JEE Advanced Chemistry Question (2026) — Solution

Step-by-step solution

In the first reaction sequence: Step 1: Aniline reacts with acetic anhydride in the presence of pyridine to undergo acetylation, forming acetanilide. Step 2: Acetanilide is treated with a nitrating mixture (conc. HNO _3 and conc. H _2 SO _4) at 288 K . The - NHCOCH _3 group is moderately activating and ortho/para directing. Due to steric hindrance, the major product P is p-nitroacetanilide. Step 3: P is hydrolyzed with H _3 O ^+ to yield p-nitroaniline. Step 4: p-nitroaniline is diazotized using NaNO _2/ HCl at 273-278 K to form the diazonium salt Q , which is p-nitrobenzenediazonium chloride. In the second reaction sequence: Step 1: Cumene (isopropylbenzene) is oxidized by O _2 followed by acid hydrolysis ( H _3 O ^+) to produce phenol and acetone. The major aromatic product is phenol. Step 2: Phenol undergoes the Kolbe-Schmitt reaction with NaOH and CO _2 to form sodium salicylate. Step 3: Acidification with H _3 O ^+ yields salicylic acid as the major product S . In the final reaction: Salicylic acid ( S ) reacts with p-nitrobenzenediazonium chloride ( Q ) in an aqueous NaOH medium. This is an electrophilic aromatic substitution (azo coupling). The phenoxide ion formed is highly activating, and coupling occurs at the para position with respect to the - OH group (position 5 of salicylic acid). The resulting product T is an azo dye, which is a highly conjugated and coloured compound. Evaluating the options: (A) Treatment of Q (p-nitrobenzenediazonium chloride) with ethanol reduces the diazonium group, yielding nitrobenzene, not an aromatic aldehyde. (Incorrect) (B) The phthalein dye test is a characteristic test for phenols. Salicylic acid ( S ) contains a phenolic - OH group with a free para position (position 5) and gives a positive phthalein dye test when heated with phthalic anhydride and conc. H _2 SO _4. (Correct) (C) P is p-nitroacetanilide, which is a mononitro compound, not a dinitro compound. (Incorrect) (D) T is an azo dye, which is a coloured compound. (Correct) Answer: S gives positive phthalein dye test.; T is a coloured compound.