# Why does this ring contraction take place in the following nucleophilic substitution? [closed]

While attempting this problem I thought that P1 would simply be Bromine getting substituted by OH through nucleophilic substitution.

P2 would give a product formed through ring expansion as a 4 membered ring would tend to rearrange to a 5 membered ring due to strain factors.(SN1 reaction(acidic conditions))

This is the mechanism I thought of:

This pointed towards option (2) but what's given is option (3).
Why does this happen?

It could be that the four-membered ring being strained rearranges, as Waylander suggests but if it would rearrange, why would it rearrange into a 3 membered ring?
Usually, this happens only when there is a stabilising factor and the only stabilising factor related to a 3 membered ring I can think of, is sigma bond resonance(Dancing resonance)

I would appreciate an explanation of what's happening here.

• Cyclobutanes are very highly strained, more strained than cyclopropanes. If they can rearrange they will. Jun 17 at 12:38
• Even into three membered rings? I thought three membered are even more strained... Does dancing resonance(sigma resonance) operate here? Jun 17 at 13:03
• Can the 'close' voters give some feedback about the post? I have edited it to make the post more clear but I would still appreciate some constructive criticcism so that I can improve. Jun 17 at 15:48
• @PrajwalTiwari dancing resonance is not an officially accepted term for cyclopropyl methyl carbocation, stick to sigma resonance in the future. Jun 18 at 3:02
• Why did this get closed though? I provided all my approaches, thoughts, even added a mechanism that I used, what else should I have provided to make this a better post? Jun 19 at 13:13

This reaction should be leaning towards SN1 because the approach of $$\ce{H2O}$$ for a back attack to the ABMO of the $$\ce{C-Br}$$ is heavily hindered since the neighborhood is congested due to the small ring of cyclobutane.
After the ring contraction the reaction would proceed in a fairly standard nucleophilic attack by $$\ce{H2O}$$ resulting in the $$P_1$$ your answer key alludes to. The overall reaction should proceed in the manner shown below:-