$N-$ substituted amide will not undergo Hoffmann bromamide reaction because formation of iso- cyanate is not possible.
(Figure) $\xrightarrow[FeB{{r}_{3}}]{B{{r}_{2}}}B\xrightarrow{Sn\,/\,HCl}C$ $\xrightarrow[HCl]{NaN{{O}_{2}}}D\xrightarrow[HBr]{CuBr}E$
