Which site is more favorable for electrophilic aromatic substitution in O-Acetotoluidide?

The answer given in my textbook says the site para to acetamide group is more favorable, however I believe the site para to methyl group should be more favorable. Here are the reasons:

1) Because of sterric inhibition of resonance, acetamide group will be slightly out of plane thereby decreasing the quality of resonance.

2) Because of hyperconjugation by 3 alpha hydrogens, electron density will increase 3 times each at sites ortho and para to methyl group. This means that methyl group due to hyperconjugation will lead to nine resonance structures as opposed to just three in acetamide.

So, because of these two reasons, I think site ortho to methyl group should be more favorable(ortho will be less favorable because of sterric hindrance).

Am I missing something or the answer given in the book is wrong?


  • $\begingroup$ During my entire high school, I never really took SIR into consideration as long as there weren't at least two methyl groups on both sides of the functional group (acetamide in this case). Also, the site para to acetamide is activated due to +M effect, which is generally dominant over hyperconjugation from methyl. $\endgroup$ – Gaurang Tandon Jan 4 '18 at 6:34
  • $\begingroup$ "Resonance effect is dominated over hyperconjugation" is just a generalisation and I don't think it should be used here. Here the hyperconjugation leads to 9 resonating structures as opposed to just three with acetamide. $\endgroup$ – Arishta Jan 4 '18 at 6:36
  • $\begingroup$ Because of SIR effect, even with one methyl group, substituted benzoic acid is more acidic than benzoic acid ; substituted aniline is less basic than aniline. This means that just one methyl group can cause significant sterric hindrance and make the bulky group go out of plane leading to decrease in quality of resonance. $\endgroup$ – Arishta Jan 4 '18 at 6:41
  • $\begingroup$ Why are you counting the number of resonating structures? We are not to find the stability of the overall molecule. We are to find which carbon of the benzene ring is most activated. We've narrowed our list to (1) para to methyl (2) para to acetamide. The 2nd position is activated by +M, while the 1st position is activated by hyperconjugation. I wish to say that the major product will be formed from 2nd position (as +M dominates over hyperconjugation). $\endgroup$ – Gaurang Tandon Jan 4 '18 at 6:44
  • $\begingroup$ "substituted benzoic acid is more acidic than benzoic acid" yes that's true. You're right then. I'm not sure of the final answer. Sorry for disturbing. $\endgroup$ – Gaurang Tandon Jan 4 '18 at 6:45

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