Exam # 3 Study Guide Lectures: 20-27Lecture 20I. Intro to AlkynesA. Have triple bondsB. Have 180° geometryC. Less reactive than alkenesD. sp hybridized orbitalsII. NamingA. Same as alkanesB. Identify the parent chainC. Drop the –ane and add –yneD. Parent chain includes both double and triple bond if both functional groups are presenti. –enyne endingIII. Alkyne reactivityA. Reactivity (and stability) order:Lecture 21IV. MechanismsA. 1 eq of HXi. Markovnikovii. Trans/EB. Excess HXi. Markovnikovii. GerminalC. 1 eq of X2 i. Markovnikovii. Trans/ED. Excess X2i. Markovnikovii. Two sets of germinal halogensE. Acid catalyzed water additioni. Leads to either a Ketone (internal alkyne) or an Aldheyde (terminal alkyne)ii. Enol stage tautomerizes into the keto stageiii. Remember that symmetric alkynes yield 100% one productiv. Asymmetric alkynes lead to a racemic mixture of productsLecture 22I. Hydroboration of alkynesA. Produces an aldehydeB. Anti-MarkovnikovC. Syn additionLecture 23I. Retrosynthesis is doing a synthesis problem backwardsLecture 24I. Delocalized electrons are elctrons shared by 3 or more atomsII. Benzene is a planar moleculeIII. Rules for drawing resonance structuresA. Only electrons moveB. Only pi electrons and lone pairs can moveC. Sigma bonds are not broken and substituents are not removedIV. Factors that decrease stabilityA. Atoms with incomplete octetsB. Negative charge on an atom that is not the most electronegative atom in the molecule or a positive charge on the most electronegative moleculeC. Charge separationREMEMBER that resonance increases stability and makes the molecule less reactive (and moreacidic )V. Criteria for aromaticityA. Must have uninterrupted pi-cloudB. Must be cyclicC. Must be planarD. Must follow Huckle rule: 4n + 2 where n is any intergeri. The number of pi electron present must be a Huckle numberLecture 25I. Hererocyclic compounds can be aromatic as long as all bonded atoms have a pi-cloudII. Antiaromatic refers to cyclic, planar compounds that do not follow Huckle’s rule. They are incredibly unstablea. Nonaromatic compounds may not be cyclic, planar, or have a cyclic pi cloudAnti-aromaticAromaticNon-aromaticIII. 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. ExampleAnti-aromaticAromaticLecture 26I. Diels-Alder ReactionsA. Always produces a six-membered ringB. Diene can attack a dienophile from either the top face or the bottom faceC. Electron rich regions do not react with each otherD. Electron poor regions react with electron rich regionsE. ExampleF. Diels Alder reaction can occur with cyclic compoundsG. The face that the diene attacks determines the stereochemistry of the moleculeThe first product occurs when top face is attacked; the second occurs when bottom face is attackedLecture 27I. Diels Alder StereochemistryA. ENDO vs. EXOi. Endo means that the substituents are facing downwards relative to the moleculeii. Exo means the substituents are facing upward relative to the moleculeiii. To visualize endo and exo, the compound formed from a diels-alder reaction must be drawn in 3Div. The Endo position is the preferred position and is formed the most rapidlyB. The stereochemistry of the diene and the dienophile don’t change during the reactioni. The inside substituents ( the circled hydrogens) always point upwards in the ring that formsii. The carbonyl groups are still trans to each otherAnother exampleFlipping the dienophile forms the endo enantiomer of the
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