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Question #44040
The Question
An organic compound (\ce{C8H8Cl2}) is treated with one equivalent of potassium phthalimide in DMF, followed by heating with hydrazine hydrate, to yield compound (\ce{C8H10ClN}) as the major product.
Compound gives a foul-smelling gas when heated with \ce{CHCl3} and alcoholic \ce{KOH}, but does \textbf{not} produce a red dye when treated with \ce{NaNO2}/\ce{HCl} at followed by alkaline 2-naphthol. Instead, the reaction of with \ce{NaNO2}/\ce{HCl} at evolves a colorless gas and forms compound .
Compound does \textbf{not} yield a yellow precipitate upon treatment with \ce{I2} and \ce{NaOH}. Vigorous oxidation of with acidic \ce{KMnO4} yields 4-chlorobenzoic acid.
Identify the structure of .
Options
- A.
\ce{1-chloro-4-(2-chloroethyl)benzene}
- B.
\ce{1-chloro-4-(1-chloroethyl)benzene}
- C.
\ce{1,4-bis(chloromethyl)benzene}
- D.
\ce{1-chloro-2-(chloromethyl)-4-methylbenzene}
Check Your Answer
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
Gabriel Phthalimide Synthesis|Aniline Resonance Pitfalls
Why This Question Is Tricky
Common Mistakes
This problem tests the boundary between aliphatic and aromatic reactivity in nucleophilic substitution.
The core trap is the 'Aryl Halide Inertness' in the Gabriel Phthalimide Synthesis; students often incorrectly assume the aryl chloride can undergo attack.
Furthermore, it integrates qualitative organic analysis (Carbylamine, Azo-dye, Iodoform tests) and side-chain oxidation to rigorously isolate the exact positional isomer, punishing those who ignore the negative iodoform test (differentiating primary vs secondary benzylic halides) or the exhaustive oxidation rules for alkylbenzenes.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Amines - Class 12 |
| Concept | Gabriel Phthalimide Synthesis |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |