= 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. \b[F contains an NH-1,2,3-triazole replacing the alkyne], with the <thiazole> and $\mathrm{(CH_2)_3CO_2Et}$ attached to its two carbon atoms. The <thiazole>'s own CO$_2$Et group remains.
The formula check is decisive:
$$
\boxed{\mathrm{C_{18}H_{16}FNO_4S+C_2H_4+HN_3=C_{20}H_{21}FN_4O_4S}.}
$$
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.
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