The Structure and Synthesis of
Alkenes
State University of New York at Albany
Alkenes:
Are hydrocarbons with carbon-carbon
double bonds.
 Are also known as olefins.
 C=C is considered to be a functional group
because it is relatively reactive.
 Double bonds (1.33 Å) are shorter than
single bonds (1.54 Å).

Elements of Unsaturation



Alkenes are unsaturated because they are capable
of adding hydrogen in the presence of a catalyst.
Alkanes are saturated because they cannot react
with anymore hydrogen.
Each pi bond of an alkene (or alkyne) or the ring of
a cyclic compound decreases the number of
hydrogen atoms in a molecular formula. This
feature is called “element or degree or unsaturation”.
Each element of unsaturation corresponds to two
fewer hydrogen atoms than the “saturated” formula.
Examples:
CH3
CH2 CH3
propane
saturated
CH3
CH
CH2
propene
one element of
unsaturation
cyclopropane
one element of
unsaturation
A molecule represented by the formula C4H8
has one element of unsaturation since it is
missing two hydrogens when compared to the
saturated alkane formula (C4H10). There are
five constitutional isomers of formula C4H8.
CH2
CH
CH2
1-butene
CH3
CH3
CH
CH
CH3
2-butene
CH3
CH3
C
CH2
isobuytlene
cyclobutane methylcyclopropane
Elements of Unsaturation with Heteroatoms




Heteroatoms are atoms different from carbon and
hydrogen.
Halogens: Halogen atoms are counted as hydrogen
atoms. E.g. C2H4F2 has no elements of unsaturation.
Oxygen atoms: should be ignored when calculating
elements of unsaturation. E.g. C2H4O has one
element of unsaturation.
Nitrogen atoms: should be counted as half of a
carbon atom. E.g. C4H9N should be treated as
C4.5H(9+2).Thus C4H9N is two hydrogens short of the
“saturated” formula, and it has one element of
unsaturation.
Nomenclature of Alkenes


Similar to that of alkanes, except the ending of the
root name corresponding to the longest continuous
chain containing the double bond is changed from
“ane” to “ene”. E.g. ethane becomes ethene,
propane becomes propene, cyclohexane becomes
cyclohexene.
When a chain contains more than 3 carbons, a
number is used give the position of the double
bond. The chain is numbered starting from the end
closest to the double bond. E.g.:
CH3
CH
CH
2-pentene
CH2
CH3

Compounds with two double bonds are
dienes, those with three double bonds are
trienes, and those with four double bonds
are tetraenes. Numbers are used to specifiy
the locations of double bonds.
CH2
CH
CH
1,3-butadiene
1,3,5,7-cyclooctatetraene
CH2

With regard to numbering the chain, the double
bond is given precedence over alkyl groups or
halogens.
Br
3-propyl-1-heptene
7-bromo-1,3,5-cycloheptatriene
CH3
CH2
CH
CH
CH3
3-methyl-1-butene

Alkenes named as substituents are called
“alkenyl” groups. Some alkenyl substituents
have common names.
CH2
methylene
CH2
CH CH2
allyl group
CH
CH2
vinyl group
Cl
3-methylenecylcohexene
allyl chloride
3-vinyl-1,5 hexadiene

Some simple alkenes have common names.
CH2
CH2
ethene
ethylene
CH3
CH
CH2
propene
propylene
ethenylbenzene
styrene
CH2
C
CH3
CH3
2-ethylpropene
isobutylene
Cis/Trans Nomenclature

When two similar groups are on the same side
of the double bond, the alkene is cis. When two
similar groups are on opposite sides of the
double bond, the alkene is trans. Note that if
either carbon of the double bond holds two
identical groups, the molecule cannot have cis
and trans forms.
cis-2-pentene
trans-2-pentene
2-methyl-2-pentene
(not cis or trans)
E/Z Nomenclature

When the cis/trans system fails to give an unambiguous name, the E/Z system is used.
Consider each carbon of the double bond
separately,
and
assign
Cahn-Ingold-Prelog
priorities to the two groups attached to each
carbon. If the two groups of high priority are on the
same side, the compound is Z; if they are on
opposite sides, it is E.
Br
CH3
C
Cl
Cl
C
CH3
C
H
Br
C
H
(Z)-1-bromo-1-chloro- (E)-1-bromo-1-chloropropene
propene
Commercial Uses of Alkanes
Used mostly in the production of polymers.
 Serve as intermediates in the synthesis of
drugs, pesticides and other valuable
chemicals including ethanol, acetic acid,
ethylene glycol and vinyl chloride.
 Ethylene is uses as a plant hormone,
accelerating the ripening of fruit.

Alkene Stability



Saytzeff’s Rule: the more highly substituted an
alkene is, the more stable it is. In other words, alkyl
groups attached to the double bonded carbons
stabilize the alkene.
The two probable factors responsible for the
stabilizing effect of alkyl groups to the double bond
are the contribution of electron density to the pi
bond by the alkyl groups, and the increased
separation of bulky groups from one another (120o
bond angle) in alkenes.
In general, trans isomers are more stable than cis
isomers because of decreased steric interactions.
Cycloalkene Stability



The presence of a ring only makes a difference in
stability if there is ring strain. Rings that are 5 carbons
or larger easily accommodate double bonds, and such
systems behave like straight chain alkenes.
The ring strain in cyclopropene and cyclobutene is
greater than that in cyclopropane and cyclobutane.
Thus cyclopropene and cyclobutene are less stable
than the analogous alkanes.
In cyclic alkenes, the cis isomer is generally more
stable than the trans. The trans isomer is rarely
observed unless the ring contains 10 or more
carbons.
Bredt’s Rule

Trans cycloalkenes are not stable unless there
are at least 8 carbon atoms in the ring. Bredt’s
rule states that a bridged bicyclic compound
cannot have a double bond at the bridgehead
position unless one of the rings contains at least
8 carbon atoms.
Bredt's rule
violation
Stable: although double
bond is at bridgehead,
system in not bridged.
Unstable: double bond is at
bridgehead and ring contains
only 6 carbons.
Polarity


Alkenes are relatively nonpolar. They are insoluble in
water, but soluble in nonpolar solvents such as
hexane. Alkenes are slightly more polar than alkanes
for two reasons: the pi bond electrons are more
polarizable, thus contributing to instantaneous dipole
moments, and the vinylic bond tends to be slightly
polar, contributing to the permanent dipole moment.
In a symmetrical trans disubstituted alkene, the sum
of the dipole moments is zero. In the analogous cis
alkene, the vector sum of the two dipoles is directed
perpendicular to the double bond. This results in a
non-zero molecular dipole. The permanent dipole
results in an increased bp.
H3C
H
C
H
C
H3C
CH3
C
H
 = 0.35 D
C
H
H
vector sum =
 = 0.33 D
cis-2-butene
bp = 4 oC
H3C
H
C
C
H
CH3
vector sum = 0
=0
trans-2-butene
bp = 1 oC
Synthesis of Alkenes by
Elimination Reactions

E2 dehydrohalogenation gives excellent yields
with bulky secondary and tertiary alkyl halides
that are poor SN2 substrates. A strong bulky
base forces second order elimination by
abstracting a proton. The bulkiness of the base
hinders second order substitution. Tertiary
halides are the best E2 substrates because they
are prone to elimination and cannot undergo
SN2 substitution.
Use of a Bulky Base

A bulky base can minimize substitution reactions
by hindering the approach to attack at carbon.
Yet, they can easily abstract a proton giving the
elimination product. Some commonly used
strong bulky bases are t-butoxide, triethylamine,
diisopropyl amine, and 2,6 dimethylpyridine.
Examples of Strong Bulky Bases
H
CH3
H3C
C
O
N
N
CH3
t-butoxide
diisopropylamine
H3C
N
CH3
2,6-dimethylpyridine
triethylamine
Example:
H Br
H
(i-Pr)2NH, heat
H
H
H
Bromocyclohexene, a 2o halide, can undergo both
substitution and elimination. E2 is favored over
SN2 by using diisopropylamine as the base.
Diisopropyl amine is too bulky to be a good
nucleophile, but it acts as a strong base to abstract
a proton.
Formation of the Hofmann Product

When non-bulky bases are used in eliminations in
which there is a choice of proton to abstract,
abstraction takes place to yield the most substituted
alkene (Saytzeff product). However, steric hindrance
may prevent a bulky base from abstracting the
proton that would lead to the most stable alkene. In
such cases, the less substituted alkene is formed
(Hofmann product).
Example:
Saytzeff product Hofmann product
H
H3C
H3C
CH3
C
C
CH2
H
Br
H
OCH2CH3
CH3CH2OH
CH3
C
CH3
CH2
C
H
C
CH3
H3C
CH3
C
C
CH2
H
Br
H
H3C
OC(CH3)3
CH3
CH2
C
H
C
CH3
28%
H
29%
CH3
C
(CH3)3COH
C
H3C
71%
H
H
H
C
H3C
H
72%
Dehydrohalogenation by the E1
Mechanism

E1 dehydrohalogenation usually takes place in a
good ionizing solvent such as an alcohol or
water, without a strong nucleophile or base to
force 2nd order kinetics. The substrate is usually
as 2o or 3o alkyl halide. This reaction is generally
accompanied by SN1 substitution because the
nucleophilic solvent can also attack the
carbocation directly.
Example:
H
H3C
CH3
C
C
CH2
H
Br
H
HOCH2CH3
H3C
CH3CH2OH
H3C
CH3
C
C
H
CH2
+
H
HOCH2CH3
CH3
C
H
H
C
CH3
H3C
H
CH3
C
C
H
O
CH3
CH2 CH3
Br
Dehalogenation of Vicinal Dibromides

Vicinal dibromides can be converted to alkenes by
reduction with either iodide ion or zinc in acetic acid.
The reaction proceeds via an E2 mechanism, taking
place through an anti-coplanar transition state.
Br H
Na I + H3C
C
C
H
Br
H3C
CH3
acetone
C
C
H
Br
C
CH3
H
C
H
+
CH3
H3C
CH3COOH
+ IBr
C
H
Br H
Zn + H3C
H
C
+
CH3
ZnBr2
NaBr
Dehydration of Alcohols

Dehydration refers to the “removal” of water. An
alcohol is protonated by an acid, and then water
leaves yielding a carbocation that loses a proton
to give an alkene.
CH3
CH3
C
CH3
H2SO4
OH
CH3
C
CH3
C
+
O
H
CH3
H
O
+
H
CH3
CH3
H2O
CH3
CH2
H CH2
+
H2O +
HSO4
CH3
CH3
CH3
H
CH3
CH3
CH3
CH3
H3O
Summary of Synthesis of Alkenes




Dehydrohalogenation by E2 mechanism: Saytzeff
and Hofmann products are possible depending on
the base used.
Dehydrohalogenation by E1 mechanism: both
substitution and elimination products are formed.
Dehalogenation of vicinal dibromides: occurs with
the treatment of vic-dibromide with I- or Zn in acetic
acid via an E2 mechanism.
Dehydration: involves conversion of the poor -OH
leaving group into the good leaving group H2O.