An organic chemistry student conducts the reaction of 2-bromobutane with sodium ethoxide carefully d — Haloalkanes and Haloarenes Chemistry Question
Question
An organic chemistry student conducts the reaction of 2-bromobutane with sodium ethoxide carefully dissolved in ethanol. At $25^\circ C$, the product mixture predictably contains a notable ratio heavily favoring 2-ethoxybutane over butenes. When the identical reaction is aggressively repeated at $80^\circ C$, the ratio overwhelmingly shifts in favour of the butene elimination products. Which of the following kinetic and thermodynamic statements structurally correct this specific, dramatic shift?
💡 Solution & Explanation
A) True. Bimolecular elimination ($E2$) generally requires physically breaking two strong sigma bonds ($C-X$ and $C-H$) and organizing them to form a pi bond simultaneously, whereas substitution primarily involves breaking only one sigma bond and forming another. Consequently, the activation energy ($E_a$) for elimination is typically strictly higher than for substitution. C) True. This explains why the activation energy is higher. B) True. According to the Arrhenius exponential relationship ($k = Ae^{-E_a/RT}$), reactions possessing higher activation energies are much more exquisitely sensitive to temperature changes. When the temperature is increased from $25^\circ C$ to $80^\circ C$, the rate constant for the sluggish $E2$ pathway increases exponentially faster than the rate constant for the $S_N2$ pathway, allowing elimination to abruptly overtake substitution. D) False. Higher temperatures typically shift reactions toward thermodynamic control, not kinetic control. Elimination is thermodynamically favored at high temperatures due to massive entropy gains.