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Question #44044
The Question
An organic compound, 2-bromopropanal, is treated with one equivalent of Potassium phthalimide in N,N-dimethylformamide (DMF) at . Instead of the expected nucleophilic substitution, the reaction predominantly yields a volatile liquid that exhibits a strong IR absorption at and rapidly decolorizes a solution of in . Identify the major product .
Options
- A.
2-Phthalimidopropanal
- B.
Prop-2-enal
- C.
2-Aminopropanal
- D.
3-(1-Bromoethyl)-2-methyloxirane-2-carbaldehyde
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
Gabriel Phthalimide Synthesis|Alpha-Hydrogen Acidity
Why This Question Is Tricky
Common Mistakes
This question tests the 'Chemical Exception' where the Gabriel Phthalimide Synthesis fails.
Students often blindly apply SN2 substitution when they see Potassium phthalimide and an alkyl halide.
The hidden variable is the dual role of the phthalimide anion as a bulky base and the substrate being a secondary alpha-halo carbonyl.
The high acidity of the alpha-hydrogen causes reversible enolate formation, but the kinetic funneling via the Curtin-Hammett principle drives the irreversible E2 elimination of the beta-hydrogen.
The thermodynamic stability of the resulting conjugated system (prop-2-enal) makes elimination the major pathway, completely bypassing the expected Gabriel synthesis.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Amines - Class 12 |
| Concept | Gabriel Phthalimide Synthesis |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |