See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: Acidity of organic compounds depends on the stability of the conjugate base after loss of a proton. More stable the conjugate base, more acidic the compound. Step 1: Identify the acidic proton in each compound. - Compound I (cyclohexanol): O-H proton. pKa ~ 16 - Compound II (cyclohexylamine): N-H proton. pKa ~ 38-40 - Compound III (phenol): O-H proton. pKa ~ 10 Step 2: Compare acidities. - Cyclohexylamine (II) has an N-H bond. Nitrogen is less electronegative than oxygen and the lone pair on the nitrogen conjugate base (cyclohexyl amide anion) is not stabilized by resonance, making it the least acidic (highest pKa ~38). - Cyclohexanol (I) has an O-H bond. Oxygen is more electronegative than nitrogen, so the alkoxide conjugate base is more stable than the amide anion. pKa ~16, making it more acidic than the amine. - Phenol (III) has an O-H bond, but the phenoxide conjugate base is stabilized by resonance delocalization into the aromatic ring. This makes phenol significantly more acidic than cyclohexanol. pKa ~10. Step 3: Order of increasing acidity (least acidic to most acidic): II (cyclohexylamine, pKa ~38) < I (cyclohexanol, pKa ~16) < III (phenol, pKa ~10) Step 4: Why other options fail. - (a) III < I < II would mean phenol is least acidic, which is wrong because phenoxide is resonance-stabilized. - (b) I < III < II would mean amine is most acidic, which contradicts the fact that N-H bonds are far less acidic than O-H bonds. - (c) III < II < I would mean cyclohexanol is most acidic and phenol is least acidic, both of which are incorrect. - (d) II < I < III correctly places cyclohexylamine as least acidic, cyclohexanol in the middle, and phenol as most acidic. Therefore, the correct answer is D.