JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

The given starting material is diazoaminobenzene (C_6H_5-N=N-NH-C_6H_5). When diazoaminobenzene is treated with an acid catalyst (HCl) and excess aniline, and warmed to 40-45^ C , it undergoes a rearrangement reaction (or dissociation followed by coupling) to form p-aminoazobenzene. The reaction sequence is as follows: 1. Diazoaminobenzene reacts with HCl to form benzenediazonium chloride and aniline. 2. The benzenediazonium ion then couples with the excess aniline at the para position to yield p-aminoazobenzene. 3. Cooling and adding acetic acid helps in precipitating the free base, p-aminoazobenzene, which is a yellow dye (often called Aniline Yellow). The structure of the yellow product (X), p-aminoazobenzene, is C_6H_5-N=N-C_6H_4-NH_2. The molecular formula of p-aminoazobenzene is C_ 12 H_ 11 N_3. Let's calculate its molar mass: Molar mass = (12 12) + (11 1) + (3 14) Molar mass = 144 + 11 + 42 = 197 g mol ^ -1 The mass of nitrogen in one mole of the compound is 3 14 = 42 g . The percentage of nitrogen in the product is: \% N = 42 197 100 21.32\% Rounding to the nearest integer, we get 21. Answer: 21