1
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Is it a kind of beckmann rearrangement? $\endgroup$ – prog_SAHIL Nov 16 '17 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 '17 at 13:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.