chapter 7 structure and synthesis of alkenesfaculty.sdmiramar.edu/choeger/chap 7 alkenes 1 (ip).pdf1...
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Chapter 7Structure and Synthesis
of Alkenes
Chapter 7-Alkenes 1 Slide 7-2
Introduction• Hydrocarbon with carbon-carbon double bonds• Sometimes called olefins, “oil-forming gas”• Planar• Pi bond is the functional group.• More reactive than sigma bond (more exposed).• Bond dissociation energies:
C=C BDE 611 kJ/molC-C BDE -347 kJ/molPi bond 264 kJ/mol
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Chapter 7-Alkenes 1 Slide 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.
A
B
C
D
side
side
endendA
B
C
D
top face
bottom face
Chapter 7-Alkenes 1 Slide 7-4
Bond Lengths and Angles
• Hybrid orbitals have more s character.• Pi overlap brings carbon atoms closer.• Bond angle with pi orbital increases.
Angle C=C-H is 121.7°Angle H-C-H is 116. 6°
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Chapter 7-Alkenes 1 Slide 7-5
Pi Bond• Sideways overlap of parallel p orbitals.• No rotation is possible without breaking the pi bond (264
kJ/mole).• Cis isomer cannot become trans without a chemical reaction
occurring.
Chapter 7-Alkenes 1 Slide 7-6
Elements of Unsaturation• A saturated hydrocarbon: CnH2n+2
• Each pi bond (and each ring) decreases the number of H’sby 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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Chapter 7-Alkenes 1 Slide 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
Chapter 7-Alkenes 1 Slide 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.
CH
HCH
HN C
H
HH
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Chapter 7-Alkenes 1 Slide 7-9
Structure for C6H7N?
• Since nitrogen counts as half a carbon, the number of H’s ifsaturated is2(6.5) + 2 = 15.
• Number of missing H’s is 15 – 7 = 8.• Elements of unsaturation is 8 ÷ 2 = 4.
Chapter 7-Alkenes 1 Slide 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 index number.• In a ring, the double bond is assumed to be between carbon 1
and carbon 2.• Place numbers in front of suffix to indicate where double
bond is.
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Chapter 7-Alkenes 1 Slide 7-11
Name These AlkenesCH2 CH CH2 CH3
CH3 C
CH3
CH CH3
CH3
CHCH2CH3
H3C
1-butenebut-1-ene
2-methyl-2-butene2-methylbut-2-ene
3-methylcyclopentene
2-sec-butyl-1,3-cyclohexadiene2-sec-butylcyclohexa-1,3-diene
3-n-propyl-1-heptene3-n-propylhept-1-ene
Chapter 7-Alkenes 1 Slide 7-12
Alkene Substituents
= CH2
methylene(methylidene)
- CH = CH2vinyl
(ethenyl)
- CH2 - CH = CH2allyl
(2-propenyl)
Name:
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Chapter 7-Alkenes 1 Slide 7-13
Common Names• Usually used for small molecules.• Examples:
CH2 CH2
ethylene
CH2 CH CH3
propylene
CH2 C CH3
CH3
isobutylene
Chapter 7-Alkenes 1 Slide 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 for rings less than 8.• Trans cycloalkenes are not stable unless the ring has at least 8
carbons (and even here trans is rare).
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Chapter 7-Alkenes 1 Slide 7-15
Name these:
C C
CH3
H
H
CH3CH2
C C
Br
H
Br
H
trans-2-pentenetrans-pent-2-ene
cis-1,2-dibromoethene
C C
Br
Cl
Br
H
What about this?
Chapter 7-Alkenes 1 Slide 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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Chapter 7-Alkenes 1 Slide 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
3,7-dichloro-(2Z, 5E)-2,5-octadiene3,7-dichloro-(2Z, 5E)-octa-2,5-diene
Chapter 7-Alkenes 1 Slide 7-18
Commercial Uses: Ethylene
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Chapter 7-Alkenes 1 Slide 7-19
Commercial Uses: Propylene
Chapter 7-Alkenes 1 Slide 7-20
Other Polymers
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Chapter 7-Alkenes 1 Slide 7-21
Stability of Alkenes• Measured by heat of hydrogenation:
Alkene + H2 → Alkane + energy• More heat released, higher energy alkene.
Chapter 7-Alkenes 1 Slide 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.
• Cis vs Trans vs terminal• Alkyl group stabilizes the double bond.• Alkene less sterically hindered.
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Chapter 7-Alkenes 1 Slide 7-23
Disubstituted Isomers• Stability: cis < geminal < trans isomer• Less stable isomer is higher in energy, has a more exothermic
heat of hydrogenation.
-116 kJTrans-2-butene
-117 kJ (CH3)2C=CH2Isobutylene
-120 kJCis-2-butene CH3C C
CH3
H H
HC C
CH3
CH3 H
Chapter 7-Alkenes 1 Slide 7-24
Relative Stabilities
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Chapter 7-Alkenes 1 Slide 7-25
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 (‘floppy’ rings).
Chapter 7-Alkenes 1 Slide 7-26
Bredt’s Rule (Bridgehead Alkenes)• A bridged bicyclic compound cannot have a double bond at
a bridgehead position unless one of the rings contains atleast eight carbon atoms.
• Examples:
Unstable.Violates Bredt’s rule
Stable. Double bondin 8-membered ring.
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Chapter 7-Alkenes 1 Slide 7-27
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-dipoleinteractions occur.
Alkyl groups are electron-donating toward the pi bond,so may have a small dipole moment.
Chapter 7-Alkenes 1 Slide 7-28
Polarity Examples
µ = 0.33 D µ = 0
cis-2-butene, bp 4°C
C C
H
H3C
H
CH3
trans-2-butene, bp 1°C
C C
H
H
H3C
CH3
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Chapter 7-Alkenes 1 Slide 7-29
Alkene Synthesis:Overview
Simple preparation involves Elimination reactions:
• E2 dehydrohalogenation (-HX)• E1 dehydrohalogenation (-HX)• Dehalogenation of vicinal dibromides (-X2)• Dehydration of alcohols (-H2O)
Chapter 7-Alkenes 1 Slide 7-30
Removing HX via E2
• Strong base abstracts H+ as X- leaves from the adjacentcarbon.
• Tertiary and hindered secondary alkyl halides give goodyields (SN2 is retarded).
• Use a bulky base if the alkyl halide usually formssubstitution products (such as isopropoxide or t-butoxide).
• Zaitsev product preferred if possible
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Chapter 7-Alkenes 1 Slide 7-31
E2 Mechanism• Lewis base approaches molecule; removal of H is more facile
than substitution:
Nu
H
H CH3
Br
HH3C
• Rate bond making = rate bond breaking• H being removed and X leaving must be periplanar;• Anti periplanar preferred to syn periplanar
Chapter 7-Alkenes 1 Slide 7-32
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
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Chapter 7-Alkenes 1 Slide 7-33
Hofmann Product• Bulky bases abstract the least hindered H+
• Least substituted alkene is major product.
Chapter 7-Alkenes 1 Slide 7-34
E2: Diastereomers
Stereospecific reaction: (S, R) produces only trans product,(R, R) produces only cis.
Ph
Br H
H CH3
Ph
≡
H
Ph CH3Br
PhHH
Ph
CH3
Ph
HBr
CH3Ph
PhH =>
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Chapter 7-Alkenes 1 Slide 7-35
E2: Cyclohexanes
Leaving groups must be trans diaxial. =>
Chapter 7-Alkenes 1 Slide 7-36
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
CH3H
Br
CH3HC C
CH3
H
H
H3C =>
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Chapter 7-Alkenes 1 Slide 7-37
Removing HX via E1
• Secondary or tertiary halides.• Formation of carbocation intermediate.• May get rearrangement.• Weak nucleophile.• Usually have substitution products, too.
=>
Chapter 7-Alkenes 1 Slide 7-38
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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Chapter 7-Alkenes 1 Slide 7-39
Dehydration Mechanism
=>
Chapter 7-Alkenes 1 Slide 7-40
Industrial Methods
• Catalytic cracking of petroleumLong-chain alkane is heated with a catalyst to produce an
alkene and shorter alkane.Complex mixtures are produced.
• Dehydrogenation of alkanesHydrogen (H2) is removed with heat, catalyst.Reaction is endothermic, but entropy-favored.
• Neither method is suitable for lab synthesis. =>
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Chapter 7-Alkenes 1 Slide 7-41
End of Chapter 7