Bài giảng Organic Chemistry - Chapter 11: Alkenes

Chapter 11: AlkenesCCDouble bondNames: endinganeene1. Find longest chain wit both Csp2 carbons in it.Rules:An octene123456784For example: ethene, propene, butene, etc.3. Name and # substituents, in alphabetical order4-Ethyl-3-methyl-3-octene2. Number the chain with C C close to terminus12345678A 3-octene (only the first of the two Csp2 carbons is named by a #) 44. Cycloalkenes123CH3CC123-Methylcyclohexeneby definitionand start such as to give substituents lowest possible numbering 5. Stereoisomers:RRRRcistransCis/trans used for 1,2-disubstituted ethenes.6. For tri- and tetrasubstituted alkenes: E, Z naming. Use R, S priority rules at each sp2-carbon separately, to find yhe higher priority group at each end. E-4-Ethyl-3-methyl-3-octene123456784Opposite sides: ESame side: ZColloquial: double bond positionRRInternalTerminal RRCH28. Substituents: Alkenyl CH2CHEthenyl (vinyl)CH2CHCH22-Propenyl (allyl)9. Exocyclic alkenes: AlkylidenecycloalkanesMethylidenecyclohexane (Methylenecyclohexane)OH ( SH)>ene123OH2-Propen-1-ol7.Structure Of The Double BondElectron richTwo components: the -bond and the -bondThe Sigma BondEtheneThe Pi BondOrbital Energies Bond is relatively weakHow Weak Is The Pi Bond?Ea = 65 kcal mol-1Typical C-C bond strength ~90 kcal mol-1(Table 3-2)Bond Strengths In AlkenesUnusually strong because C uses sp2 hybridsCC33% s character.In contrast:sp3 has 25% s characterNet effect: relatively e-withdrawingH:sp2HasAlkenyl Hydrogens Are Relatively AcidicCCHpKa ~ 44Compare CH3CH2-H pKa ~ 50+CH3LiRCHCHLiCH4+CCHHHRTherefore, in principle:Problems: regio-, stereoselectivity. Better via alkenyl organometallics:CH2CHBr+LiCH2CHLiCCBrHHRMg+CCMgBrHHRUseful: React with carbonylsDegree Of UnsaturationMolecular formula tells us how many rings and/or  bonds are present in a molecule. Reference is a saturated acyclic hydrocarbon: CnH2n+2.Simple examples:We need to determine the deviation of the molecular formula from CnH2n+2 (in increments of 2 H = degree of unsaturation). Every ring or double bond takes away 2 H, triple bond 4 H from CnH2n+2.C6H12, not C6H14.C6H10, not C6H14.Heteroatoms and Degree Of UnsaturationHalogen:-1;CHCXNitrogen:+1;CHCNRHCHCOHS, O no effect on count (still CnH2n+2 + Sx or Oy)Depends on valency of element:Steps: 1. Calculate Hsat = 2n C + 2 – nX + nN n = “number of”2. Count Hactual in given molecular formula.3. Degree of Unsaturation: (Hsat – Hactual)/2How To Calculate Degree Of UnsaturationC5H5NExamples:C10H161. Hsat = (2x10) + 2 = 222. Degree of unsaturation: (22-16)/2 = 3or..orHsat = 10 + 2 + 1 = 132. (13 - 5)/2 = 4 degrees of unsaturation:NPyridineCNor..orRelative Stability Of AlkenesMeasure heat of hydrogenation ΔHH2 of isomers, e.g., butene+H2+H2+H2ΔHH2 (kcal mol-1)-28.6-27.6-30.3InternalterminaltranscisRealtive stability:>>cat.cat.cat.CWhy? 1. Hyperconjugation: 2. Steric hindrance (strain)HC:Cis is less stable than trans because of steric hindranceGeneral order of stability:CH2 CH2RCH CH2RCH CHRcis<<RCH CHRtranstritetrasubstituted<<<CCH3CH2O-Na+, CH3CH2OH Synthesis of AlkenesE revisited. Best: E2 on RX. Regioselectivity?Saytzev-Rule Non-bulky base leads to more stable internal double bond.CH3CH2 C CH3CH3XMore stableLess stableCCHCH3CH3H3CCCH2H3CH2C+CH370%30%Small baseAlexander M. Saytzev 1841–1910)Saytzev’s Rule: The Transition StateThe transition states begin to look like product double bond.LargebaseHofmann RuleBulky base leads to less stable terminal double bondAugust Wilhelm von Hofmann (1818–1892),Recall: Stereoselectivity is the preferential formation of one stereoisomer over possible others. Here: cis or trans product? Yes, but not completely.51%18%31%++BrNa OCH3CH3OH-+Elimination Is StereoselectiveMainly transNon-bulky base: SaytzevRecall: Stereospecificity is the selective conversion of one stereoisomer of starting material to one of the product. Here: One diastereomer ofC CHX**Elimination Is Stereospecificgives only E product, the other gives only Z. Alkenes From ROH By DehydrationRprimOH::CH2 CH2 O+ H2SO4 conc., goes by E2, requires heat:+HH:H+HSO4CH2 CH2H2SO4H2O::++-+Rsec, Rtert OH + H : E1 + rearrangementsCH3CH2OH + HAcid, ΔCCHOHCC+HOHDehydrationCan be messy!+R = primary < secondary < tertiaryRelative ReactivityCH3CCCH3HHOHCH3CH2H2SO4, Δ -HOHH3CCCHH3CCH2CH354%8%CHCH3CH3CCHHCH3+ other isomers+Major Problem: RearrangementsOHα-Terpineol cajuput oil, pine oil, and petitgrain oil+Terpinolene15%Limonene9%33% H2SO4, 1 h, 100°C-H2O++28.5%Isoterpinolene18.5%γ-Terpinene15%α-TerpineneTerpenes: the scent of soap+