This reaction sequence involves organic transformations where we need to analyze the products and their chemical properties. Let's break it down step by step.

Step 1: Identify the Starting Compound and First Reaction

The starting compound is cumene (isopropylbenzene). The first reaction is oxidation with KMnO₄/KOH, which is a standard method for oxidizing alkyl side chains on aromatic rings to carboxylic acids. The isopropyl group (-CH(CH₃)₂) gets oxidized to a carboxylic acid group (-COOH).

So, cumene (C₆H₅CH(CH₃)₂) is oxidized to benzoic acid (C₆H₅COOH). Therefore, compound P is benzoic acid.

Step 2: Second Reaction - Conversion to Ester

Benzoic acid (P) reacts with SOCl₂ (thionyl chloride) to form an acid chloride. This is a common method to convert carboxylic acids to more reactive derivatives. The acid chloride then reacts with phenol in the presence of NaOH to form an ester. Phenol is activated by NaOH to form phenoxide ion, which is a good nucleophile for ester formation.

So, the product Q is phenyl benzoate (C₆H₅COOC₆H₅).

Step 3: Third Reaction - Fries Rearrangement

Phenyl benzoate (Q) undergoes Fries rearrangement with AlCl₃. This is a Lewis acid-catalyzed reaction that rearranges phenolic esters to ortho- and para-hydroxy ketones. Since the ortho position is sterically hindered, the para product is major.

So, the product R is 4-hydroxybenzophenone (p-hydroxybenzophenone).

Step 4: Fourth Reaction - Reduction with LiAlH₄

4-Hydroxybenzophenone (R) is reduced with LiAlH₄, which reduces carbonyl groups to alcohols. So, the ketone group (-CO-) is reduced to a secondary alcohol (-CH(OH)-).

The product S is 4-(hydroxy(phenyl)methyl)phenol, which is a benzhydrol derivative with a phenolic -OH group.

Analysis of the Options:

Option 1: S gives dark violet coloration with 1% aqueous FeCl₃ solution.
FeCl₃ test is used for phenols. S has a phenolic -OH group, so it should give a positive test (violet coloration). This is correct.

Option 2: R is steam volatile.
R is 4-hydroxybenzophenone. It has an intramolecular hydrogen bond between the carbonyl oxygen and the phenolic hydrogen, making it less volatile. Phenol is steam volatile due to intermolecular H-bonding, but here the intramolecular H-bonding reduces volatility. So, R is not steam volatile. This is incorrect.

Option 3: Q gives dark violet coloration with 1% aqueous FeCl₃ solution.
Q is phenyl benzoate, which is an ester. It does not have a free phenolic -OH group (it's esterified), so it will not give a positive FeCl₃ test. This is incorrect.

Option 4: S gives yellow precipitate with 2,4-dinitrophenylhydrazine.
2,4-DNPH test is for carbonyl compounds (aldehydes and ketones). S is an alcohol (reduced product) and does not have a carbonyl group, so it should not give a positive test. This is incorrect.

Final Answer:

Only option 1 is correct.

Related Topics and Formulae:

Related Topics: Oxidation of alkylbenzenes, Esterification, Fries rearrangement, Reduction of carbonyls, Qualitative tests for functional groups (FeCl₃ for phenols, 2,4-DNPH for carbonyls).

Key Reactions:
- Oxidation: $C{H}_{3}C{H}_{2}R\to COOHR$ (with KMnO₄/KOH)
- Esterification: $RCOOH+R\text{'}OH\to RCOOR\text{'}$ (via acid chloride)
- Fries Rearrangement: Phenolic ester to hydroxy ketone with AlCl₃
- Reduction: $C=O\to CHOH$ with LiAlH₄