Does the reaction of 2‐hydroxy‐6‐methyl‐5‐nitrocyclohepta‐2,4,6‐trien‐1‐one with NaOH involve ring contraction?


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> What is the product and mechanism when 2‐hydroxy‐6‐methyl‐5‐nitrocyclohepta‐2,4,6‐trien‐1‐one is reacted with
>  $\ce{NaOH}$?
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> [![2‐hydroxy‐6‐methyl‐5‐nitrocyclohepta‐2,4,6‐trien‐1‐one + NaOH](https://i.stack.imgur.com/vI8GT.png)](https://i.stack.imgur.com/vI8GT.png)
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The only reagent used is $\ce{NaOH}$ so I suppose it abstracts the acidic hydrogen from the $\ce{-OH}$ group. A nucleophilic attack at the carbonyl carbon is also possible, as it could form a tropylium like cation. But I am not sure as to how a ring contraction occurs. The final answer given is 3-methyl-4-nitrobenzoic acid. How does the formation of an acid group and the ring contraction take place?




Drawing on the suggestion of @user55119 I propose the mechanism shown below. 


Whether the first intermediate bicyclic [4.1.0] heptanone arises from an intramolecular electrocyclic process or an intramolecular aldol is unclear but either seems possible. Once the highly strained bicyclic [4.1.0] heptanone intermediate is formed, the reaction proceeds with OH- attack on the carbonyl in a process analogous to a Favorskii [mechanism here](https://www.organic-chemistry.org/namedreactions/favorsky-reaction.shtm), with the second intermediate anion stabilised by the nitro group which also directs the direction of ring opening. Finally loss of OH- in a manner similar to base-catalysed aldol and driven by re-aromatisation gives the observed product. 
[![enter image description here](https://i.stack.imgur.com/ckAO5.png)](https://i.stack.imgur.com/ckAO5.png)


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