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Author: Subject: Hydriodic reducing power
jimmyboy
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[*] posted on 6-9-2009 at 00:43
Hydriodic reducing power


Just out of sheer curiosity -- would hydriodic be strong enough to reduce benzoic acid to the alkene? -- I can't find a reference of any chemist that was crazy enough (or wasteful enough) to try it

[Edited on 6-9-2009 by jimmyboy]
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woelen
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[*] posted on 6-9-2009 at 03:45


To which alkene? Benzoic acid is based on a benzene-ring and is not easily converted to an alkene.

But in general, do not expect hydroiodic acid to be a sufficiently strong reductor to reduce organic groups like -COOH or =C=O.




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jimmyboy
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[*] posted on 6-9-2009 at 04:54


Benzene -- I have seen a few organic textbooks mention it can convert carboxylic acids to alkanes.. if it can do that then it must step through the aldehyde alcohol or alkene somewhere along the way ..

I first saw it here.. number 3
http://iit-jee-chemistry.blogspot.com/2008/01/iit-jee-revisi...

[Edited on 6-9-2009 by jimmyboy]
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Ozone
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[*] posted on 6-9-2009 at 07:08


HI can go all the way to the alkane!

In the presence of P(red) It will reduce benzophenone to give diphenyl methane (so it will do ketones). It can also reduce aliphatic hydroxyls to alkanes, eg. glycerol to propane.

See Gatterman, L. (1907), pp. 301-303 and 345-348. It can be acquired by poking around in references.

pp. 300-301 are attached.

It looks like carboxylic acids will likwewise yield the corresponding alkane, but require higher temperatures/pressures (eg. 150°C).

For brevity see:

http://www.geocities.com/nepalorganic/co_acid.htm

Even better is that the I2 can be recycled to HI using H2S!

Cheers,

O3


Gatterman_301-302.jpg - 391kB

[Edited on 6-9-2009 by Ozone]




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jimmyboy
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[*] posted on 6-9-2009 at 15:30


Nice find! thanks Oz - I was wondering if benzene acids were an exception - apparently it can tear oxygen from almost anything
I didn't realize it but the carbon in the COOH group is stuck to the benzene ring permanent - so it would reduce to toluene not benzene..

[Edited on 6-9-2009 by jimmyboy]
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[*] posted on 6-9-2009 at 23:01


The combo I2/red P can reduce many things, but is the reducing power in this case really due to HI? An interesting thing would be to try HI alone without the red P. I would be surprised if HI alone can reduce many things, but if this indeed is the case, then I have learnt something new again ;)



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chloric1
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[*] posted on 7-9-2009 at 06:11
NOPE


It seems to me that phosphonic acid would be the active component. Oxygen from the air would liberate small amounts of iodine from the HI then attack the phosphorus to form PI3 which is immediately hydrolisized to H3PO3 and HI. Any free iodine is kept in solution by HI and as long as red phosphorus exist, new phosphonic acid can be formed. Correct me if I ma wrong but does phosphonic acid give phosphine when heated in water? Phosphine might be another reactive species.



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[*] posted on 7-9-2009 at 08:01


I don't think it's that simple, either.

Sorry, the pic up-load hosed the attachment; it is small and hard to read. In there it describes the recycling of HI via hydrolysis of PI3 (in-situ) with water. This yields HI and phosphorous acid (P(OH)3).

The combination appears to be more efficient because the HI is recycled (and can thus engage in a greater-than-stoichiometric number of individual reductions).

This is over simplified, and a variety of reducing species (including perhaps PH3) exist. An interesting paper describing the regenerative role of P in HI/P reductions is attached.

Cheers,

O3

Attachment: Albouy 1997-role of P in HI reductions.pdf (423kB)
This file has been downloaded 1952 times





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[*] posted on 9-11-2009 at 16:17


hi will not reduce terminal iodoalkanes except very slugglishly like a week or mor so yes it would work but one would have to be very patient.
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