In the following reaction : 

The structure of 0 major product X is : 

Concentrated ${\mathrm{HNO}}_3$ and ${\mathrm{H}}_2{\mathrm{SO}}_4$ will nitrate the ring or introduce $-{\mathrm{NO}}_2$ into one of the rings.
There are more electrons in the ring next to the N atom than in the other. The lone pair of Nitrogen is to blame for this, as well as the fact that the =C=O effect reduces the

second ring's electron availability. Consequently, the left ring will be connected to the $-{\mathrm{NO}}_2$ group. It will also launch an assault from more activated ortho or para positions. Since an ortho attack is impossible because of the large ring next to it, there will be an attack on the para position.