= Solution
The qualitative <nucleophilicity> order is \b[$p$-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>.
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