[aldol_condensation]

Hi! Everyone  .

In my text book I had the following Question:

What is the product of Ethyl propanoate and diethyl oxalate in the presence of Sodium Ethoxide and acidic medium?

Well the answer is straightforward, ie. C2H5OOCH(CH3)COCOOC2H5. But I was wondering what would be the product if instead of taking diethyl oxalate we had taken Ethyl Methyl oxalate. The the problem, according to me, would then be that where would the enolate anion attack the Ethyl Methyl oxalate---on the carbon bearing the ethoxide group or on the carbon bearing the methoxide group---. My speculation is that there would be steric factors. I am not sure.

Please Help.
Thanks!

[OldThrashbarg]

You're using sodium ethoxide as your base for a reason.  Remember, the carbonyl carbon has a partial positive charge.  So while this reaction is going on, the nucleophilic ethoxide ion is attacking the carbonyl carbon.  So for a moment, you essentially have -OC(OEt)2R.  The carbonyl group then reforms, and one of the OEt groups falls off.  Assuming there's a large excess of NaOEt, you'll end up replacing all of the methyl groups with ethyl groups anyways.