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I know that A will most probably not happen because of high steric hindrance. However between B and C, will B become the major product? Because $\ce{-N(CH3)2}$ is less bulky compared to $\ce{C=O-NH2}$, so there will be less steric hindrance.

  • $\begingroup$ With E already twice substituted, the use of ortho, meta, para to designate the entry of the new (then third) substitutent requires to be a bit more specific. You have luck with A, Br were introduced in a position that is ortho to both $\ce{NMe2}$ and the amide. This contrasts to B, for example, were Br is ortho to $\ce{NMe2}$, and para to the amide. $\endgroup$
    – Buttonwood
    Jun 26, 2020 at 8:36

1 Answer 1


Here's the 3D structure of the substrate:


As you can see, the $\ce{-N(Me)2}$ group is clearly larger than the $\ce{-CONH2}$ group. Also notice that the $\ce{-N(Me)2}$ group is planar with the benzene ring. This is because this structure is the most favourable for $\mathrm{p-\pi}$ conjugation 1.

Your guess that A would not form is correct. The large size of the $\ce{-N(Me)2}$ group would also point to the fact that C is the major product. For a better explanation of the major product, we can take a look at the 3D structures of B and C:

Product B Product C

You can clearly see that in B, the bromine atom pushes the dimethylamine group out of it's stable configuration, where it is coplanar with the benzene ring. This requires more energy and also produces an unstable product. Therefore, C is the major product in this reaction.


  1. Novikov, V. P., et al. “Molecular Structure and Conformation of N,N-Dimethylaniline by Gas-Phase Electron Diffraction and Quantum-Chemical Calculations.” Russian Journal of General Chemistry, vol. 74, no. 8,2004, pp. 1247–53. doi:10.1007/s11176-005-0146-9.
  • 1
    $\begingroup$ This answer relies on steric hinderance to figure out which product may be formed. My anticipation was to see an answer equally considering that electronic effects by the amine (electron push) and the amide (electron withdrawing) leading 1) to a preferred direction substitution pattern by either one if the other were absent. And then 2) «because both amine and amide already are present ...» the preferred entering direction of Br as third substitutent. Vaguely I recall seeing a table where these effects of an electrophilic aromatic substitution were quantified, too. $\endgroup$
    – Buttonwood
    Jun 26, 2020 at 8:53

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