 Solution

An aromatic–aliphatic polyamide and methanol are formed. Let $\mathrm{Ar}=p\text{-}{\mathrm{C}}_6{\mathrm{H}}_4-{\mathrm{C}}_6{\mathrm{H}}_4\text{-}p$. The polymer repeat unit isEach amide linkage comes from nucleophilic acyl substitution: an amine attacks an ester carbonyl, the tetrahedral intermediate expels methoxide, and proton transfer produces methanol. There are approximately $2n$ molecules of methanol for $n$ complete repeat units in a long chain; the precise finite-chain count depends on its end groups. Removal of methanol drives this step-growth polymerization.