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Developing Lewis structures for organic molecules 1) Draw the full structure of the molecules with the connectivity suggested by the formula such that all valences are met. 2) Count the number of valance electrons in the entire molecule(equal to the group number in the Periodic Table). 3) From the number in 2, subtract the bonding electrons (two times the no. of lines in your structure). Any electrons that are unplaced are added to atoms that lack a full octet (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. *H needs only two atoms, B and Al need only six. 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. CH3NH2 CH3OH CH2O H3CCN H3CNO2 H2COH+ HCONH2 (CH3O)2CH+ H2CNN H3CSOCH3 H2NCONH2 [CH2CHCH2]+ 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 H2C H 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 H C O H bond to form a lone pair: 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) When comparing resonance structures with the same number of bonds (as in part i above) 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 cases where there are resonance structures with the same number of formal charges, the favoured structure will be that in which the negative charge is on the more electronegative atom or positive charge is 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 Properties 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 Cl C C N MgBr C H 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 Increasing e nergy We have a picture of where electrons are found in atoms: in atomic orbitals of various types. 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: H H protons In orbital terms... two overlapping 1s H atomic orbitals overlap constructively to form a new orbital with enhanced e density between the two nuclei. We can imagine the HF molecule arising from the following orbital overlap... F and H2O? F H y O x z CH4 y x y z 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 orbitals to form two bonds to the H atoms. Consider the superimposition of the 2s and 2px orbitals on the Be atom... Two possible combinations... These are called "hybrid" orbitals. In this case they are sp hybrids. 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 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. 2 3 4 Geometry Hybridization Unhybridized Ethane H H H C H C C H H H H C H H H H 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 Resulting in two π bonds, mutually perpendicular. C C C H Reactive Carbon-centered Intermediates H3C H3C H3C H2C Steric Number Hybzn: H H C C H H Carbocation H H C H H H H H Carbanion Radical Carbene Heteroatoms... H O O H N R H H Halogens... X C C N C R C N CH3OH C O C O H H2C=O H H H H C O C N H C N Hybridization and Resonance Structures Frequently, different resonance structures predict different geometries about the same atom... O O C H N CH3 C H N CH3 CH3 N is sp3 C N C N CH3 N is sp2 O C N H Remember: resonance structures show how electrons are delocalized, electrons are delocalized via π-bonds, π-bonds are formed by the overlap of atomic p orbitals. Pyrrole H H H H H N N N N N N H N H Molecular Orbital (MO) Theory - the bonding model for adults We have looked at bonds as being the result of overlapping atomic orbitals and/or hybrid atomic orbitals. MO theory is similar. We take all of the valence atomic orbitals of the atoms in a given molecule and use them to derive a set of new orbitals for a given molecule. The principles: • the number of molecular orbitals in a molecule must equal the number of valence atomic orbitals of the constituent atoms • like all orbitals, MOs can hold two electrons at most. • MOs can be delocalized - where atomic orbitals are found around an atom, molecular orbitals can be spread over the entire molecule • MOs can be σ or π • molecular orbitals can be bonding, non-bonding or anti-bonding • a chart showing the relative energies of the MOs to each other and to the AOs from which they are derived is known as an MO diagram. Consider the H2 molecule: we have ... In phase: = Out of phase: = 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 for diatomics, 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. 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 = Why are these bonding and antibonding orbitals so important, especially if the antibonding orbitals are empty? H2C CH2 Energy π∗2p LUMO 2p π2p HOMO The HOMO is the ... The LUMO is the ... For any molecule, the HOMO and LUMO are the "Frontier Molecular Orbitals". When molecules react, their bonds are rearranged. In the MO model, that means that MOs must change and the MOs that are involved are the FMOs because: MO theory is very useful for examining how electron delocalization occurs in molecules. ELECTRON DELOCALIZATION OCCURS THROUGH PI BONDS so 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... π∗2p π∗2p πΝΒΜΟ2p πΝΒΜΟ2p 2p 2p π2p π2p Benzene The benzylic system. C H H C H H Intermolecular Forces D, donor must be O or N. A, acceptor must be O or N. D Hydrogen bond A H O Ion-ion C Na CH3 O H Ion-dipole O Na H Dipole-dipole O H + C H Dispersion or London Forces H3C e e CH3 e H3C H3C e e e CH3 CH3 H3C H3C e e CH3 CH3 C Formula 1 2 3 4 5 6 7 8 9 10 16 17 18 19 Name 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 B.P. Methane Ethane Propane Butane Pentane Hexane Heptane Octane Nonane Decane Hexadecane Heptadecane Octadecane Nonadecane M.P. -162 -88.5 -42 0 36 69 98 126 151 174 287 302 317 330 18 22 29 32 CH3 CH CH3 CH3 CH CH3 CH3 1/r6 CH3 CH CH CH3 CH3 MW (g/mol) compound MP/BP 58 BP 0°C 60 BP 7°C 64.5 BP 12°C 60 BP 97°C 60 BP 116°C 60 MP 58°C O LiO2CCH3 The compounds: H2 C C H2 H3C H2 C H3C O CH3 C H O H2 C CH3 H3C CH3 H2 C C H2 OH H3CCH2 Cl 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 H O Intermolecular H 0.270 Intramolecular 0.263 O O O H H O O H O H N N 0.288 0.304 O H H N 0.310 N H N O N N NH2 H N N H O N NH O O H N HN N HN N N N H H Adenine Thymine Tender Loving Chromatography Silica gel: the solid phase adsorbent. A dilute solution of a compound or mixture is spotted onto a TLC plate with a capillary. The plate is then placed upright in a tank containing a small amound of some solvent or solvent mixture (known as the eluant) such that only the very bottom of the plate is in the solvent and that the spots are not covered. Capillary action of the adsorbent draws the eluant upward. When the eluant hits the spot, the compounds are carried with it depending on the relative strength of their interaction with the eluant or the adsorbent -- the partitioning. elution S OH S M S S S S O M S S OH OH OH OH S S The compounds: 1 C H2 H3C H2 C 4 H3C O CH3 C CH3 5 CH3 O H Hexane (non-polar) Effect of Eluant Polarity 2 LiO2CCH3 3 O H2 C H2 C H3C H2 C C H2 6 H3CCH2 OH Toluene (aromatic non-polar) Chloroform (somewhat polar) Ethyl Acetate (polar) 1 2 1 2 1 2 Ethanol (very polar) 1 CH2OH Br 2 1 2 Cl 1 2 SM Prod RM SM Prod RM SM Prod AS or DP AS DP AS or DP AS DP Applications of TLC RM