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
Alkene - Hydrocarbon With CarbonCarbon Double Bond
Also called an olefin but alkene is better
 Includes many naturally occurring materials
◦ Flavors, fragrances, vitamins

2
7.1 Industrial Preparation and
Use of Alkenes

Ethylene and propylene are the most
important organic chemicals produced
3
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
4
Example: C6H10

Saturated is C6H14
◦ 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
5
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

6
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
7
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

8
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

9
Many Alkenes Are Known by
Common Names
10
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
11
Rotation of  Bond Is Prohibitive


This prevents rotation about a carbon-carbon double
bond (unlike a carbon-carbon single bond).
Creates possible alternative structures
12
The presence of a carboncarbon 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

13
Cis, Trans Isomers Require That End
Groups Must Differ in Pairs
 180°rotation superposes
 Bottom pair cannot be superposed without breaking C=C
14
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

15
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

16
Ranking Priorities: Cahn-IngoldPrelog Rules
RULE 1
 Must rank atoms that are connected at comparison point
 Higher atomic number gets higher priority
◦ Br > Cl > S > P > O > N > C > H
17
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
18
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
19
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
20
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

21
Hyperconjugation
Electrons in neighboring filled  orbital stabilize vacant
antibonding  orbital – net positive interaction
 Alkyl groups are better than H

22
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

23
Electrophilic Addition Energy Path
Two step process
 First transition state is high energy point

24
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

25
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

26
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

27
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

28
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

29
30
Inductive stabilization of cation species
31
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 (Gº)
◦ The relative stability of the transition state (which
describes the size of the rate constant) is the activation
energy (G‡)
◦ The transition state is transient and cannot be examined
32
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
33
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
34
Competing Reactions and the
Hammond Postulate
Normal Expectation: Faster reaction gives more stable
intermediate
 Intermediate resembles transition state

35
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

36
Hydride shifts in biological
molecules
37
Learning Check:
What is the degree of unsaturation for a
compound with the formula C6H10NBr?
1.
2.
3.
4.
5.
0
1
2
3
4
Solution:
What is the degree of unsaturation for a
compound with the formula C6H10NBr?
1.
2.
3.
4.
5.
0
1
2
3
4
Learning Check:
Which of the following cannot have a
triple bond?
1.
2.
3.
4.
5.
C6H12
C10H16
C10H18
C4H6O
C4H4O
Solution:
Which of the following cannot have a
triple bond?
1.
2.
3.
4.
5.
C6H12
C10H16
C10H18
C4H6O
C4H4O
Learning Check:
What is the IUPAC name for the following
compound?
1.
2.
3.
4.
5.
(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
Solution:
What is the IUPAC name for the following
compound?
1.
2.
3.
4.
5.
(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
Learning Check:
Which of the following molecules is trans-4ethyl-5-methylhex-2-ene?
1.
3.
2.
4.
5.
Solution:
Which of the following molecules is trans-4ethyl-5-methylhex-2-ene?
1.
3.
2.
4.
5.
Learning Check:
How many isomers exist for C2H2Br2?
1.
2.
3.
4.
5.
1
2
3
4
5
Solution:
How many isomers exist for C2H2Br2?
1.
2.
3.
4.
5.
1
2
3
4
5
Learning Check:
What is the major product of the following
reaction?
HBr
2.
1.
3.
4.
5.
Solution:
What is the major product of the following
reaction?
HBr
2.
1.
3.
4.
5.
Learning Check:
Which of the following carbocations is most
stable?
1.
2.
3.
4.
5.
Solution:
Which of the following carbocations is most
stable?
1.
2.
3.
4.
5.
Learning Check:
What is the best starting material for the
following reaction?
+
?
HBr
Br
1.
3.
2.
4.
5.
Solution:
What is the best starting material for the
following reaction?
+
?
HBr
Br
1.
3.
2.
4.
5.
Learning Check:
Norethindrone, one of the first oral contraceptives,
has the structure shown below. What is its degree
of unsaturation?
H3C
OH
H
norethindrone
1.
2.
3.
4.
5.
5
6
7
8
9
O
Solution:
Norethindrone, one of the first oral contraceptives,
has the structure shown below. What is its degree
of unsaturation?
H3C
OH
H
norethindrone
1.
2.
3.
4.
5.
5
6
7
8
9
O
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?
1.
2.
3.
4.
5.
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.
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?
1.
2.
3.
4.
5.
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.
Learning Check:
What is(are) the double bond
configuration(s) in the following compound?
COOH
NC
1.
2.
3.
4.
5.
E
Z
E, Z
E, E
Z, Z
H2N
OH
CN
Solution:
What is(are) the double bond
configuration(s) in the following compound?
COOH
NC
1.
2.
3.
4.
5.
E
Z
E, Z
E, E
Z, Z
H2N
OH
CN
Learning Check:
Which of the alkenes has the most negative heat
of hydrogenation?
1.
3.
2.
4.
5.
Solution:
Which of the alkenes has the most negative heat
of hydrogenation?
1.
3.
2.
4.
5.
Learning Check:
Which statement best describes the
energetics of the reaction shown below?
C2H4 + H2 → C2H6
1.
2.
3.
4.
5.
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
Solution:
Which statement best describes the
energetics of the reaction shown below?
C2H4 + H2 → C2H6
1.
2.
3.
4.
5.
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
Learning Check:
Which of the following σ-bonds participate in
hyperconjugation?
A
E
H
1.
2.
3.
4.
5.
A and B
A and E
B and E
C and D
C and E
B
CH3
D
H
C
Solution:
Which of the following σ-bonds participate in
hyperconjugation?
A
E
H
1.
2.
3.
4.
5.
A and B
A and E
B and E
C and D
C and E
B
CH3
D
H
C
Learning Check:
Which of the following bromides is the major product of the
reaction shown below, assuming that there are no carbocation
rearrangements?
+
HBr
C13H17Br
(1 equivalent)
1.
3.
2.
4.
5.
Solution:
Which of the following bromides is the major product of the
reaction shown below, assuming that there are no carbocation
rearrangements?
+
HBr
C13H17Br
(1 equivalent)
1.
3.
2.
4.
5.
Learning Check:
How many degrees of unsaturation are
there in the molecule of prizmane?
1.
2.
3.
4.
5.
1
2
3
4
5
Solution:
How many degrees of unsaturation are
there in the molecule of prizmane?
1.
2.
3.
4.
5.
1
2
3
4
5
Learning Check:
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?
1.
2.
3.
4.
5.
4 kcal/mol
2 kcal/mol
–2 kcal/mol
–4 kcal/mol
–60 kcal/mol
Solution:
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?
1.
2.
3.
4.
5.
4 kcal/mol
2 kcal/mol
–2 kcal/mol
–4 kcal/mol
–60 kcal/mol
Learning Check:
Hyperconjugation is most important in
stabilization of which kind of species?
1.
2.
3.
4.
5.
alkanes
alkenes
carbocations
carbanions
carbon radicals
Solution:
Hyperconjugation is most important in
stabilization of which kind of species?
1.
2.
3.
4.
5.
alkanes
alkenes
carbocations
carbanions
carbon radicals
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
Cl favor rearrangements?
+ D that
1.
2.
3.
4.
5.
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
Cl favor rearrangements?
+ D that
1.
2.
3.
4.
5.
Learning Check:
Which of the marked bonds in the carbocation shown
below is the most likely to migrate to generate another
carbocation?
A
H3C
H
B
1.
2.
3.
4.
5.
A
B
C
D
E
C
E D
H
Solution:
Which of the marked bonds in the carbocation shown
below is the most likely to migrate to generate another
carbocation?
A
H3C
H
B
1.
2.
3.
4.
5.
A
B
C
D
E
C
E D
H