the-reaction-of-2-chlorobutane-with-sodium-methoxide-in-methanol-forms-2-methoxy

Question

The reaction of 2-chlorobutane with sodium methoxide in methanol forms 2-methoxybutane. When small amounts of potassium iodide are added to the reaction mixture, the rate of the formation of 2-methoxybutane increases. Explain why.

Accepted Answer

The reaction rate increases because the iodide ion is a strong nucleophile and will form a substrate with a good leaving group.

Answer

The reaction rate increases because potassium iodide is a good addition to the solvent, enhancing the nucleophilicity of the methoxide ion.

Answer

The reaction rate increases as potassium ion reacts with chloride ion, reducing its concentration at the product side and shifting the reaction to the right.

Answer

The reaction rate increases because chloride and iodide ions prefer to react, making chlorine an better leaving group.

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The reaction of 2-chlorobutane with sodium methoxide in methanol forms 2-methoxybutane. When small amounts of potassium iodide are added to the reaction mixture, the rate of the formation of 2-methoxybutane increases. Explain why.