Reaction Sequence Involving Phthalimide and Gabriel Phthalimide Synthesis

To determine the correct statements, let us analyze the reaction sequence step by step:

1. Formation of Compound $\mathbf{X}$:

   • Step (i): Tetralin ($1,2,3,4$-tetrahydronaphthalene) is subjected to strong oxidation with acidic potassium permanganate ($\text{KMnO}_4/\text{H}^+, \Delta$). The saturated alicyclic ring undergoes cleavage to yield phthalic acid (benzene-$1,2$-dicarboxylic acid): $\text{Tetralin} \xrightarrow{\text{KMnO}_4, \text{H}^+, \Delta} \text{Phthalic acid } (\text{C}_8\text{H}_6\text{O}_4)$
   • Step (ii): Reaction of phthalic acid with ammonia ($\text{NH}_3$) followed by heating leads to the elimination of two water molecules ($-2\text{H}_2\text{O}$) to form phthalimide ($\mathbf{X}$): $\text{Phthalic acid} + \text{NH}_3 \xrightarrow{\Delta, -2\text{H}_2\text{O}} \text{Phthalimide } (\mathbf{X}, \text{C}_8\text{H}_5\text{NO}_2)$
   • Compound $\mathbf{X}$ is phthalimide, which is an oxygen-containing compound (contains two carbonyl oxygens).

2. Formation of Compound $\mathbf{Y}$:

   • Phthalimide ($\mathbf{X}$) is treated with ethanolic $\text{KOH}$ to form potassium phthalimide, which then undergoes a nucleophilic substitution reaction ($S_N2$) with alkyl bromide ($\text{R-Br}$). This is the classic Gabriel phthalimide synthesis: $\text{Phthalimide } (\mathbf{X}) \xrightarrow[\text{(ii) Ethanolic KOH, (iii) R-Br}]{\text{(i) Strong heating}} N\text{-alkylphthalimide } (\mathbf{Y})$
   • Compound $\mathbf{Y}$ is $N$-alkylphthalimide ($\text{C}_8\text{H}_4\text{O}_2\text{N-R}$), which also contains oxygen atoms.

3. Formation of Compound $\mathbf{Z}$:

   • Alkaline hydrolysis of $N$-alkylphthalimide ($\mathbf{Y}$) using aqueous $\text{NaOH}$ yields sodium phthalate (aromatic compound) and a primary aliphatic amine ($\mathbf{Z}$): $N\text{-alkylphthalimide } (\mathbf{Y}) \xrightarrow{\text{NaOH}} \text{Disodium phthalate (Aromatic compound)} + \text{R-NH}_2 (\mathbf{Z})$
   • Compound $\mathbf{Z}$ is an aliphatic primary amine ($\text{R-NH}_2$).

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Evaluation of Options:

• (A) Both $\mathbf{X}$ and $\mathbf{Y}$ are oxygen containing compounds:

  • $\mathbf{X}$ (phthalimide) contains two carbonyl oxygen atoms ($\text{C}_8\text{H}_5\text{NO}_2$).
  • $\mathbf{Y}$ ($N$-alkylphthalimide) also contains two carbonyl oxygen atoms ($\text{C}_8\text{H}_4\text{O}_2\text{N-R}$).
  • Statement (A) is CORRECT.

• (B) $\mathbf{Y}$ on heating with $\text{CHCl}_3/\text{KOH}$ forms isocyanide:

  • The carbylamine test ($\text{CHCl}_3/\text{KOH}$) is given exclusively by primary amines ($\text{1}^\circ$ amines).
  • Since $\mathbf{Y}$ is an imide ($N$-alkylphthalimide) and not a primary amine, it does not give the carbylamine test.
  • Statement (B) is INCORRECT.

• (C) $\mathbf{Z}$ reacts with Hinsberg's reagent:

  • $\mathbf{Z}$ is a primary aliphatic amine ($\text{R-NH}_2$). Primary amines readily react with Hinsberg's reagent (benzenesulfonyl chloride, $\text{C}_6\text{H}_5\text{SO}_2\text{Cl}$) to form $N$-alkylbenzenesulfonamides.
  • Statement (C) is CORRECT.

• (D) $\mathbf{Z}$ is an aromatic primary amine:

  • Gabriel phthalimide synthesis yields an aliphatic primary amine ($\mathbf{Z} = \text{R-NH}_2$), not an aromatic primary amine (since aryl halides do not undergo $S_N2$ substitution with potassium phthalimide).
  • Statement (D) is INCORRECT.

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Conclusion:

The correct statements are A and C.