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Developing Lewis structures 1) Add up all the valence electrons for all atoms. For ions, add one electron for each negative charge; subtract one electron for each positive charge. 2) Connect the atoms using lines to represent bonding pairs of electrons. The central atom is usually that wich is least electronegative. (In most organic molecules, the carbons atoms make up the frame of the molecule.) 3) Any electrons that are unplaced are added to the outer atoms (usually heteroatoms in organic molecules) as lone pairs. 4) Atoms lacking octets are completed by using lone pairs from adjacent atoms to form multiple bonds. Where more than one such pair is available, then all possibilities must be drawn and evaluated: these are resonance structures. 5) Formal charges on specific atoms are assigned using the formula: FC = (no. valence electrons - no. bonds to that atom)-(no. lone pair electrons) 6) Atoms below the second row of the periodic table may have more than eight electrons. For most (neutral) organic compounds, Lewis structures can be determined simply by taking the connectivity implied, remembering that there are 4 bonds to carbon, 3 to nitrogen, 2 to oxygen and 1 to hydrogen and by filling octets. 2-aminobutane CH2O 2-bromo-3-pentanone H2CCH2 HCCH HCONH2 H3CNNN H2COH+ H3CCN benzene H3CNO2 H3CSOCH3 Rules for Drawing and Interpreting Resonance Structures Resonance structures show how electrons are delocalized within a species. When drawing and evaluating resonance structures, it is important to keep track of lone pairs. 1) Only lone pair electrons and multiple-bond electrons (pi-electrons) move from one resonance structure to another. Atoms remain in exactly the same place. O O C C H2C H H2C O OH C C H H2C H H3C H 2) From structure to structure, electron pairs move in the following ways: lone pair to from an adjacent bond: H H C H C O H bond to form a lone pair: H H O O S H3C O S CH3 H3C CH3 bond to form a new bond: H2C H H C C CH2 H2C CH2 3) Electrons move only to adjacent positions, but more than one pair of electrons can move from one structure to another. N N O N N O 4) Always check to see what happens to the formal charges on specific atoms. The net charge on each structure must be the same. 5) Structures that are identical in form are said to be degenerate. Such structures contribute the same to the overall structure. Resonce "hybrids" are often drawn of such structures. H2C H H C C CH2 H2C H = C H2C CH2 CH2 6) When resonance structures are not degenerate, use the following rules to judge which structures contribute more to the character of the species: i) Resonance structures in which second period atoms (C-F) all have octets are favoured over those with electon deficient atoms--resonance structures that have more than 8 electrons on such atoms are not valid. H H C H H C O O H H ii) Resonance structures with fewer formal charges contribute more to the character of a species. H O C H H O N C H H N H iii) In some cases there are resonance structures with the same number of formal charges. In these cases, the favoured structure will be that in which the negative charge is on the more electronegative/positive charge on the more electropositive element: N N O N N O Invalid Resonance Structures We do not show resonance structures that: i) have more than eight electrons on a second row element N N O N N O X N N O ii) have less than eight electrons on O or N (or similarly electronegative element) O C CH2 O C CH2 H H iii) have more than two atoms bearing formal charges (excepting polynitros) H H C C O O O H X O C O O iv) have any atom with a formal charge greater than +/-1. O C N O C N X 2O C N HCO2 CH3NO2 H2CN2 HCONH2 Benzene Bond Length and Bond Strength Bond Length (nm) Strength (kJ/mol) Bond Length (nm) Strength (kJ/mol) C—C 0.153 347 C—N 0.147 305 C=C 0.131 611 C=N 0.128 615 C 0.118 837 C 0.114 891 C—H 0.109 414 C—F 0.140 485 C—O 0.143 360 C—Cl 0.179 339 C=O 0.121 740 C—Br 0.197 285 C—S 0.182 C—I 0.216 218 C N Polar Covalent Bonds The periodic trend in Pauling's electronegativity... O O C C O C O H C H O C N H C O C C N H Cl C MgBr C Li VSEPR for organic chemists ...The steric number and counting to four. BeH2 BF3 CH4 H2C=CH2 HCCH H CH3CH3 CH3OH H2CO H C H2C=CHCH3 C PhNO2 C H NH CH3CN Bonding We have a picture of where electrons are found in atoms: in atomic orbitals of various types. Increasing energy The size and energy of orbitals depend on n and on the atomic number, Z. 2s 2s 2p This is a single 2p orbital. It has two lobes of different phase and a node at the nucleus. A reminder about wave addition... + 2s 1s Orbitals are wavefunctions and have wave properties: These two 2s orbitals are the same in all respects except they have different phase. 2p = - ... and about spatial components. Any vector in 2D (or 3D) can be expressed as the vector sum of two component vectors. y y x x The vector lies in the xy plane because it has some "xness" and some "yness". Chemical bonds... ... result from the sharing of electrons between atoms. electrons represented in a Lewis structure as: protons In orbital terms... two overlapping 1s H atomic orbitals We can imagine the HF molecule arising from the following orbital overlap... F and H2O? F H H y O x z CH4 y H H x H y z C H C H z H x H H Orbital hybridization BeH2 This is a linear molecule. Be has a 2s and three 2p orbitals. It needs to use two of these to form bonds to the H atoms. Consider the superimposition of the 2s and 2px orbitals on the Be atom... Two possible combinations... Note that the two hybrid orbitals share the spatial (along the x axis) and nodal properties of the atomic orbitals from which they are hybridized. H Be H The remaining atomic 2p orbitals are still there but vacant. We can consider each Be—H bond as resulting from the in phase overlap of a sp hybrid orbital on Be with the 1s atomic orbital on H. Be H BH3 - trigonal planar We need three hybrid orbitals to form the B—F bonds. Three sp2 hybrid orbitals can be formed. They have spatial components in one plane - that of the atomic p orbitals from which they are hybridized. B Where is the unhybridized orbital? H H B H CH4 - tetrahedral C C 4 sp3 hybrid orbitals. The hybridization of any atom can be determined from its geometry: Steric No. Geometry Hybridization Unhybridized 2 Linear sp 2 x 2p 3 Trigonal Planar sp2 2p 4 Tetrahedral sp3 Ethane H H H C H C C H H H H C H H H H The C—C and C—H bonds of ethane result from the overlap of atomic orbitals that point directly to each other. The resulting bonding orbital has radial symmetry. We call this type of bond a σ bond. One important aspect of this bond's symmetry is that it is possible to rotate one end of the molecule with respect to the other leading to different molecular conformations. H H C H H C C H H H H H H H C H Bonding in Ethylene H H C 12 valence electrons C H H H C C C C H H C—C σ bond. H H C The σ bond framework of ethylene. C H H H H H C C H H H H C C H H For this bond to exist, the atomic p orbitals must be parallel. One critically important result of this is that, unlike σ bonds, free rotation about the bond is no longer possible without breaking the π bond. This takes substantial energy. H3C C H C C H H H H3C CH3 H H H C C CH3 H CH3 H3C C C H H H H3C CH3 H Bonding in Acetylene H C C H 10 valence electrons. Each C atom is sp hybridized... C H C C H C H H The σ bond framework. Each C atom has TWO unhybridized atomic p orbitals... H C C H H C C C H Allene H2C C CH2 Heteroatoms... CH3OH C O C O H H2C=O H H H H C O C N H C N Molecular Orbital (MO) Theory We have looked at bonds as being the result of overlapping atomic orbitals and/or hybrid atomic orbitals. MO theory is similar. Valence atomic orbitals are mixed to give a new set of molecular orbitals. For simple diatomic molecules, MO Theory gives a simple picture. For molecules of more than two atoms, things get more complicated. In organic chemistry, MO theory is usually used to help us understand pi bonding and delocalized electron systems. Let us consider the hydrogen molecule again... In phase: = Problem: we started with two atomic orbitals (capacity four electrons). We must end up with two molecular orbitals. How? Out of phase: = We can combine other orbital types as well to give MOs of various shapes... 2p and hybrid orbitals: Out of phase In phase The π and π* orbitals Out of phase = In phase = Some combinations don't make sense We construct a picture of the electronic structure of molecules using correlation diagrams, which indicate the energy and type of molecular orbitals that are occupied and vacant. σ∗1s E 1s 1s σ1s The AOs appear on the left (and right) side of the diagram at their corresponding energy level and the MOs that result from the AOs are in the middle at their corresponding energy level. An appropriate no. of valence electrons are added to the diagram following Hund's Rule and the Pauli exclusion principle to determine which orbitals are occupied. The bond order between the two atoms is the sum of the no. of bonding electrons the sum of the antibonding electrons divided by two. The second row diatomics (Li2 to Ne2) all involve combinations of the 2s and 2p atomic orbitals giving a set of eight MOs. σ∗2pz π∗2px π∗2py 2p 2p σ2pz E π2px π2py σ∗2s 2s 2s σ2s Even for molecules as simple as CH4, MO theory becomes rather complex. It is very useful, however, for examining how electron delocalization occurs in molecules with extensive π-bonding. In these cases we look exclusively at the π-MO diagrams. H2C CH2 π∗2p = 2p π2p = H H2C C C CH2 H π∗2p π∗2p 2p π2p π2p The Allyl system H C H2C CH2 π∗2p πΝΒΜΟ2p 2p π2p Isolobal systems... Benzene Reactive Intermediates H C H C H H H H C H H H The benzylic system. C H H Intermolecular Forces In order of decreasing strength... Hydrogen bonds Dipole-dipole London dispersion or Van der Wäals interactions. C H H Hydrogen Bonding Hydrogen bonds are relatively strong non-covalent interactions. There are two components to a hydrogen bond: an O—H or N—H group (donor) and a lone pair on an electronegative atom (acceptor). Length (nm) H O O Intermolecular H 0.270 O O H O Intramolecular 0.263 O H H O O H O H N N 0.288 0.304 O H H N 0.310 N H N O O H N N N N O N NH N NH2 O H N N H HN HN N N H H Adenine Thymine Dipole-dipole attractions O H3C + C CH3 London Forces H3C e e CH3 e H3C H3C e e e CH3 CH3 H3C H3C ee C 1 2 3 4 5 6 7 8 9 10 16 17 18 19 Formula CH4 CH3CH3 CH3CH2CH3 CH3(CH2)2CH3 CH3(CH2)3CH3 CH3(CH2)4CH3 CH3(CH2)5CH3 CH3(CH2)6CH3 CH3(CH2)7CH3 CH3(CH2)8CH3 CH3(CH2)14CH3 CH3(CH2)15CH3 CH3(CH2)16CH3 CH3(CH2)17CH3 Name Methane Ethane Propane Butane Pentane Hexane Heptane Octane Nonane Decane Hexadecane Heptadecane Octadecane Nonadecane B.P. M.P. -162 -88.5 -42 0 36 69 98 126 151 174 287 302 317 330 18 22 29 32 e e CH3 CH3 CH3 CH CH3 CH CH3 CH3 CH3 CH3 CH CH CH3 CH3 Pentane B.P. 36° Isopentane B.P. 30° Neopentane B.P.10°