So A will be initial intermediate of nucleophilic attack of RNH2 on carbonyl.
But alkoxide is strong enough base to take H+ from protonated amine - this gives B.
Since you are in basic conditions with excess of amine, a small amount of C will be formed - even if only a fraction of a percent, if it reacts to give product the reaction will proceed.
Alkoxide is much better leaving group than amide (RNH- anion).