1. Chapter 24. Amines and Heterocycles
Why this Chapter?
Amines and carbonyl compounds are the most abundant and have rich chemistry
In addition to proteins and nucleic acids, a majority of pharmaceutical agents contain amine functional groups

2. Amines – Organic Nitrogen Compounds
Organic derivatives of ammonia, NH3,
Nitrogen atom with a lone pair of electrons, making amines both basic and nucleophilic
Occur in plants and animals

3. 24.1 Naming Amines
Alkyl-substituted (alkylamines) or aryl-substituted (arylamines)
Classified:
1° (RNH2), methyl (CH3NH2),
2° (R2NH),
3° (R3N)

4. Quaternary Ammonium Ions
A nitrogen atom with four attached groups is positively charged
Compounds are quaternary ammonium salts

5. IUPAC Names – Simple Amines
For simple amines, the suffix -amine is used as the parent name of the alkyl substituent
tert-Butylamine
Cyclohexylamine
Aniline
The suffix -amine can be used in place of the final -e in the parent name
1,4-Butanediamine
4,4-Dimethylcyclohexanamine

6. IUPAC Names – Amines With More Than One Functional Group
Consider the NH2 as an amino substituent on the parent molecule
4-Amino-2-butanone
2-Aminobutanoic acid
2,4-Diaminobenzoic acid

7. IUPAC Names – Multiple Alkyl Groups
Symmetrical secondary and tertiary amines are named by adding the prefix di- or tri- to the alkyl group
Triethylamine
Diphenylamine

8. IUPAC Names – Multiple, Different Alkyl Groups
Named as N-substituted primary amines
Largest alkyl group is the parent name, and other alkyl groups are considered N-substituents
N,N-Dimethylpropylamine
N-Ethyl-N-methylcyclohexylamine

9. Common Names of Heterocyclic Amines
If the nitrogen atom occurs as part of a ring, the compound is designated as being heterocyclic
Each ring system has its own parent name

10. Learning Check:
Name the following:

11. (or Ethyl methyl amine)
Solution:
Name the following:
N-methylethanamine
(or Ethyl methyl amine)
N-Ethyl-N-methylcyclohexylamine
Tricyclohexylamine
1,3-Butanediamine
N-methylpyrrolidine
Diisopropylamine

12. 24.2 Properties of Amines Bonding to N is similar to that in ammonia
N is sp3-hybridized
C–N–C bond angles are close to 109° tetrahedral value

13. Chirality Is Possible (But Not Observed)
An amine with three different substituents on nitrogen is chiral (in principle but not in practice): the lone pair of electrons is the fourth substituent
Most amines that have 3 different substituents on N are not resolved because the molecules interconvert by pyramidal inversion

14. Amines Form H-Bonds
Amines with fewer than five carbons are water-soluble
Primary and secondary amines form hydrogen bonds, increasing their boiling points

15. 24.3 Basicity of Amines
The lone pair of electrons on nitrogen makes amines basic and nucleophilic
They react with acids to form acid–base salts and they react with electrophiles

16. Relative Basicity
Amines are stronger bases than alcohols, ethers, or water
Amines establish an equilibrium with water in which the amine becomes protonated and hydroxide is produced
The most convenient way to measure the basicity of an amine (RNH2) is to look at the acidity of the corresponding ammonium ion (RNH3+)
High pKa → weaker acid and stronger conjugate base.

17. High pKa → weaker acid and stronger conjugate base.
The most convenient way to measure the basicity of an amine (RNH2) is to look at the acidity of the corresponding ammonium ion (RNH3+)
Weak Acid
Strong Base
Most simple alkylammmonium ions have pKa's of 10 to 11
High pKa → weaker acid and stronger conjugate base.
Arylamines and heterocyclic aromatic amines are considerably less basic than alkylamines (conjugate acid pKa 5 or less)
Strong Acid
Weak Base

18. Relative Basicity Stronger base Strong base Weaker Acid (pKa ~60)
Weak Acid (pKa ~40)
Strong base

19. Purification of Amines:
Amines can be separated from a mixture by reaction with acids to form soluble amine salts

20. Basicity of Amides
Amides (RCONH2) in general are not proton acceptors except in very strong acid
The C=O group is strongly electron-withdrawing, making the N a very weak base
Addition of a proton occurs on O but this destroys the double bond character of C=O as a requirement of stabilization by N

21. 24.4 Basicity of Substituted Arylamines
The N lone-pair electrons in arylamines are delocalized by interaction with the aromatic ring  electron system and are less able to accept H+ than are alkylamines
Weak Base
Weak Nucleophile
Electrons tied up in resonance aren’t as available for bonding.

22. Substituted Arylamines
Electron-donating substituents (such as CH3, NH2, OCH3) increase the basicity of the corresponding arylamine
Electron- withdrawing substituents (such as Cl, NO2, CN) decrease arylamine basicity

23. Learning Check: Without looking at a table,
Name the following compounds then
Rank the structures in each set in order of decreasing basicity (#1 = strongest) A. B.

24. Solution: Without looking at a table,
Name the following compounds then
Rank the structures in each set in order of decreasing basicity (#1 = strongest) A. 3 1 2
p-Nitroaniline
p-Aminobenzaldehyde
p-Bromoaniline B. 3 1 2
p-Methylaniline
p-(Fluoromethyl)aniline
p-(Trifluoromethyl)aniline

25. 24.5 Biological Amines
In what form do amines exist at a physiological pH of 7.3?
At pH = we expect the amount of [HA] and [A-] to be the same.
As the solution becomes more acidic (Below a pH of 10.64) we expect methylamine to be protonated. (Higher concentration of HA)
As the solution becomes more basic (pH above 10.64) we expect the acidic proton to be removed. (Higher concentration of A-)

26. Biological Amines The amino acid, alanine, at pH = 7.3 pKa = 9.69
A pH below pKa atom is protonated.
At pH above pKa proton is removed.

27. 24.6 Synthesis of Amines Reductions of Nitriles, Amides, & Nitro’s
Reactions we’ve already seen:

28. Synthesis of Amines
Arylamines are prepared from nitration of an aromatic compound and reduction of the nitro group
Reduction by catalytic hydrogenation over platinum is suitable if no other groups can be reduced
Iron, zinc, tin, and tin(II) chloride are effective in acidic solution

29. SN2 Reactions of Alkyl Halides
Ammonia and other amines are good nucleophiles
Remember: Nucleophilicity parallels Bacisity

30. Uncontrolled Multiple Alkylation
Primary, secondary, and tertiary amines all have similar reactivity, the initially formed monoalkylated substance undergoes further reaction to yield a mixture of products
Need better method to avoid mixture of products

31. Selective Preparation of Primary Amines: the Azide Synthesis
Azide ion, N3 displaces a halide ion from a primary or secondary alkyl halide to give an alkyl azide, RN3
Alkyl azides are not nucleophilic (but they are explosive)
Reduction gives the primary amine

32. Gabriel Synthesis of Primary Amines
A phthalimide alkylation for preparing a primary amine from an alkyl halide
The N-H in imides (CONHCO) can be removed by KOH followed by alkylation and hydrolysis

33. Reductive Amination of Aldehydes & Ketones
Treatment of an aldehyde or ketone with ammonia or an amine in the presence of a reducing agent
Forms an imine that is then reduced

34. Reductive Amination Is Versatile
Ammonia, primary amines, and secondary amines yield primary, secondary, and tertiary amines, respectively

35. Mechanism of Reductive Amination

36. Reducing Step
Sodium cyanoborohydride, NaBH3CN, reduces C=N but not C=O
Stable in water

37. Learning Check:
Prepare using a reductive amination.

38. Solution:
Prepare using a reductive amination.

39. Hofmann and Curtius Rearrangements
Carboxylic acid derivatives can be converted into primary amines with loss of one carbon atom by both the Hofmann rearrangement and the Curtius rearrangement

40. Hofmann Rearrangement
Amide reacts with Br2 and base
Rearranges to lose carbonyl carbon making chain 1 carbon shorter
Gives high yields of arylamines and alkylamines

41. Hofmann example:

42. Curtius Rearrangement
Heating an acyl azide prepared from an acid chloride
Migration of R from C=O to the neighboring nitrogen with simultaneous loss of a leaving group

43. Curtius Example:

44. Learning Check:
Prepare o-methylbenzylamine from a carboxyic acid using a Hofmann rearrangement.
Prepare o-methylbenzylamine from a carboxyic acid using a Curtius rearrangement.

45. Solution:
Prepare o-methylbenzylamine from a carboxyic acid using a Hofmann rearrangement.
Hofmann
Curtius
Prepare o-methylbenzylamine from a carboxyic acid using a Curtius rearrangement.

46. 24.7 Reactions of Amines
We’ve already seen: Alkylation and acylation

47. Hofmann Elimination Converts amines into alkenes
NH2 is very a poor leaving group so it converted to an alkylammonium ion, which is a good leaving group

48. Silver Oxide Is Used for the Elimination Step
Ag2O in water forms Ag(OH)2 which exchanges hydroxide ion for iodide ion in the quaternary ammonium salt, thus providing the base necessary to cause elimination

49. Orientation in Hofmann Elimination
We would expect that the more highly substituted alkene product predominates in the E2 reaction of an alkyl halide (Zaitsev's rule)
However, the less highly substituted alkene predominates in the Hofmann elimination due to the large size of the trialkylamine leaving group
The base must abstract a hydrogen from the most sterically accessible, least hindered position

50. Steric Effects Control the Orientation

51. 24.8 Reactions of Arylamines
Amino substituents are strongly activating, ortho- and para-directing groups in electrophilic aromatic substitution reactions
Reactions are controlled by conversion to amide

52. Arylamines Are Not Useful for Friedel-Crafts Reactions
The amino group forms a Lewis acid–base complex with the AlCl3 catalyst, preventing further reaction
Therefore we use the corresponding amide

53. Diazonium Salts: The Sandmeyer Reaction
Primary arylamines react with HNO2, yielding stable arenediazonium salts

54. Uses of Arenediazonium Salts
The N2 group can be replaced by a nucleophile

55. Preparation of Aryl Halides
Reaction of an arenediazonium salt with CuCl or CuBr gives aryl halides (Sandmeyer Reaction)
Aryl iodides form from reaction with NaI without a copper(I) salt

56. Aryl Nitriles and Carboxylic Acids
An arenediazonium salt and CuCN yield the nitrile, ArCN, which can be hydrolyzed to ArCOOH

57. Formation of Phenols (ArOH)
From reaction of the arenediazonium salt with copper(I) oxide in an aqueous solution of copper(II) nitrate

58. Reduction to a Hydrocarbon
By treatment of a diazonium salt with hypophosphorous acid, H3PO2

59. Mechanism of Diazonium Replacement
Through radical (rather than polar or ionic) pathways

60. Diazonium Coupling Reactions
Arenediazonium salts undergo a coupling reaction with activated aromatic rings, such as phenols and arylamines, to yield brightly colored azo compounds, ArN=NAr

61. How Diazonium Coupling Occurs
The electrophilic diazonium ion reacts with the electron- rich ring of a phenol or arylamine
Usually occurs at the para position but goes ortho if para is blocked

62. Azo Dyes
Azo-coupled products have extended  conjugation that lead to low energy electronic transitions that occur in visible light (dyes)

63. 24.9 Heterocycles
A heterocycle is a cyclic compound that contains atoms of two or more elements in its ring, usually C along with N, O, or S

64. Pyrole and Imidazole Pyrole is an amine and a conjugated diene,
however its chemical properties are not consistent with either of structural features

65. Chemistry of Pyrole
Electrophilic substitution reactions occur at C2 b/c it is position next to the N
A more stable intermediate cation having 3 resonance forms
At C3, only 2 resonance forms

66. Polycyclic Heterocycles

67. 24.10 Spectroscopy of Amines -Infrared
Characteristic N–H stretching absorptions 3300 to cm1
Amine absorption bands are sharper and less intense than hydroxyl bands
Protonated amines show an ammonium band in the range 2200 to 3000 cm1

68. Examples of Infrared Spectra

69. Nuclear Magnetic Resonance Spectroscopy
N–H hydrogens appear as broad signals without clear-cut coupling to neighboring C–H hydrogens
In D2O exchange of N–D for N–H occurs, and the N–H signal disappears

70. Chemical Shift Effects
Hydrogens on C next to N and absorb at lower field than alkane hydrogens
N-CH3 gives a sharp three-H singlet at  2.2 to  2.6

71. 13C NMR
Carbons next to amine N are slightly deshielded - about 20 ppm downfield from where they would absorb in an alkane

72. Mass Spectrometry
A compound with an odd number of nitrogen atoms has an odd-numbered molecular weight and a corresponding parent ion
Alkylamines cleave at the C–C bond nearest the nitrogen to yield an alkyl radical and a nitrogen-containing cation

73. Mass Spectrum of N-Ethylpropylamine
The two modes of a cleavage give fragment ions at m/z = 58 and m/z = 72.

74. Select the best classification for the following molecule:
1˚ aliphatic amine
2˚ aliphatic amine
3˚ aliphatic amine
aromatic amine
heterocyclic aromatic amine

75. Which of the following is not represented in the molecule shown?
aliphatic amine
aromatic amine
heterocyclic aromatic amine
secondary amine
a pyridine ring

76. Determine the IUPAC name for the following molecule:
ethylpropylphenylamine
N-phenyl-N-ethylpropanamine
N-ethyl-N-phenylpropanamine
N-ethyl-N-propylaniline
N-phenyl-N-ethylaniline

77. If the pKb of an amine is 9.54, what is the pKa of its conjugate acid?
6.46
4.46
-9.54
not enough information to determine
pKa is unrelated to pKb.

78. The major alkaloid present in tobacco leaves is nicotine, whose structure is shown below. Which will be the major ammonium ion formed when nicotine is treated with one equivalent of a strong acid? 2. 1. 1 2

79. Arrange the above from strongest to weakest base:
A, B, C
B, C, A
C, A, B
A, C, B
B, A, C

80. Amine A is more basic than amine B.
True
False

81. The equilibrium lies to the right in the above reaction:
True
False

82. Which of the following statements is true regarding the following two molecules?
Both A and B are aromatic.
Both A and B are aliphatic amines.
A is more basic than B.
B is more basic than A.
Both A and B are planar molecules.

83. Select the most acidic compound from the choices provided.1. 2. 3. 4. 5. 1 2 3 4 5

84. If a protonated amine with a pKa of 10 is placed in a solution of pH 12, the predominant form of the amine in solution will be the protonated form.
True
False

85. The following transformation can be accomplished as shown.
True
False

86. Identify the major product for the following reaction.1. 2. 3. 5. 4. 1 2 3 4 5

87. Identify the structure of A based on the products that result from a Hofmann elimination.+ 1. 2. 3. 4. 5. 1 2 3 4 5

88. b, c, d, a a, d, b, c a, b, d, c c, b, d, a c, d, b, a
Select the order of chemical shift (most downfield → most upfield) for the carbon atoms in the following molecule:
b, c, d, a
a, d, b, c
a, b, d, c
c, b, d, a
c, d, b, a

89. What is the intermediate that results from the treatment of aniline with HNO2 and H2SO4?
a diazonium salt
an aryl radical
a quaternary ammonium salt
an isocyanate
an aryl azide