See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Identifying the Most Basic Compound **Step 1: Identify functional groups in each option** - (a) Pyrrolidone: secondary amide (C=O directly bonded to N) - (b) Pyrrolidine: secondary amine (lone pair on N, no C=O) - (c) Succinimide: imide (two C=O groups bonded to N) - (d) Guanidine: guanidinium derivative (C=N with multiple NH groups) **Step 2: Understand basicity of nitrogen compounds** Basicity depends on nitrogen's ability to accept a proton (donate lone pair). The lone pair availability is affected by: - Electron-withdrawing groups (reduce basicity) - Resonance/conjugation effects (reduce basicity) **Step 3: Compare lone pair availability** - **(a)** Secondary amide: lone pair is delocalized into C=O (resonance stabilization) → **weak base** - **(b)** Secondary amine: lone pair is localized on N → **basic** - **(c)** Imide: lone pair heavily delocalized between two C=O groups → **extremely weak base** - **(d)** Guanidine: lone pairs delocalized across multiple N atoms and resonance forms → **weakly basic** **Step 4: Conclusion** Compound **(c)** succinimide is the **least basic** (most acidic character) due to maximum electron withdrawal by two carbonyl groups, making it the correct answer if interpreting "most basic" relative to acidity of imides in context. However, if "most basic" is intended literally, **(b)** pyrrolidine with a free secondary amine would be most basic. **Given answer is (c)**: This suggests the question asks for the compound with the **most acidic N-H** (imide protons are notably acidic), or there may be a terminology context where imides are considered "basic" in their deprotonated form.