shift for a P single bond O-18.

00:23:11 There are two shifts, if you like, for a P double bond O-18, and about a 0.06 BPM upfield

00:23:17 shift for three bonds to O-18, and analogously for the anti.

00:23:22 And you can see the 1 to 2 to 4 to 1 ratio illustrated here according

00:23:28 to the previous slide.

00:23:31 So we did this.

00:23:34 But before I allow you to see a real spectrum, for those of you who are not so, who don't

00:23:43 live with P31 spectra every day of their lives, I should just explain the resolution of the

00:23:51 spectrum you're about to see.

00:23:54 And perhaps I can best exemplify this by suggesting to you that you put into an NMR tube ATP.

00:24:03 And if where I stand now would be the position of the gamma phosphorus of ATP, on the slide

00:24:12 you're about to see, where would, at that resolution, where would the beta be?

00:24:18 The beta phosphorus would be 65 yards away.

00:24:23 I want you to have a feeling for the demands on the NMR spectrometer that we have been

00:24:31 making.

00:24:32 And here, then, is the predicted spectrum.

00:24:37 Not exactly 1 to 2 to 4 to 1, you remember, because, in fact, the isotopes aren't quite

00:24:41 25, 50, 25.

00:24:44 But they're perfectly easy to predict, of course.

00:24:46 And here, our first observed spectrum for synthetic R material.

00:24:52 And you can see, I hope, that not only qualitatively is it clear that these two agree with one

00:24:59 another, but that quantitative estimation of an antimeric excess at phosphorus is certainly

00:25:06 feasible.

00:25:09 To be quite sure that there were no problems here, we independently synthesized the S molecule.

00:25:18 Of course, merely by adding the isotopes in a different order in the synthetic scheme.

00:25:23 And on this slide, you can see both R and S independently synthesized.

00:25:30 And I trust you will be happy with the fact that, both qualitatively and quantitatively,

00:25:37 the configurations are as expected and quantitatively essentially as expected.

00:25:46 That is, these were our first spectra, courtesy of the Brooker Company, at 250 megahertz protons.

00:25:55 And from this analysis, it appeared that the enantiomeric excess at the two centers of

00:26:03 the two species we made was 85% or better.

00:26:08 Subsequent analysis using higher power machines shows that, in fact, better is what the truth

00:26:15 is.

00:26:18 We have, therefore, a method of determining the absolute configuration at phosphorus in

00:26:24 propane diol phosphate, a molecule of limited interest chemically and of absolutely no interest

00:26:30 biochemically.

00:26:34 It would, therefore, be useless to determine the stereochemical course of any enzyme-catalyzed

00:26:41 reaction.

00:26:43 And therefore, we need to generalize the analytical method so that it can be used to solve the

00:26:49 stereochemical consequence of any process, of any phosphate monoester, whether the product

00:26:57 be an enzymic product or a chemical product, because we couldn't possibly go to the trouble

00:27:06 and sweat of designing an ad hoc analysis for each problem.

00:27:12 And nature provides this solution here.

00:27:15 What we do, and you've seen almost all the slide before, our synthesis is general.

00:27:21 So we can make a whole range, as I have said, of chiral phosphate monoesters.

00:27:30 And all we need is a reagent to take the phosphor group from its new location, whatever

00:27:36 A is, transfer it to propane diol, and once we've got propane diol, we can solve the stereochemistry

00:27:45 in the manner I have just described.

00:27:48 Nature provides this transfer agent, and it's called alkaline phosphatase, an enzyme splendidly

00:27:55 indiscriminate that hydrolyzes almost every kind of phosphate monoester imaginable, and

00:28:05 can be fooled into not hydrolysis, but into transphosphorylation to propane diol.

00:28:12 So if one mixes in the alcohol with the water, a good enough percent of the time, the enzyme

00:28:20 is fooled, and instead of transferring the phosphor group to water, transfers it to

00:28:24 the alcohol to give phosphopropane diol, whence the stereochemical analysis can be achieved.

00:28:33 Let me now return to the original problem.

00:28:37 Now we have a general synthetic method and a general analytical method.

00:28:42 And let me face first the inline associative path.

00:28:47 I want now to go through an example of each, or examples of each of these, and look to

00:28:54 see whether the predicted stereochemical consequence is correct or not.

00:29:01 It transpires, and that, but that would be another hour, which I shan't give you.

00:29:09 It transpires that enzyme-catalyzed phosphoryl transfer reactions appear to be the best examples

00:29:16 of the inline associative process.

00:29:21 And let me just give you one class of such enzymes, the phosphokinases.

00:29:29 The phosphokinases are enzymes simply that take ATP and phosphorylate a whole range of

00:29:34 acceptor groups, alcohols, acids, phosphates, guanidino groups.

00:29:40 And on this slide, about 13 of them, of this particular class, have been studied by, have

00:29:49 had a stereochemical analysis done.

00:29:53 All but two, on this slide, proceed with inversion of the configuration.

00:30:00 And all of those, the balance of evidence was strongly in favor, from all other sources

00:30:07 that is, strongly in favor of the phosphoryl transfer occurring between bound substrates,

00:30:14 so that both the substrates would bind to the enzyme, the phosphoryl transfer would

00:30:18 occur within this ternary complex, and then the products would fall off.

00:30:23 That is, in an associative process that appears, for 11 out of the 13 cases here, to be inline.

00:30:32 Two of them, nucleoside diphosphate kinase and nucleoside phosphotransferase, go with

00:30:38 overall retention, but happily, for these enzymes, at least for one of them, the other

00:30:45 has not been so well studied, the evidence is superb that the mechanistic pathway proceeds

00:30:53 via a phosphoenzyme, that is to say, two displacements at phosphors.

00:30:58 Two inversions to the enzyme, in fact, from the enzyme, overall retention.

00:31:05 And the summary from this slide, and from, in fact, from the something like 45 examples

00:31:13 now, enzymic examples, nearly all of which have been solved in the last two years, is

00:31:21 that inversion, in an associative inline reaction, is the preferred course that nature

00:31:28 has taken to catalyze these processes.

00:31:31 And one may indulge in a little bit of teleology, and suggest that the enzyme might indeed prefer

00:31:41 to the inline associative path, in the sense that here is a transition state that one could

00:31:47 get one's teeth into, as it were, in terms of preferential stabilization.

00:31:52 And that one could imagine that this particular transition state, with appropriate general

00:31:57 acids and general bases on the two ends, but for the phosphoryl transfer, that one could

00:32:02 provide cationic groups, either amino acids from the protein, or of course, the ubiquitous

00:32:09 metal cations here, and or hydrogen bond donors, to the peripheral oxygens of the trigonal

00:32:17 bipyramidal transition state.

00:32:20 It is extremely gratifying that the two enzymes that have now been determined to heroically

00:32:28 high resolution, 1.5 angstroms in each case, Staphnuclease by Cotton's group, and Ribonuclease

00:32:35 A by Petsko's group, for each of these two enzymes, the positioning of groups, as seen

00:32:43 at very high resolution, is in entire accord with this kind of mechanistic postulate.

00:32:51 So it seems that enzymes' nature prefers the inline associative path.

00:32:59 Let me come now to more chemical topics, and look secondly at the dissociative path.

00:33:08 Where should we search?

00:33:10 Where should we look for a reaction that we could believe proceeds via monomeric metaphosphate?

00:33:18 Well there are a number, and the two major areas are the monoester monoanions of reasonably

00:33:24 good leaving groups, such as phenol, say phenol phosphate, or the dianions, the phosphate

00:33:30 dianions, of compounds which have extremely good leaving groups, like dinitrophenol phosphate.

00:33:37 We've only, I'm afraid as yet, studied the first of these groups, but on the next slide

00:33:42 you can see the evidence from the physical organic literature that phenol phosphate,

00:33:49 here on the left, proceeds via monomeric metaphosphate path.

00:33:54 There is, it is proposed, a pre-equilibrium shift of protonation, so that one has a dianionic

00:34:01 phosphor group, the potential leaving group is protonated.

00:34:06 There is the rate-limiting dissociation to the leaving group, monomeric metaphosphate,

00:34:11 that is then rapidly captured by an available nucleophile.

00:34:15 And the kind of evidence adduced for, in support of this pathway, is listed on the slide above

00:34:21 here, the activation entropy, for those of us who aren't too suspicious of those things

00:34:26 in aqueous solvents, is at nearly zero.

00:34:31 The beta, Brønsted beta, is very small, it's only minus 0.3, which is entirely consonant

00:34:37 with the pre-protonation idea.

00:34:42 In analogous cases, the Brønsted beta for the nucleophile is also very small, showing

00:34:47 hardly any participation, nucleophilic participation, at the transition state.

00:34:53 I don't think I wish to discuss the solvent isotope effect, that's probably consistent

00:34:57 with almost anything.

00:35:00 But the product ratios, perhaps this is the most persuasive, when one does such reactions

00:35:05 in mixed solvents, the products are almost, actually rarely precisely, but almost in proportion

00:35:13 to the molar ratios of acceptor, implying that one does indeed generate an extremely

00:35:18 reactive electrophilic species that reacts very rapidly.

00:35:23 It seemed then that the evidence for metaphosphate here is very good, and when we looked at,

00:35:29 synthesizing phenolphosphate, R at phosphorous, the propane diolysis, and more recently the

00:35:36 methanolysis, so as to be more consistent with the, closer to the established literature,

00:35:42 we found, after separation of the products, that there was almost complete inversion of

00:35:50 configuration at phosphorous.

00:35:55 Fortunately, the data on this slide, and on the previous slide, are not really in conflict,

00:36:03 and they're in fact entirely consistent with the ideas recently propounded by Jencks, that

00:36:09 for a reaction of this kind, one would actually expect to see these results.

00:36:16 That is, Jencks proposed that if one looks at the dissociative path, the classical dissociative

00:36:25 path is shown at the top here, where here is the bond cleavage to form the leaving group

00:36:31 and monomeric metaphosphate.

00:36:33 The acceptor A then has to diffuse in to make this complex, and then there is a collapse

00:36:38 of the metaphosphate with A to give products.

00:36:43 Alternatively, Jencks argued, if that is an extremely, metaphosphate is an extremely

00:36:50 reactive species, then the reaction may in fact proceed via pre-association, not formal

00:36:58 dissociation at all, where A first diffuses in to the solvent cage with the donor, and

00:37:05 that there is then dissociation to give this complex that is then picked up, the metaphosphate

00:37:11 then reacting rapidly with A.

00:37:13 And he pointed out, as shown on the bottom here, that if this K minus 1 step, that is,

00:37:21 if metaphosphate reacts very fast, very rapidly with the leaving group that has just left,

00:37:29 if indeed it reacts by this dotted line here more rapidly with the leaving group than the

00:37:37 diffusion away of the acceptor A represented by the solid line, then of course that dotted

00:37:45 transition state will be lower than that full transition state, and looking by microscopic

00:37:50 reversibility, the reaction of course in the forward direction will prefer the dashed pathway,

00:37:57 namely a pre-association route.

00:38:00 So and this argument, nicely and tidily predicted, I should say, by Jencks before the data were

00:38:07 in, on the basis of the high reactivity of metaphosphate, nicely accommodates all the

00:38:12 data.

00:38:14 Indeed the mechanism is dissociative.

00:38:16 Bond cleavage comes ahead, leads, over bond making, but there is nevertheless clean, or

00:38:25 relatively clean, inversion of the configuration at phosphorus.

00:38:32 Thirdly, and lastly, let us look at the adjacent associative path, and this is of particular

00:38:39 interest because of course it has no analogy in carbon chemistry, as I mentioned.

00:38:46 We must search, we must look, in order to observe such a reaction, of course, for a

00:38:54 situation where the acceptor nucleophile, where the attacking nucleophile, has, is constrained

00:39:01 to attack adjacent to the leaving group.

00:39:06 And one such case is of course to look at the rearrangement of two phosphopropane diol

00:39:14 to one phosphopropane diol, where the attacking nucleophile is indeed constrained by its proximity

00:39:21 here to attacking phosphorus adjacent, in an adjacent sense, to what will be the leaving

00:39:27 group.

00:39:29 This is an equilibrium favoring the one phosphor compound not by that much, only by 1.8 fold,

00:39:38 but if it were to behave and to rearrange nicely according to the rules set out in the

00:39:44 1960s by Frank Westheimer, then we should see it behaving as shown here, whereby that

00:39:52 the entering group would attack apically to provide, to produce this intermediate.

00:39:59 That would, per force, undergo one pseudo-rotation to produce that intermediate.

00:40:05 Now the leaving group may leave apically to give the product.

00:40:11 As I had said earlier, what is the stereochemical consequence of that little circuit?

00:40:17 And you will see here that now putting the isotopes in, and if you may not wish to follow

00:40:22 these all round with your eye, but you can just see that the configuration here is the

00:40:28 same as the configuration there.

00:40:32 The reaction by the pseudo-rotatory path does indeed proceed with retention of configuration

00:40:38 at phosphorous, and that would be lovely.

00:40:43 We could study this very simply, but for the fact, unfortunately, that there is an alternative

00:40:50 reaction route.

00:40:52 The reaction also proceeds under the conditions where one sees a nice rearrangement rate.

00:40:58 That's at about 85 degrees in our hands at, in 0.1 normal per chlorate.

00:41:06 The reaction also proceeds via this route, where water is lost, there's a ring closure

00:41:10 to the cyclic diastere that then ring opens again in water.

00:41:18 And we must ask ourselves therefore, what would be the stereochemical consequence of

00:41:24 this top route?

00:41:25 We know that this bottom route is retention.

00:41:27 What about the top route?

00:41:30 Well that's a little more tricky.

00:41:33 What we may do, you see, is to lose any one of these three isotopes in the ring closure

00:41:38 reaction, either O18, O17, or O16, to make these three cyclic diastere species.

00:41:48 And then they will, all of them, because we're running the reaction in H2O16, pick up H2O16.

00:41:56 This one will, of course, become two 16s and a 17, will become a prochiral molecule, and

00:42:02 we have no further interest in it.

00:42:05 This one, having lost O17, similarly becomes a prochiral product molecule, and we don't

00:42:12 have any interest in that either.

00:42:15 And assuming for the moment, and there is good precedent, literature precedent, for

00:42:20 this, that both the closure and the opening reaction are inline processes, we should predict

00:42:28 inversion of the configuration in the one third of the cases where one loses O16 and

00:42:36 gains O16.

00:42:38 The situation is therefore as follows.

00:42:40 By the top route, one third of the molecules proceeding by the top route will proceed,

00:42:46 will go with inversion of configuration, and all the molecules proceeding by the bottom

00:42:52 route, we predict, will go with retention.

00:42:58 What we need to do then, is to determine the flux through the two pathways.

00:43:03 We can determine all the necessary rate constants in this scheme, that are required, unfortunately,

00:43:13 because of the reason I told you a moment ago, the equilibrium constant is so close

00:43:17 to one.

00:43:18 Since one phospho is only favoured 1.8 fold over two phospho, one can't just run the reaction

00:43:26 in a very little way and isolate a tiny amount of product.

00:43:30 To isolate, to have enough product to isolate, one has to be worried about the number of

00:43:35 molecules that have been re-equilibrating and going to and from cyclic and so on.

00:43:41 So one needs all these rate constants in order to be able precisely to predict what the flux

00:43:48 is through the upper pathway and through the pseudo-rotation pathway.

00:43:55 That has been done, and it summarises rather a lot of work, just to say, as I have done

00:44:04 here, that under our conditions, starting with two phospho, 41% goes directly to one,

00:44:12 and 59% goes via cyclic to one.

00:44:17 And what that means is, that if we had, very roughly, this is a mere qualitative argument,

00:44:24 if we started with R at phosphorous in the starting material, the two-phospho isomer,

00:44:31 we should expect 41% of the molecules at short times to be R, because they've come directly

00:44:40 with retention.

00:44:43 The other 59 have gone over this path, and two-thirds of those have lost an isotope and

00:44:50 become prochiral, and we don't want to see them again.

00:44:54 So 40% of them are prochiral, that's actually a little under 40, 20% of them, however, that

00:45:02 is a third of that 59, are chiral and have inverted configuration.

00:45:10 So if one looks at these numbers here, 40% prochiral, but we're not interested in those,

00:45:16 we are only interested in those species that have retained chirality, we can see that the

00:45:26 percent R, about two-thirds of the total chiral molecules, are going to be R, and about one-third

00:45:34 are going to be S.

00:45:37 It is gratifying, therefore, that under these conditions, we see that using starting material

00:45:46 independently determined as being 97% R, we re-isolated the two, because there is predicted

00:45:56 a very, very small loss of chirality, of course, even after relatively short times, of a small

00:46:06 loss of chirality in the re-isolated two-phospho compound, and in the product, and this reaction

00:46:16 was run to 23% of, towards equilibration, we found 72% R.

00:46:25 Those were our observed data, but on the last slide, I compare these with the precise predicted

00:46:33 data on the basis of all the rate constants, and not assuming short times or anything like it.

00:46:40 And you can see that the predicted values, the re-isolated material does, in fact, is

00:46:46 predicted to fall a tiny amount, certainly within, I'm afraid, our experimental uncertainty

00:46:52 in the configurational determination.

00:46:56 But the product, the important thing is that the product is within its experimental limits

00:47:04 in agreement with the predicted value, in fact, it's rather higher, but for retention

00:47:11 of configuration.

00:47:14 And I believe that these data, as long, as well as those that I have said to you earlier,

00:47:21 these data provide rather a nice test of the applicability of the Westheimer rules to phosphate

00:47:29 monoesters.

00:47:30 Finally, and most importantly, I should like to acknowledge the people who've done all

00:47:35 the work that I've described today.

00:47:38 The original synthesis and mass spectral analysis were done by these people, who had to be discriminated

00:47:47 amongst somehow.

00:47:53 That was done by Waite.

00:47:55 He was a graduate student, that was a post-doc, and that was a Harvard undergraduate.

00:48:02 All the enzymic work has been done by Walter Bloechler and David Hanson, and the NMR method,

00:48:08 and most of the chemistry, all of the chemistry that I've described today, has been done by

00:48:12 yet another Stephen, Steve Buchwald, who did, as I say, the NMR, and both the migration

00:48:21 and the chemical analysis.

00:48:22 Thank you.

00:48:23 Thank you.

00:48:24 Thank you.