Synthesis of 4-amino-3,5-dimethylphenol from N-(2,6-dimethylphenyl)hydroxylamine using perchloric acid?

Question

I come across one reaction in which N-(2,6-dimethylphenyl)hydroxylamine is converted to 4-amino-3,5-dimethylphenol in the presence of strong acid (perchloric acid).

It,s seems rearrangement reaction, but i am unable to understand how does this rearrangement is possible. What is the reaction mechanism of rearrangement reaction?

Answer 1

N-Phenylhydroxylamines undergo what is known as the Bamberger rearrangement when subjected to aqueous acid. The mechanism is a bit strange, as it involves divalent positively charged nitrogen (very rare - in fact, it is so rare that ChemDraw keeps trying to correct my $\ce{NH}$ to $\ce{NH3}$!), but that's just how it is. Wikipedia already has what I consider to be a very good explanation, but I'll try.

The mechanism begins with protonation at oxygen. Ordinarily, one would expect the nitrogen to be the more basic heteroatom, but protonation at nitrogen is unproductive - you can protonate nitrogen, but it's a dead end. On the other hand, if you protonate on oxygen, you can lose a water molecule.

The cationic intermediate on the right is rather like a benzyl cation, but with a nitrogen instead of a carbon. It has several resonance forms:

and the only decent nucleophile in the system, water, has three choices as to where it wants to attack.

1. Attack at nitrogen just goes back the way we came.
2. Attack at the ortho positions in this case would lead to a loss of aromaticity because you can't deprotonate at the end (see the mechanism below).
3. Attack at the para position. This completes the mechanism; after a proton exchange between N and O, a deprotonation leads to rearomatisation.