Consider the following reaction scheme and choose the correct option(s) for the major products Q, R, and S.

The correct option showing the major products Q, R, and S is:

Step-by-Step Reaction Mechanism:

1. Formation of Intermediate P (Hydroboration-Oxidation of Styrene):

Styrene, ${\text{C}}_6{\text{H}}_5\text{CH}={\text{CH}}_2$, undergoes hydroboration-oxidation using (i) ${\text{B}}_2{\text{H}}_6$ followed by (ii) $\text{NaOH},{\text{H}}_2{\text{O}}_2,{\text{H}}_2\text{O}$. This reaction follows anti-Markovnikov addition of water across the double bond to yield 2-phenylethan-1-ol (or 2-phenylethanol):

$\text{P}={\text{C}}_6{\text{H}}_5-{\text{CH}}_2-{\text{CH}}_2\text{OH}$

2. Formation of Product Q:

Intermediate P (${\text{C}}_6{\text{H}}_5{\text{CH}}_2{\text{CH}}_2\text{OH}$) undergoes a sequence of three reactions:

(i) Oxidation with Jones reagent (${\text{CrO}}_3/{\text{H}}_2{\text{SO}}_4$): Primary alcohol P is oxidized to 2-phenylacetic acid (${\text{C}}_6{\text{H}}_5{\text{CH}}_2\text{COOH}$).

(ii) Hell-Volhard-Zelinsky (HVZ) Reaction (${\text{Cl}}_2,\text{ Red phosphorus}$): Chlorination takes place specifically at the $\alpha$-position of the carboxylic acid to form 2-chloro-2-phenylacetyl chloride intermediate.

(iii) Hydrolysis (${\text{H}}_2\text{O}$): The acid chloride group hydrolyzes back to the carboxylic acid, producing 2-chloro-2-phenylacetic acid:

$\text{Q}={\text{C}}_6{\text{H}}_5-\text{CH(Cl)}-\text{COOH}$

3. Formation of Product R:

Starting again from P (${\text{C}}_6{\text{H}}_5{\text{CH}}_2{\text{CH}}_2\text{OH}$):

(i) Treatment with thionyl chloride (${\text{SOCl}}_2$): Replaces the -OH group with a chlorine atom to form (2-chloroethyl)benzene (${\text{C}}_6{\text{H}}_5{\text{CH}}_2{\text{CH}}_2\text{Cl}$).

(ii) Nucleophilic substitution with $\text{NaCN}$: Cyanide replaces the chloride ion to yield 3-phenylpropanenitrile (${\text{C}}_6{\text{H}}_5{\text{CH}}_2{\text{CH}}_2\text{CN}$).

(iii) Acidic hydrolysis with heating (${\text{H}}_3{\text{O}}^{+},\Delta$): Hydrolyzes the nitrile group completely to a carboxylic acid, forming 3-phenylpropanoic acid:

$\text{R}={\text{C}}_6{\text{H}}_5-{\text{CH}}_2-{\text{CH}}_2-\text{COOH}$

4. Formation of Product S:

Compound R (${\text{C}}_6{\text{H}}_5{\text{CH}}_2{\text{CH}}_2\text{COOH}$) is treated with concentrated sulfuric acid ($\text{conc. }{\text{H}}_2{\text{SO}}_4$). Under these strong acid conditions, intramolecular Friedel-Crafts acylation takes place, where the acylium ion attacks the benzene ring to form a five-membered cyclic ketone ring (1-indanone):

$\text{S}=\text{1-indanone}$

Therefore, the correct major products Q, R, and S match the second option provided in the question image.