Animated Solution for Chemistry - Amines: Match the four starting materials (P, Q, R, S) given in List I with the corresponding reaction scheme (I, II, III, IV) provided in List - II.

## Decoding the Organic Chemistry Matrix: A Journey Through Multi-Step Synthesis

Welcome to a masterclass in organic synthesis! This problem is a beautiful, sprawling matrix match that tests your endurance and your deep understanding of reagents, protecting groups, and directing effects. We are given four distinct starting materials and four complex, multi-step reaction schemes. Our mission is to trace the chemical transformations step-by-step, deduce the final product, and match its molecular formula to the correct scheme.

Let's roll up our sleeves and dive into the chemistry.

The Acetylene Trimerization

We begin with starting material P, which is acetylene ($H-C\equiv C-H$). The first reagent in Scheme III is a red-hot iron tube at $873K$. This is a classic, high-yield reaction where three moles of acetylene undergo cyclic polymerization to form benzene ($C_6{H}_6$).

Once we have the benzene ring, we treat it with a fuming mixture of nitric and sulfuric acid ($HN{O}_3/H_{2}S{O}_4$) under heat. This powerful nitrating mixture performs a double nitration. Because the first nitro group is strongly deactivating and meta-directing, the second nitro group attaches at the meta position, yielding m-dinitrobenzene.

Next comes a brilliant piece of selectivity: the Zinin reduction. By using hydrogen sulfide in ammonia ($H_{2}S\cdot N{H}_3$), we selectively reduce only one of the two nitro groups into an amine ($-N{H}_2$), giving us m-nitroaniline.

Finally, we treat this intermediate with sodium nitrite and sulfuric acid ($NaN{O}_2/H_{2}S{O}_4$) to form a diazonium salt. Subsequent hydrolysis (warming with water) replaces the diazonium group with a hydroxyl ($-OH$) group. The final product is m-nitrophenol, which has the molecular formula $C_6{H}_{5}N{O}_3$. This perfectly matches Scheme III, meaning P maps to 3.

The Art of Protection

Next up is starting material Q, resorcinol (1,3-dihydroxybenzene). Scheme IV begins by treating resorcinol with concentrated sulfuric acid ($H_{2}S{O}_4$) at $60^{\circ }C$. This leads to disulfonation. The bulky sulfonic acid ($-S{O}_{3}H$) groups attach at the 4 and 6 positions. Why is this important? Because these groups act as temporary "blockers" or protecting groups.

When we subsequently nitrate this molecule using concentrated $HN{O}_3$ and $H_{2}S{O}_4$, the highly reactive 4 and 6 positions are already occupied. The incoming nitro group is forced to attack the sterically hindered 2-position, right between the two hydroxyl groups. This forms 2-nitroresorcinol-4,6-disulfonic acid.

In the final step, heating the molecule with dilute sulfuric acid triggers desulfonation. The sulfonic acid groups are cleaved off, leaving behind the pristine 2-nitroresorcinol. Its molecular formula is $C_6{H}_{5}N{O}_4$, which matches Scheme IV. Thus, Q maps to 4.

Taming the Amine

Let's examine starting material R, nitrobenzene. Scheme II starts with a standard reduction using tin and hydrochloric acid ($Sn/HCl$), which converts the nitro group into an amine, forming aniline.

Aniline is notoriously reactive toward electrophilic aromatic substitution and is easily oxidized by nitric acid. To tame this reactivity, we react it with acetyl chloride ($C{H}_{3}COCl$). This converts the amine into an acetamido group ($-NHCOC{H}_3$), forming acetanilide. This is a classic protection strategy.

Now, we sulfonate the ring with concentrated $H_{2}S{O}_4$. The bulky acetamido group directs the incoming sulfonic acid group primarily to the para position. With the para position blocked, subsequent nitration with $HN{O}_3$ is forced to occur at the ortho position relative to the acetamido group.

Finally, we boil the mixture with dilute sulfuric acid. This acidic hydrolysis achieves two things simultaneously: it removes the acetyl protecting group (restoring the amine) and it removes the sulfonic acid group (desulfonation). The resulting product is o-nitroaniline, with the molecular formula $C_6{H}_6{N}_2{O}_2$. This matches Scheme II, meaning R maps to 2.

Side-Chain Oxidation

Our final starting material is S, p-nitrotoluene. Scheme I utilizes alkaline potassium permanganate ($KMn{O}_4,H{O}^{-}$) followed by acidic workup. $KMn{O}_4$ is a vigorous oxidizing agent that will chew up any alkyl side chain attached to a benzene ring (provided it has benzylic hydrogens) and convert it directly into a carboxylic acid ($-COOH$). This transforms p-nitrotoluene into p-nitrobenzoic acid.

To convert this acid into an amide, we first react it with thionyl chloride ($SOC{l}_2$). This replaces the hydroxyl group of the acid with a chlorine atom, forming the highly reactive p-nitrobenzoyl chloride.

Reacting this acid chloride with ammonia ($N{H}_3$) yields the final amide, p-nitrobenzamide. Counting the atoms gives us the molecular formula $C_7{H}_6{N}_2{O}_3$, which perfectly matches Scheme I. Therefore, S maps to 1.

By systematically tracing the reagents, understanding the role of protecting groups, and keeping a close eye on directing effects, we have successfully decoded the entire matrix!