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Question #44632
The Question
Diphenyldibromomethane () is treated with an excess of Mercury(I) fluoride () in anhydrous toluene at 298 K in the dark. Considering the redox properties of the reagent and the stability of the intermediates, what is the major organic product formed?
Options
- A.
- B.
- C.
- D.
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Exception_Exposure
Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.
Concepts Behind This Question
Finkelstein and Swarts Exchange|Oxidation State Traps|E° (M²⁺/M) Anomalies
Why This Question Is Tricky
Common Mistakes
Students typically memorize the Swarts reaction as a straightforward halogen exchange (SN2) to form alkyl fluorides.
The trap lies in overlooking the disproportionation of Mercury(I) to Mercury(II) and elemental Mercury.
When paired with a sterically hindered, electronically stabilized substrate like diphenyldibromomethane in a non-polar solvent (which suppresses SN1), the mechanism shifts from kinetic SN2 control to a Single Electron Transfer (SET) pathway.
The elemental mercury reduces the substrate, leading to carbenoid formation and subsequent dimerization, making tetraphenylethene the major product rather than the expected gem-difluoride.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Haloalkanes and Haloarenes - Class 12 |
| Concept | Finkelstein and Swarts Exchange |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |