See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: The acidity of a C-H bond is determined by the stability of the carbanion formed after deprotonation, which depends on the electronegativity and polarizability of adjacent heteroatoms, as well as any resonance or inductive stabilization. Step 1 - Identify the acidic hydrogen in each compound: All four compounds have a benzylic CH at position 2 of a five-membered heterocyclic ring. The acidity of that CH depends on how well the two flanking heteroatoms stabilize the resulting carbanion. Step 2 - Compare the heteroatom effects: (a) Two oxygen atoms flank the CH in 1,3-dioxolane. Oxygen is highly electronegative but has low polarizability. It stabilizes the carbanion primarily through induction. (b) Two sulfur atoms flank the CH in 1,3-dithiolane. Sulfur is less electronegative than oxygen but is highly polarizable. The larger, more diffuse 3p orbitals of sulfur can better overlap with and stabilize the adjacent carbanion through d-orbital participation and polarizability effects. This makes sulfur a superior stabilizer of carbanions compared to oxygen in this context (as well established in dithiane/dithiolane chemistry, e.g., Corey-Seebach umpolung). Hence the C-H in the dithiolane ring is most acidic. (c) Two nitrogen atoms (NH) flank the CH. Nitrogen is less electronegative than oxygen and less polarizable than sulfur; additionally the nitrogen lone pairs donate electron density into the ring, destabilizing the carbanion. This C-H is least acidic. (d) One oxygen and one nitrogen flank the CH. This is intermediate, with the nitrogen lone pair donation reducing acidity compared to (a) and (b). Step 3 - Rank acidities: Dithiolane (b) > Dioxolane (a) > Oxazolidine (d) > Imidazolidine (c) The compound with two flanking sulfur atoms (option b) has the most acidic hydrogen because sulfur's high polarizability and ability to use d-orbitals provides superior stabilization of the resulting carbanion. Step 4 - Why other options fail: (a) Oxygen is electronegative but poor at stabilizing carbanions via polarizability compared to sulfur. (c) Nitrogen lone pairs donate into the system, actually destabilizing a carbanion at C-2. (d) Mixed O/N flanking gives intermediate acidity, still less than two sulfurs. Therefore, the correct answer is B.