Past exam of the chemistry B course of the University of Cambridge / 2024 / Paper 1 / B2 / a / Solution

In the strongly acidic nitrating mixture, the dimethylamino nitrogen is predominantly protonated. The substituent is therefore $-\mathrm{NM}{\mathrm{e}}_2{\mathrm{H}}^{+}$ during aromatic substitution, rather than the neutral electron-donating $-\mathrm{NM}{\mathrm{e}}_2$ group. Its lone pair is unavailable for conjugation. Ortho or para attack would generate a sigma-complex contributor with positive ring charge at the carbon attached to an already positive nitrogen; meta attack avoids that especially unfavorable contributor.

Sulfuric acid protonates nitric acid, and loss of water produces the nitronium ion. Aromatic attack at a meta position forms a Wheland intermediate; deprotonation restores aromaticity. The bicarbonate workup then returns the amino group to its neutral form. $A$ is 3-nitro-N,N-dimethylaniline.

There are six inequivalent aromatic carbons in this unsymmetrical meta-disubstituted ring and one carbon environment for the two equivalent N-methyl groups, giving seven carbon NMR signals. The para isomer would have only four aromatic environments plus the methyl environment. An ortho isomer could also give seven signals, so signal counting supports but does not alone establish the meta assignment; protonation-dependent directing effects are essential.