Is E1 or E2 the Correct Mechanism for Bulky Base Eliminations?

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SUMMARY

The discussion centers on the mechanisms of E1 and E2 reactions in the context of bulky base eliminations, specifically involving t-Butanol and the Hoffmann elimination. The consensus is that E1 is the correct mechanism due to the bulky alcohol, which disfavors E2 reactions. The presence of a strong base is crucial for Hoffmann elimination, and the discussion questions whether such a base is present in this scenario. Ultimately, the solution manual's preference for E1 aligns with established chemical principles.

PREREQUISITES
  • Understanding of E1 and E2 elimination mechanisms
  • Familiarity with t-Butanol as a solvent
  • Knowledge of Hoffmann elimination and its requirements
  • Concept of leaving groups in organic chemistry
NEXT STEPS
  • Study the characteristics of E1 and E2 mechanisms in detail
  • Research the role of solvents in elimination reactions, focusing on polar, non-nucleophilic solvents
  • Examine the conditions required for Hoffmann elimination, particularly the use of strong bases
  • Explore the significance of leaving groups in determining reaction pathways
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Chemistry students, organic chemists, and educators looking to deepen their understanding of elimination mechanisms and the factors influencing them.

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TL;DR Summary: E1 or E2? I did the E2 hoffman elimination because it's tbuOh so it's bulky hence does E2 with Hoffman as major product so I chose option A. But the solution manual has done E1 reaction and rearrangement so the answer marked is C. What is the correct mechanism?

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The alcohol is bulky and therefore NOT likely to be engaging in E2. Hoffman elimination has a strong base. Is there a strong base in this example?

E1 is favored by a polar,non-nucleophilic solvent like t-Butanol and good leaving groups like I. This is textbook E1.
 
chemisttree said:
The alcohol is bulky and therefore NOT likely to be engaging in E2. Hoffman elimination has a strong base. Is there a strong base in this example?

E1 is favored by a polar,non-nucleophilic solvent like t-Butanol and good leaving groups like I. This is textbook E1.
I had read that bulky bases do hoffman elimination (E2) because of less crowding.
 
Where did you read that?

Hoffmann eliminations apply to quaternary alkyl amines in presence of a strong base.

Is that the case here?
 

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