Prepare Methyl Iodide Using SN2 Mechanism

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To prepare methyl iodide using the SN2 mechanism, a methyl halide such as methyl bromide can be reacted with potassium iodide (KI) in an appropriate solvent. Iodide is recognized as the best leaving group, which raises questions about replacing it with another halide. The reaction can be represented as CH3Br + KI → CH3I + KBr, demonstrating the displacement of bromide by iodide. Using an aprotic solvent like DMSO, DMF, or acetone can enhance the reaction efficiency. Understanding the role of the solvent is crucial for successful halide substitution.
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Hi folks! :)
I have a problem with the following problem.
Take any methyl halide and prepare methyl iodide.Choose the appropriate solvent.

Now what i know is here we are going by SN2 mech and replacing a halide by iodide ion.But the problem i am facing is how can i replace a other halide by a iodide ion because my book says iodide is the best leaving group among all halides.So i am confused how will this happen. ?
 
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Yes, iodide is the best leaving group. But the solvent can make the difference. For example, the Kharasch addition in the presence of H2O2 breaks the Markonikov's rule. I'm sorry I don't have the details of this reaction, but will check into it.
 
You can do a displacement reaction with methylbromide:

CH3Br + KI ----> CH3I + KBr

You might want to use an aprotic solvent such as DMSO, DMF or acetone.
 
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