[movies]

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 pK_a difference of 8 units the deprotonation will be nearly complete if you have a full equivalent of base.  It's like mixing acetic acid (pK_a = 4.76, H₂O solvent) and water (pK_a = 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 pK_a 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 pK_a 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.

[GCT]

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.

[movies]

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?

[GCT]

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?

[movies]

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.

[GCT]

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?

[movies]

 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).

[HP]

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

[GCT]

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.  

[movies]

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.

[GCT]

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.

[movies]

First off, "original" thiourea is more acidic then methanol?  And isn't methanol a solvent in this reaction?

Yes.  The pK_a'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 reaction.

[GCT]

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.

[movies]

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.

[GCT]

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.