Animated Solution for Chemistry - Organic Chemistry: Match the compounds in List-I with the appropriate observations in List-II and choose the correct option.

Welcome to a beautiful problem on qualitative organic analysis. When faced with large, complex-looking molecules, the secret is not to get intimidated. Instead, we must act like detectives, hunting for specific functional groups that act as the chemical signatures of the molecule. Let's break down each compound and match it with its characteristic chemical test.

Compounds P and Q

Let's focus on Compound P first. If you look closely at its structure, you'll realize it's a dipeptide derivative. Notice the hydroxyl group attached directly to the benzene ring? That is a phenol. Right next to it, on the aliphatic chain, we have a free primary amine group ($-N{H}_2$).

What do these groups do? A phenolic $-OH$ gives a characteristic violet complex when treated with neutral ferric chloride ($FeC{l}_3$). Simultaneously, a free primary amine reacts with ninhydrin to give a deep purple color, famously known as Ruhemann's purple. This dual reactivity perfectly matches observation (2).

Moving on to Compound Q, we encounter another dipeptide. However, notice the nitrogen at the far left end. It is not a free amine; it is acetylated, forming an amide bond. There are absolutely no free primary amines and no phenolic groups in this entire molecule.

Because it lacks a free amine, it will not give the ninhydrin test directly. But, if we subject the molecule to complete hydrolysis, we will break the peptide and amide bonds, releasing free amino acids. These newly freed amino acids will then give a positive ninhydrin test. Furthermore, since there is no phenol present, the phthalein dye test will be negative. This perfectly aligns with observation (5).

Compounds R and S

Next up is Compound R, which is anilinium chloride ($PhN{H}_3^{+}C{l}^{-}$). This is a salt of aniline. In an aqueous solution, it exists in equilibrium with free aniline, which is a primary aromatic amine.

Primary aromatic amines are famous for undergoing a diazo coupling reaction. When reacted with phenyl diazonium salts in a mildly acidic medium, they couple to form p-aminoazobenzene, which is a brilliant yellow dye. This directly points us to observation (1).

Finally, let's examine Compound S. It is a substituted phenylhydrazine, specifically 2,6-dimethylphenylhydrazine. The key functional group here is the hydrazine moiety ($-NHN{H}_2$) at the right.

Hydrazines are classic reagents in carbohydrate chemistry. They act as strong nucleophiles, reacting with the aldehyde or ketone groups of reducing sugars like glucose to form hydrazones. This perfectly matches observation (3).

By systematically analyzing the functional groups, we have successfully decoded the puzzle. Compound P matches with 2, Q with 5, R with 1, and S with 3. This logical breakdown not only leads us to the correct answer but also reinforces the fundamental principles of organic qualitative analysis.