JEE Advanced 2021 Paper 1 Q12 Chemistry Organic Chemistry Haloalkanes & Haloarenes Hard

JEE Advanced 2021 Paper 1 · Q12 · Haloalkanes & Haloarenes

The reaction of Q with PhSNa yields an organic compound (major product) that gives positive Carius test on treatment with Na$_2$O$_2$ followed by addition of BaCl$_2$. The correct option(s) for Q is(are):

  1. A. [Figure: option A — 2,4-dinitrofluorobenzene: a benzene ring bearing two NO$_2$ groups (at positions 1 and 3 of the depicted ring, i.e. one ortho and one para to the leaving group) and F at the carbon flanked by an ortho-NO$_2$. The fluorine is the leaving group; activating NO$_2$ groups are ortho and para to F.]
  2. B. [Figure: option B — 1,3-dinitro-5-iodobenzene: benzene ring with NO$_2$ at C1 and C3 (meta to each other) and I at C5 (meta to both NO$_2$). The leaving group I has only meta NO$_2$ groups, no ortho/para EWG.]
  3. C. [Figure: option C — 3-(methylsulfanyl)-5-nitro-bromobenzene: benzene ring with MeS at C1, Br at C3 and NO$_2$ at C5. The Br has meta NO$_2$ and meta MeS only.]
  4. D. [Figure: option D — 2-(methylsulfanyl)-4-nitrobenzyl chloride: a CH$_2$Cl group attached to a benzene ring that bears NO$_2$ para to the CH$_2$Cl and MeS ortho to the CH$_2$Cl. The leaving group is a benzylic Cl, so it reacts by ordinary S$_N$2 (benzyl-chloride mechanism), giving a benzyl thioether bearing the MeS group on the ring.]
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Correct answer:A, D

Solution

The Carius test for sulfur converts an organic compound containing S to sulfate via Na$_2$O$_2$ fusion; addition of BaCl$_2$ to the resulting solution then gives a white precipitate of BaSO$_4$. To obtain BaSO$_4$, Q must give a sulfur-containing product on reaction with PhSNa.

(A) 2,4-dinitrofluorobenzene has both NO$_2$ groups ortho/para to the leaving F, so it undergoes facile aromatic nucleophilic substitution (S$_N$Ar) with PhSNa to give 2,4-dinitrothiophenyl ether (an Ar-S-Ph product). This product contains S → positive Carius test. (A) is correct.

(B) and (C): the NO$_2$ groups are only meta to the leaving group (I or Br), so they cannot activate the ring for S$_N$Ar; no thioether is formed. (B) and (C) are wrong.

(D) The leaving group is a benzylic chloride (sp$^3$ carbon, not aryl), so a standard S$_N$2 with PhS$^-$ gives a benzyl phenyl sulfide that retains its own MeS group on the ring. The product contains S → positive Carius test. (D) is correct.

Thus (A) and (D) are correct.

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