See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: Diazonium salt formation and azo coupling reaction. p-Toluidine (4-methylaniline) reacts with NaNO2/HCl at 0-5°C to form the diazonium salt, 4-methylbenzenediazonium chloride (compound A): [4-CH3-C6H4-N2+]Cl-. Step 2 - Azo Coupling with Phenol in mild basic medium: In mild basic medium (slightly alkaline, e.g., pH ~8-10), phenol is converted partially to phenoxide ion (PhO-), which is a much stronger activating group for electrophilic aromatic substitution. The diazonium ion acts as a weak electrophile and couples with the activated aromatic ring. Step 3 - Position of coupling: Phenoxide ion directs electrophiles to ortho and para positions. The para position is preferred over ortho because it is less sterically hindered. Therefore, the diazonium salt couples at the para position of phenol (i.e., para to the OH group). Step 4 - Product structure: The product is 4-hydroxyphenyl-azo-4-methylphenyl, i.e., HO-C6H4-N=N-C6H4-CH3 (para,para), where OH is at the para position of one ring and CH3 is at the para position of the other ring. This corresponds to option (b). Step 5 - Why other options fail: (a) Incorrectly places the azo group on the methyl-bearing ring with no OH group, and involves unactivated benzene as the coupling component - wrong. (c) Shows a direct C-C bond to Ph rather than an azo linkage - wrong product type. (d) Shows coupling at the ortho position to OH, which is the minor product; the major product is para-coupling. Therefore, the correct answer is B.