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Question #44639
The Question
Alcohol \textbf{X} (3,3-dimethylbutan-2-ol) is treated with concentrated aqueous at to yield an alkyl halide \textbf{Y} as the major product. Subsequently, \textbf{Y} is treated with Magnesium turnings in dry diethyl ether. During this process, the \textit{in situ} generated Grignard reagent acts as a strong, sterically hindered base and reacts with unreacted \textbf{Y} via a bimolecular elimination pathway, yielding an alkene \textbf{Z} as the major hydrocarbon product. Identify the IUPAC name of \textbf{Z}.
Options
- A.
2,3-dimethylbut-2-ene
- B.
2,3-dimethylbut-1-ene
- C.
3,3-dimethylbut-1-ene
- D.
2,3-dimethylbutane
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
Grignard Reagent Sensitivity|SN1 Carbocation Rearrangement|E2 Regioselectivity (Saytzeff vs Hoffmann)
Why This Question Is Tricky
Common Mistakes
This problem tests two sequential exceptions.
First, students must recognize the 1,2-methyl shift in the SN1 reaction, avoiding the unrearranged trap (3,3-dimethylbut-1-ene).
Second, it tests the 'Hidden Variable' of Grignard sensitivity: tertiary halides undergo auto-elimination during Grignard synthesis.
The critical trap is E2 Regioselectivity (Saytzeff vs Hoffmann).
While standard E2 on 2-bromo-2,3-dimethylbutane with a small base yields the Zaitsev product (2,3-dimethylbut-2-ene), the in situ generated tertiary Grignard is an exceptionally bulky base.
Steric hindrance forces abstraction of the primary proton, making the Hoffmann product (2,3-dimethylbut-1-ene) the major product.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Haloalkanes and Haloarenes - Class 12 |
| Concept | Grignard Reagent Sensitivity |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |