My book says that electrophilic bromination of toluene (methylbenzene) is faster than that of ethylbenzene. How is that possible?

I am doubting it because ethyl is a better $+\ce{I}$ group than methyl, so ethyl would increase the electron density in the benzene ring more than methyl. As a result, bromine, which is an electrophile, should react with ethylbenzene faster than methylbenzene.

So, why is the bromination of methylbenzene faster than ethylbenzene and please give the reaction mechanism, if possible?


2 Answers 2


First, I think it's probably important to note that the difference in $\text{+I}$ effect of a methyl vs. ethyl is very small. In principle, the ethyl is more electron-rich, however the strength of the inductive effect decays very rapidly with increasing distance. Ultimately, I suspect variance in steric factors, solvation, and electron-donating ability to the $\pi$-system of the ring by hyperconjugative overlap are the decisive factors, not the difference in induction.

If we consider sterics, the ethyl is obviously the bulkier group, and that may partially explain the difference in reactivity. The ethyl, moreso than the methyl, will hinder the approach of the electrophile, and may potentially raise the energy of the transition state and $\sigma$-complex intermediate slightly by:

(a) electrostatic repulsion toward the electrophile, particularly in the case of ortho- attack, and

(b) inhibit potential stabilization of the $\sigma$-complex intermediate by solvent molecules.

From the standpoint of electronic effects, I suspect the differences are marginal. To the extent of my understanding, the energies of $\ce{C-H}$ and $\ce{C-C}$ bonding $\sigma$ orbitals are very similar, and the efficiency of hyperconjugative overlap should consequently be similar as well. I've occasionally encountered claims that $\ce{C-H}$ bonds are slightly better $\pi$-donors, but I'm not aware of any conclusive evidence to that effect. If that is true, however, it might contribute to the observed disparities in reactivity.

For a quantitative approach, you may want to take a look at the substituent constants derived for various alkyl groups from the Hammett equation. The equation was originally developed with the aim of analyzing substituent effects on the thermodynamic equilibria of benzoic acid derivatives. However, since the reactivity of substituted benzoic acid is also affected by the mesomeric and inductive properties of ring substituents (albeit in the opposite sense from a benzene ring in an EAS reaction), I think the Hammett equation furnishes a reasonable proxy measure of at least the relative strengths of substituents as either donors or acceptors of electron density. I found one table of said substituent constants. (Keep in mind that those table values are for ionization of benzoic acid, so a negative value is actually indicative of an electron-donating effect, which is favorable for an EAS reaction but destabilizing to an acid.)

Finally, as to the reaction mechanism, I'm not going to reproduce it here. Every EAS reaction follows a very similar, schematic mechanism. For the mechanisms, take a look, for example, at the Wikipedia articles on electrophilic halogenation and electrophilic aromatic substitution more generally.


Reactivity of alkylbenzenes towards electrophillic substitution reactions is decided only on the basis of steric hinderence that alkyl groups provides to aromatic ring.if steric hinderence increases reactivity of aromatic ring decreases .steric hinderence by alkyl groups increases in the order -CH3 < 1¤ < 2¤ < 3¤ . Therefore reactivity decreases in the order CH3 > 1¤ > 2¤ > 3¤ . E.g. toluene > ethylbenzene

  • $\begingroup$ This doesn't really add much that hasn't already been said in the previous answer. $\endgroup$
    – bon
    Jan 21, 2016 at 14:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.