See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the functional groups and their hydrogens: a = carboxylic acid O-H (–COOH) b = alcohol O-H on cyclopentane ring (secondary alcohol) c = thiol S-H (–SH) d = C-H alpha to a ketone (α-hydrogen) Step 2 – Recall typical pKa values: Carboxylic acid (–COOH): pKa ≈ 4–5 (very acidic) Thiol (–SH): pKa ≈ 10–11 Alcohol (–OH): pKa ≈ 15–16 α-C-H adjacent to ketone: pKa ≈ 20 Step 3 – Compare acidity (lower pKa = more acidic): a (carboxylic acid, pKa ~4-5) >> c (thiol, pKa ~10-11) >> b (alcohol, pKa ~15-16) >> d (α-C-H, pKa ~20) Step 4 – Carboxylic acid hydrogen (a) is by far the most acidic because the resulting carboxylate anion is stabilized by resonance delocalization over two electronegative oxygen atoms, giving it the lowest pKa among all four hydrogens. Step 5 – Why other options fail: b (alcohol) is far less acidic than a carboxylic acid. c (thiol) is more acidic than an alcohol but still much less acidic than a carboxylic acid. d (α-C-H) is the least acidic of the four. Therefore, the correct answer is A.