# In Hofmann degradation of amide Me the important step is the migration of

$\begin{array}{1 1}(a)\;\text{an alkyl group without its }e^{\ominus}\text{ deficient N atom}\\(b)\;\text{an alkyl group with its electron pair to }e^{\ominus}\text{ deficient N-atom}\\(c)\;\text{an alkyl group with its electron pair to }e^{\ominus}\text{ deficient O-atom}\\(d)\;\text{an alkyl group with its electron pair to }e^{\ominus}\text{ rich N atom}\end{array}$

In the Hofmann degration ,migration of R (alkyl group with it $e^{\ominus}$ pair) to the electron deficient N-atom gives the isocyanate.
Hence (b) is the correct answer.