V. In the following reaction: CH3CH₂Br + CH3COOK → CH3COOCH₂CH3 +KBr Adding a small amount of sodium iodide increases the reaction rate significantly. Please using the reaction mechanisms, nucleophilicity and leaving group quality to explain.

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### Reaction Analysis

**Reaction:**
\[ \text{CH}_3\text{CH}_2\text{Br} + \text{CH}_3\text{COOK} \rightarrow \text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{KBr} \]

Adding a small amount of sodium iodide increases the reaction rate significantly. Please use reaction mechanisms, nucleophilicity, and leaving group quality to explain.

**Explanation:**

This reaction involves the conversion of an alkyl bromide (\(\text{CH}_3\text{CH}_2\text{Br}\)) and potassium acetate (\(\text{CH}_3\text{COOK}\)) to form an ester (\(\text{CH}_3\text{COOCH}_2\text{CH}_3\)) and potassium bromide (\(\text{KBr}\)).

1. **Nucleophilicity:**
   - In this context, sodium iodide plays a critical role by enhancing the nucleophilicity of the reaction. Iodide ions (\(I^-\)) are better nucleophiles compared to bromide ions (\(Br^-\)) due to their larger size and higher polarizability.

2. **Leaving Group Quality:**
   - The iodide ion can replace the bromide ion in the reaction, forming an alkyl iodide intermediate. This is because iodide is a better leaving group, which allows the reaction to proceed faster.

3. **Mechanism:**
   - The substitution mechanism likely follows a \(S_N2\) pathway, where the iodide ion first substitutes the bromide, enhancing the overall reactivity and allowing the acetate ion to perform nucleophilic attack more efficiently.

The addition of sodium iodide, therefore, facilitates a faster reaction by improving the quality of the leaving group and enhancing the nucleophilicity of the anion responsible for the reaction.
Transcribed Image Text:Certainly! Here's the transcription and explanation for an educational website: --- ### Reaction Analysis **Reaction:** \[ \text{CH}_3\text{CH}_2\text{Br} + \text{CH}_3\text{COOK} \rightarrow \text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{KBr} \] Adding a small amount of sodium iodide increases the reaction rate significantly. Please use reaction mechanisms, nucleophilicity, and leaving group quality to explain. **Explanation:** This reaction involves the conversion of an alkyl bromide (\(\text{CH}_3\text{CH}_2\text{Br}\)) and potassium acetate (\(\text{CH}_3\text{COOK}\)) to form an ester (\(\text{CH}_3\text{COOCH}_2\text{CH}_3\)) and potassium bromide (\(\text{KBr}\)). 1. **Nucleophilicity:** - In this context, sodium iodide plays a critical role by enhancing the nucleophilicity of the reaction. Iodide ions (\(I^-\)) are better nucleophiles compared to bromide ions (\(Br^-\)) due to their larger size and higher polarizability. 2. **Leaving Group Quality:** - The iodide ion can replace the bromide ion in the reaction, forming an alkyl iodide intermediate. This is because iodide is a better leaving group, which allows the reaction to proceed faster. 3. **Mechanism:** - The substitution mechanism likely follows a \(S_N2\) pathway, where the iodide ion first substitutes the bromide, enhancing the overall reactivity and allowing the acetate ion to perform nucleophilic attack more efficiently. The addition of sodium iodide, therefore, facilitates a faster reaction by improving the quality of the leaving group and enhancing the nucleophilicity of the anion responsible for the reaction.
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