[hanji]

So I started with 3-methoxybenzaldehyde and ended with 2-(3-methoxybenzylidene)succinic acid.    This was done as a Stobe Condensation so diethyl succinate was used.  The firststep was to treat the diethyl succinate with NaOMe and what happens is that a carbanion forms which acts as a nucleophile and attacks the carbonyl of 3-methoxybenzaldehyde.  Subsequent steps involved diluting with water, washing with ether and then acidifying the aqueous layer with conc. HCl...    I checked some mechanisms online but I'm mainly confused with how both carboxylic forms are formed in
2-(3-methoxybenzylidene) succinic acid.     Thank you in advance

[hanji]

If further explanation is needed I'll be happy to try and answer your questions....

[Honclbrif]

If you've checked the mechanism, as you've claimed, you'll see it only accounts for the creation of one carboxylate. Look at your workup. How else might an ester have been hydrolyzed?

[hanji]

Im sorry I forgot to add that after adding the 3-methoxybenzaldehyde I let it reflux for about 2 hours and then I added 5M NaOH and let it reflux overnight.  When I added the NaOH a precipitate formed which eventually redissolved.  I would go with that the NaOH hydrolyzed the ester off forming the carbxylic acid? So this means a carboxylate is not formed in this position.      So the acid workup at the end was only to protonate the previously formed carboxylate?       Thank you I appreciate the help

[hanji]

And if it's a Stobbe CONDENSATION... then shouldn't we have loss of a water molecule somewhere?

[hanji]

Sorry about my previous post ... I now realize that a carboxylate is formed in both positions...  the Na+ acts as a counter ion