07_Lecturevvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv.ppt

© 2013 Pearson Education, Inc.
Chapter 7
Lecture
Organic Chemistry,
8th
Edition
L. G. Wade, Jr.
Structure and Synthesis
of Alkenes
Rizalia Klausmeyer
Baylor University
Waco, TX

Introduction
• Alkenes are hydrocarbons with carbon–
carbon double bonds.
• Alkenes are also called olefins, meaning “oil-
forming gas.”
• The functional group of alkenes is the carbon–
carbon double bond, which gives this group its
reactivity.
Chapter 7 2

Sigma Bonds of Ethylene
Chapter 7 3

Orbital Description
• Sigma bonds around the double-bonded
carbon are sp2
hybridized.
• Angles are approximately 120º and the
molecular geometry is trigonal planar.
• Unhybridized p orbitals with one
electron will overlap, forming the double
bond (pi bond).
Chapter 7 4

Bond Lengths and Angles
• sp2
hybrid orbitals have more s character than the sp3
hybrid orbitals.
• Pi overlap brings carbon atoms closer, shortening the
C—C bond from 1.54 Å in alkanes down to 1.33 Å in
alkenes.
Chapter 7 5

Pi Bonding in Ethylene
• The pi bond in ethylene is formed by overlap of the
unhybridized p orbitals of the sp2
hybrid carbon atoms.
• Each carbon has one unpaired electron in the p orbital.
• This overlap requires the two ends of the molecule to be
coplanar.
Chapter 7 6

Cis-Trans Interconversion
• Cis and trans isomers cannot be interconverted.
• No rotation around the carbon–carbon bond is
possible without breaking the pi bond (264
kJ/mole).
Chapter 7 7

Elements of Unsaturation
• Unsaturation: A structural element that decreases
the number of hydrogens in the molecule by two.
• Also called index of hydrogen deficiency.
• Double bonds and rings are elements of
unsaturation.
Chapter 7 8

Calculating Unsaturations
• To calculate the number of unsaturations, first
find the number of hydrogens the carbons
would have if the compounds were saturated.
Then subtract the actual number of
hydrogens and divide by 2.
• This calculation cannot distinguish between
unsaturations from multiple bonds and those
from rings.
Chapter 7 9

Example: Calculate the Unsaturations
for a Compound with Formula C5H8.
• First calculate the number of hydrogen atoms for a
saturated compound with five carbons:
(2 x C) + 2
(2 x 5) + 2 = 12
• Now subtract from this number the actual number of
hydrogen atoms in the formula and divide by 2:
12 – 8 = 4 = 2 unsaturations
2 2
• The compound has two unsaturationss. They can be
two double bonds, two rings, or one double bond and
one ring.
Chapter 7 10

Elements of Unsaturation:
Heteroatoms
• Halogens replace hydrogen atoms in
hydrocarbons, so when calculating
unsaturations, count halides as hydrogen atoms.
• Oxygen does not change the C:H ratio, so
ignore oxygen in the formula.
• Nitrogen is trivalent, so it acts like half a carbon.
Add the number of nitrogen atoms when
calculating unsaturations.
Chapter 7 11

Example: Calculate the Unsaturations for
a Compound with Formula C4H7Br.
• First calculate the number of hydrogens for a
saturated compound with four carbons:
(2 x C) + 2 + N
(2 x 4) + 2 = 10
hydrogens in the formula and divide by 2.
Remember to count halides as hydrogens:
10 – 8 = 2 = 1 unsaturation
2 2
Chapter 7 12

Example: Calculate the Unsaturations
for a Compound with Formula C6H7N.
• First calculate the number of hydrogens for a
saturated compound with six carbons. Add the
number of nitrogens:
(2 x C) + 2 + N
(2 x 6) + 2 + 1 = 15
hydrogens in the formula and divide by 2:
15 – 7 = 8 = 4 unsaturations
2 2
Chapter 7 13

IUPAC Nomenclature
• Find the longest continuous carbon chain
that includes the double-bonded carbons.
• Ending -ane changes to -ene.
• Number the chain so that the double
bond has the lowest possible number.
• In a ring, the double bond is assumed to
be between carbon 1 and carbon 2.
Chapter 7 14

IUPAC and New IUPAC
Chapter 7 15

Ring Nomenclature
1-methylcyclopentene
In a ring, the double bond is assumed to
be between carbon 1 and carbon 2.
CH3
CH3
1
2 1
2
3
3-methylcyclopentene
Chapter 7 16

Multiple Double Bonds
• Give the double bonds the lowest numbers
possible.
• Use di-, tri-, tetra- before the ending -ene to
specify how many double bonds are present.
Chapter 7 17

Cis-Trans Isomers
• Also called geometric isomerism.
• Similar groups on same side of double bond, alkene is cis.
• Similar groups on opposite sides of double bond, alkene is
trans.
• Not all alkenes show cis-trans isomerism.
Chapter 7 18

Cyclic Compounds
• Trans cycloalkenes are not stable unless the
ring has at least eight carbons.
• All cycloalkenes are assumed to be cis
unless otherwise specifically named trans.
Chapter 7 19

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).
Chapter 7 20

Example
• Assign priority to the
substituents according
to their atomic number
(1 is highest priority).
• If the highest priority
groups are on opposite
sides, the isomer is E.
• If the highest priority
groups are on the same
side, the isomer is Z.
1
2
1
2
E-1-bromo-1-chloropropene
Chapter 7 21

Cyclic Stereoisomers
• Double bonds outside the ring can show
stereoisomerism.
Chapter 7 22

Stereochemistry in Dienes
• If there is more than one double bond in
the molecule, the stereochemistry of all
the double bonds should be specified.
Chapter 7 23

Commercial Uses of Ethylene
Chapter 7 24

Commercial Uses of
Propylene
Chapter 7 25

Addition Polymers
Chapter 7 26

Heat of Hydrogenation
• Combustion of an alkene and hydrogenation of an
alkene can provide valuable data as to the stability of
the double bond.
• The more substituted the double bond, the lower its
heat of hydrogenation.
Chapter 7 27

Relative Heats of
Hydrogenation
More substituted double bonds are usually more stable.
Chapter 7 28

Relative Stabilities
Chapter 7 29

Heats of hydrogenation are
usually exothermic. A larger
amount of heat given off implies
a less stable alkene, because the
less stable alkene starts from a
higher potential energy.
Chapter 7 30

Substituent Effects
• The isomer with the more substituted double bond
has a larger angular separation between the bulky
alkyl groups.
Chapter 7 31

Disubstituted Isomers
• Stability: cis < geminal < trans isomer
• The less stable isomer has a higher exothermic
heat of hydrogenation.
-116 kJ
trans-2-butene
-117 kJ
(CH3)2C=CH2
iso-butene
-120 kJ
cis-2-butene
CH3
C C
CH3
H H
H
C C
CH3
CH3 H
Chapter 7 32

Cycloalkenes
• A ring makes a major difference only if there is ring strain, either
because of a small ring or because of a trans double bond.
• Rings that are five-membered or larger can easily accommodate
double bonds, and these cycloalkenes react much like straight-
chain alkenes.
Chapter 7 33

Cyclopropene
• Cyclopropene has bond angles of about 60°,
compressing the bond angles of the carbon–carbon
double bond to half their usual value of 120°.
• The double bond in cyclopropene is highly strained.
Chapter 7 34

Stability of Cycloalkene
• Cis isomer is more stable than trans in small cycloalkenes.
• Small rings have additional ring strain.
• Must have at least eight carbons to form a stable trans double
bond.
• For cyclodecene (and larger), the trans double bond is almost
as stable as the cis.
Chapter 7 35

Bredt’s Rule
• A bridged bicyclic compound cannot have a
double bond at a bridgehead position unless
one of the rings contains at least eight carbon
atoms.
Chapter 7 36

Compound (a) is stable. Although the double bond is at a bridgehead, it is not a bridged bicyclic
system. The trans double bond is in a ten-membered ring. Compound (b) is a Bredt’s rule violation and
is not stable. The largest ring contains six carbon atoms, and the trans double bond cannot be stable in
this bridgehead position.
Compound (c) (norbornene) is stable. The (cis) double bond is not at a bridgehead carbon.
Compound (d) is stable. Although the double bond is at the bridgehead of a bridged bicyclic system,
there is an eight-membered ring to accommodate the trans double bond.
Which of the following alkenes are stable?
Solved Problem 1
Solution
Chapter 7 37

Physical Properties of Alkenes
• Low boiling points, increasing with mass.
• Branched alkenes have lower boiling points.
• Less dense than water.
• Slightly polar:
 Pi bond is polarizable, so instantaneous dipole–
dipole interactions occur.
 Alkyl groups are electron-donating toward the pi
bond, so may have a small dipole moment.
Chapter 7 38

Polarity and Dipole Moments of
Alkenes
• Cis alkenes have a greater dipole moment than trans
alkenes, so they will be slightly polar.
• The boiling point of cis alkenes will be higher than the
trans alkenes.
Chapter 7 39

Alkene Synthesis
Overview
• E2 dehydrohalogenation (-HX)
• E1 dehydrohalogenation (-HX)
• Dehalogenation of vicinal dibromides (-X2)
• Dehydration of alcohols (-H2O)
Chapter 7 40

Dehydrohalogenation by the E2
Mechanism
• Strong base abstracts H+
as double bond forms and X-
leaves from
the adjacent carbon.
• Tertiary and hindered secondary alkyl halides give alkenes in good
yields.
• Tertiary halides are the best E2 substrates because they are prone to
elimination and cannot undergo SN2 substitution.
Chapter 7 41

Bulky Bases for E2 Reactions
• If the substrate is prone to substitution, a bulky base
can minimize the amount of substitution.
• Large alkyl groups on a bulky base hinder its
approach to attack a carbon atom (substitution), yet it
can easily abstract a proton (elimination).
Chapter 7 42

Hofmann Product
Bulky bases, such as potassium tert-butoxide, abstract the
least hindered H+
, giving the less substituted alkene as the
major product (Hofmann product).
Chapter 7 43

E2 Reactions Are Stereospecific
• Depending on the stereochemistry of the alkyl halide,
the E2 elimination may produce only the cis or only
the trans isomer.
• The geometry of the product will depend on the anti-
coplanar relationship between the proton and the
leaving group.
Chapter 7 44

Stereochemistry of E2
Elimination
• Most E2 reactions go through an anti-coplanar
transition state.
• This geometry is most apparent if we view the
reaction with the alkyl halide in a Newman projection.
Chapter 7 45

Show that the dehalogenation of 2,3-dibromobutane by iodide ion is stereospecific by showing that the
two diastereomers of the starting material give different diastereomers of the product.
Rotating meso-2,3-dibromobutane into a conformation where the bromine atoms are anti and coplanar,
we find that the product will be trans-2-butene. A similar conformation of either enantiomer of the (±)
diastereomer shows that the product will be cis-2-butene. (Hint: Your models will be helpful.)
Solved Problem 2
Solution
Chapter 7 46

Don’t try to memorize your way
through these reactions. Look at
each one, and consider what it
might do. Use your models for
the ones that involve
stereochemistry.
Chapter 7 47

E2 Reactions on Cyclohexanes
• An anti-coplanar conformation (180°) can only be achieved when both the hydrogen and the
halogen occupy axial positions.
• The chair must flip to the conformation with the axial halide in order for the elimination to take place.
Chapter 7 48

Explain why the following deuterated 1-bromo-2-methylcyclohexane undergoes dehydrohalogenation
by the E2 mechanism, to give only the indicated product. Two other alkenes are not observed.
In an E2 elimination, the hydrogen atom and the leaving group must have a trans-diaxial relationship.
In this compound, only one hydrogen atom—the deuterium—is trans to the bromine atom. When the
bromine atom is axial, the adjacent deuterium is also axial, providing a trans-diaxial arrangement.
Solved Problem 3
Solution
Chapter 7 49

Debromination of Vicinal
Dibromides
• Vicinal dibromides are converted to alkenes by
reduction with iodide ion in acetone.
• Not an important reaction, but the mechanism is
similar to the E2 dehydrohalogenation reaction.
Chapter 7 50

E2 Debromination of a Vicinal
Dibromide
• E2 debromination takes place by a concerted,
stereospecific mechanism.
• Iodide ion removes one bromine atom, and
the other bromine leaves as bromide ion.
Chapter 7 51

Make a model of each compound, and
place it in the conformation where the
groups to be eliminated are anti and
coplanar. The positions of the other
groups will be near their positions in the
alkene product.
Chapter 7 52

E1 Elimination Mechanism
• Tertiary and secondary alkyl halides:
3º > 2º
• Carbocation intermediate.
• Rearrangements are possible.
• Weak nucleophiles such as water or alcohols.
• Usually have SN1 products, too, since the solvent
can attack the carbocation directly.
Chapter 7 53

E1 and SN1 Mechanisms
Chapter 7 54

Dehydration of Alcohols
• Use concentrated H2SO4 or H3PO4 and remove low-
boiling alkene as it forms to shift the equilibrium and
increase the yield of the reaction.
• E1 mechanism.
• Rearrangements are common.
• Reaction obeys Zaitsev’s rule.
Chapter 7 55

Dehydration Mechanism: E1
Step 1: Protonation of the hydroxyl group (fast equilibrium).
Step 2: Ionization to a carbocation (slow; rate limiting).
Chapter 7 56

Dehydration Mechanism: Step 3
Step 3: Deprotonation to give the alkene (fast).
Chapter 7 57

In acid-catalyzed mechanisms,
the first step is often addition of
H+
, and the last step is often
loss of H+
.
Chapter 7 58

Propose a mechanism for the sulfuric acid–catalyzed dehydration of t-butyl alcohol.
The first step is protonation of the hydroxyl group, which converts it to a good leaving group.
The second step is ionization of the protonated alcohol to give a carbocation.
Abstraction of a proton completes the mechanism.
Solved Problem 4
Solution
Chapter 7 59

Catalytic Cracking of Alkanes
• Long-chain alkane is heated with a catalyst to
produce an alkene and shorter alkane.
• Complex mixtures are produced.
Chapter 7 60

Dehydrogenation of Alkanes
• Dehydrogenation is the removal of H2 from a molecule,
forming an alkene (the reverse of hydrogenation).
• This reaction has an unfavorable enthalpy change but
a favorable entropy change.
Chapter 7 61

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