### Alkene: Structure and Reactivity

```Alkene: Structure and
Reactivity
Chapter 6
Alkenes
• An alkene (also called an olefin) is a
hydrocarbon with a carbon-carbon double
bond.
• Alkenes are present in most organic and
biological molecules.
Examples of Alkenes
H3C
OH
H
H3C
H
H
O
Testosterone
b-Carotene
Calculating Degrees of Unsaturation
• Because a double bond has fewer hydrogens
than an alkane it is refereed to as unsaturated.
One example is that of ethylene.
H
H
H
H
H
H
H
H
H
H
Ethylene
Ethylene
C2H4
C2H6
Saturated Alkane Chains
• To calculate the amount of hydrogens in a
alkane chain all you have to do is take the
number of carbons (make that n) and
multiply it by 2 then add 2 more for the
additional hydrogens at the terminals.
CnH2n+2
Degrees of Unsaturation
• There are several ways that a molecule can be
unsaturated. The first is by an double or triple
bond. The second is by the formation of a ring.
Or a combination of both.
All of these are C6H10
How to Calculate Unsaturation
• Take the original number of hydrogens found.
C6H10
• Use the calculation of saturated alkanes.
C6H2(6)+2 = 14
C6H14
• Subtract the original hydrogens from saturated
alkenes.
14 – 10 = 4 hydrogens
• For every two hydrogens (divide by 2) that are
missing 1 degree of unsaturated is found.
4 hydrogens / 2 = 2 degrees of unsaturation (answer)
Calculating Unsaturation in other
Molecules
• What if the molecules contains halogens,
oxygen, and nitrogen? There are rules for these
possibilities.
• For halogens (F, Cl, Br, I) since they bond only
once similar to hydrogen, you can consider them
a hydrogen.
• Since oxygen forms 2 bonds it doesn’t effect the
formula. IGNORE IT.
• For nitrogen, subtract the number of nitrogens
from hydrogens to get the proper saturated
number.
Summarize These Rules
• Add the number of halogens to the
number of hydrogens.
• Ignore the number of oxygens.
• Subtract the number of nitrogens from
hydrogens.
Naming Alkenes
• 1) Name the Parent hydrocarbon (THAT
INCLUDES THE ALKENE).
• 2) Number the carbon atoms in the chain with
either the alkene or branching with the lowest
number.
• 3)Write the full name. Name it accordingly with
substituents properly named and numbered.
Also remember it there are more than 1 alkene it
is diene (2), triene (3), and so forth.
Examples
1-cyclohexene
4-methyl-1,3-cyclopentadiene
CH3CH2
H3CH2CH2C
2-ethyl-1-pentene
H
H
4
3
CH3
5
CH3
1
2
1,5-dimethyl-1-cyclopentene
Older Names Still in Use
• Methylene (group)
H2C
• Vinyl (group)
H2C CH
• Allyl (group)
H2C CH CH2
• Ethylene instead of Ethene
Cis –Trans Isomers in Alkenes
• Because double bonds do not rotate freely they
are similar to ring structure because they can be
stereoisomers. They can have both a cis and
trans configuration. This occurs when the double
bond is disubstituted (not when 3 hydrogens are
bound to the double bond).
H3C
H
Cis
CH3
H
CH3
H
H
H
Not Stereoisomers
H
H3C
Trans
CH3
H
E, Z Designation
• Cis and Trans only describes disubstituted
alkanes. It does not describe when other
atoms (halogen, oxygen, nitrogen)
attaches to the double bond. The E,Z
system is used to describe when this
occurs.
• E = Trans
• Z = Cis (for me I remember CIZ)
E, Z Rules
• Rule 1. Consider Each Carbon Separately. Look at
the atoms directly attached to each and rank them
accordingly BY ATOMIC NUMBER.
• Br > Cl > S > P > O > N > C > 2H > 1H
• Rule 2. If the first atom don’t give you an answer then
move down the chain until you find a difference.
• Rule 3. Multiple bonds atoms are equivalent to the
same number of single bonds.
H
C O
H
C O
O C
C = H,O,O
O = C,C
Stability of Alkenes
• Because of steric strain the energy of cis
isomers are higher than that of the trans
isomers. This does not mean that you can not
find the cis isomer but often an equilibrium can
form with the trans more abundant.
Alkene Stability (Lower Energy)
The Heats of Hydrogenation is a determination of
the relative stabilities of cis and trans isomers
without looking at equilibrium positions. Alkyl
groups are better than H.
R
R
R
R
>
tetrasubstituted
R
H
R
R
trisubstituted
>
R
H
R
H
~
R
R
H
H
disubstituted
>
R
H
H
H
monosubstituted
Stability of Alkenes
• Stability order of alkenes is due to a combination of
two factors. The most important is the stabilizing
interactions between the C=C p bond and adjacent
substituents. This is called hyperconjugation.
Comparing Stabilities of Alkenes
• Evaluate heat given off when C=C is converted to
C-C.
• More stable alkene gives off less heat.
– Trans butene generates 5 kJ less heat than cisbutene.
Cis
H3C
H
H
CH3
H2
H
Pd
CH3
Trans
H3C
H
H2
Pd
H3C
H
H
H3C
H
H
CH3
H
H
CH3
H
H
Electrophilic Addition of HX to
Alkenes
• General reaction mechanism: electrophilic addition.
• Attack of electrophile (such as HBr) on  bond of
alkene.
• Produces carbocation and bromide ion.
• Carbocation is an electrophile, reacting with
nucleophilic bromide ion.
H
R
Br
H
R
-
Br
R
H
Nucleophile
R
H
H
H
Br
H
H
H
H
H
Electrophilic Addition Energy Path
• Two step process.
• First transition state is high energy point.
Example of Electrophilic Addition
• Addition of hydrogen bromide to
2-Methyl-propene.
• H-Br transfers proton to C=C
• Forms carbocation intermediate.
– More stable cation forms.
• Bromide adds to carbocation.
Energy Diagram for Electrophilic
Addition
• Rate determining
(slowest) step has highest
energy transition state.
– Independent of
direction.
– In this case it is the first
step in forward
direction.
– “rate” is not the same
as “rate constant”.
Carbocation Structure and Stability
• Therefore stability of carbocations: 3º > 2º > 1º > +CH3
R
R C
R
Tertiary (3o)
>
H
R C
R
>
Secondary (2o)
H
H C
R
>
Primary (1o)
H
H C
H
Methyl
Orientation of Electrophilic Addition:
Markovnikov’s Rule
• In an unsymmetrical alkene, HX reagents can
add in two different ways, but one way may be
preferred over the other.
• If one orientation predominates, the reaction is
regiospecific.
• Markovnikov observed in the 19th century that in
the addition of HX to alkene, the H attaches to
the carbon with the most H’s and X attaches to
the other end (to the one with the most alkyl
substituents).
– This is Markovnikov’s rule.
Example of Markovnikov’s Rule
• Regiospecific – one product forms where two are
possible
• If both ends have similar substitution, then not
regiospecific.
CH2
H
H
CH2
H Br
CH2
Br
BrBr
CH2
H
Not Favored
H
CH2
Br
Favored
Energy of Carbocations and
Markovnikov’s Rule
• More stable carbocation forms faster.
• Tertiary cations and associated transition states
are more stable than primary cations.
Mechanistic Source of Regiospecificity
in Addition Reactions
• If addition involves a
carbocation intermediate.
– and there are two
possible ways to add.
– the route producing the
more alkyl substituted
cationic center is lower in
energy.
– alkyl groups stabilize
carbocation.
The Hammond Postulate
• If carbocation intermediate is more stable than
another, why is the reaction through the more stable
one faster?
– The relative stability of the intermediate is related to
an equilibrium constant (DGº).
– The relative stability of the transition state .(which
describes the size of the rate constant) is the
activation energy (DG‡).
– The transition state is transient and cannot be
examined.
Transition State Structures
• A transition state is the highest energy species in a
reaction step.
• By definition, its structure is not stable enough to
exist for one vibration.
• But the structure controls the rate of reaction.
• So we need to be able to guess about its properties
in an informed way.
• We classify them in general ways and look for trends
in reactivity – the conclusions are in the Hammond
Postulate.
Statement of the Hammond Postulate
• A transition state should be similar to an
intermediate that is close in energy.
• Sequential states on a reaction path that are
close in energy are likely to be close in structure.
- G. S. Hammond
G
carbocation
Reaction
In a reaction involving a
carbocation, the
transition states look like
the intermediate.
Competing Reactions and the
Hammond Postulate
• Normal Expectation: Faster reaction gives more stable
intermediate.
• Intermediate resembles transition state.
“Non-Hammond” Behavior
• More stable intermediate from slower reaction
• Conclude: transition state and intermediate must not
be similar in this case – not common.
Take Home Message
•
•
•
•
•
•
Alkene Structures
Calculating Degrees of Unsaturation
Naming Alkenes
Alkene Stability
Markovnikov’s Rule
Transition States
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