Organic Chemistry CH7 Lecture (10/29/25
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Overview of Alkane Stability and Elimination Reactions
Context of Discussion: Continued from previous discussion about alkane stability, moving into elimination reactions.
Main Concept: Conducting elimination reactions typically requires a strong base.
Alkoxides: Preparation and Examples
Definition of Alkoxide: Conjugate base derived from an alcohol, typically characterized as RO⁻ where R is an alkyl group (the negative charge is on oxygen).
Common Alkoxides Discussed:
Ethoxide: C₂H₅O⁻ (Sodium ethoxide, Na⁺C₂H₅O⁻)
Methoxide: CH₃O⁻ (Sodium methoxide, Na⁺CH₃O⁻)
Potassium hydroxide: KOH
Making an Alkoxide:
Start with a typical alcohol.
Use a strong base to deprotonate the alcohol.
Notable strong bases include Sodium Hydride (NaH) which deprotonates alcohols by providing H⁻ ions.
Mechanism of Alkoxide Formation:
Strong base (e.g., NaH) extracts a proton (H⁺) from the alcohol, creating an alkoxide (RO⁻).
Reaction produces an alkoxide and a correspondingly charged counter ion (e.g., Na⁺).
Reaction Mechanism Involvement
Role of Sodium in Alkoxides:
Sodium acts as a counter ion balancing the charge of the alkoxide and aids the formation of the alkoxide.
Example of Alkoxide Naming:
For a methyl groups (C₁H₃) alkoxide:
Example: Sodium methoxide is Na⁺OCH₃ (alkoxide is OCH₃⁻).
For ethyl groups (C₁H₂):
Example: Sodium ethoxide is Na⁺OCH₂CH₃ (alkoxide is OCH₂CH₃⁻).
Strong Bases and Elimination Reaction Types
Reactions mediated by sterically hindered bases (e.g., potassium tert-butoxide) lead to different product outcomes. Sterically hindered bases prevent actions on more substituted internal carbons.
Types of Elimination Reactions
E2 (Bimolecular Elimination):
Identified by the presence of two reactants during the rate-determining step, specifically the base and the substrate.
E1 (Unimolecular Elimination):
The rate determined solely by substrate concentration, similar to SN1 reactions.
General Mechanism of E2 Reactions
Generic Reaction Setup:
Base attacks a beta hydrogen and donates electron density, resulting in the cleavage of C-H and C-X (leaving group).
Forming a double bond between carbons while expelling the leaving group.
Transition State Dynamics:
Discussed electron density shifts, bond formations, and how orientation affects product configurations.
Importance of Anticoplanar Configuration
Definition: Hydrogen and leaving group must be anticoplanar to allow for successful elimination.
Geometry Importance: Maintaining this orientation ensures proper hybridization transitions necessary for the formation of a double bond (sp³ to sp² hybridization).
Example of E2 Reactions with Specific Substrates
Examining 2-bromo-2-methylbutane:
Identify alpha (attached to leaving group) and beta (adjacent carbons) hydrogens.
Emphasis on pathway analysis for hydrogen abstraction, with multiple products leading to stability comparisons (di-substituted vs. tri-substituted outcomes).
Zaitsev's Rule: Small bases favor formation of more substituted alkenes (more stable).
Hoffman Product: Larger bases yield less substituted alkenes due to steric hindrance.
Bases and Their Effects on Elimination Reactions
Effect of Base on Product Formation:
Small, less sterically hindered bases tend to product more stable products.
Conversely, more sterically hindered bases produce less stable products due to spatial limitations in accessing beta protons.
Case Study: Substrate Analysis with Steric Considerations
Menthol Analogue Comparison:
Differentiation of neomenthyl chloride vs menthyl chloride as diastereomers with the same elimination reaction; analysis focuses on product identification based on anticoplanar positioning of substituents.
Conclusion and Future Learning
Identification of products from elimination reactions leads to deeper understanding through discussing transitions, base effects, and spatial orientation compliance.
Homework: Review structures of alkoxides and practice product predictions for elimination reactions.