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.

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

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

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.

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