The bimolecular elimination (E2) reaction of proceeds significantly faster than that of when treated — Haloalkanes and Haloarenes Chemistry Question
Question
The bimolecular elimination (E2) reaction of $PhCHBrCH_3$ proceeds significantly faster than that of $PhCHBrCD_3$ when treated with alcoholic KOH. What is the fundamental kinetic reason for this experimentally observed difference?
💡 Solution & Explanation
In an E2 reaction, the abstraction of the $\beta$ -proton by the base and the departure of the leaving group happen simultaneously in a single, concerted rate-determining step. Because a C-D bond has a lower zero-point energy and is therefore stronger than a C-H bond, breaking it requires a higher activation energy. This leads to a primary kinetic isotope effect ($k_H/k_D > 1$), explaining exactly why the deuterated substrate reacts at a noticeably slower rate.