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Chemistry Question 69 – JEE-MAIN 2025

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When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid (273 K) and acidified with acetic acid, the mass (g) of 0.1 mole of product formed is : (Given molar mass in g mol−1 H : 1, C : 12, N : 14, O : 16, S : 32)

Identify the type of organic reaction occurring between sulphanilic acid, nitrous acid, and 1-naphthylamine.

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Ninja Strategy•Approximate Molar Mass Estimation

Estimate the approximate molar mass of the product by counting carbons and other heavy atoms. A product formed from two aromatic rings and a sulfonic acid group will likely have a molar mass in the range of 200-400 g/mol. For 0.1 mole, the mass should be 20-40 g. Only option 2 (33 g) falls within this reasonable range.

Step 1: Diazotization of Sulphanilic Acid✦ Active

Sulphanilic acid (p-aminobenzenesulfonic acid, H2N−C6H4−SO3H) reacts with nitrous acid (HNO2) at 273 K to form p-benzenediazonium sulfonate, a zwitterionic diazonium salt. The molecular formula of this diazonium salt is C6H4N2SO3.

Step 2: Azo Coupling Reaction○ Expand

The diazonium salt (C6H4N2SO3) undergoes an azo coupling reaction with 1-naphthylamine (C10H7NH2). The coupling typically occurs at the 4-position of 1-naphthylamine (para to the amino group). The reaction forms an azo dye: SO3−−C6H4−N=N−C10H6−NH2. The molecular formula of the product is C16H12N3O3S.

Step 3: Molar Mass and Final Mass Calculation○ Expand

Using the given molar masses (H: 1, C: 12, N: 14, O: 16, S: 32), the molar mass of C16H12N3O3S is calculated:

(16×12)+(12×1)+(3×14)+(3×16)+(1×32)=192+12+42+48+32=326 g/mol

The mass of 0.1 mole of the product is 0.1 mol×326 g/mol=32.6 g. Rounding to the nearest integer among the options, the mass is 33 g.

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