Final answer:
In E2 reactions with sterically hindered bases, the more substituted alkene is usually the major product. This is due to the steric hindrance favoring elimination over substitution for sterically encumbered substrates.
Step-by-step explanation:
In an E2 reaction using a sterically hindered base, the more prominent product is the one resulting from elimination where the double bond is formed in the least hindered manner, typically leading to the more substituted alkene. This is because a bulky base is better at removing a proton leading to the more stable alkene, which in the case of E2 reactions, is often the more substituted alkene due to the extra stabilization from adjacent carbon groups (alkyl groups). The E2 mechanism occurs through a concerted transfer of electron pairs resulting in the formation of a pi bond. Steric hindrance at the base or substrate, as well as stronger bases, favor the E2 mechanism over SN2 because they inhibit backside attacks required for SN2 but promote elimination.
When tertiary substrates are present, they tend to undergo E2 reactions due to steric crowding around the reaction center which makes SN2 reactions less favorable. In contrast, primary or methyl substrates are more likely to undergo SN2 due to the lack of steric hindrance, promoting nucleophilic substitution over elimination.