1. Properties and Synthesis
Alkenes & Alkynes I
Properties and Synthesis

2. Alkenes & Alkynes Chapter 7
Physical Properties of Alkenes & Alkynes
E-Z System
Vicinal (Vic) & Geminal (Gem)
Relative Stabilities of Alkenes & Cycloalkenes
Zaitsev’s Rule & Hofmann Rule
Dehydration of Alcohols
Terminal Alkynes
Syn & Anti Additions
Hydrogenation of Alkynes
Index of Hydrogen Deficiency

3. Physical Properties Factors Intermolecular Forces (IMF)
Molecular Weight (MW)
Symmetry

4. Boiling Point Trends-General
Ethene C 2C
Propene C 3C
1-Butene C 4C
1-Pentene C 5C
1-Hexene C 6C
1-Heptene C 7C
As Molecular Weight Increases Boiling Point (BP) increases.

5. Boiling Point Trends-Branching
Branching decreases boiling point due to decrease surface area which translates into fewer surface electrons participating in building intermolecular forces (IMF).

6. Boiling Point Trends-Branching
1-Pentene C
2-Methyl-1-butene 31 0C
Basically no significant difference in BP.

7. Sharing the Burden (Concept Review)
The more the molecule can spread out the burden (Charge, Pi electrons, free radical, etc.) the more stable the molecule will be.
                               
                              
                        
                     3° 2° 1°
Methyl
most stable
next-to-most stable
next-to-least stable
least stable

8. Sharing the Burden
Ability to spread out the burden of the double bond translates into greater stability.
Pi electrons have greater flow with more Beta-Carbons.

9. Boiling Point Trends-Branching
Branching decreases boiling point due to decrease surface area which translates into fewer surface electrons participating in building intermolecular forces.
Branching of Alkenes leads to greater flow of Pi electrons which can participate in building intermolecular forces.
The two factors together cancel each other out resulting in no significant change in BP

10. Boiling Point Trends-Cis/Trans
Analyze the handout (Chart with Physical Properties of Alkenes)
Dipole Moments

11. Boiling Point Trends-Cis/Trans
Cis has a higher BP than Trans due to the stronger dipole of the Cis.

12. BP Trends-Cis/Trans vs. Terminal
Analyze the handout (Chart with Physical Properties of Alkenes) H
CH3CH2
Dipole Moments
1-Butene

13. BP Trends-Cis/Trans & Terminal
Cis & Trans have higher BP than terminal alkenes due to increase flow of Pi electrons resulting in stronger IMF.
Cis has a higher BP than Trans due to the stronger dipole of the Cis.

14. Alkane vs. Alkene Boiling Point
C C
C C
-0.5 0C C
36.1 0C C
68.7 0C C
98.4 0C C
Decreased
Decreased
Decreased
Decreased
Decreased
Decreased

15. Alkane vs. Alkene Boiling Point
Adding a double bond decreases the BP due to decreased sigma electrons participating in building IMF.
Why do Cis and Trans Butene have higher BP’s than Butane?

16. Melting Point Trends-General
Ethene C C
Propene C C
1-Butene C C
1-Pentene C C
1-Hexene C C
1-Heptene C 7C
As Molecular Weight Increases Melting Point (MP) increases, except where symmetry plays a role.

17. Melting Point Trends-Branching
No general trend
If branching increases symmetry, melting point (MP) increases.
If branching decreases symmetry, melting point decreases

18. Melting Point Trends-Cis/Trans
Analyze the handout (Chart with Physical Properties of Alkenes)
Planes of Symmetry
One Plane of Symmetry
Two Planes of Symmetry

19. Melting Point Trends-Cis/Trans
Trans has a higher MP than Cis due to the greater symmetry. Trans has two planes of symmetry, while Cis only has one plane of symmetry.

20. Melting Point Trends-Cis/Trans
Analyze the handout (Chart with Physical Properties of Alkenes) H
CH3CH2
Planes of Symmetry
1-Butene
One Plane of Symmetry
Two Planes of Symmetry No

21. Melting Point Trends-Cis/Trans
Cis & Trans have higher MP than terminal alkenes due to greater symmetry
Trans has a higher MP than Cis due to the greater symmetry. Trans has two planes of symmetry, while Cis only has one plane of symmetry.

22. Alkane vs. Alkene Melting Point
Increased
-183 0C C
-188 0C C
-138 0C C
-130 0C C
-95 0C C
-910C C
Non-Significant
Decreased
Decreased
Decreased
Decreased

23. Alkane vs. Alkene Melting Point
Adding a double bond in a central location increases symmetry and thus increases MP.
Adding a double bond in a non-central location decreases symmetry and thus decreases MP.
How does Cis and Trans Butene MP compared to the MP of Butane? How do you explain this?

24. Saturated & Unsaturated
Saturated – The carbon atoms of the molecule are complete filled with Hydrogen (No double or triple bonds)
Unsaturated-Some carbon atoms have room for additional hydrogen (Double or triple bonds present)
Polyunsaturated-Many carbon atoms have room for additional hydrogen (Many Double or triple bonds present)

25. Saturated & Unsaturated
Saturated – CH3CH2CH2CH2CH2CH2CH2CH3
Unsaturated- CH3CH=CHCH2CH2CH2CH2CH3
CH3CH=CHCH2CH=CHCH2CH3
Polyunsaturated-
CH2=CHCH=CHCH=CHCH=CH2
Fatty Acid- CH3CH2CH2CH2CH2CH2CH2COOH

26. Saturated & Unsaturated Fatty Acids
Source:

27. Saturated & Unsaturated Fats
Source:

28. Saturated & Unsaturated Fats

29. Saturated & Unsaturated Fatty Acids

30. Phospholipids
Source:

31. Cell Membrane Fluidity

32. R-S System (Concept Review)

33. E-Z System E
Entgegen (Opposite) Z
Zusammen (Together)

34. E-Z System
(Z) Together
(E) Opposite

35. E-Z System Problem 7.1 Page 284 (A-D) Answers:
(E)-1-bromo-1-chloro-1-pentene
(E)-2-bromo-1-chloro-1-iodo-1-butene
(Z)-3,5-dimethyl-2-hexene
(Z)-1-chloro-1-iodo-2-methyl-1-butene

36. Vic vs. Gem Vic CH2BrCH2Br Gem CHBr2CH3 Vicinal
Latin Vicinus meaning adjacent
Used when groups are on adjacent Carbons
Gem CHBr2CH3
Geminal
Latin Geminus meaning twins
Used when groups are on the same Carbon

37. Stabilities of Alkenes-General
Ability to spread out the burden of the double bond translates into greater stability.

38. Stabilities of Alkenes-(Cis/Trans)
Trans is more stable due to less crowding, less electron repulsion.

39. Stabilities of Alkenes-(Cis/Trans)
Cis is more stable due to less ring strain, less electron repulsion.
Ring strain electron repulsion is stronger than crowding electron repulsion.

40. Vision & Cis-Trans Stability

41. Stabilities of Alkenes-(Cis/Trans)
Gem is more stable than Vic due to greater electron flow.

42. Zaitsev Rule Zaitsev, Saytzeff, Saytseff, Saytzev
States that the major product of an elimination reaction will be the more stable alkene.
* Major Product

43. Zaitsev Rule Zaitsev, Saytzeff, Saytseff, Saytzev
States that the Major product of an elimination reaction will be the more stable alkene
Ea driven reaction (Kinetic Control)

44. Ea driven reaction (Kinetic Control)
Rate of formation of the product with the lower Ea is faster, thus it accumulates faster.

45. Ea driven reaction (Kinetic Control)

46. Hofmann Rule Opposite of Zaitsev Rule
Applies when major product of an elimination reaction is the less stable alkene.
Collision Driven Reaction

47. Hofmann Rule

48. Ea vs. Collision CH3CHBrCH(CH3)2 + NU:- E2 E2 SN2 CH3CH=C(CH3)2
CH2=CHCH(CH3)2

49. Ea vs. Collision
SN2 vs. E2
E2 (Zaitsev) vs. E2 (Hofmann)

50. Dehydration of Alcohols
Elimination Reaction
Acid Catalyzed
Products alkene or alkyne & water

51. Dehydration of 20 & 30 Alcohols
E1 Mechanism
Step 1: Protonation
Add H+ to the hydroxyl group (Acid Catalysis)
Makes a better leaving group (H2O vs. OH-)

52. Dehydration of 20 & 30 Alcohols
Step 2: Dissociation
Leaving group departs from substrate.
Carbocation formed

53. Dehydration of 20 & 30 Alcohols
Step 3: Nucleophilic Attack
Nucleophile attacks a Beta-Carbon.
Alkene or Alkyne formed

54. Acid-Catalyzed Dehydration of Secondary or Tertiary Alcohols Summary P

55. Dehydration of 10 Alcohols
E2 Mechanism
Step 1: Protonation
Add H+ to the hydroxyl group (Acid Catalysis)
Makes a better leaving group (H2O vs. OH-)

56. Dehydration of 10 Alcohols
Step 2: Nucleophilic Attack
Attacks Beta Hydrogen
Leaving group departs from substrate.
Alkene or Alkyne formed

57. Acid-Catalyzed Dehydration of Primary Alcohols Summary P. 303
Step 1
Step 2

58. Molecular Rearrangement

59. Molecular Rearrangement
Molecular Rearrangement occurs to partially stabilize the carbocation

60. Terminal Alkynes Triple bond at the end of the carbon chain
Moderate acidity
Stronger acid than terminal alkenes and alkanes
Rare in Nature

61. Terminal Alkynes- Creating a Nucleophile

62. Terminal Alkynes- Nucleophilic Attack

63. Terminal Alkynes-Advantages
Can create any size carbon chain
Can place the triple bond in any location
Can then add functional groups via the triple bond.

64. Syn & Anti Additions

65. Hydrogenation of Alkynes
Addition of a hydrogen molecule (H2) across the triple bond.
Three reactions, leading to three different products.
Platinum & H2
P-2 Catalyst (Nickel Boride)
Lithium

66. Hydrogenation of Alkynes Platinum & H2
Hydrogen adds across both Pi bonds to produce an Alkane

67. Hydrogenation of Alkynes P-2 Catalyst
Adds across only the first Pi bond
Syn Addition
Z or Cis alkene arrangement

68. Hydrogenation of Alkynes Lithium
Adds across only the first Pi bond
Anti Addition
E or Trans alkene arrangement

69. Index of Hydrogen Deficiency
Abbreviated IHD
Counts missing pairs of hydrogen
Each double bond has an IHD of 1
Each ring has an IHD of 1
Each triple bond has an IHD of 2
General formula for a straight chain alkane is RnH2n+2

70. Index of Hydrogen Deficiency

71. Index of Hydrogen Deficiency
Hydrogenation remove double and triple bonds
Hydrogenation will not effect ring structures

72. Index of Hydrogen Deficiency
Hydrogenation remove double and triple bonds
Hydrogenation will not effect ring structures

73. IHD Example Problems
P. 185 problems 4.20, 4.21

74. IHD & Functional Groups
Alkane CH3CH H
Halogen CH3CH2Br 5 H
Need to add one hydrogen for each halogen

75. IHD & Functional Groups
Alkane CH3CH H
Alcohol CH3CH2OH 6 H
Ignore Oxygen

76. IHD & Functional Groups
Alkane CH3CH H
Aldehyde CH3CHO 4 H
Contains carbonyl group (C=O)
Thus contains one double bond
Thus IHD = 1
Ignore Oxygen

77. IHD & Functional Groups
Alkane CH3CH H
Amine CH3CH2NH2 7 H
Subtract one hydrogen for each nitrogen present.

78. IHD & Functional Groups
For compounds containing halogen atoms, count the halogen atoms as hydrogen atoms.
For compounds containing oxygen, ignore the oxygen.
For compounds containing nitrogen, subtract one hydrogen for each nitrogen present.

79. Problems (p.329)
7.27
7.31
7.34
7.35
7.37
7.38
7.43
7.44
7.45
7.48 – 7.51