1. Ch #12 Alkenes and Alkynes

2. Alkene Introduction Hydrocarbon with carbon-carbon double bonds Sometimes called olefins, “oil-forming gas” General formula C n H 2n n≥2 Examples n=2 C 2 H 4

3. Common Names Usually used for small molecules. Examples: Vinyl carbons are the carbons sharing a double bond Vinyl hydrogens are the hydrogens bonded to vinyl carbons

4. IUPAC Nomenclature Parent is longest chain containing the double or triple bond. -ane changes to –ene (or -diene, -triene) for double bonds, or –yne (or –diyne, -triyne). Number the chain so that the double bond, or triple bond has the lowest possible number. In a ring, the double bond is assumed to be between carbon 1 and carbon 2.

5. Name These Alkenes

6. 1-butene

7. Name These Alkenes 1-butene 2-methyl-2-butene

8. Name These Alkenes 1-butene 2-methyl-2-butene 3-methylcyclopentene

9. Name These Alkenes 1-butene 2-methyl-2-butene 3-methylcyclopentene 2-sec-butyl-1,3-cyclohexadiene

10. Name These Alkenes 1-butene 2-methyl-2-butene 3-methylcyclopentene 2-sec-butyl-1,3-cyclohexadiene 3-n-propyl-1-heptene

11. Alkene Substituents = CH 2 methylene - CH = CH 2 vinyl - CH 2 - CH = CH 2 allyl - CH 2 - CH = CH 2 allyl Name = ?

12. Alkene Substituents = CH 2 methylene - CH = CH 2 vinyl - CH 2 - CH = CH 2 allyl - CH 2 - CH = CH 2 allyl Name = MethylenecyclohexaneName =

13. Alkene Substituents = CH 2 methylene - CH = CH 2 vinyl - CH 2 - CH = CH 2 allyl Name = MethylenecyclohexaneName = vinylcyclohexane

14. Alkyne Common Names Acetylene is the common name for the two carbon alkyne. To give common names to alkynes having more than two carbons, give alkyl names to the carbon groups attached to the vinyl carbons followed by acetylene.

15. Alkyne Examples

16. Isopropyl methyl acetylene

17. Alkyne Examples Isopropyl methyl acetylenesec-butyl Cyclopropyl acetylene

18. Cis-trans Isomerism Similar groups on same side of double bond, alkene is cis. Similar groups on opposite sides of double bond, alkene is trans. Cycloalkenes are assumed to be cis. Trans cycloalkenes are not stable unless the ring has at least 8 carbons.

19. Name these:

20. trans-2-pentene

21. Name these: trans-2-pentene

22. Name these: trans-2-pentenecis-1,2-dibromoethene

23. Which of the following show cis/trans isomers? a. 1-pentene b. 2-pentene c. 1-chloro-1-pentene d. 2-chloro-1-pentene e. 2-chloro-2-pentene

24. E-Z Nomenclature Use the Cahn-Ingold-Prelog rules to assign priorities to groups attached to each carbon in the double bond. If high priority groups are on the same side, the name is Z (for zusammen). If high priority groups are on opposite sides, the name is E (for entgegen).

25. Example, E-Z 1 2 1 2 2 1 1 2 2Z2Z 5E5E

26. 1 2 1 2 2 1 1 2 2Z2Z 5E5E 3,7-dichloro-(2Z, 5E)-2,5-octadiene

27. Physical Properties Low boiling points, increasing with mass. Branched alkenes have lower boiling points. Less dense than water. Nonpolar (Hydrophobic)

28. Alkene Synthesis Dehydrohalogenation (-HX) Dehydration of alcohols (-H 2 O) Examples: Zaitsev’s rule: The major product contains the most substituted double bond Elimination Reactions:

29. Alkene Reactions I. Addition Reactions C=C a. Hydration C-C + H-O-H C=C HO-H b. Hydrogenation C-C + H-H HH c. Halogenation + X-X Catalyst H+H+ Catalyst = Ni, Pt, Pd C-C XX Alcohol Alkane X = Cl, Br, I Dihalide Follows Markovnikov’s Rule

30. Regiospecificity Markovnikov’s Rule: The proton (H + ) of an acid adds to the carbon in the double bond that already has the most H’s. “Rich get richer.” C=C Examples: CH 3 H H HH C=C H CH 3 H + H-O-H H+H+ + H-Cl H C-C H HCl H H C-C H HO-H H CH 3 Major Products

31. Alkene Reactions (2) I. Addition Reactions (cont.) d. Hydrohalogenation C=C C-C + H-X C=C HX e. Glycol Formation + H-O-O-H C-C H-OH-OO-HO-H Alkyl halide Glycol Follows Markovnikov’s Rule

32. Alkene Reactions Step 1: Pi electrons attack the electrophile. Step 2: Nucleophile attacks the carbocation

33. Terpenes Composed of 5-carbon isopentyl groups. Isolated from plants’ essential oils. C:H ratio of 5:8, or close to that. Pleasant taste or fragrant aroma. Examples: Anise oil Bay leaves

34. Terpenes

36. head tail head tail head Geraniol (roses) Head to tail link of two isoprenes Called diterpene head tail head tail Menthol (pepermint) Head to tail link of two isoprenes another diterpene

37. Structure of Terpenes Two or more isoprene units, 2-methyl-1,3- butadiene with some modification of the double bonds. myrcene, from bay leaves =>

38. Classification Terpenes are classified by the number of carbons they contain, in groups of 10. A monoterpene has 10 C’s, 2 isoprenes. A diterpene has 20 C’s, 4 isoprenes. A sesquiterpene has 15 C’s, 3 isoprenes.

39. ALKENE REVIEW

40. Describe the geometry around the carbon–carbon double bond. a.Tetrahedral b.Trigonal pyramidal c.Trigonal planar d.Bent e.Linear

41. Answer a.Tetrahedral b.Trigonal pyramidal c.Trigonal planar d.Bent e.Linear

42. Give the formula for an alkene. a.C n H 2n-4 b.C n H 2n-2 c.C n H 2n d.C n H 2n+2 e.C n H 2n+4

43. Answer a.C n H 2n-4 b.C n H 2n-2 c.C n H 2n d.C n H 2n+2 e.C n H 2n+4

44. Name CH 3 CH=CHCH=CH 2. a.2,4-butadiene b.1,3-butadiene c.2,4-pentadiene d.1,3-pentadiene e.1,4-pentadiene

45. Answer a.2,4-butadiene b.1,3-butadiene c.2,4-pentadiene d.1,3-pentadiene e.1,4-pentadiene

46. Calculate the unsaturation number for C 6 H 10 BrCl. a.0 b.1 c.2 d.3

47. Answer a.0 b.1 c.2 d.3 U = 0.5 [2(6) + 2 – (12)] = 1

48. Name. a. Trans-2-pentene b. Cis-2-pentene c. Trans-3-methyl-2-pentene d. Cis-3-methyl-2-pentene

49. Name. a. E-2-pentene b. Z-2-pentene c. E-3-methyl-2-pentene d. Z-3-methyl-2-pentene e. Z-2-methyl-2-pentene

50. Answer a.CH 3 COOH b.CH 3 CHO c.CH 3 CH 2 OH d.HOCH 2 CH 2 OH e.CH 3 CH(OH) 2 Ethylene oxide is formed first, followed by a ring opening to form ethylene glycol.

51. a.ClCH 2 CH 2 Cl b.ClCH=CHCl c.CH 2 =CH 2 d.CH 2 =CHCl

52. Answer a.ClCH 2 CH 2 Cl b.ClCH=CHCl c.CH 2 =CH 2 d.CH 2 =CHCl Chlorine is added across the double bond, then HCl is lost.

53. a.(CH 3 ) 2 CHOH b.CH 3 CH 2 CH 2 OH c.HOCH 2 CH 2 CH 2 OH d.CH 3 CH(OH)CH 2 OH

54. Answer a.(CH 3 ) 2 CHOH b.CH 3 CH 2 CH 2 OH c.HOCH 2 CH 2 CH 2 OH d.CH 3 CH(OH)CH 2 OH Water adds by Markovnikov’s orientation across the double bond.

55. a.[CH 2 CH(CH 3 )] n b.[CH 2 CH 2 ] n c.[CH 2 =CH(CH 3 )] n d.[CH 2 =CH 2 ] n

56. Answer a.[CH 2 CH(CH 3 )] n b.[CH 2 CH 2 ] n c.[CH 2 =CH(CH 3 )] n d.[CH 2 =CH 2 ] n

57. Identify the product formed from the polymerization of tetrafluoroethylene. a.Polypropylene b.Poly(vinyl chloride), (PVC) c.Polyethylene d.Poly(tetrafluoroethylene), Teflon

58. Answer a.Polypropylene b.Poly(vinyl chloride), (PVC) c.Polyethylene d.Poly(tetrafluoroethylene), Teflon Teflon is formed from the polymerization of tetrafluoroethylene.

59. a.CH 3 C  CCH 3 b.CH 2 =CHCH=CH 2 c.CH 3 CH=CHCH 3 d.CH 3 CH 2 CH 2 CH 3

60. Answer a.CH 3 C  CCH 3 b.CH 2 =CHCH=CH 2 c.CH 3 CH=CHCH 3 d.CH 3 CH 2 CH 2 CH 3 Hydrogen adds across the double bond to form an alkane.

61. 7.15 a.(CH 3 ) 2 CHOSO 3 H b.CH 3 CH=CH 2 c.(CH 3 ) 2 C=O d.CH 3 CH 2 COOH

62. 7.15 Answer a.(CH 3 ) 2 CHOSO 3 H b.CH 3 CH=CH 2 c.(CH 3 ) 2 C=O d.CH 3 CH 2 COOH Acid dehydrates alcohols to form alkenes.

63. 7.16 Dehydration of alcohols occurs by what mechanism? a.S N 1 b.S N 2 c.E1 d.E2

64. 7.16 Answer a.S N 1 b.S N 2 c.E1 d.E2 The dehydration of alcohols occurs by an E1 mechanism.

65. 7.17 Give the products from the catalytic cracking of alkanes. a.Alkanes b.Alkenes c.Alkynes d.Alkanes + alkenes e.Alkanes + alkynes

66. 7.17 Answer a.Alkanes b.Alkenes c.Alkynes d.Alkanes + alkenes e.Alkanes + alkynes

67. 7.18 Give the products from the dehydrogenation of alkanes. a.Alkanes b.Alkenes c.Alkynes d.Alkanes + alkenes e.Alkanes + alkynes

68. 7.18 Answer a.Alkanes b.Alkenes c.Alkynes d.Alkanes + alkenes e.Alkanes + alkynes

69. 7.19 a.(CH 3 ) 3 CO -, (CH 3 ) 3 COH b.CH 3 CH 2 O -, CH 3 CH 2 OH c.NaI, acetone d.H 2, Pd

70. 7.19 Answer a.(CH 3 ) 3 CO -, (CH 3 ) 3 COH b.CH 3 CH 2 O -, CH 3 CH 2 OH c.NaI, acetone d.H 2, Pd The Hofmann product (least substituted) is favored with a bulky base.

71. 7.20 a.Pt, 500 o C b.H 2, Pt c.H 2 SO 4, 150 o C d.NaI, acetone e.NaOH

72. 7.20 Answer a.Pt, 500 o C b.H 2, Pt c.H 2 SO 4, 150 o C d.NaI, acetone e.NaOH Dehydrogenation occurs with a metal catalyst and heat.

73. End Chapter #3