Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 4 / 55 / b / i / 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 $\mathrm{RCO}-\mathrm{O}-\mathrm{CO}-\mathrm{Im}$. Acyl transfer to imidazole, with decarboxylation, produces the activated acyl imidazole $\mathrm{RCO}-\mathrm{Im}$, CO${}_2$ and imidazole.

The amine attacks the RCO carbonyl of that acyl imidazole. Its tetrahedral intermediate collapses with loss of imidazole; proton transfer gives $\mathrm{RCONH}{\mathrm{R}}^{\prime }$. 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.