Alcohols, Phenols and Ethers Part 2

Concept:The given reaction is the Kolbe-Schmitt reaction: a phenol reacts with $\mathrm{CO}_2$ in presence of $\mathrm{NaOH}$ under high pressure and temperature, followed by acidification, to form an ortho-hydroxybenzoic acid (salicylic acid).The product gives a characteristic colour with neutral $\mathrm{FeCl}_3$ confirming the presence of a phenolic $\mathrm{-OH}$ group.Explanation:First, find the molecular formula of compound (x).Vapour density = 47, so molar mass = $2 \times 47 = 94 \ \mathrm{g/mol}$.%C = 76.6, %H = 6.38, %O = $100 - (76.6 + 6.38) = 17.02$.Moles: C = $76.6/12 \approx 6.38$, H = $6.38/1 = 6.38$, O = $17.02/16 \approx 1.064$.Simplest ratio: divide by 1.064 → C = 6, H = 6, O = 1 → empirical formula $\mathrm{C_6H_6O}$.Molar mass of $\mathrm{C_6H_6O} = 12 \times 6 + 6 + 16 = 94 \ \mathrm{g/mol}$, matches.So compound (x) is phenol ($\mathrm{C_6H_5OH}$).Phenol undergoes Kolbe-Schmitt reaction: with $\mathrm{CO}_2/\mathrm{NaOH}$ at $120\,{}^{\circ}\mathrm{C}$ under high pressure, then $\mathrm{H_3O}^{+}$, gives salicylic acid (compound y, $\mathrm{2\text{-}HOC_6H_4COOH}$).Salicylic acid has both a carboxylic acid group and a phenolic OH, so it gives purple colour with neutral $\mathrm{FeCl}_3$.Now evaluate each option:A. Salicylic acid has a $\mathrm{-COOH}$ group; it dissolves in $\mathrm{NaHCO}_3$ releasing $\mathrm{CO}_2$ gas. This is correct.B. Phenol ($\mathrm{pK_a} \approx 10$) is a weaker acid than salicylic acid ($\mathrm{pK_a}$ of $\mathrm{-COOH} \approx 3$). Hence, compound x is less acidic than compound y, not more. This statement is incorrect.C. Both phenol and salicylic acid are acidic (phenolic OH and/or COOH) and will dissolve in aqueous $\mathrm{NaOH}$. Correct.D. Both are aromatic compounds with high carbon content; they burn with a sooty flame. Correct.Answer:The incorrect statement is B.