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Organic Molecular Structure, , Bonding and Structure: Part 1, , 1 Energies of Electrons in Atoms and Molecules, • Bonding, structure, shape, and reactions of organic molecules are determined primarily by the energies of, the electrons in atomic orbitals (AOs) and molecular orbitals (MOs)., • In organic chemistry, we use models, generally the simplest one that explains what we are trying to understand., • Here, we use a simple model for AOs that describes the factors that determine electron energies in atoms., Later, we will use more sophisticated models for orbitals to understand electron energies in molecules., , 1.1 Energies of Electrons in AOs, electron stabilized by, nucleus, held "tightly", close to nucleus, relatively, low energy electron, , H 1s1, , 3+, , Li, , 6+, , C 1s22s22p2, , 1s22s1, , electron in even larger 2p, A.O., but, higher positive, charge on the nucleus (6+),, thus outer electrons held, reasonably tightly by, nucleus, moderate energy, , even higher positive charge, on the nucleus (8+), but now, atomic orbital has another, negatively charged electron,, outer electrons held, reasonably tightly by, nucleus, moderate energy, , 8+, , O, , electron in larger 2s A.O.,, further from nucleus and, shielded by two 1s electrons,, not held tightly, relatively, high energy electron, , 1s22s22p4, , 1.2 Quantitative Energies of Electrons in Atoms, • Quantitative information about the relative energies of electrons is obtained from measurements of Ionization, Energies (IE), or Ionization Potentials (IP)., • The first IE/IP is the energy required to completely remove the highest energy electron from an atom or, molecule. We are interested mainly in the energies of these highest energy electrons since these are the ones, that are involved in chemical bonding and reactions., • Higher energy electrons in an atom require less energy to remove, they have smaller IEs., • First IEs for some atoms are given below in electron Volts (eV) (don’t memorize these!)., , Energy, , 13.6, eV, , higher, energy, lower, energy, , energy of an electron infinitely far from any nucleus, , 1s, , H, 1s1, , 5.4eV, , 11.3, eV, , 2s, , 2p, 1s, , Li, 2 1, 1s 2s, , 2p, , 13.6, eV, , 14.5, eV, , 2p, , 2s, , 2p, , 1s, , 2s, , C, 1s22s22p2, , 2p, , 2p, , 1s, N, – "valence" electrons, are involved in reactions/bonding, 2 2, 1s 2s 2p3, – "core" electrons, not involved in reactions/bonding, Bonding 1 : page 1, , 2p, 2p, , 2p, 2s, 1s, , O, 1s 2s22p4, 2

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• Valence electrons are in the outer shell; they are highest in energy and are involved in bonding., • Core electrons are in the inner shells; they are not involved in bonding., , Understanding the Connection Between IP/IE and Electron Energy, • The IE is a measure of the energy of the highest energy electron., • An electron that has been completely removed from an atom/molecule has a very high energy because it is not, stabilized by any nuclei. Electrons associated with an atom in an orbital are lowered in energy because they are, negatively charged and are stabilized by the positively charged nucleus., • If the energy of an electron in an atom is low (the electron is held “tightly” by the atom), more energy is, required to remove it from the atom, the energy required to ionize is large, the IE or IP is large., • If the energy of an electron in an atom is high, less energy is required to remove it from the atom, the energy, required to ionize is small, and the IE or IP is small., , very high energy electron, , energy of an electron infinitely far from any nucleus, , Energy, , smaller, less energy, 5.4eV required to remove, , larger, 13.6, eV, , more energy, required to remove, larger I.P., , higher energy electron, , 2s, , 1s, lower, energy electron H 1s1, , 1s, Li 1s22s1, , • Many factors influence atomic IEs, (orbital size, nuclear charge, orbital occupancy, etc.). However, a detailed, understanding of these factors is not necessary at this point., • What you should know for now is that electron energy decreases and IE increases, and roughly with, increasing electronegativity (i.e., left to right and from bottom to top in the periodic table):, hydrogen, , helium, , 1, , 1.0079, lithium, , beryllium, , 3, , 4, , Li Be, , 6.941, , 9.012, , sodium, , magnesium, , 11, , 12, , 22.990, , 24.306, , potassium, , calcium, , 19, , 20, , 39.098, , 40.078, , rubidium, , 37, , 85.468, , 87.62, , 2, , increasing electronegativity, DECREASING electron energy, INCREASING IP, , H, , He, carbon, , nitrogen, , oxygen, , fluorine, , 5, , 6, , 7, , 8, , 9, , B, , Na Mg, scandium, , titanium, , 21, , 22, , 44.956, , 47.867, , 50.942, , strontium, , yttrium, , zirconium, , 38, , 39, , 40, , 91.224, , Sc Ti, , Rb Sr, , Y, , 88.906, , vanadium, , C, , N, , O, , F, , 10.811, , 12.0107, , 14.007, , 15.999, , 18.998, , aluminium, , silicon, , phosphorus, , sulfur, , chlorine, , 13, , 14, , 15, , 16, , 17, , 26.912, , 28.086, , 30.974, , 32.067, , 35.453, , Al, , K Ca, , Si P, , S, , Cl, , chromium, , manganese, , iron, , cobalt, , nickel, , copper, , zinc, , gallium, , germanium, , arsenic, , selenium, , bromine, , 24, , 25, , 26, , 27, , 28, , 29, , 30, , 31, , 32, , 33, , 34, , 35, , 51.996, , 54.938, , 55.845, , 39.098, , 58.693, , 63.546, , 65.39, , 69.723, , 72.61, , 74.922, , 78.96, , 79.904, , niobium, , molybdenum, , technetium, , ruthenium, , rhodium, , palladium, , silver, , cadmium, , indium, , tin, , antimony, , tellurium, , iodine, , 41, , 42, , 43, , 44, , 45, , 46, , 47, , 48, , 49, , 50, , 51, , 52, , 53, , 92.906, , 95.94, , [98.91], , 101.07, , 85.468, , 106.42, , 107.87, , 112.41, , 114.818, , 118.71, , 121.760, , 127.60, , 126.904, , 23, , V, , increasing, electronegativity, DECREASING, electron energy, INCREASING IP, , 4.0026, , boron, , Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br, , Zr Nb Mo Tc Ru Rh Pd Ag Cd, , In Sn Sb Te, , I, , neon, , 10, , Ne, 20.180, argon, , 18, , Ar, 39.948, krypton, , 36, , Kr, 83.80, xenon, , 54, , Xe, 131.29, , 1.3 Energies of Electrons in Molecules, for example, Hydrogen, • We can also measure energies of electrons in molecules as IEs., • The energies of electrons in molecules are lower than the energies of electrons in atoms., , Energy, , energy of an electron that is infinitely far from any nucleus, IP ~ 13.6, , H atom, 1s atomic orbital, , IP ~ 15.4, lower energy, IN A BOND, , H2 molecule, σ molecular orbital, , • The IE of the electron in the hydrogen molecule is larger than that in the hydrogen atom., Bonding 1 : page 2

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• The energy of the electrons in the molecule must be lower than in the atom., • Why? Because in the electrons in the molecule are in a bond—this is really important!, , each electron, stabilized by, two nuclei, , electron stabilized, by one nucleus, H atom, 1s atomic orbital, , H2 molecule, covalent bond, σ molecular orbital, , • In the atom, the negatively charged electron is stabilized by one positively charged nucleus; in the, molecule, the negatively charged electrons are stabilized by two positively charged nuclei., • We have just learned why hydrogen exists as the molecule H2 not as hydrogen atoms, making the, molecule lowers the energy of the electrons by getting them into a bond:, , H•, , H, H, molecule, , + •H, atoms, , • This is our first very simple chemical reaction. Two hydrogen atoms react to form a hydrogen molecule, the, electrons in the hydrogen atoms are chemically reactive, they want to do a chemical recation to form H2., We have already learned two critical concepts for understanding organic chemistry:, 1. Forming bonds stabilizes (lowers the energies of) electrons., 2. Higher energy electrons are more chemically reactive (because they want to form bonds)., , 2 Bonding in Molecules: The Lewis Structure Model, • Lewis structures represent a model for bonding in organic molecules, they are not, always 100% accurate, but they are useful because they are simple, and we tend to, use the simplest model wherever possible., , 2.1 Atomic Valence and Lewis Structures, Making a Molecule of Hydrogen (H2) as a Simple Example, • Each hydrogen atom makes one bond, because in doing so it lowers the energy of, its one valence electron., • Hydrogen can only make one bond because it only has one valence electron., • Making one bond fills the first shell of the hydrogen atom with two electrons., , 1 valence electron, H, 1s, H electron configuration, , H•, , shared electrons: covalent bond, H, , H, , •H, , Lewis "dot" structure, , H, H, Lewis or Kekule, structure, , • Note the use of curved arrow notation, “move or use an electron from each atom to make a new bond.”, , Making a Molecule of CH4 as an Example, 2 core, electrons, , H, , 4 valence electrons, , 1s2 2s22p2, , C, , H, , C, , shares 2, H, , shares 8, filled second shell H, , H, H, , C, H, , H, , H C H, , H, Lewis, or Kekule, Lewis "dot" structure, H, structure, • After making four bonds, a carbon atom’s second outer shell is “full” with eight shared electrons, it obeys the, “filled shell,” sometimes called the “octet rule,” for second row elements., Bonding 1 : page 3, C electron configuration

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• The hydrogens also have a “full"” first shell with two electrons., • The filled shell “rule” and the pictures of the Lewis dot structures are models of organic structures., , Making a Molecule of NH3 as a Simple Example, 2, 2 core, core, electrons, electrons, electrons 55 valence, valence electrons, N, , H, , N, , 1s2 2s22p3, , H no more bonds, , N electron configuration, , H, , H, , shares 8, X, shares 2, shares 2, octet rule, H N H, H HN N H H, shares 8, HH, H, Lewis, or Kekule, Lewis "dot" structure, structure, , • Nitrogen has five valence electrons, but it can only lower the energy of three electrons in bonds before the, outer electron shell is filled with eight electrons. Attempting to make another bond “overfills” the shell, violating the, filled shell “rule.”, • Nitrogen in this structure therefore has two nonbonding electrons in the outer shell., • Lewis “dot” structures are tedious to draw, we will use them rarely., Unfortunately, there is more than one use of the word Valence., 1. The outer shell is the valence shell, the electrons in the outer shell are the valence electrons., 2. The number of bonds an atom normally makes without violating the “filled shell rule” is the normal valence., • The normal number of bonds, the “Normal Rules of Valence” for different atoms are obtained from the atomic, configuration and the filled shell rule., Do not memorize this table! Learn the normal valences by working with and building organic structures., , atom, 1st shell, , 2nd shell, , 3rd shell, , electron, configuration, , # valence, electrons, , maximum possible, electrons in outer shell, , normal valence, (normal # of bonds), , H, , 1s, , 1, , 2, , 11, , B, , 1s2 2s22p1, , 3, , 8, , 33 only 3 valence electrons, , C, , 1s2 2s22p2, , 4, , 8, , 44, , N, , 1s2 2s22p3, , 5, , 8, , O, , 1s2 2s22p4, , 6, , 8, , F, , 1s2 2s22p5, , 7, , 8, , 33 octet "rule", 22 octet "rule", 11 octet "rule", , P, , 1s2 2s22p63s23p3, , 5, , 8, , 33 or, or 5!, 5!, , S, , 1s2 2s22p63s23p4, , 6, , 8, , 2,4, 2, 4,oror6!6!, , violates, octet rule!, , • The number of valence electrons is the number of electrons in the highest energy shell., • The Normal Valence is the number of electrons that can be used to make a bond before the highest energy, shell is filled, it is equal to the number of bonds an atom “normally” makes., • Boron has only three valence electrons and so it can make only three bonds, it does not have enough electrons, to fill the shell, even if all electrons are involved in bonding., • Nitrogen has five valence electrons, but after making three bonds the valence shell is full and the remaining two, electrons cannot make bonds, therefore, the normal valence of nitrogen is also three., • Oxygen has six valence electrons, after making two bonds the valence shell is full, the normal valence is two., • The filled shell or octet rule doesn’t really work for third row elements, for example, phosphorus and sulfur., Example Problem: Draw two different Lewis structures for C2H6O. The Normal Rules of Valence require:, • Four bonds to each tetravalent carbon (each carbon wants to make four bonds)., • Two bonds to each divalent oxygen (each oxygen wants to make two bonds)., • One bond to each monovalent hydrogen (each hydrogen wants to make one bond)., Bonding 1 : page 4

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• To generate Lewis structures, we will learn to assemble molecules using common organic structural motifs, for, example, a carbon at the end of a chain usually has three hydrogen atoms bonded to it, -CH3., , different structures, (isomers), H, , H, , O, C, C, H, H, H, H, , H, , H, , H, , C, H, , C, H, , O, H, , same structure, drawn, H, differently, , O H, , H, C, , C, , H, , H, , H, , • Lewis/Kekule structures indicate connectivity of atoms, the orientation or the direction the bonds point doesn’t, matter (at least for now)., • Different Lewis (Kekule) structures for a specific molecular formula are isomers, more on isomers later., , 2.2 Condensed Structures, • In condensed structures, the order of atom connectivity/bonding is implied by the “written” order of the atoms,, but the bonds are not explicitly shown, they are implied., • This means that the normal valences for the atoms in the structure must be assumed., Example Problem 1: Convert the provided condensed formula into a Lewis/Kekule structure., , carbon with 3 H's connected to, carbon with 2 H's connected to, carbon with 2 H's connected to, oxygen with 1 H, , CH3CH2CH2OH, condensed, , H, , H, , H, , H, , C, , C, , C, , H, , H, , H, , O, , Lewis, , H, , • Obeying the normal rules of valence can generate only one possible Lewis structure., • Nonbonding electrons are shown in Lewis structures., Example Problem 2: Convert the provided condensed formula into a Lewis/Kekule structure., • Obeying the normal rules of valence for C (four bonds), H (one bond), and O (two bonds) can only generate, one possible Lewis structure., , parenthesis means repeated unit, parenthesis means "off" the main chain, , CH3C(CH3)2CH(OH)CO2CH3, , functional group, H, H C H, H, H C, C, only possible, H H C H, connectivity order, H, , O, C, H, , H, , O, C, , O, , "double bond", , 2 bonds to C, , H, C H, H, , -CO2CH3, , =, , C, , H, O, , O, , C H, H, , • Note the carbon to oxygen double bond, which is required to satisfy the normal rules of valence for all atoms., • Note the two different uses of parentheses., 1. Parentheses are used to indicate repeating units along the main chain, for example, (CH2)3 in the earlier, structure., 2. Parentheses can be used to indicate a part of the structure that comes “off” the main chain, for example, (OH), in the earlier structure., Bonding 1 : page 5

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• “Rules” for condensed structures are not strictly observed, for example, the parentheses around the “OH” in the, earlier structure may be omitted if it is considered obvious that it is “off” the main chain., Example Problem 3: Convert the provided condensed formula into a Lewis/Kekule structure., , parenthesis, means 2 -CH2- groups in a row, , CH3(CH2)2CCCOCH2CO2H, , acid functional group, H, H C, H, , H H, C, , C, , H, , H, , C, , C, , O, , H, , C, , C, H, , O, C, , O, , must be a triple bond here, H, , • Note the use of parentheses in this case to indicate repeating units. The two uses of parentheses are easily, distinguishable using the normal rules of valence, only one will “work” in a particular context., , 2.3 Line-Angle Structures, • Line-angle or skeletal structures are most commonly used in organic chemistry., • The lines show bonds between carbon atoms; hydrogen atoms are not included unless they are part of a, functional group (see later for definition)., Example 1, , H, H, O, H, C, C H, H C, C, C H H, H, H C H, O C H, H, H, Lewis structure, H, , (CH3)3CCOCH2OCH3, condensed, , the "kink", indicates the, position of a, carbon atom, , O, , O, , Line angle, (skeletal), , • There is an atom (carbon, unless otherwise specified) at the “end” of each line, each line is a covalent bond., Example 2, Draw one example of Lewis (Kekule), condensed and line-angle structures for C4H6O., , Lewis structure, H, , H, C, H 1, , C, 2, , O, C 4, C, H, H 3 H, , Condensed structure, , Line Angle structure, O, , 1 2 3 4, CH2CHCH2CHO, 1, , 2, , 3, , 4, , H, , this H is part of, the aldehyde, functional group, , • Where reasonable, draw the angles roughly correct for the molecular shape, see later., • Which kind of structure to draw (condensed, Lewis, line angle, etc.) depends upon the context, we will use, mainly line angle or even a mixture of line angle and Lewis in one structure., , Bonding 1 : page 6

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3 Functional Groups, 3.1 Anatomy of an Organic Structure, • Organic molecules consist of a carbon/hydrogen (hydrocarbon) “skeleton” that mainly determines the size and, shape of the molecule., • Added to this skeleton are the functional groups, which generally include atoms that are more electronegative, than carbon, such as O, N, S, and so on., • Chemistry takes place at the functional groups. When we start to discuss reactions, we will divide the reactions, into those that are characteristic of the different functional groups., • Line-angle or skeletal structures are most commonly used, especially for larger molecules., The main structural features of a typical organic molecule are the following:, Nonbonding electrons are not included in most of the structures in this section for clarity, , O, , HO, , functional groups, , functional groups, , OH, , O, cortisone, anti-inflammatory, , functional groups, , carbon/hydrogen "skeleton", , O, , • The alkane or alkyl part of a molecule is the “skeleton” or “backbone” of the molecule., • Alkanes or alkyl groups consist of chains (or rings) of carbon atoms connected together by single C–C and an, appropriate number of C–H bonds. C–C and C–H bonds tend to be strong and relatively unreactive., H H, , H, H, , an example alkane (heptane), only C and H, no double bonds, , C, O, , C, H, , C, , H, C, , H, , an example 3 carbon, alkyl chain in a molecule, , H, , R stands for any alkyl chain, e.g., R–OH could be:, , an alkyl chain, , H3C–OH or CH3CH2OH, or, OH, etc., , R-OH =, , NH, , N, , bupivacaine, an, epidural anesthetic, , O, , • Carbon or nitrogen atoms are characterized by the number of alkyl chain (-R) or aryl (aromatic) groups (-Ar,, see below) attached to them., • Atoms with one substituent are primary (1°)., • Atoms with two substituents are secondary (2°)., • Atoms with three substituents are tertiary (3°)., • Atoms with four substituents are quaternary (4°)., • H is not counted as a substituent in this context., , H is, not a, substituent, , H, H, H, , H, , R, , C, , C, R, , Primary or 1°, carbon, , R, , R, , R, , H, R, , Secondary or 2°, carbon, Bonding 1 : page 7, , C, H, , R, , Tertiary or 3°, carbon, , R, , C, R, , R, , Quaternary or 4°, carbon

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3.2 Important Functional Groups, Alkene Functional Group, • Carbon–carbon double bond:, , OH, C, , alkene, , diene, , alkene, , C, , linalool, used in the perfume industry, , NOT aromatic (see next section), does not have alternating double/single bonds, , Aromatic Functional Group, • Alternating C–C and C=C bonds in a ring—be careful to distinguish from alkene:, , stands for any aromatic ring system, e.g. Ar–OH could be:, O, aromatic, OH, OH, Cl, aspirin, or, etc., , Ar, , OH, O, O, , this is aromatic because it has alternating single/, double bonds in a (large) ring, an aromatic ring, can have more than 6 carbon atoms!, , Alkyne Functional Group, • Carbon–carbon triple bond:, , HO, , alkyne, mestranol, the estrogen used in many oral, contraceptives, , C, , C, , O, , Amine Functional Group, • Contain a nitrogen with at least one alkyl or aryl group, here R1, R2, and so on, stands for any alkyl chain that, may or may not be the same:, , R1, R or Ar, R2, , 3° amine, , N, R3, , tertiary (3°) amine, , N, triethylamine, fishy smell!, , 2° amine, N, H, , coniine, the, poison in hemlock, , HO, O, , S, , O, , 1° amine, NH2, , taurine, supposedly active ingredient in, energy drinks e.g. Red Bull, , Ether and Epoxide Functional Group, • An ether is an oxygen between two alkyl or aryl groups., • For the specific case where the oxygen is part of a 3-membered ring the functional group is an epoxide., , (R, Ar), , O, , O, , ether, , (R, Ar), , 4,5-benzo[a]pyrene oxide,, highly carcinognic, , O, , tetrahydrofuran, THF, a common, organic solvent, , O, , Bonding 1 : page 8, , epoxide

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Alcohol Functional Group, • Oxygen with one R or Ar group and one hydrogen., • Hydrogen atoms are not included in line-angle structures unless they are part of a functional group, the alcohol, functional group provides an example of this:, , 3° carbon, O, , (R, Ar), , bisabolol is used to aid in the, transfer of drugs through the skin, , OH, , H, , H is PART of the functional group!, , 3° alcohol, , Halide Functional Group, • Aryl or alkyl group with fluoride, chloride, bromide, and iodide:, , R or Ar, R, , R, , F, , Cl, , R, , chloride, Br, , R, , Cl, , Cl, , chloride, , chloride, , I, , O, bromide, , chloride, Cl, , Br, , Cl, , N, H, , bromide, H, N, , Br, , O, , 6,6'-dibromoindigo, a component, of a natural purple dye, , a polychlorinated biphenyl (PCB), many industrial, uses but toxic, bioaccumulates in animals, , Ketone Functional Group, • C=O double bond with two alkyl or aryl groups:, , O, (R, Ar), , C, , ketone, (R, Ar), , H 3C, , O, C, , acetone, the simplest, ketone, common organic, solvent, nail-polish remover, , CH3, , Aldehyde Functional Group, • C=O double bond with one R or Ar group and one H., • Hydrogen atoms are not included in line-angle structures unless they are part of a functional group, the, aldehyde functional group provides another example of this:, , O, O, (R, Ar), , C, , aldehyde, vanillin, main extract fron, vanilla bean, , H, H, , HO, , H is PART of the, functional group!, , OMe, , Carboxylic Acid Functional Group, • C=O with one R or Ar group and one -OH., • Hydrogen atoms are not included in line-angle structures unless they are part of a functional group, the, carboxylic acid functional group provides another example of this., , (R, Ar), , O, C, , carboxylic acid, , O, OH, , OH, NOT ketone, NOT alcohol!, , sorbic acid, food, preservative, , H is PART of the functional group!, , Bonding 1 : page 9

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Ester Functional Group, • C=O double bond with -OR or -OAr group:, , ester, , O, (R, Ar), , C, , O, , methyl paraben, food, preservative, cosmetics, additive, anibacterial/fungal, , HO, , O (R, Ar), , O, NOT ketone, NOT ether!, , Amide Functional Group, • C=O with -NR2 (R or Ar):, , 3° amide, , O, C, , N, , H, N, , R, , O, , 2° amiDe, , N, , OH, , N, , R, 3° amide, , 3° amiNe, , O, , Cl, loperamide, anti-diarrhea drug, , Acid Chloride Functional Group, • C=O double bond with -Cl group:, , (R, Ar), , O, C, , acid chloride, , Cl, , O, , acetyl chloride, many useful, reactions, , Cl, , Nitrile Functional Group, • Aryl or alkyl group with carbon–nitrogen triple bond:, , O, , nitrile, , N, (Ar, R) C, , C, , N, , N, , citalopram,, antidepressant drug, take after first midterm, , (Ar,R) means aryl or alkyl, , How Functional Groups Are Represented?, Here are some functional groups incorporated into line-angle and condensed structures that you will see and, need to understand:, nitrile, aldehyde, alcohol, carboxylic acid, amide, and amine, , O, , O, H, or, , O, OH, or, , or, , CHO, , H, , CO2H, , O, N, H, , N, , H, C, , or, , O, , OH, NH, Bonding 1 : page 10, , N H, or, , NC, , NH2

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Example problems: Indicate all functional groups, do not include alkanes., , ester, (not ketone or ether), , N, , O, , ether, , N, , amine, , alcohol, , amine, , OH, , aromatic (not alkene), , ether, , O, Heroin, , H, , aromatic (not alkene), , O, , O, ketone, , O, O, , alkene, , ester, (not ketone or ether), , O, , aromatic, fluoride CF3, , testosterone, , alkene, , Fluoxetine (Prozac), , •We like to minimize the amount of memorization in this course, but functional groups have names that can’t, be worked out, you just have to know them., • You can think of learning the functional groups a couple of ways, by memorization, or by using them., • There is nothing wrong with some memorization in any course, things that are memorized are recalled more, quickly and accurately, but memorizing names is boring and not the most efficicent use of your time!, • We suggest you learn the functional groups by working the homework problems, solve as many problems, as you can with this list of functional groups, don’t use flashcards!, • This way you will learn many of the names of the functional groups by doing, and the small number left over, that you have not learned yet, those you can memorize!, , 4 Structural Isomers, • Isomers are different compounds with the same molecular formula (we have already seen some of these)., • We meet two kinds of isomers in this course: structural isomers and stereoisomers, to be discussed later., • Structural isomers differ in the order in which the atoms are connected (connectivity of the atoms), the order in, which the atoms are bonded together are different., • The physical and chemical properties of structural isomers are different; they are different molecules., Example 1: Structural isomers for C4H10, , C4H10, , TWO isomers are, , H3C, , CH2 CH2 CH3, , CH, , CH3, , CH3, isobutane b.p. = -11.7°, , butane b.p. = -0.5°, , possible, , H3C, , Example 2: Structural isomers for C5H12, , 4, 3, CH2 CH3, , C5H12, , THREE isomers, are possible, , H3C CH2 CH2 CH2 CH3, , (5) 2, H3C CH, , 1 CH3, , 4 CH3, 3 CH2, , CH3, 1, 2 CH, , (5) CH3, , CH3, H3C C CH3, CH3, , same structure NOT isomers, • Note that for now, the direction in which the bonds “point” is irrelevant, structural isomers are generated by, connecting atoms together in a different order only., • There are thus three different structural isomers with the molecular formula C5H12., Bonding 1 : page 11

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Example 3: Generate all of the structural isomers for C6H14., • A useful strategy is often to start with the longest possible chain, and progressively “branch”:, , CH3, , CH2 CH2 CH2 CH2 CH3, CH3, , same, structure, , CH3, , CH2 CH2 CH, CH3, , CH3, H3C CH, , CH3 generates a new isomer, , CH2 CH2 CH3, , CH3, , CH2 CH, , CH3, CH2 CH3, , CH3, , CH2 C CH3, CH3, , generates a new isomer, , CH3, CH3, , CH, CH3, , CH CH3, , • We find that there are five possible structural isomers., • The direction in which the bonds “point” doesn’t matter (for now), the structures are defined only by the order in, which the atoms are bonded together., Example 4: How many different structural isomers are shown below?, , A, , CH3, , CH3, 1, , CH2 CH, 2, 3, bond, rotation, , 2 bond rotations, converts A into B, , CH2 CH3, 4, 5, , 1, CH3, , 2, 3, CH2 CH, , 4, 5, CH2 CH3, , CH3, , CH3, , B, CH2 CH CH2, 2 3 4, 5 CH, H 3C 1, 3, , H 3C, , 1 2, CH2, , 5 CH, 3, 1, 2, 3, 4, CH3 CH2 CH CH2, CH3, , 3, HC CH3, 5, 4, H3C CH2, , • These are all the same structure, none of them are isomers., • Rotation around single C–C bonds does not generate a new chemical structure, just the same structure drawn a, different way (later we will see that these different ways of drawing the structures are called conformers)., , 5 Formal Charges, • The Lewis structure model attempts to obey the filled shell/octet rule for all atoms by getting as many electrons, as possible into bonds (remember, forming bonds lowers electron energy)., • Sometimes, this can result in structures that disobey the normal rules of valence, resulting in an atom “owning”, either more or less electrons than its normal number of valence electrons., • This results in a mismatch of protons and electrons, which results in the atom having a formal charge., Example 1, , # of valence electrons, O=6, CH3NO2, N=5, , H, H C, H, , O, N, O, , O sees 8, owns 6 electrons, neutral, N sees 8 electrons, owns 4, positive charge, O sees 8, owns 7 electrons, negative charge, , • This is the only reasonable way to draw a Lewis structure for CH3NO2 that gets as many valence electrons into, bonds as possible., Bonding 1 : page 12

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• The “filled shell/octet rule” is obeyed for all atoms, but the normal rules of valence are not (there are four bonds, to the nitrogen instead of three, and one bond to the lower oxygen instead of two)., • The central nitrogen “sees” eight electrons, that is, the ones that are shared in the four bonds, and “owns” four, electrons (half of the shared electrons), but is required to “own” five valence electrons to be neutral., • It “owns” one fewer electron than required to balance the positive charge on the nucleus, its charge is thus +1., • The lower oxygen “sees” eight electrons (six nonbonding and the shared pair in the bond), it “owns” seven, electrons (the nonbonding ones and half the shared pair), but it requires six valence electrons to be neutral., • This oxygen thus “owns” one extra electron than would be required to balance the positive charge on the, nucleus, its charge is thus −1., , The Calculation of Formal Charge Using a Formula, Formal charge = # valence electrons − (# nonbonding electrons) − 1/2 (# electrons in bonds), Example 2:, , H, , valence nonbonding bonding, , H C, C = (4) - (0) - 1/2 (8) = 0, (zero formal charge), , H, , O, , N = (5) - (0) - 1/2 (8) = +1, , N, O, , O = (6) - (6) - 1/2 (2) = –1, valence nonbonding, , bonding, , • Later, we will learn better ways to determine formal charges that do not use this formula, but it may be useful to, use the formula at least for now as a way of making progress in understanding structures., Example 3, , Na 5 valance electrons : 5 electrons to be NEUTRAL, Na formally "owns" 1 electron for each bond = 4, Na has zero non-bonding electrons, "owns" one eless lectron than needed for neutrality : POSITIVE CHARGE, , H, C, , CH2N2, H, , Na, , Nb, Nb = (5) - (4) - 1/2 (4) = –1 (one negative charge), , valence, , nonbonding, , bonding, , • The “filled shell/octet” rule is obeyed in all cases, but the normal rules of valence are not., • Formal charges are associated with structures in which the normal rules of valence are not obeyed., , # nonbonding, , zero formal charge, , Charge = 4 - 2 - 1/2 (4) = 0, , C, H, , H, , # valence, , # bonding, , • This is methylene (you do not have to know this), it is very reactive (its lifetime is less than 0.0000000001, second in solution), but we can still draw a valid Lewis structure for it., • The carbon atom has only two bonds but no charge, not all atoms that disobey the normal rules of, valence have a charge!, , 6 Quantum Mechanical Description of Atomic Orbitals, • The Lewis structure model works well but doesn’t give a proper description of exactly where the electrons are in, the orbitals, we need a much better understanding of what atomic and molecular orbitals really look like so, that we will know where the electrons are in atoms and molecules., • We also need to learn that electron properties are more complex than indicated by simple Lewis structures., • Quantum Mechanics provides a more realistic description of electrons in orbitals., Bonding 1 : page 13

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An Apparent Problem: Quantum Mechanics shows that it is impossible to say exactly where an electron is in, an orbital, only the probability of finding the electron at a particular point in space can be obtained., • However, this is not really a problem, since it is a natural consequence of the wave nature of electrons. Without, understanding this wave nature of electrons we cannot properly understand orbitals and bonding., • Quantum Mechanics describes atomic and MOs in terms of a wave function equation., • The wave function is given the Greek letter capital ψ (pronounced Psi), and for AOs has the form:, , wavefunction, symbol, , quantum numbers, Ψ = f ( n, l, m, s, r, ...), , is a function of, these parameters (and more), , distance from nucleus, Here, n, l, m, and s are the quantum numbers you encountered in general chemistry:, • n = principal quantum number, • l = angular momentum quantum number, • m = magnetic quantum number, • s = spin quantum number, • Another parameter that is important for us is the distance of the electron from the nucleus, r., • Wave functions exhibit wave behavior, that is, just like any wave they can be positive, negative, or zero., • Solving the wave equation for the different quantum numbers gives the different AOs., • The probability of finding an electron at a particular distance from the nucleus, P(r), is given by the value of the, wave function squared, this is our answer to where are the electrons in an orbital!, , P(r) = Ψ (r) 2, • Quantum Mechanics is highly mathematical, however, we do it in pictures, as a plot, or a pictorial, representation of the wave function versus distance from the nucleus, which is much easier!, At this point, you should be confused since all of this is difficult to understand without examples., , 6.1 2p Atomic Orbital, • From general chemistry, we know the values of the atomic quantum numbers n, l, m, and so on . . ., , Ψ = f (n, l, m, s, r, ...), , in this case n = 2, l = 1, m = 0, +1, -1, , • n = 2 means the second shell., • l = 1 in this case means a p AO and not an s AO., • m = 0, +1, −1 means that there are three p AOs, px, py, and pz., • Let’s start with the familiar “hourglass” shape for the 2p AO:, , probability zero here, WRONG!!, , nucleus, , r, , r, , distance from, nucleus, , probability largest here, orbital shape should NOT be interpreted to, illustrate a "figure of 8" movement of electrons, , orbital shape SHOULD be interpreted to illustrate, the probability of finding the electron, , • The “electron cloud” picture from general chemistry relates the idea of electron density to the probability of, finding the electron, the higher the probability of finding the electron, the “denser” the apparent electron cloud., "cloud" picture of, the electron density, , density zero here, density largest here, , • We can derive the shape of the wave function for the 2p AO by making a plot of probability of finding the, electron, P(r) = wave function squared, as a function of distance from the nucleus, r., Bonding 1 : page 14

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• Then, take the square root to get a plot of the wave function., picture of P(r) = Ψ2, , probability zero here, NOT a node, , 10%, r, , never goes to zero, plot of Ψ, looks like a wave!, , r, , picture of Ψ, shows phase information, , 0, , r, , never reaches zero, here, , goes through zero, 1 PLANE node!, , Ψ, , 0, , r, , AND, Ψ, , need to consider BOTH, wavefunctions, , r, picture of Ψ, shows the OTHER phasing situation that, must also be considered, , max away from nucleus, , P(r) = Ψ2, , 0, , r, 1 node, , • The orbital shape given by the wave function defines an arbitrary “boundary” beyond which the probability of, finding the electron is small., • The wave function squared gives the probability of finding the electron as a function of distance from the, nucleus (P(r)). This probability plot contains no phase information since a probability (and indeed any number, that is squared) can only be positive or zero, never negative., • The plot of the wave function looks like a wave, it gives phase information associated with the wave, that is,, the positive and negative regions due to the wave behavior., • For the 2p AO, the wave function has one node at the nucleus, that is, the electron is never at the nucleus., • There are different kinds of nodes, the p AO has a plane node., PLANE NODE, , • The wave function is the square root of the probability plot, thus we don’t know which phase is positive and, which is negative, which is why we need to consider possibilities, shown in both plots and both pictures. Positive, and negative are arbitrary in wave functions, thus we distinguish the different phases by shading (or coloring), in the picture rather than assigning an absolute positive or negative sign., • Understanding this wave nature is essential for building MOs!, , Bonding 1 : page 15

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6.2 2s Atomic Orbital, in this case n = 2, l = 0, m = 0, , Ψ = f (n, l, m, s, r, ...), , • The quantum numbers tell us about the wave nature of the orbitals, the number of nodes is the value of the, principal quantum number n − 1, therefore, a 2s AO (n = 2) must have one node., , zero electron density, (sperical node), , WRONG!!, , • The familiar spherical orbital should not be interpreted to illustrate a circular movement of electrons., • The “cloud” picture of electron density reveals a spherical region within the orbital where the electron density is, zero, that is, where the wave function must have a node., , picture of P(r) = Ψ2, no phase information here, , P(r) = Ψ2, , highest closest to the nucleus (r = 0), , 10%, , r, , never goes to zero here, r, , electron probability zero here, , Ψ, goes through zero, ONE (spherical) node, , plot of Ψ, looks like a wave!, , 0, , never goes to zero, , picture of Ψ shows phase, information, , ONE spherical node, , change in "phase" at the node, Ψ, Again, we need to consider both, wavefunctions, picture of Ψ, shows the other phasing situation that, must also be considered, , r, , AND, r, ONE spherical node, , need BOTH pictures, , • The wave function squared gives the probability of finding the electron as a function of distance from the, nucleus (P(r)). The probability plot contains no phase information since a probability (in fact any number, squared) can only be positive or zero, never negative., • The wave function for the 2s AO has a spherical node that surrounds the nucleus, the probability of finding the, electron this distance from the nucleus is zero at all distances from the nucleus defined by the sphere., Bonding 1 : page 16

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z, , z, , x, , y, , spherical, node, x, , y, , • The wave function is the square root of the probability plot, thus we don’t know which phase is positive and, which negative, which is why we need to consider possibilities, shown in both plots and both pictures. Positive, and negative are arbitrary in wave functions, thus we distinguish the different phases by shading (or coloring), rather than positive or negative signs. The colors are arbitrary, just like shading and nonshading., , 6.3 1s Atomic Orbital, in this case n = 1, l = 0, m = 0, , Ψ = f (n, l, m, s, r, ...), , • The quantum numbers tell us about the wave nature of the orbitals, the number of nodes is the value of the, principal quantum number n − 1; therefore, a 1s AO (n = 1) must have zero nodes., • The familiar spherical orbital should not be interpreted to illustrate a circular movement of electrons., • The “cloud” picture of electron density, zero nodes in this case., , electron, “cloud” picture, , WRONG!!, , picture of P(r) = Ψ2, no phase information here, , P(r) = Ψ2, plot of Ψ, looks sort of like a wave!, , highest closest to the nucleus (r = 0), , r, , positive, r, , never goes to zero, , 10%, 0, , AND, , need to consider BOTH, wavefunctions, , r, , 0, Ψ, , picture of Ψ shows phase, information, , never goes to zero, , 10%, , r, , never goes to zero, Ψ, , r, negative, , 0, , r, , picture of Ψ, shows the OTHER phasing situation that, must also be considered, , Bonding 1 : page 17

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6.4 So, Why Are There Orbitals?, • The wave nature of electrons is an odd concept, it doesn't seem to make sense in our macroscopic world;, however, the wave nature of electrons is the most important factor that determines their properties!, • If we accept that electrons have wave nature, there must be orbitals, there is no other explanation for orbitals., • Consider the sound waves related to vibration of a string:, , a plucked string, , fundamental, harmonic, , boundary, , only, CERTAIN, waves, allowed, , boundary, standing waves, , • Only the fundamental wave and its harmonics are observed, because the boundary conditions imposed by the, length of the string requires these to be standing waves, only certain standing waves are allowed., • Now consider the electron waves defined by the attraction of the negatively charged electron and a, positively charged nucleus:, standing wave, (orbital )"allowed", only, CERTAIN, STANDING waves, allowed, wave (orbital), NOT "allowed", the "ends" don't meet!, • Because of the boundary conditions of the charge on the nucleus and the requirement for proper overlap of the, wave structure where it “meets,” only certain standing waves are allowed., • The allowed standing waves represent the allowed ways that electrons can be distributed around a nucleus and, their associated allowed energies., • These standing waves are the orbitals., • If the electron does not have wave nature, then there are no standing waves, there are no orbitals, and, there is no other way to explain the existence of orbitals without taking the wave nature of the electrons into, account., , only, CERTAIN, waves are allowed, these are the orbitals, , 1s, , 2s, , 2pz, , 2px, , 2py, , 7 The MO Theory Model of Bonding, 7.1 Introduction to Localized MO Theory: The Simple Example of H2, • Electrons in atoms are in AOs, electrons in molecules are in MOs., • MOs can be constructed by combining the familiar AOs into MOs., • A completely accurate MO theory would take all of the AOs associated with all of the atoms in the molecule,, which is impossible to do without complicated math, instead, we will use an approximate method to generate, localized MOs., Bonding 1 : page 18

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• The approximate method is called a Linear Combination of Atomic Orbitals (LCAO)., • The LCAO model generates approximate MOs that are localized on individual bonds. Localized MO theory, “pretends” that MOs are localized between the two atoms that “make” the bond. This is the “next level” up from, Lewis structures in models of bonding and structure., • LCAO is very quantitative, but we will do it qualitatively, using pictures., The LCAO “Rules” to make a pictorial localized MO that describes a bond between two atoms are the following:, 1. Choose the appropriate two AOs to make the bond (one from each atom)., 2. Draw pictures of the two wave functions that correspond to these AOs., 3. Make two new MOs by combining (overlapping) the two AOs both in-phase and out-of-phase., • This is why we need the phase information contained in the wave functions to make MOs., • Note that when the AOs are combined in a molecule, we must get both in-phase and out-of-phase overlap, of the wave functions, which generates two new orbitals in the molecule (orbitals can’t disappear, two AOs, must make two MOs)., • Let’s look at molecular hydrogen as an (easy) example:, H, , +, , H H, , H, , molecular orbital, , atomic orbitals, , • Overlap the wave function of one 1s AO from each hydrogen atom to make the new MOs., , Bonding, H 1s A.O. Ψ, , x, , x, , Anti-Bonding, H 1s A.O. Ψ, , H 1s A.O. Ψ, , combine (overlap) IN PHASE, , x, , x, , H 1s A.O. Ψ, , combine (overlap) OUT OF PHASE, , • When the AOs are close enough that overlapping (combining) in-phase occurs, then overlapping (combining), out-of-phase also happens simultaneously, it is an unavoidable consequence of the wave nature of the orbitals., • The two situations are called bonding (in-phase) and antibonding (out-of-phase)., • This is how the two AOs are combined to make the two MOs in molecular hydrogen., • Each hydrogen atom provides one electron each to make the bond., • One of these two new MOs will contain the two valence electrons in the new covalent bond., Energy diagram for formation of localized MOs, overlap wavefunctions OUT of phase, Energy, , ANTI-BONDING, σ* M.O., , =, , x, , DESTRUCTIVE INTERFERENCE, , x, , x, , bring, together, , x, , 1s A.O., , or, 1s A.O., x, , bring, together, , σ M.O., BONDING, x, , x, , = negative phase, (arbitrary), , =, , x, , x, , Bonding 1 : page 19, , x, , = positive phase, (arbitrary), = nucleus, , = electron, , CONSTRUCTIVE, INTERFERENCE, overlap wavefunctions IN phase

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• Bringing the atoms together to make a bond results in the overlap of the AOs., • Because of the wave nature of the AOs, when they overlap, they do so “in-phase” and also “out-of-phase,” at, the same time, whether we like it or not!, • Overlap of the AO wave functions in-phase results in constructive interference of the wave structures to, generate a new wave structure, which is largest where constructive overlap is greatest, that is, between the, nuclei. The result is a localized σ bonding MO (σ pronounced as sigma). This is where the two electrons “go.”, • An electron in the bonding σ–MO is lower in energy than an electron in either of the two AOs that were used to, construct it, both electrons are thus lowered in energy, this is the definition of bonding., • Corresponding overlap of the wave functions out of phase results in the destructive interference of the wave, structures to generate another new wave structure. This new structure has a node where the destructive, overlap occurs, that is, between the nuclei, the result is a localized σ*–antibonding MO. The electrons do not, “go” there, since they would be higher in energy., Antibonding Orbitals: Antibonding orbitals must form if bonding orbitals are formed, the combination of two, AOs must generate two MOs, orbitals can’t disappear., There is nothing wrong with having antibonding MOs with no electrons in them! Think about a hydrogen, atom. It has a 2p AO, for example, it just doesn’t have any electrons in it, it is an empty orbital the atom has., Antibonding orbitals are empty orbitals that molecules have. Antibonding orbitals become important later when, we come to chemical reactions., , σ, , x, , BOTH are symmetrical, with respect to the, internuclear axis, , x, , Ψ for σ-bonding M.O., , x, , σ∗, , Ψ for σ∗-ANTIbonding M.O., , • Both of the orbitals are symmetrical with respect to the internuclear axis (they are both the same above and, below the internuclear axis), which is why they are called σ orbitals., • This picture also gives us the real explanation for why electrons form bonds., • The conventional explanation is that in the MOs the charges are “balanced,” each electron “sees” two nuclei, in, essence this an electrostatic argument, but it doesn’t really make sense because in the MO the nuclei are very, close to each other too!, , H•, lower volume, higher kinetic energy, , +, , H•, , H–H, larger volume, lower kinetic energy, , +, , Ψ for σ-bonding M.O., , Ψ for 1s A.O., , • The best explanation is that the new MO is larger (has a larger volume) and is less curved than the AOs. An, electron moves more slowly in a larger volume orbital, the electrons thus have lower kinetic energy in the, bonding MO., Now, square the wave function to give the probability of finding the electrons in the MO., , H, σ M.O., , H, , x, , x, , Ψ, , square, x, , x, , Highest electron density between nuclei,, fits our understanding of "balanced" charges, , Ψ2, Bonding 1 : page 20

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square, σ* M.O., , x, , x, , x, , ZERO electron density between nuclei, no, shielding of nuclei, high in energy, , x, , Higher Energy Antibonding M.O., , Ψ2, , Ψ, , 7.2 Valence Shell Electron Pair Repulsion Determines (Almost) Everything, • Before we get to the more complex MOs, we need to learn how electrons are arranged in organic molecules., , Everything Starts with Valence Shell Electron Pair Repulsion, • The orbitals you learned in general chemistry for atoms are appropriate for isolated atoms., • But most atoms aren’t isolated like that, they are associated with other atoms, for example, in molecules, where, they are surrounded by other atoms, electrons and protons, that is, other charges and are in bonds., • The AOs respond and change as a result of making bonds to the surrounding atoms., • When an atom is in a molecule, the electrons around the atom arrange themselves to make bonds and keep out, of each other’s way, that is, they minimize electron repulsion., • The AOs the atom had when it was an isolated atom are don’t work when in a molecule., , valence, 1s, 2s and 2p, C atomic orbitals, work here, isolated carbon atom, 1s2 2s2 2p2, , "dashed bond", points away from viewer, , H, , these atomic, orbitals can’t, work here, , C, H, , H, , H, , "wedged bond", points towards viewer, , carbon atom in a molecule, CH4, methane, • This means that we can’t use the normal AOs in LCAO to make the MOs. We need to use rearranged, (hybridized) AOs and it is the electron repulsion that allows us to work out how to do this., • Valence Shell Electron Pair Repulsion (VSEPR) takes all of this into account!, • New Notation: The wedged and dashed bonds above are a method of describing a three-dimensional structure, on a two-dimensional piece of paper that is very useful for the tetrahedral geometry (see below)., • VSEPR Accounts for the locations of the surrounding atoms, and the bonding and nonbonding electron pairs., • Determines what each atom needs to “do,” and where in space the electrons around each atom need to go., VSEPR thus determines hybridization, how the AOs change, which determines how the MOs are built., , VSEPR with Four Valence Electron Domains, • The following structures have four sets or domains of Valence Shell Electron Pairs around C, N, and O atoms., • The four sets or domains of electron pairs avoid each other as much as possible to minimize electron repulsion., the 2s and 2p valence shell atomic orbitals won’t work for these C, N and O atoms, they are surrounded by other atoms (charges) - 4 electron domains, the atomic orbitals in the C, N and O must respond by “hybridizing”, , R, ~109°, , ••, N, , R, B R, ~109° R, R, 4 electron domains, 4 electron domains, in the 4 single bonds, 3 bonding domains, 1 non-bonding domain, tetrahedral electron geometry, tetrahedral electron geometry, all four angles are the same, three angles are the same, tetrahedral molecular geometry trigonal pyramidal molecular geometry, A R, , C, , R, , Bonding 1 : page 21, , ••, , non-bonding electrons, R, ~109°, O, C, ••, R, , 4 electron domains, 2 in bonds, 2 non-bonding, tetrahedral electron, geometry, one bond angle, bent molecular geometry

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• The locations of atoms (nuclei) can be well defined, but the locations of electrons cannot., • Each structure has four electron domains, all four have tetrahedral electron geometry due to VSEPR., • The molecular geometry of structure A (above) can be defined by the positions of the atoms as tetrahedral., • For structure B, the molecular geometry can only describe the positions of the four atoms: trigonal pyramidal., • For structure C, the molecular geometry can only describe the positions of the H–O–H atoms, thus bent., , VSEPR with Three Valence Electron Domains, • The following structures have three sets or domains of valence electrons around the indicated C and N atom, • The three electron domains separate from each other as much as possible due to electron repulsion., the 2s and 2p valence shell atomic orbitals won’t work for these C and N atoms, the atomic orbitals must respond to the 3 new electron domains by “hybridizing”, , H, , H 120°, A, , C, , C, , H, , H, , 120° 3 electron, domains, , HH, C, , H, H, C, C H, B H ~120°, 3 electron, H C, C, H domains, H N, , 2 domains in C-H single bonds, 1 domain in N-C single bond, 1 domain in a C=C double bond, 1 domain in a N=C double bond, all 3 angles roughly the same, 1 non-bonding domain, trigonal planar molecular geometry, bent molecular geometry, • The electron geometry at each of the earlier circled atoms is trigonal planar., • The molecular geometry at carbon in A (above) is defined by the positions of four atoms as trigonal planar., • For structure B, the molecular geometry at nitrogen can only describe the positions of the C–N=C atoms: bent., • Knowing there are three electrons domains around an atom helps in building the MOs., , VSEPR with Two Valence Electron Domains, • The following structures have two sets or domains of valence electrons around the indicated C and N atoms., • The two electron domains separate from each other as much as possible due to electron repulsion., the 2s and 2p valence shell atomic orbitals won’t work for these C and N atoms, the atomic orbitals must respond to the 2 new electron domains by “hybridizing”, , 180°, H, , C, , C, , H, , 2 electron, domains, , H, , C, , 2 electron, domains, , N, , 1 domain in C-H single bond, 1 domain in N non-bonding pair, 1 domain in a C–C triple bond, 1 domain in a C–C triple bond, angle = 180°, the geometry at N can't be defined, linear molecular geometry, • The electron geometry at each of the earlier circled atoms is linear., • The molecular geometry at carbon of structure A (above) is defined by the positions of the two atoms as linear., • The molecular geometry at the nitrogen cannot be defined., , Practice Using VSEPR: VSEPR analysis of geometries and angles considers one atom at a time in a molecule., • For the circled atoms, give the electron geometry, the molecular geometry and the relevant bond angles., , H H ~120°, H, 3 electron domains, C, O, C, C, H, ~120° trigonal planar electron geometry, H ~120°C H, trigonal planar molecular geometry, H H, , Bonding 1 : page 22, , H, , ~120°, H, 3 electron domains, trigonal planar electron geometry, trigonal planar molecular geometry, H, , C

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H, 4 electron domains, tetrahedral electron geometry, tetrahedral molecular geometry H, , H, N, H, , C, C, CH3, H ~120°, 3 electron domains, trigonal planar electron geometry, bent molecular geometry, , all, H angles, ~109°, , 7.3 Methane as a Simple Organic Molecular Structure, Everything Starts with VSEPR, • There are four C–H bonds around the central carbon atom, there are thus four electron domains that need to, be as far apart from each other as possible to lower the total electron energy., • The lowest energy molecular geometry is tetrahedral at the central carbon atom., H, H, H, ~1.1Å, ~109°, CH4 H C H, 4 electron domains, C, H, H, C, tetrahedral, H, H, H, H, methane, H, • The central carbon thus needs four identical AOs to construct four identical C–H bonds., • These four AOs must be separated by ~109° in order to generate the correct structure and geometry., • The Valence Shell electrons in atomic carbon are associated with the 2s and the three 2p AOs., • Methane cannot use these orbitals to make the four identical C–H bonds since these four orbitals are obviously, not identical and they are not separated by angles of ~109°., , these orbitals are at 90°, they can’t make these bonds, , 1s2, , 2s2 2p2, , valence electrons, , 2px, , 2py, , 2s, , 2 electrons 1 electron, , 2pz, , 1 electron, , X, , carbon, , H, , H, , C, , H, , 0 electrons, , ~109°, , H, , • Hybridize (mix together) the four different valence AOs that carbon has in an isolated atom to make four new, identical hybridized (mixed) AOs that can make the required four identical bonds., • Each new hybrid orbital consists of an equal mixture of a 2s, a 2px, a 2py and a 2pz orbital. Mixing the four 2s,, 2p, 2p, and 2p AOs must make four new hybridized sp3 orbitals (orbitals can’t disappear!)., , 4 valence orbitals the isolated atom has, , four identical sp3 hybrid (mixed) orbitals, , Energy, , valence orbitals the carbon NEEDS in methane (molecule), 2px, , 2py, , 2pz, , Ψ, , mix, ALL 4, , 2s, , sp3, , sp3, , sp3, , sp3, , and, , Ψ, Ψ, , EACH sp3 = (25% 2s + 25% 2px + 25% 2py + 25% 2pz), sp3 = (25% 2s + 75% p), , Ψ, , • All of the valance orbitals are mixed, 2s, 2px, 2py, and 2pz, not just the orbitals that are occupied, we are mixing, orbitals, not the electrons., Bonding 1 : page 23

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• Each orbital is one part s and three parts p: that is, we have four new hybridized sp3 AOs that are used to, make four identical C–H bonds., , H, , C, , 90°, , 109°, , these orbitals, become these, orbitals, , isolated carbon atom, 2s and three 2p orbitals, , H, , C, H, , H, , carbon atom in CH4, four sp3 hybrid orbitals separated by 109°, , • The sp3 notation comes not from the number of electrons, but from the number of orbitals that are mixed, together, three 2p orbitals are mixed with one 2s orbital., • An sp3 AO has 25% 2s character and 75% 2p character; therefore, an electron in an sp3 hybrid AO is lower in, energy than one in a 2p AO, but significantly higher in energy than one in a 2s AO., • Carbon will use one of its four valence electrons in each of the sp3 hybridized orbitals to make the four bonds., • Each C–H bond requires one sp3 hybrid AO from C and one 1s AO from H., • Combine a 1s AO from H with an sp3 AO from C according to the pictorial LCAO “rules” to make a localized MO, for each bond., , cancel, Energy, , Ψ, Anti-Bonding σ* M.O., , Ψ, , out of phase overlap makes antibonding M.O., DEstructive inteference, H, , x, , node, , C sp3, x, , Bonding σ M.O., , H 1s, , AND, , Ψ, , C, , H, , Ψ, , H, , H, Ψ σ-bonding, M.O., , in-phase overlap makes bonding M.O., CONstructive inteference, , Ψ, larger due to addition, , • The hydrogen 1s AO is lower in energy than the carbon sp3 AO., • The node in the MO is at the carbon nucleus, because there was one there in the sp3 orbital., • Both of the MOs are symmetrical with respect to the internuclear axis, therefore bonding MO is the σ orbital, and, the antibonding MO is the σ* orbital., • The shading (phasing) of the orbitals is arbitrary, a diagram can also be drawn with opposite shading:, , Energy, , σ* M.O., , this diagram is EXACTLY, Ψ σ* M.O., EQUIVALENT to the one above, the, phasing (shading) is arbitrary!, anti-Bonding, H, 2 nodes, , C sp3, x, , Ψ σ M.O. Bonding, , H 1s, σ M.O. Bonding, Bonding 1 : page 24, , AND, , H, , C, , H, , H, Ψ σ-bonding, M.O.

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• Why doesn't the hydrogen rehybridize? The H atom only has to accommodate one bond, this is a much smaller, perturbation than the carbon experiences, hybridization is not necessary for atoms that only make one bond., Summary of what we just did:, , sp3 A.O. H, C H, 1s A.O. H, H, , H, C H, H, , H, , pictures of the Ψ for the A.O.s, overlapping IN PHASE, (these do NOT exist in the molecule), , H, H, , picture of Ψ2 for the σ-bonding, M.O., "where" the electrons are in, the C-H bond, , picture of the Ψ for the σ-bonding, M.O. made by combining the A.O.s, IN PHASE, , sp3 A.O. H, C H, 1s A.O. H, H, , C H, H, , H, , H, , C H, H, , H, , H, , picture of Ψ2 for the σ*-ANTIbonding, M.O., "where" the electrons WOULD BE, if they were in the anti-bonding M.O., , picture of the Ψ for the σ*ANTIbonding M.O. made by, combining the A.O.s OUT OF PHASE, , pictures of the Ψ for the A.O.s, overlapping OUT OF PHASE, (these do NOT exist in the molecule), , C H, H, , • The orbital is the “largest” between the C and H, consistent with shared electrons between the two atoms., • When the AOs overlap, they do so to make the strongest bond they can, the best overlap of the 1s AO is with, the larger lobe of the sp3 AO., , 7.4 Structures with Double Bonds: Ethylene as a Simple Example, •Start With VSEPR: There are three electron domains around each carbon atom that need to get as far apart, from each other as possible, the geometry is this trigonal planar and the bond angles are all ~120°., • An additional consideration is that the two bonds in the C=C double bond cannot be the same, two pairs of, electrons cannot occupy the same area of space due to electron repulsion., , ~120°, , H, , C2H4, , 3 similar σ, bonds, , Ethylene, , C, H, , H, , 3 electron domains, trigonal planar geometry (VSEPR), sp2 hybridized, , C, H, , ~120°, , 1 "extra" bond, , • To make three similar bonds that are separated by ca. 120°, each carbon needs three AOs on that are the same, and that are separated by ca. 120°., • The carbon also needs to make one bond that is different, the second of the bonds in the C=C double bond., • The valence AOs of an isolated carbon (2s, 2px, 2py, and 2pz) cannot do this, certainly not the 120° bond, angles, the carbon will need to hybridize (mix) its valence orbitals to make new hybridized valence orbitals., , four valence orbitals the carbon NEEDS in ethylene (molecule), 3 sp2 + 1 p, , 4 valence orbitals atom has, Energy, 2p, , 2p, , 2p, , Ψ, , mix, , each is 33% s + 66% p, Ψ, , THESE 3, 2s, , Ψ, , sp2, , sp2, , MAKES, , sp2 three sp2 hybrid, orbitals, , used to make 3 σ-bonds, Bonding 1 : page 25, , and, , 2p, , Ψ, , one unhybridized, p orbital, used to make, π-bond

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• Combine (hybridize or mix) one 2s and two 2p to make three new hybridized sp2 orbitals., • Orbitals can’t disappear, mixing three AOs must make three new AOs., • An sp2 AO has 33% 2s character and 66% 2p character, therefore, an electron in an sp2 hybrid AO is lower in, energy than one in a 2p AO, but significantly higher in energy than one in a 2s AO, and is lower in energy than, one in a sp3 hybrid AO., , these orbitals, become these, orbitals, , C, , 90°, isolated carbon atom, 2s and three 2p orbitals, , H, , H, , 120°, , C, , C, H, , H, , carbon atom in CH2CH2, 3 sp2 hybrid orbitals and 1 unhybridized p orbital, , C–H Bonds, , • Combine a 1s AO on H with an sp2 AO on C to generate the localized MOs associated with the C–H bonds., • Draw the wave function squared of the appropriate localized MO to indicate the locations of the electrons., , 1s H, , H, C, , sp2, , C, , H, , H, , H, C, , C, , node on C H, , H, , pictures of the Ψ for the A.O.s, overlapping IN PHASE, (these do NOT exist in the molecule), , H, , H, C, , C, , node on C H, , H, , H, Ψ2, , picture of the Ψ for the σ-bonding, M.O. made by combining the, A.O.s IN PHASE, , picture of, for the σ-bonding, M.O., "where" the electrons are in, the C-H bond, , C–C σ Bond, , • Combine an sp2 AO on C with an sp2 AO on C. Draw the wave function squared of the localized MO., , H, , H, C, , C, , H, , H, , H, C, H, , H, , pictures of the Ψ for the A.O.s, overlapping IN PHASE, (these do NOT exist in the molecule), , C, , H, , H, C, , nodes here, H, , H, , C, H, , picture of Ψ2 for the σ-bonding, M.O., "where" the electrons are in, the C-C sigma bond, , picture of the Ψ for the σ-bonding, M.O. made by combining the, A.O.s IN PHASE, , The Second C–C Bond, • What about the other (“extra”) bond between the carbons? Use the one p AO on each carbon that is “left over”, after the hybridization to make a “pi” bond (π bond)., • The p AOs on each carbon are orthogonal to the plane that contains the sp2 hybrid AOs, to see them properly, we need to look at the molecules from the “side.”, , H, looking, from the "top", , C, H, , p A.O., , 3 sp2 hybrid A.O.s, , H, p A.O., , C, , H, , H, C, H, , H, , C, H, , looking from the side, , need to look from the SIDE, , • Now, we can see how the second bond in the double bond is built from the two p AOs., , Bonding 1 : page 26, , looking, from the "side"

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H, , H, C, H, , H, , H, , C, , C, , C, , H, , H, , pictures of the Ψ for the A.O.s, overlapping IN PHASE, (these do NOT exist in the molecule), , H, , H, C, , C, , H, , H, , picture of the Ψ for the π-bonding, M.O. made by combining the A.O.s, IN PHASE, , H, , picture of Ψ2 for the π-bonding, M.O., "where" the electrons are in, the C-C π-bond, , 2 PLANE nodes, (higher energy orbital), Energy, , Anti-Bonding, , C C, , unsymmetrical with respect to, internuclear axis: π*-M.O., , π* M.O., , C or C, , C, p A.O., , p A.O., π M.O., Bonding, , unsymmetrical with respect to, internuclear axis: π-M.O., , C C, , 1 PLANE node, The following diagram shows the location of all of the electrons are in the C=C double bond (in drawings of, wave functions squared), that is, the σ and the π bonds that together make up the double bond., , 2 electrons in, C–C π M.O., above and below, plane, , H, , H, , H, , C, , C, , H, , 2 electrons in, C–C σ M.O., in the plane, , • All six atoms in (horizontal plane) the π bond is above and below the plane., • The electrons in the π and σ MOs occupy different regions of space., • The electrons in the π MO are further from nuclei, held less tightly, and are higher in energy., • The π bond is weaker than the σ bond (electrons are higher in energy in the π bond)., , 7.5 Structures with Triple Bonds: Acetylene as a Simple Example, • Start with VSEPR: There are two electron domains around each carbon atom that need to get as far apart from, each other as possible, the molecular geometry is linear and the bond angles are all ~180°., • An additional consideration is that the three bonds in the C–C triple bond cannot be the same, three pairs of, electrons cannot occupy the same area of space due to electron repulsion., • To make two similar bonds that are separated by 180°, each carbon needs two AOs separated by 180°., , C2H2, Acetylene, , ~180° 2 domains of electrons, sp, C, C, H, H, sp, linear, Lewis structure, , • The carbon also needs to make two π bonds that are different., Bonding 1 : page 27

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• The valence AOs that an isolated carbon atom (2s, 2px, 2py, and 2pz) will not be able to do this, the carbon will, need to hybridize (mix) its valence AOs to make new hybridized valence AOs., • Combine (hybridize or mix) one 2s and one 2p to make two new hybridized sp orbitals., • Orbitals can’t disappear, mixing two AOs must make two new AOs., • An sp AO has 50% 2s character and 50% 2p character; therefore, an electron in an sp hybrid AO has an energy, that is exactly between those of electrons in 2p and 2s AOs, and, has an energy that is lower than those of, electrons in sp2 or sp3 hybrid AOs., , 4 valence orbitals atom has, , four valence orbitals the carbon NEEDS in acetylene (molecule), 2 sp + 2 p, , Energy, 2p, , 2p, , 2p, , each is 50% s + 50% p, , mix, , 2s, , Ψ and, sp, two sp hybrid orbitals, , sp, , THESE 2, , 2p, , Ψ, , 2p, , two unhybridized, p orbitals, used to make, 2 π-bonds, , used to make 2 σ-bonds, , C–H σ Bond, • Combine a 1s AO on H with an sp AO on C to generate the localized MOs associated with the C–H σ bond., • Draw the wave function squared of the appropriate localized MO to indicate the locations of the electrons., , H, , C, , sp + 1s, , sp + 1s, , sp 1s, C, H, , H, , overlapping atomic orbitals, do not exist in the molecule, , C, , H, , C, , H, , C, , H, , C, , Ψ2 for C-H σ-bonding M.O., , Ψ for C-H σ-bonding M.O., , C–C Triple Bond, There are two electrons in a localized σ MO constructed by combining a 1s AO from H and an sp hybrid AO on, each C., , sp + sp, , sp + sp, H, H, C, C, overlapping atomic orbitals, do not exist in the molecule, , H, , C, , sp + sp, , C, , H, , H, , C, , C, , H, , Ψ2 for C-C σ-bonding M.O., , Ψ for C-C σ-bonding M.O., , • There are two electrons in one π MO constructed from a 2 p AO on each C (top and bottom, wave function, squared), it is easy to draw the first MO because it can be drawn in the plane of the paper., , Ψ for C-C, π-bonding M.O. #1, , H, , C, , C, , H, , H, , C, , C, , H, , Ψ2 for C-C, π-bonding M.O. #1, , • There are two electrons in a second π MO constructed from a 2 p AO on each C (one lobe on each “side,” wave, function squared), this second π MO is at right angles to the first one, it’s difficult to draw!, , Ψ for C-C, π-bonding M.O. #2, , H, , C, , C, , H, , H, , Bonding 1 : page 28, , C, , C, , H, , Ψ2 for C-C, π-bonding M.O. #2

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π-M.O. #1, H, , C, , molecular model showing the 2 πbonds in acetylene, H, , C, , π-bond 1, π-bond 2, , π-M.O. #2, , the bonds are orthogonal (90°) to, minimize the overlap of the 2 sets, of electrons in the 2 π-bonds, , 7.6 The Electrons in the Molecule Methanol, for example, • Start with VSEPR, , H, , CH3OH, , H, , methanol, , 4 electron domains, tetrahedral electron geometry, sp3 hybridized, bent moelcular geometry, , O, , C, , H, H ~109°, , • There are four electron domains around the oxygen, one in the O–H bond, one in the O–C bond and two, associated with the two nonbonding (lone) pairs., The electron domain geometry is tetrahedral, the molecular geometry is bent., (A detailed consideration of the local symmetries of the orbitals for the two nonbonding pairs of electrons tells us, that they can’t actually be exactly the same, they can’t have the same energy, but that is an advanced concept, and in this class we will assume that they are the same and that the oxygen atom is truly sp3 hybridized), Combine an sp3 hybrid AO from C and an sp3 hybrid AO from O to make the MOs associated with the C–O bond:, , C(sp3), , H, H, , O, , C, , O(sp3), , H, , H, , C(sp3), , H, , H, , A.O.s used to form the M.O., (these do not exist in the molecule), , O, , C, , O(sp3), H, , H, C(sp3), , H, , H, Ψ2, , Ψ for C-O σ-bonding M.O., , O, , C, , O(sp3), H, , H, , for C-O σ-bonding M.O., , • The bonding MO is formed by in-phase overlap of the AO.s., • There is no phase information in the wave function squared picture because a probability can only be positive or, zero, never negative, the wave function squared is not a wave, it does not have nodes., • There is zero probability of finding the electrons in the bonds at the C and O nuclei because there was a node at, these nuclei in the original hybrid sp3 AOs. The MO is larger toward the oxygen (see later)., To draw the wave function for the localized MO that describes the O–H bond, we need to include phase, information in the wave function:, , H, H, , 3, , sp A.O. on O and 1s A.O. on H, , O, , C, , H, H, , • Draw the wave function for the nonbonding electrons on the oxygen atom, we need to include phase information, in the wave function:, , oxygen hybrid orbitals, , sp3 A.O. on O, H, O, C, H H, , sp3 A.O. on O, H, , sp3, sp3, sp3, sp3, , • The table explains exactly how the oxygen atom uses all of the hybrid orbitals., Bonding 1 : page 29, , O-C σ-bond, O-H σ-bond, non-bonding pair #1, non-bonding pair #2

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7.7 The Electrons in the Molecule Chloroform, for Example, • VSEPR is a bit different this time, because the Cl makes only one bond., • Hybridization does not occur when only one bond is made, the perturbation to the AOs that results by bringing, only one new atom to make a bond is not large enough to rehybridize the AOs, the Cl atom does not hybridize., , H, CH3Cl, chloroform, , H, , C, , sp3, , H, , only ONE BOND, not hybridized!, 3s, 3px, 3py, 3pz, , Cl, , • The carbon makes four bonds, and the hybridization is therefore sp3 (as usual)., • Because the chlorine only makes one bond, it does not hybridize, it uses a simple p AO to make the σ bond., , H, Ψ for σ∗-antibonding M.O., , H, , C(sp3) + Cl (p), , C, , H, Cl, , H, , H, , C, H, , Cl, , Ψ for σ-bonding M.O., C(sp3) + Cl (p), , 7.8 Summary of Hybrid AOs, Energy, , Atomic Orbital, , what they really, look like, , what we probably what we will draw, because it is easier, should draw, , 2p, , (2)sp3 25%s+75%p, (2)sp2 33%s+66%p, (2)sp 50%s+50%p, energy midway between, s and p orbitals, , 2s, , • The hybrid orbitals are really 2sp, 2sp2 etc., but we usually drop the 2 (which is why it is in parenthesis above)., • The actual hybrid orbitals actually have quite complex shapes, as shown in the table., • With increasing p-character, the energy of electrons in the hybrid orbitals increases., • The sp AO is 50% s + 50% p, its energy is midway between those of the pure s and p AOs., Let’s talk about the antibonding orbitals. At this point, many students are confused about these. What is, the point of antibonding orbitals, what are they used for?, We haven’t used antibonding orbitals much yet, but you need to understand that they form, and that they form as a, direct consequence of quantum mechanics and the wave nature of the electrons. If electrons have wave nature,, there has to be antibonding orbitals, you can’t have bonding orbitals without also having antibonding orbitals. By, accepting that antibonding orbitals form, you are embracing the concept of quantum and wave mechanics for, electrons. The wave nature of electrons is counterintuitive, yet without wave properties there are no orbitals, and, without orbitals there is no you, no me, no world, no planet, nothing!, Antibonding orbitals are similar to a p AO on a hydrogen atom. Does a hydrogen atom have a p AO? In a hydrogen, atom, there are no electrons in the p AO, but you could promote an electron from the 1s orbital into a 2p orbital using, Bonding 1 : page 30

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light energy. And so, in this sense, a hydrogen has a p AO, there just aren’t normally any electrons in it. It is the, same with molecules, molecules have lots of orbitals that don’t have all electrons in them, e.g. antibonding orbitals., But how are they used? Well, a stable molecule not doing anything doesn’t use them, they are empty. But, a, molecule undergoing a chemical reaction is different. Reacting molecules that are making and breaking bonds by, donating and accepting electrons. So, if a molecule needs to accept electrons in a reaction, they can’t go into the, bonding orbitals, since they are full. The only orbitals that are empty are the antibonding orbitals. And, just as, electrons in bonding orbitals make bonds, putting electrons into antibonding orbitals helps to break bonds, exactly, what has to happen in a chemical reaction! Later, we will look at reactions that can only be understood by, considering the antibonding orbitals and their wave functions., Quantum mechanics is counterintuitive, students should be confused by it! There is a famous quote by Niels Bohr,, “Those who are not shocked when they first come across quantum theory cannot possibly have understood it”; and, a more scary one by Richard Feynman, “It is safe to say that nobody understands quantum mechanics.”, , 7.9 Putting It All Together: From VSEPR/Hybridization to MOs, We will now do an exercise where we work through an entire complex organic structure and learn how VSPER,, hybridization, and the construction of MOs works., First, fill in the missing angles using VSEPR to determine molecular geometry from the number of “domains” of, electrons around each atom, in turn this determines the hybridization of each atom:, , N, , sp2, , sp, 180°, C, , H ~120°, , O, , C, , C, , C, , sp, ~109°, , H, N, , C, H, , H, , sp3 sp2, , H, , ~109°, H, , C, , H, , sp3, , H, , Second, give the hybridized AOs used to make the indicated bonds or hold the nonbonding electrons., , sp2 A.O., , p(C) + p(N), N, , n, , π∗, H, C, , C, , C, H, , H, , C, C, , H σ, , n sp3 A.O., , O, N, C, H, , H, , H, , H, , sp3(C) + sp2(C), , Third, draw pictures of the localized wave functions for the indicated electrons and give the appropriate AOs., • Construct localized MOs by in-phase (bonding) and out-of-phase (antibonding) combination of the AOs., , sp2 A.O., Ψn, π∗, Ψ, N, , p(C) + p(N), , H, C, , C, , n, , O, , HΨ, , C, , C, , C, , H H, , H, , Ψ σ, sp3(C) + sp2(C), , N, C, H, , Bonding 1 : page 31, , H, , H, , sp3 A.O.

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• We now have a very detailed understanding of exactly which orbitals the important electrons are in, and also, by, extension, an idea of their relative energies, which in turn gives us information on their chemical reactivity—higher, energy electrons tend to be more chemically reactive., , 7.10 Connection to Electron Energies, Example Problem: Give the relative energies of the indicated electron pairs and draw a picture of the appropriate, localized wave function squared superimposed on the structure (everything starts with VSEPR!):, , H A. non-bonding electrons, H, C, O, B. σ-bonding electrons, H, H, H, C, C. non-bonding electrons, C, C, C, C, H, H H, H, O, Energy, D. σ-bonding electrons, C. non-bonding electrons on atom with formal negative charge clearly highest, A. non-bonding electrons higher than any bonding electrons, D. σ-bond "made from" 2 sp3 A.O.s, larger more "p-like" orbital, higher energy electrons, B. σ-bond "made from" 2 sp A.O.s, smaller more "σ-like" orbital, lower energy electrons, , Example Problem: Give the relative energies of the indicated electron pairs and the orbital compositions, that is,, which AOs are used to make the localized MOs and which AOs contain the nonbonding electrons:, , H, C. π-bonding electrons, , H, , 2 p A.O.s on C, H, Energy, , C, , C, C, , H, C, , OH, , C, , H, , C, H, , A. non-bonding electrons sp3 A.O., , H, , NH2, , H, , B. σ-bonding electrons sp3 A.O.s on C and O, D. σ-bonding electrons sp3 A.O.s on C and C, E. non-bonding electrons sp3 A.O., , E. non-bonding electrons on less electronegative atom, highest, A. non-bonding electrons on more electronegative atom, lower, C. π-bond electrons higher than most σ-bond electrons, D. σ-bond involving less electronegative atom, higher than σ-bond electrons B, B. σ-bond involving more electronegative atom, lower than σ-bond electrons D, , • Electron energies are related to strengths of bonds, bond dissociation energies (see later)., • Note, number of nonbonding pairs is not important! Nonbonding electrons are lower in energy on oxygen, compared to those on nitrogen because oxygen is more electronegative than nitrogen, even though oxygen has, two nonbonding pairs. This is because the two nonbonding pairs “avoid” each other, they occupy different regions, of space. This is different from the situation where two electrons occupy the same orbital. In that case, the energy, goes up because the electrons can’t avoid each other., , Bonding 1 : page 32