k -oc(ch3)3 is a great base for elimination reactions, but it does have one issue: if there are two or more possible products, the less substituted double bond is the major product. this is the opposite of what zaitsev says. why is this the case with k -oc(ch3)3? group of answer choices negative oxygens are strong bases. potassium gets involved in the reaction. the tertiary butyl group has too much steric hindrance for the oxygen to get to the more substituted b-hydrogen. less substituted double bonds are more stable than more substituted double bonds all of these answers are correct
