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Question #44222
The Question
Consider the solvolysis of the optically active alkyl halide (2S,3S)-2-chloro-3-methylpentane in \ce{H2O} at 298 K. The reaction proceeds via an initial carbocation which undergoes a rapid 1,2-shift to form a more thermodynamically stable carbocation , ultimately yielding the major product . The global enthalpy path shifts due to the superior hyperconjugative stabilization of .
Let and be the number of hyperconjugable -hydrogens in and , respectively. Identify the major product , its optical activity, and the difference in hyperconjugable hydrogens () that provides the thermodynamic driving force () for the shift.
Options
- A.
3-methylpentan-3-ol; Optically Inactive;
- B.
3-methylpentan-2-ol; Optically Active;
- C.
3-methylpentan-3-ol; Optically Active;
- D.
2-methylpentan-2-ol; Optically Inactive;
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Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Mechanism_Trap
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Concepts Behind This Question
Enthalpy Definition|SN1 Carbocation Rearrangement|Hyperconjugation Logic
Why This Question Is Tricky
Common Mistakes
This question tests the intersection of thermodynamic control in intermediates (Hammond Postulate) and stereochemistry.
Students often rote-learn that reactions of chiral substrates lead to 'racemization' (and thus optical inactivity via a racemic mixture), but fail to recognize when a rearrangement completely destroys the chiral center, yielding a fundamentally achiral molecule.
Furthermore, it forces the quantitative evaluation of hyperconjugation (-hydrogen counting) as the primary thermodynamic driver for the enthalpy shift (), overriding simple inductive effect assumptions.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Thermodynamics - Class 11 Part 1 |
| Concept | Enthalpy Definition |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |