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Question #44920

Difficulty: 8/10 | 🟢 Free

The Question

Consider three isomeric amines A\mathbf{A}, D\mathbf{D}, and F\mathbf{F} with the molecular formula \ceC7H9N\ce{C7H9N}.

  1. A\mathbf{A} cannot be prepared by Gabriel phthalimide synthesis. It reacts with Hinsberg's reagent to form a precipitate insoluble in aqueous \ceNaOH\ce{NaOH}. Treatment of A\mathbf{A} with \ceNaNO2/HCl\ce{NaNO2/HCl} at 273 K273\text{ K} yields a yellow oil C\mathbf{C}.
  2. D\mathbf{D} can be prepared by Gabriel phthalimide synthesis. Treatment of D\mathbf{D} with \ceNaNO2/HCl\ce{NaNO2/HCl} at 273 K273\text{ K} produces brisk effervescence of a diatomic gas and a major organic product E\mathbf{E}.
  3. F\mathbf{F} cannot be prepared by Gabriel phthalimide synthesis, but its Hinsberg product is completely soluble in aqueous \ceNaOH\ce{NaOH}. Treatment of F\mathbf{F} with \ceNaNO2/HCl\ce{NaNO2/HCl} at 273 K273\text{ K} yields a water-soluble salt G\mathbf{G} which does not evolve gas. Warming the aqueous solution of G\mathbf{G} to 298 K298\text{ K} yields H\mathbf{H}.

Identify C\mathbf{C}, E\mathbf{E}, and H\mathbf{H} (assuming F\mathbf{F} is the para-isomer).

Options

  • A.

    C\mathbf{C}: NN-methyl-NN-nitrosoaniline, E\mathbf{E}: Phenylmethanol, H\mathbf{H}: 4-methylphenol

  • B.

    C\mathbf{C}: 4-nitroso-NN-methylaniline, E\mathbf{E}: Chlorophenylmethane, H\mathbf{H}: 4-methylphenol

  • C.

    C\mathbf{C}: NN-methyl-NN-nitrosoaniline, E\mathbf{E}: 2-methylphenol, H\mathbf{H}: Phenylmethanol

  • D.

    C\mathbf{C}: 4-nitroso-NN-methylaniline, E\mathbf{E}: Phenylmethanediazonium chloride, H\mathbf{H}: 4-methylbenzenediazonium chloride

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Diazotization Temperature Sensitivity|Gabriel Phthalimide Synthesis|Hinsberg Reagent Selectivity

Why This Question Is Tricky

Common Mistakes

This problem tests the boundaries of amine synthesis and qualitative analysis.

The primary trap is the kinetic/thermodynamic distinction in diazotization: students often assume all diazonium salts are stable at 273 K, forgetting that aliphatic diazonium salts (from D) decompose instantly due to the lack of resonance stabilization, whereas aromatic ones (from F) require warming to 298 K to decompose.

A secondary trap involves the Hinsberg test and nitrosation: students frequently confuse the electrophilic aromatic substitution of tertiary amines with the N-nitrosation of secondary amines (forming the yellow oil C), or misidentify the solubility of sulfonamides in alkali.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterAmines - Class 12
ConceptDiazotization Temperature Sensitivity
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree