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

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    $\begingroup$ 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? $\endgroup$ Jan 28 '14 at 9:22
  • $\begingroup$ Yes.This is not possible.$Br^+$ and$Cl^+$ cation cannot exist as free ions leaving the aromatic ring in electrophilic substitution $\endgroup$ Jan 28 '14 at 14:03
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    $\begingroup$ @KamleshPatel please substantiate your answers with sources, facts, reasonings… $\endgroup$
    – F'x
    Jan 28 '14 at 20:17
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    $\begingroup$ I'm voting to close this question as off-topic because it is based on wrong premises. $\endgroup$ May 8 '15 at 6:59
  • $\begingroup$ onlinelibrary.wiley.com/doi/10.1002/hlca.19720550633/abstract $\endgroup$
    – 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.

  • $\begingroup$ No.I meant $Br^+$ and $Cl^+$. $\endgroup$ Jan 28 '14 at 7:59
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    $\begingroup$ @DevgeetPatel, apologies, I came to the wrong conclusion. That said, I think your question requires clarification/rephrasing. The halogens do not form isolated cations under typical reaction conditions, and certainly not when acting as leaving groups. In electrophilic aromatic substitution reactions, the halogens react with Lewis acid catalysts to form highly reactive zwitterionic complexes. While these complexes may contain halogen atoms with positive formal charges, at no point do said atoms act as cationic leaving groups. Given that, it's unclear to me what you're really asking. $\endgroup$
    – Greg E.
    Jan 28 '14 at 8:33
  • $\begingroup$ If possible(hypothetical case considered) which one will be better? $\endgroup$ Jan 28 '14 at 8:38

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