Exam 1 Study Guide CHEM 215Resonance:-Only electrons change-Formal charge: # of valence e – nbe – ½(bonding electrons)-The least number of formal charges/ where negative charges are on the most electronegative atomsMajor contributorConstitutional Isomers/Structural Isomers-Different molecules with the same molecular formula, but different connectivity/skeletons [CH3OCH3 vs. CH3CH2OH]Conformational Isomers/Conformers-Same molecule with different rotations about single sigma bondsStereoisomers-Different molecules with the same structure and connectivity but different orientation in 3D space-Either enantiomers or diastereomers-May be chiral or achiral-# of Stereoisomers = 2n, where n is the number of stereocentersStereocenter-The location on a molecule that confers stereochemistry-The location on a connectivity/skeleton where stereoisomers differ in their 3D orientation-May or may not be a chiral center= chiral site = chiral carbon-Sp3 tetrahedral carbon bound to 4 chemically distinct groups-Has R/S absolute stereoconfigurationChiral Compound-Has one or more chiral centers, but lacks an internal plane of symmetry-Is optically active-Has exactly one enantiomerEnantiomers-Non-superimposable mirror images-Every chiral compound has exactly ONE-Have opposite optical activities-A 50/50 racemic mixture is optically inactive-All chiral centers mirror across an external plane of symmetry, but achiral stereocenters remain the same-Must not contain an internal plane of symmetryDiastereomers-Stereoisomers that are NOT enantiomers-May be chiral or achiralChiral Diastereomer-Usually mirrors some, but not all chiral centersAchiral Diastereomers:-have internal planes of symmetry-E/Z (across a double bond)-cis/trans (on a ring)Meso-optically inactive-has an even number of chiral centers that cancel each other out across an internal plane of symmetry-it is superimposable with its mirror imagewhy they aren’t enantiomersSN2: -one stereoisomeric product-inversion of stereochemistry-good nucleophile-bad leaving group-aprotic solvent-steric hindrance is a factorSN2:-enantiomeric at the Carbon in question-good leaving group-bad nucleophile (don’t want a fight—need to form C+)-polar, protic solvent—gain/lose protons (R-OH, R-NH2, R-SH)E1:-2 stereoisomeric products-usually E/Z at the alpha-beta bond-weak Base (SN1 has a nucleophile)-strong LG-polar-protic solvent-heatZeitsev’s Rule—more substituted alkenes are more stableE2: -one stereoisomeric product-strong Base-bad LG-aprotic solvent-heatHydration: -reagents: H2O and catalytic amount of H2SO4-Add an OH group by first protonating the double bound according to markovnikov’s ruleDehydration:-reagents: H2O and H2SO4 and heat (takes energy to break bonds)-protonotate the –OR group to convert them into good LG-with heat present, ELIMINATE the LG to yield an alkenePolymerization:-reagents: H2SO4 and Hexane-add a nucleophile to quench the reaction (nucleophile like H2O)Hydride Shifts-in order to give the most stabilized C+-resonance stabilized>tertiary>secondary>primary-can also shift –CH3 and –CH2CH3Hydroboration:-reagents: 1) BH3 2)H2O2 and NaOH (‘work up’)-syn addition of H/OH-yields an anti-markovnikov R-OH-driven by sterics—can also use 9-BBN instead which gives different steric outlookAddition of Any Group 1 Solid Metal and Any Liquid Amine-Na, K, Li and NH3 CH3NH2-yields an anti-addition of H/HHalogenation:-Anti-addition of X/X or X/OR-pi bond attacks the electrophilic halogen to form a halonium ion ring-nucleophile attacks assymetric halonium at more substituted side-alsoHOCl, HOI, Br2, Cl2, I2, but NOT F (most electronegative)Ozonolysis:-reagents: O3 and a “Work up”-cut double bond in half and tack an O on either side-reductive work up Zn+H2O or CH3SCH3 (add just an -H)-oxidative work up H2O2 and NaOH (add an -OH)Oxidation of Osmium -reagents: 1) OsO4 2)Na2SO3/H2O-syn addition of OH/OH +enantiomeric form-cis diolFormation of an Epoxide-reagent: any peroxy acid (mcpba)Substitution:-reagents: 1) Na+ and NH2- 2) CH3Br-deprotonate to form a carbanion-add CH3Redox Reactions:-reduction of simplest double bond-each substituent has an oxidation number associated with it that determines, in part, the oxidation number of the carbon.-considerations for oxidation number assignment:1) if an atom has no formal charge, sum of oxidation numbers=02) the sign (+ or -) and the magnitude of the oxidation number of a substituent is dictated by its relative electronegativity and the number of bonds it has to the carbon in questionTrends:-more electronegative than Carbon: Nitrogen, Oxygen, Fluorine, Chlorine, -1 for every single bond, double bond is -2, triple bond is -3-less electronegative than carbon: H is +1 (if we were using Boron, we could see a +2 or +3 based on the number of bonds)-we use oxidation numbers to keep track of the oxidation state of carbon and to determine whether or not a reaction is oxidative or reductive-Oxidize primary alcoholsaldehydescarboxylic acids-oxidize secondary alcoholsketones-tertiary alcohols cannot be oxidizedGeneral Oxidation Scheme:1) Protonate oxygen in –OH group to make it a better leaving group, or use TsCl 2) Elimination with base –remove beta hydrogen while LG leaves-2 options for reagents: 1) high valence metals [ Cr 6+ ]2) purely organic moleculesAqueous Cr 6+[acid catalyzed] -will turn primary all the way to carboxylic and secondary to ketones-allows for selectivity-Jones Reagent: CrO3, H2SO4-Chromate: Na2CrO4, H2SO4-Dicrhomate: Na2Cr2O7, H2SO4Anhydrous Cr 6+[not acid catalyzed]-allows you to stop oxidation at the aldehyde-CrO3, pyridine-ClCrO3 -, PCC +-Cr2O7 2-, PDC 2+**key intermediate**-Chromate EsterSwern Reagents:1)DMSO and Oxalylchloride2) Et3N (methylamine)**will oxidize to aldehydeSubstitution Reactions with R-OH- -OH- not good leaving groups- Primary alcohols will undergo Sn2 with a good nucleophile- Tertiary alcohols will undergo Sn1 with a good nucleophile- Secondary can undergo both Sn1 and Sn2o Could do a hydride shift with Sn1o Could form a tosylate with Sn1-converts –OH into a good leaving groupAlcohols as Nucleophiles-Intermolecular Reaction-Best Conditions: -sterically least-hindered-Strong base that is not a good nucleophile (NaH)-Intramolecular Reactions-3,5,6 membered rings are favored-if our –OH and LG are not in proper positions, get Elimination -Requirements:-nucleophile must be able to approach the ephile from the opposite side(this is why