14. Conjugated Compounds
and Ultraviolet Spectroscopy
Conjugated and Nonconjugated
Dienes
 Compounds can have more than one double or triple bond
 If they are separated by only one single bond they are
conjugated and their orbitals interact
 The conjugated diene 1,3-butadiene has properties very
different from those of the nonconjugated diene, 1,5pentadiene
2
Conjugated compounds: common in nature
 Extended conjugation leads to absorption of visible light, producing color
Conjugated hydrocarbon with many double bonds = polyenes
Lycopene is a conjugated polyene responsible for red color in tomatoes
3
Learning Check:
 Circle any conjugation in the following:
A.
C.
B.
CH2 CH C N
H2C
D.
E.
O
C
O
CH3
CH3
C
CH2
F.
H
C
CH2
4
Solution:
 Circle any conjugation in the following:
A.
C.
B.
CH2 CH C N
H2C
Nonconjugated
D.
E.
O
C
O
CH3
CH3
C
CH2
F.
H
C
CH2
Nonconjugated
5
14.1 Conjugated Dienes:
Preparation: Typically by elimination of allylic halide
 Specific industrial processes for large scale production of
commodities by catalytic dehydrogenation and dehydration
6
Stability of Conjugated Dienes:
 Conjugated dienes are more stable than nonconjugated
based on heats of hydrogenation
 Hydrogenating 1,3-butadiene produces 16 kJ/mol less
heat than 1,4-pentadiene
7
Stability of Conjugated Dienes:
 Conjugation makes more stable
 Substitution makes more stable
8
Description of 1,3-Butadiene
The single bond between the conjugated double bonds is
shorter and stronger than sp3
Partial Double Bond Character
9
Molecular Orbital Description
Isolated Double bond
Conjugated Double bonds
The bonding -orbitals are made from 4 p orbitals that provide
greater delocalization and lower energy than in isolated C=C
•electrons in 1,3-butadiene are delocalized over the  bond system
Delocalization leads to stabilization
10
14.2 Electrophilic Additions to Conjugated
Dienes: Allylic Carbocations
 Remember: addition of electrophile to C=C
 Markovnikov regiochemistry via more stable carbocation
11
Carbocations from Conjugated Dienes
 Now: Addition of H+ gives delocalized 2o allylic carbocation
12
Addition Products to Delocalized Carbocation
 Nucleophile can add to either cationic site
The transition states for the
two possible products are
not equal in energy
13
Another Example:
14
Learning Check:
 Give the likely products from reaction of 1 equiv HCl with
2-methyl-1,3-cyclohexadiene.
15
Solution:
 Give the likely products from reaction of 1 equiv HCl with
2-methyl-1,3-cyclohexadiene.
16
14.3 Kinetic vs. Thermodynamic Control
 At completion, all reactions are at equilibrium and the
relative concentrations are controlled by the
differences in free energies of reactants and products
(Thermodynamic Control)
 If a reaction is irreversible or if a reaction is far from
equilibrium, then the relative concentrations of
products depends on how fast each forms, which is
controlled by the relative free energies of the
transition states leading to each (Kinetic Control)
17
Kinetic and Thermodynamic Control:
Example
 Addition to a conjugated diene at or below room temperature normally
leads to a mixture of products in which the 1,2 adduct predominates
over the 1,4 adduct
 At higher temperature, product ratio changes and 1,4 adduct
predominates
18
19
_____ Kinetic control
_____ Thermodynamic control
20
14.4 The Diels-Alder Cycloaddition
Reaction
 Conjugate dienes can combine with alkenes to form six-membered cyclic
compounds
 Ring formation involves no intermediate (concerted formation of 2 bonds)
 Discovered by Otto Paul Hermann Diels and Kurt Alder in Germany in the 1930’s
21
View of the Diels-Alder Reaction
 Woodward and Hoffman showed to be an example of the general
class of pericyclic reactions
 Involves orbital overlap, change of hybridization and electron
delocalization in transition state
 The reaction is called a cycloaddition
Diene
Dieneophile
22
14.5 Characteristics of Diels-Alder Reaction
 The alkene component is called a dienophile


C=C is conjugated to an electron withdrawing group, such as C=O or
CN
Alkynes can also be dienophiles
23
Stereospecificity of the Diels-Alder
Reaction
 The reaction is stereospecific,

There is a one-to-one relationship between stereoisomeric
reactants and products
24
Endo vs Exo positions
Endo and Exo denote relative sterechemistry of groups in bicyclic
systems
Substituent on one bridge is exo if it is anti (trans) to the larger of
the other two bridges and endo if it is syn (cis) to the larger of the
other two bridges
25
Regiochemistry of the Diels-Alder Reaction
 Reactants align to produce endo (rather than exo) product
26
Example:
Endo product formed
27
Conformations of Dienes in the
Diels-Alder Reaction
 The two double bonds
in the diene are “cis”
or “trans” to each
other about the single
bond (being in a plane
maximizes overlap)
 These conformations
are called s-cis and strans (“s” stands for
“single bond”)
 Dienes react in the s-
cis conformation in
the Diels-Alder
reaction
28
 Not all dienes can rotate to be s-cis
29
Learning Check:
 Which can form an s-cis diene?
30
Solution:
 Which can form an s-cis diene?
31
Learning Check:
 Which would be good Diels-Alder dienophiles?
32
Solution:
 Which would be good Diels-Alder dienophiles?
33
14.6 Diene Polymers: Natural and
Synthetic Rubbers
 Conjugated dienes can be polymerized
 The initiator for the reaction can be a radical, or an acid
 Polymerization: 1,4 addition of growing chain to
conjugated diene monomer
34
Natural Rubber
 A material from latex, in plant sap
 In rubber repeating unit has 5 carbons and Z stereochemistry of all
C=C
 Gutta-Percha is natural material with E in all C=C
 Looks as if it is the head-to-tail polymer of isoprene (2-methyl-1,3butadiene)
35
Synthetic Rubber
 Chemical polymerization of isoprene does not produce rubber
(stereochemistry is not controlled)
 Synthetic alternatives include neoprene, polymer of 2-chloro-
1,3-butadiene
 This resists weathering better than rubber
36
Vulcanization
 Natural and synthetic rubbers are too soft to be used in products
 Charles Goodyear discovered heating with small amount of sulfur
produces strong material
 Sulfur forms bridges between hydrocarbon chains (cross-links)
37
14.7 Structure Determination in Conjugated
Systems: UV Spectroscopy
 Conjugated compounds can absorb light in the ultraviolet region of the
spectrum
 The electrons in the highest occupied molecular orbital (HOMO) undergo a
transition to the lowest unoccupied molecular orbital (LUMO)
 The region from 2 x 10-7m to 4 x 10-7m (200 to 400 nm) is most useful in
organic chemistry
 A plot of absorbance (log of the ratio of the intensity of light in over light
transmitted) against wavelength in this region is an ultraviolet spectrum –
38
Ultraviolet Spectrum of 1,3-Butadiene
 Example: 1,4-butadiene has four  molecular orbitals with the lowest
two occupied
 Electronic transition is from HOMO to LUMO at 217 nm (peak is
broad because of combination with stretching, bending)
39
Quantitative Use of UV Spectra
 Absorbance for a particular compound in a specific solvent at a
specified wavelength is directly proportional to its concentration
 You can follow changes in concentration with time by recording
absorbance at the wavelength
 Beers’ law: A




= ecl
“A” = absorbance
“e” is molar absorptivity (extinction coefficient)=amount of UV
light absorbed
“c” is concentration in mol/L
“l” is path of light through sample in cm
40
14.8 Interpreting UV Spectra: The
Effect of Conjugation
 max: wavelength where UV absorbance for a compound is greatest
 Energy difference between HOMO and LUMO decreases as the
extent of conjugation increases
 max increases as conjugation increases (lower energy)
 1,3-butadiene: 217 nm, 1,3,5-hexatriene: 258 nm
41
Substituents on  system increase max
1,3-butadiene:
H2C=CH-CH=CH2
217 nm
42
Calculations of max
 The max of the   * transition for compounds with < 4 conjugated
double bonds can be calculated using Woodward-Fieser rules.
O
Start with a base number:

CH2 CH CH CH2
max = 217 nm
C C

C H

max = 210 nm
O

C C

C R

max = 215 nm
To the base add:
1. 30 for each extra conjugated double bond
2. 5 each time a conjugated double bond is an exocyclic double bond
3. 36 for each conjugated double bond frozen s-cis
4. 5 for each alkyl group or halogen bonded to conjugated system of polyene
5. 10 for an -substituent of a conjugated alkehyde or ketone
6. 12 for a -substituent of a conjugated aldehyde or ketone
43
Example:
 Calculate expected max for following:
Base
= 217
3 alkyl substituents @ 5 each = 15
Calculated
Observed
= 232
= 232
44
Learning Check:
 Calculate expected max for following:
O
To the base add:
1. 30 for each extra conjugated double bond
2. 5 each time a conjugated double bond is an exocyclic double bond
3. 36 for each conjugated double bond frozen s-cis
4. 5 for each alkyl group or halogen bonded to conjugated system of polyene
5. 10 for an -substituent of a conjugated alkehyde or ketone
6. 12 for a -substituent of a conjugated aldehyde or ketone
45
Solution:
 Calculate expected max for following:
O
Base
= 217
Additional db bd = 30
Locked s-cis
= 36
2 alkyl substituents = 10
Base
 substituent
 substituent
Exocyclic db bd
= 215
= 10
= 12
= 5
Calculated
= 242
Calculated
= 293
Observed
= 241
Observed
= 293
46
14.9 Conjugation, Color and the
Chemistry of Vision
 Visible region is about 400 to 800 nm
 Extended systems of conjugation absorb in visible region
 Visual pigments are responsible for absorbing light in eye and
triggering nerves to send signal to brain
47
-Carotene,
11 double
bonds in
conjugation,
max = 455 nm
48
Vision:
Vision:
Beta-Carotene
H3C
CH3
CH3
H3C
CH3
CH3
CH3
CH3
H3C
CH3
Liver Enzymes
Vitamin A
H3C
CH3
CH3
CH3
OH
H3C
CH3
CH3
CH3
O
H
CH3
Retinol
CH3
Retinal
CH3
H3C
CH3
CH3
CH3
H3C
O
CH3
CH3
H
CH3
Retinal
O
H
NH2
CH3
NH2
Dissociation triggers Opsin
nerve impulse
which gives visual image
Opsin
CH3
H3C
H3C
CH3
CH3
11
10
CH3
trans
CH3
H
CH3
H
13
12
N
N
CH3
Rhodopsin
Rhodopsin = opsin protein +
retinal as a prosthetic group
CH3
Metarhodopsin II
Light
FIND THE MAN IN THE COFFEE BEANS
Vision:
Vision:
 Doctors have concluded that if you find the man in the
coffee beans in 3 seconds, the right half of your brain is
better developed than most people. If you find the man
between 3 seconds and 1 minute, the right half of the
brain is developed normally. If you find the man
between 1 minute and 3 minutes, then the right half of
your brain is functioning slowly and you need to eat more
protein. If you have not found the man after 3 minutes,
the advice is to look for more of this type of exercise to
make that part of the brain stronger!!!
Vision:
Stare at the eye of the red parrot while you count slowly to 20, then look
immediately at one spot in the empty bird cage. The faint, ghostly image of a
bluegreen bird will appear in the cage.
Try the same thing with the green cardinal. A faint magenta bird will appear
in the cage.
The ghostly birds you see here are called afterimages. An afterimage is an
image that stays with you even after you have stopped looking at an object.
The back of your eye is lined with light sensitive cells, called cones, which are
sensitive to certain colors of light. When you stare at the red bird, your redsensitive cones adapt to the light and lose their sensitivity. When you shift your
gaze to the white background of the bird cage, you see white (minus red)
where the red-sensitive cells have become adapted. White light minus red light
is blue-green light. That's why the afterimage you see is blue-green and in the
shape of a parrot. The same thing happens when you stare at the green bird,
but this time it's the green-sensitive cones that adapt. White minus green light
is magenta light, so you see the afterimage as a magenta cardinal.
Look at the flag for
about 20 seconds,
then look down to the
white space below.
Notice how the
ghostly image of the
familar "stars and
stripes" appears. This
afterimage occurs
because red, white
and blue are the
complementary colors
of cyan,black and
yellow.
Vision:
Which of the following is not conjugated?
20%
20%
20%
20%
3
4
20%
2.
1.
O
3.
5.
4.
O
1
2
5
How many π (pi) bonding molecular orbitals are
present in 1,3,5-hexatriene ?
20%
1.
2.
3.
4.
5.
20%
20%
20%
3
4
20%
2
3
4
5
6
1
2
5
Arrange the following alkenes in order of least
stable to most stable.
20%
A
1.
2.
3.
4.
5.
B
20%
20%
20%
3
4
20%
C
A<B<C
B<C<A
B<A<C
A<C<B
C<A<B
1
2
5
Predict the expected major organic product of the
following reaction.
HBr
20%
20%
20%
20%
3
4
20%
°C
55 ÞC
1.
2.
Br
Br
3.
Br
4.
Br
5.
Br
1
2
5
Predict the expected major organic product of the
following reaction.
O
20%
Toluene
+
20%
20%
20%
3
4
20%
O
heat
O
1.
2.
O
H
O
O
O
O
H
3.
O
H O
O
4.
H
O
H O
5.
O
O
O
H
1
O
O
2
5
Which of the following dienes would be expected to
react fastest in a Diels-Alder reaction?
20%
1.
20%
20%
20%
3
4
20%
2.
3.
4.
5.
1
2
5
Which of the following dienophiles would be expected
to react fastest in a Diels-Alder reaction?
20%
1.
20%
20%
20%
3
4
20%
2.
OH
O
3.
Cl
5.
4.
OCH3
O
1
2
5
Which of the following dienes will not participate in a
Diels-Alder reaction?
25%
1.
2.
3.
4.
1
25%
2
25%
3
25%
4
Select the most appropriate name for the
following polymer:
20%
F
20%
20%
20%
3
4
20%
n
1.
2.
3.
4.
5.
poly(methyl fluoride)
poly(ethyl fluoride)
poly(vinyl fluoride)
polyfluoromethane
polyfluoroethane
1
2
5
UV/Visible spectroscopy is based on _____
excitation.
25%
25%
25%
25%
electronic
2. rotational
3. nuclear
4. vibrational
1.
1
2
3
4
If one knows the molar absorptivity for a molecule, then
one can determine its concentration by taking an
ultraviolet/visible absorption spectrum.
50%
50%
True
2. False
1.
1
2
Which of the following wavelengths can be
classified as UV light?
20%
1.
2.
3.
4.
5.
20%
20%
20%
3
4
20%
900 nm
725 nm
500 nm
450 nm
none of these
1
2
5
Which of the following statements is not true regarding
ultraviolet/visible spectroscopy?
1.
2.
3.
4.
20%
As the extent of conjugation in a
molecule increases, λmax increases.
As the extent of conjugation in a
molecule increases, the HOMOLUMO gap increases.
The ultraviolet/visible region of the
electromagnetic spectrum ranges
from 200 – 800 nm.
The amount of UV light absorbed by
a compound can be expressed by
20%
20%
20%
3
4
20%
its molar absorptivity, ε.
5.
As the path length of the sample
increases, its absorbance increases.
1
2
5
Which of the following compounds is most likely to absorb in the
visible region of the electromagnetic spectrum?
20%
1.
20%
20%
20%
3
4
20%
2.
3.
4.
5.
O
O
1
2
5
The electronic transition that occurs for conjugated
molecules in ultraviolet/visible spectroscopy is:
25%
1.
2.
3.
4.
25%
25%
25%
σ → π*
σ → σ*
π → π*
σ* → π*
1
2
3
4