Why are fluorides more reactive in nucleophilic aromatic substitutions than bromides?

Question

In nucleophilic aromatic substitution reactions, why do fluorides react faster than bromides? Ordinarily bromide is a better leaving group than fluoride, e.g. in $\\mathrm{S_N2}$ reactions, so why isn't this the case here? The only thing I can think of is that fluorine is more electron-withdrawing (via the inductive effect), which could stabilise the Meisenheimer complex formed as an intermediate.

Answer 1

What it comes down to is the initial nucheophilic attack on the aromatic ring. Bromine has an electronegativity only slightly larger than carbon (2.96 vs. 2.66 respectively). Thus, there is very little partial positive charge on the electrophilic carbon.

Fluorine, on the other hand, is significantly more electronegative than carbon (3.98 vs. 2.66 respectively). The result is that there is a much larger partial positive charge on the substituted carbon due to the significantly more polar C-F bond.

We can safely assume that elimination from the Meisenheimer complex is quite fast in both cases, so the rate determining step must be the initial nucleophilic attack. Electron rich nucleophiles (hydroxide, for example) will react more quickly with the more electropositive carbon (the fluorinated one). Therefore, the more favorable the initial attack, the faster the reaction. Taken together, we can now understand how aryl fluorides react more quickly in SnAr reactions compared to aryl bromides.