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Question #44375
The Question
Consider the following reaction sequence starting from 2-methylbut-2-ene (Compound ):
Compound is subjected to elimination under two different conditions: Condition I: Condition II:
Identify the major products and .
Options
- A.
= 3-methylbut-1-ene; = 2-methylbut-2-ene
- B.
= 2-methylbut-1-ene; = 2-methylbut-2-ene
- C.
= 2-methylbut-2-ene; = 3-methylbut-1-ene
- D.
= 2-methylbut-2-ene; = 2-methylbut-2-ene
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
Hydroboration-Oxidation Regiochemistry|E2 Regioselectivity (Saytzeff vs Hoffmann)
Why This Question Is Tricky
Common Mistakes
This problem tests the student's ability to track regiochemistry through a multi-step sequence, specifically linking anti-Markovnikov installation to subsequent elimination preferences.
The first trap is the HBO step: students might incorrectly predict Markovnikov addition or carbocation rearrangement, leading to a tertiary alcohol (which would yield 2-methylbut-1-ene with a bulky base, as seen in Option B).
The second trap is the E2 elimination of the secondary tosylate.
Students must recognize that the choice of base dictates the regioselectivity (Saytzeff vs.
Hoffmann).
A bulky base (t-BuOK) suffers from steric clash during internal proton abstraction, favoring the less substituted kinetic product (Hoffmann).
A small base (EtONa) favors the more stable, substituted alkene (Saytzeff).
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Alcohols, Phenols, and Ethers - Class 12 |
| Concept | Hydroboration-Oxidation Regiochemistry |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |