2
$\begingroup$

Ring expansion of vinyl cyclobutane -

enter image description here

Could someone please help me with a detailed mechanism for the above conversion?

It is the acid-catalyzed rearrangement of 4,7,7-trimethylbicyclo[3.2.0]hept-3-en-6-one to 3,3-dimethyl-6-methylidenebicyclo[2.2.1]heptan-2-one.

I couldn't find the mechanism anywhere, and couldn't propose a good one either. Please help.

P.S. As one of the users in the comments is interested in seeing my best effort, I'll explain my thought process- As the carbonyl oxygen is the most nucleophilic, it is the first to be protonated in acidic medium. Furthermore, this species has a resonating structure (less contributing, but the only one that can lead to a ring expansion, I believe) – one obtained by shifting the π bond towards oxygen – a carbocation. After this, I'm clueless. I think, that the expansion could have something to do with the hydrogen at alpha position to the sp² hybridised carbon atom in the five membered ring – as the product has an exocyclic double bond.

$\endgroup$
0

1 Answer 1

6
$\begingroup$

You are correct about protonating the carbonyl group. [The first of your products is the same as the reactant.] The cyclobutanone 2 was prepared from a mixture of geranic acids. (J. J. Beereboom, J. Org. Chem., 1965, 30, 4230). The structures of 2, 4 and 5 were determined by Beereboom. I am providing the most reasonable mechanism. Cyclobutanone 2 must have a more stable cis ring juncture owing to its method of formation or exposure to p-TsOH. The trans isomer of 2 cannot give the trans isomer of 3 because of strain. Tertiary carbocation 3 arises by addition of the double bond to the protonated carbonyl. Collapse of the carbocation produces bicyclic ketone 4, which in the presence of p-TsOH isomerizes to the exocyclic isomer 5. In general, 1,1-disubstituted double bonds are less stable than tri-substituted ones. In this case, ring strain may be present in the endocyclic alkene 4.

![enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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