See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: In an acid-base equilibrium, the backward reaction is favoured when the products are a stronger acid and stronger base than the reactants. Equivalently, equilibrium lies to the LEFT (backward reaction favoured) when the conjugate acid on the product side is a stronger acid than the original acid on the reactant side — i.e., when the base on the left is weaker than the base on the right, or when the pKa of the acid being deprotonated is higher than the pKa of the conjugate acid formed. Step 1 - Analyze each reaction in the forward direction: (a) EtO(-) (conjugate acid: EtOH, pKa ~16) deprotonates cyclohexanol (pKa ~16-17). These pKa values are very similar, so equilibrium is roughly balanced, but EtOH and cyclohexanolate are comparable; forward reaction is slightly favoured or neutral. (b) KH (hydride, conjugate acid: H2, pKa ~35) deprotonates EtOH (pKa ~16). Since H2 is a much weaker acid than EtOH, the forward reaction is strongly favoured (equilibrium lies far to the right). (c) Me3CO(-) (conjugate acid: tert-butanol, pKa ~19) deprotonates H2O (pKa ~15.7). Here the conjugate acid tert-BuOH (pKa ~19) is weaker than H2O (pKa ~15.7), meaning Me3CO(-) is a stronger base than HO(-). The forward reaction deprotonates a weaker acid (H2O, pKa 15.7) with a stronger base (Me3CO(-)); actually, since pKa of H2O (15.7) < pKa of t-BuOH (19), t-BuO(-) is a stronger base than OH(-), so the forward reaction is favoured. (d) Naphthalen-1-olate anion (O(-) on naphthalene, conjugate acid: 1-naphthol, pKa ~9.4) reacts with CH3OH (pKa ~15.5). In the forward direction, 1-naphthol would be produced and methoxide (MeO(-)) would be produced. The conjugate acid of 1-naphthoxide is 1-naphthol (pKa ~9.4), which is a much stronger acid than MeOH (pKa ~15.5). This means 1-naphthoxide is a much weaker base than methoxide. Therefore, the backward reaction — where methoxide deprotonates 1-naphthol — is strongly favoured. The equilibrium lies to the LEFT, meaning the backward reaction is favoured. Step 2 - Why other options fail: (a) EtO(-) and cyclohexanolate have similar pKa values (~16), so neither forward nor backward is strongly favoured. (b) KH + EtOH: H2 (pKa ~35) >> EtOH (pKa ~16), so forward reaction is overwhelmingly favoured, not backward. (c) Me3CO(-) + H2O: t-BuOH (pKa ~19) > H2O (pKa ~15.7), so forward reaction is favoured. (d) 1-Naphthoxide + CH3OH: 1-naphthol (pKa ~9.4) is a much stronger acid than MeOH (pKa ~15.5), so the backward reaction is strongly favoured. Therefore, the correct answer is D.