Consider the following reaction sequence

Compound (y) develops characterstic colour with neutral $$FeCl_{3}$$ solution.
Identify the INCORRECT statement from the following for the above sequence.

To find the molecular formula of x, we use the percentage composition and vapor density:

• Molecular Mass: $$2 \times \text{Vapour Density} = 2 \times 47 = 94 \text{ g/mol}$$.
• Empirical Calculation: * C: $$76.6\%$$ of $$94 \approx 72 \text{ g}$$ (which is 6 atoms of Carbon).
  • H: $$6.38\%$$ of $$94 \approx 6 \text{ g}$$ (which is 6 atoms of Hydrogen).
  • O: Remaining mass is $$94 - (72 + 6) = 16 \text{ g}$$ (which is 1 atom of Oxygen).
• Result: The formula is $$C_6H_6O$$. Compound (x) is Phenol.

Phenol undergoes the Kolbe-Schmitt Reaction when treated with $$CO_2$$ and $$NaOH$$ under high pressure, followed by acidification.

• Product (y): Salicylic acid ($$o$$-hydroxybenzoic acid).
• Identification Test: Salicylic acid contains a phenolic $$-OH$$ group, which is why it develops a characteristic violet color with neutral $$FeCl_3$$.

Based on the properties of Phenol (x) and Salicylic acid (y):

1. Solubility in $$NaHCO_3$$: Correct. Salicylic acid contains a carboxylic acid group ($$-COOH$$), which is strong enough to react with sodium bicarbonate and evolve $$CO_2$$ gas. Phenol does not react with $$NaHCO_3$$.
2. Relative Acidity: Option B is INCORRECT. Compound (x) (Phenol) is less acidic than compound (y) (Salicylic acid). The presence of the electron-withdrawing $$-COOH$$ group significantly increases the acidity of the molecule.
3. Combustion: Correct. Both are aromatic compounds (high carbon-to-hydrogen ratio) and typically burn with a sooty flame.
4. Solubility in $$NaOH$$: Correct. Both are acidic (phenol is weakly acidic; salicylic acid is more acidic) and will dissolve in a strong base like $$NaOH$$.