Chemetix
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HI/P Halogenation agent - reductant?
While perusing Vogel on the synthesis of haloalkanes, the transformation of alcohols to alkyl iodides features the use of HI/P commonly. Fair enough I
say, that makes sense. i.e the use of HBr with isopropyl alcohol to form isopropyl bromide is given pg 561 5th edition.
Similar reaction pg 568 on the preparation of butyl iodide, where butanol and 2P + 5I2 gives 10 mol of the alkyl halide.
So why does the reaction on pg 755 with 4- benzylidene-2-phenyloxazole-5-one and HI/P yield phenylalanine? Why, for example, does it become a
reductant with a well known secondary alcohol (that shall not be named) to form a secondary amine, and not a halogenation(oxidation) reaction?
[Edited on 19-6-2020 by Chemetix]
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draculic acid69
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Water free Vs not water free maybe. You don't use water in the alcohol to alkyl iodide do you?
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karlos³
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Not maybe, it is purely the water content.
The alkyl iodide gets reduced to the alkane when the reaction takes place in the presence of water.
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Boffis
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No P is required in the conversion of alcohols to iodoalkanes, just HI. However, you can generated HI in situ from iodine, phosphorus and water. P can
also be added to remove iodine liberated from the HI by atmospheric oxidation, this occurs readily.
In the second example HI is being used as a reducing agent but in the process iodine is liberated. Since iodine is an oxidising agent you need
sometime to remove the iodine and drive the reaction forward. In effect P is the reducing agent and HI is merely a catalyst since it is constantly
regenerated and it not incorporated into the final product.
I suspect the difference is in the ratio of HI/P
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Racconized
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What the others have said, the hi/p reduction is based on hydrolysis of the iodo intermediate and regeneration of hydrogen iodine catalyst. While the
red P and iodine halogenenation is done in water free conditions, I have done both reactions iodopropane and mandelic acid reduction. And the only
difference is that water is present in the reduction and the halogenation is done without water, only using the reactant alcohol as the solvent (I
used it stoichiometrically for iodopropane)
In the halogenation of alcohols, Phosphorus triodide is the actual reactant, which is formed by red phosphorus and iodine.
Edit: It seems not everything I wrote is 100% correct, apparently HI is perfectly capable of turning primary alcohols to alkyl iodines, in which case
water would be present. So take my reply with a grain of salt, seems like the post above mine gives a good explanation
[Edited on 19-6-2020 by Racconized]
[Edited on 19-6-2020 by Racconized]
[Edited on 19-6-2020 by Racconized]
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Racconized
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Edit 4th: Proposed mechanism seems to depend on weather hydrogen iodide is used in catalytic amounts or in stoichiometrically (or excess) according
to: Dobmeier, M.; Herrmann, J. M.; Lenoir, D.; König, B. Beilstein J. Org. Chem. 2012, 8, 330–336. doi:10.3762/bjoc.8.36
(pretty much what Boffis said, already, the linked study/paper is only if you wanted something more in depth.)
Can't post picture due to copyright, but I guess just use sci-hub
also sorry for my previous incorrect and unfounded attempt at an explanation.
[Edited on 20-6-2020 by Racconized]
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dextro88
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Actualy the whole mechanism of the reaction of HI/P is the creation of the phosphorous tribriomide as the guy recconized said, wich can react with
solutions even of water and alcohols, and actualy this is making the iodine so vigoursly reacting and heat donating, i think methyl iodide for example
can be made in solution of water and methanol for example, as every liberated molecule of phosphorous triiodide will react with one molecule of the
solution its just the concentration that will propose the speed of halogenation, and its concentration will depend the speed or possiblity of forming
complexes with another iodine atoms and removing them. Its all one pot, depend on conditions.
Best regards Dex.
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