[Benzene]

Lithium Aluminum Hydride is used to reduce Nitroalkenes. 4 equivalents of Lithium Aluminum Hydride are used, and I have read a common impurity is the oxime impurity.

I have included a sketch of what I think happens, in short.
1. Hydride Attacks Nitrogen, Oxygen takes Hydrogen.
2. Hydroxide leaves, N=O bond forms
3.conjugate addition by Hydride, deprotonated oxime

then I have some other things involving AlH₃, but this scheme would account for the oxime impurity.

[zsinger]

I would consult Dr. Lennox, as he is basically correct in every word that exits his mouth.  I like how you showed where the equivalents of hydride are used, however I'm somewhat sure the mechanism is not 100% correct, but surely close.  I don't quite follow your electron pushing.
           Cheers.
                         -Zack

[zsinger]

Also, is OH- a good leaving group? 

[Benzene]

Yah sorry, the lighting in my house is not great, and sometimes my drawings aren't good. I drew it again.

my mechanism is direct addition of the hydride.
oxygen removes proton on positively charged Nitrogen, oxygen double forms, kicks out a hydroxide.
conjugate addition by the hydride, forming the deprotonated oxime.
negatively charged oxygen pairs with AlH₃ and Hydride attacks Nitrogen Oxime, creating a N-O-AlH₃ transition state
Hydride quickly attacks Nitrogen kicking off AlH₃O^- Li⁺
and creating the boxed product

[Benzene]

Couldn't ^-OH leave in an extremely basic solution.   ?

[orgopete]

Couldn't ^-OH leave in an extremely basic solution.   ?

Yes, in an aldol condensation, but not here. If an -NOH bond were to form, it would react with LAH to give H2 plus -NO(-). I'd start with conjugate addition. I'd also look up an example of this reaction to learn more about it, how it was done, and the stoichiometry. You might compare this reduction to the reduction of a carboxylic acid.

[Benzene]

The paper I have uses a large excess of LiAlH₄ in tertrahydrofuran (anhydrous). The ratio of LiAlH₄ to Substrate is 5:1.

I know carboxylic acid reductions require aqueous work-ups at a couple points on  the way to the alkane. I included a drawing of LiAlH₄ reduction of carboxylic acid.

The reduction with large excess of LiAlH₄ doesn't require H₃O⁺. This is the reason why I am confused, it seems like hydroxides must be involved?

[orgopete]

I suggested the reduction of a carboxylic acid as it too involves the expulsion of an O(-). I think this is done by an aluminum, see http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/c_acid/reaktionen_organoli_carbons.vlu.html. I wrote a similar mechanism in my book, A Guide to Organic Chemistry Mechanisms.

This mechanism requires more work to be done. The formation of trivalent aluminum must be done two times. First to capture the oxygen and then for elimination of that oxygen. If you were to write out all steps of borohydride or aluminum hydride reductions, you would have written steps showing exchanges between tri and tetravalent atoms.

When I researched my mechanism, I had begun to wonder about the efficiency of these steps. Could aluminum capture another oxygen? The reactions I saw often involved longer reaction times, intermediate additions of reagent, and stoichiometric excesses. I thought this was consistent with a greater reluctance to eliminate the O(-) by aluminum. Indeed, I think you will find the preferred reduction of a COOH is by BH3 or convert the acid into an ester.