Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: bharathi on July 20, 2005, 03:44:59 AM
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Hi all,
I am doing condensation of diethyl-2-ethyl-2-phenylmalonate with thiourea. This reaction is proceeding only in the presence of sodium methoxide in methanol.
I have no idea on the mechanism by which sodium methoxide is causing the reaction.
Is this dry base acting as a catalyst / base / nucleophile??
Can this not be substitued by alcoholic KOH
Could someone help me out in outlining the mechanism for this condensation.
Thanks in advance,
Bharathi
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diethyl-2-ethyl-2-phenylmalonate ???
methoxide deprotonates the alpha carbonyl hydrogen to form a synthetic equivalent of an enolate (the central carbon in malonate)
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There aren't any alpha protons in the malonate compound.
So the product you are getting is the cyclic urea, right, where the nitrogens have displaced the EtO groups to make a 6-membered ring? It's possible that the methoxide trans-esterifies with the ethoxide first and the smaller MeO groups are sterically small enough to allow the reaction to proceed.
A reagent like KOH needs some water as well, since KOH isn't very soluble in neat alcohol. If you had water you might get a lot of ester saponification and once you have the carboxylate you can't attack the carbonyl to make the ring.
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Or quite possibly the methoxide serves to remove the proton from positively charged nitrogen in the tertiary intermediate formed by the nucleophilic addition of a theourea nitrogen to the carbonyl, thus the reaction proceeds afterwards.
diethyl-2-ethyl-2-phenylmalonate...how did you deduce the structure?
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If the methoxide was just a base then you would theoretically only need a catalytic amount because you produce an equivalent of ethoxide from the condensation. I suppose it's possible that it is just shuttling protons though.
Dialkyl malonates always imply two esters, the other groups have to be in the middle. So dimethyl malonate is two methyl esters with a CH2 in the middle.
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what do you mean...the elimination occurs after the addition doesn't it....perhaps you're saying that the addition occurs, the positive charge on the nitrogen remaining, the elimination of ethoxide, then ethoxide returns to abstract the proton from the positively charged nitrogen???
In the end. you do need a base to stabilize the positively charged nitrogen, my guess is that this is methoxide, I just don't think that elimination of ethoxide will occcur appreciably when another good leaving group state of RNH2+ is present.
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in reference to the original question by bharathi
the point being, you need a base to start the reaction, my guess is that it's methoxide from the start
I'm not quite sure if the ethoxide formed will take over as the proton remover or whether it will react with methanol...don't have time to consider this aspect
signing off ;)
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The mechanism probably does start by deprotonating the urea with methoxide, but you only need a catalytic amount of base because after the addition to the ester you eliminate an equivalent of ethoxide which can then deprotonate another equivalent of urea.
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yeah, I'm not absolutely sure about what happens afterwards, it's just that I've never heard of the mechanism you're proposing...to be honest, it sounds a bit strange.
althhough from what I remember, ethanoate is certainly more reactive then methanoate
Perhaps you would consider the proposal that ethanoate reacts with metanol, to product more methanoate?
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Perhaps you would consider the proposal that ethanoate reacts with metanol, to product more methanoate?
Yes, that's what I first suggested.
As for the rest of the mechanism, the attached picture shows what I mean. My point is that you really only need one molecule of MeONa to initiate the process since you ultimately produce a molecule of MeOH, EtOH, and NaOEt in the reaction. The addition NaOEt can then deprotonate another molecule of urea and start the process over again.
I suppose it might be advantageous to add more base since the imide proton in the compound I labled A is probably more acidic than the amide protons; it is a Curtin-Hammet situation though.
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Not to sound rude, but please read the whole post
Note that it's thiourea...not sure if that makes a difference at all.
That's not really what I had in mind....I thought that the nitrogen of the thiourea would attack directly and form a positively charged tertiary intermediate (neg. charge on oxygen also of course), with the positive charge on the nitrogen. Just recently performed the fisher esterification in the lab; an anion is required to deprotonate the tertiary intermediate.
Your's is essentially the same, but slightly different, in that the methoxide deprotonates the thiourea before the tertiary intermediate adduct. I'm suggesting that it would be more significant that methoxide deprotonate the tertiary adduct since the positive charge is greater. If I'm not mistaken, doesn't your first product (due to the intramolecular rearrangement) technically represent a resonance hybrid? There will be a positive charge on the nitrogen, not as great as in the adduct. Right?
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My mistake about the thiourea. I fixed it above. Shouldn't make much of a difference.
Yes, the deprotonated thiourea would be a resonance hybrids, I just drew them that way to look more like traditional enolates.
The problem with the thiourea attacking directly is that there would be a positively charged intermediate when the whole reaction is under basic conditions. That is usually unlikely. Also, the pKa of thiourea is 21 (DMSO solvent) and that of methanol is 29 (DMSO solvent). That difference is more than enough to ensure that the equilibrium lies almost entirely with the deprotonated thiourea and methanol.
The case of the Fisher esterification is different because it is typically acid catalyzed. In an acid catalyzed reaction then positively charged intermediates are much more likely than negatively charged intermediates.
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Thank you Sirs, for the inputs.
Is sodium methoxide for this reaction irreplacable?
I am concerned of the storage and handling w.r.t. process & product safety.
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bharathi, why do you want to switch bases?
My mistake about the thiourea. I fixed it above. Shouldn't make much of a difference.
Yes, the deprotonated thiourea would be a resonance hybrids, I just drew them that way to look more like traditional enolates.
The problem with the thiourea attacking directly is that there would be a positively charged intermediate when the whole reaction is under basic conditions. That is usually unlikely.
it happens, plain and simple as that, there's nothing preventing the direct attack
Also, the pKa of thiourea is 21 (DMSO solvent) and that of methanol is 29 (DMSO solvent). That difference is more than enough to ensure that the equilibrium lies almost entirely with the deprotonated thiourea and methanol.
but note that it is the protonated form of thiourea
I'm not trying to be an ass, it's just that all of this is a bit interesting to me, it's really the question of whether ethanoate serves as the proton remover in the long run or methanoate. I'm moving towards the methanoate route, since it is the less complicated of the two, I just don't know exactly why ethanote would take over...and if it should, the dyanmics of the reaction should be a bit more complicated I imagine....for e.g. the reaction would propagate exponentially with time according to the ethanoate route, if I'm note mistaken. Also, in general, what happens to the methanoate as ethanoate takes over?
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Isnt't it classic Biginelli multicomponent reaction ???
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but note that it is the protonated form of thiourea
No, it's not. It's just plain old thiourea. Protonated thiourea would be H2NCSNH3+ When you have a pKa difference of 8 units the deprotonation will be nearly complete if you have a full equivalent of base. It's like mixing acetic acid (pKa = 4.76, H2O solvent) and water (pKa = 14), it exists mostly as H3O+ and acetate ions, right? The same thing will occur with thiourea and methoxide: mostly deprotonated thiourea and methanol. The deprotonated form will be a lot more nucleophilic too because the adjacent thiocarbonyl group can only stabilize one of the lone pairs on the nitrogen at a time.
Which base does the deprotonating doesn't make much of a difference, but if you only added a catalytic amount of methoxide to start the reaction then ethoxide would probably carry on the reaction after all the methoxide had turned into methanol. The pKa difference between methanol and ethanol is pretty small (~0.72 units, in water solvent) so if there is only a small amount of methanol then that equilibrium won't play a large role. You typically need a pKa difference of at least 2 units to get a significant amount of deprotonation. Also, if the ethoxide which is formed has a choice between deprotonating methanol or thiourea, it'll pick the thiourea because the thiourea is so much more acidic than methanol.
All in all, I don't think that the kinetics would change that much by changing the base, since in both cases the initial thiourea/alkoxide equilibrium will be largely shifted towards the deprotonated thiourea/alkanol side. There might be a change if there was a significant amount of trans esterification because the attack on the ester is likely a relatively slow step and sterics might play a role.
Also, "methanoate" refers to a HCO2- ion and "ethanoate" to a acetate anion. I think what you mean to say is "methoxide" and "ethoxide."
Isnt't it classic Biginelli multicomponent reaction ???
Not quite, the Biginelli reaction is acid catalyzed and typically uses an aldehyde, not an ester. It's very similar though.
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I'll take this discussion a bit slower,
When the nitrogen of thiourea attacks a carbonyl carbon of malonate, the thiourea in its tetrahedral intermediate adduct will be in its protonated form.
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I'll take this discussion a bit slower,
When the nitrogen of thiourea attacks a carbonyl carbon of malonate, the thiourea in its tetrahedral intermediate adduct will be in its protonated form.
I don't agree. There won't be a significant amount of H2NCSNH2 in solution because it will be deprotonated by the methoxide; the predominant species would be H2NCSNH-. The latter will be a more the nucleophilic species as well, and therefore more likely to be the attacking species.
There will be one hydrogen left on the nitrogen that attacked the ester, but the other proton will be gone before the attack.
Just so we're clear, you are proposing a mechanism like the one I have attached to this post, right?
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yes,
what is the relative pka for thiourea and methoxide (methanoate) again?
Surely the protonated form which I suggested will be more acidic though... and say that for your sakes that thiourea was deprotonated from the start, would you say that this goes to completion, probably not right?
Thus if any of the original (remaining) thiourea, were to encounter the carbonyl carbon, surely the kinetics would favor the deprotonation of the thiourea adduct with the positive charge on the nitrogen of thiourea over the resonance hybrid of an unprotonated thiourea right?
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Surely the protonated form which I suggested will be more acidic though... and say that for your sakes that thiourea was deprotonated from the start, would you say that this goes to completion, probably not right?
So what if it's more acidic? For that acid/base reaction to occur the nucleophilic attack must happen first, during which time the acid/base chemistry of the starting materials will already be going. Acid/base reactions happen very, very fast, certainly faster than the attack of a poor nucleophile (thiourea) on an electron rich, sterically hindered carbonyl group! And why would there be a reason to think that it wouldn't go to completion? If you argument is simply that there are two resonance forms of the deprotonated thiourea then you have neglected the two resonance forms of "regular" thiourea! The same forces are at play. The key difference in nucleophilicity is that in the deprotonated form only one of the two lone pairs on the nitrogen can be delocalized into the thiocarbonyl group at a time, so there is always one exposed to perform the nucleophilic addition.
I'm certain that if the thiourea did attack as you suggest that it would be rapidly deprotonated and go along about the course of the reaction, but my point is that the mechanism you propose will be kinetically much slower, and therefor contribute much less to the observed rate.
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And why would there be a reason to think that it wouldn't go to completion?
You do know that most acid/base reactions don't go to completion unless one of them is a strong base/acid right?
you have neglected the two resonance forms of "regular" thiourea
I was referring to the "regular" conformation, sorry for the confusion
I'm certain that if the thiourea did attack as you suggest that it would be rapidly deprotonated and go along about the course of the reaction, but my point is that the mechanism you propose will be kinetically much slower, and therefor contribute much less to the observed rate.
perhaps
It's just that the mechanism you propose seems a bit complicated and "unncessary." Another issue which has been disregarded for a while here, actually it was the original point of our discussion, was on whether ethoxide or methoxide would maintain itself as the base in the reaction. My thoughts were on methoxide yours was the situation where methoxide acted as somewhat of a catalyst in relasing the ethoxide. I would imagine that they would have a concept established for such an interesting and nonsimplist mechanism. My intuition tells me that the rate of reaction according to your proposal would have a exponentially increasing rate of reaction, something I've never heard of but would be more than happy to accept if it were true.
What is the actual mechanism of this reaction anyways?
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You do know that most acid/base reactions don't go to completion unless one of them is a strong base/acid right?
Oh, you mean the acid/base part? Sure, that will be an equilibrium.
It's just that the mechanism you propose seems a bit complicated and "unncessary." Another issue which has been disregarded for a while here, actually it was the original point of our discussion, was on whether ethoxide or methoxide would maintain itself as the base in the reaction. My thoughts were on methoxide yours was the situation where methoxide acted as somewhat of a catalyst in relasing the ethoxide. I would imagine that they would have a concept established for such an interesting and nonsimplist mechanism. My intuition tells me that the rate of reaction according to your proposal would have a exponentially increasing rate of reaction, something I've never heard of but would be more than happy to accept if it were true.
I never like to apply Occhem's Razor to chemistry discussions because I think chemists simplify things a lot anyway. And no, I don't think a pre-existing equilibrium between an acid and a base is an unnecessary consideration.
I think you have misinterpreted the point I was making about the production of ethoxide in the course of the reaction. Throughout the reaction the amount of alkoxide (that is, methoxide and ethoxide) would remain constant and therefore won't change the kinetics of the reaction. However, the make-up of the alkoxide mixture (methoxide + ethoxide) would change from being mostly methoxide to being mostly ethoxide by the end of the reaction. For each molecule of alkoxide that initiates the reaction you end up producing another equivalent of alkoxide at the end of the reaction. There would be no exponential growth of the concentration of alkoxide throughout the reaction. If there were an exponential growth of alkoxide concentration, then surely that would have some effect on the rate. However, I doubt that deprotonation is the rate determining step of the reaction.
I suppose I was eroneous in calling methoxide a catalyst, since it is not reproduced when the reaction is completed. I should have said that methoxide could act as an initiator (in sub-stoichiometric amounts).
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I think in this basic madia the equilibrium thiourea-isothiourea is in isoform advantage and the sugested reaction mechanism is most possble...
Really intrigue mechanism ;)
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I suppose I was eroneous in calling methoxide a catalyst, since it is not reproduced when the reaction is completed. I should have said that methoxide could act as an initiator
an initiator? I'm not familiar with this concept, although I'm not disregarding it at all.
Don't you think that the ethoxide would react with methanol, afterall, that's how acid base reactions occur...a molecule such as ethoxide is surrounded by methanol, proton transfer would occur. This is why I believe methoxide is the base in the course of the reaction.
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an initiator? I'm not familiar with this concept, although I'm not disregarding it at all.
Yeah, a reagent that only requires a small (sub-stoichiometric) amount in order to initiate the reaction. It differs from a catalyst in that it is consumed in the initiation process. Other similar species then carry out the bulk of the reaction turnover. A good example of this is a radical initiator such as AIBN.
Don't you think that the ethoxide would react with methanol, afterall, that's how acid base reactions occur...a molecule such as ethoxide is surrounded by methanol, proton transfer would occur. This is why I believe methoxide is the base in the course of the reaction.
As I mentioned above, ethoxide is a slightly stronger base than methoxide, so they might react to some small extent, but in the presence of a stronger acid (thiourea) the resultant ethoxide would most likely react with that instead of methanol. Also, if methanol were used in sub-stoichiometric amounts, the concentration of ethanol would quickly overtake the concentration of methanol since in the course of the reaction two equivalents of ethanol are produced (one from each ester). Once the concentration of ethanol exceeded that of methanol it would be increasingly less likely for an ethoxide to encounter a methanol molecule before encountering a thiourea or another ethanol molecule first.
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I guess it's time to start putting this argument to a close, partly since the op has abandoned it.
Yeah, a reagent that only requires a small (sub-stoichiometric) amount in order to initiate the reaction. It differs from a catalyst in that it is consumed in the initiation process. Other similar species then carry out the bulk of the reaction turnover. A good example of this is a radical initiator such as AIBN.
there's actually a established paradigm for this? Note that wth radical reactions, you also have the chain reaction, and termination step. Don't know why, the "initiator" concept sounds a bit weird to me; it sounds more of a extremely inefficient genetic experiment
As I mentioned above, ethoxide is a slightly stronger base than methoxide, so they might react to some small extent, but in the presence of a stronger acid (thiourea) the resultant ethoxide would most likely react with that instead of methanol. Also, if methanol were used in sub-stoichiometric amounts, the concentration of ethanol would quickly overtake the concentration of methanol since in the course of the reaction two equivalents of ethanol are produced (one from each ester). Once the concentration of ethanol exceeded that of methanol it would be increasingly less likely for an ethoxide to encounter a methanol molecule before encountering a thiourea or another ethanol molecule first.
First off, "original" thiourea is more acidic then methanol? And isn't methanol a solvent in this reaction?
-it seems much more efficient that methoxide be the base throughout the course of the reaction. Methoxide reacts with thiourea, releases ethoxide, which reacts with a surrounding methanol to furnish more methoxide, and so on. In your case, methoxide becomes somewhat of a spectator ion. You've got reactions going both ways. My assertion is that anytime methanol is the solvent, and the released ethoxide reacts with its surrounding methanol, the kinetics will drastically favor the methoxide route. Whereas in your route, the ethoxide is essentially getting "tugged" along, as it reacts with methanol along the way, releasing methoxide here and there (don't you see where this is all pointing?). The kinetics just seem perfect with the methoxide route.
why wait at all for the ethoxide to be relasead, when you have methoxide right at hand?
-Ethoxide would actually have to be in 2 times in stoichiometric equivalence (at least) to thiourea, that is two moles of ethoxide are needed for every thiourea to achieve your proposed final product.
-seems there are plenty of disadvantages in considering ethoxide as a candidate base.
I guess we can agree to disagree. I have not disregarded your proposal at all, and I'll be more than happy to hear more about the topic from you, particularly if it includes a specific reference (article, advanced text, website) to this particular reaction.
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First off, "original" thiourea is more acidic then methanol? And isn't methanol a solvent in this reaction?
Yes. The pKa's are 21 and 29 respectively, as I said above. Yes methanol is solvent, but that doesn't make it more acidic.
The observable kinetics of the reaction won't be affected in any measurable way when using methoxide or ethoxide to do the deprotonation because the acid/base step with thiourea is not rate limiting and probably not even rate determining. My only point about methoxide as base in the first place is that you could use less than a full equivalent of methoxide. If the ethoxide that is produced goes and reacts with methanol then it just doesn't matter for the rate because either alkoxide will react extremely fast with the much more acidic thiourea. It's not a question of "waiting" for thiourea to react, its the fact that a by-product of the reaction is another equivalent of alkoxide. Think about the reaction of a mole of substrate and thiourea with one molecule of methoxide kicking things off. Theoretically, that is all that you would need.
Look at the complete, balanced equation for the reaction (attached) assuming no further equilibrium with the alcohols and the alkoxide. I don't know why you think there would have to be two full equivalents of base, since one is produced in each condensation step.
Ethoxide must at some point be used as a base to deprotonate something.
The original reference from this reaction is J. Am. Chem. Soc. 1935, 57, 1961-1963, but there is no mechanistic discussion. The reaction is, however, analogous to the Claisen condensation (http://www.organic-chemistry.org/frames.htm?http://www.organic-chemistry.org/namedreactions/claisen-condensation.shtm) reaction.
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This isn't analogous to the claisen condensation at all. As you can see the with the claisen condensation, the base is actually the conjugate of the solvent, and that's exactly what I was suggesting.
Ethoxide must at some point be used as a base to deprotonate something.
my point exactly, it's ethoxide interacts with its surrounding solvent. if you're suggesting that the kinetics of an ethoxide/thiourea is actually faster in relevance to rate constants, than one can make the same assertion with the methoxide/thiourea argument, in particular since methoxide is so much more abundant if it is actually used as a solvent. This would be particularly important since one would want to drive the reaction forward. There will always be much more methoxide present than ethoxide.
you know something, I'm starting to think that perhaps the original post was in error, perhaps a ethoxide/ethanol solvent was used, or supposed to have been used.
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you know something, I'm starting to think that perhaps the original post was in error, perhaps a ethoxide/ethanol solvent was used, or supposed to have been used.
If you look in the JACS paper I cited above, then you'll see that they did use ethoxide/ethanol. I suspect that the modification the original poster referred to has to do with the availability of methoxide/methanol for a lab course or something. It's easier to come by dry methanol than dry ethanol.
I think that our argument has been very semantic. I don't think that we are too far from agreement here.
But note that the solvent is methanol, not methoxide. You can't use sodium methoxide as a solvent because it's a solid salt. When in methanol solvent, there would likely be more methoxide than ethoxide provided that the equilibrium between ethoxide and methanol adjusts faster than the ethoxide reacts with thiourea. Kinetically, however, if there are many, many more molecules of methanol than thiourea (the case with MeOH solvent) then the ethoxide/methanol equilibrium probably does occur prior to deprotonation of thiourea.
I know that a Claisen condensation usually uses the same alkoxide as in the ester, but I see no reason why a slight change in the solvent/alkoxide reagent (e.g. MeO-/MeOH instead of EtO-/EtOH) would significantly change the mechanism. I think that the alkoxide is usually matched to the ester in order to negate the effect of trans-esterification.
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I know that a Claisen condensation usually uses the same alkoxide as in the ester, but I see no reason why a slight change in the solvent/alkoxide reagent (e.g. MeO-/MeOH instead of EtO-/EtOH) would significantly change the mechanism. I think that the alkoxide is usually matched to the ester in order to negate the effect of trans-esterification.
Yeah, well evidently, according to the op, the reaction progressed with the methanol/methoxide solvent, although I wish I could actually try it out for myself. One could use the ethanol/ethoxide solvent of course...I was just interested in the complications that might arise in using a different solvent, after all, it is somewhat of a nonstandard procedure and I guess people get interested when an alternative is proposed. Perhaps I was overly obssessed with the idea though, but it is somewhat in accordance with my current enrollment in organic lab II (having the final test this tuesday).
I'll check out that paper you referred to sometime, I'm not particularly close to a library at the moment.
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I'll check out that paper you referred to sometime, I'm not particularly close to a library at the moment.
I think it's a general case in organic chemistry that a number of slightly different conditions will all succesfully carry out the reaction, just one particular set of reagents usually works best for a particular substrate or one set of reagents is just more convenient to use.
I actually wouldn't recommend wasting your time looking up that paper. There is practically nothing about the synthesis (except the procedure), it's mostly about the pharmacology of the products.