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In the reaction below, why does acetic anhydride not react with the -OH group to form an ester instead?

Reaction scheme

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The amino group is more nuclephilic than the -OH. For this reason the reaction will proceed first with the amine. If an excess of acetic anhydryde is used the acetylation of both moieties would happen.

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  • $\begingroup$ The amino group cannot be more nucleophilic because it hardly has an available electron pair. The hydroxy group does. $\endgroup$ – Jan May 8 '16 at 15:54
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The reaction is likely with one equivalent of acetic anhydride and under thermodynamic control.

Kinetically, the nucleophilic attack of phenol on acetic anhydride should be a lot faster than that of aniline. Aniline’s lone pair is taking part in resonance across the phenyl ring so it is barely able to attack. Phenol has two lone pairs on oxygen, though, of which only one is able to resonate with the aromatic system. Under kinetic control I would thus expect acylation on oxygen.

However, there is a second process which renders us thermodynamic control: The initial product can react with a second molecule in a transacylation reaction.

$$\ce{RHN-C6H4-OAc + H2N-C6H4-OR <=>> RHN-C6H4-OH + HAcN-C6H4-OR}$$

Both the oxygen of a free phenol and the nitrogen of a free aniline (the latter slower, but still) can attack a corresponding acylated atom. In overall thermodynamic stability, the acetylamine is more stable than the acetate (and is also kinetically more inert) so after equilibration has taken place the acetamide will prevail.

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