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Question #44934

Difficulty: 8/10 | 🟢 Free

The Question

Compound A\mathbf{A} (\ceC9H10Br2\ce{C9H10Br2}) is treated with 1 equivalent of potassium phthalimide in DMF at 350 K350\text{ K}, followed by heating with aqueous \ceNaOH\ce{NaOH} to yield compound B\mathbf{B} (\ceC9H12BrN\ce{C9H12BrN}). Compound B\mathbf{B} is then treated with \ceNaNO2\ce{NaNO2} and \ceHCl\ce{HCl} at 273 K273\text{ K}, followed by the addition of phenol in dilute \ceNaOH\ce{NaOH}. No azo dye is formed; instead, a clear solution with the evolution of a colorless gas is observed, yielding compound C\mathbf{C} (\ceC9H11BrO\ce{C9H11BrO}) as the major organic product. Compound C\mathbf{C} on vigorous oxidation with acidic \ceKMnO4\ce{KMnO4} yields a monobasic acid D\mathbf{D} (\ceC7H5BrO2\ce{C7H5BrO2}). Identify the correct structure of starting material A\mathbf{A}.

Options

  • A.

    1-Bromo-4-(3-bromopropyl)benzene

  • B.

    1-(2-Bromoethyl)-4-(bromomethyl)benzene

  • C.

    1,4-Dibromo-2-propylbenzene

  • D.

    1-Bromo-2-(2-bromoethyl)-4-methylbenzene

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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

Ammonolysis Complexity|Gabriel Phthalimide Synthesis

Why This Question Is Tricky

Common Mistakes

The problem tests the fundamental limitation of the Gabriel Phthalimide Synthesis: the phthalimide anion is a bulky nucleophile that exclusively undergoes SN2S_N2 reactions with aliphatic halides and is completely inert towards aryl halides.

Students often rote-learn Gabriel synthesis as a universal primary amine generator and fall into the trap of attempting SNArS_NAr.

Furthermore, the problem tests the differentiation between aliphatic and aromatic primary amines using the diazonium coupling test, and the exhaustive oxidation of alkyl side chains on benzene rings where any length of alkyl chain with benzylic hydrogens is cleaved to a single carboxyl group, differentiating mono-alkylated from di-alkylated rings.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterAmines - Class 12
ConceptAmmonolysis Complexity
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree