Question 7

Concept: This reaction sequence is based on the oxidative cleavage of substituted benzenes, followed by the Gabriel Phthalimide Synthesis, which is a standard method for the preparation of pure primary aliphatic amines.

Step-by-step Reaction Breakdown:

• Step 1 (Formation of X): Tetralin undergoes vigorous oxidation with acidic $KMnO_4$ to form Phthalic acid (benzene-1,2-dicarboxylic acid). When phthalic acid is treated with $NH_3$ and heated ($\Delta, -2H_2O$), it forms Phthalamide ($\mathbf{X}$). (Note: The explicit loss of $2H_2O$ indicates the formation of phthalamide).
• Step 2 (Formation of Y): * (i) Strong heating of Phthalamide ($\mathbf{X}$) causes the loss of an $NH_3$ molecule, resulting in ring closure to form Phthalimide.
  • (ii) Reaction with ethanolic $KOH$ forms Potassium phthalimide (a strong nucleophile).
  • (iii) Reaction with an alkyl halide ($R-Br$) results in an $S_N2$ substitution, yielding $N$-alkylphthalimide ($\mathbf{Y}$).
• Step 3 (Formation of Z): Alkaline hydrolysis (using $NaOH$) of $N$-alkylphthalimide ($\mathbf{Y}$) yields Sodium phthalate (the aromatic compound) and a primary aliphatic amine, $R-NH_2$ ($\mathbf{Z}$).

Evaluation of Options:

• (A) Both X and Y are oxygen containing compounds: True. $\mathbf{X}$ (Phthalamide) and $\mathbf{Y}$ ($N$-alkylphthalimide) both contain carbonyl oxygen atoms.
• (B) Y on heating with $CHCl_3/KOH$ forms isocyanide: False. $\mathbf{Y}$ is an imide, not a primary amine. Only primary amines give the positive isocyanide (carbylamine) test.
• (C) Z reacts with Hinsberg's reagent: True. $\mathbf{Z}$ ($R-NH_2$) is a primary amine, which reacts with Hinsberg's reagent (Benzenesulfonyl chloride, $C_6H_5SO_2Cl$) to form an $N$-alkylbenzenesulfonamide that is soluble in alkali.
• (D) Z is an aromatic primary amine: False. The Gabriel phthalimide synthesis cannot be used to prepare aromatic primary amines because aryl halides do not undergo nucleophilic substitution ($S_N2$) with the phthalimide anion under normal conditions. Thus, $\mathbf{Z}$ is strictly an aliphatic primary amine.

Final Answer: Options (A) and (C) are correct.