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Author: Subject: Imidazole Experiment?
Pomzazed
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[*] posted on 19-11-2008 at 06:59
Imidazole Experiment?


I've just acquired around 100g of pure imidazole (>98%). It has a distinctive hospital-like smell.

Just wonder if there're any interesting experiment with this if you could have pointed me out?




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JohnWW
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[*] posted on 19-11-2008 at 09:47


This should be in the Organic Chemistry section. Because imidazole is aromatic, like pyrrole, it lacks the usual basicity of amines. It is actually weakly acidic, forming an anion (also aromatic) with strong bases by losing the H+ from the -NH= nitrogen. Imidazole undergoes aromatic electrophilic substition, so you could try nitrating some of the stuff to obtain 4-nitroimidazole, which may be an energetic compound. It also undergoes electrophilic ubstitution with Br2 or Cl2 to form 4-bromoimidazole. These 4-substituted products are also 5-substituted imidazoles because of tautomeric shift of the H on one N atom to the other N atom.
(Sorry - confused imidazole with indole, due to another post of mine this morning about indole derivatives).

[Edited on 20-11-08 by JohnWW]
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Pomzazed
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[*] posted on 19-11-2008 at 11:10


Well, i know it is an aromatic compound.
But at first hand I'm thinking about experimenting with its salt and complex properties first, before shifting to messing with organic reactions.

Do you know the molar ratio it form complex with Cu(II) in aqueous media? Is the [Cu(imd)4]2+ in analogy to [Cu(NH3)4]2+ too crowded in ligand number? so i can calculate the correct amount to use.

Google search only reveals other Cu2+/imidazole containing indicator complex etc.

Quote:
so you could try nitrating some of the stuff to obtain 4-nitroindole, which may be an energetic compound. It also undergoes electrophilic ubstitution with Br2 or Cl2 to form 4-bromoindole. These 4-substituted products are also 5-substituted imidazoles because of tautomeric shift of the H on one N atom to the other N atom.

How can it become an indole? the cyclic core of imidazole and indole is different!

PS. does imidazole even withstand the basic acid-nitrator media without decomposing due to the oxidative power of the nitrate?

[Edited on 20-11-2008 by Pomzazed]




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Nerro
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[*] posted on 19-11-2008 at 16:09


If you couple two sidechains to the N atoms you can deprotonate the C between the N's to form a stable carbene which can serve as a very solidly connected ligand!



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chemrox
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[*] posted on 19-11-2008 at 18:37
imidazole book


Imidazole also has amphoteric properties due to the two N's in the aromatic ring. I d/l'd a book on them as a class of compounds:

Attachment: M.R.Grimmett - Imidazole and benzimidazole synthesis.djvu (2.9MB)
This file has been downloaded 250 times





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Pomzazed
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[*] posted on 22-11-2008 at 10:39


Update #1

I'm trying to crystalize out the imidazole copper complex as sulfate salt by mixing 10mmol of imidazole and 2.5mmol of copper sulfate with water as solvent.

They both dissolve readily and form a deep blue complex in the same fashion as copper-ammonia one, then i let it stand to crystalize out.

What i have notice is i got both deep blue crystal of the complex!
BUT!
There's very fine precipitate which is olive green! and this contaminate my crystal, so i try again and mix 5mmol/1.25mmol ratio and watch it.

At first the solution is clear deep blue, but after 10 min, the olive green precipitate started to form and gather at the bottom! so i filter them out, and see again, and the precipitate slowly forms again, and then the complex started to crystallize out.

I separate the precipitate and try washing with water, which doesnt dissolve at all but give turbid green instead.
The precipitate is so fine that holding the beaker will make one in the bottom go up into solution.

Any idea what this is?
- copper hydroxide/carbonate from contacting with air?
- redox product between Cu(II) and imidazole?
- copper imidazolide?
- ideas here please!




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