1. PN. ROZAINI ABDULLAH

3.  Able to UNDERSTAND the hybridization theory.  Able to DISCUSS organic reactions and their mechanisms.  Able to EXPLAIN the basic concepts theoretically and APPLY the knowledge of the physical and chemical properties of each functional group.

5. An atom consists of electrons, positively charged protons, and neutral neutrons. Electrons form chemical bonds. Atomic number: numbers of protons in its nucleus. Mass number: the sum of the protons and neutrons of an atom Isotopes have the same atomic number but different mass numbers. The atomic weight: the average weighted mass of its atoms. Molecular weight: the sum of the atomic weights of all the atoms in the molecule.

6. AN ORBITAL The volume of space around the nuclues in which an electron is most likely to be found.

7. The s Orbitals

8.  Node is the part of an orbital in which there is zero probability of finding an electron. Why node occur? Because the electron have the both particle like and wave like properties. What is node?

9. S orbital P orbitals

10. 10 The p Orbitals Teardrop shape Doorknob shape

11.  At the first energy level, the only orbital available to electrons is the 1s orbital, but at the second level, as well as a 2s orbital, there are also orbitals called 2p orbitals.  A p orbital is rather like 2 identical balloons tied together at the nucleus.

12. Quantum mechanics uses the mathematical equation of wave motions to characterize the motion of an electron around a nucleus. Wave functions or orbitals tell us the energy of the electron and the volume of space around the nucleus where an electron is most likely to be found. The atomic orbital closer to the nucleus has the lowest energy. Degenerate orbitals have the same energy. The Distribution of Electrons in an Atom

14. The ground-state electronic configuration describes the orbitals occupied by the atom’s electrons with the lowest energy

15. 15 The Aufbau principle: an electron always goes to the available orbital with the lowest energy The Pauli exclusion principle: only two electrons can occupy one atomic orbital and the two electrons have opposite spin. Hund’s rule: electrons will occupy empty degenerated orbitals before pairing up in the same orbital The following principles determine which orbitals electrons occupy:

16.  Combines the tendency of atoms to fill their octets by sharing electrons (the Lewis model) with their wavelike properties assigning electrons to a volume of space call orbital.  According to MO theory, covalents bond results when atomic orbitals combine to form molecular orbital.  MO Theory is like Atomic Orbital Theory.  MO theory also have their own specific sizes, shapes and energy.

17. Molecular orbitals belong to the whole molecule.  bond: formed by overlapping of two s orbitals. Bond strength/bond dissociation: energy required to break a bond or energy released to form a bond.

18. 18

19. Potential energy H + H Internuclear distance HH weak net attraction at long distances

20. Potential energy H + H Internuclear distance HH HH H H attractive forces increase faster than repulsive forces as atoms approach each other

21. Potential energy H + H H2H2H2H2 Internuclear distance 74 pm HH HH H H -436 kJ/mol maximum net attraction (minimum potential energy) at 74 pm internuclear distance

22. 1s1s1s1s 1s1s1s1s H H 2 H atoms: each electron "feels" attractive force of one proton H 2 molecule: each electron "feels" attractive force of both protons H H

23. Potential energy H + H H2H2H2H2 Internuclear distance 74 pm HH HH H H -436 kJ/mol repulsive forces increase faster than attractive forces at distances closer than 74 pm

24. 24

25.  The bond length : the new covalent covalent bond.  Example: the bond length of H 2 is 0.74Å.  Bond strengh = Bond dissociation energy, the energy required to break the a bond or energy level release when a bond is formed.

26. 26 In-phase overlap forms a bonding MO; out-of-phase overlap forms an antibonding MO:

27.  The bonding molecular orbital is lower in energy and therefore more stable than individual atomic orbitals.  The anti bonding molecular orbital, with less electron density between the nuclei is less stable. It need higher energy than orbital orbitals.  Remember: “ The strongest covalent bond are formed by electrons that occupy the molecular orbitals with the lowest energy”.

28. 28 Sigma bond (  ) is formed by end-on overlap of two p orbitals: A  bond is stronger than a  bond

29. 29 Pi bond (  ) is formed by sideways overlap of two parallel p orbitals:

30.  This theory states that whatever a molecular or ion can be represented by two or more Lewis structures that differ only in the positions of electron, 2 things will be true: 1) none of these structures are that we called resonance contributors, will be correct representation for the molecule or ion. 2) The actual molecule or ion will be represented by a hybrid of these structure.

31.  LOCALIZED – electron are shared between two atoms in a covalent bond or unshared electron belonging to a single atom.  DELOCALIZED – electron share by several nuclei.

32.  Step 1: The connectivity must be the same in all resonance structures The Lewis formulas above are not resonance forms of the same compound. ARE THIS THE SAME COMPOUND OR DIFFERENT??? WHAT IS THE ANSWER???

33.  Step 2: Each contributing structure must have the same number of electrons and same net charge.  Example: All structures have 18 electrons and a net charge of 0.

34.  Step 3: Calculate formal charges on the first structure. Formal Charge = Number of valence electron – (number of lone pair electron + 1/2 of bonding electron) None of the atoms possess a formal charge in this Lewis structure. Formal Charge for O = 6 – [4 +( ½ x 4)] = 0 Formal Charge for C = 4 – [0 + (½ x 8) ]= 0 Formal Charge for N = 5 – [2 +( ½ x 6) ]= 0

35.  Step 4: Calculate formal charges on the second and third structures. These structures have formal charges; these are less stable Lewis structures. Formal Charge for O = 6 – [6 +( ½ x 2)] = -1 Formal Charge for C = 4 – [0 + (½ x 6) ]= 1 Formal Charge for N = 5 – [2 +( ½ x 6) ]= 0 Formal Charge for O = 6 – [6 +( ½ x 2)] = -1 Formal Charge for C = 4 – [0 + (½ x 8) ]= 0 Formal Charge for N = 5 – [0 +( ½ x 8) ]= 1

36.  Resonance structure exist on paper.  In writing resonance, we are allowed to move electrons only.  All the structures must be proper Lewis structure.  Greater number of covalent bond, greater the stability since more atom will have complete octets.  The structure with the least (or no) separation of charge is more stable.  Other things being equal a structure with negative charge is more electronegativity element will be more stable.  Resonance form that are equivalent, have no different stability and contribute equally.

39. 39 Intermolecular Forces Intermolecular forces are exist between molecules. Functional groups determine the type and strength of these interactions interactions that. There are several types of intermolecular interactions. Ionic compounds contain oppositely charged particles held together by extremely strong electrostatic inter-actions. These ionic inter- actions are much stronger than the intermolecular forces present between covalent molecules.

40. 40 Covalent compounds are composed of discrete molecules. The nature of the forces between molecules depends on the functional group present. There are three different types of interactions, shown below in order of increasing strength:  van der Waals forces  dipole-dipole interactions  hydrogen bonding

41. 41 van der Waals Forces van der Waals forces are also known as London forces. They are weak interactions caused by momentary changes in electron density in a molecule. They are the only attractive forces present in nonpolar compounds. Even though CH 4 has no net dipole, at any one instant its electron density may not be completely symmetrical, resulting in a temporary dipole. This can induce a temporary dipole in another molecule. The weak interaction of these temporary dipoles constitutes van der Waals forces.

42. 42 All compounds exhibit van der Waals forces. The surface area of a molecule determines the strength of the van der Waals interactions between molecules. The larger the surface area, the larger the attractive force between two molecules, and the stronger the intermolecular forces. Figure 3.1 Surface area andvan der Waals forces

43. 43 van der Waals forces are also affected by polarizability. Polarizability is a measure of how the electron cloud around an atom responds to changes in its electronic environment. Larger atoms, like iodine, which have more loosely held valence electrons, are more polarizable than smaller atoms like fluorine, which have more tightly held electrons. Thus, two F 2 molecules have little attractive force between them since the electrons are tightly held and temporary dipoles are difficult to induce.

44. 44 Dipole-Dipole Interactions Dipole—dipole interactions are the attractive forces between the permanent dipoles of two polar molecules. Consider acetone (below). The dipoles in adjacent molecules align so that the partial positive and partial negative charges are in close proximity. These attractive forces caused by permanent dipoles are much stronger than weak van der Waals forces.

45. 45 Hydrogen bonding typically occurs when a hydrogen atom bonded to O, N, or F, is electrostatically attracted to a lone pair of electrons on an O, N, or F atom in another molecule. Hydrogen Bonding

46. 46 Note: as the polarity of an organic molecule increases, so does the strength of its intermolecular forces.

47.  Theory in developing the bonds in selected examples such as methane, ethane, ethene, ethyne, metyl cation.

48. HybridBondingShapeRotationBond sp3 ( σ )Head- head TetrahedralFreeSingle sp 2 ( π ) Side waysTriangularRigidDouble sp ( π ) SidewaysLinearRigidTriple

49. Methane Ethene

50.  Formula molecule: CH 4  Carbo electron configuration : 1s 2 2s 2 2p x 1 2p y 1 How can it have bonds to four hydrogens?

51. Electron configuration of carbon at in most stable state. An electron is promoted from the 2s orbital to the vacant 2p orbital. The 2s orbital and the three orbital 2p orbital are combine to give a set of four equal-energy sp 3 hybridized orbitals, each of which contains one electron.

52.  The sp3 hybrid orbitals are arrange in tetrahedral fashion around the carbon.  Each orbital contains one electron and can form a bond with a hydrogen atom to give a tetrahedral methane molecule. sp 3 109.5 o

53.  Each methyl group consists of an sp3- hybridized carbon attached to three hydrogens by sp 3 -1s σ bonds.  Overlap remaining hall-filled orbital of one carbon with that of the other generates a σ bond a bond between them.  Here the third kind of σ bond, one that has as its basis the overlap of two sp 3 -hybridized orbitals.

54. In general, you can expect that carbon will be sp3- hybridized when it is directly bonded to four atoms

55. Electron configuration of carbon at in most stable state. An electron is promoted from the 2s orbital to the vacant 2p orbital. 2sp 2 2p The 2s orbital and the three orbital 2p orbital are combine to give a set of three equal-energy sp 2 hybridized orbitals. One of the 2p orbital remains uncharge.

56. In general, you can expect that carbon will be sp 2 hybridized when it is directly bonded to three atoms. The half-filled 2p orbitals have their axes perdicular to frame of σ bonds of the molecule and overlap in a side-by-side manner that called pi( π ) bond.

57. Electron configuration of carbon at in most stable state. An electron is promoted from the 2s orbital to the vacant 2p orbital. 2sp 2 2p The 2s orbital and the three orbital 2p orbital are combine to give a set of two equal-energy sp 2 hybridized orbitals. Two of the 2p orbital remains uncharge.

58. In general, you can expect that carbon will be sp- hybridized when it is directly bonded to two atoms.

59. 1.14 Bruice: Organic Chemistry, © 2011 Pearson Education, Inc. Answer: E

60.  Organic reactions and their mechanisms  Bronstead-Lowry and Lewis acid and bas  Acid-base reactions  Heterolysis of bonds to carbon and the strength of acids and bases: Ka and pKa.

61.  Substitution reactions in which an atom or group of atoms is replaced by a different atom or group of atoms.  In this familiar example, OH ion replaces Cl of methylchloride.

62.  Addition reactions in which the elements of one compound or molecule are added to another compound or molecule.  A familiar example is the addition of H 2 to an alkene in the presence of a catalyst.

63.  Elements of a simple compound are removed, or eliminated from the compound undergoing reaction (sometimes referred to as the substrate).  Small molecules expelled in elimination reactions include H 2 O, HX, HCN and NH 3.  An elimination reaction results in the formation of a multiple bond or a ring in the product. An example of an elimination reaction is HBr from 2-bromo-2-methylpropane, under the influence of base, to form 2-methylpropene.

64.  A molecule undergoes a reorganization of its constituent parts.  Example: heating the alkene with a strong acid causes the formation of another isomeric alkene.  Not only the position of double bond and a hydrogen atom changed, but a methyl group has moved from one to another.

65.  Any species that has a hydrogen can potentially act as an acid and any compound possessing a lone pair can potentially act as base.

66.  Brønsted-Lowry Acids and Bases A Brønsted-Lowry ACID is a PROTON DONOR. A Brønsted-Lowry BASE is a PROTON ACCEPTOR. H + = proton Examples of Brønsted–Lowry acids and bases

67. Some molecules contain both hydrogen atoms and lone pairs and thus, can act either as acids or bases, depending on the particular reaction. An example is the addictive pain reliever morphine.

68. Which of the following molecules is not a BrØnsted-Lowry acid? HBr NH 3 CCl 4 BrØnsted-Lowry acid must contain a proton. Which of the following is not a BrØnsted-Lowry base? BrØnsted-Lowry bases must have a lone pair or pi bond to donate.

69. Water can behave as Acid because it has a proton that it can donate. Base bacause it has a lone pairs That can accept a proton.

70. A Brønsted-Lowry acid base reaction results in the transfer of a proton from an acid to a base. In an acid-base reaction, one bond is broken, and another one is formed. The electron pair of the base B: forms a new bond to the proton of the acid. The acid H—A loses a proton, leaving the electron pair in the H—A bond on A.

71. The movement of electrons in reactions can be illustrated using curved arrow notation. Because two electron pairs are involved in this reaction, two curved arrows are needed. Loss of a proton from an acid forms its conjugate base. Gain of a proton by a base forms its conjugate acid. A double reaction arrow is used between starting materials and products to indicate that the reaction can proceed in the forward and reverse directions. These are equilibrium arrows.

73. Label the acid, base, conjugate acid and conjugate base in the following reaction. A B CB CA Use curved arrows to show the movement of electron pairs.

74. Acid strength is the tendency of an acid to donate a proton. The more readily a compound donates a proton, the stronger an acid it is. Acidity is measured by an equilibrium constant. When a Brønsted-Lowry acid H—A is dissolved in water, an acid- base reaction occurs, and an equilibrium constant can be written for the reaction.

75. Because the concentration of the solvent H 2 O is essentially constant, the equation can be rearranged and a new equilibrium constant, called the acidity constant, K a, can be defined. It is generally more convenient when describing acid strength to use “pK a ” values than K a values.

77. Anything that stabilizes a conjugate base A: ¯ makes the starting acid H—A more acidic. Four factors affect the acidity of H—A. These are: Element effects Inductive effects Resonance effects Hybridization effects

78. No matter which factor is discussed, the same procedure is always followed. To compare the acidity of any two acids: oAlways draw the conjugate bases. oDetermine which conjugate base is more stable. oThe more stable the conjugate base, the more acidic the acid.

79. Element Effects—Trends in the Periodic Table. Across a row of the periodic table, the acidity of H—A increases as the electronegativity of A increases. Positive or negative charge is stabilized when it is spread over a larger volume.

80. Down a column of the periodic table, the acidity of H—A increases as the size of A increases. Size, and not electronegativity, determines acidity down a column. The acidity of H—A increases both left-to-right across a row and down a column of the periodic table. Although four factors determine the overall acidity of a particular hydrogen atom, element effects—the identity of A—is the single most important factor in determining the acidity of the H—A bond.

81. An inductive effect is the pull of electron density through  bonds caused by electronegativity differences between atoms. In the example below, when we compare the acidities of ethanol and 2,2,2-trifluoroethanol, we note that the latter is more acidic than the former.

82. The reason for the increased acidity of 2,2,2-trifluoroethanol is that the three electronegative fluorine atoms stabilize the negatively charged conjugate base.

83. When electron density is pulled away from the negative charge through  bonds by very electronegative atoms, it is referred to as an electron withdrawing inductive effect. More electronegative atoms stabilize regions of high electron density by an electron withdrawing inductive effect. The more electronegative the atom and the closer it is to the site of the negative charge, the greater the effect. The acidity of H — A increases with the presence of electron withdrawing groups in A.

84. Resonance is a third factor that influences acidity. In the example below, when we compare the acidities of ethanol and acetic acid, we note that the latter is more acidic than the former.

85. When the conjugate bases of the two species are compared, it is evident that the conjugate base of acetic acid enjoys resonance stabilization, whereas that of ethanol does not.

86. Resonance delocalization makes CH 3 COO ¯ more stable than CH 3 CH 2 O ¯, so CH 3 COOH is a stronger acid than CH 3 CH 2 OH. The acidity of H—A increases when the conjugate base A: ¯ is resonance stabilized.

87. Electrostatic potential plots of CH 3 CH 2 O ¯ and CH 3 COO ¯ below indicate that the negative charge is concentrated on a single O in CH 3 CH 2 O ¯, but delocalized over both of the O atoms in CH 3 COO ¯.

88. The final factor affecting the acidity of H—A is the hybridization of A. Let us consider the relative acidities of three different compounds containing C—H bonds The higher the percent of s-character of the hybrid orbital, the closer the lone pair is held to the nucleus, and the more stable the conjugate base.

91. Which is the strongest base? H 2 0 (pk a = 15.7) NH 3 (pk a = 38) CH 4 (pk a = 58) CH 4 Draw the products and determine the direction of the equilibrium.

92. Which of these bases is strong enough to deprotonate CH 3 COOH (pk a = 4.8) ? pk a = 2535-7 Both acetylene and hydrogen will work (must have higher pk a ) Not using pk a, which is the stronger acid? H 2 S HBr HBr, b/c Br is further to the right and further down in the periodic table.

93. Which is the stronger acid? F 2 CHCH 2 COOH OR Cl 2 CH 2 COOH F 2 CHCH 2 COOH is the most acidic b/c acids are stabilized by electron withdrawing groups and F is more electronegative than Cl Which is the stronger acid and why? B/c the C-H bond involves sp 2 hybridized orbital instead of sp 3

94. The Lewis definition of acids and bases is more general than the BrØnsted- Lowry definition. A Lewis acid is an electron pair acceptor. A Lewis base is an electron pair donor. Lewis bases are structurally the same as BrØnsted-Lowry bases. Both have an available electron pair—a lone pair or an electron pair in a  bond. A BrØnsted-Lowry base always donates this electron pair to a proton, but a Lewis base donates this electron pair to anything that is electron deficient.

95. A Lewis acid must be able to accept an electron pair, but there are many ways for this to occur. All Br Ø nsted-Lowry acids are also Lewis acids, but the reverse is not necessarily true. oAny species that is electron deficient and capable of accepting an electron pair is also a Lewis acid. Common examples of Lewis acids (which are not Br Ø nsted-Lowry acids) include BF 3 and AlCl 3. These compounds contain elements in group 3A of the periodic table that can accept an electron pair because they do not have filled valence shells of electrons.

96. Any reaction in which one species donates an electron pair to another species is a Lewis acid-base reaction. In a Lewis acid-base reaction, a Lewis base donates an electron pair to a Lewis acid. Lewis acid-base reactions illustrate a general pattern in organic chemistry. Electron-rich species react with electron-poor species. In the simplest Lewis acid-base reaction one bond is formed and no bonds are broken. This is illustrated in the reaction of BF 3 with H 2 O. H 2 O donates an electron pair to BF 3 to form a new bond.

97. A Lewis acid is also called an electrophile. When a Lewis base reacts with an electrophile other than a proton, the Lewis base is also called a nucleophile. In this example, BF 3 is the electrophile and H 2 O is the nucleophile. electrophile nucleophile Electrophile: a species that can act as a Lewis Acid or electron pair acceptor. Example: Carbocation. Nucleophile: An atom that has an unshared electron pair which can be used to form A bond to carbon. Nucleophile is Lewis Base.

98. Two other examples are shown below. Note that in each reaction, the electron pair is not removed from the Lewis base. Instead, it is donated to an atom of the Lewis acid and one new covalent bond is formed.

99. Which is not a Lewis base? Lewis base must have an electron pair to donate. Which is not a Lewis acid? None, all are Lewis acids b/c they either contain an unfilled valence shell or an acidic proton.