What are the mechanisms for the reduction of organic compounds employing red phosphorus and iodine?

Question

I was recently studying oxidation and reduction reactions.

I found red phosphorus and iodine reduce many different organic compounds.

Out of interest I would like to know the mechanism for these reactions. A concerted path is preferred.

Answer 1

I found Red P and Iodine reduce many different organic compounds.

Actually, it is typically used for the deoxygenation in benzylic positions, using benzylic alcohols or benzoins as starting materials.

Aliphatic alcohols yield iodoalkanes, propargylic alcohols often undergo elimination under these conditions whereas allylic alcohols are typically converted to complex mixtures without synthetic value.

The deoxygenation is often carried out with catalytic amounts of iodine but an excess of red phosphorous. In principle, catalytic amounts of $\ce{NaI}$ plus one equivalent of $\ce{H3PO3}$ in water would work as well.

When using iodine and red phosphorous, hydrogen iodide is generated in situ via the formation and subsequent hydrolysis of $\ce{PI3}$.

The formal replacement by a proton is not fully understood and is disputed whether the formation of a benzylic iodide is really involved.

In princpiple, elimination from water from the protonated starting material would yield a stabilized benzylic cation. This might get reduced by a iodide to a benzylic radical and, in a second step to a very shortlived benzylic anion which is instantaneously protonated.

Both reduction steps yield iodine radicals from iodide and in summary regenerate iodine.