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Alkyl Iodides are formed by reaction of alcohols with iodine and red phosphorus. I already am well aware that this forms $\ce{PI3}$ in situ, which acts similarly to $\ce{PBr3}$ and allows alkyl iodides to be prepared from alcohols.

However, when the same reaction is performed on ephedrine, the hydroxyl group is reduced rather than iodinated. What determimes whether iodination or reduction occurs?

I have a few theories, one being temperature may be kept lower to form the alkyl iodide. Or is it to do with the benzylic position of the hydroxyl group? Or perhaps the red phosphorus is in excess to favor reduction? Any incite will be helpful.

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