Before You Begin
Question #44633
The Question
An optically active alkyl halide, (2, 3)-2-bromo-3-methylpentane, is subjected to two different sets of reaction conditions:
Condition I: Treatment with 1.0 equivalent of \ce{NaI} in anhydrous acetone at 298 K yields major product P. Condition II: Treatment with 1.0 equivalent of \ce{NaOMe} in methanol at 330 K yields major product Q.
Which of the following correctly identifies the stereochemistry of P and Q?
Options
- A.
P is (2, 3)-2-iodo-3-methylpentane; Q is ()-3-methylpent-2-ene
- B.
P is (2, 3)-2-iodo-3-methylpentane; Q is ()-3-methylpent-2-ene
- C.
P is (2, 3)-2-iodo-3-methylpentane; Q is ()-3-methylpent-2-ene
- D.
P is (2, 3)-2-iodo-3-methylpentane; Q is ()-3-methylpent-2-ene
Check Your Answer
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
Finkelstein and Swarts Exchange|E1/E2 Solvent and Base Effects|SN2 Stereochemical Inversion
Why This Question Is Tricky
Common Mistakes
This question tests the competition between S2 and E2 pathways dictated by solvent and reagent choices, alongside rigorous stereochemical analysis.
The Finkelstein condition (polar aprotic solvent) isolates the S2 pathway, testing the student's ability to invert only the reacting chiral center.
The E2 condition (strong base, protic solvent, heat) tests the anti-periplanar requirement on a chiral substrate.
Students often misapply the Newman projection rotations, incorrectly predicting the ()-alkene.
The hidden variable is the pre-existing anti-periplanar geometry in the staggered conformation, which directly dictates the formation of the ()-isomer.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Haloalkanes and Haloarenes - Class 12 |
| Concept | Finkelstein and Swarts Exchange |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |