Chapter 10
Alkenes
Introduction
Structure and Synthesis of Alkenes
• Hydrocarbon with carbon-carbon
double bonds
• Sometimes called olefins, “oilforming gas”
Functional Group
• Pi (π) bond is the functional group.
• More reactive than sigma bond.
• Bond dissociation energies:
– C=C BDE 611 kJ/mol
– C-C BDE -347 kJ/mol
– Pi bond
264 kJ/mol
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.
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°
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.
Elements of Unsaturation
• A saturated hydrocarbon: CnH2n+2
• Each pi bond (and each ring) decreases the
number of H atoms by two.
• Each of these is an element of unsaturation.
• To calculate: find number of H atoms if it were
saturated, subtract the actual number of H atoms,
then divide by 2.
• Or use formula :
½(2C+2-H)
where C = number of carbons
H = number of hydrogens
Heteroatoms
• Halogens take the place of hydrogens, so
add their number to the number of H atoms.
• Oxygen doesn’t change the C:H ratio, so
ignore oxygen in the formula.
• Nitrogen is trivalent, so it acts like half a
carbon.
E.g.:
CH3-CH=CH-CBr3
CH2COH
C4H5Br
C2H4O
(1 element of unsaturation)
Propose a Structure:
for C5H8
• First calculate the number of elements of
unsaturation.
• Remember:
– A double bond is one element of unsaturation.
– A cyclic ring is one element of unsaturation.
– A triple bond is two elements of unsaturation.
Structure for C6H7N?
• Since nitrogen counts as half a carbon, the
number of H atoms if saturated is
2(6.5) + 2 = 15.
• Number of missing H atoms is 15 – 7 = 8.
• Element of unsaturation is 8 ÷ 2 = 4.
IUPAC Nomenclature
• Parent chain is the 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.
• A compound with two double bonds is a
diene, three C=C is a triene and four C=C
is a tetraene.
Name These Alkenes
CH2
CH3
CH CH2
1-butene
but-1-ene
CH3
C CH CH3
CHCH2CH3
H3C
2-sec-butyl-1,3-cyclohexadiene
2-sec-butylcyclohexa-1,3-diene
CH3
2-methyl-2-butene
2-methylbut-2-ene
CH3
3-methylcyclopentene
3-methylcyclopent-1-ene
3-n-propyl-1-heptene
3-n-propylhept-1-ene
Br
7-bromo-1,3,5-cycloheptatriene
7-bromo-cyclohepta-1,3,5-triene
Alkene Substituents
= CH2
methylene
(methylidene)
- CH = CH2
vinyl
(ethenyl)
Name: methylidenecyclohexane
- CH2 - CH = CH2
allyl
(2-propenyl)
Common Names
• Usually used for small molecules.
• Examples:
CH3
CH2
CH2
CH2
ethylene
CH2
CH CH3
C CH3
isobutylene
propylene
2-methylpropene
ethenylbenzene
styrene
H2C
CH3
H
CH2
isoprene
2-methylbuta-1,3-diene
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.
Name these:
H
CH3
Br
C C
CH3CH2
Br
C C
H
trans-2-pentene
trans-pent-2-ene
H
H
cis-1,2-dibromoethene
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).
Example, E-Z
1
H3C
2
H
Cl
1
CH CH3
1
Cl
C C 1
C C
CH2
H
H
2
2
2
2Z
5E
3,7-dichloro-(2Z, 5E)-2,5-octadiene
3,7-dichloro-(2Z, 5E)-octa-2,5-diene
Commercial Uses of Ethylene
Commercial Uses of Propylene
=>
Other Polymers
Stability of Alkenes
• Measured by heat of hydrogenation:
Alkene + H2 → Alkane + energy
• More heat released, higher energy alkene.
Substituent Effects
• Zaitsev’s rule : More substituted alkenes are more stable.
H2C=CH2 < R-CH=CH2 < R-CH=CH-R < R-CH=CR2 < R2C=CR2
unsub. < monosub. < disub.
< trisub.
< tetrasub.
• Alkyl group stabilizes the double bond by two factors:
(i) They are e- donating, thus contributing e- density to π bond
(ii) Bulky substitutuents like alkyl groups are best situated
as far apart as possible
• Alkenes are less sterically hindered as compared to alkanes.
Disubstituted Isomers
• Stability: trans isomer > geminal > cis
• Less stable isomer is higher in energy, has
a more exothermic heat of hydrogenation.
Cis-2-butene
CH3
C C
H
Isobutylene
Trans-2-butene
CH3
H
(CH3)2C=CH2
H
CH3
-120 kJ
C C
CH3
H
-117 kJ
-116 kJ
Relative Stabilities
Cycloalkene Stability
• Cis isomer more stable than trans.
• Small rings have additional ring strain,
making more reactive than a thpical C=C.
• 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.
• In acyclic alkenes, the trans isomers are
usually more stable.
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 dipoledipole interactions occur.
– Alkyl groups are electron-donating toward the pi
bond, so may have a small dipole moment.
Polarity Examples
H3C
H
H3C
H
H
H
propene
μ = 0.35D
CH3
H
cis- 2-butene
Vector sum =
μ = 0.33D
H3C
H
H
CH3
tran-2-butene
vector sum = 0
μ=0
Alkene Synthesis Overview
•
•
•
•
•
•
E2 dehydrohalogenation (-HX)
β ELIMINATIONS
E1 dehydrohalogenation (-HX)
Dehalogenation of vicinal dibromides (-X2)
Dehydration of alcohols (-H2O)
Catalytic cracking of petroleum
industry method
Dehydrogenation of alkanes
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.
• Is a reliable synthetic reaction especially
if the alkyl halide is a poor SN2 substrate
• Usually better for synthetic purposes
because the E1 has a more competing
reactions.
Some Bulky Bases
(CH3)2CH
H3C
N
..
CH3
2,6-dimethylpyridine
(CH3)2CH
N:
H
diisopropylamine
Hofmann Product
• Bulky bases abstract the least hindered H+
• Least substituted alkene is major product,
known as a Hofmann Product.
E2: Cyclohexanes
Leaving groups (both Br atoms in this case)
must be trans diaxial.
E2: Vicinal Dibromides
• Remove Br2 from adjacent carbons.
• Bromines must be anti-coplanar (E2).
• Use NaI in acetone, or Zn in acetic acid.
Br
CH3
H
Ph
H
+ I-Br + Br-
CH3
Br
H
H
Ph
Removing HX via E1
•
•
•
•
•
Secondary or tertiary halides
Formation of carbocation intermediates
Carbocations may rearrange
Use a weak nucleophile
Usually have substitution products
Dehydration of Alcohols
• Reversible reaction
• Use concentrated sulfuric or phosphoric
acid, remove (by distillation) low-boiling
alkene as it forms.
• Protonation of OH converts it to a good
leaving group, H2O+
• Carbocation intermediate, resembles E1
• Protic solvent (H2O) or HSO4- removes
adjacent H+
Dehydration Mechanism
Step 1: protonation of the hydroxyl group (fast equilibrium)
Step 2: Ionization to a carbocation (slow, rate-limiting)
Step 3: deprotonation to give the alkene (fast)
Industrial Methods
• Catalytic cracking of petroleum
– Long-chain alkanes are heated with a catalyst
to produce alkenes and shorter-chain alkanes.
– Complex mixtures are produced.
• Dehydrogenation of alkanes
– Hydrogen (H2) is removed with heat, catalyst.
– Reaction is endothermic (enthalpy unfavored),
but entropy-favored.
• Neither method is suitable for lab synthesis
Reactions of Alkenes
Reactivity of C=C
• Electrons in pi bond are loosely held.
• Electrophiles are attracted to the pi
electrons.
• Carbocation intermediate forms.
• Nucleophile adds to the carbocation.
• Net result is addition to the double bond.
Electrophilic Addition
• Step 1: Pi electrons attack the electrophile.
• Step 2: Nucleophile attacks the carbocation.
• Because the carbon atoms of a double bond are both trigonal
planar, the elements of X and Y can be added to them from the
same side or from opposite sides.
Types of Additions
Addition of HX (1)
Protonation of double bond yields the most
stable carbocation. Positive charge goes to the
carbon that was not protonated.
Addition of HX (2)
CH3
CH3 C CH CH3
H Br
CH3
CH3 C CH CH3
+
H
_
Br
CH3
CH3 C CH CH3
+
H
+ Br
CH3
CH3 C CH CH3
Br H
_
Regiospecificity
• Markovnikov’s Rule: The proton of an
acid adds to the carbon in the double
bond that already has the most H’s. “Rich
get richer.”
• More general: In an electrophilic addition
to an alkene, the electrophile adds in
such a way as to form the most stable
intermediate.
• HCl, HBr, and HI add to alkenes to form
Markovnikov products.
Alkenes
Hydrohalogenation—Reaction Stereochemistry
• Recall that trigonal planar atoms react with reagents from two
directions with equal probability.
• Achiral starting materials yield achiral products.
• Sometimes new stereogenic centers are formed from
hydrohalogenation:
A racemic mixture →
Alkenes
Hydrohalogenation—Reaction Stereochemistry
• The mechanism of hydrohalogenation illustrates why two
enantiomers are formed. Initial addition of H+ occurs from either
side of the planar double bond.
• Both modes of addition generate the same achiral carbocation.
Either representation of this carbocation can be used to draw the
second step of the mechanism.
Alkenes
Hydrohalogenation—Reaction Stereochemistry
• Nucleophilic attack of Cl¯ on the trigonal planar carbocation also
occurs from two different directions, forming two products, A
and B, having a new stereogenic center.
• A and B are enantiomers. Since attack from either direction
occurs with equal probability, a racemic mixture of A and B is
formed.
Alkenes
Hydrohalogenation—Reaction Stereochemistry
• Hydrohalogenation occurs with syn and anti addition of HX.
• The terms cis and trans refer to the arrangement of groups in a
particular compound, usually an alkene or disubstituted
cycloalkene.
• The terms syn and anti describe stereochemistry of a process—
for example, how two groups are added to a double bond.
• Addition of HX to 1,2-dimethylcyclohexene forms two new
stereogenic centers, resulting in the formation of four
stereoisomers (2 pairs of enantiomers).
Hydrohalogenation—Reaction Stereochemistry
Hydrohalogenation—Summary
Free-Radical
Addition of HBr
• In the presence of peroxides, HBr adds to
an alkene to form the “anti-Markovnikov”
product.
• Only HBr has the right bond energy.
• HCl bond is too strong.
• HI bond tends to break heterolytically to
form ions.
Free Radical Initiation
• Peroxide O-O bond breaks easily to form
free radicals.
R O O R
heat
R O
+
O R
• Hydrogen is abstracted from HBr.
R O
+ H Br
R O H
+ Br
Electrophile
Propagation Steps
• Bromine adds to the double bond.
H
H
+
H
.Br
H
H3C
H3C
.
H
Br
H
• Hydrogen is abstracted from HBr.
C C
Br
+
H Br
C C
+
Br
Br H
Electrophile
Anti-Markovnikov ??
CH3
CH3 C CH CH3
CH3
CH3 C CH CH3
Br
+
Br
X
CH3
CH3 C CH CH3
Br
2o
• Tertiary radical is more stable, so that
intermediate forms faster.
3o
Hydration of Alkenes
+
H
C C
alkene
+ H2O
H OH
C C
alcohol
• Reverse of dehydration of alcohol
• Use very dilute solutions of H2SO4 or
H3PO4 to drive equilibrium toward
hydration.
Mechanism for Hydration
H
H
C C
+
+
H2O
H
+
H O H
H
+
C C
+
H2O
H
+
H O H
C C
+
C C
+
H O H
C C
H
H O
+
H2O
C C
+
H3O
+
Orientation for Hydration
• Markovnikov product is formed.
H
H
H3C
+
H3C
..
H2O:
CH3
..
H2O:
+
O
..
H
CH3
H
H3C
CH3
+
CH3
H
..
H3C
H
+
CH3
H H
O+
CH3
H3C
CH3
H
HO
H3C
CH3
CH3
+ H3O+
Indirect Hydration
• Oxymercuration-Demercuration
– Markovnikov product formed
– Anti addition of H-OH
– No rearrangements
• Hydroboration
– Anti-Markovnikov product formed
– Syn addition of H-OH
Oxymercuration (1)
• Reagent is mercury(II) acetate,
Hg(OAc)2. which dissociates slightly to
form +Hg(OAc).
• +Hg(OAc) is the electrophile that attacks
the pi bond.
Note: OAc is the abbrev. of COOCH3
H3C
CH3
H3C
CH3
+
Hg(OAc)2
H2O
H3C
HO
CH3
HgOAc
NaBH4
H3C
HO
CH3
H
Oxymercuration (2)
The intermediate is a cyclic mercurinium
ion, a three-membered ring with a positive
charge.
Oxymercuration (3)
• Water approaches the mercurinium ion from
the side opposite the ring (anti addition).
• Water adds to the more substituted carbon to
form the Markovnikov product.
Demercuration
Sodium borohydride, a reducing agent,
replaces the mercury with hydrogen.
Advantages and Disadvantages of
Oxymercuration-Demercuration
Advantages
• Gives better yields than direct acidcatalysed hydration
• Avoids the possibiliti of rearrangements
• Does not involve harsh (acidic) conditions
Disadvantages
• Highly toxic
Alkoxymercuration-Demercuration
• Follows the reaction pathway or
mechanism of oximercurationdemercuration
• If the nucleophile is an alcohol, ROH,
(instead of water, HOH) the product is an
ether
• Markovnikov orientation of addition
Hydroboration
• Borane, BH3, adds a hydrogen to the
most substituted carbon in the double
bond.
• The alkylborane is then oxidized to the
alcohol which is the anti-Markovnikov
product.
Borane Reagent
• Borane exists as a dimer, B2H6,
in equilibrium with its monomer.
• Borane is a toxic, flammable, explosive gas.
• Safe when complexed with tetrahydrofuran,
THF.
Stoichiometry
H3C
CH3
3
H3C
CH3
+
BH3
THF
H3C
CH3
B
H
3
H3C
H2O 2, -OH
CH3
3
H
Borane prefers least-substituted carbon
due to steric hindrance as well as charge
distribution.
OH
Mechanism
• The electron-deficient borane adds to
the least-substituted carbon.
• The other carbon acquires a positive charge.
• H adds to adjacent C on same side (syn).
Oxidation to Alcohol
• Oxidation of the alkyl borane with basic
hydrogen peroxide produces the alcohol.
• Orientation is anti-Markovnikov.
Addition of Halogens
• Cl2, Br2, and sometimes I2 add to a double
bond to form a vicinal dibromide.
• Anti addition, so reaction is stereospecific.
Mechanism for Halogenation
• Pi electrons attack the bromine molecule.
• A bromide ion splits off.
• Intermediate is a cyclic bromonium ion.
H3C
H3C
+
CH3
CH3
Br
+
Br-Br
H3C
H3C
CH3
CH3
bromonium ion
Mechanism (2)
Halide ion approaches from side opposite
the three-membered ring. (resembles SN2)
H3C
H3C
Br
CH3
CH3
+
Br-Br
+
CH3
H3C
H3C
Br CH3
..
:Br:
..
CH3
H3C
CH3
CH3 Br
Examples of
Stereospecificity
Br
H
H 3C
Br
H
CH3
H
CH3
H
CH3
Br
1800
H
CH3
H
CH3
Br Br
Br
CH3
Br
R
H
H
R
Br
CH3
H 3C
H
Br
Br
Br
H
CH3
H
CH3
1800
CH3
CH3
H
H
H
CH3
Br Br
Br
CH3
H
S
H
R
CH3
Br
Br
Test for Unsaturation
• Add Br2 in CCl4 (dark,
red-brown color) to an
alkene in the presence
of light.
• The color quickly
disappears as the
bromine adds to the
double bond.
• “Decolorizing bromine”
is the chemical test for
the presence of a
double bond.
Formation of Halohydrin
• If a halogen is added in the presence of
water, a halohydrin is formed.
• Water is the nucleophile, instead of
halide.
• Product is Markovnikov and antistereochemistry.
Regiospecificity
The most highly substituted carbon has
the most positive charge, so nucleophile
attacks there.
Hydrogenation
•
•
•
•
•
Alkene + H2 → Alkane
Conditions: At room T & 1 atm
Catalyst required, usually Pt, Pd, or Ni
Finely divided metal, heterogeneous
Syn addition, syn stereochemistry
Epoxidation
• Alkene reacts with a peroxyacid (sometimes called peracids) ,
a carboxilic acid that has an extra O atom in –O-O-(peroxy)
linkage to form an epoxide (also called oxirane).
• Usual reagent is peroxybenzoic acid (m-chloroperoxybenzioc
acid, MCPBA) for its desirable solubility properties: The
peroxyacid dissolves, then the spent acid precipitates out of
solution.
R
R
OH
O
a carboxylic acid
O
O
OH
a peroxy acid
Mechanism
One-step, concerted electrophilic reaction.
Several bonds break and form simultaneously.
Epoxide Stereochemistry
No rotation around the double-bonded
carbons, so cis or trans stereochemistry is
retained.
m-chloroperoxybenzoic acid
precipitate
Opening the Epoxide Ring
• Most epoxides are easily isolated as stable
products if the solution is not too acidic.
• Acid catalyzed.
Any moderately strong acid protonates the
epoxide.
• Water attacks the protonated epoxide (back-side
attack), opening the ring.
• Anti orientation, trans diol is formed.
One-Step Reaction
• To synthesize the glycol without
isolating the epoxide, use aqueous
peroxyacetic acid or peroxyformic acid.
• The reaction is stereospecific.
Epoxides vs Glycol
• Epoxidation reagents can be chosen to
favour either epoxide or glycol.
• Peroxyacetic acid is used in strongly acidic
water solutions to yield glycols.
(through the protonation of epoxide)
• Peroxybenzoic acids are weak acids that can
be used in NON nucleophlic solvents such as
CCl4 to yield epoxides.
• Refer p357, figure 8.8 for more examples
Syn Hydroxylation
of Alkenes
• Alkene is converted to a cis-1,2-diol
• Two reagents:
– Osmium tetroxide (expensive, highly toxic &
volatile), followed by hydrogen peroxide
or
– Cold, dilute aqueous potassium
permanganate, followed by hydrolysis with
base
Mechanism with OsO4
Concerted syn addition of two oxygens to
form a cyclic ester.
Stereospecificity
If a chiral carbon is formed, only one
stereoisomer will be produced (or a pair of
enantiomers).
Permanganate hydroxylation
• Syn stereochemistry
• Slightly reduced yields in most cases as
compared to OsO4.
• Provides a simple chemical test for the
presence of an alkene, colour changes fr
deep purple to murky, opague brown colour of
MnO2.
Oxidative Cleavage
• Both the pi and sigma bonds break.
• C=C becomes C=O.
• Two methods:
– Warm or concentrated or acidic KMnO4.
– Ozonolysis.
• Used to determine the position of a
double bond in an unknown alkene.
Cleavage with MnO4
-
• Permanganate is a strong oxidizing
agent.
• Glycol initially formed is further oxidized.
• Disubstituted carbons become ketones.
• Monosubstituted carbons become
carboxylic acids (fr aldehydes).
• Terminal =CH2- becomes CO2.
Examples
CH3
H3C
O
CH3
CH3
KMnO4
(warm, concd.)
CH2
+
(warm, concd.)
CH3
OH
O
KMnO4
H3C
COOH
COOH
O
+
CO2
Ozonolysis
• Reaction with ozone forms an ozonide.
• Ozonides are not isolated, but are treated
with a mild reducing agent like Zn or
dimethyl sulfide (CH3)2S.
• Milder oxidation than permanganate.
• Products formed are ketones or aldehydes.
• A common method to determine the
position of double bonds in alkenes
Ozonolysis Mechanism
• Formation of ozonide, then
reduction with dimethyl sulfide (CH3)2S).
Ozonolysis Example
CH3
CH3
(1) O 3
H3C
(2) (CH3)2S
CH2
H3C
CHO
CHO
O
+
H2C
O
-
Comparison between MnO4
cleavage & ozonolysis
• Both MnO4- cleavage & ozonolysis break
C=C bond & replace it with carbonyl (C=O)
groups.
• In MnO4 cleavage, any aldehyde products
are further oxidized to COOH.
• In ozonolysis-reduction, aldehyde products
are generated in the DMSO reduction step
& are not oxidized.
Alkenes
Alkenes in Organic Synthesis
Suppose we wish to synthesize 1,2-dibromocyclohexane from
cyclohexanol.
To solve this problem we must:
Alkenes
Alkenes in Organic Synthesis
Working backwards from the product to determine the starting
material from which it is made is called retrosynthetic analysis.
Predict the MAJOR products of the following reactions,
and give the structures of any intermediates. Include
stereochemistry where appropriate.
(a)
H3C
(b)
(c)
(e)
H3C
HCl
(1) Hg(OAc)2.H2O
(2) NaBH4
CH3
CH3
CH3
(f)
HCl
+
H , H 2O
ROOR
(g)
CH3CO3H
CH3
OsO4
H2O 2
(d)
(h)
CH3
(1) BH3.THF
-
(2) H2O2, OH
CH3
-
KMnO4, OH
(cold, dil.)
Show how you would synthesize each compound using
methylenecyclohexane as your starting material.
CH2
(a)
(d)
CH3
(b)
OH
O
(g)
OH
(e)
Cl
Br
CH3
OCH3
(c)
OH
(f)
OH
OH