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Question #44511
The Question
Consider the following reaction sequence:
- \ce{4,4-dimethylpent-2-yne} is treated with aqueous \ce{H2SO4} in the presence of \ce{HgSO4} at 298 K to give the major product \textbf{P}.
- \textbf{P} is then treated with potassium \textit{tert}-butoxide (\ce{t-BuOK}) in \textit{tert}-butanol (\ce{t-BuOH}) at 355 K (reflux), followed by the addition of one equivalent of methyl iodide (\ce{CH3I}) to yield the major product \textbf{Q}.
Identify the IUPAC name of the major product \textbf{Q}.
Options
- A.
5,5-dimethylhexan-3-one
- B.
3,4,4-trimethylpentan-2-one
- C.
2,2,4-trimethylpentan-3-one
- D.
2,2-dimethylhexan-3-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
Alkyne Hydration Tautomerization|Enolate Regioselectivity|Reaction Intermediates Stability
Why This Question Is Tricky
Common Mistakes
The problem tests two critical exceptions to standard heuristic rules.
First, in the hydration of the unsymmetrical alkyne, hyperconjugation overrides the inductive effect, directing nucleophilic attack to the methyl-substituted carbon rather than the tert-butyl-substituted one.
Second, and more importantly, it traps students who blindly apply the rule that 'thermodynamic conditions yield the more substituted enolate'.
Due to extreme A(1,3) allylic strain and steric hindrance caused by the bulky tert-butyl group, the internal enolate is highly destabilized.
Thus, the terminal (less substituted) enolate is both the kinetic and thermodynamic product, leading to terminal alkylation.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Hydrocarbons - Class 11 Part 2 |
| Concept | Alkyne Hydration Tautomerization |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |