1. 7. Alkenes: Structure and Reactivity
Why this Chapter?
C-C double bonds are present in most organic and biological molecules
To examine consequences of alkene stereoisomerism
To focus on general alkene reaction: electrophilic addition

2. Alkene - Hydrocarbon With Carbon-Carbon Double Bond
Also called an olefin but alkene is better
Includes many naturally occurring materials
Flavors, fragrances, vitamins

3. 7.1 Industrial Preparation and Use of Alkenes
Ethylene and propylene are the most important organic chemicals produced

4. 7.2 Calculating Degree of Unsaturation
Relates molecular formula to possible structures
Degree of unsaturation: number of multiple bonds or rings
Formula for a saturated acyclic compound is CnH2n+2
Each ring or multiple bond replaces 2 H's

5. Example: C6H10 Saturated is C6H14 This has two degrees of unsaturation
Therefore 4 H's are not present
This has two degrees of unsaturation
Two double bonds?
or triple bond?
or two rings
or ring and double bond

6. Degree of Unsaturation With Other Elements
Organohalogens (X: F, Cl, Br, I)
Halogen replaces hydrogen
C4H6Br2 and C4H8 have one degree of unsaturation
Organoxygen compounds (C,H,O) - if connected by single bonds
These don't affect the total count of H's

7. Organonitrogen compounds
Nitrogen has three bonds
So if it connects where H was, it adds a connection point
Subtract one H for equivalent degree of unsaturation in hydrocarbon

8. Summary - Degree of Unsaturation
Count pairs of H's below CnH2n+2
Add number of halogens to number of H's (X equivalent to H)
Ignore oxygens (oxygen links H)
Subtract N's - they have two connections

9. 7.3 Naming of Alkenes Name the parent hydrocarbon
Number carbons in chain so that double bond carbons have lowest possible numbers
Rings have “cyclo” prefix

10. Many Alkenes Are Known by Common Names

11. 7.4 Cis-Trans Isomerism in Alkenes
Carbon atoms in a double bond are sp2-hybridized
Three equivalent orbitals at 120º separation in plane
Fourth orbital is atomic p orbital
Combination of electrons in two sp2 orbitals of two atoms forms  bond between them
Additive interaction of p orbitals creates a  bonding orbital
Subtractive interaction creates a  anti-bonding orbital
Occupied  orbital prevents rotation about -bond
Rotation prevented by  bond - high barrier, about 268 kJ/mole in ethylene

12. Rotation of  Bond Is Prohibitive
This prevents rotation about a carbon-carbon double bond (unlike a carbon-carbon single bond).
Creates possible alternative structures

13. The presence of a carbon- carbon double bond can create two possible structures
cis isomer - two similar groups on same side of the double bond
trans isomer - similar groups on opposite sides
Each carbon must have two different groups for these isomers to occur

14. Cis, Trans Isomers Require That End Groups Must Differ in Pairs
180°rotation superposes
Bottom pair cannot be superposed without breaking C=C

15. 7.5 Sequence Rules: The E,Z Designation
Neither compound is clearly “cis” or “trans”
Substituents on C1 are different than those on C2
We need to define “similarity” in a precise way to distinguish the two stereoisomers
Cis, trans nomenclature only works for disubstituted double bonds

16. E,Z Stereochemical Nomenclature
Priority rules of Cahn, Ingold, and Prelog
Compare where higher priority groups are with respect to bond and designate as prefix
E -entgegen, opposite sides
Z - zusammen, together on the same side

17. Ranking Priorities: Cahn-Ingold-Prelog Rules
Must rank atoms that are connected at comparison point
Higher atomic number gets higher priority
Br > Cl > S > P > O > N > C > H

18. Extended Comparison RULE 2
If atomic numbers are the same, compare at next connection point at same distance
Compare until something has higher atomic number
Do not combine – always compare

19. Dealing With Multiple Bonds:
RULE 3
Substituent is drawn with connections shown and no double or triple bonds
Added atoms are valued with 0 ligands themselves

20. 7.6 Stability of Alkenes
Cis alkenes are less stable than trans alkenes
Compare heat given off on hydrogenation: Ho
Less stable isomer is higher in energy
And gives off more heat
tetrasubstituted > trisubstituted > disubstituted > monosusbtituted
hyperconjugation stabilizes

21. Comparing Stabilities of Alkenes
Evaluate heat given off when C=C is converted to C-C
More stable alkene gives off less heat
trans-Butene generates 5 kJ less heat than cis-butene

22. Hyperconjugation
Electrons in neighboring filled  orbital stabilize vacant antibonding  orbital – net positive interaction
Alkyl groups are better than H

23. 7.7 Electrophilic Addition of Alkenes
General reaction mechanism: electrophilic addition
Attack of electrophile (such as HBr) on  bond of alkene
Produces carbocation and bromide ion
Carbocation is an electrophile, reacting with nucleophilic bromide ion

24. Electrophilic Addition Energy Path
Two step process
First transition state is high energy point

25. Electrophilic Addition for preparations
The reaction is successful with HCl and with HI as well as HBr
HI is generated from KI and phosphoric acid

26. 7.8 Orientation of Electrophilic Addition: Markovnikov’s Rule
In an unsymmetrical alkene, HX reagents can add in two different ways, but one way may be preferred over the other
If one orientation predominates, the reaction is regiospecific
Markovnikov observed in the 19th century that in the addition of HX to alkene, the H attaches to the carbon with the most H’s and X attaches to the other end (to the one with the most alkyl substituents)
This is Markovnikov’s rule

27. Example of Markovnikov’s Rule
Addition of HCl to 2-methylpropene
Regiospecific – one product forms where two are possible
If both ends have similar substitution, then not regiospecific

28. Markovnikov’s Rule (restated)
More highly substituted carbocation forms as intermediate rather than less highly substituted one
Tertiary cations and associated transition states are more stable than primary cations

29. 7.9 Carbocation Structure and Stability
Carbocations are planar and the tricoordinate carbon is surrounded by only 6 electrons in sp2 orbitals
The fourth orbital on carbon is a vacant p-orbital
The stability of the carbocation (measured by energy needed to form it from R-X) is increased by the presence of alkyl substituents

31. Inductive stabilization of cation species

32. 7.10 The Hammond Postulate
If carbocation intermediate is more stable than another, why is the reaction through the more stable one faster?
The relative stability of the intermediate is related to an equilibrium constant (DGº)
The relative stability of the transition state (which describes the size of the rate constant) is the activation energy (DG‡)
The transition state is transient and cannot be examined

33. Transition State Structures
A transition state is the highest energy species in a reaction step
By definition, its structure is not stable enough to exist for one vibration
But the structure controls the rate of reaction
So we need to be able to guess about its properties in an informed way
We classify them in general ways and look for trends in reactivity – the conclusions are in the Hammond Postulate

34. Examination of the Hammond Postulate
A transition state should be similar to an intermediate that is close in energy
Sequential states on a reaction path that are close in energy are likely to be close in structure - G. S. Hammond

35. Competing Reactions and the Hammond Postulate
Normal Expectation: Faster reaction gives more stable intermediate
Intermediate resembles transition state

36. 7.11 Mechanism of Electrophilic Addition: Rearrangements of Carbocations
Carbocations undergo structural rearrangements following set patterns
1,2-H and 1,2-alkyl shifts occur
Goes to give more stable carbocation
Can go through less stable ions as intermediates

37. Hydride shifts in biological molecules

38. Learning Check:
What is the degree of unsaturation for a compound with the formula C6H10NBr? 1 2 3 4

39. Solution:
What is the degree of unsaturation for a compound with the formula C6H10NBr? 1 2 3 4

40. Which of the following cannot have a triple bond?
Learning Check:
Which of the following cannot have a triple bond?
C6H12
C10H16
C10H18
C4H6O
C4H4O

41. Which of the following cannot have a triple bond?
Solution:
Which of the following cannot have a triple bond?
C6H12
C10H16
C10H18
C4H6O
C4H4O

42. What is the IUPAC name for the following compound?
Learning Check:
What is the IUPAC name for the following compound?
(E)-2,5-dimethylhex-2-ene
(Z)-2,5-dimethylhex-2-ene
1-isobutyl-2-methylprop-1-ene
1-isopropyl-2-methylbut-2-ene
2,5-dimethylhex-2-ene

43. What is the IUPAC name for the following compound?
Solution:
What is the IUPAC name for the following compound?
(E)-2,5-dimethylhex-2-ene
(Z)-2,5-dimethylhex-2-ene
1-isobutyl-2-methylprop-1-ene
1-isopropyl-2-methylbut-2-ene
2,5-dimethylhex-2-ene

44. Which of the following molecules is trans-4-ethyl-5-methylhex-2-ene?
Learning Check:
Which of the following molecules is trans-4-ethyl-5-methylhex-2-ene? a b c d e 1. 2. 4. 3. 5.

45. Which of the following molecules is trans-4-ethyl-5-methylhex-2-ene?
Solution:
Which of the following molecules is trans-4-ethyl-5-methylhex-2-ene? a b c d e 1. 2. 4. 3. 5.

46. How many isomers exist for C2H2Br2?
Learning Check:
How many isomers exist for C2H2Br2? 1 2 3 4 5

47. How many isomers exist for C2H2Br2?
Solution:
How many isomers exist for C2H2Br2? 1 2 3 4 5

48. What is the major product of the following reaction?
Learning Check:
What is the major product of the following reaction? 2. 1. 3. 4. 5. a b c d e

49. What is the major product of the following reaction?
Solution:
What is the major product of the following reaction? 2. 1. 3. 4. 5. a b c d e

50. Which of the following carbocations is most stable?
Learning Check:
Which of the following carbocations is most stable? 1. 2. 3. 4. 5. a b c d e

51. Which of the following carbocations is most stable?
Solution:
Which of the following carbocations is most stable? 1. 2. 3. 4. 5. a b c d e

52. What is the best starting material for the following reaction?
Learning Check:
What is the best starting material for the following reaction? 1. 2. 5. 3. 4. a b c d e

53. What is the best starting material for the following reaction?
Solution:
What is the best starting material for the following reaction? 1. 2. 5. 3. 4. a b c d e

54. Learning Check:
Norethindrone, one of the first oral contraceptives, has the structure shown below. What is its degree of unsaturation? 5 6 7 8 9

55. Solution:
Norethindrone, one of the first oral contraceptives, has the structure shown below. What is its degree of unsaturation? 5 6 7 8 9

56. Learning Check:
Consider the reaction between HCl and ethene (the reactants). The carbocation intermediate and the Cl– may revert to the reactants or go on to form the product, chloroethane. Which statement about this process is correct?
The formation of chloroethane from the carbocation and Cl– is the rate limiting process.
The structure of the transition state for conversion of the carbocation to chloroethane resembles the product.
The carbocation intermediate (and Cl–) has energy that is intermediate between the reactants and the products.
The intermediate may bond with Cl– on either of its two carbons.
Based on the Hammond postulate, the carbocation and Cl– form the chloroethane faster than they revert to reactants.

57. Solution:
Consider the reaction between HCl and ethene (the reactants). The carbocation intermediate and the Cl– may revert to the reactants or go on to form the product, chloroethane. Which statement about this process is correct?
The formation of chloroethane from the carbocation and Cl– is the rate limiting process.
The structure of the transition state for conversion of the carbocation to chloroethane resembles the product.
The carbocation intermediate (and Cl–) has energy that is intermediate between the reactants and the products.
The intermediate may bond with Cl– on either of its two carbons.
Based on the Hammond postulate, the carbocation and Cl– form the chloroethane faster than they revert to reactants.

58. Learning Check:
What is(are) the double bond configuration(s) in the following compound? E Z
E, Z
E, E
Z, Z

59. Solution:
What is(are) the double bond configuration(s) in the following compound? E Z
E, Z
E, E
Z, Z

60. Which of the alkenes has the most negative heat of hydrogenation?
Learning Check:
Which of the alkenes has the most negative heat of hydrogenation? 1. 2. 5. 4. 3. a b c d e

61. Which of the alkenes has the most negative heat of hydrogenation?
Solution:
Which of the alkenes has the most negative heat of hydrogenation? 1. 2. 5. 4. 3. a b c d e

62. Learning Check:
Which statement best describes the energetics of the reaction shown below? C2H H2 → C2H6
exothermic with positive activation energy
exothermic with negative activation energy
endothermic with positive activation energy
endothermic with negative activation energy
cannot be determined without knowing BDE’s

63. Solution:
Which statement best describes the energetics of the reaction shown below? C2H H2 → C2H6
exothermic with positive activation energy
exothermic with negative activation energy
endothermic with positive activation energy
endothermic with negative activation energy
cannot be determined without knowing BDE’s

64. Which of the following σ-bonds participate in hyperconjugation?
Learning Check:
Which of the following σ-bonds participate in hyperconjugation?
A and B
A and E
B and E
C and D
C and E

65. Which of the following σ-bonds participate in hyperconjugation?
Solution:
Which of the following σ-bonds participate in hyperconjugation?
A and B
A and E
B and E
C and D
C and E

66. Learning Check:
Which of the following bromides is the major product of the reaction shown below, assuming that there are no carbocation rearrangements? 1. 2. 5. 4. 3. a b c d e

67. Solution:
Which of the following bromides is the major product of the reaction shown below, assuming that there are no carbocation rearrangements? 1. 2. 5. 4. 3. a b c d e

68. Learning Check:
How many degrees of unsaturation are there in the molecule of prizmane? 1 2 3 4 5

69. Solution:
How many degrees of unsaturation are there in the molecule of prizmane? 1 2 3 4 5

70. Learning Check: 4 kcal/mol 2 kcal/mol –2 kcal/mol –4 kcal/mol
The heat of formation of 2-methylpropane is –32 kcal/mol. The heat of formation of butane is –30 kcal/mol. The heat of hydrogenation of 2-methylpropene is –28 kcal/mol. What is the heat of formation of 2-methylpropene?
4 kcal/mol
2 kcal/mol
–2 kcal/mol
–4 kcal/mol
–60 kcal/mol

71. Solution: 4 kcal/mol 2 kcal/mol –2 kcal/mol –4 kcal/mol –60 kcal/mol
The heat of formation of 2-methylpropane is –32 kcal/mol. The heat of formation of butane is –30 kcal/mol. The heat of hydrogenation of 2-methylpropene is –28 kcal/mol. What is the heat of formation of 2-methylpropene?
4 kcal/mol
2 kcal/mol
–2 kcal/mol
–4 kcal/mol
–60 kcal/mol

72. Learning Check:
Hyperconjugation is most important in stabilization of which kind of species?
alkanes
alkenes
carbocations
carbanions
carbon radicals

73. Solution:
Hyperconjugation is most important in stabilization of which kind of species?
alkanes
alkenes
carbocations
carbanions
carbon radicals

74. Learning Check:
If D = 2H reacts in the same way as 1H, what would be the most likely product of the following addition run under conditions that favor rearrangements? 1. 2. 3. 4. 5. a b c d e

75. Solution:
If D = 2H reacts in the same way as 1H, what would be the most likely product of the following addition run under conditions that favor rearrangements? 1. 2. 3. 4. 5. a b c d e

76. Learning Check:
Which of the marked bonds in the carbocation shown below is the most likely to migrate to generate another carbocation? A B C D E

77. Solution:
Which of the marked bonds in the carbocation shown below is the most likely to migrate to generate another carbocation? A B C D E