CHEM 2211 1st Edition Lecture 25Outline of Last LectureI. Delocalized ElectronsII. Bonding in benzeneIII. Drawing Resonance structuresIV. Factors that decrease stabilityV. Delocalization adds stabilityVI. Criteria for AromaticityOutline of Current LectureI. Delocalization EnergyII. Criteria for Aromatic compoundsIII. Aromaticity of Heterocyclic compoundsIV. AntiaromaticityV. Drawing molecular orbitals and Frost DiagramsCurrent LectureI. Delocalization EnergyA. Aromatic compounds like benzene have very large delocalization energiesII. Criteria for Aromatic compoundsIII. Aromaticity of Heterocyclic compoundsA. Heterocyclic compounds are compounds with one or more atoms that isn’t carbonB. Heterocyclic atoms must have a pi orbital in order to fulfill the aromaticity rulesC. Most common hetero atoms are N, O, and SD. ExamplesExample of Heterocyclic compoundsIV. AntiaromaticityA. Characterized by instabilityB. Is considered to be antiaromatic if the compound fulfills the first rule but not the second rule for aromaticityC. Molecuels that do not adhere to either rule are non-aromaticD. ExampleAnti-aromaticAromaticNon-aromaticV. Drawing Molecular Orbitals (MO’s) and frost diagramsA. Molecular orbitals demonstrate pi bonding and illustrate the pi system in a compoundB. The number of MO’s is dependent upon the number of pi bonds in a compoundi. Each MO has a node. The number of nodes is the orbital number minus oneii. Nodes separate out of sync orbitalsiii. Orbitals that are in sync with each other are shaded on the same side of the balloons and vice versa for out of sync orbitalsC. Example1, 3-butenei. Red line = nonbonding lineii. Blue dashed line = nodesiii. Purple numbers = number of carboniv. Blue numbers = orbital levelD. When drawing Frost diagrams make sure that the molecule has a point facing downwardi. Each carbon is represented by a line (level)ii. Each set of levels is bisected by the non-bonding lineiii. The bottom half contains the bonding orbitals and the top half contains the antibonding orbitalsE.
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