1. © E.V. Blackburn, 2011 The chemistry of organic compounds

2. © E.V. Blackburn, 2011 What is “organic chemistry” and why is it so important? Organic chemistry is the study of the compounds of carbon. Think about how organic compounds affect our daily life: Our clothes – natural and synthetic fibers Our medicines Our food – carbohydrates, proteins, triglycerides Oils, perfumes, paints, plastics, detergents, etc.

3. © E.V. Blackburn, 2011 What is “organic chemistry” and why is it so important? alizarin: the first naturally occurring dye to be synthesized (1868) indigo: used to dye blue jeans ethanol: a fermentation product

4. © E.V. Blackburn, 2011 “Vital force” theory Friedrich Wöhler (1828): A slight problem! The ammonium cyanate was synthesized from bones.......but the Kolbe synthesis of acetic acid in 1845 put the theory to rest!

5. © E.V. Blackburn, 2011 Vitamin B 12 The synthesis of vitamin B 12 was finally completed in 1972 by Woodward and Eschenmoser after 10 years of work and the assistance of roughly one hundred graduate students.

6. © E.V. Blackburn, 2011 “A production of amino acids under possible primitive earth conditions” S.L. Miller, Science, 117, 528 (1953) amino acids including glycine and alanine

7. © E.V. Blackburn, 2011 Proteins E. coli contains ~5,000 different chemical compounds of which 3,000 are proteins. Man contains ~2,000,000 different proteins. Biologists believe that there are in excess 10,000,000 of proteins which take part in the process of life! http://www.wisegeek.com/how-many-proteins- exist.htm In order to give you an idea of the vastness of organic chemistry, we will look at proteins, our final topic of study in CHEM 263!

8. © E.V. Blackburn, 2011 Angiotensin II Angiotensin II is a blood pressure regulating hormone. It contains 8 amino acid residues. Its structure is actually: Asp-Arg-Val-Tyr-Ile-His-Pro-Phe. It is possible to arrange these in 40,320 different ways only one of which corresponds to the hormone!

9. © E.V. Blackburn, 2011 Organic chemistry A logical subject based on fundamental principles: Can we predict the reactivity of a group of atoms based on what we’ve learned in school chemistry courses? The answer is YES! You know a great deal about acetone so let us look at it.

10. © E.V. Blackburn, 2011 Acetone Here is acetone! But…where are the carbons and hydrogens?

11. © E.V. Blackburn, 2011 Acetone Here is acetone! But…where are the carbons?

12. © E.V. Blackburn, 2011 Acetone But…where are the hydrogens?

13. © E.V. Blackburn, 2011

14. The chemistry of acetone

15. © E.V. Blackburn, 2011 Acetone - a base! Remember the curved-arrow convention. The tail points to the electron source and the head to the electron destination.

16. © E.V. Blackburn, 2011 Acetone - a Lewis acid! 

17. © E.V. Blackburn, 2011 Models of chemical bonding These models are based on the premise that atoms react to produce the electronic configuration of a noble gas.

18. © E.V. Blackburn, 2011 Models of chemical bonding: the ionic bond Consider the Li - F bond. Li1s 2 2s 1 - gives He configuration on loss of e - F1s 2 2s 2 2p 5 - gives Ne configuration on gain of e - “Ions” are the structural units of ionic compounds and have strong electrostatic forces holding them together.

19. © E.V. Blackburn, 2011 When atoms of similar electronegativities are bonded, complete electron transfer cannot take place. The noble gas configurations are attained by sharing electrons: Models of chemical bonding: the covalent bond Electronegativity - the ability of an atom to attract electrons. In the covalent bond, the atoms share electrons. The structural unit is the “molecule.”

20. © E.V. Blackburn, 2011 Lewis structures 1. Find the total number of valence electrons of all of the atoms. 2. Use pairs of electrons to form bonds between all bonded pairs of atoms. 3. Distribute the remaining electrons to give each hydrogen a duet and atoms of the second period an octet. e.g. CH 4 e.g. C 2 H 4

21. © E.V. Blackburn, 2011 Lewis structures (CH 3 ) 2 CHCH 2 OH 1.Find the total number of valence electrons of all of the atoms. 4 x 4 + 10 x 1 + 6 = 32 2. Use pairs of electrons to form bonds between all bonded pairs of atoms. 28 electrons

22. © E.V. Blackburn, 2011 Lewis structures (CH 3 ) 2 CHCH 2 OH 3. Distribute the remaining electrons to give each hydrogen a duet and atoms of the second period an octet.

23. © E.V. Blackburn, 2011 Formal charge formal = valence - unshared electrons - 0.5 x shared electrons charge electrons in bonded atom in bonded atom in free atom It is often necessary to include a formal charge when we draw a structure. Thus in drawing the hydronium ion we need to know where the charge is located. H 3 O + - for oxygen, formal charge = 6 - 2 - 0.5 x 6 = +1 What is the formal charge on nitrogen in the ammonium ion?

24. © E.V. Blackburn, 2011 Formal charge for atom 1 = 6 - 6 (non-bonded e - ) - 0.5 x 2 = -1 Formal charge for atom 2 = 6 - 2 - 0.5 x 6 = +1 Formal charge for atom 3 = 6 - 4 - 0.5 x 4 = 0 formal = valence - unshared electrons - 0.5 x shared electrons charge electrons in bonded atom in bonded atom in free atom

25. © E.V. Blackburn, 2011 Resonance Consider the carbonate ion, CO 3 2-. We can draw three equivalent structures: In reality the ion is perfectly symmetric. All C-O bond lengths are identical and the negative charge is delocalized over the three oxygens. The structure is a hybrid of these three contributing structures.

26. © E.V. Blackburn, 2011 The theory of resonance The molecule is a hybrid of all the contributing structures and cannot be adequately represented by any one of these structures. Whenever a molecule can be represented by 2 or more structures which differ only in the arrangement of their electrons, there may be resonance:

27. © E.V. Blackburn, 2011 The theory of resonance

28. © E.V. Blackburn, 2011 The theory of resonance The hybrid is more stable than any of the contributing structures. This increase in stability is called the resonance energy. Resonance is important when these structures are of about the same stability. For example, However:

29. © E.V. Blackburn, 2011 Benzene

30. © E.V. Blackburn, 2011 Benzene

31. © E.V. Blackburn, 2011 Benzene

32. © E.V. Blackburn, 2011 Benzene

33. © E.V. Blackburn, 2011 How to draw resonance structures

34. © E.V. Blackburn, 2011 How to draw resonance structures

35. © E.V. Blackburn, 2011 How to draw resonance structures - 1,3-dienes

36. © E.V. Blackburn, 2011 How to draw resonance structures - benzene

37. © E.V. Blackburn, 2011 How to draw resonance structures

38. © E.V. Blackburn, 2011 Draw all the resonance structures of: How to draw resonance structures

39. © E.V. Blackburn, 2011 Quantum mechanics Schrödinger calculated mathematical expressions which describe the motion of electrons. These “wave equations” give the energy levels available to electrons as well as the relative probability of finding an electron associated with a given energy level at any point in space. Orbitals are 3-dimensional representations of these probabilities. In order to understand covalent bonding, we must start by studying the electronic structure of individual atoms.

40. © E.V. Blackburn, 2011 Atomic orbitals Atomic orbitals are described by three quantum numbers: The most important is the principal quantum number, n. It governs the energy of an orbital. It can be equal to any positive integer except for zero. The second is the angular momentum quantum number, l, whose value depends on n: l = 0, 1, 2, …. n-1. It determines the shape of the orbital. The third is the magnetic quantum number, m l, which governs the orientation of the orbital relative to the three axes. m l = -l, -l + 1 … 0 … l - 1, l.

41. © E.V. Blackburn, 2011 Atomic orbitals Orbitals are classified (named) according to their values of n and l using a number and a letter. The number represents the value of n and the letter represents the value of l. Electrons in an orbital having l = 0 are called s electrons. Electrons in an orbital having l = 1 are called p electrons. Electrons in an orbital having l = 2 are called d electrons.

42. © E.V. Blackburn, 2011 Atomic orbitals n = 1, l = 0, m l = 0one 1s orbital n = 2, l = 0, m l = 0one 2s orbital n = 2, l = 1, m l = -1, 0, or +1three 2p orbitals!

43. © E.V. Blackburn, 2011 Atomic orbitals

44. © E.V. Blackburn, 2011 1s 2s 2p

45. © E.V. Blackburn, 2011 Phase signs When the value of a wave equation is calculated for a particular point in space relative to the nucleus, the result may be a positive number, a negative number, or zero.   node

46. © E.V. Blackburn, 2011 Electron configurations The Pauli exclusion principle: Orbitals can accommodate a maximum of two electrons but only if they are of opposite spin. Hund’s rule: One electron is placed in each degenerate orbital before adding a second electron to an orbital. The aufbau principle: Orbitals of lowest energy are filled first. The electronic configuration of carbon is therefore 1s 2 2s 2 2p 1 2p 1

47. © E.V. Blackburn, 2011 Linear combination of atomic orbitals Representation of the formation of the H-H bond by the sharing of electrons by the two hydrogens and overlap of their singly occupied atomic orbitals.

48. © E.V. Blackburn, 2011 Molecular orbitals no interaction maximum stability The H-H bond

49. © E.V. Blackburn, 2011  bond formation

50. © E.V. Blackburn, 2011 Antibonding molecular orbitals When two atomic orbitals combine, they form two molecular orbitals. We have met the bonding molecular orbital. Here the atomic orbitals combine by addition. Thus orbitals of the same phase sign overlap. The antibonding molecular orbital is formed by interaction of orbitals of opposite phase sign:

51. © E.V. Blackburn, 2011 Energy diagram for the hydrogen molecule

52. © E.V. Blackburn, 2011 The structure of methane Ground state electronic configuration of carbon:  

53. © E.V. Blackburn, 2011 The structure of methane Promote an electron to give four half filled atomic orbitals: 

54. © E.V. Blackburn, 2011 Hybrid atomic orbitals Orbital hybridization is a mathematical approach that involves the combination of individual orbital wave functions to obtain wave functions for new orbitals. These orbitals have, in varying proportions, the properties of the original orbitals taken separately. What does this mean for methane?

55. © E.V. Blackburn, 2011 sp 3 hybrid orbitals  The four orbitals are mixed (hybridized) to give four new sp 3 hybrid orbitals which are oriented at angles of 109.5 o and are more directional in character:

56. © E.V. Blackburn, 2011

57. Ethane - C 2 H 6

58. © E.V. Blackburn, 2011 sp 2 hybrid orbitals The electronic configuration of boron is 1s 2 2s 2 2p 1. What is the structure of BF 3 ? 

59. © E.V. Blackburn, 2011 Structure - ethylene - C 2 H 4

60. © E.V. Blackburn, 2011 sp hybrid orbitals sp hybridization leads to a linear structure. BeH 2 is an example of such a molecule. The 2s orbital and one of the 2p orbitals are hybridized.

61. © E.V. Blackburn, 2011 Acetylene - C 2 H 2

62. © E.V. Blackburn, 2011 NH 3 Ammonia is pyramidal and its bond angles (H-N-H) are 107 o.

63. © E.V. Blackburn, 2011 H2OH2O Oxygen has 8 valence electrons in the water molecule: Its bond angle is 104.5 .

64. © E.V. Blackburn, 2011 Hybrid orbitals and carbon Using sp, sp 2 and sp 3 hybrid orbitals of carbon, we can construct the carbon chains and rings of organic compounds. The compound below is hystrionicotoxin, a non-protein based toxin that has been isolated from the skin of the “Poison Dart Frog”. The Golden Poison Dart frog may have enough toxin to kill 10 adult humans. Identify the hybrid orbitals to form the C-C, C-H, C-O and O-H bonds in this compound. Predict bond angles.

65. © E.V. Blackburn, 2011 Functional groups – structural families Organic compounds are divided into “families” based on the groupings of atoms within the molecule. Lets now look at the “families” that we will encounter in CHEM 261/263.

66. © E.V. Blackburn, 2011 Hydrocarbons These are compounds containing only carbon and hydrogen atoms. This “clan” is divided into smaller “family” units: alkanes alkenes alkynes aromatic compounds

67. © E.V. Blackburn, 2011 Alkanes Alkanes do not have multiple bonds between carbon atoms. They are therefore “saturated” compounds.

68. © E.V. Blackburn, 2011 Alkenes Alkenes have a carbon – carbon double bond. They are therefore said to be “unsaturated” compounds.

69. © E.V. Blackburn, 2011 Alkynes Alkynes have a carbon – carbon triple bond and are therefore also “unsaturated”.

70. © E.V. Blackburn, 2011 Benzene – a representative aromatic hydrocarbon We will study this group of unstaurated cyclic hydrocarbons in CHEM 263. The six  electrons are said to be “delocalized” over the six carbons of the benzene ring.

71. © E.V. Blackburn, 2011 Alkyl groups These groups are named by replacing the -ane suffix of the corresponding alkane by -yl, hence “alkyl”. CH 3 - methyl (Me-) CH 3 CH 2 - ethyl (Et-) CH 3 CH 2 CH 2 - propyl (Pr-) An alkyl group is the structure obtained when a hydrogen atom is removed from an alkane. “R” is used as a general symbol to represent any alkyl group.

72. © E.V. Blackburn, 2011 Phenyl and benzyl groups “Ar” is used as a general symbol to represent any aromatic ring system. Ar- Ar-CH 2 -

73. © E.V. Blackburn, 2011 Functional groups A group of atoms and their associated bonds that has about the same chemical reactivity whenever it occurs in different compounds.

74. © E.V. Blackburn, 2011 Alkyl halides R-X alkyl halide R = alkyl group (substituted or unsubstituted)

75. © E.V. Blackburn, 2011 carbon classification a “secondary” carbon is bonded to two carbon atoms a “tertiary” carbon is bonded to three carbon atoms a “primary” carbon is bonded to one other carbon

76. © E.V. Blackburn, 2011 Alcohols and phenols

77. © E.V. Blackburn, 2011 R = alkyl or aryl

78. © E.V. Blackburn, 2011

79. The “hottest” compound This is probably capsaicin, the spicy constituent of peppers (capsicum annuum), cayenne pepper, chili peppers and other capsicum species. Identify the functional groups.

80. © E.V. Blackburn, 2011 Bond Properties

81. © E.V. Blackburn, 2011 Bond dissociation energies Energy is absorbed when bonds are broken. This energy is called the bond dissociation energy, D. Energy is released when bonds are formed.

82. © E.V. Blackburn, 2011 Modes of bond breaking

83. © E.V. Blackburn, 2011 Breaking of bonds to carbon

84. © E.V. Blackburn, 2011 Bond dissociation energies  H for bond homolysis at 25C

85. © E.V. Blackburn, 2011 Bond dissociation energies

86. © E.V. Blackburn, 2011  H = + 438 + 243 - 351 - 432 = -102 kJ

87. © E.V. Blackburn, 2011 Polar bonds and polar molecules A polar bond is a covalent bond between two atoms of differing electronegativity.

88. © E.V. Blackburn, 2011 Dipole moments This uneven electron distribution is measured by a quantity called the dipole moment, , measured in units of debye. The direction of bond and molecule dipoles are often indicated using an arrow:  = 1.75D A polar molecule is one in which the centres of positive and of negative charge do not coincide.

89. © E.V. Blackburn, 2011 Dipole moments

90. © E.V. Blackburn, 2011 Physical properties Physical properties such as melting point, boiling point and solubility are dependant on bond polarity and bond type (ionic or covalent).

91. © E.V. Blackburn, 2011 Melting point The temperature at which the thermal energy of the structural units is great enough to overcome the intracrystalline forces that hold them together.

92. © E.V. Blackburn, 2011 Melting points v structural unit ionic compounds High mp. Why? NaCl has a mp = 801C. Each sodium ion is surrounded by six chloride ions.....the crystal is very strong and the structure rigid. covalent compounds Lower mp. Why? The structural unit is the molecule. What are the intermolecular forces of attraction?

93. © E.V. Blackburn, 2011 Dipole-dipole attractions The attraction of the positive ends of polar molecules by the negative ends of other polar molecules. Polar compounds therefore have higher mp and bp than non-polar compounds of comparable molecular mass.

94. © E.V. Blackburn, 2011 Hydrogen bonding Hydrogen bonding is an example of a dipole - dipole interaction in which a hydrogen atom acts as a “bridge” between two very electronegative atoms. This attractive force is ~20 kJ/mol. Ethanol (CH 3 CH 2 OH) and dimethyl ether (CH 3 OCH 3 ) have the same molecular formula. However ethanol has a boiling point of 78.3 o C whereas that of the ether is -24.8 o C.

95. © E.V. Blackburn, 2011 van der Waals forces Methane has no net dipole moment. However when two molecules come in close contact, their electron clouds repel one another. The result is an induced dipole, albeit a small one. The electron deficient part of one molecule attracts the electron rich part of another. The dipoles continually change but the net result is an attraction between the molecules.

96. © E.V. Blackburn, 2011 Criteria of purity All scientists observe systems, make observations and draw conclusions from their results. However, we must know for certain exactly what is under scrutiny. Otherwise any results obtained are simply a worthless jumble of irreproducible facts. Even a sample taken from a labeled bottle is no assurance of purity!

97. © E.V. Blackburn, 2011 The peroxide effect

98. © E.V. Blackburn, 2011 The peroxide effect

99. © E.V. Blackburn, 2011 Melting point - used to indicate degree of purity and in identification of a solid - recrystallize to constant melting point - a sharp melting point range (<1C) between appearance of drops of liquid within the sample to the disappearance of the last trace of solid - mixed melting points. Salt is often placed on an icy sidewalk. Why?

100. © E.V. Blackburn, 2011 Boiling points Pure liquids which distil without decomposition will have sharp, constant boiling points and will leave no residue on distillation to dryness. However, boiling points are very pressure dependent making comparisons unreliable. Some liquid mixtures pose problems – azeotropic mixtures.

101. © E.V. Blackburn, 2011 Refractive index Using an Abbé refractometer, the index can be measured and compared with that of an authentic sample or with literature values. An excellent method of evaluating the purity of a liquid.

102. © E.V. Blackburn, 2011 Thin layer chromatography How can chemists prove that a compound is pure or identical with an authentic sample using chromatography? We can’t! If more than one spot or peak is present, we know that the sample is impure. The presence of one spot or peak is an indication of purity (or identity) but it can never constitute a proof of purity.

103. © E.V. Blackburn, 2011 Elemental analysis

104. © E.V. Blackburn, 2011 Mass spectrometry - used to determine molecular weights - used to determine molecular formulas - used in structure elucidation

105. © E.V. Blackburn, 2011 Infrared spectroscopy - bonds vibrate with characteristic frequencies. - absorption of IR energy occurs only when there is a match between the wavelength of the radiation and the wavelength of the bond vibration. - the position of the IR absorption depends on the strength of the bond, the masses of the bonded atoms and the type of vibration observed

106. © E.V. Blackburn, 2011 Bond vibrations stretching vibrations bending vibrations

107. © E.V. Blackburn, 2011 The 5000 to 1250 cm -1 region Absorptions between 5000 and 1250cm -1 can usually be assigned to a specific bond and a specific vibrational mode.

108. © E.V. Blackburn, 2011 The finger print region This region is therefore called the “finger print region” of the IR spectrum. Absorptions between 1250 and 675cm-1 are associated with more complex vibrational and rotational modes and are often characteristic of the molecule as a whole.

109. © E.V. Blackburn, 2011

115. C7H8OC7H8O

116. Examining a spectrum ~3400 cm -1 : O-H and N-H stretch ~3100 cm -1 : sp and sp 2 C-H stretch ~2900 cm -1 : sp 3 C-H stretch ~2750 cm -1 : aldehyde C-H stretch ~2200 cm -1 : C≡C and C≡N stretch ~1700 cm -1 : C=O stretch ~1600 cm -1 : C=C stretch of alkenes and arenes ~1200 cm -1 : C-O stretch (limited utility)

117. © E.V. Blackburn, 2011 Degree of unsaturation Degree of unsaturation = (2N C - N X + N N – N H + 2)/2 N C = number of carbons N X = number of halogens N N = number of nitrogens N H = number of hydrogens

118. © E.V. Blackburn, 2011 C7H6O2C7H6O2

119. C7H5NC7H5N

120. C7H9NC7H9N