The Chemistry of Carboxylic Acids and their Derivatives
Brown's Chemistry Advanced Level Pre-University Chemistry Revision Study
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chemistry of carboxylic acids and derivatives
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Selective reduction of
carboxylic acids and use of the products in organic synthesis routes
and a unique
carboxylic acid oxidation!
6.5.1 Reduction of a carboxylic acid to a
primary aliphatic alcohol
tetrahydridoborate(III) NaBH4, (sodium borohydride), is not a powerful enough reducing agent to reduce
tetrahydridoaluminate(III), LiAlH4, (lithium aluminium tetrahydride), is a
powerful reducing agent than NaBH4, and in ether solvent, readily
reduces carboxylic acids to primary alcohols.
The LiAlH4 effectively
releases a hydride ion,
powerful nucleophile - electron pair donor, which can attacks the
δ+ carbon of polarised carbonyl bond
(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
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
or benzoic acid to
C6H5COOH + 4[H] ==> C6H5CH2OH
+ 4[H] ===>
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
e.g. propenoic acid H2C=CH-COOH
which has two functional groups -
alkene and carboxylic acid
+ [H] == NaBH4 ==> No reaction,
4[H] == LiAlH4 ==> H2C=CH-CH2OH
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
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 double bond.
+ H2 == Ni ==> H3C-CH2-COOH
Only the alkene group is
reduced with a nickel catalyst, giving propanoic acid with only one
This means you
can selectively reduce either functional groups or you need two
reductions to form propan-1-ol.
now has the chemistry of primary alcohols
A unique oxidation
Carboxylic acids are usually quite stable
Think of the end product of oxidising
alcohols and aldehydes with acidified potassium
dichromate(VI) - carboxylic acids.
However, in terms of lower members of the
simple aliphatic carboxylic acids, there is one clear
Apart from a carboxylic acid with an aldehyde group as or in
a side chain, the first in this homologous series of aliphatic carboxylic
acids, methanoic acid (HCOOH) is the only carboxylic acid that is
the only one easily
oxidised, acting as a reducing agent, 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) However, it does not give a yellow-orange precipitate with
2,4-dinitrophenylhydrazine - carboxylic acids themselves do not usually undergo
nucleophilic addition - elimination reactions like aldehydes do.
In tests (i) and (ii) methanoic acid is oxidised to
carbon dioxide and water.
HCOOH + [O] ===> CO2 + H2O
So methanoic acid acts as a
reducing agent and readily oxidised to carbon
dioxide and water by these two reagents.
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 !!!
So methanoic acid
behaves like an aldehyde, the left of this particular
carboxylic acid molecule has a
grouping, identical to an aldehyde group.
Methanoic acid esters have the same
'end' grouping e.g. ethyl methanoate,
so I assume they can also give a positive 'aldehyde'
Ethanoic acid CH3COOH, cannot behave in this
way, the left H is replaced by an alkyl group, so there
is no equivalent of an aldehyde group present in the
You can't 'rearrange' any of the other carboxylic
acids to be both aldehyde and carboxylic acid.
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