JEE Advanced 2022 — Paper 1 — CHEM-I solution

JEE Advanced2022Paper 1CHEM-I

Consider the following reaction.

On estimation of bromine in 1.00 g of R using Carius method, the amount of AgBr formed (in g) is ________.

[Given: Atomic mass of H = 1, C = 12, O = 16, P = 31, Br = 80, Ag = 108]

Your answer

Solution

Red phosphorous with Br $_{2}$ generates PBr $_{3}$ in situ, which converts a primary benzylic –OH into –Br:

$3 p-Br-C_{6} H_{4} -CH_{2} OH + PBr_{3} \to 3 p-Br-C_{6} H_{4} -CH_{2} Br + H_{3} PO_{3}$

So R = p-bromobenzyl bromide, C $_{7}$ H $_{6}$ Br $_{2}$ , with two Br atoms per molecule. Molar mass = $7 (12) + 6 (1) + 2 (80) = 84 + 6 + 160 = 250$ g/mol.

Moles of R in 1.00 g = $\frac{1}{250}$ .

Carius method: each Br becomes one AgBr. So moles of AgBr = $2 \times \frac{1}{250} = \frac{2}{250}$ .

Molar mass of AgBr = 108 + 80 = 188 g/mol.

Mass of AgBr = $\frac{2 \times 188}{250} = 1.504 \approx 1.50$ g.

← Back to full paper

Practice more CHEM-I

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →