JEE Advanced 2025 — Paper 1 — CHEM-III solution

JEE Advanced2025Paper 1CHEM-III

The reaction sequence given below is carried out with 16 moles of X. The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of S produced is _______.

Use: Atomic mass (in amu): H = 1, C = 12, O = 16, Br = 80

Your answer

Solution

Tracking the moles through each step, starting from 16 moles of X:

Step 1 (100% yield, Wurtz then acetal hydrolysis): two molecules of X couple to one P. Moles of P $= 16/2 = 8$ mol. P is biphenyl-2,2'-dicarbaldehyde (M = 210).

Step 2 (50%, intramolecular Cannizzaro): one CHO is reduced to CH $_{2}$ OH and the other oxidised to COOH on the same biphenyl. Q has 4 mol.

Step 3a (50%, soda-lime decarboxylation): removes the COOH, leaving 2-(hydroxymethyl)biphenyl R. Moles of R available for the final coupling $= 4 \times 0.5 = 2$ mol.

Step 3b (50%, PBr $_{3}$ ): R → T (2-bromomethylbiphenyl). Moles of T $= 2 \times 0.5 = 1$ mol.

Final step (50%, Williamson ether synthesis): R (deprotonated by NaH) + T → S = bis(biphenyl-2-yl-methyl) ether. Each S requires one R and one T; T is limiting (1 mol).

Moles of S $= 1 \times 0.5 = 0.5$ mol.

Molecular formula of S: (C $_{6}$ H $_{5}$ )–C $_{6}$ H $_{4}$ –CH $_{2}$ –O–CH $_{2}$ –C $_{6}$ H $_{4}$ –(C $_{6}$ H $_{5}$ ) $=$ C $_{26}$ H $_{22}$ O. Molar mass $= 26 \times 12 + 22 \times 1 + 16 = 312 + 22 + 16 = 350$ g/mol.

Mass of S $= 0.5 \times 350 = 175$ g.

$∴$ Mass of S $= 175.00$ g.

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