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Magazine Chapter 8: Problem 11
How does each of the following changes affect the rate of an \(E2\) reaction? a. tripling [RX] b.halving [B:] c. changing the solvent from \(CH_3OH\) to \(DMSO\) d. changing the leaving group from \(I^-\) to \(Br^-\) e. changing the base from \(^-OH\) to \(H_2O\) f. changing the alkyl halide from \(CH_3CH_2Br\) to \((CH_3)_2CHBr\)
Short Answer
Expert verified
a) Rate increases 3x; b) Rate decreases 2x; c) Rate increases; d) Rate decreases; e) Rate decreases; f) Rate increases.
Step by step solution
01
Understand E2 Reaction Rate Equation
An E2 reaction is bimolecular and the rate law is given by: \( \text{Rate} = k [\text{RX}] [\text{B:}] \), where \([\text{RX}]\) is the concentration of the alkyl halide and \([\text{B:}]\) is the concentration of the base. The reaction rate is directly proportional to the concentrations of both the alkyl halide and the base.
02
Analyze Impact of Tripling [RX]
Given the rate equation, tripling the concentration of \([\text{RX}]\) would multiply the reaction rate by 3. Thus, the rate of the reaction increases by a factor of 3.
03
Analyze Impact of Halving [B:]
According to the rate equation, halving the concentration of \([\text{B:}]\) would divide the reaction rate by 2. Therefore, the reaction rate is decreased by a factor of 2.
04
Effect of Solvent Change from Methanol to DMSO
Changing the solvent from methanol, a protic solvent, to DMSO, an aprotic solvent, generally increases the rate of E2 reactions. Aprotic solvents do not solvate the base as much, thus enhancing its effectiveness in promoting elimination.
05
Effect of Changing the Leaving Group from I- to Br-
In E2 reactions, better leaving groups stabilize the transition state more effectively. Since \(I^-\) is a better leaving group than \(Br^-\) due to its larger size and weaker bond strength with carbon, replacing \(I^-\) with \(Br^-\) will decrease the reaction rate.
06
Effect of Changing Base from Hydroxide Ion to Water
\(^-OH\) is a stronger base compared to \(H_2O\). Replacing \(^-OH\) with \(H_2O\), a much weaker base, will significantly decrease the reaction rate as the reaction relies on a strong base to abstract a proton.
07
Effect of Changing Alkyl Halide from Ethyl Bromide to Isopropyl Bromide
The rate of E2 reactions can be affected by the structure of the alkyl halide. Increasing steric bulk, as when changing from \(CH_3CH_2Br\) to \((CH_3)_2CHBr\), typically increases susceptibility to E2 reactions due to the increased availability of beta hydrogens, thereby causing an increase in the reaction rate.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rate Equation
Understanding the rate equation is crucial in predicting how different factors will impact the speed of a reaction, especially for an E2 reaction, which is a type of bimolecular elimination reaction. The rate equation for an E2 reaction is given by:\[ \text{Rate} = k [\text{RX}] [\text{B:}] \]Here,
- \([\text{RX}]\) stands for the concentration of the alkyl halide.
- \([\text{B:}]\) is the concentration of the base involved in the reaction.
- \(k\) is the rate constant, specific to the reaction conditions.
Alkyl Halides
Alkyl halides play a significant role in E2 reactions. They consist of an alkyl group bonded to a halogen. In the context of E2 reactions:- **Structure**: The structure of the alkyl halide can affect the reaction's mechanism and speed. Bulky alkyl groups often favor E2 reactions due to steric hindrance, which makes nucleophilic substitution (SN2) less favorable.- **Example Change**: When changing from ethyl bromide \(CH_3CH_2Br\) to isopropyl bromide \((CH_3)_2CHBr\), the latter increases the steric bulkiness, enhancing the likelihood of an E2 mechanism by providing more accessible beta hydrogens, increasing the reaction rate.
Solvent Effects
The choice of solvent can significantly influence the rate of an E2 reaction. E2 reactions generally prefer aprotic solvents:- **Protic vs Aprotic**: Protic solvents, like methanol \(CH_3OH\), can stabilize ions through hydrogen bonding. Aprotic solvents, like DMSO, do not solvate the base as strongly, allowing it to be more available and active in the reaction.- **Impact on Rate**: Switching from a protic solvent to an aprotic one like DMSO enhances the base's effectiveness, thus increasing the reaction rate by promoting the elimination process more efficiently.
Leaving Groups
Leaving groups are responsible for departing from the molecule during the reaction, and their ability to leave affects the reaction rate:- **Good Leaving Groups**: A good leaving group is essential for a fast E2 reaction, as it stabilizes the transition state.- **Comparison**: Iodide \(I^-\) is a better leaving group than bromide \(Br^-\) because of its larger size and weaker bond with carbon, allowing it to detach more easily. Replacing \(I^-\) with \(Br^-\) results in a decrease in the reaction rate because the transition state is less stabilized.
Base Strength
The base used in an E2 reaction significantly impacts the reaction rate. Strong bases readily abstract protons to facilitate the elimination process:- **Base Strength**: A strong base, such as \(^-OH\), is more effective at removing protons compared to a weaker base like water \(H_2O\).- **Effect on Rate**: Substituting \(^-OH\) with \(H_2O\) diminishes the reaction speed considerably. This happens because the weaker base \(H_2O\) is much less capable of removing a proton, thus slowing down the overall elimination process.
