[bharathi]

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

[GCT]

diethyl-2-ethyl-2-phenylmalonate

methoxide deprotonates the alpha carbonyl hydrogen to form a synthetic equivalent of an enolate (the central carbon in malonate)

[movies]

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.

[GCT]

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?

[movies]

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.

[GCT]

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.

[GCT]

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

[movies]

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.

[GCT]

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?

[movies]

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.

[GCT]

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?

[movies]

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 pK_a 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.

[bharathi]

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.

[GCT]

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 pK_a 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?  

[HP]

Isnt't it classic Biginelli multicomponent reaction