Carboxylic acids and their derivatives on reaction with hydroxylamine $\ce{NH2OH}$ and hydrazine $\ce{N2H4}$ give hydroxamic acids and acid hydrazides, respectively, through acyl nucleophilic subsitution.

Why don't they give the same reaction as aldehydes and ketones (i.e. removal of $\ce{H2O}$) giving formation of oximes as the acids also have oxo ($\ce{=O}$) group?


1 Answer 1


For simplicity, I will consider the reaction of acetic acid with ammonia, because the same principles apply. The two alternatives are: $$ \underbrace{\ce{CH3C(O)NH2}}_{\mathrm{amide}} + \ce{H2O} \leftarrow \ce{CH3C(O)OH} + \ce{NH3}\rightarrow \underbrace{\ce{CH3C(NH)OH}}_{\mathrm{imidic \ acid}} + \ce{H2O} $$ where the group connected via a double bond has been set in parentheses. The two products are connected by protonation/deprotonation in a tautomeric equilibrium, so the question boils down to thermodynamic stability. For most cases, we expect the C-O double bond/C-N single bond combination to be more stable than the C-N double bond/C-O single bond combination.

  • $\begingroup$ Is it a kind of beckmann rearrangement? $\endgroup$
    – prog_SAHIL
    Nov 16, 2017 at 12:45
  • $\begingroup$ @prog_SAHIL I clarified the answer. No, it is not a Beckmann rearrangement, because no new C-N bonds are formed - they only change from double to single or vice versa. It is a simple matter of protonation and deprotonation, just like keto/enol. $\endgroup$
    – TAR86
    Nov 16, 2017 at 13:10

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