An advanced organic chemist is heavily evaluating the following five distinct negatively charged ion — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

To strictly physically maximize $S_N2$ on a moderately hindered $2^\circ$ substrate and entirely avoid $E2$, the chosen reagent must fiercely be a very fast, excellent nucleophile but a decidedly very weak base. Strong bases far too eagerly abstract exposed protons to form unwanted alkenes. Let's meticulously evaluate the available ions: 1) $F^-$: Noticeably poor nucleophile in typical protic solvents due to immense hydrogen bonding, and a weak/moderate base. Not ideal. 2) $CH_3O^-$: Highly strong nucleophile, remarkably strong base (highly unstable conjugate of the weak acid $CH_3OH$). This ion strongly, fiercely promotes $E2$ on $2^\circ$ halides. 3) $I^-$: Excellent, incredibly fast nucleophile (very large, highly polarizable), very incredibly weak base (highly stable conjugate of the very strong acid $HI$). Absolutely ideal for $S_N2$. 4) $CN^-$: Fast, excellent nucleophile, physically weak base (stable conjugate of $HCN$). Extremely ideal for $S_N2$. 5) $NH_2^-$: Massively, powerfully strong base (highly unstable conjugate of the incredibly weak acid $NH_3$). Massively promotes $E2$. Only $I^-$ and $CN^-$ perfectly physically fit the highly precise chemical profile required to strictly force substitution on a secondary halide. Total count = 2.