Nucleophilic Substitution and Carius Detection Test for Organic Substrate

To determine which structure(s) of compound $\mathbf{Q}$ yield a major organic product that gives a positive Carius test for sulfur, let us analyze the chemical reactions involved:

1. Principle of the Carius Test for Sulfur

When an organic compound containing sulfur is oxidized using sodium peroxide ($\text{Na}_2\text{O}_2$), the sulfur atom is converted into sulfate ions ($\text{SO}_4^{2-}$). Subsequent addition of barium chloride ($\text{BaCl}_2$) results in the formation of a white precipitate of barium sulfate ($\text{BaSO}_4$): $\text{Organic Sulfur} \xrightarrow{\text{Na}_2\text{O}_2} \text{SO}_4^{2-} \xrightarrow{\text{BaCl}_2} \text{BaSO}_4 \downarrow \text{ (white precipitate)}$

Thus, for $\mathbf{Q}$ to give a positive Carius test after reaction with $\text{PhSNa}$, the major organic product formed must contain sulfur.

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2. Analysis of the Options

• Option (A): 1-fluoro-2,4-dinitrobenzene

  • The fluorine atom is activated towards Nucleophilic Aromatic Substitution ($S_N\text{Ar}$) because the strongly electron-withdrawing nitro groups ($-\text{NO}_2$) are located at the ortho (position 2) and para (position 4) positions relative to the fluorine atom.
  • The thiolate nucleophile ($\text{PhS}^-$) attacks the ring, displacing $\text{F}^-$ via a Meisenheimer complex intermediate: $\text{C}_6\text{H}_3(\text{NO}_2)_2\text{F} + \text{PhSNa} \longrightarrow \text{C}_6\text{H}_3(\text{NO}_2)_2\text{SPh} + \text{NaF}$
  • The product, (2,4-dinitrophenyl)(phenyl)sulfide, contains a thioether linkage ($-\text{SPh}$) and therefore gives a positive Carius test for sulfur.
  • Option (A) is correct.

• Option (B): 1-iodo-3,5-dinitrobenzene

  • The nitro groups are located at the meta positions relative to the iodine atom.
  • Meta-nitro groups cannot stabilize the Meisenheimer complex intermediate by resonance. Therefore, nucleophilic aromatic substitution ($S_N\text{Ar}$) does not occur, and no sulfur-containing substitution product is formed.
  • Option (B) is incorrect.

• Option (C): 1-bromo-3-(methylthio)-5-nitrobenzene

  • The bromine atom is meta to both the $-\text{NO}_2$ and $-\text{SMe}$ groups, meaning it is not activated for $S_N\text{Ar}$ substitution by $\text{PhS}^-$.
  • Since no nucleophilic substitution occurs to give the desired major product, Option (C) is incorrect.

• Option (D): 4-(methylthio)-3-nitrobenzyl chloride

  • The compound contains a benzylic chloride ($-\text{CH}_2\text{Cl}$) group.
  • Benzylic halides undergo rapid $S_N2$ nucleophilic substitution with $\text{PhS}^-$: $\text{Ar-CH}_2\text{Cl} + \text{PhSNa} \longrightarrow \text{Ar-CH}_2\text{-SPh} + \text{NaCl}$
  • The resulting organic product contains sulfur atoms (both $-\text{SMe}$ and $-\text{SPh}$) and undergoes oxidation with $\text{Na}_2\text{O}_2$ followed by precipitation with $\text{BaCl}_2$ to give a positive Carius test.
  • Option (D) is correct.

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Conclusion

The correct options representing potential structures of $\mathbf{Q}$ are (A) and (D).