See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Solution: Basicity Order Analysis ## Option (c) is correct: Pyridine > Piperidine > Morpholine > Thiomorpholine **Key Principle:** Basicity depends on: 1. Lone pair availability on nitrogen 2. Electron density at nitrogen 3. Resonance/inductive effects ### Analysis of Option (c): **Pyridine (aromatic):** $pK_b \approx 8.8$ (strongest base) - Although aromatic, the lone pair is *not* in the aromatic π-system; it occupies an sp² orbital perpendicular to the ring - This lone pair is relatively accessible for protonation - Aromatic stabilization of the conjugate acid actually *increases* basicity **Piperidine (saturated):** $pK_b \approx 2.9$ - Saturated cyclic secondary amine with sp³ nitrogen - Lone pair is more available than aromatic competitors - However, electron density on N is reduced by adjacent electron-withdrawing atoms in subsequent compounds **Morpholine (oxygen in ring):** $pK_b \approx 5.7$ - Oxygen is highly electronegative and withdraws electron density from nitrogen via the $\sigma$-bond - This reduces lone pair availability and basicity **Thiomorpholine (sulfur in ring):** $pK_b \approx 6.3$ (weakest base) - Although sulfur is less electronegative than oxygen, it is larger and more polarizable - Poor orbital overlap with nitrogen reduces electron donation efficiency - Least basic of the four ### Why others fail: - **(a)** Incorrect: Secondary amines are more basic than primary; charged $SMe_2$ group is electron-withdrawing - **(b)** Incorrect: Alkene conjugation (*sp²* carbon) reduces lone pair availability; quaternary ammonium $N^+$ has no lone pair - **(d)** Incorrect: Bicyclic strain and geometric constraints reduce basicity; bicyclic compound should be stronger than $Et_3N$ **Answer: (c)**