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Question #44037
The Question
Compound is 5-oxohexanamide. is treated with (1 equivalent) and (4 equivalents) in an aqueous medium. The reaction mixture is initially kept at and then heated to to yield an intermediate . Under the reaction conditions, spontaneously undergoes intramolecular cyclization with the loss of a water molecule to form the major product . Identify the major product .
Options
- A.
2-Methyl-1-pyrroline
- B.
6-Bromo-5-oxohexanamide
- C.
Piperidine-2,6-dione
- D.
5-Methyl-2-pyrrolidone
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Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Mechanism_Trap
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Concepts Behind This Question
Hoffmann Bromamide Rearrangement|Haloform Reaction Logic|Alpha-Hydrogen Acidity
Why This Question Is Tricky
Common Mistakes
The problem tests the student's ability to evaluate competitive kinetics and site-selectivity based on pKa differences.
The molecule contains both a primary amide and a methyl ketone.
Students often blindly apply the Haloform reaction to methyl ketones in alkaline Br2.
However, the pKa of a primary amide (~15) is significantly lower than the alpha-protons of a methyl ketone (~20).
Consequently, the amide is deprotonated and N-brominated much faster than the rate-limiting enolization of the ketone.
With only 1 equivalent of Br2, the reagent is entirely consumed by the amide, leading exclusively to the Hoffmann bromamide degradation.
The resulting primary amine then undergoes spontaneous intramolecular condensation with the intact ketone to form a 5-membered cyclic imine (2-methyl-1-pyrroline).
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Amines - Class 12 |
| Concept | Hoffmann Bromamide Rearrangement |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |