See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

## Analysis of Solubility in NaOH Solution **Key Concept:** Compounds dissolve in $NaOH$ if they contain acidic groups (N-H or S-H) that can be deprotonated to form water-soluble anionic salts. --- ### Examining Each Compound: **(A) Phthalimide (Isoindole-1,3-dione)** - Contains **N-H group** (secondary imide) - $NaOH$ deprotonates: $\text{Phthalimide-NH} + NaOH \rightarrow \text{Phthalimide-N}^- Na^+ + H_2O$ - Forms soluble sodium salt ✓ **(B) Benzoisothiazole-3(2H)-one-1,1-dioxide** - Contains **N-H group** (secondary amide in heterocycle) - $NaOH$ deprotonates the N-H to form soluble anionic salt ✓ **(C) N-Methylbenzenesulfonamide** - Contains $N-CH_3$ (tertiary nitrogen, **no N-H**) - Cannot be deprotonated by $NaOH$ - Remains insoluble ✗ **(D) Benzenesulfonamide** - Contains **N-H group** (secondary sulfonamide) - $NaOH$ deprotonates: $\text{PhSO}_2NH + NaOH \rightarrow \text{PhSO}_2N}^- Na^+ + H_2O$ - Forms soluble sodium salt ✓ --- **Answer: A, B, D** — all contain acidic N-H groups that form water-soluble anionic salts with $NaOH$.