See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

**Analysis of Acid Strength and Reaction Feasibility** **Step 1: Identify the $K_a$ values** - Alcohols: $K_a = 10^{-17}$ (weakest acid) - Water: $K_a = 10^{-14}$ (intermediate acid) - Phenol: $K_a = 10^{-10}$ (strongest acid) **Step 2: Evaluate Reaction (I): $C_6H_5O^\circ + H_2O \rightarrow C_6H_5OH + OH^\circ$** Phenoxide ion ($C_6H_5O^-$) acting as a base accepts proton from water. For this to occur, **phenol must be a stronger acid than water**. Since $K_a(\text{phenol}) = 10^{-10} > 10^{-14} = K_a(\text{water})$ ✓ However, phenol is still a weak acid. Water is the base here, and the equilibrium **favors the left side** (reactants). This reaction is **not favorable** ✗ **Step 3: Evaluate Reaction (II): $C_2H_5O^\circ + H_2O \rightarrow C_6H_5OH + OH^\circ$** Ethoxide ion ($C_2H_5O^-$) acting as a base accepts proton from water. For this reaction, **alcohol must be a stronger acid than water**. Since $K_a(\text{alcohol}) = 10^{-17} < 10^{-14} = K_a(\text{water})$ Alcohol is a **much weaker acid** than water, so ethoxide is a **much stronger base** than hydroxide. Water readily donates its proton to ethoxide. This reaction **proceeds forward** ✓ **Conclusion:** Only Reaction (II) is possible because ethoxide is a strong enough base to abstract a proton from water. **Answer: (B) only II**