Alkene Reactions Cheat Sheet, Cheat Sheet of Organic Chemistry

Download Alkene Reactions Cheat Sheet and more Cheat Sheet Organic Chemistry in PDF only on Docsity! CH3 CH3 CH3 Br CH3 Br H CH3 H CH3 H Br CH3 CH3 CH3 CH3 X OH CH3 CH3 BH2 CH3 OH H H BH2 H CH3 CH3 CH3 Hg(OAc) OR CH3 H CH3 H OH X+ CH3 CH3 CH3 X X X+ Hg(OAc) β-Mercury- substituted carbocation H2SO4 or H3PO4 or NaHSO4 Which upon reaction with: Reagent Yields Halonium Ion "X+" Carbocation Halonium Ion What adds? HOOH/NaOH Hg(OAc)+ "X+" To give the more stable: Radical Substrate (NaBH4 reduction replaces Hg(OAc) with H) Radical Hydrohalogenation: X2/H2O Organoborane Halogenation: Hydration: Halohydrin Formation: Hydroboration/Oxidation: Oxymercuration: Summary of Alkene Reactions Hydrohalogenation: HBr/peroxides Carbocation B2H6 Br• ROHHg(OAc)2 X– Br–H+ H+ HBr HBr X2 H2O H2O ORGANIC CHEMISTRY SYSTEMATIC NOMENCLATURE OF BICYCLIC COMPOUNDS Bicyclic compounds require the breaking of two carbon-carbon bonds to convert them to open chained compounds--containing no rings. Bicyclic compounds are named by prefixing bicyclo- to the name of the parent hydrocarbon. The name of the parent hydrocarbon is obtained by counting the total number of carbon atoms in all of the rings of the compound. Thus, because norbornane (below) contains seven carbon atoms in the two rings of the molecule it is a bicycloheptane. Carbon atoms shared by both rings are referred to as bridgehead carbons. 1 2 3 4 5 6 7 1 2 3 4 5 6 7 norbornane bicyclo[2.2.1]heptane The numbers of carbon atoms between bridgehead carbons in the molecule are specified by counting from the bridgehead carbon and listing each of the numbers in brackets in decreasing order prior to the name of the hydrocarbon. Thus, norbornane is bicyclo[2.2.1]heptane. The following are additional examples of bicycloheptanes: 1 2 3 4 5 6 7 1 2 3 4 5 6 7 bicyclo[3.1.1]heptane bicyclo[3.2.0]heptane Guide to Displacement Reactions Reactivity Type of carbon SN1 (carbocation intermediate) E1 (carbocation intermediate) SN2 (concerted) E2 (concerted) 1 ° X X • good LG • good Nu: • polar aprotic solvent • strong, bulky bases 2 ° • good LG• poor Nu: • polar protic solvent • competes with SN1 when base is present • good LG • good Nu: (weaker base than HO–) • polar aprotic solvent • competes with SN2 • favored by bulky bases stronger than HO– 3 ° • good LG• any Nu: • polar protic solvent • competes with SN1 when base is present X • strong base Nucleophilicity Classification Nucleophiles Rel. React. Excellent I–, HS–, RS– >105 Good Br–, HO–, RO–, CN–, N3– 104 Fair NH3, Cl–, F–, RCO2– 103 Weak H2O, ROH 1 Very Weak RCO2H 10–2 Leaving Groups Classification Leaving Groups Excellent ROSO2–, H2O, ROH, N2 Good I–, Br–, Cl–, NR3, RCO2– Poor HO–, RO–, F–, CN–, NH2 HBr CH3 Give all possible products, including stereochemistry! H2SO4 (dil.) HBr peroxides Br2 CCl4 CH3 CH3 OH Br + enantiomers CH3 Br CH3 Br CH3 Br Br + enantiomer OH H H OH + enantiomer + enantiomer OH H H OH + enantiomer + enantiomer Br H H Br + enantiomer + enantiomer Br H H Br + enantiomer + enantiomer Br H H Br + enantiomer + enantiomer Br H H Br + enantiomer + enantiomer Br Br Br Br + enantiomer OH Br Br Br Br Br Br Br Reagent Substrate CH3 CH3 CH3 CH3OH Cl Give all possible products, including stereochemistry! Cl2 H2O 1. Hg(OAc)2 2. NaBH4/OH– 1. BH3 2. H2O2/OH– 1. OsO4 2. H2O2/OH– + enantiomer CH3 OH OH OH OH + enantiomer + enantiomer OH Cl + enantiomer OH H H OH + enantiomer + enantiomer + enantiomer H OH + enantiomer HO OH OH Cl + enantiomer OH H H OH + enantiomer + enantiomer + enantiomer H OH HO OH Cl OH HO Cl OH OH OH OH OH HO OH Reagent Substrate LEWIS STRUCTURES OFTEN DO NOT ADEQUATELY REPRESENT ACTUAL STRUCTURES: Cl N O O Remember... Implies unequal bonding of N to two oxygens! Experimentally, they are the same length! Linus Pauling (Feb. 28, 1901-Aug. 19, 1994) introduced the concept of... Resonance: When a molecule can be represented by 2 or more Lewis structures where thearrangement of atoms is the same, but electrons are shifted Resonance arrow (double headed). Do not under any circumstances confuse with an equilibrium arrow! The small curved arrows are "electron pushing" arrows. They represent the figurative movement of an electron pair. This is for bookkeeping purposes only! Cl N O O 2 1 ...RESONANCE HYBRID 2 Cl N O O 22 1 THE ACTUAL MOLECULE IS MORE STABLE THAN ANY INDIVIDUAL RESONANCE FORM IMPLIES. IN FACT, THE MORE RESONANCE FORMS ONE CAN DRAW OF A SPECIES, THE MORE STABLE IT IS. OZONE, O3, APPEARS TO HAVE A SEPARATION OF CHARGE, IMPLYING SEVERE INSTABILITY... 11 Neither Lewis structure by itself represents reality. The structures are not rapidly interconverting. Rather, the actual structure is a mix of the two (or more) resonance forms and is called a... RESONANCE THEORY ...THE RESONANCE HYBRID SHOWS THAT BOTH TERMINAL OXYGENS SHARE THE NEGATIVE CHARGE, RESULTING IN GREATER STABILITY RULES OF RESONANCE THEORY Cl N O O 3. RESONANCE FORMS FOR A GIVEN MOLECULE NEED NOT CONTRI- BUTE EQUALLY TO THE OVERALL STRUCTURE (THE RESONANCE HYBRID) SINCE THEY MAY DIFFER IN ENERGY AND 1. ATOMIC POSITIONS REMAIN THE SAME 2. FOLLOW THE OCTET RULE are not resonance forms!!! ...is not! are resonance forms!!! are valid resonance forms but... O O O Separation of charge indicates a higher energy resonance form... more stable contributes more less stable contributes less O O O more stable ( on O) less stable ( on C) Formal negative charge on electropositive elements indicates a higher energy resonance form, as well as positive charge on electronegative elements... O OO (kind of like blue and yellow mixing to make green) 4. INSURE THAT ALL RESONANCE FORMS HAVE THE SAME NET CHARGE, THE SAME NUMBER OF ELECTRONS, AND THE SAME NUMBER OF UNPAIRED ELECTRONS N C O C N O N C O N C O N O C C H H H H H N O C C H H H H H N O C C H H H H H C C H HH H C C H HH H N O C C H H H H H N O C C H H H H H A B A B Orbital Hybridization Linear Carbon (Atom) • Only two other atoms or • One atom and one lone pair surrounding carbon (atom) {one s + one p} two sp orbitals ORIENT THEM: 180° apart Two p orbitals left over σ bonds are formed by the sp hybrids and π bonds by the p orbitals Example Trigonal Planar Carbon (Atom) • Only three other atoms or • Two atoms and one lone pair or • One atom and two lone pairs (rare) surrounding carbon (atom) H C N {one s + two p} three sp2 orbitals ORIENT THEM: 120° apart Tetrahedral Carbon (Atom) • Four other atoms or • Three atoms and one lone pair or • Two atoms and one lone pair surrounding carbon (atom) {one s + three p} four sp3 orbitals ORIENT THEM: 109.5° apart Example Hydrogen cyanide No p orbitals left over σ bonds are formed by the sp3 hybrids—no π bonds C C Methane (side view) H H H H (top view) Ethylene H H C HH One p orbital left over σ bonds are formed by the sp2 hybrids and a π bond by the p orbital Example C C H H H H