The thermodynamic rate of the reaction is extraordinarily sensitive to steric hindrance surrounding — Haloalkanes and Haloarenes Chemistry Question
Question
The thermodynamic rate of the $S_N2$ reaction is extraordinarily sensitive to steric hindrance surrounding the electrophilic carbon. Consequently, which of the following listed alkyl bromides will undergo $S_N2$ displacement at a substantially slower rate than simple ethyl bromide ($CH_3CH_2Br$)?
💡 Solution & Explanation
Ethyl bromide is a $1^\circ$ alkyl halide. A) Methyl bromide is utterly unhindered and reacts vastly *faster* than ethyl bromide. B) Isopropyl bromide ($2^\circ$) features significant $\alpha$ -branching, shielding the backside and slowing the rate. C) Neopentyl bromide ($1^\circ$) has extreme steric bulk at the adjacent $\beta$ -carbon, almost entirely blocking the backside approach, making its $S_N2$ rate practically zero. D) tert-Butyl bromide ($3^\circ$) has massive $\alpha$ -branching that completely prevents $S_N2$ attack, slowing the relative rate to zero.