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Question #45095
The Question
Consider the following two reaction sequences performed on 2-methylcyclohexan-1-one:
Sequence 1:
- LDA (1.05 eq), THF,
- , Major Product P
Sequence 2:
- , ,
- , , DMF, Major Product Q
Identify the major products P and Q.
Options
- A.
\textbf{P} = 6-methylcyclohex-2-en-1-one; \textbf{Q} = 2-methylcyclohex-2-en-1-one
- B.
\textbf{P} = 2-methylcyclohex-2-en-1-one; \textbf{Q} = 6-methylcyclohex-2-en-1-one
- C.
\textbf{P} = 6-methylcyclohex-2-en-1-one; \textbf{Q} = 6-methylcyclohex-2-en-1-one
- D.
\textbf{P} = 2-methylcyclohex-2-en-1-one; \textbf{Q} = 2-methylcyclohex-2-en-1-one
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
Tautomerism Conditions|Enolate Regioselectivity|Reaction Intermediates Stability
Why This Question Is Tricky
Common Mistakes
This question tests the 'Kinetic vs Thermodynamic Control' trap in enolate chemistry.
Students often assume that the most stable product always forms, failing to recognize that LDA at low temperatures strictly yields the kinetic enolate due to the inability of the intermediate to equilibrate (insufficient thermal energy).
Conversely, acid-catalyzed halogenation proceeds via the thermodynamic enol.
A secondary trap lies in the IUPAC nomenclature: after the kinetic sequence forms the double bond at the less substituted side, the ring must be re-numbered to give the alkene the lowest possible locants, shifting the methyl group's position number from 2 to 6.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Organic Chemistry – Some Basic Principles and Techniques - Class 11 Part 2 |
| Concept | Tautomerism Conditions |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |