# Which is better leaving group [closed]

How can I compare the leaving group ability of $\ce{Br^+}$ and $\ce{Cl^+}$ ions for the electrophilic substitution reactions for aromatic compounds?

## closed as off-topic by Klaus-Dieter Warzecha, user15489, Gowtham, Burak Ulgut, FreddyMay 11 '15 at 4:16

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• I've never heard about halonium atoms as leaving groups in electrophilic aromatic substitutions. Would you mind to provide a sketch of a reaction you're thinking of? – Klaus-Dieter Warzecha Jan 28 '14 at 9:22
• Yes.This is not possible.$Br^+$ and$Cl^+$ cation cannot exist as free ions leaving the aromatic ring in electrophilic substitution – Kamlesh Patel Jan 28 '14 at 14:03
• @KamleshPatel please substantiate your answers with sources, facts, reasonings… – F'x Jan 28 '14 at 20:17
• I'm voting to close this question as off-topic because it is based on wrong premises. – Klaus-Dieter Warzecha May 8 '15 at 6:59
• onlinelibrary.wiley.com/doi/10.1002/hlca.19720550633/abstract – Mithoron May 8 '15 at 11:21

I presume you're referring to $\ce{Br-}$ and $\ce{Cl-}$ in aromatic nucleophilic substitution reactions. In that case, $\ce{Br-}$ is typically the superior leaving group to $\ce{Cl-}$, due to the former's larger size lending it greater polarizability (i.e., the ability to disperse the negative charge over a greater area). In general, when comparing elements in the same group, the dominant factor in determining anion stability is the larger size as one moves down the group. This largely accounts for the order of acidity of the halogen-containing binary acids (i.e., $\ce{HI} > \ce{HBr} > \ce{HCl}$), the greater acidity of thiols over alcohols, etc. Solvent effects are relevant, since smaller anions are more easily and thoroughly solvated in protic solvents. However, as far as I'm aware, this effect is rarely decisive, and $\ce{Br-}$ usually remains the better leaving group.
That said, the rate-determining step in nucleophilic aromatic substitution reactions is not the loss of the leaving group, but rather the formation of the negatively charged Meisenheimer complex resulting from the initial nucleophilic attack. This intermediate is extremely high in energy due to the temporary loss of aromaticity it entails. Thus, the reaction is largely controlled by kinetics, and therefore dependent upon the stability of the intermediates and transition states. $\ce{Cl-}$ is more electronegative than $\ce{Br-}$, and therefore better stabilizes the negative charge in the intermediate complex by induction ($-I$ effect). Consequently, we can expect the chlorobenzene to be more reactive in practice under most conditions, with the disparity in leaving group quality being less important. A cursory search yielded one reference confirming this order, but unfortunately I don't have access at the moment to more authoritative primary sources.
In summary, while $\ce{Br-}$ is the superior leaving group, its loss is not the rate-determining step. Hence, chlorobenzenes will in fact be more reactive due to chlorine's superior ability to inductively stabilize the intermediate anionic complex.
• No.I meant $Br^+$ and $Cl^+$. – Devgeet Patel Jan 28 '14 at 7:59