A few years back I remember reading that it was an exceptionally powerful (yet simple) reducing system but I can't seem to find the page(s) I read
that on anymore. Can anybody give me some info about it?Sauron - 17-9-2007 at 19:50
Al and formic acid? That will certainly dissolve and liberate H2, and make aluminum formate.
To reduce what exactly? Selectivity?stygian - 17-9-2007 at 19:53
To reduce an amide to an N-methyl. Not many of your more common agents seem capable of that. I was looking specifically for its reaction towards
alcohols and/or esters (formic esters, hoping that they would not be reduced or could be protected in some way)
[Edited on by stygian]Sauron - 18-9-2007 at 00:29
You mean to reduce a -C(O)NH-Me to -CH2-NH-Me is that right?
Or is this a reductive amination, sat of a formamide with methylamine under dehydrating conditions like toluene and a Dean Stark trap??
No Al involved. That's a Leuckart reaction. As commonly used to prepare N-methylformamide, and I'm sure you know what the next step is, don't you? And
the end product?stygian - 18-9-2007 at 08:14
The archetypical ammonium formate leuckart amination produces the N-formyl derivative, which is commonly then hydrolysed off. Instead, I aim to
reduce said -C(O)H to -CH3.
I read of the HCOOH/Al reduction on few (more than one, but admittedly not many) pages on the web. The difficulty in finding any mention makes me
wonder if it was even correct. It seemed at the time like a fine alternative to BH3*THF. Maybe it's wishful thinking that an agent to reduce the
formyl group would leave an ester hopefully untouched.Dr. Beaker - 18-9-2007 at 14:57
A rough guess:
Al indeed dissolved in formic acid to produce Al(HCOO-)3. Al+3 acts as a lewis acid and activates the carbonyl\amide. HCOO- is known as a hydride
synthon ansd it reduces carbonyl\amide:
" HCOO- = H- + CO2"
HCOO- + RCONHR + H2O ---> HCO3- + RCHOHNHR
alkoxide takes proton from water. formate oxidized to CO2 that reacts with the OH- from the deprotonated water
RCHOHNHR ---> H2O + RHC=NR
vicinal amino alcohol losses water.
RHC=NR + HCOO- + H2O ---> HCO3- + RCH2NHR
amide takes proton from water
VwalaSauron - 18-9-2007 at 16:23
Googling "formic acid-aluminum reduction" gets us nothing pertinent.guy - 18-9-2007 at 16:25
This Leuckart reaction looks intersting. How good are the yields usually?Organikum - 19-9-2007 at 02:09
Quote:
Originally posted by guy
This Leuckart reaction looks intersting. How good are the yields usually?
If you dont know exactly what you are doing then you will get a lot of the famous "red tar" and up to no product.Dr. Beaker - 19-9-2007 at 09:27
That's what scifinder brought in
I know RhIII is not Al+3 but the principal is similar - formate as a hydride source
Aqueous Transformation of a Metal Diformate to a Metal Dihydride Carbonyl Complex Accompanied by H2 Evolution from the Formato Ligands. Ogo,
Seiji; Nishida, Hiromi; Hayashi, Hideki; Murata, Yusuke; Fukuzumi, Shunichi. Department of Material and Life Science, Graduate School of
Engineering, Osaka University, Suita, Osaka, Japan. Organometallics (2005), 24(20), 4816-4823. Publisher: American Chemical Society, CODEN:
ORGND7 ISSN: 0276-7333. Journal written in English. CAN 143:415091 AN 2005:904478 CAPLUS (Copyright (C) 2007 ACS on SciFinder (R))
Abstract
[RhIII(tacn)(HCOO)2](OTf) (2, tacn = 1,4,7-triazacyclononane, OTf- = CF3SO3-) reacts with HCOONa at pH 6-7 at 80 in H2O to form a Rh dihydride
CO complex, [RhIII(tacn)(H)2(CO)](OTf) (5), with H2 evolution from the formato ligands of 2. This is the 1st example of H2 evolution from H atoms of
formato ligands of a metal diformate complex. Complex 2 was synthesized from the reaction of [RhIII(tacn)(H2O)3](OTf)3 (1) with HCOONa in H2O.
Recrystn. of 2 with HCOONa in H2O gave crystals of [RhIII(tacn)(HCOO)3] (3) whose structure was unequivocally detd. by x-ray anal. The key
intermediate in the transformation from 2 to 5 is a Rh hydride formate complex, [RhIII(tacn)H(HCOO)](OTf) (4), which was obtained by heating of an aq.
soln. of the Rh diformate complex 2 with 10 equiv of DMSO at pH 6-7 in H2O. Complex 4 was well characterized by isotopic labeling measurements of 1H
NMR, electrospray ionization mass spectrometry (ESI-MS), and GC.Mardec - 20-9-2007 at 13:43
Sorry for the kinda off topic question, but what is so great about N-Methyl?.. (I am kinda new to organic..)
And I will look into it, it sounds familiar Al+some acid reduction. Should have some files on it.
