C6H5−CONH2 + Br2/NaOH → C6H5−NH2 (Aniline) (A)
(CH3CO)2O/pyridine → C6H5−NH−O||C−CH3 (N-phenylethanamide)
When benzamide reacts with Br2/NaOH, it undergoes Hoffmann degradation reaction to form aniline (A) as a product. When this aniline reacts with acetic anhydride in the presence of pyridine, it forms phenylethanamide as a product.