JEE Advanced 2025 · Paper 1 · Chemistry Q13

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ChemistryNumerical+4Question 45 of 48

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

X = 2-(2-bromophenyl)-1,3-dioxolane. X → [(i) Na, dry ether; (ii) H3O+] → P (molar mass 210, 100 %) → [(i) NaOH, Δ; (ii) H3O+] → Q (50 %) → [(i) NaOH, CaO; (ii) Δ] → R (50 %) → [PBr3, (C2H5)2O] → T (50 %). T then reacts with R in the presence of NaH, (C2H5)2O to give S (50 %).

Your answer

Answer and worked solution

Official answer: 175

X → P (100 %). Na in dry ether couples two aryl bromides (Wurtz-Fittig). $\ce{H3O+}$ then removes the acetal protecting groups. P is biphenyl-2,2′-dicarbaldehyde, $\ce{C14H10O2}$ , $M=210$ ✓. From 16 mol X: 8 mol P.
P → Q (50 %). NaOH with heat causes an intramolecular Cannizzaro reaction. Q is 2′-(hydroxymethyl)biphenyl-2-carboxylic acid: 4 mol.
Q → R (50 %). NaOH/CaO with heat decarboxylates it. R is 2-biphenylylmethanol: 2 mol.
R → T (50 %). $\ce{PBr3}$ gives 2-(bromomethyl)biphenyl, T: 1 mol.
T + R → S (50 %). NaH makes the alkoxide of R, and a Williamson synthesis with T gives the ether $(\ce{C6H5C6H4CH2})_2\ce{O}$ : 0.5 mol.

Mass of S. $M(\ce{C26H22O})=312+22+16=350$ , so
$0.5\times350=175\ \text{g}.$

Topics:Named reactions Haloarenes Ethers Carboxylic acids

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JEE Advanced 2025 · Paper 1, Chemistry question 13