Sodium
tetrahydridoborate(III) NaBH4, (sodium borohydride), is not a powerful enough reducing agent to reduce
carboxylic acids.
Lithium
tetrahydridoaluminate(III), LiAlH4, (lithium aluminium tetrahydride), is a
much more
powerful reducing agent than NaBH4, and in ether solvent, readily
reduces carboxylic acids to primary alcohols.
The LiAlH4 effectively
releases a hydride ion,
:H-, a
powerful nucleophile - electron pair donor, which can attacks the
δ+ carbon of polarised carbonyl bond
δ+C=Oδ-.
(electronegativity of O > C).
The reaction must be carried with a
dry solvent such as ethoxyethane ('ether') because LiAlH4 reacts rapidly with water
(and ethanol too).
The reaction is complex and goes
through various stages and can be summarised
as:
RCOOH + 4[H]
===> RCH2OH + H2O (R = H, alkyl or aryl)
The initial product must then be
hydrolysed with water to release the primary alcohol.
e.g. propanoic acid to
propan–1–ol: CH3CH2COOH + 4[H] ==> CH3CH2CH2OH
+ H2O
or benzoic acid to
phenylmethanol:
C6H5COOH + 4[H] ==> C6H5CH2OH
+ H2O
+ 4[H] ===>
+ H2O
So, LiAlH4 (not
NaBH4) readily reduces the
carbonyl group (>C=O) in carboxylic acids and derivatives to the
primary alcohol functional group.
As far as I know,
metal/acid reducing agents like Zn(s)/HCl(aq) is not powerful enough to reduce carboxylic acids.
H2(g)/Ni(s) will NOT reduce
carboxylic acids, but there are other specialised catalysts that can effect
this reduction using hydrogen gas in the chemical industry.
RCOOH + 2H2
===> RCH2OH + H2O (R = H, alkyl or aryl)
A simple example
to illustrate what is, and is not, reduced in terms of the functional
groups.
e.g. propenoic acid H2C=CH-COOH
('acrylic acid'),
which has two functional groups -
alkene and carboxylic acid
(i) H2C=CH-COOH
+ [H] == NaBH4 ==> No reaction
(ii)
H2C=CH-COOH +
4[H] == LiAlH4 ==> H2C=CH-CH2OH
+ H2O
The alkene is NOT
reduced, but the carbonyl group is, giving prop-2-en-1-ol.
via the negative hydride ion nucleophile attacking the polarised
carbonyl bond.
Nucleophilic attack:
-H:
==> δ+C=Oδ-
This product is an unsaturated primary alcohol,
still with two functional groups.
The >C=C< alkene group is NOT reduced by
LiAlH4 because the attacking nucleophile is
essentially a negative hydride ion (:H-)
which would be repelled by the high electron density of the pi electron cloud of the
non-polar C=C bond.
(iii)
H2C=CH-COOH
+ H2 == Ni ==> H3C-CH2-COOH
Only the alkene group is
reduced, giving propanoic acid with only one
functional group.
This means you
can selectively reduce either functional groups or you need two
reductions to form propan-1-ol.
The molecule
now has the chemistry of primary alcohols
6.5.2
A unique oxidation
Apart from a carboxylic acid with an aldehyde group as or in
a side chain, methanoic acid is the only carboxylic acid that is so easily
oxidised it gives the following results with the following reagents:
(i) It gives a silver mirror with ammoniacal silver
nitrate (Tollen's reagent).
(ii) It gives a red-brown copper(I) oxide (Cu2O)
with blue Fehling's solution.
(iii) It does not give a yellow-orange precipitate with
2,4-dinitrophenylhydrazine - carboxylic acids do not usually undergo
nucleophilic addition - elimination reactions.
Explanation
In tests (i) and (ii) methanoic acid is oxidised to
carbon dioxide and water.
HCOOH + [O] ===> CO2 + H2O
When you write the formula of methanoic acid, HCOOH, the
'usual' abbreviated structural formula, the explanation isn't as obvious
until you write it another way ...
i.e. HOCHO is now a formula of a
hydroxy-aldehyde, hence its ease of oxidation, giving positive results
for two the simple tests for aldehydes !!!
You can't 'rearrange' any of the other carboxylic
acids to be both aldehyde and carboxylic acid.
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INDEX of all carboxylic acids
and derivatives notes
All Advanced Organic
Chemistry Notes
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