In the production of methamphetamine from reducing epinephrine with #"HI"# in the presence of red phosphorus, what is the role of #"HI"#, and could #"HCl"# be used instead? What is an accepted mechanism?

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.

We could suppose that the #"HI"# is concentrated, and the polar protic solvent could be water. In those conditions, iodine is actually the strongest nucleophile of all the halogens in polar protic solvents. If we didn't have enough #"HI"#, elimination would be more likely.

One potential problem is that chloride is a worse leaving group than iodide because the pKa of #"HI"# is lower than that of #"HCl"#. 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.

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.

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"#. Therein lies the "driving force".

I did find a schematic of what's probably going on, except it's with an alcohol:

In the end, what you have essentially happening for OUR reaction is:

• The iodide acting as a nucleophile (mixture of #"S"_N1# and #"S"_N2#), except with no initial protonation necessary.
• At some point, #"R"cdot# and #"I"cdot# form. An #"I"^(-)# reacts radically with #"R"cdot# to form #"R"^(-)# and #"I"cdot#
• The #"R"^(-)# gets protonated. The two radical iodides probably reform #"I"_2#, as #"I"_2# is depicted in the above diagram to be "leaving" (produced in) the reaction to react with more red phosphorus.

One could draw out the final two aforementioned steps like so:

Ultimately you have reduced the alkyl halide into an alkane.

I personally have never heard of this reaction before (it's fairly new; 1982), and apparently it's quite dangerous (as are many radical reactions, such as peroxide radicals...), so maybe that's why I haven't been taught this reaction.

I wouldn't worry too much about the mechanism. In my opinion, no good professor in their right mind would put this on an exam. :)