Hi! Can I please ask all the possible products for this reaction mechanism?

Question

Reaction mechanism of 1-Bromo-2-Butene with CH3OH. I'm aware it can have two products but how and why?

Answer 1

First of all, methanol is a highly ionizing solvent being in the polar protic category of solvents that typically assist in carbocation formation following the SN1 route. The carbon bearing the bromine leaving group is primary allylic in this case (in the absence of the double bond, the reaction pathway would be SN2 for the saturated primary bromide). As such, one can construct a more stable resonance formed secondary allylic carbocation by participation of the double bond. Methanol can then act as a weak nucleophile and attack both allylic carbocations to form intermediate oxonium ions. After deprotonation, one would end up with two different ether products, namely3-methoxy-1-butene (major) and 1-methoxy-2-butene (minor).