Final answer:
When CH3-Br reacts with a strong bulky base, it undergoes an E2 elimination reaction forming an alkene, with the base abstracting a proton and the bromide ion leaving, influenced by steric hindrance that makes substitution reactions less favorable.
Step-by-step explanation:
When CH3-Br reacts with a strong bulky base, it typically undergoes an E2 elimination reaction. In this bimolecular reaction, the base abstracts a proton (–H) from the carbon atom adjacent to the carbon bonded to bromine, simultaneously the carbon-bromine bond breaks, and the bromide ion serves as the leaving group. This leads to the formation of an alkene as the double bond is established between the two carbons as a result of the elimination process. Strong bulky bases favor E2 mechanisms because they hinder substitution reactions (such as SN2) due to steric hindrance; they are less able to approach the electrophilic carbon atom directly to displace the leaving group.The reaction rate for this process is represented by the equation: rate = k" [CH3 Br][OH-], indicating a second-order reaction. One might infer that, with a strong and bulky base, this reaction would proceed efficiently towards the elimination product given the base's strength and the favorable reaction conditions for E2 processes.