JEE Main 2026 — Apr 2, Shift 2 — Chemistry solution

JEE Main2026Apr 2, Shift 2Chemistry

Consider the following reactions sequence

When the product (P) is subjected to Carius analysis using AgNO $_{3}$ , $1.0$ g of the product (P) will produce _________ g of the precipitate of AgBr. (Nearest integer)

(Given : molar mass in g mol $^{- 1}$ C : $12$ , H : $1$ , O : $16$ , N : $14$ , Br : $80$ , Ag : $108$ )

Your answer

Solution

Step (i): Sn/HCl reduces $-$ NO $_{2}$ in $p$ -nitrotoluene to $-$ NH $_{2}$ , giving $p$ -toluidine after basification.

Step (ii): Acetic anhydride acetylates the amine to $p$ -methylacetanilide, converting the strongly activating $-$ NH $_{2}$ to the more moderate $-$ NHCOCH $_{3}$ (this protects against over-bromination).

Step (iii): Br $_{2}$ /AlBr $_{3}$ brominates ortho to $-$ NHCOCH $_{3}$ (still strongly $o, p$ -directing) and ortho to $-$ CH $_{3}$ , giving a single Br on the ring.

Step (iv): Acidic hydrolysis cleaves the acetyl group, regenerating the free amine.

Product P is $2$ -bromo- $4$ -methylaniline (C $_{7}$ H $_{8}$ BrN), molar mass $= 7 (12) + 8 (1) + 80 + 14 = 186$ g/mol.

Each mole of P yields one mole of AgBr (molar mass $188$ ) in Carius analysis.

Mass of AgBr from $1.0$ g of P: $\frac{1.0}{186} \times 188 \approx 1.011$ g, i.e. $\approx 1$ g.

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