Final answer:
The E2 mechanism is a concerted, bimolecular elimination reaction where a base removes a proton and a leaving group departs simultaneously, forming a double bond. It is especially common with tertiary alkyl halides due to steric hindrance that makes S₂ reactions less likely. The reaction rate depends on both the substrate and the base.
Step-by-step explanation:
The E2 mechanism is a type of elimination reaction where a base abstracts a proton, and in the same concerted step, a leaving group (usually a halide) departs. This results in the formation of a double bond. The key characteristic of an E2 reaction is that it is bimolecular, meaning the rate of reaction depends on the concentration of both the substrate (the alkyl halide) and the base. Importantly, E2 occurs in a single, concerted step, making it a concerted mechanism. Steric hindrance affects the reactivity in E2 in the order: methyl < primary < secondary < tertiary. In particular, tertiary alkyl halides are prone to react via the E2 mechanism due to the difficulty that nucleophiles face in approaching the crowded tertiary centers, making substitution reactions less favourable.
When comparing E2 to other elimination mechanisms like E1, the primary distinction is that E1 occurs in a step-wise fashion and involves a unimolecular rate-determining step; the formation of a carbocation intermediate. For tertiary substrates, which cannot form stable carbocations, E2 becomes the preferred mechanism. Polar protic solvents do not favor S₂ and E2 reactions, as they can stabilize the intermediates or transition states involved in competing SN1 and E1 mechanisms.