1. Ch #12
Alkenes and Alkynes

2. Alkene Introduction Hydrocarbon with carbon-carbon double bonds
Sometimes called olefins, “oil-forming gas”
General formula CnH2n n≥2
Examples
n=2 C2H4

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

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

11. Alkene Substituents = CH2 methylene - CH = CH2 vinyl - CH2 - CH = CH2
allyl
- CH2 - CH = CH2
allyl
Name = ?

12. Alkene Substituents = CH2 methylene - CH = CH2 vinyl - CH2 - CH = CH2
allyl
- CH2 - CH = CH2
allyl
Name = Methylenecyclohexane
Name =

13. Alkene Substituents = CH2 methylene - CH = CH2 vinyl - CH2 - CH = CH2
allyl
Name = Methylenecyclohexane
Name = 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. Alkyne Examples
Isopropyl methyl acetylene

17. Alkyne Examples Isopropyl methyl acetylene
sec-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. Name these:
trans-2-pentene

21. Name these:
trans-2-pentene

22. Name these:
trans-2-pentene
cis-1,2-dibromoethene

23. Which of the following show cis/trans isomers. a. 1-pentene. b
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 2 1 1 2 1 2 2Z 5E

26. Example, E-Z 1 2 2 1 1 2 1 2 2Z 5E
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 Elimination Reactions: Dehydrohalogenation (-HX)
Dehydration of alcohols (-H2O)
Examples:
Zaitsev’s rule: The major product contains the most substituted double bond

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

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.” H
O-H
Examples: H H H+
C-C H H
C=C
+ H-O-H H
CH3 H
CH3
Major Products H Cl H H
C-C
C=C
+ H-Cl H H H
CH3 H
CH3

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

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

35. Terpenes

36. Terpenes head tail head head tail head tail tail head Geraniol (roses)
Head to tail link of two isoprenes
Called diterpene
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 40

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

42. Give the formula for an alkene.
a. CnH2n-4
b. CnH2n-2
c. CnH2n
d. CnH2n+2
e. CnH2n+4 42

43. Answer
a. CnH2n-4
b. CnH2n-2
c. CnH2n
d. CnH2n+2
e. CnH2n+4 43

44. Name CH3CH=CHCH=CH2. a. 2,4-butadiene b. 1,3-butadiene
c. 2,4-pentadiene
d. 1,3-pentadiene
e. 1,4-pentadiene 44

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

46. Calculate the unsaturation number for C6H10BrCl.
d. 3 46

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

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

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 49

50. Answer a. CH3COOH b. CH3CHO c. CH3CH2OH d. HOCH2CH2OH e. CH3CH(OH)2
Ethylene oxide is formed first, followed by a ring opening to form ethylene glycol. 50

51. a. ClCH2CH2Cl
b. ClCH=CHCl
c. CH2=CH2
d. CH2=CHCl 51

52. Answer a. ClCH2CH2Cl b. ClCH=CHCl c. CH2=CH2 d. CH2=CHCl
Chlorine is added across the double bond, then HCl is lost. 52

53. a. (CH3)2CHOH
b. CH3CH2CH2OH
c. HOCH2CH2CH2OH
d. CH3CH(OH)CH2OH 53

54. Answer a. (CH3)2CHOH b. CH3CH2CH2OH c. HOCH2CH2CH2OH d. CH3CH(OH)CH2OH
Water adds by Markovnikov’s orientation across the double bond. 54

55. a. [CH2CH(CH3)]n
b. [CH2CH2]n
c. [CH2=CH(CH3)]n
d. [CH2=CH2]n 55

56. Answer a. [CH2CH(CH3)]n b. [CH2CH2]n c. [CH2=CH(CH3)]n d. [CH2=CH2]n56

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

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

59. a. CH3CCCH3
b. CH2=CHCH=CH2
c. CH3CH=CHCH3
d. CH3CH2CH2CH3 59

60. Answer a. CH3CCCH3 b. CH2=CHCH=CH2 c. CH3CH=CHCH3 d. CH3CH2CH2CH3
Hydrogen adds across the double bond to form an alkane. 60

61. 7.15
a. (CH3)2CHOSO3H
b. CH3CH=CH2
c. (CH3)2C=O
d. CH3CH2COOH 61

62. 7.15 Answer a. (CH3)2CHOSO3H b. CH3CH=CH2 c. (CH3)2C=O d. CH3CH2COOH
Acid dehydrates alcohols to form alkenes. 62

63. 7.16 Dehydration of alcohols occurs by what mechanism?
a. SN1
b. SN2
c. E1
d. E2 63

64. 7.16 Answer
a. SN1
b. SN2
c. E1
d. E2
The dehydration of alcohols occurs by an E1 mechanism. 64

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

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

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

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

69. 7.19 a. (CH3)3CO-, (CH3)3COH b. CH3CH2O-, CH3CH2OH c. NaI, acetone
d. H2, Pd 69

70. 7.19 Answer a. (CH3)3CO-, (CH3)3COH b. CH3CH2O-, CH3CH2OH
c. NaI, acetone
d. H2, Pd
The Hofmann product (least substituted) is favored with a bulky base. 70

71. 7.20 a. Pt, 500o C b. H2, Pt c. H2SO4, 150o C d. NaI, acetone e. NaOH71

72. 7.20 Answer a. Pt, 500o C b. H2, Pt c. H2SO4, 150o C d. NaI, acetone
e. NaOH
Dehydrogenation occurs with a metal catalyst and heat. 72

73. End Chapter #3