Step 1 (Identification of Reactions and Functional Groups): [0.5 Mark]

• The reaction of (A) with $\text{Br}_2/\text{KOH}$ (Hofmann Bromamide Degradation) indicates that (A) is an amide. Given the formula $\text{C}_3\text{H}_7\text{NO}$, (A) is propanamide: $\text{CH}_3\text{CH}_2\text{CONH}_2$.
• The degradation product (B) is an amine with one carbon less: ethanamine ($\text{CH}_3\text{CH}_2\text{NH}_2$).

Step 2 (Step-by-Step Chemical Deductions): [0.25 Mark]

• Formation of (C): (B) reacts with $\text{CHCl}_3/\text{alc. KOH}$ (Carbylamine test) to produce a foul-smelling isocyanide. $$\text{CH}_3\text{CH}_2\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH} \rightarrow \text{CH}_3\text{CH}_2\text{NC} + 3\text{KCl} + 3\text{H}_2\text{O}$$ Compound (C) is ethyl isocyanide ($\text{CH}_3\text{CH}_2\text{NC}$).
• Formation of (D): (B) reacts with Hinsberg's reagent ($\text{C}_6\text{H}_5\text{SO}_2\text{Cl}$) to form a sulfonamide. $$\text{CH}_3\text{CH}_2\text{NH}_2 + \text{C}_6\text{H}_5\text{SO}_2\text{Cl} \rightarrow \text{C}_6\text{H}_5\text{SO}_2\text{NHCH}_2\text{CH}_3 + \text{HCl}$$ Compound (D) is N-ethylbenzenesulfonamide. Since it contains an acidic hydrogen on the nitrogen atom, it is soluble in alkali.

Step 3 (Final Concluded Answer): [0.25 Mark]

The structures are as follows:

• (A): $\text{CH}_3\text{CH}_2\text{CONH}_2$ (Propanamide)
• (B): $\text{CH}_3\text{CH}_2\text{NH}_2$ (Ethanamine)
• (C): $\text{CH}_3\text{CH}_2\text{NC}$ (Ethyl isocyanide)
• (D): $\text{C}_6\text{H}_5\text{SO}_2\text{NHCH}_2\text{CH}_3$ (N-ethylbenzenesulfonamide)

💡 EL10 Concept Insight: Think of this like a chemical "detective game": the Hofmann reaction chops off a carbon atom to turn an amide into an amine, the Carbylamine test acts as a "stinky" fingerprint for primary amines, and the Hinsberg test checks if the amine has a hydrogen "handle" left to dissolve in base.