This chapter discusses the structure, properties, and synthesis of alkenes. Key points include:
- Alkenes contain a pi bond between two carbons as the functional group. This pi bond is more reactive than a sigma bond.
- Common methods for alkene synthesis are E2 elimination of hydrogen halides, E1 elimination of hydrogen halides, removal of vicinal dibromides, and dehydration of alcohols.
- Alkene stability is affected by substitution - more substituted alkenes are more stable. Cis isomers are generally more stable than trans.
- IUPAC nomenclature is used to systematically name alkenes based on
2. Chapter 7 2
Functional Group
• Pi bond is the functional group.
• More reactive than sigma bond.
• Bond dissociation energies:
C=C BDE 146 kcal/mol
C-C BDE -83 kcal/mol
Pi bond 63 kcal/mol
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3. Chapter 7 3
Orbital Description
• Sigma bonds around C are sp2
hybridized.
• Angles are approximately 120 degrees.
• No nonbonding electrons.
• Molecule is planar around the double bond.
• Pi bond is formed by the sideways overlap of
parallel p orbitals perpendicular to the plane
of the molecule.
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4. Chapter 7 4
Bond Lengths and Angles
• Hybrid orbitals have more s character.
• Pi overlap brings carbon atoms closer.
• Bond angle with pi orbitals increases.
Angle C=C-H is 121.7°
Angle H-C-H is 116. 6° =>
5. Chapter 7 5
Pi Bond
• Sideways overlap of parallel p orbitals.
• No rotation is possible without breaking
the pi bond (63 kcal/mole).
• Cis isomer cannot become trans
without a chemical reaction occurring.
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6. Chapter 7 6
Elements of
Unsaturation
• A saturated hydrocarbon: CnH2n+2
• Each pi bond (and each ring) decreases the
number of H’s by two.
• Each of these is an element of unsaturation.
• To calculate: find number of H’s if it were
saturated, subtract the actual number of H’s,
then divide by 2.
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7. Chapter 7 7
Propose a Structure:
• First calculate the number of elements of
unsaturation.
• Remember:
A double bond is one element of unsaturation.
A ring is one element of unsaturation.
A triple bond is two elements of unsaturation. =>
for C5H8
8. Chapter 7 8
Heteroatoms
• Halogens take the place of hydrogens, so add
their number to the number of H’s.
• Oxygen doesn’t change the C:H ratio, so
ignore oxygen in the formula.
• Nitrogen is trivalent, so it acts like half a
carbon.
C
H
H
C
H
H
N C
H
HH
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9. Chapter 7 9
Structure for C6H7N?
• Since nitrogen counts as half a carbon,
the number of H’s if saturated is
2(6.5) + 2 = 15.
• Number of missing H’s is 15 – 7 = 8.
• Elements of unsaturation is 8 ÷ 2 = 4.
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10. Chapter 7 10
IUPAC Nomenclature
• Parent is longest chain containing the
double bond.
• -ane changes to -ene. (or -diene, -triene)
• 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.
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11. Chapter 7 11
Name These Alkenes
CH2 CH CH2 CH3
CH3 C
CH3
CH CH3
CH3
CHCH2CH3
H3C
1-butene
2-methyl-2-butene
3-methylcyclopentene
2-sec-butyl-1,3-cyclohexadiene
3-n-propyl-1-heptene
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13. Chapter 7 13
Common Names
• Usually used for small molecules.
• Examples:
CH2 CH2
ethylene
CH2 CH CH3
propylene
CH2 C CH3
CH3
isobutylene
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14. Chapter 7 14
Cis-trans Isomerism
• Similar groups on same side of double
bond, alkene is cis.
• Similar groups on opposite sides of
double bond, alkene is trans.
• Cycloalkenes are assumed to be cis.
• Trans cycloalkenes are not stable
unless the ring has at least 8 carbons.
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15. Chapter 7 15
Name these:
C C
CH3
H
H
CH3CH2
C C
Br
H
Br
H
trans-2-pentene cis-1,2-dibromoethene
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16. Chapter 7 16
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).
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17. Chapter 7 17
Example, E-Z
C C
H3C
H
Cl
CH2
C C
H
H
CH CH3
Cl1
2
1
2
2Z
2
1
1
2
5E
(2Z, 5E)-3,7-dichloro-2,5-octadiene
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21. Chapter 7 21
Stability of Alkenes
• Measured by heat of hydrogenation:
Alkene + H2 → Alkane + energy
• More heat released, higher energy alkene.
=>
30.3 kcal
27.6 kcal
22. Chapter 7 22
Substituent Effects
• More substituted alkenes are more stable.
H2C=CH2 < R-CH=CH2 < R-CH=CH-R < R-CH=CR2 < R2C=CR2
unsub. < monosub. < disub. < trisub. < tetra sub.
• Alkyl group stabilizes the double bond.
• Alkene less sterically hindered.
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23. Chapter 7 23
Disubstituted Isomers
• Stability: cis < geminal < trans isomer
• Less stable isomer is higher in energy, has
a more exothermic heat of hydrogenation.
27.6 kcalTrans-2-butene
28.0 kcal(CH3)2C=CH2Isobutylene
28.6 kcalCis-2-butene CH3
C C
CH3
H H
H
C C
CH3
CH3 H
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24. Chapter 7 24
Cycloalkene Stability
• Cis isomer more stable than trans.
• Small rings have additional ring strain.
• Must have at least 8 carbons to form a
stable trans double bond.
• For cyclodecene (and larger) trans
double bond is almost as stable as the
cis.
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25. Chapter 7 25
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.
• Examples:
Unstable.
Violates Bredt’s rule
Stable. Double bond
in 8-membered ring.
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26. Chapter 7 26
Physical Properties
• 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.
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27. Chapter 7 27
Polarity Examples
µ = 0.33 D µ = 0
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cis-2-butene, bp 4°C
C C
H
H3C
H
CH3
trans-2-butene, bp 1°C
C C
H
H
H3C
CH3
29. Chapter 7 29
Removing HX via E2
• Strong base abstracts H+
as X-
leaves
from the adjacent carbon.
• Tertiary and hindered secondary alkyl
halides give good yields.
• Use a bulky base if the alkyl halide
usually forms substitution products.
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30. Chapter 7 30
Some Bulky Bases
C
CH3
H3C
CH3
O
_
tert-butoxide
(CH3CH2)3N :
triethylamine
=>
N
H
CH(CH3)2
CH(CH3)2
diisopropylamine
N CH3H3C
2,6-dimethylpyridine
31. Chapter 7 31
Hofmann Product
• Bulky bases abstract the least hindered H+
• Least substituted alkene is major product.
CH3 C
H
H
C
CH3
Br
CH2
H
CH3CH2O
CH3CH2OH
_
C C
CH3
CH3H
H3C
C C
H
HH3C
CH3CH2
71% 29%
72%28%
C C
H
HH3C
CH3CH2
C C
CH3
CH3H
H3C_
CH3CH2OH
CH3 C
H
H
C
CH3
Br
CH2
H
(CH3)3CO
=>
32. Chapter 7 32
E2: Diastereomers
Stereospecific reaction: (S, R) produces only
trans product, (R, R) produces only cis.
Ph
Br H
H CH3
Ph
≡
H
Ph CH3
Br
PhH
H
Ph
CH3
Ph
H
Br
CH3Ph
PhH
=>
33. Chapter 7 33
E2: Cyclohexanes
Leaving groups must be trans diaxial. =>
34. Chapter 7 34
E2: Vicinal Dibromides
• Remove Br2 from adjacent carbons.
• Bromines must be anti-coplanar (E2).
• Use NaI in acetone, or Zn in acetic acid.
I
-
Br
CH3
H
Br
CH3H
C C
CH3
H
H
H3C
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35. Chapter 7 35
Removing HX via E1
• Secondary or tertiary halides
• Formation of carbocation intermediate
• Weak nucleophile
• Usually have substitution products too
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36. Chapter 7 36
Dehydration of
Alcohols
• Reversible reaction
• Use concentrated sulfuric or phosphoric
acid, remove low-boiling alkene as it
forms.
• Protonation of OH converts it to a good
leaving group, HOH
• Carbocation intermediate, like E1
• Protic solvent removes adjacent H+
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38. Chapter 7 38
Industrial Methods
• Catalytic cracking of petroleum
Long-chain alkane is heated with a catalyst to
produce an alkene and shorter alkane.
Complex mixtures are produced.
• Dehydrogenation of alkanes
Hydrogen (H2) is removed with heat, catalyst.
Reaction is endothermic, but entropy-favored.
• Neither method is suitable for lab synthesis
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