Chem 230 Lecture 9 Page- 1 Isomerization of alkenes.
65.9 kcal/mole. This is the energy it takes to break a pi bond
* Instructor shows how to think about a pi-bond breaking.
- compare this to the C-C bond energy, ~83 kcal/mole.
- the C=C bond energy is between 146 and 151 kcal/mole.
65.9 + 83 = 148 kcal/mole.
The barrier of 65.9 kcal/mole is easily enough to isolate two separate rotamers.
- Remember that the barrier to ethane rotation was only ~3 kcal/mole
* We said that the ethane rotamers quickly interchange at room temperature
* What is the origin of the barrier in ethane versus ethane?
- In the case of 1,2-disubstituted ethene derivatives we don’t call these rotamers, we call
them geometrical isomers.
Nomenclature of alkenes.
- cis and trans
and
cis-2-butene and trans-2-butene
- if disubstituted with different substituents we apply (Z) zusammen or (E) entgegan.
Steps to naming Alkenes
(1) Decide if there are isomer issues such as cis/trans or E/Z. Apply Cahn-Ingold-Prelog rules.
(2) Find the longest chain that contains the double bond(s).
when two alkenes are present include both double bonds in the longest chain.
(3) Follow the rules for alkanes.
- minimize the number that refers to the position of the double bond.
Examples.
cis-3-hexene
- draw a structure for the name above
(Z)-2-chloro-3,5-dimethyl-2-heptene
- draw a structure for the name above
Cl
F
[name the structure above: (1E)-1-fluoro-2-ethyl-1,3-butadiene]
Highest priority groups on the same side of the alkene then (Z)
Learn the rules below; you will need them again.
Cahn-Ingold-Prelog Priority System
See text for more details.
(1) heaviest atoms have higher priority
(2) if you can't decide with the first atom in the group then go to the second
(3) in the case of unsaturation expand the group.
Chem 230 Lecture 9 Page- 2 -
More substituted alkenes are more stable than less substituted alkenes.
* compare the isomeric hexenes
alkene
* 1-hexene
* cis-3-hexene
* trans-3-hexene
* 2-methyl-2-pentene
* 2,3-dimethyl-2-butene
∆Hf
-10.0
-11.2
-12.1
-16.0
-16.6
- question is why?
The usual answer is hyperconjugation.
Instructor explains hyperconjugation.
1,2-interactions
Torsional strain We have seen this term before in the context of rotamers of alkanes.
* The gain in stability from substitution drops off in the Table above as the alkene
is more substituted.
Structural constraints due to the double bond
-alkenes at bridgehead positions
* decalin has a bridgehead
A
B
C
D
E
Decalin and some dehydrodecalins
-There is no problem with the double bonds in the structures above.
- The two dehydrodecalins in the center are at bridgehead positions
Problematic Bridgehead for double bonds
* [2.2.1]-bicycloheptane and [2.2.2]-bicyclooctane
* [2.2.1]-bicycloheptene and [2.2.2]-bicyclooctene
- Why are some of these structures problematic for the double bonds they contain?
- Answer: The p orbitals cannot overlap therefore there is no π-bond.
* Instructor showed this by diagramming the molecular orbitals.
nomenclature of bicycloalkanes
•
•
•
•
based on "bridgehead" positions
[m.n.p] bicycloalkane
m, n, p are the numbers of carbons in each bridge, listed in decreasing order
carbon numbers for substituents start at a bridgehead and traverse larger bridges in order
Br
6-bromobicyclo[3.2.1]octane
Chem 230 Lecture 9 Page- 3 and
H
H
H
H H
H
H H
H
HH
H
bicyclo[4.2.1]nona-2,4-diene, a conjugated diene
bicyclo[4.2.1]nona-2,7-diene, a non-conjugated diene