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Question #44017
The Question
Thymine (\ce{5-methylpyrimidine-2,4(1H,3H)-dione}) (s) is treated with one equivalent of sodium hydride (\ce{NaH}) (s) in anhydrous DMF (l) at 298 K, followed by the addition of one equivalent of methyl iodide (\ce{CH3I}) (l). Which of the following is the major organic product formed?
Options
- A.
\ce{1,5-dimethylpyrimidine-2,4(1H,3H)-dione}
- B.
\ce{3,5-dimethylpyrimidine-2,4(1H,3H)-dione}
- C.
\ce{5-ethylpyrimidine-2,4(1H,3H)-dione}
- D.
\ce{2-methoxy-5-methylpyrimidin-4(3H)-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
Enolate Regioselectivity|Nitrogenous Base Keto-Enol Tautomerism
Why This Question Is Tricky
Common Mistakes
The trap tests the student's ability to differentiate between standard ketone enolate chemistry and nucleobase reactivity.
Students often erroneously predict C-alkylation (due to vinylogous alpha-protons) or O-alkylation (due to high charge density on oxygen).
The hidden mechanism relies on recognizing the ambident nature of the thymine anion, applying HSAB theory for soft-soft N-alkylation, and utilizing the Curtin-Hammett principle where the less sterically hindered N1 position dictates the major kinetic product despite rapid N1/N3 anion equilibration.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Aldehydes, Ketones, and Carboxylic Acids - Class 12 |
| Concept | Enolate Regioselectivity |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |