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Question #44934
The Question
Compound () is treated with 1 equivalent of potassium phthalimide in DMF at , followed by heating with aqueous to yield compound (). Compound is then treated with and at , followed by the addition of phenol in dilute . No azo dye is formed; instead, a clear solution with the evolution of a colorless gas is observed, yielding compound () as the major organic product. Compound on vigorous oxidation with acidic yields a monobasic acid (). Identify the correct structure of starting material .
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 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 .
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
| Subject | Chemistry |
| Chapter | Amines - Class 12 |
| Concept | Ammonolysis Complexity |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |