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Part 6.
The Chemistry of Carboxylic Acids and their Derivatives
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Brown's Chemistry Advanced Level Pre-University Chemistry Revision Study
Notes for UK KS5 A/AS GCE IB advanced level organic chemistry students US
K12 grade 11 grade 12 organic chemistry preparation of carboxylic acids
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Part 6.3
Synthetic routes and methods of preparing aliphatic and aromatic carboxylic acids
Sub-index for this page
(some overlap with
section 6.6)
6.3.1
Oxidation of primary
alcohols (aliphatic/aromatic acids)
6.3.2
Oxidation of aldehydes
(to aliphatic/aromatic acids)
6.3.3
Oxidation of aromatic
hydrocarbons to make aromatic acids
6.3.4
Hydrolysis of nitriles
(aliphatic/aromatic acids)
6.3.5
Hydrolysis of esters
6.3.6
The iodoform reaction
6.3.7
The hydrolysis of
amides
6.3.8
Examples of synthetic
routes to substituted carboxylic acids
6.3.1 Oxidation of primary alcohols (aliphatic or aromatic
acids)
When oxidised they form aldehydes and further oxidation gives
the relatively stable carboxylic acid e.g.
-
===>
-
===>
-
 ===>
-
 ===>
 Oxidation of primary alcohols to carboxylic acids
The technique illustrated on the right
(diagram PD5) is called heating under
reflux, a method which enables a reaction to be carried out at a higher
temperature than room temperature to speed up the reaction AND retain the
solvent (reaction medium e.g. water) and any volatile reactant or product
(e.g. an alcohol/aldehyde/ketone).
As the mixture boils, the vapours of the
solvent or volatile reactant/product are condensed back into the flask in
the vertical condenser, so any volatile reactant is used up and no volatile
product lost (at least at this stage in a preparation!).
Refluxing the alcohol with excess of the potassium dichromate(VI)/sulfuric acid
mixture.
The oxidation is ...
primary alcohol
==> carboxylic acid
2Cr2O72-(aq)
+ 3RCH2OH(aq) + 16H+(aq) ===>
3RCOOH(aq) + 4Cr3+(aq) + 11H2O(l)
oxidation half-reaction:
RCH2OH(aq) + H2O(l)
==>
RCOOH(aq) + 4H+(aq) + 4e-(aq) (R = alkyl or aryl)
reduction half reaction:
Cr2O72–(aq) + 14H+(aq)
+ 6e– ===> 2Cr3+(aq) + 7H2O(l)
The orange dichromate(VI) ion is
reduced to the green chromium(III) ion.
Examples using
simplified symbol equations:
ethanol ==> ethanoic acid
CH3CH2OH
+ 2[O] ===> CH3COOH + H2O
propan-1-ol
(1-propanol, n-propyl alcohol, n-propanol) ==> propanoic
acid
CH3CH2CH2OH
+ 2[O] ===> CH3CH2COOH + H2O
Aromatic primary alcohols (NOT phenols)
can be similarly oxidised e.g. phenylmethanol to benzoic acid.
+ 2[O] ===>
+ H2O
 Oxidation
of primary alcohols with alkaline potassium manganate(VII)
Heating a
primary alcohol with a aqueous sodium hydroxide and potassium manganate(VII)
mixture under reflux (diagram PD2) will
give the sodium salt of the carboxylic acid and it is not possible to
isolate the intermediate aldehyde.
However, the acid/dichromate(VI) method
under reflux is better, because the carboxylic acid is less liable to further
degradative oxidation. The complex reaction can be summarised as:
RCH2OH(aq) +
NaOH(aq) + 2[O] ==>
RCOO-Na+(aq) + 2H2O(l)
(R = alkyl or aryl)
After removing the
excess KMnO4/MnO2 the weak acid is freed from its
sodium salt by adding strong dilute hydrochloric acid.
RCOO-(aq) + H+(aq)
==> RCOOH(aq/l/s)
TOP OF PAGE
and sub-index
6.3.2 Oxidation of aldehydes (aliphatic or aromatic acids)
You can use exactly the same method
described in 6.3.1 using a mixture potassium dichromate(VI) and moderately
concentrated sulfuric acid.
aldehyde ==>
carboxylic acid
Cr2O72-(aq)
+ 3RCHO(aq) + 8H+(aq) ===> 3RCOOH(aq)
+ 2Cr3+(aq) + 4H2O(l) (R =
alkyl or aryl)
oxidation half-reaction:
RCHO(aq) + H2O(l) ===> RCOOH(aq)
+ 2H+(aq) + 2e-(aq)
reduction half reaction:
Cr2O72–(aq) + 14H+(aq)
+ 6e– ===> 2Cr3+(aq) + 7H2O(l)
The orange dichromate(VI)
ion is reduced to the green chromium(III) ion.
Examples using
simplified symbol equations:
ethanal
==> ethanoic acid:
CH3CHO + [O] ===>
CH3COOH
propanal
(propionaldehyde) ==> propanoic acid (propionic
acid)
CH3CH2CHO
+ [O] ===> CH3CH2COOH
butanal
(butyraldehyde) ==> butanoic acid (butyric acid)
CH3CH2CH2CHO
+ [O] ===> CH3CH2CH2COOH
TOP OF PAGE
and sub-index
6.3.3 Oxidation of aromatic hydrocarbons to make aromatic
acids
Acidified potassium dichromate(VI) might
oxidise alkyl benzene compounds to benzoic acid, but if it does, it will be
much slower than with the alkaline manganate(VII) method
described below - NaOH/KMnO4 is a much stronger oxidising agent.
Overall change is
represented by the equations for the conversion:
===>
C6H5CH3 + 3[O]
===> C6H5COOH
+ H2O
Aromatic
are not easily oxidised and longish reflux times are necessary (illustrated,
fig. PD2 right).
Hydrocarbons are
difficult to oxidise with typical organic oxidising agents compared to
compounds like alcohols.
However, aromatic hydrocarbons with an alkyl side
chain can be oxidised with strong reagents such as aqueous potassium
manganate(VII)/sodium hydroxide.
Whatever the length of
the alkyl group on a benzene ring it gets whittled down to carbon of
the carboxylic acid group e.g. propylbenzene ends up as benzoic acid, the
same product obtained by oxidising the shorter methylbenzene.
The
more stable aromatic benzene ring is left intact.
The overall process for
producing benzoic acid from methylbenzene can be summarised ..
C6H5CH3
(l) + NaOH(aq) + 3[O]
===> C6H5COO-Na+
(aq)
+ 2H2O(l)
After removing the
excess KMnO4/MnO2 with a reducing agent, the weak
acid, benzoic acid is freed from
its sodium salt by adding dilute hydrochloric acid.
C6H5COO-(aq) + H+(aq)
===>
C6H5COOH
and you can oxidise the three
isomeric dimethyl benzene to make the three isomeric aromatic
dicarboxylic acids:
(i)
benzene-1,2-dicarboxylic acid from 1,2-dimethylbenzene
 + 6[O] ===>
 + 2H2O
(i)
benzene-1,3-dicarboxylic acid from 1,3-dimethylbenzene
 + 6[O] ===>
 + 2H2O
(i)
benzene-1,4-dicarboxylic acid from 1,2-dimethylbenzene
 + 6[O] ===>
+ 2H2O
After recrystallisation of the
products e.g. using aqueous ethanol, you can determine the melting
point and confirm the identity from data tables of melting
points of carboxylic acids.
A sharp melting point
indicates a reasonable pure sample of the prepared acid.
The product should also give
effervescence of carbon dioxide (limewater test) when added to
sodium hydrogen carbonate solution - a simple test for a
carboxylic acid.
RCOOH + NaHCO3
===> RCOO-Na+ + H2O +
CO2
TOP OF PAGE
and sub-index
 6.3.4 Hydrolysis of nitriles
to give aliphatic or aromatic carboxylic acids
The triple bonded nitrile group C≡N
is hydrolysed to the carboxylic acid group COOH.
The nitrogen ends up as ammonia or the
ammonium ion. If the nitrile is refluxed with dilute
hydrochloric/sulfuric acid (strong acids) or sodium hydroxide (strong base - alkali) the
corresponding carboxylic acid or its sodium salt is formed.
The hydrolysis with pure water is to
slow, but the reaction is speeded up by a strong acid or strong alkali
and reflux conditions. Strictly speaking all the reactants and products
should be suffixed by (aq)
(a) Nitriles
synthesised from halogenoalkane:
RX + KCN ===>
RCN + KX
Nucleophilic substitution reaction where X = Cl, Br or I and R =
H, alkyl or aryl.
The
nucleophilic substitution reaction between halogenoalkanes (haloalkanes) and potassium cyanide
(i) Equations for the dilute mineral acid hydrolysis of a nitrile to give the
free (weaker) acid
In this case converting
propanenitrile to propanoic acid or its salt, sodium propanoate
+ 2H2O + H+
+ NH4+
Here the free acid and an ammonium
ion are formed. (more detailed structured formula hydrolysis equation)
+ 2H2O + H+
+ NH4+
(less detailed structured formula hydrolysis equation)
+ 2H2O + H+
+ NH4+
(skeletal formula hydrolysis equation)
(ii) Equations for the alkaline hydrolysis of a nitrile to give the sodium salt
(if aqueous sodium hydroxide is used), in the equations you write out the
product as the carboxylate anion.
In this case converting
propanenitrile to its salt, e.g. sodium propanoate
+ H2O + OH-
+ NH3
Here the carboxylate anion
(propanoate ion) and free ammonia are formed.
(structured formula hydrolysis equation)
+ H2O + OH-
+ NH3
(abbreviated structured formula hydrolysis equation)
+ H2O + OH-
+ NH3
(skeletal formula hydrolysis equation)
The acid, here propanoic acid, is set
free by the addition of a strong mineral acid e.g. dilute sulfuric acid.
CH3CH2COO-(aq) + H+(aq)
===> CH3CH2COOH
 (iii)
The hydrolysis of
2-methylpropanenitrile
2-methylpropanenitrile
=== hydrolysis ===> free 2-methylpropanoic acid or its salt
(CH3)2CHC≡N
+ 2H2O + H+
(CH3)2CHCOOH + NH4+
(acid hydrolysis, free acid)
(CH3)2CHC≡N
+ H2O + OH-
(CH3)2CHCOO- + NH3
(alkaline hydrolysis, salt of acid)
(b) Hydroxynitriles prepared from aldehydes and
ketones by addition of hydrogen cyanide
R2C=O
+ HCN ===> R2C(OH)CN (R = H, alkyl or aryl)
Aldehydes
and ketones and nucleophilic addition of hydrogen cyanide
The general equation for the hydrolysis
of a hydroxy nitrile to a hydroxy-carboxylic acid.
R2C(OH)CN
+ 2H2O
===> R2C(OH)COOH
+ NH3
The nitrile is converted to a carboxylic acid by hydrolysis.
The reaction is slow, so it is speeded up by heating the
nitrile with the hydrolysis reagent (e.g. dilute sulfuric acid or aqueous
sodium hydroxide) under reflux.
(a) Hydrolysis
equation with acid - good yield under reflux conditions
(i)
+ H+ + 2H2O ===>
 + NH4+
2-hydroxypropanenitrile +
hydrogen ion ===> 2-hydroxypropanoic acid + ammonium
ion
Note that this produces the free acid.
(ii)
+ H+ + 2H2O ===>
 + NH4+
2-hydroxy-2-methylpropanenitrile + hydrogen ion ===> 2-hydroxy-2-methylpropanoic acid + ammonium
ion
Note that this produces the free acid.
(iii)
 + H+
+ H2O ===> +
NH4+
2-hydroxy-2-methylbutanenitrile
+ water ===> 2-hydroxy-2-methylbutanoic acid +
ammonia
Note that this produces the free acid.
(b) Hydrolysis
with alkali - good yield under reflux conditions
(i)
+ OH- + H2O ===>
 + NH3
2-hydroxypropanenitrile +
hydroxide in ===> 2-hydroxypropanoate ion + ammonia
Note that this produces the salt of the acid and the ammonia
would be boiled off under reflux conditions.
e.g. from sodium hydroxide you would get sodium
2-hydroxypropanoate.
You add dilute mineral acid (sulfuric or hydrochloric) to
free the acid - strong acid displacing a weak one.
(ii)
+ OH- + H2O ===>
 + NH3
2-hydroxy-2-methylpropanenitrile + hydroxide in ===> 2-hydroxy-2-methylpropanoate
ion + ammonia
Note that this produces the salt of the acid and the ammonia
would be boiled off under reflux conditions.
e.g. from sodium hydroxide you would get sodium
2-hydroxy-2-methylpropanoate.
You add dilute mineral acid (sulfuric or hydrochloric) to
free the acid - strong acid displacing a weak one.
(iii)
+
OH- + H2O ===>
 + NH3
2-hydroxy-2-methylbutanenitrile
+ water ===> 2-hydroxy-2-methylbutanoate ion +
ammonia
Note that this produces the salt of the acid and the ammonia
would be boiled off under reflux conditions.
e.g. from potassium hydroxide you would get potassium
2-hydroxy-2-methylbutanoate
You add dilute mineral acid (sulfuric or hydrochloric) to
free the acid - strong acid displacing a weak one.
The acid, here 2-hydroxy-2-methylbutanoic acid, is set free by
the addition of a strong mineral acid e.g. dilute sulfuric acid.
CH3CH2C(CH3)(OH)COO-(aq) + H+(aq)
===> CH3CH2C(CH3)(OH)COOH
Note on R/S isomers (optical stereoisomers)
If the nitrile is derived from aldehydes from ethanal (CH3CHO)
onwards and all unsymmetrical ketones (e.g. butanone CH3COCH2CH3),
R/S isomers are formed on equal probability basis - a racemic mixture
(racemate). In other words, if the resulting nitriles from
these carbonyl compounds are hydrolysed, you will also get a racemic (50 :
50)mixture of the mirror image forms of the hydroxynitrile and hence a
racemic mixture of the acid obtained by the hydrolysis of the nitrile.
TOP OF PAGE
and sub-index
 6.3.5 Hydrolysis of esters
If an ester is warmed with
a dilute aqueous strong acid or alkali (much faster than pure water), it changes back into the
original carboxylic acid and alcohol from which the ester was made.
This reaction is called hydrolysis and this
particular hydrolysis is also called saponification if alkali is used
(i) Hydrolysis
with acid (dilute sulfuric or hydrochloric acids)
ethyl ethanoate + water
===> ethanoic acid + ethanol
+ H2O ===>
+
(ii) Hydrolysis
with alkali (dilute sodium hydroxide)
CH3COOCH2CH3
+ OH- ===> CH3COO-Na+
+ CH3CH2OH
With alkali, you have to add a strong dilute mineral acid
(hydrochloric or sulfuric) to free the weak organic carboxylic acid.
CH3COO-Na+
+ H+ ===> CH3COOH
(Note that esterification is the
opposite of the hydrolysis: ethanoic acid + ethanol
==>
ethyl ethanoate + water)
Hydrolysis of esters and soap production
One type of soap is made by hydrolysing triglyceride
esters from natural plant oils or fats using sodium/potassium hydroxide.
For example hydrolysing a saturated fat molecule - a
triglyceride ester e.g. using sodium hydroxide
 forming
sodium palmitate
Another hydrolysis of a fat molecule might yield
the salt sodium stearate,
sodium stearate
On adding a strong dilute mineral acid like sulfuric
or hydrochloric acid, the free long chain fatty acid can be extracted,
stearic acid
The soap is the sodium or potassium salt of a long
chain saturated or unsaturated fatty acid.
Soaps are made by hydrolysing esters from naturally occurring plant oils
and animal fat
TOP OF PAGE
and sub-index
6.3.6
The iodoform reaction
This reaction is given by all secondary alcohols with the 2-ol structure -R-CHOH-CH3
and all ketones with the 2-one
structure R-CO-CH3 ('methyl ketones') to
yield carboxylic acids. R
= alkyl or aryl
To make the reagent, sodium hydroxide solution
is added to a solution of iodine in potassium iodide solution until most
of the colour has gone.
You can also use a mixture of
potassium iodide dissolved in sodium chlorate(I) solution.
The organic compound is warmed with this
solution.
It is a complicated reaction, in which
the hydrogen atoms of the methyl group of the CH3CO grouping are
replaced by iodine atoms. The C=O group becomes part of the carboxylic acid
group.
The sodium hydroxide/iodine reagent
causes the C-C bond of the CH3CO grouping to break in the process
of forming the pale yellow
precipitate of CHI3 (triiodomethane, 'iodoform'),
that smells like an antiseptic.
The following equilibrium is found in the
NaOH/I2 reagent:
I2 + 2OH-
IO- + I- H2O
The iodate(I) ion, IO-,
substitutes I for H forming the I3CCO grouping.
The electron withdrawing effect of the
three iodine atoms weakens the C-C sigma bond allowing for the formation of
CHI3.
For R = alkyl or aryl e.g. CH3,
CH2CH3, C6H5 etc., the overall equations
are:
The simple 'molecular' general equation:
CH3COR + 3I2
+ 4NaOH ===> CHI3 + RCOONa +
3NaI + 3H2O
The general ionic equation with state symbols:
CH3COR(aq) + 3I2(aq)
+ 4OH-(aq) ===> CHI3(s) + RCOO-(aq) +
3I-(aq) + 3H2O(l)
The weak carboxylic acid must be freed by adding a strong
dilute mineral acid e.g. hydrochloric or sulfuric/
RCOO-(aq) +
H+(aq) ===>
RCOOH(aq/s)
Examples of conversions
(i) butan-2-ol OR butanone will yield
propanoic acid
CH3CH2CH(OH)CH3
OR
CH3CH2COCH3
===> CH3CH2COOH
For the full equations R would be CH3CH2
(ii) 1-phenylethanol would give benzoic acid
===>
For the full equations R would be C6H5
This would not normally be considered a useful synthetic
route to prepare carboxylic acids.
TOP OF PAGE
and sub-index
6.3.7 The hydrolysis of amides
Amides are another derivative of carboxylic acids that
can be hydrolysed back to the parent carboxylic acid.
If amides are refluxed with strong acid
(hydrochloric or sulfuric) or alkali potassium/sodium hydroxide, the product
is the original carboxylic acid or its salt.
(a) Acid
Hydrolysis of amides
e.g. the formation of the free weaker carboxylic acid
RCONH2(aq) + HCl(aq) + H2O(l)
===> RCOOH(aq) + NH4Cl(aq)
and the more correct ionic
equation
RCONH2(aq)
+ H+(aq) + H2O(l)
===> RCOOH(aq) + NH4+(aq)
e.g. propanamide hydrolysed to propanoic acid
(aq)
+ H+(aq) + H2O(l)
===>
(aq) + NH4+(aq)
(b) Alkaline hydrolysis of amides
e.g. the formation of the salt of the
parent carboxylic acid.
RCONH2(aq)
+ NaOH(aq) ===> RCOONa(aq)
+ NH3(aq/g)
and the more correct ionic equation
RCONH2(aq) + OH-(aq) ===> RCOO-(aq)
+ NH3(aq/l)
e.g. hydrolysis of benzamide to benzoic acid
(aq) + OH-(aq)
===>
 (aq)
+ NH3(aq/l)
The carboxylic acid is freed
by adding a stronger mineral acid
RCOO-(aq)
+ H+(aq) ===> RCOO-(aq)
e.g.
(aq)
+ H+(aq) ===>
(aq)
TOP OF PAGE
and sub-index
6.3.8 Examples of synthetic
routes to substituted carboxylic acids
Just a few brief notes to give some ideas of how synthetic
sequences can be put together from reactions you are likely to have, or
will, study.
(a) Chlorination - chlorinated carboxylic acids
You can chlorinate carboxylic acids in the same way as
for alkanes e.g. chlorine plus uv light.
e.g. chlorine + ethanoic acid
===> chloroethanoic acid + hydrogen chloride
CH3COOH + Cl2
===> ClCH2COOH + HCl
The chlorination may continue to give
dichloroethanoic acid and trichloroethanoic acid.
Cl2CHCOOH and
Cl3COOH
It is a free radical chain substitution
reaction mechanism.
For propanoic acid, there are two possible isomers,
2-chloropropanoic acid or 3-chloropropanoic acid.
CH3CH2COOH + Cl2
===> {CH3CHClCOOH or ClCH2CH2COOH}
+ HCl
(b) Hydroxy carboxylic acid formation
You can hydrolyse the chlorinated acids by refluxing
with sodium hydroxide.
e.g. chloroethanoic acid + sodium hydroxide
===> hydroxyethanoic acid + sodium chloride
ClCH2COOH + NaOH ===>
HOCH2COOH + NaCl
CH3COOH + OH-
===> HOCH2COOH + Cl-
This will a nucleophilic substitution
reaction mechanism.
With excess alkali, the carboxylic acid is formed as
the sodium salt (not shown) and freed by adding a stronger mineral
acid (HCl or H2SO4)
Similarly for the chloropropanoic acids you
2-hydroxypropanoic acid and 3-hydroxypropanoic acid, though you just get
one product from each isomer (no or involved!).
CH3CHClCOOH + OH-
===> CH3CH(OH)COOH + Cl-
ClCH2CH2COOH + OH-
===> HOCH2CH2COOH + Cl-
(b) Amine formation - amino acid synthesis
You can react the chlorinated carboxylic acid with
ammonia under high pressure and temperature to insert the amine group
substituent forming a synthetic route to amino acids.
e.g. chloroethanoic acid + ammonia
===> aminoethanoic acid
ClCH2COOH + 2NH3
===> H2NCH2COOH + NH4Cl
This will also be nucleophilic substitution
reaction mechanism.
Similarly for the chloropropanoic acids you
2-aminopropanoic acid and 3-aminopropanoic acid you just get one product
from each isomer.
CH3CHClCOOH + 2NH3
===> CH3CH(NH2)COOH + NH4Cl
ClCH2CH2COOH + 2NH3
===> H2NCH2CH2COOH + NH4Cl
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