HI/P reduction

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The HI/P reduction (or HI+P reduction) is a very common and useful reduction process, which uses hydroiodic acid and elemental phosphorus to reduce organic compounds, like alcohols to their respective alkanes, in two steps. Heinrich Kiliani described this method over 140 years ago.

Due to its great effectiveness and easy to produce, this reduction is sadly used by meth cooks to produce methamphetamine from pseudoephedrine, which has tainted the reputation of this procedure.


The process involves two steps:

  • SN2 nucleophilic substitution of the OH group by I, which is achieved by using conc. hydroiodic acid
  • Reduction of the alkyl iodide by hydriodic acid

For the first step, hydroiodic acid converts the hydroxy compound to iodide.[1]

This reduction process is known to be powerful enough to remove oxygen from any organic compound and if reaction conditions are harsh enough it can even remove nitrogen as well. The reduction of terminal alkyl iodide however is longer and has a lower yield.

The exact mechanism reaction for this reaction is not well described in literature and the information is relative scarce.


To an iodide salt, which can be NaI or KI add small amounts of conc. Phosphoric acid and a small amount of red phosphorus. Concentration of H3PO4 between 70-90% is indicated to give a good performance, as it allows temperatures which are sufficient to reduce almost everything without the need of a pressure vessel. Iodine is produced as side product. Since the maximum concentration of hydroiodic acid can be 57%, any excess hydrogen iodide will boil off. Add the HI solution to the compound you want to reduce. The reaction takes place in aqueous medium. Reflux the solution until the compound has been converted to alkyl iodide.

R-OH + HI → R-I + H2O
HI + O2 → I2 + H2O

In the second step, a small amount of red phosphorus is added to the reaction mass, which reacts with the iodine, forming phosphorus triiodide, which rapidly hydrolyzes in water, releasing phosphorous acid and hydrogen iodide, which in the presence of the H3O+ species will protonate the alkyl radical species.

2 P + 3 I2 → 2 PI3
PI3 + H2O → H3PO3 + HI
H2O + HI ⇌ H3O+ + I-
R-I + HI + H3O+ → R-H + I2 + H2O


The HI/P reduction can be used to reduce various OH containing compounds, or halocarbons to their respective alkanes.

Carboxylic acids can also be reduced to their respective alkanes.[2]


The reaction will produce phosphine as side product, which is very toxic and very flammable.


  1. http://www.beilstein-journals.org/bjoc/articles/8/36
  2. http://www.sciencemadness.org/talk/files.php?pid=161627&aid=8575

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