Chapter 8: Problem 15
How does each of the following changes affect the rate of an E1 reaction? a. doubling [RX] d. changing the leaving group from \(\mathrm{Cl}^{-}\) to \(\mathrm{Br}\) b. doubling [B:] e. changing the solvent from DMSO to \(\mathrm{CH}_{3} \mathrm{OH}\) c. changing the halide from \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) to \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}\)
Short Answer
Step by step solution
Understanding E1 Reaction Kinetics
Effect of Doubling [RX]
Effect of Doubling [B:]
Effect of Changing the Leaving Group
Effect of Changing the Solvent
Effect of Changing the Substrate Structure
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Unimolecular Elimination
Leaving Group Stability
- \(\text{Br}^-\) is a better leaving group than \(\text{Cl}^-\) because of its larger size and greater ability to distribute and stabilize the negative charge across its volume.
- When the leaving group is changed from \(\text{Cl}^-\) to \(\text{Br}^-\), the rate of the E1 reaction increases, as a more stable leaving group results in a quicker formation of the carbocation intermediate.
Substrate Concentration
- Doubling the concentration of substrate \([\text{RX}]\) effectively doubles the speed at which the reaction takes place.
- This distinct relationship mirrors the rate law for E1 reactions: \[ \text{Rate} = k[ ext{RX}] \], indicating that altering substrate concentration can directly influence the overall time needed for the reaction to complete.
Carbocation Stability
- Tertiary carbocations, which are attached to three alkyl groups, are more stable than secondary, and far more stable than primary carbocations.
- In E1 reactions, a substrate that forms a more stable carbocation will generally react faster. For instance, when changing from \((\text{CH}_3)_3\text{CBr}\) to \(\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}\), the former leads to a tertiary carbocation which is more stable, thus accelerating the reaction.