Animated Solution for Chemistry - Organic Chemistry: Consider the following reaction scheme and choose the correct option(s) for the major products Q, R and S.

This JEE Advanced organic chemistry problem is a beautiful symphony of sequential reactions. It tests your ability to track functional group transformations, regioselectivity, and intramolecular cyclizations. Let's embark on this molecular journey step by step.

Setting the Stage with Hydroboration

The reaction scheme kicks off with Styrene ($Ph-CH=C{H}_2$) reacting with diborane ($B_2{H}_6$) followed by alkaline hydrogen peroxide ($NaOH,H_2{O}_2,H_{2}O$). This is the classic Hydroboration-Oxidation sequence.

Unlike acid-catalyzed hydration which follows Markovnikov's rule and involves carbocation intermediates (prone to rearrangement), hydroboration-oxidation is a concerted process. It adds water across the double bond in an anti-Markovnikov fashion. The boron atom attaches to the less sterically hindered terminal carbon, which is subsequently replaced by a hydroxyl group.

Thus, Styrene is smoothly converted into our first intermediate, P: 2-phenylethanol ($Ph-C{H}_2-C{H}_2-OH$).

Oxidation and Alpha-Halogenation

Now, the scheme splits. Let's follow the upper path to find product Q. Intermediate P is treated with Chromium trioxide in sulfuric acid ($Cr{O}_3,H_{2}S{O}_4$), commonly known as Jones reagent. This is a robust oxidizing agent that takes primary alcohols all the way up the oxidation ladder to carboxylic acids.

The resulting phenylacetic acid is then subjected to Chlorine and Red Phosphorus followed by water. This specific cocktail triggers the Hell-Volhard-Zelinsky (HVZ) reaction. The HVZ reaction is highly selective; it exclusively halogenates the $\alpha$-carbon (the carbon directly adjacent to the carboxyl group) by proceeding through an enol intermediate of the acyl halide.

This gives us product Q: $\alpha$-chloro-phenylacetic acid.

Chain Elongation via Substitution

Returning to intermediate P, the lower path aims to synthesize product R. First, 2-phenylethanol is reacted with Thionyl chloride ($SOC{l}_2$). This is an excellent method to convert alcohols into alkyl chlorides, often proceeding via an $S_{N}i$ mechanism with retention of configuration, though stereochemistry isn't the primary concern here.

Next, Sodium Cyanide ($NaCN$) is introduced. The cyanide ion is a strong nucleophile that executes a classic $S_{N}2$ attack, displacing the chloride leaving group. This step is strategically crucial because it performs a one-carbon homologation—it extends the carbon chain by one atom.

Finally, the nitrile is subjected to acidic hydrolysis with heat ($H_3{O}^{+},\Delta$). This vigorously converts the nitrile group into a carboxylic acid.

We have successfully synthesized product R: 3-phenylpropanoic acid.

Intramolecular Cyclization to S

The final transformation takes product R and treats it with concentrated sulfuric acid ($H_{2}S{O}_4$). When a carboxylic acid is heated with a strong mineral acid, it can be protonated and lose a molecule of water to generate a highly reactive acylium ion ($R-C^{+}=O$).

In our molecule, this powerful electrophile is tethered to an electron-rich benzene ring by a two-carbon chain. The proximity is perfect for an intramolecular Friedel-Crafts acylation. The acylium ion attacks the ortho position of the benzene ring, closing the loop.

Because of the tether length, the attack geometrically favors the formation of a stable, strain-free five-membered ring fused to the aromatic system.

This elegant cyclization yields product S: 1-indanone.

Comparing our meticulously derived structures for Q, R, and S with the given options, we find a perfect match with Option (B). This problem is a masterclass in organic synthesis, seamlessly weaving together addition, oxidation, substitution, and electrophilic aromatic substitution into a single, cohesive narrative.