= Solution
\b[H is the fused pyridone analogue of G: replace the pyrone ring oxygen by NH, retaining its carbonyl and the remote carboxylic acid.] Ammonia adds to the electron-poor pyrone system, opening its lactone/enol-ether framework into an amino-containing unsaturated acyl intermediate. Proton transfers permit intramolecular closure through nitrogen and loss of water, giving the conjugated pyridone/lactam ring. Acid work-up restores the neutral acid and the preferred neutral lactam representation.
The net atom change is $+\mathrm{NH}-\mathrm O$, so
$$
\boxed{\mathrm{C_{16}H_{12}FNO_4S\longrightarrow C_{16}H_{13}FN_2O_3S},}
$$
exactly matching H. Merely forming a carboxamide from the terminal acid would remove a different oxygen and leave the pyrone oxygen intact; it would not explain the characteristic pyrone-to-pyridone ring transformation.
\Image[/past-exam-of-the-chemistry-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2-heterocyclic-products.png]
{title=Original structures of the thiazole triflate, triazole, fused pyridone and acylated pyrrole}
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