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
- 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.
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