 Solution

The qualitative nucleophilicity order is $p$-OMe $>$ $p$-Me $>$ H $>$ $p$-Cl $\gg$ $p$-NO${}_2$, matching the product-yield order at the stated stopping time. Electron donation raises the availability of the aniline lone pair for attack. A para methoxy group donates by resonance, while methyl is a weaker donor. Chlorine withdraws inductively and gives an aniline less nucleophilic than the unsubstituted compound. A para nitro group strongly withdraws by resonance: electron flow from nitrogen through the ring into the nitro group stabilizes a charge-separated form and greatly reduces nucleophilicity.

The key arrows for the nitroaniline comparison move the nitrogen lone pair into its own aromatic ring, shift the ring $\pi$ bonds toward the para nitro group, and shift a nitro N=O bond toward oxygen. The methoxy donor produces the opposite electronic bias. These are finite-time yields, so they support a kinetic explanation qualitatively; they are not by themselves measured rate constants or equilibrium constants.