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Author: Subject: LAH reductions with H2SO4: rationale?
flyingbanana
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[*] posted on 3-12-2009 at 02:10
LAH reductions with H2SO4: rationale?


In PIHKAL, several nitrostyrene reductions with LAH are performed with the aid of 100% (fuming) sulfuric acid. Why? I haven't seen anything similar done in common literature for other compounds (and if there are, references would be welcome).
Would it have anything to do with nucleophilic mechanism of hydride attack? If so, then would an alternative reduction employing borane based reagents be more effective?
Either way, adding sulfuric acid to a fully loaded solution of LAH does not seem like something I would ever want to do.
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Nicodem
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[*] posted on 3-12-2009 at 08:34


H2SO4 is added because sometimes the chemist prefers do the reduction with alane ethereates instead of lithium tetrahydridoaluminate. Their chemoselectivity differs. For example, LiAlH4 can often dehalogenate aryl halides while alane can not. Also, some functional groups reduce faster with alane so the rationale can also be in increasing the reaction rate.
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flyingbanana
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[*] posted on 4-12-2009 at 18:17


Thanks for pointing me to the actual reducing agent. Opened up a whole new search space.
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Barium
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[*] posted on 6-12-2009 at 07:41


Alane can be generated from LAH or SAH (sodium aluminum hydride) and, pretty much, any strong acid; HCl, methanesulfonic acid or strongly acidic cation-exchange resins.
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smuv
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[*] posted on 6-12-2009 at 09:33


...just to add, lewis acids such as AlCl3 or ZnCl2 will also produce alane.



"Titanium tetrachloride…You sly temptress." --Walter Bishop
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Barium
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[*] posted on 7-12-2009 at 02:32


If AlCl3 is used to generate alane, one needs to use about a threefold excess in order to supress the formation of AlH2Cl and AlHCl2. As alane cleaves THF, solutions needs to be used quite promptly. Triethylamine can be used as a lewis base to form the AlH3-NEt3 complex which is as strong reducing agent as AlH3 alone but can be stored in THF for a month without cleavage of the ether.
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smuv
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[*] posted on 7-12-2009 at 18:42


Yes, a three fold excess of LAH to AlCl3 is required (just to clarify the wording of your post implies the opposite). Brauer outlines this procedure.

The mixed hydrides can be useful though for some reductions, such as the reduction of certain ketones to the hydrocarbon.

ex. J. Am. Chem. Soc., 1958, 80 (11), pp 2896–2898




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[*] posted on 7-12-2009 at 22:05


So alane reduction of 4-bromo-2,5-dimethoxynitroalkene would yield the halogenated aminoalkane? Pretty awesome!
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[*] posted on 8-12-2009 at 08:18


Yes you are absolutely right Smuv. Even though I read and re-read the text before posting, I still managed to fuck it up. Is this a sign to start using other solvents than benzene and chloroform? :o
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