Bamberger Rearrangement — Why no ortho- products?
I was researching the Nitrobenzene -> Phenylhydroxylamine -> p-Aminophenol pathway. The final step is achieved using a Bamberger Rearrangement.
I would post a diagram of the reaction, but the spam filter has banned me on multiple occasions for posting URLs so I'm going to avoid that.
Basically, the -OH group gets protonated, and pops off as water, leaving the nitrogen positive. The pi electrons on the ring move to form a double
bond with the nitrogen, which creates a positive formal charge on the ring.
Now, we can draw resonance structures that show that the positive charge is able to exist in the ortho and para positions.
Next, all diagrams show water attacking the carbocation while it's in the para position (at which point water picks up the excess hydrogen and the
product tautomerizes to p-aminophenol).
However, as the carbocation is able to exist in the ortho position, why water not attack there? I don't believe it's due to steric hindrance, as the
-NH and water groups are not that big, are they? If anything, the ortho position would just make it easier for the tautomerization, as the nitrogen
can easily pick up a hydrogen from the OH2 group, no?
I can't seem to find anything on the internet describing this process, except one paper. However, that paper looks at this reaction in a
trifluoroacetic acid which forms a complex of some sort, and this is not applicable when done in water.
What's going on here?
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