Chapter 11 Alkenes and IR I.Alkene Nomenclature A.Unsaturation 1)Alkanes: C n H 2n+2 2)Alkenes: C n H 2n 3)Degree of Unsaturation a)Tells us how many rings.

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Chapter 11 Alkenes and IR I.Alkene Nomenclature A.Unsaturation 1)Alkanes: C n H 2n+2 2)Alkenes: C n H 2n 3)Degree of Unsaturation a)Tells us how many rings and double bonds in molecule b)H sat = 2C + 2 – X + N (Ignore O, S) c)Degree of Unsaturation = (H sat – H act )/2 d)Example: C 5 H 8 NOCl i.H sat = 2(5) + 2 – = 12 ii.(H sat – H act )/2 = (12 – 8)/2 = 2 degrees of unsaturation

B.Nomenclature 1.Common Names end with –ylene a.Ethylene b.Propylene 2.IUPAC: Replace –ane of an alkane with –ene of an alkene a.Ethene b.Propene 3.Alkenes follow alkane nomenclature, with double bond location numbered closest to end a.1-butene b.2-butene c.Cylclohexene 4.Substituents named as prefixes with lowest numbers a.3-methyl-1-pentene b.3-methylcyclohexene

5.Disubstituted Alkenes can be cis or trans streoisomers a.cis-2-butene b.trans-2-butene c.Cycloalkenes cis unless large 6.Tri- or Tetra-substituted Alkenes can be E or Z stereoisomers a.Use priorities from R/S nomenclature b.Assign 1-2 on each carbon c.Move from to trace out an E or Z 7.Alcohols have priority over alkene in numbering: Alkenol 8.Alkene substituents are named alkenyl Z-1-bromo-1,2-difluoroethene E-1-chloro-3-ethyl-4-methyl-3-heptene 2-propen-1-ol Z-5-chloro-3-ethyl-4-hexen-2-ol ethenylcyclohexane2-propenyl- trans-1-propenyl-

II.Pi-bonding in Alkenes A.The  -bond 1)sp 2 hybridization results in 120 o angles 2)H1s-Csp 2 overlap gives the CH  -bonds 3)Csp 2 —Csp 2 overlap gives the C—C  -bond 4)Cp—Cp overlap gives the C—C  -bond B.Bond Strength 1)Bond strength is proportional to orbital overlap 2)The  -bond in ethene is very strong because of good overlap 3)The  -bond in ethene is fairly weak because of poor overlap 4)Overall, the double bond is stronger than a C—C single bond 5)The weak  -bond will be the reactive part of the molecule

6)Orbital and Energy level diagrams for ethene

7)Thermal Isomerization tells us the  -bond energy a.cis/trans interconversion must go through broken  -bond T.S. b.E a = 65 kcal/mol should be about the same as the  -bond strength c.The  -bond is slightly stronger than alkane due to better sp 2 overlap d.C—H bonds are also stronger than in alkanes (110 kcal/mol) 8)Radical H-atom abstraction doesn’t occur in alkenes because of the strong C-H bonds. The chemistry is dominated by the weak  -bond.

III.Physical properties of Alkenes A.Boiling points are about like alkanes B.Melting points depend on the isomer 1)cis-alkenes have a U-shape that disrupts packing in the solid, giving lower temperatures (Vegetable oils have cis-alkenes) 2)trans-alkenes have melting points close to the alkanes C.Polarization 1)Alkenes are more polar than alkanes due to more e-withdrawing sp 2 hybrid orbitals (more s-character draws e- closer to nucleus) 2)cis-alkenes are more polar than trans-alkenes due to their shape D.Acidity of alkenes > alkanes, again because of the greater s-character of sp 2 hybrid orbitals. 1)Ethane pK a = 50 2)Ethene pK a = 44

IV.NMR of Alkene  -electrons deshield hydrogens 1)Alkane H 1.0 ppm 2)Alkene H 5-6 ppm 3)Spectrum of an alkene

B.Coupling in Alkenes depends on the isomer C. 13 C NMR of alkenes gives peaks at ppm due to deshielding