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

Leaving Groups

Organic Chemistry

8. Elimination Reactions

Leaving Groups

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3:56 minutes

Problem 50b

Textbook Question

For which of the following reactions would you expect elimination to be more favored than substitution?
(b)

Verified step by step guidance

Step 1: Analyze the structure of the alkyl halides in both reactions. The first structure contains a tertiary alkyl bromide, while the second structure contains a secondary alkyl bromide. Tertiary alkyl halides are more prone to elimination reactions due to steric hindrance, which makes substitution less favorable.

Step 2: Consider the reaction conditions. The presence of NaOH in H2O suggests a basic environment. In such conditions, elimination reactions (E2 mechanism) are often favored, especially for tertiary alkyl halides, as the base can abstract a proton from a β-carbon to form an alkene.

Step 3: Evaluate the competition between substitution and elimination. For the tertiary alkyl bromide, steric hindrance around the carbon bearing the leaving group (Br) makes nucleophilic attack difficult, favoring elimination. For the secondary alkyl bromide, substitution (SN2 mechanism) is more competitive due to less steric hindrance.

Step 4: Consider the role of the solvent (H2O). Water is a polar protic solvent, which can stabilize the transition state for elimination reactions. This further supports elimination being favored for the tertiary alkyl bromide.

Step 5: Conclude that elimination is more favored for the reaction involving the tertiary alkyl bromide compared to the secondary alkyl bromide, due to steric hindrance and the basic reaction conditions promoting the E2 mechanism.

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Key Concepts

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

Elimination vs. Substitution Reactions

In organic chemistry, elimination reactions involve the removal of atoms or groups from a molecule, resulting in the formation of a double bond, while substitution reactions involve replacing one atom or group with another. The preference for one type of reaction over the other can depend on factors such as the structure of the substrate, the nature of the nucleophile or base, and the reaction conditions.

Zaitsev's Rule

Zaitsev's Rule states that in elimination reactions, the more substituted alkene is generally the favored product. This is because more substituted alkenes are typically more stable due to hyperconjugation and the inductive effect. Understanding this rule helps predict the outcome of elimination reactions, especially when comparing different substrates.

Sterics and Nucleophilicity

Steric hindrance refers to the crowding around a reactive center in a molecule, which can influence the pathway of a reaction. In the context of elimination versus substitution, bulky bases favor elimination due to their inability to approach the substrate for substitution. Additionally, the strength and size of the nucleophile or base can significantly affect whether substitution or elimination is favored in a given reaction.

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