An organic chemist compiles a list of anionic species to precisely evaluate their leaving group abil — Haloalkanes and Haloarenes Chemistry Question
Question
An organic chemist compiles a list of anionic species to precisely evaluate their leaving group abilities in $S_N2$ nucleophilic substitution reactions: (i) $F^-$ (ii) $Cl^-$ (iii) $Br^-$ (iv) $I^-$ (v) $CH_3COO^-$ (acetate) (vi) $OH^-$ (vii) $CF_3SO_3^-$ (triflate) How many of these anions are considered strictly better leaving groups than the bromide ion ($Br^-$) under standard polar aprotic conditions?
💡 Solution & Explanation
A good leaving group is a weak, stable base (the conjugate base of a strong acid). The order of basicity among halide ions is $F^- > Cl^- > Br^- > I^-$. Therefore, the leaving group ability is the reverse: $I^- > Br^- > Cl^- > F^-$. Among the halides, only $I^-$ is a better leaving group than $Br^-$. Looking at the oxygen-based leaving groups, $OH^-$ and $CH_3COO^-$ are relatively strong bases and poor leaving groups. However, the triflate ion ($CF_3SO_3^-$) is exceptionally stable due to extensive resonance across three oxygen atoms and the powerful $-I$ effect of the $CF_3$ group. It is the conjugate base of triflic acid (a superacid), making it a much better leaving group than even the halides. Thus, the two species better than $Br^-$ are $I^-$ and $CF_3SO_3^-$.