clearly_not_atara
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[theory] anhydrous metal halides by Wurtz coupling
Usually anhydrous metal halides are prepared from either the halogens or the anhydrous hydrogen halides, which is annoying because they are not
supplied to consumers and they are volatile, which often causes them to boil during reaction. Instead halogens are supplied as haloimides, and
"deprotecting" the halogen is usually associated with significant losses.
However, benzyl chloride can be prepared directly from TCCA:
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&...
It stands to reason that a similar reaction will proceed with brominated hydantoins which are the active ingredient in "pool bromine", since HBr is
more acidic than HCl and the attendant radical formation from Br2 proceeds more easily.
We also know that benzyl bromide reacts with magnesium to produce magnesium bromide and bibenzyl. This paper mentions a Wurtz coupling with aluminum
and benzyl chloride, which should give an AlCl3 byproduct:
http://pubs.acs.org/doi/abs/10.1021/ol202733v
The Wurtz coupling is usually an undesirable side-reaction in the preparation of benzylic organometallics: Google readily finds hundreds of ways to
prevent it from happening, but papers rarely discuss performing it intentionally except under strange conditions (e.g. in water). However there are
reports of performing the Wurtz coupling even with manganese, iron, copper or chromium, in addition to the traditional substrates magnesium, zinc,
lithium, and sodium. Using the preparation directly from available forms of the halogen may give significantly higher whole-process yields.
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Bobymartinez
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I think the Wurtz coupling will occur when the halogen compound is in excess compared to the metal so if you have large amount of BnBr compared to the
aluminium you should obtain the anhydrous chloride/bromide and the coupling product (maybe aluminium amalgam would also work).
Why not using directly TCCA to react with aluminium ? I don't exactly know the conditions though...
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tsathoggua1
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You can dehydrate hydrated metal chlorides (speaking of the transition series here) by reflux in SOCl2, the reaction products being gaseous is a handy
thing and drives the reaction to the right. The other sulfur halides would probably work too.
Question-will this replace other halogens (ignoring F of course) if performed on a hydrated metal bromide or iodide? Or if both hydrolysis of the
sulfur halide/SOCl2 and halogen exchange take place, can kinetic control be used to ensure that primarily the result is that of the hydrolysis
product?
[Edited on 14-3-2017 by tsathoggua1]
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clearly_not_atara
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| Quote: | | Why not using directly TCCA to react with aluminium ? |
Separating AlCl3 from the byproduct aluminium cyanurate is likely to be anywhere from annoying to impossible. By contrast if benzyl chloride is used
to oxidize aluminium the byproduct bibenzyl is soluble in organic solvents and can be removed by evaporation if necessary.
| Quote: | | You can dehydrate hydrated metal chlorides (speaking of the transition series here) by reflux in SOCl2, the reaction products being gaseous is a handy
thing and drives the reaction to the right. The other sulfur halides would probably work too. |
I suspect that anyone who is trying to prepare metal halides starting from pool chlorine is going to have a difficult time obtaining SOCl2.
It's my understanding that usually the reaction n*HX + MCln >> n*HCl + MXn favors the products, because chlorine has the shortest covalent
bonding radius and hence forms more stable bonds with hydrogen. SOCl2 reacts with water of crystallization to form SO2 + HCl which generally will not
react with a metal bromide. However if SOBr2 is used on a metal chloride, HCl gas may be released.
One example of the above property is the iodide process for refining titanium, where TiCl4 + HI >> TiI4 + HCl is used to prepare titanium
tetraiodide from the chloride.
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tsathoggua1
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Well the difficulty in for most, obtaining SOCl2 is why the other sulfur halides were suggested, they are less of a shit to prepare. Still somewhat
constipated but not the chilli-est turd in the rectum.
What about refluxing over CaO in a chlorinated solvent (dried in the process, conveniently)would that work?
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Bobymartinez
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| Quote: | | Separating AlCl3 from the byproduct aluminium cyanurate is likely to be anywhere from annoying to impossible. |
I don't think it will be that hard to separate aluminium chloride from byproducts. AlCl3 sublimation point is only 183°C and is usually purified that
way before a Friedel-Craft. Juste use a cold finger to condense everything.
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clearly_not_atara
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Quote: Originally posted by Bobymartinez  |
I don't think it will be that hard to separate aluminium chloride from byproducts. AlCl3 sublimation point is only 183°C and is usually purified that
way before a Friedel-Craft. Juste use a cold finger to condense everything. |
Are you trolling me? Do you have any idea why this is considered to be such a difficult transformation? Do you understand what a "double salt" is? Has
it not occurred to you that cyanurate is a Lewis base?
Haven't you figured out that if it were that easy there wouldn't already be ten years of threads about anhydrous metal halides? Is it not clear why
the laziest possible idea has already been tried and failed? | Quote: | What about refluxing over CaO in a chlorinated solvent (dried in the process, conveniently)would that work?
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I think you'll get a calcium aluminate. You may be able to use dry CaCl2 if it's available.
[Edited on 14-3-2017 by clearly_not_atara]
[Edited on 14-3-2017 by clearly_not_atara]
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