1
$\begingroup$

What is the product and mechanism when 2‐hydroxy‐6‐methyl‐5‐nitrocyclohepta‐2,4,6‐trien‐1‐one is reacted with $\ce{NaOH}$?

2‐hydroxy‐6‐methyl‐5‐nitrocyclohepta‐2,4,6‐trien‐1‐one + NaOH

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?

$\endgroup$
  • 2
    $\begingroup$ It may be worthwhile drawing the dicarbonyl tautomer and thinking of a possible C-H deprotonation, and also considering the mechanism of the Favorskii rearrangement. $\endgroup$ – Waylander Apr 9 '20 at 10:58
  • 2
    $\begingroup$ Electrocyclization would get you to the Favorskii intermediate. Alternatively, the @Waylander diketone may undergo a benzylic acid rearrangement. $\endgroup$ – user55119 Apr 9 '20 at 19:13
5
$\begingroup$

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, 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

$\endgroup$
  • $\begingroup$ I no longer have Scifinder access to see if this is a literature example. If anyone who does would care to post the reference, please do. $\endgroup$ – Waylander Apr 10 '20 at 11:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.