Question 7

**Working out the sequence** - Tetralin (1,2,3,4‑tetrahydronaphthalene) with hot acidic KMnO₄: both benzylic CH₂ groups are oxidised, degrading the saturated ring to give phthalic acid (benzene‑1,2‑dicarboxylic acid). - Phthalic acid + NH₃, Δ, −2H₂O → **X = phthalimide** (cyclic imide, C₆H₄(CO)₂NH). - Phthalimide: heating, then ethanolic KOH gives potassium phthalimide; alkylation with R–Br gives **Y = N‑alkylphthalimide**. - Y + NaOH (hydrolysis) → sodium phthalate (the aromatic compound) + **Z = R–NH₂**, a primary aliphatic amine. This is the Gabriel phthalimide synthesis. **Checking the options** - (A) X (phthalimide) and Y (N‑alkylphthalimide) both contain C=O oxygens → correct. - (B) Carbylamine (CHCl₃/KOH) reaction needs a primary amine; Y is an imide with no N–H → incorrect. - (C) Z = RNH₂ (primary amine) reacts with Hinsberg's reagent (benzenesulfonyl chloride) to give an alkali‑soluble sulfonamide → correct. - (D) Z is an aliphatic primary amine, not aromatic → incorrect.

No written reasoning was returned for this item.

No written reasoning was returned for this item.