06-06-02 19:17
No 318282
http://www.chemistry.ccsu.edu/glagovich/
In this link they are reducing nitro compounds to hydroxlamines with Zn %50 Etoh in the prescence of ammonium chloride. Is this a standard pratice. It appears that they are using it for some kind of test purposes so swim might conclude that it is a valid method.
R-NO2 + H4 >-----Zn/AmCl--------> R-NH2OH + H20
In the case o 3,4MDP as variable r them would then the product not be MDOH. Which then if desired could be thermaly degraded to MDA.( Not that it Would make a difference)
Said product could be reduced
(Chief Bee)
06-06-02 19:25
No 318284
It is a reasonably standard practice, I've seen it several times before, and also for the synthesis of N-hydroxy-phenethylamines.
(Hive Addict)
06-06-02 19:59
No 318293
Is it feasable then that this could be used to reduce MdP2NP. Has any one tried anything like this. Swim would assume that solubility might be an issue. Would running the rxn with ETOH/AA solve the pseudonitrite solubility without adversly affecting the rxn due to the PH. This would probably facilitate the rxn with the extra H+ ions from the acid. Use 4 moles of acid per mole of nitro to supply the 4H+ that it seems the rxn calls for. Or Is that calculation uneducated.
VL_
(Chief Bee)
06-06-02 20:27
No 318302
No, you must use the saturated nitro compound, not the phenylnitropropene (reduce it with NaBH4 first).
(Hive Bee)
06-06-02 21:11
No 318316
SWIM is very interested in this.RHODIUM,SWIM wants to make sure he understands. Your just saying the compound needs to be MDP2NO2P(SWIM doesnt think he wrote this correctly but its suposed to be a NO2 compound). Correct? Maybe thats what V_L meant also?
I think i like it!
(Hive Addict)
06-06-02 21:44
No 318328
Yes but it has to be reduced first to the saturated state.
VL_
(Newbee)
06-07-02 00:25
No 318360
ah ha a moment of clarity, if the nitro alkENE is used rather than the the nitro alkANe then I bet the product is the oxime. thinking goes like this it would reduce the nitro group to the N-hydroxyamine and the alkene group would be unreduced there for one would have a vinyl hydroxylamine, the enol form of the oxime to which it would 'tauteromize'. the oxime could be useful....
tis late and i lak sleep
(Hive Addict)
06-07-02 00:38
No 318361
How could one nitrosaturate the alkane? What is the 3,4 md nitroalkane?
VL_
(Chief Bee)
06-07-02 01:31
No 318377
Obia: Correct.
VL: Huh? Nitrating an alkane is not feasible, but reducing a phenyl-2-nitropropene with NaBH4 gives the phenyl-2-nitropropane (the saturated nitroalkane) which in turn can be reduced to the N-hydroxyamphetamine with Zn/NH4Cl.
(Hive Addict)
06-07-02 02:00
No 318381
yes the question was wasit possible somehow to nitrate the ankene.
(Chief Bee)
06-07-02 03:04
No 318408
You can nitrate an alkene (such as isosafrole) to the phenylnitropropene using either the pseudonitrosite route or NaNO2/I2/HOCH2CH2OH (but you probably already knew that).
(Official Hive Translator)
08-25-02 22:37
No 348912
(Rated as: excellent)
Check this out, this is from Patent GB990092.
Personally for me, this piece of info is very reassuring as to the fact that the NH4OH/Zn/Uru Ni system, described by Rhodium and Zooligan in Post 270027 (Rhodium: "Zn/NiCl2 reduction of oxime/nitro/nitriles/ketone", Chemistry Discourse) holds a great promise for reducing nitropropenes.
Whatever direction the reduction dominantly takes place (two possible ways: via nitroalkane or from nitroalkene to oxime) - that system should work on both, and at the same time the presence of ammonia suppresses side reactions and favors formation of the amines.
Just a thought,
Antoncho
(Hive Bee)
08-26-02 20:26
No 349240
I have understood that the reason why acidic reaction conditions are better when reducing nitroalkenes is because the dimer formation through Michael addition is suppressed. Well, at least with catalytic hydrogenations. Comments?
(Official Hive Translator)
08-27-02 14:40
No 349610
Dear Moo (or someone else), could you please elaborate a little bit for the ignorant?
From what i've learned, Michael addition is the addition of an enolized ketone to a nucleophil, such as double bond. In this case, i guess, a nucleophil is meant to bee the end aminogroup, while the electropositive part is.... what? The 'acidic' carbon attached to nitrogroup of the starting nitropropene? In this case, what does an alkali base have to do w/all this business?
Or - another possibility - a base abstracts a proton from nitropropene and it condenses w/itself?
Or am i altogether wandering in the dark??
Antoncho
(Stranger)
08-28-02 06:56
No 349941
Atoncho
Nitroethenes and nitropropenes are electronically analogous to alpha,beta-unsaturated ketones(the classical Michael acceptor). The benzyllic carbon is electrophillic and is subject to attack by suitable nucleophiles, which in the case of nitrostyrene reductions could either be the fully reduced amine or the deprotonated nitroethane(electronically analogous to an enolate). Both nitroethenes and nitropropenes are good michael acceptors, and a deprotonated nitroethane is a good nucleophile, but a deprotonated nitropropane is usually so sterically hindered that it is not very reactive as a nucleophile, and dimerization is therefore much less of a problem in nitropropene reductions than it is in nitroethene reductions.