What would be the order of electron density in the benzene ring in the following compounds?

Methyl Benzene

Ethyl Benzene

t-butyl benzene

I would say that t-butyl benzene would have the highest electron density as the alkyl group attached to it is bulkier than methyl and ethyl thus it will exert a greater +I effect pushing the electrons into the ring making it electron rich. Using same analogy ethyl benzene would have a greater electron density in the ring compared to methyl benzene.

However that is not the case. Something counter intuitive seems to be happening here. The order is methyl benzene >ethyl benzene >t-butyl benzene.

I think this phenomenon can somehow be explained using hyper conjugation, but i don't get it how?

  • $\begingroup$ The differences are small, but consistent with hyperconjugation arguments. If we look at the relative rates of electrophilic nitration as a measure of electron density we find benzene=1, toluene=24, tert-butylbenzene=15.7. See this earlier Q & A. $\endgroup$ – ron Dec 28 '16 at 15:48

Well, tetrahydrocannabinol, you said it correct only the correct reason/explanation is hyperconjugation only. More the number of (alpha) hydrogen atoms on a C=C [isolated as in alkenes or in resonance as in benzene], more the possibility of hyperconjugation.

In benzene, this will shift the electron density from C-H bond of CH3 group towards the ring, making it electron rich.

Contrary to what could have been thought from +I effect of corresponding alkyl groups on the ring, hyperconjugation is more in CH3.

Since hyperconjugation involves complete shifting of electron from valence shell of atoms to the other, it is more "effective" that INDUCTIVE which just pushes electron little more towards more EN (more correctly, towards group with more -I) atom. Hyperconjugation in alkyl benzenes

  • $\begingroup$ First of all: C-C bonds are able to engage in hyperkonjugation, so "NO hyperconjugation possible" is just wrong. And Second: the drawing is terribly wrong, since no bonds are broken in case of hyperconjugation, and for hyperconjugation you need a certain orientation of the orbitals to each other, for a methyl group only one CH bond can really engage in hyperconjugation, the other two aren't aligned very well. $\endgroup$ – DSVA Dec 28 '16 at 18:13
  • $\begingroup$ @DSVA I will agree with your ‘first’. Obviously, as ron commented on the question, tert-butylbenzene has the highest electron density so there must be hyperconjugation. Your ‘second’, however, is not correct. The depiction as shown by Che Mistry shows resonance structures connected by mesomery and not reactions. Understand the difference between reaction and mesomery arrows. (However, I’ll admit that the difference could be made clearer in the answer.) $\endgroup$ – Jan Dec 28 '16 at 22:55
  • $\begingroup$ @Jan even if this would be a mesomeric structure (I never seen an actual sigma bond shifted while describing mesomeric structures) it doesn't make sense and is a mesomeric structure that shouldn't be used since it's extremly far away from the actual structure. Hyperconjugation is too weak to justify such a form. $\endgroup$ – DSVA Dec 28 '16 at 23:51
  • $\begingroup$ @DSVA Feel free to downvote all answers in which I use these resonance forms to explain hyperconjugation. $\endgroup$ – Jan Dec 28 '16 at 23:54
  • $\begingroup$ @Jan well, I like to learn, any high level textbook or paper doing it this way? It just doesn't make sense to me, I mean the same way I could write H- "ethene" H+ as resonance structure for ethane, but that's just not very useful at all. $\endgroup$ – DSVA Dec 29 '16 at 0:01

It's simply due to hyperconjugation effect. As hyperconjugation effect is considered stronger than inductive effect therefore before seeing the inductive effect of methyl, ethyl and t butyl, we should check it's hyperconjugation via the number of alpha hydrogens. In the first case, it's 3, In the second case, it's 2' and In the third case, it's 0. Hence 3 > 2 > 1

  • $\begingroup$ Both C-C and C-H bonds can act as hyperconjugative donors. So your reasoning is not justified. $\endgroup$ – Tan Yong Boon Jan 9 '18 at 5:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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