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Question #45104
The Question
Consider the E2 elimination reaction of an isomer of 1-bromo-4-(\textit{tert}-butyl)-2-methylcyclohexane, specifically where the methyl group is \textit{cis} to the \textit{tert}-butyl group, and the bromo group is \textit{trans} to the methyl group. The reaction is carried out using a high concentration of sodium ethoxide () in ethanol () at . What is the major elimination product formed?
Options
- A.
5-(\textit{tert}-butyl)-1-methylcyclohex-1-ene
- B.
\textit{cis}-5-(\textit{tert}-butyl)-3-methylcyclohex-1-ene
- C.
\textit{trans}-5-(\textit{tert}-butyl)-3-methylcyclohex-1-ene
- D.
4-(\textit{tert}-butyl)-1-methylcyclohex-1-ene
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
Bond Angle and Strain|E1/E2 Solvent and Base Effects
Why This Question Is Tricky
Common Mistakes
This problem tests the 'Conformational Lock' exception in E2 eliminations.
Students typically apply Zaitsev's rule to predict the most substituted alkene (Thermodynamic control).
However, the hidden variable is the strict stereoelectronic requirement for anti-periplanar geometry in the transition state.
Because the bulky tert-butyl group locks the molecule, the required ring flip places the C2 methyl group in an axial position, removing the anti-periplanar beta-hydrogen necessary for Zaitsev elimination.
This forces the reaction to proceed via the C6 hydrogen, exclusively yielding the kinetically forced Hofmann product.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Organic Chemistry – Some Basic Principles and Techniques - Class 11 Part 2 |
| Concept | Bond Angle and Strain |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |