Author Topic: Sb(III) and As(III) instead of P(III)  (Read 5028 times)

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stratosphere

  • Guest
Sb(III) and As(III) instead of P(III)
« on: December 20, 2003, 12:33:00 PM »
from what i understand P(III) present as either PI3 or hypophosphorous acid is involved in a catylitic cycle to convert I2 to HI in the (in)famous RPI reduction.

i would guess the chemistry of Sb(III) and As(III) would be very similiar, so my question is would these work in place of P(III) compounds?

arsenic(III) oxide and antimony (III) oxide are available for ceramic glazing, fire retardants, as well as for mineral collectors.
my understanding is that they are soluable in acidic solutions (with non-oxidizing acids) as well as in basic solutions. of course acidic being of principle interest in this case.

Rhodium

  • Guest
Antimony & Arsenic are no Phosphorus substitutes
« Reply #1 on: December 20, 2003, 06:36:00 PM »
As, which is the most similar to P in its chemistry, and thus most likely to work as a substitute (I have no idea), is so hideously toxic that it is absolutely unfit for labwork.

Also, as the As2O3 available to you is already oxidized, you first would need to reduce it to its elemental state for it to work as intended. Thus, scrap that idea completely - these two obstacles are not to be overcome by non-professionals.


xtaldoc

  • Guest
Antimony & Arsenic compounds as subs for P
« Reply #2 on: December 20, 2003, 09:09:00 PM »
Swix did some looking around on the off-chance that there was some possible merit to the notion, but can find no data except that which agrees with Rhodium's prior statement.  I would like to add that Antimony compounds are frequently the most noxious smelling compounds that exist..... Nothing seems to suggest that this would bee a potential route worth investigation.  Take heart though, an inquisitive mind is my favorite kind of mind.

fierceness

  • Guest
if something that was already oxidized would...
« Reply #3 on: December 20, 2003, 11:12:00 PM »
if something that was already oxidized would work, you'd be able to use phosphoric acid which is readibly available.. unfortunately, you can't.

Organikum

  • Guest
already told was
« Reply #4 on: December 21, 2003, 10:28:00 AM »
salts of tin + iodine = HI (stannic and stannous, Ref: Gmelin)
Make a tin-salt: plumbers soldering tin (NO electronics soldering tin! Lead!) + strong acid, let react till tin is gone, boil excess acid away (cough, cough): tin-salt (HCl is most easy).

H3PO4 aka phosphoric acid + NaI or KI = HI (look for SHORTY´s posts on this, forget Argox)

lugh

  • Guest
Numerous Reducing Agents Have Been Used
« Reply #5 on: December 21, 2003, 05:07:00 PM »
In Journ Pharm Chem 13 421 (1848) E. Soubeiran used dilute sulfuric acid, sulphites, arsenious acid and stannous salts to reduce iodine to hydriotic acid in the presence of water. The article isn't presently available, it will be located as soon as possible  ;)


stratosphere

  • Guest
Rhodium>> Your right that the As2O3 or...
« Reply #6 on: December 22, 2003, 08:57:00 AM »
Rhodium>>
Your right that the As2O3 or Sb203 avaliable are oxidized, ut only to the +3 state, this might not bee a problem as evidenced by the usefulness of hypophorous acid in this reduction and it contains phosphorous in its +3 state.

regarding the toxcitity, that is a real concern. i for one would be a little leary buying some tweak off the street that was made via a arsenic route, but much as working with mercury im sure adequate procedures for purification could be devoloped.

stratosphere

  • Guest
if something that was already oxidized ...
« Reply #7 on: December 22, 2003, 12:53:00 PM »
if something that was already oxidized would...   Bookmark  Reply    

if something that was already oxidized would work, you'd be able to use phosphoric acid which is readibly available.. unfortunately, you can't


but phosphoric is oxidized to a +5 state, you can use (HO)3P which is only oxidized to its +3 state

SHORTY

  • Guest
H3PO4
« Reply #8 on: December 22, 2003, 02:37:00 PM »
Phosphoric Acid is H3PO4.

Im pretty sure that Perphosphoric Acid is H3PO5.

We already know that H3PO3 (phosphorous acid) and H3PO2 (hypophosphorous acid) can bee used instead of rp.

H3PO4 can bee used to make HI but not insitu and not with iodine.


spectralshift

  • Guest
We already know that H3PO3 (phosphorous acid)...
« Reply #9 on: December 22, 2003, 11:09:00 PM »
We already know that H3PO3 (phosphorous acid) and H3PO2 (hypophosphorous acid) can bee used instead of rp.

H3PO4 can bee used to make HI but not insitu and not with iodine.



Exactly. there r lots of nice P substitutes available today, tomorrow probably not (ready made and basically ready made anyway).

iodine is much more rare than phosphorous...no one here at the hive has broken through the barrier though, as with phosphorous, it's too hard. wa...

There's some really good popular kelp concentrates and nutrients around I think, worth a play.

stratosphere

  • Guest
ideally the iodine can be recycled in this...
« Reply #10 on: December 24, 2003, 08:30:00 AM »
ideally the iodine can be recycled in this reaction, of course in practicality lets say you recover 90% of it per reaction, that would take you about 6 reaction cycles before you lost half of it.
so iodine shouldn't bee a major problem.

lets say old macdonald had a farm, and needed to castrate a few hundred pigs, a huge jug of iodine tincture wouldn't bee so out of place , in fact you might take an educational visit to your local farm supply store. although they might bee suspicious if you have a mohawk and black eye shadow on or something of the such.

what i see as more of the limitation of this process is the (psuedo)ephedrine, both in terms of their cost, the fact that stores limit purchases, and extracting the pills from all the denaturants they put in.

spectralshift

  • Guest
Hey! stop concerning yourself with my offtopic
« Reply #11 on: December 24, 2003, 05:59:00 PM »
Hey! stop concerning yourself with my offtopic rants it's about you wanting a P substitute, which is no problem at all. You have a hypo precursor, a h3po3 precursor, and sodium sulfite has also been discussed a bit.

biosynth and reductive alkylation for pseudoephedrine.

imvho It makes more sense than P2P and reductive amination for the bees who have been specialising in HI reductions, because they know the rxn very well, they know sudo very well. anyway thats how i see it.

ragnaroekk

  • Guest
and still the question is
« Reply #12 on: December 25, 2003, 08:13:00 AM »
Why not HCl ?

HCl + NaI = HI + NaCl

Whats wrong with this? Use stochiometric amounts and distill away the water and then the 57% constant boiling HI..... NaCl is left back. Some excess HCl wouldnt hurt as HCl was often used as solvent for HI reductions.

btw. the correct production of HI from H3PO4 calls for 100% phosphoric acid which is crystalline at roomtemperature and melts at about 55C as I remember. It can be prepared from H3PO4 of any concentration by boiling all water away at >150C and let it cool down.
This produces almost waterfree HI - this way the fabolous HI of density 1,9 aka about 70% aka fuming HI can be prepared.

RaGNArOeKk


Osmium

  • Guest
> HCl + NaI = HI + NaCl > Whats wrong...
« Reply #13 on: December 25, 2003, 08:51:00 AM »
> HCl + NaI = HI + NaCl
> Whats wrong with this?

Maybe this reaction isn't taking place?

> Use stochiometric amounts and distill away the water and then
> the 57% constant boiling HI.....

Well don't forget to tell your reaction that it is the HI you want to distill out, otherwise you might end up distilling HCl...  :P

Being able to write down such a reaction:
> > HCl + NaI => HI + NaCl
doesn't automatically mean it will happen.

What if I write down the following instead?
HI + NaCl ---> HCl + NaI

Do you think that I could make HCl that way?


Organikum

  • Guest
HI + NaCl ---> HCl + NaI = BULLSHIT
« Reply #14 on: December 25, 2003, 12:12:00 PM »
HI is not decomposed by NaCl.

Reference: Gmelin

And look here:

Post 323629 (missing)

(Rhodium: "Good start!", Stimulants)


Rhodium proposes:
KI + HCl -> HI + KCl


Osmium

  • Guest
What makes you think that this is not a ...
« Reply #15 on: December 25, 2003, 02:51:00 PM »
What makes you think that this is not a equilibrium reaction?

When I dissolve NaCl in HI (or NaI in HCl, same thing), then I end up with a solution of ions in water, something most bees do not seem to comprehend.

Distilling such a mixture results in the least volatile component to distill first, NOT the weaker acid! In this case this will be HCl, not HI. Since their boiling points are rather close together you will most likely end up a mixture of both acids distilling over, and certainly not azeotropic HI (or any concentration close to it).

> HI + NaCl ---> HCl + NaI = BULLSHIT

It's exactly the same bullshit as NaI + HCl ----> HI + NaCl.


biotechdude

  • Guest
Con. HI - Many Qu's
« Reply #16 on: December 26, 2003, 12:18:00 AM »
"...btw. the correct production of HI from H3PO4 calls for 100% phosphoric acid which is crystalline at roomtemperature and melts at about 55C as I remember. It can be prepared from H3PO4 of any concentration by boiling all water away at >150C and let it cool down.
This produces almost waterfree HI - this way the fabolous HI of density 1,9 aka about 70% aka fuming HI can be prepared...."

Are you referring to HI production using Iodide salt (KI, NaI) and H3PO4?  In situ? Or distilling off as it is being produced?  How is waterless HI handled, what is its physical properties?

What are the advantages of 1.9D [70%] HI?  Quicker reaction? Stronger Reduction Force?  Less needed?  Any concerns with side reactions or arizidines?  Will it become standard [57%] after 10 mins in a rxn (will the P-recycler be able to keep up and maintain 1.9D [70%] HI?)


Organikum

  • Guest
dissoziation and hydrates
« Reply #17 on: December 26, 2003, 06:14:00 AM »

When I dissolve NaCl in HI (or NaI in HCl, same thing), then I end up with a solution of ions in water, something most bees do not seem to comprehend



When Rhodium dissolves NaI (KI) in HCl he gets hydroiodic acid and NaCl (KCl). PERHAPS also hydrochloric acid and NaI (KI) stay present forming an equilibrium with the former two compounds.
But:
Hydroiodic acid is nothing else than the hydrated form of HI aka hydrogen iodide and so not dissoziated in form of H+ and I+ ions bathing in a swimmingpool of H2O but a certain amount of H+ and I+ ions are engaged in a heavy groupsexorgy with a certain amount of water molecules - the most stable gangbangscenario is the hydrate forming the so called "constant boiling hydroiodic acid" with about 57% concentration. (One molecule HI on eleven H2O as I remember, or better twentythree H+ ions gangbanging eleven O+ and one I+ ?)
The same can be told similar on the hydrochloric acid and on the hydrobromic acid.

So: Nobody ends up with these bathing ion circus but there is hydrohalogenic acid formed - and thats what we want.

And this way seems to feasible in reductions as Miescher et al tell in their article, as they use KI and HCl for generation of hydroiodic acid (they add some phosphorus for regeneration ONLY whats obvious from the used and regained amounts).
They even use only an catalytic amount of KI (about 20%) so the formation of hydroidic acid by the dil.hydrochloric acid seems to proceed quite well.

So also if the distillation of the constant boiling hydroiodic acid is not possible - what may be or maybe not, I dont know for sure, the hydroiodic acid formed by the action of HCl on NaI or KI may be used "as is" as the HCl doesnt hurt the reaction - no it dehydrates the hydroidic acid and pushes the reaction forward this way.

HCl-aqueous aka muriatic acid aka hydrochloric acid was the solvent of choice in the old days when industry used this type of reduction widely - before 1920.
But they also used pressure vessels very often.

In short:
equilibrium HI(aqueous) <-> HCl(aqueous):   maybe.
NaCl (KCl) + HI <-> HCl + NaI (KI):   never

The picture of the dissoziated ions is misleading as we talk here about new compounds formed with specific properties: Hydrates of hydrogenhalides.


Osmium

  • Guest
Bollocks. Maybe you should try to read some...
« Reply #18 on: December 26, 2003, 09:31:00 AM »
Bollocks. Maybe you should try to read some beginner chemistry books instead of Gmelin.


tpower9s2003

  • Guest
equali<--->briums
« Reply #19 on: December 27, 2003, 01:27:00 PM »
SWIT thought of using HCl for HI a while back in regards to I2 tinc extraction where HCl is often added to KI or NaI - to UTFSE will see that even Rhodium agreed this was an equilibrium rxn "KI + HCl <---> HI + NaCL". Osmium is absolutely right and HCl+iodide salt will not work so well, if at all. This rxn is not a decomposition...that would be a one-way arrow.   

And then think about which acid is easier to concentrate sufficiently... HCl or H3PO4? Getting the HCl concentrated enough for the resulting HI to come over at 57% or more (if that rxn even worked) isn't simply boiling the water off, which can be done with H3PO4 given good ventilation. You'd be real lucky to find OTC or reasonble cost HCl more than 37%... And there is no reason to use HI of anything other than 57%, esp. for freebase rxns. Drier rxns are less likely to get equal results to long wet reflux at least in regards to standard HI+P rxns. The Shorty version of Argox is probly the easiest method to HI so why even bother with dilute muriatic from the nearest Depot.


Organikum

  • Guest
H3PO4
« Reply #20 on: December 28, 2003, 09:01:00 AM »
is of course preferable for preparation and for subsequent dehydration of the hydroiodic acid in the reaction so no phosphorus is available (or H3PO2/3).

Nevertheless I cannot agree in Osmium being "absolutely" right. Naming others names cannot replace references and information.
Naming names is usually the argument of the dumb.
And the fat ones I want to add.

have a nice day.
ORG

Rhodium

  • Guest
Read this clarification
« Reply #21 on: December 28, 2003, 09:55:00 AM »
Read this clarification I made to the above post the last time that was overinterpreted:

Post 459244

(Rhodium: "Behavior of ions in solution", Stimulants)



Osmium

  • Guest
> Nevertheless I cannot agree in Osmium...
« Reply #22 on: December 28, 2003, 10:12:00 AM »
> Nevertheless I cannot agree in Osmium being "absolutely" right.

That's ok, you don't need to agree. Doesn't change the facts though.

> Naming others names cannot replace references and information.

Did I call you names in this thread?

> Naming names is usually the argument of the dumb.
> And the fat ones I want to add.

Congratulations for the nice example for dumbness you so generously provided. Couldn't have said it better myself.


ragnaroekk

  • Guest
why I asked
« Reply #23 on: December 29, 2003, 10:28:00 AM »
"Reduction of Phytalonsäure to Phytalid-Carbonsäure
With potassiumiodide in diluted hydrochloric acid. 19,4g Phytalonsäure (0,1mol) have been boiled together with 8,3g KI (25% of theory) and 6g phosphorus for 4 hours in 12ccm conc. HCl and 22ccm....."
The rest is description of the workup, yields had been 78% of theory, 2,1g phosphorus were regathered.

Sorry I dont know the english names of the compounds.
Phosphorus produces no HI from KI as far as I know and the amount would be to low to do all the HI production and recycling if I didnt miscalculated it.

This is translated from this reference:

https://www.thevespiary.org/rhodium/Rhodium/djvu/miescher.djvu



Thats why I asked.

Please look at this post also:

Post 445996 (missing)

(ragnaroekk: "Nobody really read this it seems", Stimulants)

where I repost parts of an older thread where it is also stated that HCl + KI works for making HI (Crankenstein and Prickleberry)

I am a Bollock.
RaG


Osmium

  • Guest
> where it is also stated that HCl + KI...
« Reply #24 on: December 29, 2003, 11:33:00 AM »
> where it is also stated that HCl + KI works for making HI

Dissolving KI in aq. HCl produces a solution containing K(+), H(+), Cl(-) and I(-). Saying that the solution contains "HI" is misleading since the H(+) doesn't know that it is supposed to produce HI.
HI and HCl are very strong acids and can be considered to be 100% dissociated in aqueous solution.


ragnaroekk

  • Guest
?
« Reply #25 on: December 29, 2003, 03:07:00 PM »
This was already stated by Rhodium.
And it does by no way answer my question.
It explains plain nothing.

Why does it work in the translated reference?
And another question brought up (not by me) was this hydrate thingie?
If I understand it right, it is asked: Why do completely dissociated ions form quite stable constant boiling hydrates?


Rhodium

  • Guest
They form azeotropes/complexes.
« Reply #26 on: December 30, 2003, 05:05:00 AM »
They form something which can be seen as azeotropes/complexes.


ragnaroekk

  • Guest
?
« Reply #27 on: December 30, 2003, 09:40:00 AM »
Chief, isnt that exactly what I wrote?
...constant boiling hydrates.
....azeotropes/complexes.

I would define a hydrate as a complex - perhaps I am wrong, but for sure a constant boiling mixture of two compounds is an azeotrope.

Isnt it this way that HI dissoziates into H+ and I- ions which are as dissociated ions associated to a certain number of H2O molecules? (or H+ and O- ions if the water is also dissociated? Serious question, no teasing) This way things would fit under one hat - the dissociated ions of the HI form complexes with watermolecules (the ratio varying with temperature and concentration) called hydrates?

If this is right, the explanaition of the reactionmechanism of HI reductions should regard this and should be interpreted from the view of the hydrated HI and not from completely dissociated ions complexed with nothing.

Festers HBr dehydration with HCl
The HI dehydration with H2SO4 or GAA or HCl

Damn, far away from the question: NaI + HCl = HI yes or no, or partly, this would provide a working theory for halogenations with haloacids which could be made high yielding and also adjustable so that the halogenation might stop at a certain point and not affect all possible parts of a molecule (semiselective?).
I believe this being useful.


Rhodium

  • Guest
"why do complexes form"?
« Reply #28 on: December 30, 2003, 03:42:00 PM »
I simply said why hydrates form, as you asked it - and as you said in the later post, it is a kind of complex.

Is your question now "why do complexes form"? If so, then that is too complicated for me to answer - it's an electron/orbital thing.