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ALKENES
Alkene Nomenclature
Alkenes
Alkenes are hydrocarbons that contain a
carbon-carbon double bond
also called "olefins"
characterized by molecular formula CnH2n
said to be "unsaturated"
Alkene Nomenclature
H2C
CH2
Ethene
or
Ethylene
(both are acceptable
IUPAC names)
H2C
CHCH3
Propene
(Propylene is
sometimes used
but is not an acceptable
IUPAC name)
Alkene Nomenclature
H2C
CHCH2CH3
1-Butene
1) Find the longest continuous chain that
includes the double bond.
2) Replace the -ane ending of the unbranched
alkane having the same number of carbons
by -ene.
3) Number the chain in the direction that gives
the lowest number to the doubly bonded
carbon.
Alkene Nomenclature
H2C
CHCHCH2Br
CH3
4) If a substituent is present, identify its position
by number. The double bond takes
precedence over alkyl groups and halogens
when the chain is numbered.
The compound shown above is
4-bromo-3-methyl-1-butene.
Alkene Nomenclature
H2C
CHCHCH2OH
CH3
4) If a substituent is present, identify its position
by number. Hydroxyl groups take
precedence over the double bond when the
chain is numbered.
The compound shown above is
2-methyl-3-buten-1-ol.
Alkenyl groups
methylene
H2C
vinyl
H2C
CH
allyl
H2C
CHCH2
isopropenyl
H2C
CCH3
Cycloalkene Nomenclature
Cyclohexene
1) Replace the -ane ending of the cycloalkane
having the same number of carbons by -ene.
Cycloalkene Nomenclature
CH3
CH2CH3
1) Replace the -ane ending of the cycloalkane
having the same number of carbons by -ene.
2) Number through the double bond in the
direction that gives the lower number to the
first-appearing substituent.
Cycloalkene Nomenclature
CH3
6-Ethyl-1-methylcyclohexene
CH2CH3
1) Replace the -ane ending of the cycloalkane
having the same number of carbons by -ene.
2) Number through the double bond in the
direction that gives the lower number to the
first-appearing substituent.
Structure and Bonding in
Alkenes
Structure of Ethylene
bond angles:
H-C-H = 117°
H-C-C = 121°
bond distances:
C—H = 110 pm
C=C = 134 pm
planar
Bonding in Ethylene





• Framework of  bonds
• Each carbon is sp2
hybridized
Bonding in Ethylene
•
Each carbon has a
half-filled p orbital
Bonding in Ethylene
•
Side-by-side overlap of halffilled p orbitals gives a  bond
Isomerism in Alkenes
Isomers
Isomers are different compounds that
have the same molecular formula.
Isomers
Constitutional isomers
Stereoisomers
Isomers
Constitutional isomers
different connectivity
Stereoisomers
same connectivity;
different arrangement
of atoms in space
Isomers
Constitutional isomers
Stereoisomers
consider the isomeric alkenes of
molecular formula C4H8
H
CH2CH3
C
H
1-Butene
H3C
H
H
C
C
H
C
H3C
H
2-Methylpropene
CH3
C
H3C
H
H3C
C
C
H
cis-2-Butene
H
C
CH3
trans-2-Butene
H
CH2CH3
C
H
1-Butene
H3C
H
C
C
H
H3C
C
H
2-Methylpropene
CH3
C
H
H3C
C
Constitutional isomers
H
cis-2-Butene
H
CH2CH3
C
H3C
C
C
H
H
H
1-Butene
C
H3C
H
2-Methylpropene
H
H3C
C
Constitutional isomers
H
C
CH3
trans-2-Butene
Stereoisomers
H3C
CH3
C
H
H
H3C
C
C
H
cis-2-Butene
H
C
CH3
trans-2-Butene
Stereochemical Notation
cis (identical or
analogous substituents
on same side)
trans (identical or
analogous substitutents
on opposite sides)
Figure 5.2
Interconversion of stereoisomeric
alkenes does not normally occur.
Requires that component of double
bond be broken.
cis
trans
Figure 5.2
cis
trans
Naming Stereoisomeric
Alkenes by the E-Z Notational
System
Stereochemical Notation
CH2(CH2)6CO2H
CH3(CH2)6CH2
C
H
C
Oleic acid
H
cis and trans are useful when substituents are
identical or analogous (oleic acid has a cis
double bond)
cis and trans are ambiguous when analogies
are not obvious
Cl
Example
Br
C
H
C
F
What is needed:
1)
2)
systematic body of rules for ranking
substituents
new set of stereochemical symbols other
than cis and trans
The E-Z Notational System
E : higher ranked substituents on opposite sides
Z : higher ranked substituents on same side
higher
C
lower
C
The E-Z Notational System
E : higher ranked substituents on opposite sides
Z : higher ranked substituents on same side
lower
C
C
higher
The E-Z Notational System
E : higher ranked substituents on opposite sides
Z : higher ranked substituents on same side
higher
C
lower
lower
C
higher
Entgegen
The E-Z Notational System
E : higher ranked substituents on opposite sides
Z : higher ranked substituents on same side
higher
C
lower
lower
C
higher
Entgegen
higher
C
lower
higher
C
lower
Zusammen
The E-Z Notational System
Question: How are substituents ranked?
Answer:
higher
C
lower
They are ranked in order of
decreasing atomic number.
lower
C
higher
Entgegen
higher
C
lower
higher
C
lower
Zusammen
The Cahn-Ingold-Prelog (CIP) System
The system that we use was devised by
R. S. Cahn
Sir Christopher Ingold
Vladimir Prelog
Their rules for ranking groups were devised in
connection with a different kind of
stereochemistry—one that we will discuss in
Chapter 7—but have been adapted to alkene
stereochemistry.
Table 5.1 CIP Rules
(1) Higher atomic number outranks lower
atomic number
Br > F
Cl > H
higher Br
C
lower
F
Cl
higher
H
lower
C
Table 5.1 CIP Rules
(1) Higher atomic number outranks lower
atomic number
Br > F
Cl > H
higher Br
C
lower
F
Cl
higher
H
lower
C
(Z )-1-Bromo-2-chloro-1-fluoroethene
Table 5.1 CIP Rules
(2) When two atoms are identical, compare the
atoms attached to them on the basis of their
atomic numbers. Precedence is established
at the first point of difference.
—CH2CH3 outranks —CH3
—C(C,H,H)
—C(H,H,H)
Table 5.1 CIP Rules
(3) Work outward from the point of attachment,
comparing all the atoms attached to a
particular atom before proceeding further
along the chain.
—CH(CH3)2 outranks —CH2CH2OH
—C(C,C,H)
—C(C,H,H)
Table 5.1 CIP Rules
(4) Evaluate substituents one by one.
Don't add atomic numbers within groups.
—CH2OH outranks —C(CH3)3
—C(O,H,H)
—C(C,C,C)
Table 5.1 CIP Rules
(5) An atom that is multiply bonded to another
atom is considered to be replicated as a
substituent on that atom.
—CH=O outranks —CH2OH
—C(O,O,H)
—C(O,H,H)
(A table of commonly encountered substituents ranked according to
precedence is given on the inside back cover of the text.)
Physical Properties of Alkenes
Dipole moments
H
What is direction of
dipole moment?
Does a methyl group
donate electrons to the
double bond, or does it
withdraw them?
H
C
C
H
H
=0D
H3C
H
C
H
C
H
 = 0.3 D
Dipole moments
 = 1.4 D
H
H
C
H
H
C
C
H
Cl
C
H
H
=0D
H3C
H
C
H
C
H
 = 0.3 D
Chlorine is
electronegative
and attracts
electrons.
Dipole moments
 = 1.4 D
Dipole moment H
of 1C
chloropropene
is equal to the H
sum of the
dipole
moments of
vinyl chloride H C
3
and propene.
C
H
H
C
Cl
H3C
H
C
C
H
H
C
H
 = 0.3 D
 = 1.7 D
Cl
Dipole moments
 = 1.4 D
H
H
C
Therefore, a
methyl group
donates
electrons to
the double
bond.
C
H
Cl
H3C
H
C
C
Cl
H
H3C
H
C
H
C
H
 = 0.3 D
 = 1.7 D
Alkyl groups stabilize sp2 hybridized
carbon by releasing electrons
R—C+
is more stable than
H—C+
R—C .
is more stable than
H—C .
R—C
is more stable than
H—C
Relative Stabilities of Alkenes
Double bonds are classified according to
the number of carbons attached to them.
H
R
C
C
H
H
H
R
R'
R
C
R'
monosubstituted
C
C
H
disubstituted
H
H
R
C
C
H
disubstituted
H
C
R'
disubstituted
Double bonds are classified according to
the number of carbons attached to them.
R"
R
C
R'
R"
R
C
C
H
trisubstituted
R'
C
R"'
tetrasubstituted
Substituent effects on alkene stability
Electronic
disubstituted alkenes are more stable
than monosubstituted alkenes
Steric
trans alkenes are more stable than cis alkenes
Fig. 5.4 Heats of
combustion of C4H8
isomers.
2717 kJ/mol
+ 6O2
2710 kJ/mol
2707 kJ/mol
2700 kJ/mol
4CO2 + 8H2O
Substituent effects on alkene stability
Electronic
alkyl groups stabilize double bonds more than H
more highly substituted double bonds are more
stable than less highly substituted ones.
Problem 5.8
Give the structure or make a molecular model of
the most stable C6H12 alkene.
C
C
Problem 5.8
Give the structure or make a molecular model of
the most stable C6H12 alkene.
H3C
CH3
C
H3C
C
CH3
Substituent effects on alkene stability
Steric effects
trans alkenes are more stable than cis alkenes
cis alkenes are destabilized by van der Waals
strain
van der Waals strain
due to crowding of
cis-methyl groups
cis-2-butene
Figure 5.5
cis and trans-2-Butene
trans-2-butene
Fig. 5.5
cis and trans-2-butene
van der Waals strain
due to crowding of
cis-methyl groups
cis-2-butene
trans-2-butene
Van der Waals Strain
Steric effect causes a large difference in stability
between cis and trans-(CH3)3CCH=CHC(CH3)3
cis is 44 kJ/mol less stable than trans
CH3 H3C
H3C
H3C
C
C
C
H
CH3
C
H
CH3
Cycloalkenes
Cycloalkenes
Cyclopropene and cyclobutene have angle
strain.
Larger cycloalkenes, such as cyclopentene
and cyclohexene, can incorporate a double
bond into the ring with little or no angle strain.
Stereoisomeric cycloalkenes
cis-cyclooctene and trans-cyclooctene
are stereoisomers
cis-cyclooctene is 39 kJ/ mol more stable
than trans-cyclooctene
H
H
cis-Cyclooctene
H
H
trans-Cyclooctene
Stereoisomeric cycloalkenes
trans-cyclooctene is smallest trans-cycloalkene
that is stable at room temperature
cis stereoisomer is more stable than trans
through C11 cycloalkenes
cis and trans-cyclododecene are approximately
equal in stability
H
H
trans-Cyclooctene
Stereoisomeric cycloalkenes
trans-cyclooctene is smallest trans-cycloalkene
that is stable at room temperature
cis stereoisomer is more stable than trans
through C11 cycloalkenes
cis and trans-cyclododecene are approximately
equal in stability
cis-Cyclododecene
trans-Cyclododecene
Stereoisomeric cycloalkenes
trans-cyclooctene is smallest trans-cycloalkene
that is stable at room temperature
cis stereoisomer is more stable than trans
through C11 cycloalkenes
cis and trans-cyclododecene are approximately
equal in stability
When there are more than 12 carbons in the
ring, trans-cycloalkenes are more stable than cis.
The ring is large enough so the cycloalkene behaves
much like a noncyclic one.
Preparation of Alkenes:
Elimination Reactions
-Elimination Reactions Overview
•dehydrogenation of alkanes:
 H; Y = H
•dehydration of alcohols:
 H; Y = OH
•dehydrohalogenation of alkyl halides:
 H; Y = Br, etc.
H

C
C

Y
C
C
+
H
Y
Dehydrogenation
• limited to industrial syntheses of ethylene,
propene, 1,3-butadiene, and styrene
• important economically, but rarely used in
laboratory-scale syntheses
CH3CH3
CH3CH2CH3
750°C
750°C
H2C
CH2 + H2
H2C
CHCH3 + H2
Dehydration of Alcohols
Dehydration of Alcohols
CH3CH2OH
OH
H2SO4
160°C
H2C
CH2 +
H2O
+
H2O
H2SO4
140°C
(79-87%)
CH3
H3C
C
CH3
OH H2SO4
heat
H3C
C
H3C
CH2
(82%)
+
H2O
R'
Relative
Reactivity
R
C
OH
tertiary:
most reactive
R"
R'
R
C
OH
H
H
R
C
H
OH
primary:
least reactive
Regioselectivity in Alcohol Dehydration:
The Zaitsev Rule
Regioselectivity
H2SO4
HO
+
80°C
10 %
•
90 %
A reaction that can proceed in more
than one direction, but in which one
direction predominates, is said to be
regioselective.
Regioselectivity
CH3
CH3
OH
H3PO4
CH3
+
heat
84 %
•
16 %
A reaction that can proceed in more
than one direction, but in which one
direction predominates, is said to be
regioselective.
The Zaitsev Rule
• When elimination can occur in more than one
direction, the principal alkene is the one formed
by loss of H from the  carbon having the
fewest hydrogens.
R
R
OH
C
C
H
CH3
CH2R
three protons on this  carbon
The Zaitsev Rule
• When elimination can occur in more than one
direction, the principal alkene is the one formed
by loss of H from the  carbon having the
fewest hydrogens.
R
R
OH
C
C
H
CH3
CH2R
two protons on this  carbon
The Zaitsev Rule
• When elimination can occur in more than one
direction, the principal alkene is the one formed
by loss of H from the  carbon having the
fewest hydrogens.
R
R
OH
C
C
H
CH3
CH2R
only one proton on this  carbon
The Zaitsev Rule
• When elimination can occur in more than one
direction, the principal alkene is the one formed
by loss of H from the  carbon having the
fewest hydrogens.
R
R
OH
C
C
H
CH3
R
CH2R
C
CH2R
R
C
CH3
only one proton on this  carbon
The Zaitsev Rule
Zaitsev Rule states that the elimination
reaction yields the more highly substituted
alkene as the major product.
The more stable alkene product
predominates.
Stereoselectivity
in
Alcohol Dehydration
Stereoselectivity
H2SO4
+
heat
OH
(25%)
(75%)
• A stereoselective reaction is one in which
a single starting material can yield two or more
stereoisomeric products, but gives one of them
in greater amounts than any other.
The Mechanism of the
Acid-Catalyzed Dehydration of
Alcohols
A connecting point...
• The dehydration of alcohols and the reaction
of alcohols with hydrogen halides share the
following common features:
• 1) Both reactions are promoted by acids
• 2) The relative reactivity decreases in the
order tertiary > secondary > primary
These similarities suggest that carbocations
are intermediates in the acid-catalyzed
dehydration of alcohols, just as they are in
the reaction of alcohols with hydrogen halides.
Dehydration of tert-Butyl Alcohol
CH3
H3C
C
CH3
OH
H2SO4
H3C
C
heat
CH2
+
H2O
H3C
•first two steps of mechanism are identical to
those for the reaction of tert-butyl alcohol with
hydrogen halides
Mechanism
Step 1:
Proton transfer to tert-butyl alcohol
H
..
+
(CH3)3C O : + H O
..
H
H
fast, bimolecular
H
+
(CH3)3C O :
H
+
H
tert-Butyloxonium ion
:O:
H
Mechanism
Step 2:
Dissociation of tert-butyloxonium ion
to carbocation
H
+
(CH3)3C O :
H
slow, unimolecular
H
+
(CH3)3C
+
tert-Butyl cation
:O:
H
Mechanism
Step 3:
Deprotonation of tert-butyl cation.
H
H3C
+C
H
+
:O:
H
CH2
H3C
fast, bimolecular
H
H3C
C
H3C
CH2
+
H
+
O:
H
Carbocations
are intermediates in the acid-catalyzed
dehydration of tertiary and secondary alcohols
Carbocations can:
•react with nucleophiles
•lose a -proton to form an alkene
(Called an E1 mechanism)
Dehydration of primary alcohols
CH3CH2OH
H2SO4
160°C
H2C
CH2 +
H2O
•A different mechanism from 3 o or 2 o alcohols
•avoids carbocation because primary
carbocations are too unstable
•oxonium ion loses water and a proton in a
bimolecular step
Mechanism
Step 1:
Proton transfer from acid to ethanol
H
..
CH3CH2 O : + H O
..
H
H
fast, bimolecular
H
+
CH3CH2 O :
H
Ethyloxonium ion
H
+
:O:
H
Mechanism
Step 2:
Oxonium ion loses both a proton and
a water molecule in the same step.
H
H
+
: O : + H CH2 CH2 O :
H
H
slow, bimolecular
H
+
:O
H
H
H
+
H2C
CH2
+
:O:
H
Mechanism
Step 2:
H
+
:O
H
Oxonium ion loses both a proton and
a water molecule in the same step.
H
H
+
: O : + H CH2 CH2 O :
H
H
Because rate-determining
step is bimolecular,
thisbimolecular
slow,
is called the E2 mechanism. H
H
+
H2C
CH2
+
:O:
H
Rearrangements in Alcohol
Dehydration
Sometimes the alkene product does not
have the same carbon skeleton as the
starting alcohol.
Example
OH
H3PO4, heat
+
3%
+
64%
33%
Rearrangement involves alkyl group migration
CH3
CH3 C
CHCH3
+
CH3
3%
• carbocation can
lose a proton as
shown
• or it can undergo a
methyl migration
• CH3 group migrates
with its pair of
electrons to
adjacent positively
charged carbon
Rearrangement involves alkyl group migration
CH3
CH3
CH3 C
CHCH3
+
97%
CH3
+
C
CHCH3
CH3
CH3
3%
• tertiary
carbocation;
more stable
Rearrangement involves alkyl group migration
CH3
CH3
CH3 C
CHCH3
+
97%
CH3
+
C
CH3
CH3
3%
CHCH3
Another rearrangement
CH3CH2CH2CH2OH
H3PO4, heat
CH3CH2CH
12%
CH2
+
CH3CH
CHCH3
mixture of cis (32%)
and trans-2-butene (56%)
Rearrangement involves hydride shift
CH3CH2CH2CH2
H
+
O:
H
CH3CH2CH
CH2
oxonium ion can
lose water and a
proton (from C-2) to
give1-butene
doesn't give a
carbocation directly
because primary
carbocations are
too unstable
Rearrangement involves hydride shift
CH3CH2CH2CH2
H
+
O:
H
CH3CH2CH
CH2
CH3CH2CHCH3
+
hydrogen migrates
with its pair of electrons
from C-2 to C-1 as
water is lost
carbocation formed
by hydride shift is
secondary
Hydride shift
H
CH3CH2CHCH2
+
O:
H
H
+
CH3CH2CHCH2 +
H
H
: O:
H
Rearrangement involves hydride shift
CH3CH2CH2CH2
H
+
O:
CH3CH2CHCH3
+
H
CH3CH2CH
CH2
CH3CH
CHCH3
mixture of cis
and trans-2-butene
Carbocations can...
•react with nucleophiles
•lose a proton from the -carbon to form
an alkene
•rearrange (less stable to more stable)
(alkyl shift or hydride shift)
Dehydrohalogenation of
Alkyl Halides
-Elimination Reactions Overview
•dehydrogenation of alkanes:
 H; Y = H
•dehydration of alcohols:
 H; Y = OH
•dehydrohalogenation of alkyl halides:
 H; Y = Br, etc.
H
C

C Y
C
C
+
H
Y
-Elimination Reactions Overview
•dehydrogenation of alkanes:
industrial process; not regioselective
•dehydration of alcohols:
acid-catalyzed
•dehydrohalogenation of alkyl halides:
consumes base
H
C

C Y
C
C
+
H
Y
Dehydrohalogenation
A useful method for the preparation of alkenes
Cl
NaOCH2CH3
ethanol, 55°C
(100 %)
likewise, NaOCH3 in methanol, or KOH in ethanol
Dehydrohalogenation
When the alkyl halide is primary, potassium
tert-butoxide in dimethyl sulfoxide is the
base/solvent system that is normally used.
CH3(CH2)15CH2CH2Cl
KOC(CH3)3
dimethyl sulfoxide
CH3(CH2)15CH
(86%)
CH2
Regioselectivity
KOCH2CH3
Br
+
ethanol, 70°C
29 %
71 %
follows Zaitsev's rule:
more highly substituted double bond
predominates
Stereoselectivity
KOCH2CH3
ethanol
Br
+
(23%)
(77%)
•more stable configuration
of double bond predominates
Stereoselectivity
Br
KOCH2CH3
ethanol
+
(85%)
•more stable configuration
of double bond predominates
(15%)
Mechanism of the
Dehydrohalogenation of Alkyl Halides:
The E2 Mechanism
Facts
• (1) Dehydrohalogenation of alkyl
halides exhibits second-order kinetics
first order in alkyl halide
first order in base
rate = k[alkyl halide][base]
implies that rate-determining step
involves both base and alkyl halide;
i.e., it is bimolecular (second-order)
Facts
• (2) Rate of elimination depends on
halogen
weaker C—X bond; faster rate
rate: RI > RBr > RCl > RF
implies that carbon-halogen bond
breaks in the rate-determining step
The E2 Mechanism
•concerted (one-step) bimolecular process
•single transition state
C—H bond breaks
 component of double bond forms
C—X bond breaks
The E2 Mechanism
R
.. –
O
.. :
H
C
C
: X:
..
Reactants
The E2 Mechanism
R
.. –
O
.. :
H
C
C
: X:
..
Reactants
The E2 Mechanism
R
–
..
O
..
Transition state
H
C
C
–
: X:
..
The E2 Mechanism
R
..
O
..
H
C
C
.. –
: X:
..
Products
Anti Elimination in E2
Reactions
Stereoelectronic Effects
Isotope Effects
Stereoelectronic effect
Br
KOC(CH3)3
(CH3)3COH
(CH3)3C
cis-1-Bromo-4-tertbutylcyclohexane
(CH3)3C
Stereoelectronic effect
(CH3)3C
trans-1-Bromo-4-tertbutylcyclohexane
Br
(CH3)3C
KOC(CH3)3
(CH3)3COH
Stereoelectronic effect
cis
Br
KOC(CH3)3
(CH3)3COH
(CH3)3C
Rate constant for
dehydrohalogenation of
cis is 500 times
greater than that of trans
(CH3)3C
Br
(CH3)3C
trans
KOC(CH3)3
(CH3)3COH
Stereoelectronic effect
cis
Br
KOC(CH3)3
(CH3)3COH
(CH3)3C
H H
(CH3)3C
H that is removed by base must be
anti periplanar to Br
Two anti periplanar H atoms in cis
stereoisomer
Stereoelectronic effect
trans
H
Br
H
(CH3)3C
KOC(CH3)3
(CH3)3COH
H H
(CH3)3C
H that is removed by base must be anti
periplanar to Br
No anti periplanar H atoms in trans stereoisomer;
all vicinal H atoms are gauche to Br
Stereoelectronic effect
cis
more reactive
trans
less reactive
Stereoelectronic effect
An effect on reactivity that has its
origin in the spatial arrangement of
orbitals or bonds is called a
stereoelectronic effect.
The preference for an anti periplanar
arrangement of H and Br in the
transition state for E2
dehydrohalogenation is an example of a
stereoelectronic effect.
Isotope effect
Deuterium,D, is a heavy isotope of
hydrogen but will undergo the same
reactions. But the C-D bond is stronger
so a reaction where the rate involves
breaking a C-H (C-D) bond, the
deuterated sample will have a reaction
rate 3-8 times slower.
In other words comparing the two rates,
i.e. kH/kD = 3-8 WHEN the rate
determining step involves breaking the
C-H bond.
A Different Mechanism for
Alkyl Halide Elimination:
The E1 Mechanism
Example
CH3
CH3
CH2CH3
C
Br
Ethanol, heat
H3C
CH3
H2C
+
C
CH2CH3
(25%)
H
C
C
CH3
H3C
(75%)
The E1 Mechanism
1. Alkyl halides can undergo elimination in
absence of base.
2. Carbocation is intermediate
3. Rate-determining step is unimolecular
ionization of alkyl halide.
4. Generally with tertiary halide, base is
weak and at low concentration.
CH3
Step 1
CH3
CH2CH3
C
: Br:
..
slow, unimolecular
CH3
C
CH3 + CH2CH3
.. –
: Br :
..
CH3
Step 2
CH3
C
+
CH2CH3
– H+
CH3
CH2
+
C
CH3
CH2CH3
C
CH3
CHCH3