Animated Solution for Chemistry - Organic Chemistry: In the following reaction sequence Q, R, S and T are the major products. The correct statement(s) about Q, R, S and T is (are)

Imagine you are an organic chemist tasked with building a complex heterocyclic molecule from a simple aliphatic salt. This problem takes us on exactly that epic journey. We begin with sodium butyrate, a four-carbon carboxylate salt.

The first step is Kolbe's electrolysis. When an aqueous solution of sodium butyrate is electrolyzed, the carboxylate anion migrates to the anode, loses an electron, and undergoes decarboxylation. This releases carbon dioxide gas and leaves behind a highly reactive propyl radical ($C_3{H}_7^{\bullet }$). Two of these propyl radicals rapidly dimerize to form a stable six-carbon alkane: hexane ($C_6{H}_{14}$).

But we don't stop at a straight chain. By passing hexane over a vanadium pentoxide ($V_2{O}_5$) catalyst at a scorching $500^{\circ }C$ and high pressure, we force the molecule to undergo aromatization. The chain folds onto itself, shedding hydrogen atoms to form the highly stable, conjugated ring of benzene. This is our intermediate Q.

Now that we have our aromatic core, it's time to decorate it. We introduce phthalic anhydride and anhydrous aluminum chloride ($AlC{l}_3$). This sets the stage for a classic Friedel-Crafts acylation. The Lewis acid activates the anhydride, generating a powerful electrophile. The benzene ring attacks, opening the anhydride ring. The result is o-benzoylbenzoic acid, our intermediate R. Notice how this molecule has two distinct functional groups: a ketone and a carboxylic acid.

To selectively modify the carboxylic acid, we treat R with phosphorus pentachloride ($PC{l}_5$). This aggressive chlorinating agent converts the $-COOH$ group into an acid chloride ($-COCl$).

Next comes a stroke of chemical elegance: the Rosenmund reduction. By using hydrogen gas with a palladium catalyst poisoned by barium sulfate ($Pd/BaS{O}_4$), we can selectively reduce the highly reactive acid chloride down to an aldehyde, without over-reducing it to an alcohol or touching the ketone group. This yields o-benzoylbenzaldehyde, our intermediate S.

Compound S is perfectly primed for a condensation reaction. It possesses both a ketone and an aldehyde group in close proximity. When we introduce hydrazine ($N{H}_{2}N{H}_2$) and apply heat, the two nitrogen atoms attack the two carbonyl carbons. This double condensation releases two molecules of water and stitches the molecule shut, forming a new six-membered ring containing two adjacent nitrogen atoms. This fused heterocyclic system is a phthalazine derivative (specifically, 1-phenylphthalazine), which is our final product T.

With our molecular cast fully assembled, let's evaluate the given statements:

Option (A): Compound S is o-benzoylbenzaldehyde. Because it contains an aldehyde group, it will readily reduce ammoniacal silver nitrate (Tollen's reagent) to metallic silver, producing a beautiful silver mirror. This statement is correct. Option (B): Compound Q is benzene. When benzene is exposed to excess chlorine gas under ultraviolet (UV) light, it undergoes a radical addition reaction, breaking its aromaticity to form benzene hexachloride ($C_6{H}_{6}C{l}_6$), commonly known as gammaxane or BHC. This statement is correct. Option (C): Compound T is a phthalazine derivative. Because its ring structure contains atoms other than carbon (specifically, nitrogen), it is by definition a heterocyclic compound. This statement is correct. Option (D): If we subject compound R (o-benzoylbenzoic acid) to acid-catalyzed cyclization, it forms anthraquinone. However, treating anthraquinone with zinc amalgam and hydrochloric acid (Clemmensen reduction) completely reduces the carbonyl groups to methylene groups, yielding anthracene, not 9,10-dihydroxyanthracene. Therefore, this statement is incorrect.

The correct options are (A), (B), and (C).