2017 H2 Chem P1 Q25 — Hydroxyl Compounds Solution

The answer is D (2, 1).
Mandelic acid, \(\text{C}_6\text{H}_5\text{CH(OH)CO}_2\text{H}\), has an alcohol OH and a carboxylic acid \(\text{CO}_2\text{H}\). Sodium reacts with both, 1 mol Na per OH group:
\(\text{ROH} + \text{Na} \rightarrow \text{RO}^-\text{Na}^+ + \tfrac{1}{2}\text{H}_2\) and \(\text{RCO}_2\text{H} + \text{Na} \rightarrow \text{RCO}_2^-\text{Na}^+ + \tfrac{1}{2}\text{H}_2\), so 2 mol Na react.
Saligenin has a phenol OH and a primary alcohol \(\text{CH}_2\text{OH}\). Aqueous NaOH reacts with the acidic phenol, \(\text{ArOH} + \text{NaOH} \rightarrow \text{ArO}^-\text{Na}^+ + \text{H}_2\text{O}\), but not with the alcohol, which is not acidic enough. So 1 mol NaOH reacts.
• A is wrong: it counts only one group of mandelic acid reacting with Na (the alcohol OH also reacts) and says saligenin does not react with NaOH (its phenol does).
• B is wrong: the saligenin value is right, but sodium reacts with the alcohol OH of mandelic acid as well as the \(\text{CO}_2\text{H}\), so 2 mol, not 1.
• C is wrong: the mandelic acid value is right, but the phenol group of saligenin is acidic enough to react with aqueous NaOH, so 1 mol, not 0.