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JEE Advanced 2026Paper 2
Exam
JEE Advanced
Year
2026
Paper
Paper 2
Subject
Chemistry
Chapter
Practical Chemistry

JEE Advanced 2026 · Paper 2 · Chemistry · Practical Chemistry

Passage: Consider the following reaction sequence in which J , K ,...

Passage: Consider the following reaction sequence in which J\textbf{J}, K\textbf{K}, L\textbf{L} and M\textbf{M} are the major products.
Passage: Consider the following reaction sequence in which J , K , L and M are the major products. Given: Atomic mass
Given:
Atomic mass (in amu): H : 11, C : 1212, N : 1414, O : 1616, S : 3232, Br : 8080, Ba : 137137

Question: In sulphur estimation by Carius method, the amount of BaSO4_4 formed from 3.793.79 g of M\textbf{M} is ____ g.

Correct answer

0

Step-by-step solution

Step 1: The starting material is mm-xylene (1,3-dimethylbenzene). It undergoes Friedel-Crafts acylation with chloroacetyl chloride (ClCH2COCl\text{ClCH}_2\text{COCl}) in the presence of anhydrous AlCl3\text{AlCl}_3 to form 2-chloro-1-(2,4-dimethylphenyl)ethanone.

Step 2: The chloro ketone reacts with NaI\text{NaI} (Finkelstein reaction) to form 2-iodo-1-(2,4-dimethylphenyl)ethanone.

Step 3: This iodo compound undergoes Williamson ether synthesis with sodium 3-nitrophenoxide to form compound J\textbf{J}.
The molecular formula of J\textbf{J} is C16H15NO4\text{C}_{16}\text{H}_{15}\text{NO}_4.

Step 4: Compound J\textbf{J} is reduced by NaBH4\text{NaBH}_4, which converts the ketone group to a secondary alcohol. Subsequent reaction with PBr3\text{PBr}_3 replaces the hydroxyl group with a bromine atom, yielding compound K\textbf{K}.
The molecular formula of K\textbf{K} is C16H16BrNO3\text{C}_{16}\text{H}_{16}\text{BrNO}_3.
Molar mass of K=(16×12)+(16×1)+80+14+(3×16)=350 g/mol\textbf{K} = (16 \times 12) + (16 \times 1) + 80 + 14 + (3 \times 16) = 350 \text{ g/mol}, which matches the given data.

Step 5: Compound K\textbf{K} reacts with sodium thiophenoxide (PhSNa\text{PhSNa}) in an SN2\text{S}_\text{N}2 reaction, replacing the bromide with a phenylthio group (-SPh\text{-SPh}) to form compound M\textbf{M}.
The molecular formula of M\textbf{M} is C16H16(SC6H5)NO3⇒C22H21NO3S\text{C}_{16}\text{H}_{16}(\text{SC}_6\text{H}_5)\text{NO}_3 \Rightarrow \text{C}_{22}\text{H}_{21}\text{NO}_3\text{S}.
Molar mass of M=350−80(for Br)+109(for SPh)=379 g/mol\textbf{M} = 350 - 80 (\text{for Br}) + 109 (\text{for SPh}) = 379 \text{ g/mol}.

Step 6: Calculate the moles of compound M\textbf{M} used in the Carius method.
Moles of M=3.79 g379 g/mol=0.01 mol\text{Moles of } \textbf{M} = \dfrac{3.79 \text{ g}}{379 \text{ g/mol}} = 0.01 \text{ mol}.

Step 7: Since each molecule of M\textbf{M} contains exactly one sulphur atom, the moles of sulphur present is 0.01 mol0.01 \text{ mol}. In the Carius method, all sulphur is quantitatively converted to barium sulphate (BaSO4\text{BaSO}_4).
Moles of BaSO4 formed=0.01 mol\text{Moles of BaSO}_4 \text{ formed} = 0.01 \text{ mol}.

Step 8: Calculate the mass of BaSO4\text{BaSO}_4 formed.
Molar mass of BaSO4=137(Ba)+32(S)+4×16(O)=233 g/mol\text{BaSO}_4 = 137 (\text{Ba}) + 32 (\text{S}) + 4 \times 16 (\text{O}) = 233 \text{ g/mol}.
Mass of BaSO4=0.01 mol×233 g/mol=2.33 g\text{Mass of BaSO}_4 = 0.01 \text{ mol} \times 233 \text{ g/mol} = 2.33 \text{ g}.

Answer: 2.332.33
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