Question

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?

Answer 1

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:

http://chemistry.stackexchange.com/

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. :)