Step 1: Hoffmann Bromamide Degradation reaction applies to aliphatic or aromatic primary amides to produce a primary amine with one carbon atom less than the parent amide. Step 2: Among the given options, Propanamide (\( CH_3{-}CH_2{-}CONH_2 \)) is an aliphatic primary amide which undergoes Hoffmann Bromamide Degradation to give ethylamine (\( CH_3CH_2NH_2 \)). Step 3: While Benzamide (option B) also gives Hoffmann degradation, the question’s image shows an aliphatic amide in option D (Propanamide) which exactly fits the typical example for the reaction. Step 4: Therefore, Propanamide is the correct answer.