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Question #44405
The Question
An optically active haloalkane , \ce{CH3CH(Br)C(CH3)3}, is treated with potassium \textit{tert}-butoxide in \textit{tert}-butanol at to yield a major organic product . Subsequent treatment of with anhydrous \ce{HI} yields compound as the major product. Which of the following correctly identifies and ?
Options
- A.
= 2-(\textit{tert}-butoxy)-3,3-dimethylbutane ; = 2-iodo-3,3-dimethylbutane
- B.
= 3,3-dimethyl-1-butene ; = 2-iodo-3,3-dimethylbutane
- C.
= 3,3-dimethyl-1-butene ; = 2-iodo-2,3-dimethylbutane
- D.
= 2,3-dimethyl-2-butene ; = 2-iodo-2,3-dimethylbutane
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Mechanism_Trap
Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.
Concepts Behind This Question
Williamson Ether Synthesis Constraints|Ether Cleavage Patterns|E1/E2 Solvent and Base Effects
Why This Question Is Tricky
Common Mistakes
This problem tests three critical mechanistic traps.
First, it traps students who default to Williamson Ether Synthesis (SN2) instead of recognizing that a bulky base with a hindered halide exclusively drives E2 elimination.
Second, it traps students who incorrectly apply carbocation rearrangements to the concerted E2 pathway, falsely predicting the Zaitsev alkene (2,3-dimethyl-2-butene).
Finally, it tests the electrophilic addition of HI to the correct terminal alkene, requiring the recognition of a 1,2-methanide shift to yield the rearranged alkyl iodide.
Option C is the only path that correctly applies E2 kinetics followed by carbocation thermodynamics.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Alcohols, Phenols, and Ethers - Class 12 |
| Concept | Ether Cleavage Patterns |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |