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Question #44426
The Question
An optically active ether (\ce{C6H14O}) is treated with one equivalent of cold anhydrous \ce{HI} to yield an alkyl iodide and an alcohol . Compound produces a yellow precipitate when treated with \ce{I2} and aqueous \ce{NaOH}. When is heated with excess concentrated \ce{HI}, it produces and a rearranged alkyl iodide as the major product. Which of the following correctly identifies and ?
Options
- A.
= 2-methoxy-3-methylbutane; = 2-iodo-2-methylbutane
- B.
= 2-methoxy-3-methylbutane; = 2-iodo-3-methylbutane
- C.
= 2-methoxy-2-methylbutane; = 2-iodo-2-methylbutane
- D.
= 2-methoxypentane; = 2-iodopentane
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
Iodoform Reaction Requirements|Ether Cleavage Patterns
Why This Question Is Tricky
Common Mistakes
This problem tests the regioselectivity of ether cleavage and the stability of intermediates.
A common trap is assuming ether cleavage with \ce{HI} always follows an \ce{S_{N}1} pathway, which would lead to cleavage at the more substituted carbon and produce methanol (failing the iodoform test).
By correctly identifying the \ce{S_{N}2} pathway for a secondary-primary ether, the student finds the iodoform-positive alcohol.
Furthermore, the problem tests the hidden variable of carbocation rearrangement (1,2-hydride shift) when the resulting alcohol reacts with excess hot \ce{HI}, distinguishing the final thermodynamic product from the kinetic one.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Alcohols, Phenols, and Ethers - Class 12 |
| Concept | Iodoform Reaction Requirements |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |