= Solution
In the strongly acidic nitrating mixture, the dimethylamino <nitrogen> is predominantly protonated. The substituent is therefore $\mathrm{-NMe_2H^+}$ during aromatic substitution, rather than the neutral <electron>-donating $\mathrm{-NMe_2}$ 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. \b[$A$ 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 \b[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.
Back to article page
