Identification of Reaction Products in Cumene to Ether Synthesis

To determine the correct statement(s), let us analyze the chemical transformation sequence step by step:

1. Formation of Compound $\mathbf{P}$: $\text{Cumene } (\text{C}_9\text{H}_{12}) \xrightarrow[\text{ii) } \text{H}_3\text{O}^+]{\text{i) } \text{O}_2} \mathbf{P}$

   • Reaction of cumene (isopropylbenzene) with $\text{O}_2$ followed by acidic hydrolysis is the industrial process for manufacturing phenol.
   • Thus, compound $\mathbf{P}$ is Phenol ($\text{C}_6\text{H}_5\text{OH}$).

2. Formation of Compounds $\mathbf{Q}$ and $\mathbf{R}$: $\mathbf{P} \xrightarrow{\text{CHCl}_3/\text{NaOH}} \mathbf{Q} \text{ (major)} + \mathbf{R} \text{ (minor)}$

   • This is the Reimer-Tiemann reaction. Treatment of phenol with chloroform ($\text{CHCl}_3$) and base ($\text{NaOH}$) introduces a formyl group ($-\text{CHO}$) onto the aromatic ring.
   • The major product $\mathbf{Q}$ is Salicylaldehyde (2-hydroxybenzaldehyde, ortho isomer).
   • The minor product $\mathbf{R}$ is 4-Hydroxybenzaldehyde (para isomer).

3. Formation of Compound $\mathbf{S}$: $\mathbf{Q} \xrightarrow[\text{PhCH}_2\text{Br}]{\text{NaOH}} \mathbf{S}$

   • Phenol/Phenolic $-\text{OH}$ in salicylaldehyde ($\mathbf{Q}$) reacts with $\text{NaOH}$ to form the phenoxide anion, which acts as a nucleophile in a Williamson ether synthesis reaction with benzyl bromide ($\text{PhCH}_2\text{Br}$).
   • The free phenolic $-\text{OH}$ group is etherified to a benzyloxy group ($-\text{OCH}_2\text{Ph}$).
   • Thus, compound $\mathbf{S}$ is 2-(benzyloxy)benzaldehyde.

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Evaluation of Options:

• Option (A): $\mathbf{R}$ exhibits steam volatility

  • Salicylaldehyde ($\mathbf{Q}$, ortho isomer) exhibits intramolecular hydrogen bonding, making it steam volatile.
  • 4-Hydroxybenzaldehyde ($\mathbf{R}$, para isomer) exhibits intermolecular hydrogen bonding, leading to higher boiling points and rendering it non-steam volatile.
  • Therefore, Option A is incorrect.

• Option (B): $\mathbf{Q}$ yields a dark violet coloration upon reaction with $1\%$ aqueous $\text{FeCl}_3$ solution

  • Compound $\mathbf{Q}$ (salicylaldehyde) contains a free phenolic $-\text{OH}$ group. Free phenols react with aqueous $\text{FeCl}_3$ solution to form colored coordination complexes (dark violet coloration).
  • Therefore, Option B is correct.

• Option (C): $\mathbf{S}$ forms a yellow precipitate with 2,4-dinitrophenylhydrazine

  • Compound $\mathbf{S}$ retains the aldehyde group ($-\text{CHO}$) from salicylaldehyde. Aldehydes react readily with 2,4-DNP (2,4-dinitrophenylhydrazine) to give a characteristic yellow-orange hydrazone precipitate.
  • Therefore, Option C is correct.

• Option (D): $\mathbf{S}$ yields a dark violet coloration upon reaction with $1\%$ aqueous $\text{FeCl}_3$ solution

  • In compound $\mathbf{S}$, the phenolic hydroxy group is protected as a benzyl ether ($-\text{OCH}_2\text{Ph}$). Since $\mathbf{S}$ no longer contains a free phenolic $-\text{OH}$ group, it will not give a positive $\text{FeCl}_3$ test.
  • Therefore, Option D is incorrect.

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Conclusion:

The correct options are B and C.