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What are the products when $\ce{HClO4}$ attacks this compound? structure of compound

I expect $\ce{HClO4}$ to protonate the ketone, forming an (aromatic?) carbocation. I don't know where to go from here. Answer suggests formation of butanoic acid and production of $\ce{CO2}$. Unsure what really happens here. All hints appreciated.

This question appeared in the JEE 1997.

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    $\begingroup$ That is one very strained system, so once it protonates it is going to rearrange $\endgroup$
    – Waylander
    Commented Apr 21, 2020 at 18:38
  • $\begingroup$ @Waylander thanks for your insight - it would be amazing if you could elaborate yourself further; what rearrangement do you see happening? Do you see ring cleavage? $\endgroup$
    – McSuperbX1
    Commented Apr 21, 2020 at 18:40
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    $\begingroup$ I wonder where the extra oxygens come from if you are getting butanoic acid and CO2? Presumably the perchloric acid is an aq solution. $\endgroup$
    – Waylander
    Commented Apr 21, 2020 at 19:53
  • $\begingroup$ See this prior thread on StackExchange at chemistry.stackexchange.com/questions/31670/… $\endgroup$
    – AJKOER
    Commented Apr 21, 2020 at 23:57
  • $\begingroup$ AJKOER, I've already seen that question as well as the accepted answer - but I'm unable to see how anything from there could possibly help me with this question.... :( @Waylander If we assume that the acid is an aqueous solution, what do you feel may happen? I'm unable to see what potential difference it could make... $\endgroup$
    – McSuperbX1
    Commented Apr 22, 2020 at 6:44

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Assuming that your end products are correct I think this reaction would proceed in a similar manner to the way it will proceed if you use hot $\ce{KMnO4}$ as a reagent.

The reason I think that is because $\ce{HClO4}$ is practically similar to $\ce{KMnO4}$ structure and is an even stronger oxidizing agent ($\ce{Cl^{+7}}$ is much more unstable (and hence much more willing to oxidize) than $\ce{Mn^{+7}}$ ion) which should be able to imitate the behavior of hot $\ce{KMnO4}$ reagent (presuming that the greater oxidative strength compensates the level of reactivity $\ce{KMnO4}$ shows at a higher temperature)

If the similarity is established then the reaction would proceed in this manner giving the products matching your answer:-

enter image description here

regarding the formation of aromatic cyclopropenium ion, I think cleavage of alkene relieves the angle strain which provides a higher stability than aromaticity.

N.B. :- Here I have presumed that $\ce{HClO4}$ can behave similar to hot $\ce{KMnO4}$ however I was unable to find a source for that.

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  • $\begingroup$ Here is the source chemistry.stackexchange.com/a/131374/106370 $\endgroup$
    – Rishi
    Commented Jul 8, 2021 at 6:53
  • $\begingroup$ @Rishi it is for $\ce{HIO4}$ and it doesn't suggest that $\ce{HIO4}$ can do oxidative cleavage of alkenes. $\endgroup$
    – napstablook
    Commented Jul 8, 2021 at 6:59

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