# Which is the major product of bromination of 3-(dimethylamino)benzamide, ortho or para?

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.

• 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. – Buttonwood Jun 26 at 8:36

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: