1. Organic
Chemistry
William H. Brown &
Christopher S. Foote

2. Amines
Chapter 22

3. Structure & Classification
Amines are classified as
1°, 2°, or , 3° amines: amines in which 1, 2, or 3 hydrogens of NH3 are replaced by alkyl or aryl groups

4. 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

5. Structure & Classification
heterocyclic amine: an amine in which nitrogen is one of the atoms of a ring

6. Structure & Classification
Example: classify each amino group by type

7. Structure & Classification
Aliphatic amines: replace the suffix -e of the parent alkane by -amine

8. Nomenclature
The IUPAC system retains the name aniline

9. Nomenclature
Among the various functional groups discussed in the text, -NH2 is one of the lowest in order of precedence

10. 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

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

12. Chirality of Amines
if we 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 3 different groups bonded to N is chiral and exists as a pair of enantiomers and, in principle, can be resolved

13. Chirality of Amines
in practice, however, they cannot be resolved because they undergo pyramidal inversion, which converts one enantiomer to the other

14. Chirality of Amines
pyramidal inversion is not possible with quaternary ammonium ions, and their salts can be resolved

15. 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 ( =0.9) is less than that between O and H ( = 1.4)

16. NMR Spectroscopy 1H-NMR 13C-NMR
the chemical shift of amine hydrogens is variable, and may be found in the region  0.5 to 5.0 depending on the solvent, concentration, and temperature
they generally appear as singlets
13C-NMR
carbons bonded to nitrogen are generally shifted approximately 20 ppm downfield relative to their signal in an alkane of comparable structure

17. IR Spectroscopy
1° and 2° amines show N-H stretching absorption in the region cm-1
1° amines show two bands in this region
2° amines show only one band in this region

18. Basicity
All amines are weak bases, and aqueous solutions of amines are basic

19. Basicity
it is more common to discuss the basicity of amines by reference to the acid ionization constant of the corresponding conjugate acid
for any acid-conjugate base pair

20. Basicity
using values of pKa, we can compare the acidities of amine conjugate acids with other acids

21. Basicity-Aliphatic Amines

22. Basicity-Aromatic Amines

23. Basicity-Aromatic Amines
aromatic amines are considerably weaker bases than aliphatic amines

24. 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

25. Basicity-Aromatic Amines
resonance delocalization of the electron pair on nitrogen by interaction with the pi system of the aromatic ring decreases basicity

26. Basicity-Aromatic Amines
the greater electron-withdrawing inductive effect of the sp2 carbon of an aromatic amine compared with the sp3 carbon of an aliphatic amine also decreases basicity
Electron-releasing, 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

27. Basicity-Aromatic Amines
4-nitroaniline is a weaker base than 3-nitroaniline

28. Basicity-Aromatic Amines
Heterocyclic aromatic amines are weaker bases than heterocyclic aliphatic amines

29. 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)

30. Basicity-Aromatic Amines
Imidazole

31. 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

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

33. Preparation We have already covered these methods
nucleophilic ring opening of an epoxide by ammonia and amines (11.9B)
addition of ammonia and 1° and 2° amines to aldehydes and ketones to give an imine followed by reduction of the imine to an amine (16.10)
reduction of an amide by LiAlH4 (18.11B)
reduction of a nitrile to a 1° amine (18.11C)
Hofmann rearrangement of a 1° amide (18.12)
nitration of an arene followed by reduction of the NO2 group to a 1° amine (21.1B)

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

35. Preparation via Azides
Alkylation of azide ion

36. Preparation via Azides
alkylation of azide ion

37. Reaction with HNO2
Nitrous acid is a weak acid, most commonly prepared by treating aqueous NaNO2 aqueous H2SO4 or HCl
In its reactions with amines, it
participates in proton-transfer reactions
is a source of the nitrosyl cation, NO+, a weak electrophile

38. Reaction with HNO2 NO+ is formed in the following way
we study the reactions of HNO2 with 1°, 2°, and 3° aliphatic and aromatic amines

39. 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, as such, participates in EAS
2° aliphatic and aromatic amines react with NO+ to give N-nitrosoamines

40. 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

41. 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

42. 1° RNH2 with HNO2
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 1 and/or 2

43. 1° RNH2 with HNO2
Tiffeneau-Demjanov reaction: treatment of a -aminoalcohol with HNO2 gives a ketone and N2

44. 1° RNH2 with HNO2 reaction with NO+ gives a diazonium ion
concerted loss of N2 and rearrangement followed by proton transfer gives the ketone

45. 1° ArNH2 with HNO2
The -N2+ group of an arenediazonium salt can be replaced in a regioselective manner by these groups

46. 1° ArNH2 with HNO2
A 1° aromatic amine can be converted to a phenol

47. 1° ArNH2 with HNO2
Problem: what reagents and experimental conditions will bring about this conversion?

48. 1° ArNH2 with HNO2
Problem: Show how to bring about each conversion

49. Hofmann Elimination
Hofmann elimination: thermal decomposition of a quaternary ammonium hydroxide to give an alkene
Step 1: formation of a 4° ammonium hydroxide

50. Hofmann Elimination
Step 2: thermal decomposition of the 4° ammonium hydroxide

51. Hofmann Elimination
Hofmann elimination is regioselective - the major product is the least substituted alkene
Hofmann’s rule: any -elimination that occurs preferentially to give the least substituted alkene as the major product is said to follow Hofmann’s rule

52. 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 alkyl halides

53. 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

54. Cope Elimination
Cope elimination shows syn stereoselectivity but little or no regioselectivity
mechanism: a cyclic flow of electrons in a six-membered transition state

55. Prob 22.24
From each pair, select the stronger base.

56. Prob 22.25
Calculate the ratio of morpholine to its conjugate acid at pH 7.0. At what pH are the concentrations of morpholine and its conjugate acid equal?

57. Prob 22.26
Which of the nitrogens of pyridoxamine is the stronger base? Explain.

58. Prob 22.27
Which of the nitrogens in epibatidine is the stronger base? How many stereocenters are present in epibatidine?

59. Prob 22.29
Complete each acid-base reaction and predict whether equilibrium lies to the left or right.

60. Prob (cont’d)
Complete each acid-base reaction and predict whether equilibrium lies to the left or right.

61. Prob (cont’d)
Complete each acid-base reaction and predict whether equilibrium lies to the left or right.

62. Prob 22.30
Provide an explanation for fact that quinuclidine is a considerably stronger base than triethylamine.

63. Prob 22.31
Devise a chemical procedure to separate a mixture of these three compounds and recover each in pure form.

64. Prob 22.33
Show how to convert each compound to benzylamine.

65. Prob 22.34
Propose a structural formula for acetylcholine chloride.

66. Prob 22.35
Show how the hydroxylated compound can be transformed into an alkyldiazonium ion, an active carcinogen, in the presence of an acid catalyst.

67. Prob 22.36
Analyze the mechanism of each rearrangement, and list their similarities and differences.

68. Prob 22.37
Propose a mechanism for this conversion. Account for the stereospecificity of the reaction.

69. Prob 22.38
Propose structural formulas for compound A and B.

70. Prob 22.39
Propose a structural formula for C10H16.

71. Prob 22.41
Propose a mechanism for pyrolysis of this ester.

72. Prob 22.43
Show how propranil can be synthesized from benzene and propanoic acid.

73. Prob 22.44
Show to bring about each step in this synthesis.

74. Prob 22.45
Show how to bring about this synthesis.

75. Prob 22.46
Show how to bring about the conversion of phenol to 4-methoxybenzylamine.

76. Prob 22.47
Show how to prepare methylparaben from toluene.

77. Prob 22.48
Show how to synthesize this 3° amine from benzene.

78. Prob 22.49
Show how to prepare N-methylmorpholine from the named starting materials.

79. Prob 22.50
Given this retrosynthetic analysis, propose a synthesis for tridemorph, a systemic agricultural fungicide.

80. Prob 22.51
Propose a mechanism for this example of a Ritter reaction. What would be the product of a Ritter reaction using acetonitrile?

81. Prob 22.52
Show how each of these diamines can be prepared from acetonitrile.

82. Prob 22.53
Given this retrosynthetic analysis, propose a synthesis for the intravenous anesthetic propofol.

83. Prob 22.54
Given this retrosynthetic analysis, propose a synthesis for propoxyphene.

84. Amines
End Chapter 22