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Question #44300
The Question
Consider the solvolysis of the following alkyl chlorides in 80% aqueous ethanol at 298 K: \begin{itemize} \item[(I)] 2-Chloro-2-methylpropane \item[(II)] 1-Chloroadamantane \item[(III)] 2-Chloro-3,3-dimethylbutane \end{itemize} Evaluate the following statements regarding their thermodynamic and kinetic profiles: \begin{itemize} \item[(P)] In the gas phase, the ionization energy of the 1-adamantyl radical is lower than that of the -butyl radical, yet I solvolyzes significantly faster than II in solution. \item[(Q)] The higher activation energy for the solvolysis of II compared to I is primarily due to steric exclusion, which prevents optimal solvent-shell organization and results in a less exothermic solvation enthalpy for the 1-adamantyl cation. \item[(R)] The solvolysis of III predominantly yields 2,3-dimethylbutan-2-ol (and its corresponding ethyl ether) via a 1,2-methyl shift, as the initial ionization produces a less stable secondary carbocation. \item[(S)] The initial heterolytic cleavage of III is kinetically faster than that of I due to the thermodynamic driving force of the subsequent rapid rearrangement to a tertiary carbocation. \end{itemize} Which of the following combinations of statements is CORRECT?
Options
- A.
P, Q, and R only
- B.
P and Q only
- C.
Q, R, and S only
- D.
P, R, and S only
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Numerical_Precision
Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.
Concepts Behind This Question
Successive Ionization Energies|Solvent Effects on Reactivity|SN1 Carbocation Rearrangement
Why This Question Is Tricky
Common Mistakes
This question tests the critical distinction between gas-phase thermodynamic stability (Ionization Energy) and solution-phase kinetic reactivity (Solvation Energy) in mechanisms.
Students often fall into the trap of assuming that a lower gas-phase IE or greater intrinsic carbocation stability always translates to a faster rate.
The 'Hidden Variable' here is the steric hindrance to solvation: bulky, rigid systems like 1-adamantyl cannot be effectively solvated, drastically reducing the exothermic solvent-assisted stabilization energy and increasing the activation barrier.
Additionally, it tests the kinetic vs. thermodynamic control of rearrangements, ensuring students recognize that a subsequent fast rearrangement (1,2-methyl shift) does not accelerate the initial rate-determining ionization step beyond that of a direct tertiary halide.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Classification of Elements and Periodicity in Properties - Class 11 Part 1 |
| Concept | Successive Ionization Energies |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |