Consider the following six rigorous experimental combinations of pure alkyl halides and specific rea — Haloalkanes and Haloarenes Chemistry Question
Question
Consider the following six rigorous experimental combinations of pure alkyl halides and specific reagents thoroughly dissolved in polar aprotic solvents: 1. 1-bromobutane + $CH_3S^-Na^+$ 2. 2-bromo-2-methylpropane + $CH_3O^-Na^+$ 3. 1-bromo-2,2-dimethylpropane + $CH_3O^-Na^+$ 4. 2-bromobutane + $(CH_3)_3CO^-K^+$ 5. 1-bromobutane + $(CH_3)_3CO^-K^+$ 6. Bromomethane + $HO^-Na^+$ Exactly how many of these specific reaction mixtures will predominantly and fundamentally yield an elimination product ($E2$) rather than a standard substitution product ($S_N2$)?
💡 Solution & Explanation
We must evaluate each substrate/reagent pairing: 1) $1^\circ$ halide + good nucleophile/weak base $\rightarrow$ predominantly $S_N2$. 2) $3^\circ$ halide + strong base ($CH_3O^-$) $\rightarrow$ exclusively $E2$ because backside attack is physically completely blocked. 3) Neopentyl halide ($1^\circ$ but extremely sterically hindered) + strong base $\rightarrow$ highly resistant to $S_N2$ due to intense $\beta$ -branching blocking the backside, but vitally, it lacks any $\beta$ -hydrogens. It literally cannot undergo $E2$. It slowly undergoes substitution or rearranged $S_N1/E1$ under different conditions. 4) $2^\circ$ halide + bulky strong base ($t-BuO^-$) $\rightarrow$ predominantly $E2$ (the bulky base forces elimination over substitution). 5) $1^\circ$ halide + massively bulky strong base ($t-BuO^-$) $\rightarrow$ predominantly $E2$. While $1^\circ$ inherently favors $S_N2$, $t-BuO^-$ is so spectacularly bulky that it acts almost exclusively as a base, attacking the highly exposed $\beta$ -hydrogens rather than navigating to the carbon. 6) Methyl halide + strong base/nuc $\rightarrow$ exclusively $S_N2$ (cannot eliminate, it lacks a $\beta$ -carbon). The predominant $E2$ mixtures are 2, 4, and 5. Total = 3.