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Question #44499
The Question
An aqueous solution of potassium 2,2-dimethylpropanoate () is subjected to Kolbe electrolysis using platinum electrodes at 298 K. Which of the following represents the major hydrocarbon product(s) formed at the anode?
Options
- A.
2,2,3,3-tetramethylbutane
- B.
2-methylpropane and 2-methylpropene
- C.
2,2-dimethylpropan-1-ol
- D.
2,2-dimethylbutane
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
Kolbe’s Electrolysis Mechanism|Steric Hindrance (B-Strain)|Hyperconjugation Logic
Why This Question Is Tricky
Common Mistakes
The question tests the limits of the standard Kolbe electrolysis coupling reaction.
Students typically memorize that yields and will reflexively choose 2,2,3,3-tetramethylbutane (Option A).
The hidden variable is the steric bulk of the intermediate tert-butyl radical.
When radicals are highly branched (tertiary), the steric hindrance (B-strain) prevents the two radical centers from approaching close enough to form a stable C-C bond.
Consequently, the reaction pathway shifts entirely from coupling to disproportionation, where one radical acts as a hydrogen atom abstractor, yielding an alkane and an alkene.
Failing to account for this steric inhibition of coupling is the primary trap.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Hydrocarbons - Class 11 Part 2 |
| Concept | Kolbe’s Electrolysis Mechanism |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |