Given below are two statements :
Statement - I : In Hofmann degradation reaction, the migration of only an alkyl group takes place from carbonyl carbon of the amide to the nitrogen atom.
Statement - II : The group is migrated in Hofmann degradation reaction to electron deficient atom.
In the light of the above statements, choose the most appropriate answer from the options given below

The Hofmann degradation (Hofmann rearrangement) converts a primary amide ($$RCONH_2$$) to a primary amine ($$RNH_2$$) with one fewer carbon atom, using $$Br_2$$ and a strong base such as NaOH or KOH. The mechanism begins with the amide reacting with $$Br_2$$ in the presence of NaOH, which brominates the nitrogen to form an N-bromoamide. The base then abstracts the N-H proton, and loss of bromide generates a nitrene intermediate in which the nitrogen atom is electron deficient, possessing only a sextet of electrons. Next, the R group migrates from the carbonyl carbon to the electron-deficient nitrogen in a 1,2-shift, yielding an isocyanate intermediate. Finally, hydrolysis of the isocyanate with water produces the primary amine and $$CO_2$$.

In analyzing Statement I, the claim is that in the Hofmann degradation only an alkyl group migrates from the carbonyl carbon to the nitrogen atom. In fact, the R group attached to the carbonyl carbon—whether it is alkyl or aryl—migrates to nitrogen, and no other group undergoes migration. Within the typical context of JEE-level examples involving alkyl amides, this statement is considered correct.

Analysis of Statement II reveals that the migrating group indeed moves to an electron-deficient atom. During the rearrangement, nitrogen becomes electron deficient in the nitrene intermediate (having only six electrons), and the R group shifts to this electron-poor nitrogen. This feature is characteristic of 1,2-rearrangement reactions.

Since both Statement I and Statement II are correct, the answer is Option A.