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Question #44361
The Question
Consider the reaction of (4R)-1-(2-bromophenoxy)-4-chloropentane with excess sodium amide (\ce{NaNH2}) in liquid ammonia at 195 K. Identify the major organic product formed.
Options
- A.
(2S)-5-(3-aminophenoxy)pentan-2-amine
- B.
(2S)-5-(2-bromophenoxy)pentan-2-amine
- C.
(2R)-5-(3-aminophenoxy)pentan-2-amine
- D.
(2S)-5-(2-aminophenoxy)pentan-2-amine
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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
Benzyne Intermediate (Elimination-Addition)|SN2 Stereochemical Inversion|C-X Bond Polarization and Polarizability
Why This Question Is Tricky
Common Mistakes
This problem tests the boundary between standard substitution and the Benzyne mechanism.
The primary trap is assuming only the alkyl chloride reacts (Option B).
The secondary trap involves misapplying electrophilic aromatic substitution rules (+M effect) to benzyne regioselectivity, leading to the ortho product (Option D).
In reality, benzyne nucleophilic addition is strictly controlled by inductive (-I) effects because the intermediate carbanion is orthogonal to the pi system, directing attack to the meta position.
Finally, the problem tests stereochemical inversion and IUPAC nomenclature, as the principal functional group changes, reversing the carbon chain numbering while preserving the R to S inversion logic.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Haloalkanes and Haloarenes - Class 12 |
| Concept | Benzyne Intermediate (Elimination-Addition) |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |