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Question #44042
The Question
An optically active primary amine () with an -configuration is synthesized in high yield from an amide using at . An attempt to synthesize via the Gabriel phthalimide synthesis using the appropriate chiral alkyl bromide and potassium phthalimide in DMF at fails, yielding a hydrocarbon () as the major product. Identify , , and .
Options
- A.
= -2-phenylpropanamide; = -(1-bromoethyl)benzene; = styrene
- B.
= -2-phenylpropanamide; = -(1-bromoethyl)benzene; = styrene
- C.
= -2-phenylpropanamide; = -(1-bromoethyl)benzene; = styrene
- D.
= -2-phenylpropanamide; = -(1-bromoethyl)benzene; = ethylbenzene
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Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Mechanism_Trap
Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.
Concepts Behind This Question
Gabriel Phthalimide Synthesis|Hoffmann Bromamide Rearrangement
Why This Question Is Tricky
Common Mistakes
This question tests the 'Lethal Collision' trap between nucleophilic substitution and molecular rearrangement.
Students often mechanically apply the Gabriel synthesis to any primary amine without considering the steric limitations of the alkyl halide precursor (secondary halides undergo E2 elimination with the bulky phthalimide base).
Furthermore, it rigorously tests the stereochemical outcomes of both reactions: retention of configuration in the Hoffmann rearrangement versus inversion of configuration in the SN2 pathway, requiring precise Cahn-Ingold-Prelog (CIP) priority tracking.
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 |