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One synthesis of quinolones begins with the formation of an ethyl ethoxymethylenemalonate, as seen in this Organic Syntheses paper.

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I've been asked if the malonate derivative would be formed if methylmalonate was treated with trimethyl orthoformate with a catalytic amount of $\mathrm{H^+}$, without heating. The nature of the solvent isn't mentioned, but I guess the reaction is carried out in a non-aqueous medium with an ion-exchange resin, since I think trimethyl orthoformate acts as a water trap.

My hunch is that there shouldn't be a significant quantity of product formed. We're dealing with an activated methylene, why not simply use a base and then carry on with our addition? I can't imagine how, under normal circumstances, the elimination product ethyl ethoxymethylenemalonate will be formed.

How can this reaction happen in a mechanistically reasonable way?

CHM
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  • You are using triethyl orthoformate, not the trimethyl orthoester. That's where the pendant group is coming from. If it had been just a water trapping agent, I presume the OrgSyn people would have had the sense to put the thing in an arrow instead on the left side of the reaction. –  Apr 27 '12 at 10:41
  • Ethyl, methyl, propyl, etc. doesn't matter here. I can devise a mechanism, but I'd need to provide heat for it to work, since it consists of an elimination. I'm just wondering if there's some unknown-to-me chemistry at work for the elimination to proceed without heating. – CHM Apr 27 '12 at 19:38

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Ok, I will try my hand on the mechanism. In my opinion the reaction should work with a catalytic amount of $\ce{H+}$ since the key step for the reaction is the hydrolysis of the orthoester. This is an entropically very favourable reaction since from one reactant molecule you get two product molecules.

The resulting molecule is extremely reactive. So it can easily react with the enolized malonate. The enolization might be facilitated by the $\ce{ZnCl2}$ and/or the cat. $\ce{H+}$. (I'm not entirely sure whether the $\ce{ZnCl2}$ really plays the role I've given it below in this reaction. So, if anyone knows good arguments against it, please let me know.)

The protonated product can then take part in an $\mathrm{E}1$ elimination reaction to yield ethyl ethoxymethylenemalonate. The carbocation which results when $\ce{EtOH}$ leaves the molecule is stabilized by the neighboring ethoxy group. The hydrogen on the central carbon atom is very acidic, since it is placed between three carboxylate groups, so it can even be removed by rather non-basic molecules, e.g. by an $\ce{EtOH}$, a malonate or an othoester molecule.

So much for the acid catalysed mechanism. Now for the question, why one doesn't simply use a base to facilitate the addition of the activated methylene compound. Well, this is due to the fact, that orthoesters are quite stable against basic hydrolysis. So, under basic conditions you don't get the carbonyl component of the addition reaction.

Philipp
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  • This looks plausible. I'm just surprised to see E1 occur without heat. Thank you sir. – CHM May 12 '12 at 16:26
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    @CHM: Ok, I will try to explain why I think it works. The rate determining step of the E1 mechanism is when the leaving group leaves the molecule thereby forming a carbocation. Since this step is very much facilitated (i.e. the activation barrier is significantly lowered) by the stabilization of the intermediate carbocation by the ethoxy group, less heat than normal will be needed to let the reaction happen at a reasonable rate. So room temperature might be enough. – Philipp May 12 '12 at 17:17