JEE Advanced 2026 · Paper 2 · Chemistry Q17

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ChemistryNumerical+2Question 53 of 54

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Question Stem for Question Nos. 17 and 18

Consider the following reaction sequence in which J, K, L and M are the major products.

Given:
Atomic mass (in amu): H : 1, C : 12, N : 14, O : 16, S : 32, Br : 80, Ba : 137

m-Xylene with (1) ClCOCH₂Cl, anhyd. AlCl₃; (2) NaI, heat; (3) sodium 3-nitrophenoxide gives J. J with (1) NaBH₄, (2) PBr₃ gives K (molar mass 350 g/mol). K with NH₃ (excess) gives L (a nitrogen-containing compound); K with PhSNa gives M (a sulphur-containing compound).

The volume of 1 M aqueous $\ce{H2SO4}$ required to completely neutralize the ammonia evolved from 5.72 g of L in Kjeldahl's method of nitrogen estimation is ____ mL.

Your answer

Answer and worked solution

Official answer: 10

The sequence.

m-Xylene undergoes Friedel-Crafts acylation with $\ce{ClCOCH2Cl}$ at C4, giving $\ce{ArCOCH2Cl}$ (Ar = 2,4-dimethylphenyl).
NaI (Finkelstein) gives $\ce{ArCOCH2I}$ .
3-Nitrophenoxide ( $\ce{Ar'O-}$ ) displaces iodide: J $=\ce{ArCOCH2OAr'}$ .
$\ce{NaBH4}$ reduces C=O to $\ce{CH(OH)}$ , and $\ce{PBr3}$ converts it to $\ce{CHBr}$ : K $=\ce{ArCHBrCH2OAr'}$ , $\ce{C16H16BrNO3}$ , $M=350$ ✓.

L. $\ce{NH3}$ substitutes Br: $\ce{ArCH(NH2)CH2OAr'}$ , $\ce{C16H18N2O3}$ , $M=286$ .

Kjeldahl. The method converts amine nitrogen to $\ce{NH3}$ but not nitro nitrogen. So each L gives one $\ce{NH3}$ . $5.72/286=0.020$ mol $\ce{NH3}$ needs 0.010 mol $\ce{H2SO4}$ , i.e. 10 mL of 1 M acid.

Topics:Qualitative and quantitative analysis of organic compounds Aromatic hydrocarbons and electrophilic substitution SN1 and SN2

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JEE Advanced 2026 · Paper 2, Chemistry question 17