organic chemistry - HI/P reduction mechanism

I recently answered a question about Breaking Bad's initial methamphetamine production method (i.e. the reduction of (pseudo)ephedrine). The reaction is as follows:

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It is infamously known as the $\ce{HI/P}$ reduction and was (and still is somewhat) a real plague in the United States until cough medicine containing ephedrine came under tightened control. That for an introduction, now the question: what is actually the reaction mechanism of the $\ce{HI/P}$ reduction?

The first step seems like a simple $\ce{S_{N}2}$ substitution of $\ce{I^{-}}$, catalyzed by the protonation of the alcohol group by hydroiodic acid (making it a far better leaving group). But what is the mechanism after that? How is phosphorus involved? Does the $\ce{HI/P}$ reduction have any legal use?

Answer

According to the source mentioned in the comments to your question, the first step is indeed nucleophilic substitution of the OH group by $\ce{I-}$, faciliated by protonation of the alcohol. For the second step ($\ce{HI}$ reduction), a radical species was found as an intermediate, and therefore a reduction by single electron transfer (SET) with oxidation of $\ce{I-}$ to $\ce{I2}$ is proposed. Red phosphorus is required for the regeneration of $\ce{HI}$ from $\ce{I2}$. $\ce{P_{red}}$ reacts with $\ce{I2}$ to the phosphorus iodides $\ce{PI3}$ and $\ce{PI5}$, which are subsequently hydrolyzed to $\ce{H3PO3}$/$\ce{H3PO4}$ and $\ce{HI}$, which is reused in further reduction steps. Catalytic amounts of $\ce{HI}$ are therefore sufficient to run the reaction in the presence of phosphorus. However, a too low concentration of $\ce{I-}$ will lead to elimination instead of substitution in the first reaction step.

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The SET mechanism of the reduction is not described in detail in the article, so the following is a bit speculative. As a resonance-stabilized radical is involved, homolytic cleavage of the $\ce{C-I}$ bond is one of the first steps, which gives the benzylic radical $\ce{R.}$ and an iodine radical $\ce{I.}$. A single electron is transferred from $\ce{I-}$ to $\ce{R.}$, forming the carbanion $\ce{R-}$ and $\ce{I.}$, which recombines with another iodine radical to $\ce{I2}$. Protonation of $\ce{R-}$ yields the final product.