Both cases would form a tetrahedral intermediate, but the RO- would always be preferentially kicked off:
There are other factors than leaving group ability which decide whether a substitution takes place, such as how well the intermediate forms. The amide is so bad an electrophile that the immediate is very unlikely to form.
So for a sucessful attack we require:
1. A good enough electrophile
2. A good enough nucleophile
3. A leaving group on the defending molecule which is better at leaving than the attacking nucleophile.
Leaving group ability is a good guide to how good a nucleophile something is - good nucleophiles are bad leaving groups. Hence a good nucleophile will have a high pKaH. Another way to think about it is that something which wants to form a bond to hydrogen will be likely to want to form a bond to carbon.
A reaction particularly suited to substitution will have more relaxed reaction conditions. Eg. Will react faster, at a lower temperature, and without needing a catalyst.
For example, below are four carboxylic acid derivatives with different leaving groups, each attacked by H2O. Try to relate the reaction conditions to the table above:
The difficulty of attacking an amide is due to n -> pi* overlap. This can be depicted as resonance:
Or spacial overlap of the filled n and empty π* orbital:
Or as an MO diagram:
Now we can see why an amide is such a bad nucleophile - because resonance raises the LUMO of the molecule.
This MO approach also gives another explanation for why an amide is a worst base than an amine - a lower lone pair energy makes it less favorable to donate them into a proton's empty 1s orbital
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