I saw a question on how to predict the product of a reaction between (hydroxymethyl)cyclobutane and HBr. The correct answer was given as 1-bromocyclopentane, however, I do not understand why a ring expansion would occur in the reaction.

My attempt: I would think that the final product would by (bromomethyl)cyclobutane. This is due to the fact that HBr would first protonate the alcohol, causing it to become a good leaving group. Afterwards, an $\ce{S_N2}$ reaction between this protonated alcohol and $\ce{Br-}$ would take place, as the substrate is a primary alcohol, where no ring expansion/carbocation formation could occur as the steps are concerted, resulting in (bromomethyl)cyclobutane.

Why is this mechanism wrong, and why does a ring expansion occur instead?

  • $\begingroup$ Reduce ring strain. Also secondary carbocation is more stable than primary, so there's a double driving force for rearrangement via ring expansion. There might be a duplicate somewhere, I'm not sure. $\endgroup$ – orthocresol Jul 24 '17 at 5:55
  • $\begingroup$ @orthocresol Would the reaction not occur in a concerted SN2 mechanism, in where there is no chance of C+ rearrangement? The rearrangement you suggested would involve a SN1 mechanism, which would not occur due to the fact that the alcohol is primary, right? Or am I misunderstanding something here? $\endgroup$ – Teoc Jul 24 '17 at 5:58
  • 3
    $\begingroup$ True. The rearrangement can still occur in the same step as loss of H2O. $\endgroup$ – orthocresol Jul 24 '17 at 5:59
  • $\begingroup$ Plus intramolecular is faster than intermolecular with bromide $\endgroup$ – Beerhunter Jul 24 '17 at 11:48
  • $\begingroup$ chemistry.stackexchange.com/questions/29559/… $\endgroup$ – Mithoron Jul 24 '17 at 11:53

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