In the end, what you have essentially happening for OUR reaction is: I did find a schematic of what's probably going on, except it's with an alcohol: Red phosphorus acts as basically a catalyst it reacts with the #"I"_2# that forms after #"HI"# reacts, to form #"PI"_3# and #"PI"_5#, which are then hydrolyzed to form #"H"_3"PO"_4#, #"H"_3"PO"_3#, and more #"HI"#. That implies that it would be a great idea to somehow have a driving force to get more hydroiodic acid to drive the equilibrium forward. Hence, if the reaction were to use exactly-as-needed quantities and 100% of the reactants reacted, iodide would rather leave than attack the electrophilic center and stay on there. ![]() One potential problem is that chloride is a worse leaving group than iodide because the pKa of #"HI"# is lower than that of #"HCl"#. If we didn't have enough #"HI"#, elimination would be more likely. ![]() In those conditions, iodine is actually the strongest nucleophile of all the halogens in polar protic solvents. ![]() We could suppose that the #"HI"# is concentrated, and the polar protic solvent could be water. I had to do a bit of research on this (finding this and this), but as it turns out, iodine acts as a nucleophile in this circumstance.
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