= Solution
Write Im for an imidazolyl group. The acid protonates one imidazole nitrogen, and its carboxylate attacks the central <carbonyl> of <carbonyldiimidazole>. Collapse of the <tetrahedral intermediate> expels imidazole and gives the mixed anhydride $\mathrm{RCO-O-CO-Im}$. Acyl transfer to imidazole, with decarboxylation, produces the activated acyl imidazole $\mathrm{RCO-Im}$, CO$_2$ and imidazole.
The amine attacks the \b[RCO <carbonyl>] of that acyl imidazole. Its <tetrahedral intermediate> collapses with loss of imidazole; proton transfer gives $\mathrm{RCONHR'}$. The <carbonyl> originally belonging to CDI leaves as CO$_2$, while the product amide <carbonyl> is the original acid <carbonyl>. Imidazole serves both as a leaving group and a proton-transfer base.
\Image[/past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-4-carbonyl-routes.png]
{title=Activated acyl transfer, hemiaminal opening and carbene formation}
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