The final major product of the following reaction is:

The reaction sequence involves treatment of an amide with $$Br_2$$ in an aqueous alkaline medium ($$KOH$$/$$NaOH$$). This combination is the classical Hofmann bromamide degradation (also called the Hofmann rearrangement).

Step 1 - Formation of N-bromoamide
The amide first undergoes base-catalysed deprotonation to give the anion, which is rapidly brominated by $$Br_2$$ to form an $$N$$-bromoamide.

Step 2 - Rearrangement to an isocyanate
On warming, the $$N$$-bromoamide loses $$Br^-$$. A concerted 1,2-migration (the “Hofmann rearrangement”) shifts the acyl $$R-CO-$$ group onto nitrogen, generating an isocyanate $$R-N=C=O$$ and releasing $$Br^-$$.

Step 3 - Hydrolysis of the isocyanate
Because the reaction medium is aqueous, the isocyanate is immediately attacked by water. Proton transfers followed by decarboxylation give a primary amine that contains one carbon fewer than the starting amide.

Thus an amide $$RCONH_2$$ is converted into the primary amine $$RNH_2$$.

Among the given options, Option D shows that primary amine. Hence the final major product is the amine obtained after Hofmann degradation.

Therefore, the correct choice is:
Option D which is: the corresponding primary amine ($$RNH_2$$)