Question

When N,N-Dimethylaniline is treated with bromine, ortho and para products are observed. However, when N,N-Dimethylaniline is treated with a mixture of nitric and sulfuric acid, only the meta product is observed. Explain these results. (Hint: what can happen to the nitrogen atom in the presence of the strong acids

Answer 1

See explanation below

Step-by-step explanation:

To get a better understanding watch the picture attached.

In the case of the reaction with Bromine, the -N(CH₃)₂ is a strong ring activator, therefore, it promotes a electrophilic aromatic sustitution, so, in the mechanism of reaction, the lone pair of the Nitrogen, will move to the ring by resonance and activate the ortho and para positions. That's why the bromine wil go to the ortho and para positions, mostly the para position, because the -N(CH₃)₂ cause a steric hindrance in the ortho position.

In the case of the reaction with HNO₃/H₂SO₄, the acid transform the -N(CH₃)₂ in a protonated form, the anilinium ion, which is a deactivating of the ring, and also a strong electron withdrawing, so, the electrophile will go to the meta position instead.

Hope this helps.