Chapter 3: Alkenes and Alkynes
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© R. Spinney 2013
General Properties
Alkenes contain double bonds and alkynes triple
bonds. Both classes of compounds are
hydrocarbons, containing only C and H atoms.
– a double bond consists of 1 s and 1 p bond,
– a triple bond consists of 1 s and 2 p bonds.
They are also “unsaturated” as they contain p
bonds and are also known as olefins.
General Properties (cont’d)
They contain less H atoms that the corresponding
alkane, generic chemical formulas are;
CnH2n+2
CnH2n
CnH2n-2
alkanes
alkenes
alkynes
Every p bond results in the loss of a pair of H atoms.
General Properties (cont’d)
With multiple double (or triple) bonds three
possible arrangements arise: cumulated,
conjugated or isolated (non-conjugated).
Conjugated are especially important as the p
bonds can interact.
Cumulated
Conjugated
Isolated
C=C=C
C=C-C=C
C=C-C-C=C
C=C=C=C
C≡C-C≡C
C≡C-C-C-C≡C
Bonding in Alkenes
•
•
•
•
•
Alkenes are sp2 hybridized
Trigonal planar – bond angle ~120°
3 s and 1 p bond (or 2 single and 1 double)
C=C double bond ~ 1.34 Å
The p bond lock the geometry to planar
Cis – Trans Isomerism
The double bond in an alkene is rigid, that is it
will not rotate freely. Therefore substituents on
the carbon atoms will produce geometric
isomers the same as on a cycloalkane ring.
Cis – Trans Isomerism (cont’d)
If the two non-hydrogen atoms or groups are on the same side of the
double bond it is a cis- arrangement, on opposite sides a transarrangement, i.e.
Note that they have different physical properties, this is because their
dipole moments are different.
Cis – Trans Isomerism (cont’d)
Under normal conditions the two isomers will not interconvert
since you would have to break the p bond, however using light
or heat it is possible to interconvert the two isomers. i.e.
C l
C l
heat or
lig h t
C l
•
•
C l
C l
C l
•
•
•
C l
C l
•
C l
C l
Chemical Reactivity
The chemical reactivity of alkenes arise
from the p electrons. The p bond is
weaker than the s bond so these
electron react first. The reagent will add
across the double bond so these are
addition reactions.
Electrophilic Addition Mechanism
The basic mechanism is the same for all
reagents, a two step reaction where an
electrophile (E) add to the p bond in the first
step creating a carbocation intermediate. A
nucleophile (Nu) then adds to the carbocation in
the second step.
Nu
+
E
E
E
+
Nu-
Nu
Carbocations
Are electron deficient and have an
empty p orbital (sp2 hybridized). Not all
carbocations are equally stable so there
are predictable patterns for which ones
will form.
C H3
C H3
H3C
C
+
3°
C H3
>
H
C
+
2°
C H3
H
>
H
C
H
+
C H3
>
1°
Note: this is the same order of stability as carbon radicals.
H
C
+
H
m e th y l
Carbocations (cont’d)
The order of carbocation stability arise from
three sources.
1)
2)
Inductive electron donation: the
electrons in C-C s bonds will be pulled
closer to the C+ helping to minimize the
charge. Note: this does not work for C-H
bonds.
Hyperconjugation: this is a orbital
interaction between adjacent C-H bonds
that can overlap the empty p orbital of
the C+, this again helps to minimize the
charge on the C+.
C H3
C
H3C
+
H
Carbocations (cont’d)
3) Resonance: a carbocation immediately adjacent to a p
system (double bond, triple bond or aromatic ring) can be
stabilized by resonance. This lowers the energy by
spreading the charge over more atoms, i.e.
Carbocations (cont’d)
One final complexity for carbocations: rearrangement.
Carbocations are susceptible to 1,2-hydride shifts, where an
adjacent H atom (and its bonding electrons) shift to the C+ to
produce a more stable carbocation, i.e.
+
H
+
H
This is also possible for methyl groups. These shifts are
common when a tertiary, allylic or benzylic carbocation is
produced.
Halogenation
• Addition of Cl2 or Br2 across the double
bond to product 1,2-dihalide alkane.
• Halogen is usually dissolved in chloroform
or carbon tetrachloride.
• Reaction is rapid at room temperatures
• Addition of bromine is a common chemical
test for the presence of double bonds as
the red colored bromine solution turns
colorless when it reacts with an alkene
• Reaction relies on the polarizability of the
halogen bonds
Halogenation (cont’d)
B
r B
r
In the first step the p electrons act as a
nucleophile attacking the bromine displacing
a bromide ion and forming the cyclic bromonium
cation intermediate.
+
B
r
In the second step the nucleophilic bromide ion
attacks the side of the bromonium ion away from
the bromine atom opening the highly strain cyclic
structure and producing the 1,2 dibromo
hydrocarbon.
Note this is a trans addition since the two Br
atoms add to opposite side of the double bond.
B
r-
B
r
B
r
Symmetry and Addition Reactions
The halogenation of ethene is a
very symmetry reaction as both
the reagent (Br2) and the
substrate (ethene) are symmetric.
In this situation there can only be
a single product. What about
asymmetric reactions? Two
products are possible now, which
regioisomer will form?
Markovnikov’s Rule
Fortunately there is a simple rule of thumb to
predict which product will form: Markovnikov’s
rule, which states that when an unsymmetrical
reagent adds to a double bond the electrophilic
part of the reagent adds to the carbon with the
most hydrogen atoms on it.
Markovnikov’s Rule (cont’d)
Why? This mode of addition will always produce
the most stable carbocation intermediate, i.e.
consider the addition of H+ to propene…
H
1 ° c a rb o c a tio n - a n ti-M a rk o v n ik o v p ro d u c t
+
H+
+
H
2 ° c a rb o c a tio n - M a rk o v n ik o v p ro d u c t
Hydration
• Addition of water across the double bond
• Product is an alcohol
• Requires an acid as a catalyst as water is not
acidic enough to produce the electrophile
Hydration (cont’d)
H+
First step is the protonation of the alkene in a
Markovnikov orientation to generate the most
stable carbocation.
+
H
O
H
The second step is the attack of the nucleophilic
water molecule on the carbocation
H +H
O
Finally an acid/base reaction deprotonates the
alkyloxonium ion to form the alcohol.
H
OH
O
H
Hydrohalogenation
• Addition of HX across a double bond to produce a
Markovnikov halide alkane
• Reactivity: HI > HBr > HCl > HF (parallels acidity)
• HBr needs to be used in the dark and under an
inert atmosphere to prevent a free radical
addition process that produces the antiMarkovnikov product.
• Note: this same mechanism applies to other acids
such as H2SO4
Hydrohalogenation (cont’d)
First step in the Markovnikov addition of the
electrophilic acidic proton to produce the most
stable carbocation intermediate.
+
H
+
The second step is the nucleophilic attack of the
halide anion on the carbocation intermediate
generating the alkyl halide product.
Br-
Br
Thermodynamics & Equilibrium
For a generic chemical reaction:
aA + bB ⇌ cC + dD
We can define the where the equilibrium lies by:
[C ]c [ D]d
K eq 
[ A]a [ B]b
Thermodynamics & Equilibrium
The size of the equilibrium constant, Keq tells us
which direction of the reaction is favored.
Greater than 1, the products and the reaction
proceeds from left to right.
The larger Keq is the further towards the
products the equilibrium lies.
Thermodynamics & Equilibrium
The equilibrium can be shifted, this is Le Châtelier’s
Principle. Adding a reactant or removing a product will
help “drive” the reaction forward, i.e.
O
+
O H
O H
O
O
+
Adding ethanol, or removing water will drive this
reaction to the right.
H 2O
Thermodynamics & Equilibrium
So what determines if a reaction proceeds to the right or left?
Spontaneous chemical processes release heat as a product. These reactions
are exothermic (the enthalpy, DH < 0).
Endothermic (DH > 0) processes can be made to work, but you have to
provide heat to “drive” the reaction.
Thermodynamics & Equilibrium
The equilibrium constant and enthalpy tell us
which direction a reaction proceeds in, and
whether it is spontaneous or not, but neither
tells us how fast it will go. For that we need to
use Transition State Theory.
This theory states that the reactants will
proceed through a high energy “Transition
State” (TS) and then proceed to the products.
Thermodynamics & Equilibrium
The TS is the point where the molecules are colliding and bonds
are being broken and reformed. It is commonly represented in a
“reaction coordinate diagram”.
Thermodynamics & Equilibrium
In contrast to an intermediate, the TS has no real life time. You
cannot isolate the molecules in the TS. Currently the TS state is
studied primarily by computational chemical models, although
new laser spectroscopy techniques have a time frame short
enough to provide information on the TS.
Thermodynamics & Equilibrium
So how does this work?
We can explain the reaction of
propene with a proton based on a
reaction coordinate diagram. The
reason the 1° carbocation is less
stable than the 2° is because it has
a TS higher in energy and so is
harder to form.
Thermodynamics & Equilibrium
The reaction of HBr with ethene can be
explained in a similar fashion.
TS 1 is high E as we are in the process of
breaking the H-Br bond but have not yet
formed the new C-H bond.
The intermediate carbocation is lower in E as
the new C-H bond has formed.
TS2 is higher in E as the C-Br bond is in the
process of forming.
Finally the product has the lowest E so the
reaction is exothermic and spontaneous
providing there is sufficient E to form TS1.
Thermodynamics & Equilibrium
Catalysts act by
providing an alternate
pathway for the
reaction which has a
lower activation
barrier. This applies to
enzymatic and nonenzymatic catalysts.
Thermodynamics & Equilibrium
Thermal effects change the energy of the reactants. If the
temperature is raised the reactants have a higher starting energy
and hence have a low barrier to cross, the Ea. Note this has no
effect of the energy at the TS or the products.