Mechanism for the synthesis of 2-pyridone from pyridine N-oxide I'm specifically interested in the pathway from pyridine N-oxide, where there is some kind of reaction and rearrangement with acetic anhydride.

I'm assuming there is first a substitution (by the negative oxygen) at the carbonyl. The -OAc group can then attack at the alpha-position on the ring.

We could continue and say that the loss of H and the -OAc group on nitrogen happens next to restore aromaticity, but then we're stuck with that OAc group instead of the carbonyl.

From what I could find, it doesn't seem that there's a radical mechanism here, but if there is some intramolecular rearrangement, I cannot seem to find it.
Pyridine N-oxide itself is generally fairly resistant to nucleophilic attack as the ring is already 'electron rich' and doesn't want to accept additional electron pairs. 

When treated with acetic anhydride, however, the carbons at the 2 position become electrophilic (the acetate is electron withdrawing making the C=N bond more like an iminium), hence it is possible to have acetate attack, with subsequent elimination to re-aromatise the ring, leaving us with a 2-substituted pyridine. 

If the acetate is hydrolysed (water, or mild aqueous base), the 2-hydroxy pyridine is formed, which may tautomerise to the more stable 2-pyridnone. 


  


It's worth noting that this mechanism differs significantly from the Boekelheide reaction (the initial pyridine has an additional methyl group at the 2-position), whereby the acetate bound to nitrogen is migrated by means of a [3,3] sigmatropic rearrangement. This rearrangement isn't possible/favourable in this case, as the [3,3] requires a 6 membered transition state, whereas without the methyl group only a 5 membered TS exists.