See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

To rank these anions in decreasing order of basic strength, we assess how well each anion holds its negative charge (stability of conjugate acid) and the electronegativity/hybridization of the atom bearing the charge. Concept: Basic strength is inversely related to the stability of the anion (i.e., the more stable the anion, the weaker the base). We compare the conjugate acids: - (A) CH3CH2C≡CH: terminal alkyne, pKa ≈ 25. The carbon is sp-hybridized, which provides moderate stabilization. - (B) CH3CH2SH: thiol, pKa ≈ 10–11. Sulfur is larger and more polarizable, stabilizing the negative charge well, making S⁻ a weaker base than O⁻ or C≡C⁻. - (C) CH3CH2CO2H: carboxylic acid, pKa ≈ 4–5. The carboxylate anion is highly stabilized by resonance (delocalization over two oxygens), making it the weakest base. - (D) CH3CH2OH: alcohol, pKa ≈ 16. The alkoxide is on sp3 oxygen with no resonance stabilization, making it a stronger base than S⁻ or CO2⁻. Ranking conjugate acid pKa values (higher pKa = stronger base): - A (pKa ≈ 25) > D (pKa ≈ 16) > B (pKa ≈ 10) > C (pKa ≈ 5) Therefore decreasing order of basic strength: A > D > B > C This matches option (c). Why other options fail: - (a) B > A > D > C: Incorrect; S⁻ is not more basic than C≡C⁻ or O⁻. - (b) D > A > B > C: Incorrect; alkoxide (D) is not more basic than alkynide (A) since pKa of alkyne (~25) > pKa of alcohol (~16). - (d) A > D > C > B: Incorrect; carboxylate (C) is more stable (weaker base) than thiolate (B), so B should come before C. Therefore, the correct answer is C.