I cant seem to figure this out. If a benzene ring has 2 substituents 1) OH group 2) COOH group and we are attacking the electrophile carbon in the carbonyl CH3C=OOC=OCH3 (acet. anhydride), why is the O from OH rather than the O from the OH in COOH used? Both cant go through resonance, so what other differences would there be.

Imagine attack by PhO- and RCO2-. The negative charge is more effectively delocalised across the two oxygens in the carboxylic acid, than the -ve charge in PhO- which is less effectively delocalised within the ring. In this case the larger effective negative charge on the phenoxide will be the most important factor in determining nucleophilicity

That would be true if you did this reaction in presence of at least 2 Eq of a suitable base; but when both OH and COOH aren't deprotonated, does the same explanation still hold?

 

Sometimes these chemoselective reactions are difficult to rationalise. o-aminophenol, for instance, reacts with some electrophiles at the oxygen in presence of Et3N (because you have O- and NH2 competing), and at the nitrogen in presence of pyridine (because you have OH and NH2 competing).

 

Here, with both groups still in the acid form, I think it's more a matter of electron density.

There is a thermodynamic consideration here too as the ester will be the favoured product over the mixed anhydride