See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: In acid-base equilibria, the reaction is favoured in the direction that goes from the stronger acid to the weaker acid (and stronger base to weaker base). The backward reaction is favoured when the products are stronger acids/bases than the reactants, i.e., when the equilibrium lies to the left. For each option, we compare the acidity of the acids on each side: (a) Reactant acid: acetylene (HC≡CH, pKa ≈ 25); Product acid: ethane (H3C-CH3, pKa ≈ 50). Acetylene is a stronger acid than ethane, so the forward reaction (deprotonation of acetylene by ethyl carbanion) is favoured. Forward reaction is favoured. (b) Reactant acid: trifluoroacetic acid (CF3COOH, pKa ≈ 0); Product acid: ethanol (HOCH2CH3, pKa ≈ 16). CF3COOH is a much stronger acid than ethanol, so forward reaction is favoured. (c) Reactant acid: protonated ethanethiol (CH3CH2SH2+, very strong acid, pKa ≈ -5 to 0); Product acid: protonated ethanol (CH3CH2OH2+, pKa ≈ -2.4). Both are strong acids but the protonated thiol is stronger (sulfur is less electronegative, so S-H bond is weaker and the conjugate base is more stable). The forward reaction converts a stronger acid to a comparably strong acid - this option is less clearly backward-favoured. (d) Reactant acid: anilinium ion (PhNH3+, pKa ≈ 4.6); Product acid: protonated phenol / phenyoxonium (PhOH2+, pKa ≈ -6). The product acid (PhOH2+) is a much stronger acid (much lower pKa) than the reactant acid (PhNH3+, pKa ≈ 4.6). Since equilibrium favours formation of the weaker acid, and the product side has a much stronger acid (PhOH2+), the backward reaction is strongly favoured. In other words, PhOH (pKa ≈ 10) is a weaker acid than PhNH3+ (pKa ≈ 4.6), meaning aniline (PhNH2) is a weaker base than phenoxide... Let us reconsider: The forward reaction is: PhNH3+ (acid, pKa 4.6) + PhOH (acid, pKa 10) → PhNH2 (conjugate base of pKa 4.6) + PhOH2+ (conjugate acid, very strong acid). Equilibrium favours the side with the weaker acid. PhOH (pKa 10) is weaker than PhNH3+ (pKa 4.6). So the stronger acid is PhNH3+ on the left. The reaction should go forward to produce the weaker acid (PhOH). Wait - the proton transfer is FROM PhNH3+ TO PhO-? No: PhNH3+ + PhOH → PhNH2 + PhOH2+. Here PhNH3+ donates a proton to PhOH (which acts as base). But PhOH is a very weak base. The conjugate acid PhOH2+ has pKa ≈ -6, making it a very strong acid - much stronger than PhNH3+ (pKa 4.6). Equilibrium favours the side with the weaker acid. Since PhNH3+ (pKa 4.6) is a weaker acid than PhOH2+ (pKa -6), the backward reaction is favoured. The products contain a stronger acid (PhOH2+) than the reactants (PhNH3+), so the backward reaction is favoured. Therefore, the correct answer is D.