I'm currently working on a chemistry project in which I need to convert ethyl methyl carbonate to ethyl formate.

ethyl methyl carbonate ‐> ethyl formate

I only have a simple background in organic chemistry (and I haven't seen anything about reaction mechanisms yet), but I was guessing water could be used in this reaction, to produce $\ce{CH3OH}$ and replace the $\ce{-OCH3}$ group by an $\ce{-OH}$ group. Then the next step would be to remove the oxygen from the $\ce{-OH}$ group. I've seen that compounds such as $\ce{LiAlH4}$ can do this, but I'm looking for a catalyst to make the process more economically viable, in case the proposed synthetic path is correct.

Does anyone know how to perform the conversion, or even by following another synthetic path? I've got no idea whether this process is correct. I'm also trying to keep the process as low energy-consuming as possible.

I agree it would be better to start from ethanol and formic acid, but I have to start with this molecule.

  • 2
    $\begingroup$ This is a carbonate derivative, using water will fragment your molecule:$$\ce{EtO-CO-OMe + 2H2O -> EtOH + MeOH + "H2CO3"\\ <=> EtOH + MeOH + CO2 ^ + H2O}$$ You need to reduce the central carbon with "$\ce{H-}$" and you also need a soft way to reduce. LAH is the best choice here. $\endgroup$ May 16, 2014 at 16:45
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    $\begingroup$ I'm not sure LAH is the right choice here; it's an aggressive reagent and would probably reduce the carbonyl moiety all the way to an alcohol. A more selective reagent (e.g., sterically hindered, electrophilic hydride donors such as DIBAH) is probably called for. There's still the problem of selecting which alkoxy group will leave, however. $\endgroup$
    – Greg E.
    May 16, 2014 at 18:08
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    $\begingroup$ There's also the brute force approach: reduce the carbonate ester all the way down to methanol with excess LAH, oxidize the methanol to formic acid, then esterify with ethanol. The nice thing is that the molecule you have already contains within its skeleton two molecules of methanol and one molecule of ethanol (if reduced with LAH and then quenched), which could be recovered, separated, and reused. $\endgroup$
    – Greg E.
    May 16, 2014 at 18:19
  • 5
    $\begingroup$ Economically, the conversion of ethylmethylcarbonate (98%, 100mL: 212.50 Euro) to ethylformate (97%, 500mL: 44.30 Euro) - prices from Sigma Aldrich, 2014-05-17 - is pointless. $\endgroup$ May 17, 2014 at 5:23
  • $\begingroup$ Most likely any reduction of the molecule will result in a mixture of different reduction products, unless total reduction to alcohols is done, so, it is extremely impractical. Moreover, formate will reduce more readily than the original molecule. This said, direct approach is the last thing to consider. $\endgroup$
    – permeakra
    Jun 26, 2014 at 9:09

1 Answer 1


First use $\ce{EtO-/EtOH}$, then lithium-tri-tert-butoxy-aluminium hydride. This will convert it to diethyl carbonate and as the reducing agent is hindered you can manipulate reaction conditions to get the desired product.

Or another way is to hydrolyse it in basic medium to get $\ce{H2CO3}$ and then evaporate $\ce{H2O}$ and $\ce{CO2}$ then after separating it will open a large number if ways as you have got ethyl alcohol and methyl alcohol.


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