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Question #44417
The Question
When (1-methoxyethyl)cyclobutane is heated with excess concentrated \ce{HI} at 373 K, a reaction sequence occurs leading to the formation of two major alkyl iodide products. Identify these major products.
Options
- A.
Iodomethane and 1-iodo-1-methylcyclopentane
- B.
Iodomethane and (1-iodoethyl)cyclobutane
- C.
Iodomethane and 1-iodo-2-methylcyclopentane
- D.
Iodomethane and 1-ethyl-1-iodocyclobutane
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
Oxidation State Traps|Ether Cleavage Patterns|SN1 Carbocation Rearrangement
Why This Question Is Tricky
Common Mistakes
The core trap lies in the carbocation rearrangement cascade following ether cleavage.
While students may correctly identify the initial cleavage to yield iodomethane and an alcohol, the subsequent reaction of the alcohol presents a bifurcation.
The 'Chemical Exception' tested is the thermodynamic preference for ring expansion (relieving cyclobutane strain) over a simple 1,2-hydride shift that would immediately yield a tertiary carbocation but retain the strained four-membered ring.
Furthermore, it tests the recognition that the initial ring expansion yields a secondary carbocation, which must undergo a further 1,2-hydride shift to reach the global minimum (tertiary cyclopentyl cation) before halide attack.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Alcohols, Phenols, and Ethers - Class 12 |
| Concept | Oxidation State Traps |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |