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Ch. 9 - Substitution and Elimination Reactions of Alkyl Halides
Bruice - Organic Chemistry 8th Edition
Bruice8th EditionOrganic ChemistryISBN: 9780135213711Not the one you use?Change textbook
Chapter 10, Problem 71c,d

Explain how each of the following changes affect the rate of the reaction of 1-bromobutane with ethoxide ion in DMF.
c. The alkyl halide is changed to 1-chlorobutane.
d. The alkyl halide is changed to 2-bromobutane.

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1
Step 1: Understand the reaction mechanism. The reaction of 1-bromobutane with ethoxide ion in DMF (dimethylformamide) is an example of an SN2 reaction. SN2 reactions are bimolecular nucleophilic substitution reactions where the rate depends on both the concentration of the nucleophile and the substrate.
Step 2: Analyze the effect of changing the alkyl halide to 1-chlorobutane. In SN2 reactions, the leaving group plays a critical role. Bromide (Br⁻) is a better leaving group than chloride (Cl⁻) because it is larger and more polarizable, making it more stable as a free ion. Replacing 1-bromobutane with 1-chlorobutane will decrease the reaction rate because chloride is a less effective leaving group.
Step 3: Analyze the effect of changing the alkyl halide to 2-bromobutane. The structure of the alkyl halide also affects the SN2 reaction rate. 1-bromobutane is a primary alkyl halide, while 2-bromobutane is a secondary alkyl halide. SN2 reactions proceed more readily with primary alkyl halides because there is less steric hindrance around the carbon undergoing substitution. Changing to 2-bromobutane will decrease the reaction rate due to increased steric hindrance.
Step 4: Consider the solvent's role. DMF is a polar aprotic solvent, which stabilizes the nucleophile (ethoxide ion) without stabilizing the leaving group. This makes it an ideal solvent for SN2 reactions. The solvent remains constant in this problem, so its effect does not change.
Step 5: Summarize the effects. Changing the alkyl halide to 1-chlorobutane decreases the reaction rate due to a poorer leaving group. Changing the alkyl halide to 2-bromobutane decreases the reaction rate due to increased steric hindrance around the reactive carbon.

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Key Concepts

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

Nucleophilicity

Nucleophilicity refers to the ability of a nucleophile to donate an electron pair to an electrophile, facilitating a chemical reaction. In the context of the reaction between 1-bromobutane and ethoxide ion, the strength of the nucleophile (ethoxide) and the leaving group (bromide or chloride) significantly influence the reaction rate. Generally, stronger nucleophiles and better leaving groups enhance the reaction rate.
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Leaving Group Ability

The leaving group ability is a critical factor in nucleophilic substitution reactions. A good leaving group can stabilize the negative charge after it departs, making the reaction more favorable. In this case, bromide is a better leaving group than chloride, which means that changing from 1-bromobutane to 1-chlorobutane would likely decrease the reaction rate due to the poorer leaving group.
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Substitution Mechanism

The substitution mechanism, particularly whether the reaction follows an SN1 or SN2 pathway, affects the reaction rate. 1-bromobutane typically undergoes an SN2 reaction due to its primary carbon structure, while 2-bromobutane can undergo both SN1 and SN2 mechanisms. The steric hindrance in 2-bromobutane can slow down the SN2 pathway, potentially leading to a different rate of reaction compared to 1-bromobutane.
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