[123456789]

Bromocyclopentane is reacted with CH₃O^- in a solvent of ether. Will the reaction mechanism be Sn2 or E2? I know the answer is both, around 50-50 for each. However, what I was thinking is that for Sn2 reaction, the methoxide will need to attack from the "backside" (ie from inside of the cyclopentane ring), which will not be that easy compared to E2, where the methoxide does not need to get into the centre, as it is going to have electron cloud repulsion and bulkiness nonetheless. So why is this train of thoughts wrong?

Also, how much does the size of the cyclo ring matter? (will E2 be more favoured if methoxide is reacted with Bromocyclobutane instead?)

Thank you. (and I really think this is a very very helpful forum-salute to all:D)

[fledarmus]

What I have been telling a lot of people recently - build a model! And remember the puckered-ring shape of a saturated cyclopentane ring. What is the trajectory of your CH₃3O^- to get the back-side attack needed for an S_N2 reaction? How do the Br and H on the ring need to line up to get the orientation required for an E2 reaction and how much will this change the ring strain?

[rrr_asd]

In polar solvents the dominant mechanism is Sn1.
I think, cyclopentyl carbocation will be stable.

[orgopete]

I have been quoting Brown, Foote, and Iverson in suggesting that nucleophiles with pKa of greater than 12 will give elimination reactions. So this would fit that example. Yet, this secondary bromide is suggested to give 50% substitution. Can someone give a reference to this reaction?

I certainly cannot explain why the B, F, and I rule of thumb should hold. Indeed, one might expect that anions should be better nucleophiles. Why should a strong base or any strong base have an increased affinity for a proton to give an elimination?