Organic Chemistry

1
Amines
Chapter 23
2
Structure Classification

• Amines are classified as
• 1, 2, or , 3 amines Amines in which there are
  1, 2, or 3 alkyl or aryl groups.

3
Structure Classification

• Amines are further divided into aliphatic,
  aromatic, and heterocyclic amines
• Aliphatic amine An amine in which nitrogen is
  bonded only to alkyl groups.
• Aromatic amine An amine in which nitrogen is
  bonded to one or more aryl groups.

4
Structure Classification

• Heterocyclic amine An amine in which nitrogen is
  one of the atoms of a ring.

5
Structure Classification

• Example Classify each amino group by type.

6
Nomenclature

• Aliphatic amines replace the suffix -e of the
  parent alkane by -amine.

7
Nomenclature

• The IUPAC system retains the name aniline.

8
Nomenclature

• Among the various functional groups, -NH2 is one
  of the lowest in order of precedence.

Amine vs alcohol
Amine vs acid
9
Nomenclature

• Common names for most aliphatic amines are
  derived by listing the alkyl groups bonded to
  nitrogen in one word ending with the suffix
  -amine.

10
Nomenclature

• When four groups are bonded to nitrogen, the
  compound is named as a salt of the corresponding
  amine.

11
Chirality of Amines

• Consider the unshared pair of electrons on
  nitrogen as a fourth group, then the arrangement
  of groups around N is approximately tetrahedral.
• An amine with three different groups bonded to N
  is chiral and exists as a pair of enantiomers
  and, in principle, can be resolved.

12
Chirality of Amines

• In practice, however, they cannot be resolved
  because they undergo inversion, which converts
  one enantiomer to the other.

13
Chirality of Amines

• Pyramidal inversion is not possible with
  quaternary ammonium ions, and their salts can be
  resolved.

14
Physical Properties

• Amines are polar compounds, and both 1 and 2
  amines form intermolecular hydrogen bonds.
• N-H- - -N hydrogen bonds are weaker than O-H- -
  -O hydrogen bonds because the difference in
  electronegativity between N and H (3.0 - 2.1
  0.9) is less than that between O and H (3.5 -
  2.1 1.4).

Using bp as an indication of H bonding
Increasing strength
15
Basicity

• All amines are weak bases, and aqueous solutions
  of amines are basic.
• It is common to discuss their basicity by
  reference to the acid ionization constant of the
  conjugate acid.

16
Basicity

• Using values of pKa, we can compare the acidities
  of amine conjugate acids with other acids.

17
Basicity-Aliphatic Amines

• Aliphatic Amines
• note that pKa pKb 14

Stronger bases
18
Basicity-Aromatic Amines
Weaker bases
Intermediate
19
Basicity-Aromatic Amines

• Aromatic amines are considerably weaker bases
  than aliphatic amines.

20
Basicity-Aromatic Amines

• Aromatic amines are weaker bases than aliphatic
  amines because of two factors
• Resonance stabilization of the free base, which
  is lost on protonation.

21
Basicity-Aromatic Amines

• The greater electron-withdrawing inductive effect
  of the sp2-hybridized carbon of an aromatic amine
  compared with that of the sp3-hybridized carbon
  of an aliphatic amine.
• And note the effect of substituents
• Electron-releasing groups, such as alkyl groups,
  increase the basicity of aromatic amines.
• Electron-withdrawing groups, such as halogens,
  the nitro group, and a carbonyl group decrease
  the basicity of aromatic amines by a combination
  of resonance and inductive effects.

22
Example Basicity-Aromatic Amines

• 3-nitroaniline is a stronger base than
  4-Nitroaniline.

Cannot do this kind of resonance in 3 nitroaniline
23
Basicity-Aromatic Amines

• Heterocyclic aromatic amines are weaker bases
  than heterocyclic aliphatic amines.

24
Basicity-Aromatic Amines

• In pyridine, the unshared pair of electrons on N
  is not part of the aromatic sextet.
• Pyridine is a weaker base than heterocyclic
  aliphatic amines because the free electron pair
  on N lies in an sp2 hybrid orbital (33 s
  character) and is held more tightly to the
  nucleus than the free electron pair on N in an
  sp3 hybrid orbital (25 s character).

25
Basicity-Aromatic Amines

• Imidazole Which N lone pair is protonated? The
  one which is not part of the aromatic system.

26
Basicity-Guanidine

• Guanidine is the strongest base among neutral
  organic compounds.
• Its basicity is due to the delocalization of the
  positive charge over the three nitrogen atoms.

27
Reaction with Acids

• All amines, whether soluble or insoluble in
  water, react quantitatively with strong acids to
  form water-soluble salts.

28
Reaction with acids

• Separation and purification of an amine and a
  neutral compound.

29
Preparation

• We have already covered these methods
• nucleophilic ring opening of epoxides by ammonia
  and amines.
• addition of nitrogen nucleophiles to aldehydes
  and ketones to form imines
• reduction of imines to amines
• reduction of amides to amines by LiAlH4
• reduction of nitriles to a 1 amine
• nitration of arenes followed by reduction of the
  NO2 group to a 1 amine

30
Preparation

• Alkylation of ammonia and amines by SN2
  substitution.
• Unfortunately, such alkylations give mixtures of
  products through a series of proton transfer and
  nucleophilic substitution reactions.

polyalkylations
31
Preparation via Azides

• Alkylation of azide ion.

Overall Alkyl Halide ? Alkyl amine
32
Example Preparation via Azides

• Alkylation of azide ion.

Note retention of configuration, trans ? trans
33
Reaction with HNO2

• Nitrous acid, a weak acid, is most commonly
  prepared by treating NaNO2 with aqueous H2SO4 or
  HCl.
• In its reactions with amines, nitrous acid
• Participates in proton-transfer reactions.
• A source of the nitrosyl cation, NO, a weak
  electrophile.

34
Reaction with HNO2

• NO is formed in the following way.
• Step 1 Protonation of HONO.
• Step 2 Loss of H2O.
• We study the reactions of HNO2 with 1, 2, and
  3 aliphatic and aromatic amines.

35
Tertiary Amines with HNO2

• 3 Aliphatic amines, whether water-soluble or
  water-insoluble, are protonated to form
  water-soluble salts.
• 3 Aromatic amines NO is a weak electrophile
  and participates in Electrophilic Aromatic
  Substitution.

36
Secondary Amines with HNO2

• 2 Aliphatic and aromatic amines react with NO
  to give N-nitrosamines.

carcinogens
Mechanism
37
RNH2 with HNO2

• 1 aliphatic amines give a mixture of
  unrearranged and rearranged substitution and
  elimination products, all of which are produced
  by way of a diazonium ion and its loss of N2 to
  give a carbocation.
• Diazonium ion An RN2 or ArN2 ion

38
1 RNH2 with HNO2

• Formation of a diazonium ion.
• Step 1 Reaction of a 1 amine with the nitrosyl
  cation.
• Step 2 Protonation followed by loss of water.

39
1 RNH2 with HNO2 (Aliphatic)

• Aliphatic diazonium ions are unstable and lose N2
  to give a carbocation which may
• 1. Lose a proton to give an alkene.
• 2. React with a nucleophile to give a
  substitution product.
• 3. Rearrange and then react by Steps 1 and/or 2.

40
1 RNH2 with HNO2

• Tiffeneau-Demjanov reaction Treatment of a
  ?-aminoalcohol with HNO2 gives a ketone and N2.

41
Mechanism of Tiffeneau-Demjanov

• Reaction with NO gives a diazonium ion.
• Concerted loss of N2 and rearrangement followed
  by proton transfer gives the ketone.

Similar to pinacol rearrangement
42
Pinacol Rearrangement an example of
stabilization of a carbocation by an adjacent
lone pair.
Overall
43
Mechanism
Reversible protonation.
Elimination of water to yield tertiary
carbocation.
This is a protonated ketone!
1,2 rearrangement to yield resonance stabilized
cation.
Deprotonation.
44
1 Primary Amines with HNO2 (Aromatic)

• The -N2 group of an arenediazonium salt can be
  replaced in a regioselective manner by these
  groups.

45
1 ArNH2 with HNO2

• A 1 aromatic amine converted to a phenol.

46
1 ArNH2 with HNO2

• Problem What reagents and experimental
  conditions will bring about this conversion?

47
1 ArNH2 with HNO2

• Problem Show how to bring about each conversion.

48
Hofmann Elimination

• Hofmann elimination Thermal decomposition of a
  quaternary ammonium hydroxide to give an alkene.
• Step 1 Formation of a 4 ammonium hydroxide.

49
Hofmann Elimination

• Step 2 Thermal decomposition of the 4 ammonium
  hydroxide.

50
Hofmann Elimination

• Hofmann elimination is regioselective - the major
  product is the least substituted alkene.
• Hofmanns rule Any ?-elimination that occurs
  preferentially to give the least substituted
  alkene as the major product is said to follow
  Hofmanns rule.

51
Hofmann Elimination

• The regioselectivity of Hofmann elimination is
  determined largely by steric factors, namely the
  bulk of the -NR3 group.
• Hydroxide ion preferentially approaches and
  removes the least hindered hydrogen and, thus,
  gives the least substituted alkene.
• Bulky bases such as (CH3)3CO-K give largely
  Hofmann elimination with haloalkanes.

52
Cope Elimination

• Cope elimination Thermal decomposition of an
  amine oxide.
• Step 1 Oxidation of a 3 amine gives an amine
  oxide.
• Step 2 If the amine oxide has at least one
  ?-hydrogen, it undergoes thermal decomposition to
  give an alkene.

53
Cope Elimination

• Cope elimination shows syn stereoselectivity but
  little or no regioselectivity.
• Mechanism a cyclic flow of electrons in a
  six-membered transition state.