A chemist heats 2-bromo-2-methylpropane with aqueous KOH. Why does increasing the temperature shift the reaction preference from substitution toward the elimination product?
Q1. A chemist heats 2-bromo-2-methylpropane with aqueous KOH. Why does increasing the temperature shift the reaction preference from substitution toward the elimination product?
Answer: Elimination is favored because it is more sensitive to temperature increases.
Explanation: High temperature and strong bases favor elimination over substitution because elimination has a higher activation energy and increases entropy.