Explaining the Catalysis of SN2 Reactions by Iodide Ions

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Iodide ions catalyze SN2 reactions of alkyl chlorides and bromides primarily due to their effectiveness as leaving groups, which lowers the activation energy of the reaction. The reaction pathway involving iodide is more favorable because the activation energy for both steps with iodide is lower than that of the direct reaction with hydroxide. Additionally, iodide is a weaker base compared to chloride and bromide, facilitating its ability to displace these halides. The presence of sodium or potassium iodide serves to solubilize iodide in the reaction medium, with cost considerations influencing the choice of these salts over others like lithium. Overall, the catalytic role of iodide enhances the reaction rate significantly.
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The addition of sodium or potassium iodide catalyzes many SN2 reactions of alkyl chlorides or bromides. Explain.

I think the reason why it catalyzes many SN2 reactions, has to do with the fact that it is the best leaving group of all the alkyl halides. But why else? Also, is the potassium and sodium the group that brings the iodide into the solution? Or do we use them, because they are cheaper than let's say lithium. Ignore my tangent, and please explain my initial question.
 
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The question comes from Williamson's Macroscale and Microscale Organic Experiments. Please respond if you have any ideas.
 
Any ideas?
 
I'll say the main reason is because the activation energy of both steps in
I- + R-Cl ->R-I + Cl- and R-I + OH-->ROH + I-
is less than the activation energy in
R-Cl + OH- ---> R-OH + Cl-
So the rate of the reaction where I- is a catalyst is more.
 
You mean, because iodide is a weaker base than chlorine and bromine it can displace the alkyl chloride or bromides?
 
I'll say the main reason is because the activation energy of both steps in
I- + R-Cl ->R-I + Cl- and R-I + OH-->ROH + I-
is less than the activation energy in
R-Cl + OH- ---> R-OH + Cl-
also relevant here is that Cl- is further stabilized by water, than I-
 
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