Organic Chemistry Notes by Jim Maxka
Chapter 14: Dienes and Conjugation
Topics
jim.maxka@nau.edu
Dienes: Naming and Properties
Conjugation
1,2 vs 1,4 addition and the stability of the allyl cation
Diels Alder Reaction
Simple Orbital Picture of dienes
Color and light absorption
Connections:
H2C
H2C
CH2
CH3
+
H3C
H3C
X
CH3
H3C
H3C
X
CH2
H3C
X
Cl
CH2
CH3
+
H3C
H2C
X
Cl
Cl
H3C
Cl
+
Cl
H3C
O
Cl
Cl
Cl
H
O
+
O
O
H
O
O
color with organic chemistry
H3C
CH3
17
CH3
16
1
2
19
7
6
3
5
4
CH3
18
18'
20
9
8
H3C
CH3
11
10
13
12
15
14
14'
15'
β,β-carotene
12'
13'
CH3
20'
10'
11'
8'
9'
CH3
19'
4'
5'
3'
6'
2'
7'
1'
H3C
16'
CH14-1
CH3
17'
Organic Chemistry Notes by Jim Maxka
Dienes
Naming: Draw the following:
1,3-Butadiene
1,3,5-Hexatriene
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Isoprene (2-Methyl-1,3-Butadiene)
Properties
Dienes have a new type of conformation. s-cis and s-trans.
Draw 1,3-butadiene in the s-Cis (syn) and in the s-Trans (anti) forms:
Label the hybridization of each C above
Make sure you understand the difference between conformation and configuration!!!!
The simplest possible diene is not butadiene.
C’s. The simplest diene would have
Butadiene has
Draw and name the simplest diene.
What is the hybridization of the C’s in allene (common name).
Conjugation:
Are the above molecules (butadiene and allene) conjugated? Why or why not?
Draw 1,4-pentadiene
Draw 1,3,5-hexatriene
Are these molecules conjugated? Why or why not?
Below are a ketone and the allyl cation:
O
CH
C
CH
H2C
+
CH2
H2C
CH3 and
And
Are these molecules conjugated? Why or why not?
What is a good definition of a conjugated molecule?
?
One way to show the stability in a conjugated molecule is to draw resonance structures that go over a long
path of the molecule:
Worked Example
O
O
-
CH
CH
C
+
C
H2C
H2C
CH3
CH3
CH14-2
Organic Chemistry Notes by Jim Maxka
jim.maxka@nau.edu
Thermochemistry: Stability of alkenes
Zaitsev and Markovnikov’s rules are based on the most substituted C with p-orbitals is the most stable.
Consider alkenes first.
Which is more stable 1-butene or -2-butene (cis/trans)? Why?
Zaitsev’s rule states that the most substituted alkene is the the most stable. Why?
Zaitsev’s rule is related to Markovnikov’s rule. For Zaitsev’s rule, we speak about substitution of p orbitals
in bonds called
bonds. For Markovnikov’s rule, we speak about substitution of p orbitals:
H
H
H
<
R
H
Substitution
Stability
R
<
R
H
H
10
0 (methyl)
R
R
R
30
20
Least stable
H
C
H
<
Most stable
H
R
C
C
R
R
For a simple graphic of hyperconjugation check out
http://chemistry.boisestate.edu/rbanks/organic/resonance.html
Thermochemistry-
The standard method is to compare a standard reaction for the same chemical structure, but
varying the position of the double bond.
CnH2n+2 (alkane)
CnH2n (alkene) + H2 (Pt)
Substitution on alkene: The more C’s attached to the alkene.
H3C
CH3
–30.3 kcal/mole
H2C
E
least subst-alkene H2
most subst-alkene H2
∆H
H3C
CH2
–28.5 kcal/mole
∆H
H3C
alkane
–26.9 kcal/mole
H3C
H2
H3C
Pt
CH3
H3C
CH3
H3C
To a rough approximation, we see that each alkyl substituent on a double bond stabilizes the
alkene about
kcal/mole.
CH14-3
Organic Chemistry Notes by Jim Maxka
jim.maxka@nau.edu
Cations
Simple Markovnikov theory will predict one product of the reaction of propene + HBr
+
CH2
CH2
H
H3C
Br
or
+
H3C
CH
Br
CH3
H3C
One of the propryl cations above is more stable than the other. Which one?
But conjugation is a more powerful stabilizing influence. One of the cations below is exceptionally stable.
Which one? Why? Draw the resonance structure in the box.
CH3
+
CH2
+
HC
HC
or
CH2
CH3
20 C+
10 C +
What substitution is the carbocation?
Why is it still so stable?
Identify the allyl H’s in the molecule below:
H
H
H
H
H
H
Below are allyl cations
Practice: Which systems below are allyl?
CH
+
CH
+
CH
+
CH
+
Electrophilic Addition of Conjugated Dienes
Review
In the electrophilic addition of an alkene with HBr, the reaction is a two step reaction.
+
CH
H
H3C
Br
CH3
H3C
or
CH2
CH2
H3C
Br
-
+
CH2
Step 1 is the formation of the C+ intermediate – the slow step.
Step 2 is the combination of Br with the C+ intermediate – the fast step. Why?
CH14-4
Organic Chemistry Notes by Jim Maxka
jim.maxka@nau.edu
Label the reaction diagram below:
B
D
∆G
C
A
∆G
A=
B=
C=
D=
0
0
1,2 vs 1,4 addition: In the reaction of propene with HBr as shown above, the stability of 2 > 1 C+ is the
key to the outcome of the reaction. For the reaction of conjugated dienes, the _ _ _ _ _ _ _ C+ is the key.
Let’s follow the course of the reaction of HBr with 1,3-butadiene.
Step 1: HBr reacts with 1,3-butadiene. Which intermediate is stabilized?
H
H2C
H
CH
Br
+
Br
or
CH2
-
+
CH2
H2C
H2C
H
What type of cation can be formed?
Draw resonance.
-
Step 2: What happens next? Rearrangement of the allyl cation or reaction with Br: ?
Follow both pathways:
-Reaction with Br :
H
H
H2C
2
1
H2C
+
CH
Br
-
Br
What is the type of this product?
--
Rearrangement followed by reaction with Br .
H
H2C
+
Br
-
CH2
+
H2C
CH
This is called the
H
H
CH2
Br
CH2
CH2
4
1
addition.
CH14-5
Organic Chemistry Notes by Jim Maxka
jim.maxka@nau.edu
If there were no barrier to the formation of either the 1,2 or the 1,4 addition, there would be mostly 1,4
product. Why?
more heat
hotter-more kinetic energy
HBr
Reaction
Recap
0
The intermediate in the 1,2 addition is what type of cation? _ _ _ _ _ _ _ c cation
Draw it:
0
The intermediate in the 1,4 addition is what type of cation? _ _ _ _ _ _ _ c cation
Draw it:
What is the relationship between these two structures?
Which one is more stable? Why?
What are thermodynamic conditions?
What are kinetic conditions?
Reaction with dihalogens
The reaction is similar to HBr but now we make 1,2 and 1,4 dibromides:
Br
CH 2
H 2C
+
Br Br
Br
+
-
CH 2
Practice for 1,2 vs 1,4 electrophilic addition
1. What would be the HBr addition products of 1,3-hexadiene?
2. Why is there no cis/trans for 1,3-butadiene? What about 1,3-hexadiene?
3. Is 2,3-hexadiene conjugated? Explain. Cis/trans here?
4. 2,3-hexadiene has enantiomers. Can you draw them? Why is this unusual.
5. Which is favored by high T, 1,2 or 1,4? Which is favored by low T, 1,2 or 1,4?
CH14-6
Organic Chemistry Notes by Jim Maxka
jim.maxka@nau.edu
Diels Alder Reaction
Orbital Model of Conjugation
Pi Bonds are areas for electron density outside of the normal sigma bond space.
Bonding model of pi bond in ethylene: First consider each p orbital independent.
H
H
H
H
H
H
H
H
H
H
H
H
Individual p orbitals
in-phase combination
out-of-phase combination
Note this the in-phase combination of the two p orbitals is bonding (pi bond), the out-of-phase combination
is antibonding (pi*) and higher in energy.
If we build a bonding correlation diagram, we can see that since each atomic p orbital brings 1 e to the
molecule, we have
e total for the molecule. Fill in all the electrons for the ethylene system below.
H
H
H
H
Antibonding (1 node) LUMO
2 p orbitals | |
H
H
H
H
Bonding (0 nodes) HOMO
What is the significance of a node?
The HOMO (Highest Occupied)
Is this molecular orbital electron rich or poor?.
The LUMO (Lowest Unoccupied)
Is this molecular orbital electron rich or poor?.
Now Consider the p orbital arrangement for butadiene. How many nodes* and bonding molecular orbitals
can be formed? Color the orbitals. Count the number of electrons. Which is HOMO, which is LUMO?
H
H
H
H
H
H
Bonding?
Nodes*
Bonding?
H
H
H
H
H
H
Nodes*
H
H
H
Bonding?
H
H
H
Nodes*
H
H
Nodes*
Bonding?
*Node= Nodal plane. What is a good definition of node?
Diels Alder Reaction is a completely electronic reaction between orbitals. Formally,
CH14-7
Organic Chemistry Notes by Jim Maxka
jim.maxka@nau.edu
EWG
EWG
C
C
C
C
EWG
EWG
Diene
1,3-Butadiene
New sigma bonds
New pi bond
Still cis arrangement
of EWG.
Dienophile
Note cis arrangement
of EWG groups
Notes: (1) There must be a diene and a dienophile.
(2) The reactivity of the diene is increased by EWG (electron withdrawing groups.
(3) EWG are highly EN groups (without reactive LP) or groups with oxygens and
nitrogens (highly EN) positioned by one atom away. The electrons are pulled away
creating a positive charge next to the reactive group.
O
O
Y
Y
Pick the following systems that have electron withdrawing groups:
O
O
H
CH3
H
CH
H
O
H
C
COCH3
O
H
CH3
H
CH
H
O
O
COCH3
O
O
C
CH
C
O
CH3
H3C
H
CH
N
C
C
N
H3C
CH3
The molecules that do not have EWG have EDG (electron donating groups).
The Reaction
Identify the diene and the dienophile and predict the product of the following Diels-Alder Reactions. Note
the EWG and EDG. Diene reactivity is increased by EDG; Dienophile reactivity is increased by EWG.
Cl
H3C
Cl
Some reactions can make bicyclic products:
N
C
C
N
CH14-8
Organic Chemistry Notes by Jim Maxka
jim.maxka@nau.edu
Sorting out Regiochemistry of Diels Alder
When bicyclic molecules are formed there may be the possibility of up or down attachment. The reaction
below can be endo or exo – ONLY the endo is formed:
O
H
C
O
H
C
O
O
CH2
C
O
H
C
up
O
exo
H
CH2
H
O
C
H
O
down
endo
C
O
There are a couple of ways of looking at the exclusion of the exo product:
(1) Sterics
(2) Electronics
Not all reactions have endo/exo:
Conformational Requirements for Diels Alder
The diene must by in the s-cis conformation for the reaction to occur.
The following molecule does not react by Diels Alder. Why?
O
HC
HC
O
Anything that stops the formation of the s-cis conformation will inhibit or slow down the reaction. This is
not to say that the a diene written in the s-trans form will not react.
CH14-9
Organic Chemistry Notes by Jim Maxka
Consider the reaction below. It does occur. Draw the product and explain:
jim.maxka@nau.edu
CN
H2C
+
H3C
CH2
CN
s-trans
More on the Electronic Requirements of Diels Alder
Now let’s go through an explanation of electronic requirements of the Diels Alder:
For the Diels Alder reaction to work, the diene needs
groups and the dienophile needs
groups.
The reason for this has to do with the orbitals that interact. For the HOMO of the diene to become more
reactive, that is more electron donating, it needs EDG. What is the EDG group here?
For practice color in the p-lobes to give the phases of the the HOMO of the diene and the LUMO of the
dienophile.
H3C
C
CH2
EWG
CH
HC
HC
CH2
EWG
Dienophile
Diene
The dienophile reacts from the LUMO. EWG promote reaction in the LUMO because they pull electron
density out of the molecule bringing the molecule closer in energy to the diene’s HOMO. Similarly, the
EDG pump energy into the diene. This brings the orbitals closer together and increases reactivity.
pump electrons in
HOMO
electron rich
empty
H
H
R
H
H
H
H
H
LUMO
electron poor
pull electrons out
H
Diels Alder Practice:
What dienes and dienophiles would react to give the following D A products.
CH2
H3C
Cl
Cl
H
CHO
H
Cl
CHO
CH14-10
Organic Chemistry Notes by Jim Maxka
Molecular color
The EM Spectrum:
X ray
E
wavelength
Far UV
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Near UV
Visible
IR
Microwave
The colorwheel concept. Light absorbed is not transmitted.
blue absorbed = what color?
red absorbed = what color?
The phenomena of absorption and emission
in
Absorbtion
out
in
Emission
out
pi*
hv
relax
pi
Ground State
Excited State
hv
Ground State
The longer the conjugation length, the longer the wavelength and the lower the energy of absorption of
light.
Beta-Carotene – from carrots What color?
What color would ethylene be?
redder color
butadiene?
benzene?
Ultraviolet
http://www.nist.gov/kinetics/spectra/1,3-butadiene_spectra.htm
CH14-11
Organic Chemistry Notes by Jim Maxka
jim.maxka@nau.edu
Why should nylon materials not be exposed to direct sunlight?
O
R
NH
(CH 2)n
NH
O
(CH2)n
O
NH
(CH2)n
NH
O
(CH2)n
R
CH14-12
Organic Chemistry Notes by Jim Maxka
Do you know the use of the compound called PABA?
jim.maxka@nau.edu
O
H2N
OH
Draw a resonance structure starting from the LP on N.
What is this phenomenon called?
HN
N
NH
SO 3-
N
Na
+
HN
N
N
N
SO 3-
N
+
Na
NH
N
N
O
O
FWA-1 (Disodium 4,4'-bis[(4-anilino-6-morpholino-1,3,5-triazin-2-yl)amino]stilbene-2,2'disulphonate (CAS-No. 16090-02-1)) is a Fluorescent Whitening Agent (FWA) mainly used
(more than 90%) in household detergents in concentrations ranging from 0.05 to 0.15%. It is also
used to a far lesser extent (less than 10% in total) in textiles and paper. FWA-1 behaves like
colorless direct cotton dyes. Its specific molecular structure is responsible for the high affinity to
cotton.
What is the role of sulfate groups?
Is this a UV absorber?
http://web.jrc.cec.eu.int/asso-eaeme/P94-95.htm
http://www.heraproject.com/ExecutiveSummary.cfm?ID=171
http://www.heraproject.com/files/23-F-04-HERA-FWA1(Version%203_1%20).pdf
http://www.photonics.com/spectra/applications/XQ/ASP/aoaid.325/QX/read.htm
Summaries
1. Name dienes and draw dienes from names.
2. Understand the difference between cis/trans and s-cis/s-trans of conjugated dienes.
3. Find conjugated regions in molecules. Determine when dienes are conjugated or not.
4. Determine when a cation or alkene is stabilized by conjugation or hyperconjugation.
5. Which is stronger stabilization? Conjugation or hyperconjugation.
6. Recognize allylic cations.
7. Predict the products of 1,2 and 1,4 addition of a halo acid.
8. Determine when a reaction will lead to kinetic or thermodynamic products based on heat or cold.
9. Be able to write molecular orbital diagrams for simple p-orbital systems.
10. Differentiate between pi and pi* orbitals.
11. Determine which orbitals will be filled and which ones will be unfilled.
12. Be able to identify EWG and EDG.
13. Determine the reactivity of the Diels Alder reaction in terms of diene and dienophile.
14. Determine the reactivity of the Diels Alder reaction in terms of EWG and EDG.
15. Determine the reactivity of the Diels Alder reaction in terms of stability of the s-cis conformation of
the diene.
16. Predict whether the Diels Alder reaction will produce endo or exo products. Be able to draw these
products. Be able to identify these products.
17. Understand the difference between color absorption and emission.
18. Recognize how conjugation affects color of organic molecules.
19. What is the difference between UV and VIS absorption?
CH14-13