Transcript Chapter 20: Carboxylic Acids and Nitriles

Chapter 20:
Carboxylic Acids and Nitriles
The Importance of Carboxylic Acids
(RCO2H)

Abundant in nature from oxidation of aldehydes
and alcohols in metabolism
◦ Acetic acid, CH3CO2H, - vinegar
◦ Butanoic acid, CH3CH2CH2CO2H (rancid butter)
◦ Long-chain aliphatic acids from the breakdown of fats

Carboxylic acids present in many industrial
processes and most biological processes

An understanding of their properties and reactions
is fundamental to understanding organic chemistry
2
The Importance of Carboxylic Acids
(RCO2H)

They are the starting materials from which
other acyl derivatives are made
20.1 Naming Carboxylic Acids & Nitriles
Carboxylic Acids, RCO2H
 If derived from open-chain alkanes, replace the terminal -e
of the alkane name with -oic acid
 The carboxyl carbon atom is C #1

Propanoic acid
4-Methylpentanoic acid
3-Ethyl-6-methyloctanedioic acid
4
Alternative Names


Compounds with CO2H bonded to a ring are named
using the suffix -carboxylic acid
The CO2H carbon is not itself numbered in this system
trans-4-Hydroxycyclohexanecarboxylic acid
1-cyclopentenecarboxylic acid
5

Common Names:
Names
accepted
by IUPAC
Carboxylic acids some of the 1st compound
studied so some of their common names
are still used. Especially for those with
biological interest.
Common Names:
Nitriles, RCN


Closely related to carboxylic acids named by adding -nitrile
as a suffix to the alkane name, with the
nitrile carbon numbered C #1
4-Methylpentanenitrile
8
Nitriles, RCN

Complex nitriles are named as derivatives of carboxylic
acids.
◦ Replace -ic acid or -oic acid ending with -onitrile
Acetonitrile
(From acetic acid)
Benzonitrile
(From benzoic acid)
2,2-Dimethylcyclohexanecarbonitrile
(From 2,2-dimethylcyclohexanecarboxylic acid)
9
Learning Check:

Give IUPAC names for the following:
Solution:

Give IUPAC names for the following:
3-Methylbutanoic
acid
4-Bromopentanoic
acid
2-ethylpentanoic
acid
2,4-Dimethylpentanenitrile
cis-4-hexenoic acid
cis-1,3cyclopentanedicarboxylic
acid)
20.2 Structure & Properties
Carboxyl carbon sp2 hybridized:
planar with bond angles of approximately 120°
12
20.2 Structure & Properties:

Carboxylic acids form hydrogen bonds,
 exist as cyclic dimers held together by 2 hydrogen bonds

Strong hydrogen bonding causes much higher boiling points
than the corresponding alcohols
1-Propanal
(MW = 58)
O
Boiling
Point
CH3CH2C H
49oC
1-Propanol
(MW = 60)
CH3CH2CH2-OH
97oC
Ethanoic Acid (Acetic acid)
MW = 60
O
CH3C OH
118oC
Dissociation of Carboxylic Acids

Carboxylic acids are proton donors toward weak and strong
bases, producing metal carboxylate salts, RCO2 +M

Carboxylic acids with more than six carbons are only
slightly soluble in water, but their conjugate base salts are
water-soluble
14
Acidity Constant and pKa

Carboxylic acids transfer a proton to water to give H3O+ and
carboxylate anions, RCO2, but H3O+ is a much stronger acid
pKa = ~5

pKa = 15.7
(acts as a base here)
pKa = -1.7
Lower pKa means stronger
acid so equilibrium favors
starting materials
The acidity constant, Ka,, is about 10-5 for a typical
carboxylic acid (pKa ~ 5)
15
Substituent Effects on Acidity
Electron withdrawing (electronegative) substituents (like F, Cl, Br, I) promote
formation of the carboxylate ion so raise the Ka (lower the pKa)
The acidity
constant, Ka,, is
about 10-5 for a
typical carboxylic
acid (pKa ~ 5)
Ka,, ~10-15 to -16 for
alcohols and
water (pKa ~ 15
to 16)
CH3OH
H2O
15.5
15.7
CH3CH2OH
15.9
16
Resonance Effects on Acidity
Phenols
Alcohols
Carboxylic acid
Inorganic acids
17
Resonance Effects on Acidity
Alcohols weaker acids since O- not stabilized by resonance
C=O Bond length
(120 pm)
Negative charge spread over O-C-O
C-O Bond length
(134 pm)
Carboxylic acids more acidic since O- stabilized by resonance
C-O Bond
lengths same
(127 pm)
20.3 Biological Acids and the
Henderson-Hasselbalch Equation

If pKa of given acid and the pH of the medium are known,
% of dissociated and undissociated forms can be
calculated using the Henderson-Hasselbalch eqn
19
Learning Check:

Calculate the % of acid form (undissociated = HA) and
carboxylate ion form (dissociated = A-) present in a 0.0020
M propanoic acid solution at pH = 5.30. (pKa = 4.87)
Solution:

Calculate the % of acid form (undissociated = HA) and
carboxylate ion form (dissociated = A-) present in a 0.0020
M propanoic acid solution at pH = 5.30. (pKa = 4.87)
HA
O
CH3CH2C OH
A-
+ H2O
pH - pKa = log [A-]
[HA]
5.30 - 4.87 = 0.43 = log [A-]
[HA]
100.43 = 2.69 = [A-]
[HA]
1
O
CH3CH2C O
+
H3O
2.69
x 100 = 73 % [A-]
2.69 + 1
and
1
x 100 = 27 % [HA]
2.69 + 1
20.4 Substituent Effects on Acidity
Electron withdrawing (electronegative) substituents (like F, Cl, Br, I) promote
formation of the carboxylate ion so raise the Ka (lower the pKa)
They stabilize the carboxylate anion by induction
22
Aromatic Substituent Effects
electron-donating
electron-withdrawing
(activating) group (like OCH3)
( activating) groups (like -NO2)
decreases acidity by
destabilizing the
carboxylate anion
increase acidity by
stabilizing the
carboxylate anion
23
Aromatic Substituent Effects
24
Learning Check:
Rank the following in order of increasing acidity:
(#1 is least acidic) (Don’t look at a table of pKa data to help you.)
A.
O
O
C
C
OH
O
H3C
B.
O
C
C
OH
C
HO
O2N
O
OH
C
OH
Cl
O
OH
O
CH3
C
OH
OH
Solution:
Rank the following in order of increasing acidity:
(#1 is least acidic) (Don’t look at a table of pKa data to help you.)
A.
O
O
C
C
OH
O
H3C
3
C
OH
O
C
HO
4
O
C
OH
1
O
OH
CH3
C
OH
O2N
2
3
C
OH
Cl
2
B.
O
1
OH
20.5 Preparation of Carboxylic Acids:
From Oxidation of Benzylic Carbons

Oxidation of a substituted alkylbenzene with KMnO4 or
Na2Cr2O7 gives a substituted benzoic acid

1° and 2° alkyl groups can be oxidized, but 3° are not
1o
CH2CH3
H3C
H3C
3o
O
C OH
KMnO4
H3C
C
CH3
CH CH3
CH3 2o
H3C
3o
C
CH3
C OH
O
27
From Oxidative Cleavage of Alkenes

Oxidative cleavage of an alkene with KMnO4 gives a
carboxylic acid if the alkene has at least one vinylic hydrogen
CH3CH2
H
C C
CH3
CH3
CH3CH2
H3O+
H
KMnO4
H
H3O+
C C
H
KMnO4
CH3CH2
OH
C O
+
O C
CH3
CH3
CH3CH2
OH
C O
+
O C
HO
OH
CO2
28
From Oxidation of 1o Alcohols & Aldehydes

Oxidation of a 1o alcohols or aldehydes with CrO3 in
aqueous acid
29
Hydrolysis of Nitriles

Hot acid or base yields carboxylic acids
O
+
CH3
CH2
CH2
C N
H3O
or
1) NaOH
2) H3O+
CH3
CH2
CH2
C OH
30
Halides  Nitriles  Carboxylic Acids


Conversion of an alkyl halide to a nitrile (by an SN2 with
cyanide ion) followed by hydrolysis produces a carboxylic
acid with one more carbon (RBr  RCN  RCO2H)
Best with 1o halides
because competing
elimination reactions
occur with 2o or 3o
alkyl halides
31
Carboxylation of Grignard Reagents

Grignard reagents react w/ dry CO2 to yield a metal carboxylate
◦ Limited to alkyl halides that can form Grignard reagents
The organomagnesium
halide adds to C=O of
carbon dioxide
Protonation by addition of aqueous
HCl in a separate step gives the free
carboxylic acid
32
Halides  Nitriles  Carboxylic Acids
Halides  Grignard  Carboxylic Acids
20.6 Reactions of Carboxylic Acids:
An Overview
Like ketones,
Carboxylic acids
transfer a
proton to a base
to give anions,
which are good
nucleophiles in
SN2 reactions
carboxylic acids
undergo
addition of
nucleophiles to
the carbonyl
group
NaOH
1) LiAlH4
2) H3O+
Carboxylic acids undergo substitutions reactions characteristic of
neither alcohols nor ketones
34
Learning Check:
Prepare 2-phenylethanol from benzyl bromide.
OH
CH2 Br
CH2
CH2
Solution:
Prepare 2-phenylethanol from benzyl bromide.
CH2 Br
1) NaCN
2) H3O+, heat
OH
CH2
CH2
3) LiAlH4
4) H3O+
1) LiAlH4
2) H3O+
Na C N
O
CH2
C N
+
H3O
heat
CH2
C OH
20.7 Chemistry of Nitriles


Nitriles and carboxylic acids both have a carbon atom
with three bonds to an electronegative atom, and contain
a  bond
C’s of nitriles and carboxylic acids are electrophilic
d-
d-
d+
d+
d+
d-
37
Preparation of Nitriles:
Dehydration of Amides


Reaction of primary amides RCONH2 with SOCl2 or
POCl3 (or other dehydrating agents)
Not limited by steric hindrance or side reactions (as is the
reaction of alkyl halides with NaCN)
38
Mechanism:
Dehydration of Amides

Nucleophilic amide oxygen atom attacks SOCl2 followed
by deprotonation and elimination
Nucleophilic amide
oxygen atom
attacks SOCl2
Deprotonation of
acidic H on N
Elimination of the SO2
and Cl leaving groups
39
Reactions of Nitriles


RCN is strongly polarized and with an electrophilic carbon
Attacked by nucleophiles to yield sp2-hybridized imine anions
40
Hydrolysis: Nitriles  Carboxylic Acids

Hydrolyzed in with acid or base catalysis to a carboxylic
acid and ammonia or an amine
41
Mechanism:
Hydrolysis of Nitriles

Nucleophilic addition of
hydroxide to CN bond

Protonation gives a
hydroxy imine, which
tautomerizes to an
amide

A second hydroxide adds
to the amide carbonyl
group and loss of a
proton gives a dianion

Expulsion of NH2 gives
the carboxylate
42
Mechanism:
Step 4: Hydrolysis of Amides
Reduction: Nitriles  1o Amines
◦Reduction of a nitrile with LiAlH4 gives a primary amine
Nucleophilic addition
of hydride ion (H-) to
the polar CN bond,
yields an imine anion
The C=N bond undergoes a second
nucleophilic addition of hydride (H-) to give
a dianion, which is protonated by water
44
Reaction of Nitriles with Organometallic
Reagents

Grignard reagents add to give an intermediate imine anion
that is hydrolyzed by addition of water to yield a ketone
45
Reactions of Nitriles: Overview
Learning Check:
Prepare 2-methyl-3-pentanone from a nitrile.
Solution:
Prepare 2-methyl-3-pentanone from a nitrile.
A couple of possibilities:
20.8 Spectroscopy of Carboxylic Acids
and Nitriles
Infrared Spectroscopy


O–H bond of the carboxyl group gives a very broad
absorption 2500 to 3300 cm1
C=O bond absorbs sharply between 1710 and 1760 cm1
Free carboxyl
groups absorb at
1760 cm1
Commonly encountered
dimeric carboxyl groups
absorb in a broad band
centered around 1710 cm1
49
Infrared Spectroscopy
O–H bond of the
carboxyl group gives
a very broad
absorption 2500 to
3300 cm1
Commonly encountered
dimeric carboxyl groups
absorb in a broad band
centered around 1710 cm1
IR of Nitriles


Nitriles show an intense CN bond absorption near 2250
cm1 for saturated compounds and 2230 cm1 for
aromatic and conjugated molecules
This is highly diagnostic for nitriles
51
C-13 NMR



Carboxyl 13COOH signals are at d165 to d185
Aromatic and ,b-unsaturated acids are near d165 and
saturated aliphatic acids are near d185
13C  N signal d115 to d130
52
Proton NMR


The acidic CO2H proton is a singlet near d 12
When D2O is added to the sample the CO2H proton is
replaced by D causing the absorption to disappear from
the NMR spectrum
53
Which of the following is not an acyl derivative?
20%
1.
20%
20%
20%
3
4
20%
2.
O
O
O
Cl
3.
H
N
4.
O
O
O
O
5.
O
O
1
2
5
Learning Check:
Name the following:
O
A.
O
4.
benzoic acid
1-cyclohexenoic acid
2-cyclohexenoic acid
1-cyclohexenecarboxylic acid
5.
2-cyclohexenecarboxylic acid
1.
2.
3.
B.
OH
1.
2.
3.
4.
5.
Br
OH
3-bromo-4-methylbenzoic acid
4-methyl-3-bromobenzoic acid
phthalic acid
4-carboxy-2-bromotoluene
5-carboxy-2-methyl-1bromobenzene
Solution:
Name the following:
O
A.
O
4.
benzoic acid
1-cyclohexenoic acid
2-cyclohexenoic acid
1-cyclohexenecarboxylic acid
5.
2-cyclohexenecarboxylic acid
1.
2.
3.
B.
OH
1.
2.
3.
4.
5.
Br
OH
3-bromo-4-methylbenzoic acid
4-methyl-3-bromobenzoic acid
phthalic acid
4-carboxy-2-bromotoluene
5-carboxy-2-methyl-1bromobenzene
Learning Check:
Name the following:
A.
B.
CN
OH
O
1.
2.
3.
4.
5.
2-propylhexanoic acid
4-propylhexanoic acid
2-butylpentanoic acid
4-carboxyoctane
5-carboxyoctane
1.
2.
3.
4.
5.
pentylpropylnitrile
4-cyanooctane
5-cyanooctane
2-propylhexanenitrile
2-butylpentanenitrile
Solution:
Name the following:
A.
B.
CN
OH
O
1.
2.
3.
4.
5.
2-propylhexanoic acid
4-propylhexanoic acid
2-butylpentanoic acid
4-carboxyoctane
5-carboxyoctane
1.
2.
3.
4.
5.
pentylpropylnitrile
4-cyanooctane
5-cyanooctane
2-propylhexanenitrile
2-butylpentanenitrile
Learning Check:
The structure for oleic acid is shown here. What is
the IUPAC name of this compound?
O
OH
1.
2.
3.
4.
5.
cis-octadecenoic acid
(Z)-octadec-9-enoic acid
(Z)-octadecyl-9-en-1-oic acid
(Z)-octadecen-9-oic acid
(Z)-9-octadecylenoic acid
Solution:
The structure for oleic acid is shown here. What is
the IUPAC name of this compound?
O
OH
1.
2.
3.
4.
5.
cis-octadecenoic acid
(Z)-octadec-9-enoic acid
(Z)-octadecyl-9-en-1-oic acid
(Z)-octadecen-9-oic acid
(Z)-9-octadecylenoic acid
Learning Check:
Why do carboxylic acids boil at higher temperatures than
most molecules of the same molecular weight?
1.
2.
3.
4.
5.
They have more oxygen atoms than most other
molecules.
They can participate in hydrogen bonding.
They are polar molecules.
They have more oxygen atoms than most other
molecules; and they can participate in hydrogen
bonding.
They can participate in hydrogen bonding; and they are
polar molecules.
Solution:
Why do carboxylic acids boil at higher temperatures than
most molecules of the same molecular weight?
1.
2.
3.
4.
5.
They have more oxygen atoms than most other
molecules.
They can participate in hydrogen bonding.
They are polar molecules.
They have more oxygen atoms than most other
molecules; and they can participate in hydrogen
bonding.
They can participate in hydrogen bonding; and they are
polar molecules.
Learning Check:
Rank the following molecules in acidity from least
acidic to most acidic.
Cl
O
O
OH
Cl
A
OH
B
1.
2.
3.
4.
5.
A, C, B
B, A, C
C, B, A
B, C, A
C, A, B
O
OH
Cl
C
Solution:
Rank the following molecules in acidity from least
acidic to most acidic.
Cl
O
O
OH
Cl
A
OH
B
1.
2.
3.
4.
5.
A, C, B
B, A, C
C, B, A
B, C, A
C, A, B
O
OH
Cl
C
Learning Check:
Rank the following molecules in acidity from most acidic
to least acidic.
OH
OH
O
OCH3
HO
A
B
1.
2.
3.
4.
5.
C
A, C, B, D
C, B, D, A
B, C, A, D
A, B, C, D
C, B, A, D
D
Solution:
Rank the following molecules in acidity from most acidic
to least acidic.
OH
OH
O
OCH3
HO
A
B
1.
2.
3.
4.
5.
C
A, C, B, D
C, B, D, A
B, C, A, D
A, B, C, D
C, B, A, D
D
Learning Check:
Chloroacetic acid is a stronger acid than acetic acid.
Which is the best explanation?
H3C COOH
H3C COO
H2C COOH
Cl
H2C COO
Cl
acetic acid
acetate
chloroacetic acid
chloroacetate
1.
More resonance structures can be drawn for chloroacetic acid than for acetic acid.
2.
More resonance structures can be drawn for chloroacetate ion than for acetate
ion.
3.
Because of its high electronegativity, chlorine is able to donate electrons to the
chloroacetate ion by the inductive effect, thereby stabilizing this ion.
4.
Because of its high electronegativity, chlorine is able to withdraw electrons from
the chloroacetate ion by the inductive effect, thereby stabilizing this ion.
5.
Chlorine is larger than hydrogen and can better hold a negative charge.
Solution:
Chloroacetic acid is a stronger acid than acetic acid.
Which is the best explanation?
H3C COOH
H3C COO
H2C COOH
Cl
H2C COO
Cl
acetic acid
acetate
chloroacetic acid
chloroacetate
1.
More resonance structures can be drawn for chloroacetic acid than for acetic acid.
2.
More resonance structures can be drawn for chloroacetate ion than for acetate
ion.
3.
Because of its high electronegativity, chlorine is able to donate electrons to the
chloroacetate ion by the inductive effect, thereby stabilizing this ion.
4.
Because of its high electronegativity, chlorine is able to withdraw electrons from
the chloroacetate ion by the inductive effect, thereby stabilizing this ion.
5.
Chlorine is larger than hydrogen and can better hold a negative charge.
Learning Check:
Which is the stronger acid?
O
O
OH
OH
Cl
p-Chlorobenzoic acid
p-methylbenzoic acid
Solution:
Which is the stronger acid?
O
O
OH
OH
Cl
p-Chlorobenzoic acid
p-methylbenzoic acid
Learning Check:
What sequence of reagents will accomplish the following?
O
OH
1.
2.
H3O+
CO2
Mg
HBr
Et2O
Br2
Mg
h
Et2O
CO2
H3O+
3.
Br2
CO2
H3O+
h
4.
HBr
5.
CO2
H3O+
Mg
Et2O
CO2
H3O+
Solution:
What sequence of reagents will accomplish the following?
O
OH
1.
2.
H3O+
CO2
Mg
HBr
Et2O
Br2
Mg
h
Et2O
CO2
H3O+
3.
Br2
CO2
H3O+
h
4.
HBr
5.
CO2
H3O+
Mg
Et2O
CO2
H3O+
Learning Check:
Which of the following cannot be performed with a nitrile?
1.
2.
O
CN
O
CN
OH
NH2
3.
CN
NH2
4.
5.
CN
Br
CN
O
Solution:
Which of the following cannot be performed with a nitrile?
1.
2.
O
CN
O
CN
OH
NH2
3.
CN
NH2
4.
5.
CN
Br
CN
O
Learning Check:
Determine the product of the following sequence:
Br
1.
NaCN
H3O+
DMSO
heat
3.
2.
O
OH
NH2
NH2
O
O
4.
5.
O
OH
OH
CN
Solution:
Determine the product of the following sequence:
Br
1.
NaCN
H3O+
DMSO
heat
3.
2.
O
OH
NH2
NH2
O
O
4.
5.
O
OH
OH
CN
Learning Check:
Determine the product of the following:
O
CH3MgBr
SOCl2
NH2
1.
H3O+
Et2O
2.
O
O
3.
5.
NH2
HO
CN
4.
NHCH3
Solution:
Determine the product of the following:
O
CH3MgBr
SOCl2
NH2
1.
H3O+
Et2O
2.
O
O
3.
5.
NH2
HO
CN
4.
NHCH3