Answer:
see explanation below
Explanation:
In the first case, we have a reaction where we have the 3-chloro-3-methylpentane reacting with t-butoxide. The t-butoxide is a very voluminous base, so the strength of substracting a hydrogen atom is reduced. Therefore, the reaction taking place here will be an E2 but instead of substracting the hydrogen from the carbons 2 or 4, it will substract it from the methyl group, cause it has less steric hindrance there and the reaction will go faster.
In the second case, the sodium ethoxide is a strong base, so it will rapidly substract an atom of hydrogen from carbon 2 or 4 to form the (Z) - 3 - methyl - 2- pentene and the substitution product.
Look picture for mechanism and products.
