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Question #44363
The Question
An optically active alkyl halide, (2S, 3R)-2-bromo-3-phenylbutane, is treated with a concentrated solution of sodium ethoxide (\ce{NaOEt}) in ethanol at . Which of the following represents the major organic product formed?
Options
- A.
(Z)-2-phenylbut-2-ene
- B.
(E)-2-phenylbut-2-ene
- C.
Racemic 2-ethoxy-2-phenylbutane
- D.
3-phenylbut-1-ene
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
Optical Activity and Enantiomers|SN1 Carbocation Rearrangement|E1/E2 Solvent and Base Effects
Why This Question Is Tricky
Common Mistakes
Students often fall into two traps: assuming an SN1 mechanism due to the benzylic potential (leading to a rearranged racemic ether) or assuming that E2 elimination will yield the thermodynamically most stable alkene (Zaitsev's rule leading to the E-isomer).
The hidden variable is the strict stereospecificity of the E2 mechanism.
The fixed (2S, 3R) stereocenters dictate an anti-periplanar transition state that exclusively forms the sterically hindered (Z)-alkene, demonstrating that kinetic stereochemical requirements override thermodynamic stability.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Haloalkanes and Haloarenes - Class 12 |
| Concept | Optical Activity and Enantiomers |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |