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enter image description here

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:

enter image description here

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.

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    $\begingroup$ Cyclobutanes are very highly strained, more strained than cyclopropanes. If they can rearrange they will. $\endgroup$
    – Waylander
    Jun 17, 2021 at 12:38
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    $\begingroup$ Even into three membered rings? I thought three membered are even more strained... Does dancing resonance(sigma resonance) operate here? $\endgroup$
    – Prajwal Tiwari
    Jun 17, 2021 at 13:03
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    $\begingroup$ 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. $\endgroup$
    – Prajwal Tiwari
    Jun 17, 2021 at 15:48
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    $\begingroup$ @PrajwalTiwari dancing resonance is not an officially accepted term for cyclopropyl methyl carbocation, stick to sigma resonance in the future. $\endgroup$
    – napstablook
    Jun 18, 2021 at 3:02
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    $\begingroup$ 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? $\endgroup$
    – Prajwal Tiwari
    Jun 19, 2021 at 13:13

1 Answer 1

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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.

Once cation formation occurs the reaction will interestingly, result in a ring contraction because the carbocation formed would be very stable due to this kind of resonance.

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:-

enter image description here

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