Question

Give the decreasing order of reactivity of the following amides in Hofmann bromamide rearrangement `(H.B.R.)` reaction.
(a) (I) `PhCONH_2`
(II)
(III)
(IV)
(V)
(VI)
(b) (I) `PhCONH_2`
(II)
(III)
(IV)
(V) .

Answer 1

`EDG` at `o-` and `p-positions` of a migrating group accelerates the reaction.
`ED` order of `Me gt Et gt i-Pt gt t-Bu` [due to `H.C` (hyper conjugation)]. In `Me, Et, i-Pr`, and `t-Bu`, the `H.C` structures, respectively, are `3, 2, 1` and no `H.C`)
So the decreasing order of reactivity for `H.B.R.` is :
(a) `(II) gt (III) gt (IV) gt (V) gt (I) gt (VI)`
(b) `ED` order of `-OMe[(+R) and (-I)] gt (-Me)` (Three `H.C` structures and `(+ I)]`.
`EW` order of `(-NO_2) [(-I and (-R)] gt (-Cl) (-I)` So the decreasing order of reactivity for `H.B.R.` is :
`underset(("More EDG favour"))(((II)gt(III)gt(I))/(EDG)) ("Standard") ((V)gt (VI))/(EWG)`