Asked by Jenn
3. Predict the relative rates for the reaction of butadiene with ethylene, methyl propenoate and maleic anhydride
All Answers 1
Answered by
GPT-5 mini
AI
Order of reactivity (fastest → slowest):
maleic anhydride >> methyl propenoate (methyl acrylate) > ethylene.
Why:
- Diels–Alder rates are controlled by the HOMO(diene)–LUMO(dienophile) gap. Electron‑withdrawing substituents on the dienophile lower its LUMO and so increase the orbital overlap and rate.
- Maleic anhydride is a very strong electron‑withdrawing, cyclic dienophile (and is locked in the cis geometry required for the reaction), so it reacts fastest and often at or near room temperature.
- Methyl propenoate (methyl acrylate) is moderately activating (ester group), so it reacts faster than unsubstituted ethylene but much more slowly than maleic anhydride.
- Ethylene has no activating group (highest LUMO) and is the least reactive, requiring high temperature/pressure to cycloadd to butadiene.
(Also note maleic anhydride typically shows a strong endo preference due to secondary orbital interactions.)
maleic anhydride >> methyl propenoate (methyl acrylate) > ethylene.
Why:
- Diels–Alder rates are controlled by the HOMO(diene)–LUMO(dienophile) gap. Electron‑withdrawing substituents on the dienophile lower its LUMO and so increase the orbital overlap and rate.
- Maleic anhydride is a very strong electron‑withdrawing, cyclic dienophile (and is locked in the cis geometry required for the reaction), so it reacts fastest and often at or near room temperature.
- Methyl propenoate (methyl acrylate) is moderately activating (ester group), so it reacts faster than unsubstituted ethylene but much more slowly than maleic anhydride.
- Ethylene has no activating group (highest LUMO) and is the least reactive, requiring high temperature/pressure to cycloadd to butadiene.
(Also note maleic anhydride typically shows a strong endo preference due to secondary orbital interactions.)
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.