Table of contents

1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

8. Elimination Reactions

E1 Reaction

Organic Chemistry

8. Elimination Reactions

E1 Reaction

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3:14 minutes

Problem 46b

Textbook Question

Provide a mechanism for the following E1 reactions.
(b)

Verified step by step guidance

Step 1: Identify the reaction type. This is an E1 elimination reaction, which proceeds via a two-step mechanism involving the formation of a carbocation intermediate.

Step 2: Protonation of the leaving group. In the presence of ethanol (EtOH), the iodine atom acts as the leaving group. The bond between the carbon and iodine breaks, forming a carbocation intermediate. The iodine departs as I⁻.

Step 3: Formation of the carbocation intermediate. The carbon attached to the iodine becomes positively charged after the iodine leaves. This carbocation is stabilized by hyperconjugation and inductive effects from the surrounding alkyl groups.

Step 4: Rearrangement of the carbocation (if necessary). In this case, the carbocation is already tertiary and stable, so no rearrangement occurs.

Step 5: Elimination of a proton to form the double bond. A base (ethanol or another molecule in the solution) abstracts a proton from a β-hydrogen (a hydrogen on a carbon adjacent to the carbocation). This results in the formation of a double bond, yielding the alkene product.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

E1 Reaction Mechanism

The E1 (unimolecular elimination) reaction mechanism involves two main steps: the formation of a carbocation intermediate followed by the loss of a leaving group to form a double bond. This mechanism typically occurs in polar protic solvents and is characterized by the rate-determining step being the formation of the carbocation, which is influenced by the stability of the carbocation formed.

Carbocation Stability

Carbocation stability is crucial in E1 reactions, as more stable carbocations lead to faster reaction rates. Stability is influenced by factors such as the degree of substitution (tertiary > secondary > primary) and resonance effects. In the provided reaction, the formation of a stable carbocation from the cyclohexane structure is essential for the subsequent elimination step to occur efficiently.

Role of Solvent

The choice of solvent plays a significant role in E1 reactions. Polar protic solvents, like ethanol (EtOH), stabilize the carbocation intermediate and facilitate the departure of the leaving group. This stabilization lowers the activation energy required for the reaction, making the E1 pathway more favorable compared to other mechanisms, such as E2, which may require a different solvent environment.

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