Reaction Sequence Involving Lactone Reduction and Oxidation Reactions

To determine the correct statements, we analyze each step of the given reaction sequence:

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1. Identification of Compound $\mathbf{P}$

The starting material is a $6$-membered cyclic lactone ($4$-benzyltetrahydro-$2H$-pyran-$2$-one).

Reduction of this lactone using lithium aluminium hydride ($\text{LiAlH}_4$) cleaves the ester linkage to yield a primary diol, 3-benzylpentane-1,5-diol ($\mathbf{P}$):

$\text{Lactone} \xrightarrow{\text{LiAlH}_4} \mathbf{P} : \text{HO}-\text{CH}_2-\text{CH}_2-\underset{\substack{| \\text{CH}_2\text{Ph}}}{\text{CH}}-\text{CH}_2-\text{CH}_2-\text{OH}$

• Symmetry and Optical Activity: The C-3 carbon of $\mathbf{P}$ is bonded to:

  1. A hydrogen atom ($-\text{H}$)
  2. A benzyl group ($-\text{CH}_2\text{Ph}$)
  3. Two identical $2$-hydroxyethyl groups ($-\text{CH}_2\text{CH}_2\text{OH}$)

  Since two identical substituents are attached to the central C-3 atom, the molecule possesses a plane of symmetry ($\sigma$). Consequently, $\mathbf{P}$ is achiral and optically inactive.

  $\implies$ Option A is incorrect.

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2. Identification of Compound $\mathbf{Q}$

Oxidation of diol $\mathbf{P}$ using Jones reagent ($\text{CrO}_3 / \text{H}_2\text{SO}_4$) oxidizes both primary alcohol groups ($-\text{CH}_2\text{OH}$) to carboxylic acid groups ($-\text{COOH}$):

$\mathbf{P} \xrightarrow{\text{CrO}_3/\text{H}_2\text{SO}_4} \mathbf{Q} : \text{HOOC}-\text{CH}_2-\underset{\substack{| \\text{CH}_2\text{Ph}}}{\text{CH}}-\text{CH}_2-\text{COOH}$

• Reaction with $\text{NaHCO}_3$: Compound $\mathbf{Q}$ is a dicarboxylic acid ($3$-benzylpentanedioic acid). Carboxylic acids react with aqueous sodium bicarbonate ($\text{NaHCO}_3$) to evolve carbon dioxide gas ($\text{CO}_2$), producing brisk effervescence:

  $\text{R-COOH} + \text{NaHCO}_3 \rightarrow \text{R-COONa} + \text{H}_2\text{O} + \text{CO}_2\uparrow$

  $\implies$ Option C is correct.

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3. Identification of Compound $\mathbf{R}$

Heating dicarboxylic acid $\mathbf{Q}$ with soda lime ($\text{NaOH}$ and $\text{CaO}, \Delta$) results in complete decarboxylation, removing both carboxylic acid groups as $\text{Na}_2\text{CO}_3$:

$\mathbf{Q} \xrightarrow{\text{NaOH, CaO}, \Delta} \mathbf{R} : \text{PhCH}_2-\text{CH}(\text{CH}_3)_2$

• Nature of Compound $\mathbf{R}$: Compound $\mathbf{R}$ is isobutylbenzene, which is an substituted aromatic alkane, not an alkyne.

  $\implies$ Option D is incorrect.

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4. Identification of Compound $\mathbf{S}$

Heating diol $\mathbf{P}$ with concentrated $\text{H}_2\text{SO}_4$ at $443\text{ K}$ ($170^\circ\text{C}$) undergoes acid-catalyzed double dehydration (elimination) to yield an unsaturated compound containing carbon-carbon double bonds ($\text{C}=\text{C}$), such as 3-benzylpenta-1,4-diene:

$\mathbf{P} \xrightarrow{\text{H}_2\text{SO}_4, 443\text{ K}} \mathbf{S} : \text{CH}_2=\text{CH}-\underset{\substack{| \\text{CH}_2\text{Ph}}}{\text{CH}}-\text{CH}=\text{CH}_2 + 2\text{H}_2\text{O}$

• Baeyer's Test: Alkenes/dienes contain $\text{C}=\text{C}$ double bonds and react with cold, dilute, alkaline $\text{KMnO}_4$ solution (Baeyer's reagent) to form glycol derivatives with the decolorization of the purple $\text{KMnO}_4$ solution. Thus, $\mathbf{S}$ gives a positive Baeyer's test.

  $\implies$ Option B is correct.

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Conclusion:

The correct statements are B and C.