# Ring expansion in the reaction of (hydroxymethyl)cyclobutane and HBr

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?

• 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. – orthocresol Jul 24 '17 at 5:55
• @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? – Cyclohexanol. Jul 24 '17 at 5:58
• True. The rearrangement can still occur in the same step as loss of H2O. – orthocresol Jul 24 '17 at 5:59
• Plus intramolecular is faster than intermolecular with bromide – Beerhunter Jul 24 '17 at 11:48
• chemistry.stackexchange.com/questions/29559/… – Mithoron Jul 24 '17 at 11:53