CuReUS
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humanity against hydrides
most of us know hydrides are difficult to work with.They are expensive,not OTC,difficult(if not impossible) to make at home ,watched and one must be
very careful working if them(if you don't want to convert your flask into a flamethrower. )
So I am starting this thread to request all the members to post information here whenever they come across novel reagents that can be used instead of
hydrides,so that future members will be benefited when they are searching for alternate reagents.
I also request the moderators consider making this thread a sticky,so that it can be easily found.
since I started this thread,let me do the honours
http://www.ncbi.nlm.nih.gov/pubmed/12596148
preparation of SmI2
http://pubs.acs.org/doi/abs/10.1021/jo300135v
preparation using just I2,with video demonstration
http://www.jove.com/video/4323/preparation-use-samarium-diio...
[Edited on 26-5-2015 by CuReUS]
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ledob86
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Hi !
Begin to read : Reductions in Organic Chemistry, Hudlicky (it's only one
example)
If you are located in europe there is a good powdered metals supplier (cheap) > metallpulver24.de
S3chemicals sell LiAlH4 more easily than NaBH4 ... just have to attest on paper what you want to do with. Onyxmet is probably another source.
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Pumukli
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Good topic but SmI2 sounds almost as exotic as hydrides (for an amateur at least).
Then comes the question: what kind of compounds would you like to reduce? Because hydrides are versatile and can reduce this and that and even that
other one too, if you tinker a bit with the reaction.
On the other hand there is a thread on SM e.g. on using copper activated aluminum instead of aluminium amalgam for some quite specific reactions.
For us, amateurs, maybe the best was to explore the limits of electrochemical reductions. Electricity is cheap, OTC, high current power supplies are
also inexpensive nowadays, the anode materials can be varied - but can be cheap and OTC too, the workup would be easier in most cases, etc. etc.
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adk
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Why not use Vitride / Synhydrid / Red-Al?
http://en.wikipedia.org/wiki/Sodium_bis%282-methoxyethoxy%29...
Almost identical to LiAlH4 in terms of reactivity. Mechanism slightly different but useful for reducing very many compounds in good yield due to high
solubility in non polar solvents.
It is a hydride, but many compounds simply can not be reduced in good yield by other means.
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byko3y
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>many compounds simply can not be reduced in good yield by other means
"Some compounds" actually is an amide.
Dissolving metals (Al, Zn, Fe, Mg) can perform many jobs of hydrides, urea peroxide can reduce ketones and dithionite can reduce unsaturated compounds
just like NaBH4 do.
I want to emphasize that the reduction of amides by LAH is not selective, any other group will be reduced first and then the amide. You can activate
amide by converting it into chloriminium with phosphorus chlorides (you can even alkylate the amide carbon with grignard, AFAIK), and then reduce it
with dissolving metal, but I suggest avoiding the need of amide reduction.
You Need silanes to selectively reduce amide, but they are expensive and hard to obtain.
Ordering LAH instantly draws attention to you, amateur chemists doesn't wants this. And you can't make hydrides at home.
And I don't know no source of samarium at my region.
[Edited on 27-5-2015 by byko3y]
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CuReUS
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Quote: Originally posted by Pumukli  |
Then comes the question: what kind of compounds would you like to reduce? Because hydrides are versatile and can reduce this and that and even that
other one too, if you tinker a bit with the reaction. |
it doesn't have to be a specific group.what I am asking is if anyone finds out an alternate reagent to reduce something which is generally reduced by
hydrides,they should post it here.
| Quote: | | For us, amateurs, maybe the best was to explore the limits of electrochemical reductions. Electricity is cheap, OTC, high current power supplies are
also inexpensive nowadays, the anode materials can be varied - but can be cheap and OTC too, the workup would be easier in most cases, etc. etc.
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yes,you are right.but the main problem of electrochemical reactions is that the yield doesn't cross 40-50% for most reactions
| Quote: Originally posted by adk |
It is a hydride, but many compounds simply can not be reduced in good yield by other means. |
don't tell me it can't be done.you should be ashamed of yourself,being a chemist and saying that something is not possible.
| Quote: Originally posted by byko3y |
Dissolving metals (Al, Zn, Fe, Mg) can perform many jobs of hydrides, urea peroxide can reduce ketones and dithionite can reduce unsaturated compounds
just like NaBH4 do. |
very good.this is exactly the kind of thing I want.now please provide the necessary papers.(especially UHP)
| Quote: | | And I don't know no source of samarium at my region. |
IIRC,one member had mentioned that Sm could be obtained from magnets.But it can be easily bought from chinese suppliers.and its not that
expensive-5$/10g
[Edited on 27-5-2015 by CuReUS]
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byko3y
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Okay, samarium seems to cost 25$ per kg, often it's like 50-60$, but still it's relatively hard to obtain in many regions, and there's no use in
magnets, because neodymium is extremely expensive (800$ per kg).
But anyway - thanks for the idea. Reduction with SmI2/H2O/THF looks very sexy.
I was talking about thiourea dioxide (misstaken with UHP):
http://www.sciencedirect.com/science/article/pii/S0040403901...
http://www.sciencedirect.com/science/article/pii/01437208948...
http://www.sciencedirect.com/science/article/pii/S0040403900...
The problem with electroreduction is that it's very specific to a particular compound, and not to its functional group. You can reduce one compound
with one electrode, while its homolog requires another electrode for a successfull reduction. And substituting aromatic group for aliphatic almost
always leads to completely different products.
Non-selectivity is another problem (related to the first one). You can reduce aldehydes, but ofter they become coupled (pinacol coupling). Same
problem with carboxylic acids oxidation to alcohols vs alkene vs kolbe coupling.
You can electroreduce salicylic acid to salicylic aldehyde, but you cannot reduce benzoic acid to benzaldehyde - you will get benzyl alcohol instead.
This problem is similar to ligated nickel coupling, so basically when you develop a reduction or coupling of a single compound - your knowledge is
barely usefull for similar coumpounds.
UPD: reduction with samarium is similar to electroreduction and other dissolving metal reduction, and can be used interchangeably https://en.wikipedia.org/wiki/Mark%C3%B3%E2%80%93Lam_deoxyge... This leads to a need of many equivalents of SmI2.
The problem is not in the cost of samarium, but the cost of iodine. For reduction of 5 moles (~1 kg of some random 200 g/mole compound) you will need
20 moles of samarium iodide, which is 5 kg of iodine.
[Edited on 28-5-2015 by byko3y]
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CuReUS
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| Quote: Originally posted by byko3y |
The problem is not in the cost of samarium, but the cost of iodine. For reduction of 5 moles (~1 kg of some random 200 g/mole compound) you will need
20 moles of samarium iodide, which is 5 kg of iodine. |
first of all,why would a home chemist need to reduce 1kg of any substance,unless they are involved in nefarious activities.I don't think any honest
home chemist would need to work with more than 50-100g of reagents at a time.And that's a lot.
And I am sure that if LiAlH4 had been used to reduce 1kg of that same compound,it would have cost 10 times more,the workup wouldn't have
been clean,an explosion might have occurred and your purchase would have been flagged by the DEA 
hope this makes you happy
from http://www.google.co.in/patents/US6211392
| Quote: | | catalytic amounts of samarium diiodide can be used for the synthesis of ferrocenyl-1,3-butadienes. Magnesium can be used to promote the recycle of
samarium diiodide in the reactions. By using magnesium to improve the recycling of samarium diiodide, a molar ratio of the samarium diiodide to
ferrocenecarbonyl is in a range of about 0.01-10. |
hit me if I am wrong,but if Mg can be used,Al can be used as well.(I know you
can't do this for all reactions,but still)
[Edited on 28-5-2015 by CuReUS]
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byko3y
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You need something like of 1 eq. of LAH per 2 moles of reduced compound. In my region 1kg (26 moles) of LAH costs 250$, while 1 kg (4 moles) of I2
costs 100$.
In case of LAH vs SmI2 for 5 moles of reduced compound we need 5-10 moles of LAH, which costs 50-100$, or 10-20 moles of I2, which costs 250-500$.
As you can see, LAH is a relatively cheap reagent, 1 kilo of it costs a lot mainly because of its low molar mass (28g/mole).
Of course, Zn or Al are much cheaper (3-10$ per kg), so their usage can be less expensive than LAH, but Sn(tin) and SnCl2 is more expensive, close to
the price of LAH: 10-15 moles of SnCl2, 20$ per kg, 225g/mole, 2-3 kg needed for the reaction, total cost 40-60$ - something close to LAH cost.
This price is reasonable only if your compound is expensive (5 moles of it should cost at least 300$), so those reagents are usually used for small
scale preparations and production of illegal substances.
UPD: nice catch on Mg+SmI2, I'd be glade to know more http://www.researchgate.net/publication/273437366_Aza-Reform...
But why do we need samarium in the first place, when we have a dirt cheap magnesium, which is extremely powerfull both as grignard reagent and
magnesium methoxide http://pubs.acs.org/doi/pdf/10.1021/jo035191d
[Edited on 28-5-2015 by byko3y]
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CuReUS
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thanks to you.I was thinking that the cost could be reduced if some reducing agent could convert Sm+3 back to Sm+2 and boom,I
found it.Too bad we can't make a complete catalytic cycle like the Wacker
oxidation,as there is no H2 in the atmosphere
| Quote: | | But why do we need samarium in the first place, when we have a dirt cheap magnesium, which is extremely powerfull both as grignard reagent and
magnesium methoxide |
but it has only limited use,the reduction of peroxides.Sm on the other hand can reduce many things.
| Quote: | | hit me if I am wrong,but if Mg can be used,Al can be used as well.(I know you
can't do this for all reactions,but still) |
do you think Al could be used ? because I checked the electrochemical series for the reduction potential of Sm+3 to Sm+2 but I
couldn't find it.Obviously, you would have to use something with a more negative reduction potential than Sm+3 to Sm+2.Do you
think Al is powerful enough ?
Also,Mg cannot be used as a co-reductant when your substrate has halides or carbonyl groups.So I was thinking,If Al could not be used,maybe we could
use La? La is very powerful,won't interfere with other groups(hopefully) and it can be obtained OTC from lighter flint
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