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Comparing the nucleophilicity of 2 Bases:

  • Potassium Methoxide
  • Potassium t-BuO

Electrophile: 2-bromoheptane

Prediction:

t-BuO will produce a greater ratio of the Hoffman product to the Zaitsev and Methoxide will produce a greater ratio of the Zaitsev to the Hoffman.

More Abundant: Hofman, 1-heptene, 30.249%

Less Abundant: Zaitsev, 2-heptene, 19.047%

RATIO: 1.59 : 1

Most abundant - 1-heptene

2nd most abundant - 2-heptene

3rd most abundant - Unreacted product

More Abundant: Zaitsev, 2-heptene, 16.590%

Less Abundant: Hofman, 1-heptene, 4.685%

RATIO: 3.54 : 1

Most abundant - Unreacted Product

2nd most abundant - 2-heptene

3rd most abundant- 1-heptene

  • Steric hindrance of a base does impact the ratio of the products in an E2 reaction
  • Size: t-BuO>KCH3O
  • t-BuO - Hofman
  • KCH3O - Zaitsev
  • This is in agreement with what we have learned in lecture/lab!!

Results: Reactants, 2-bromoheptane

Results: t-BuO + 2-bromoheptane

Procedure

Unreacted

  • 2-bromoheptane + KCH3O (in methanol)
  • 2-bromoheptane + t-butoxide
  • Equal moles
  • Reflux - 30 minutes
  • Separatory funnel
  • 2 water washes
  • Methylene Chloride
  • GC-MS

  • Expected Result: 1-heptene major product (Hofman)
  • Fragmentation pattern: 98, 83, 70, 56, 41

Results: Potassium Methoxide + 2-bromoheptane

Major Product: Unreacted Product

Expected Result: 2-heptene

Fragmentation pattern: 69, 55, 41.

33.751%

How will the relative steric hindrance of a base impact the products of an E2 reaction?

Conclusions

Steric Hinderance in E2 Reactions

Myfanwy Adams and Shreya Thatai

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