The Hydrolysis of an Imide A Way to Synthesize Primary Amines

Of the two potential leaving groups in the tetrahedral intermediate, OH is the weaker base and therefore the one more likely to be eliminated, thereby reforming the amide. 

Occasionally, an NH2 is eliminated. When this happens, the carboxylic acid that is formed immediately loses a proton. Since this step is irreversible (the negatively charged carboxylate ion is not reactive), it disturbs the equilibrium and drives the reaction toward products. 

In strongly basic solutions, the reaction is second order in hydroxide ion. That is, two equivalents of hydroxide ion participate in the reaction. The second equivalent of hydroxide ion removes a proton from the initially formed tetrahedral intermediate. 

Now the possible leaving groups are NH2 and it is eliminated and the carboxylate ion is formed. 

Notice that one equivalent of hydroxide ion is not a catalyst (it is not regenerated) but the second equivalent is regenerated, so it is a catalyst. 

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