White Yeti - 18-11-2012 at 12:15
This may be a rather fussy question, but is there a practical reaction which allows for the reduction of an acid group to an aldehyde? I've done some
searching and came up with nothing but reductions of acid groups all the way down to alcohols.
tatapouette - 18-11-2012 at 12:28
There are some methods that describe the reduction of the acid to the aldehyde.
But, often, it seems to work better on the whiteboard than in the flask...
The usual method is indirect and consists in :
- converting caboxylic acid to the methyl ester ;
- reducing to the primary alcohol ;
- smoothly oxidize to the aldehyde.
This usually is found better, even if in three steps, because :
- the first two steps are usually easy, clean and good yielding ;
- there are lots of methods (and papers) for the the third step.
EDIT : oops, my bad... There are some methods for reducing carboxylic acid derivatives more or less easily obtained (acyl chloride etc) to the
aldehyde. Starting strictly from the carboxylic acid might be more exotic. I guess the lability of the proton is the main problem here.
[Edited on 18-11-2012 by tatapouette]
[Edited on 18-11-2012 by tatapouette]
[Edited on 18-11-2012 by tatapouette]
kristofvagyok - 18-11-2012 at 12:47
There bis a LOT reference on this kind of reductions, use the search engine.
I would suggest 2 method:
1: DIBALH. A bit expensive, but great to work with.
2: convert the RCOOH to RCOCl and reduce it with Pd/BaSO4. This is a selective method used widely, it is called Rosemund Reduction.
Sydenhams chorea - 18-11-2012 at 12:58
Na(Hg) works too
SM2 - 18-11-2012 at 13:21
Yeti, were you looking for an electronic explanation as to why it is possible to oxidize an alcohol to the aldehyde/ketone stage? You can easily over
do it and take things a step further to the COOH acid. Then again, some alcohols are tougher, and some are very vulnerable. Lots of factors can be
involved in such a seemingly simple reaction because your end product is highly reactive, and is best taken up and away as it is formed.
Dr.Bob - 19-11-2012 at 06:33
I've tried to do the selective partial reduction of acids to aldehydes. Even the best working reaction usually gives multiple products, which is why
I agree with tatapouette. I just try to avoid needing the aldehyde if possible. For example, rather than do a reductive amination of an aldehyde
and amine, I would make the amide and then reduce, which works pretty well, in most cases. Not all routes work this way, but aldehydes are some of
the most challenging compounds to work with, as they are hard to make, often hard to purify, and most do not store well, as they oxidize, dimerize,
trimerize, polymerize, and degrade faster than most other chemicals.
PHILOU Zrealone - 20-11-2012 at 14:39
It doesn't work for all organic acids because of the roughness/harshness of the reaction conditions,
but the fusion (dry distillation) of Calcium formate in excess with the Calcium carboxylate of the organic acid will give you the aldehyde...It works
also with Baryum or Strontium.
It is based on the same kind of reaction as:
Ca(O2C-CH3)2 -heat-> CaCO3 + CH3-CO-CH3
Ca(O2C-CH3)2 + Ca(O2C-CH2-CH3)2 -heat-> CaCO3 + CH3-CO-CH3 + CH3-CO-CH2-CH3 + CH3-CH2-CO-CH2-CH3
Ca(O2C-CH3)2 + Ca(O2CH)2 -heat-> CaCO3 + CH3-CO-CH3 + CH3-CH=O + CH2=O
Playing with the proportions allows one to favourise one of the reaction products.
Distillation must be used to separate the products.
In some cases of diacids one can get cyclocetons.
Hexan-1,6-dioic acid leads to cyclopentanone and heptan-1,7-dioic acid to cyclohexanone.
Succunic acid provides 1,4-cyclohexadione.