Exam 3 Study Guide Lectures 20 27 Lecture 20 I Intro to Alkynes A Have triple bonds B Have 180 geometry C Less reactive than alkenes D sp hybridized orbitals II Naming A Same as alkanes B Identify the parent chain C Drop the ane and add yne D Parent chain includes both double and triple bond if both functional groups are present i enyne ending III Alkyne reactivity A Reactivity and stability order Lecture 21 IV Mechanisms A 1 eq of HX i Markovnikov ii Trans E B Excess HX i Markovnikov ii Germinal C 1 eq of X2 i Markovnikov ii Trans E D Excess X2 i Markovnikov ii Two sets of germinal halogens E Acid catalyzed water addition i Leads to either a Ketone internal alkyne or an Aldheyde terminal alkyne ii Enol stage tautomerizes into the keto stage iii Remember that symmetric alkynes yield 100 one product iv Asymmetric alkynes lead to a racemic mixture of products Lecture 22 I Hydroboration of alkynes A Produces an aldehyde B Anti Markovnikov C Syn addition Lecture 23 I Retrosynthesis is doing a synthesis problem backwards Lecture 24 I II III IV Delocalized electrons are elctrons shared by 3 or more atoms Benzene is a planar molecule Rules for drawing resonance structures A Only electrons move B Only pi electrons and lone pairs can move C Sigma bonds are not broken and substituents are not removed Factors that decrease stability A Atoms with incomplete octets B Negative charge on an atom that is not the most electronegative atom in the molecule or a positive charge on the most electronegative molecule C Charge separation REMEMBER that resonance increases stability and makes the molecule less reactive and more acidic V Criteria for aromaticity A Must have uninterrupted pi cloud B Must be cyclic C Must be planar D Must follow Huckle rule 4n 2 where n is any interger i The number of pi electron present must be a Huckle number Lecture 25 I II Hererocyclic compounds can be aromatic as long as all bonded atoms have a pi cloud Antiaromatic refers to cyclic planar compounds that do not follow Huckle s rule They are incredibly unstable a Nonaromatic compounds may not be cyclic planar or have a cyclic pi cloud Non aromatic Anti aromatic III Aromatic Drawing Molecular Orbitals MO s and frost diagrams A Molecular orbitals demonstrate pi bonding and illustrate the pi system in a compound B The number of MO s is dependent upon the number of pi bonds in a compound i Each MO has a node The number of nodes is the orbital number minus one ii Nodes separate out of sync orbitals iii Orbitals that are in sync with each other are shaded on the same side of the balloons and vice versa for out of sync orbitals C Example 1 3 butene i ii iii iv Red line nonbonding line Blue dashed line nodes Purple numbers number of carbon Blue numbers orbital level D When drawing Frost diagrams make sure that the molecule has a point facing downward i Each carbon is represented by a line level ii Each set of levels is bisected by the non bonding line iii The bottom half contains the bonding orbitals and the top half contains the antibonding orbitals E Example Aromatic Anti aromatic Lecture 26 I Diels Alder Reactions A Always produces a six membered ring B Diene can attack a dienophile from either the top face or the bottom face C Electron rich regions do not react with each other D Electron poor regions react with electron rich regions E Example F Diels Alder reaction can occur with cyclic compounds G The face that the diene attacks determines the stereochemistry of the molecule The first product occurs when top face is attacked the second occurs when bottom face is attacked Lecture 27 I Diels Alder Stereochemistry A ENDO vs EXO i Endo means that the substituents are facing downwards relative to the molecule ii Exo means the substituents are facing upward relative to the molecule iii To visualize endo and exo the compound formed from a diels alder reaction must be drawn in 3D iv The Endo position is the preferred position and is formed the most rapidly B The stereochemistry of the diene and the dienophile don t change during the reaction i The inside substituents the circled hydrogens always point upwards in the ring that forms ii The carbonyl groups are still trans to each other Another example Flipping the dienophile forms the endo enantiomer of the product
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