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Question #44629
The Question
Consider the reaction of 2,2-dimethyl-3,3-dideuterooxirane with excess potassium \textit{tert}-butoxide in \textit{tert}-butanol at , followed by mild aqueous workup. Which of the following represents the major organic product?
Options
- A.
\ce{H2C=C(CH3)-CD2OH}
- B.
\ce{(CH3)3C-O-CD2-C(CH3)2OH}
- C.
\ce{(CH3)3C-O-C(CH3)2-CD2OH}
- D.
\ce{(CH3)2CD-CD=O}
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
E1/E2 Solvent and Base Effects|Ether Cleavage Patterns|Steric Hindrance in SN2
Why This Question Is Tricky
Common Mistakes
Students typically memorize that base-catalyzed epoxide opening strictly follows SN2 regioselectivity at the less hindered carbon.
The hidden variable here is the 'neopentyl-like' steric hindrance at the primary carbon combined with the extreme bulkiness of the tert-butoxide base, which completely suppresses the SN2 pathway.
Instead, the strong base triggers an E2 elimination by abstracting a proton from the adjacent methyl group, leading to an allylic alcohol.
The isotopic labeling (CD2) acts as a trap to ensure the student correctly identifies which C-O bond breaks during the elimination.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Haloalkanes and Haloarenes - Class 12 |
| Concept | E1/E2 Solvent and Base Effects |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |