 Solution

The acidic ethanolic step converts the terminal carboxylic acid to its ethyl ester, leaving the pre-existing thiazole ester. Azide then adds across the alkyne by 1,3-dipolar cycloaddition, followed by protonation at work-up. F contains an NH-1,2,3-triazole replacing the alkyne, with the thiazole and $(\mathrm{C}{\mathrm{H}}_2{)}_3\mathrm{C}{\mathrm{O}}_2\mathrm{Et}$ attached to its two carbon atoms. The thiazole's own CO${}_2$Et group remains.

The formula check is decisive:One can represent the cycloaddition with cyclic movement of the azide resonance-form electrons and the alkyne $\pi$ electrons into the two new C–N bonds. Alkali azide initially gives a triazolide; aqueous work-up provides the NH. Interchanging formal 4/5 labels for the carbon substituents in an N-unsubstituted triazole does not establish a separate permanently fixed N-substituted regioisomer.