Lecture from Monday
Topics covered:
- Functional groups
- ChemDraw, Chem3D (computer modeling)
- Hybridization picture of H2C=C=CH2 (allene)
- Drawing resonance structures (Klein handout)
- Assessing “good” vs. “bad” resonance structures
- Introduction to Conjugation (Part 1 of 2)
Ch14C, 4/9/12
Made using ChemDraw and Chem3D
Programs available on computers in UCLA Science Learning Center (Young 4335 and others)
Intro tutorial to how to use ChemDraw: http://www.chem.umass.edu/~samal/267/chemdraw.pdf
Vollhardt, Figure 14-4 (p. 618)
IUPAC name:
propanediene
Common name: allene
•So, the hydrogens of allene are NOT in the same plane, but are instead in two
 planes.
•This is not something we might have been able to predict just from looking at
the Lewis structure
Which Resonance Contributor Represents Reality?
Your friend asks you to describe what a nectarine is because he’s never seen or eaten one… (Klein, Sec 2.1)
“True” structure:
Peach
Plum
Nectarine
The carbon monoxide case:
C
O
C
O
Neither contributor fully represents CO
Resonance hybrid: A weighted average or blend of resonance contributors; the
most accurate representation of the electronic structure of a molecule.
Which Resonance Contributor Represents Reality?
Once upon a study break...
X
Fantasy creatures
Neither fully represents reality
Real creature
A unicorn-dragon hybrid?
The carbon monoxide case:
C
O
X
C
O
Neither contributor fully represents CO
Resonance hybrid: A weighted average or blend of resonance contributors; the
most accurate representation of the electronic structure of a molecule.
Drawing the Resonance Hybrid
Example: Draw the resonance hybrid for acetate ion, CH3CO2-.
O
O
O
1. Draw contributors
CH3
O
CH3
O
2. Draw the features that are the same for all contributors
• Sigma and pi bonds, lone pairs, and formal charges
3. Add features that are not the same for all contributors
•Partial (shared) pi bonds shown as ---•Partial (shared) charges shown as d+ or d-
CH3
O
O
CH3
CH3
O
O
d-
d+
O
d-
Resonance hybrid
Do All Contributors Have Equal Importance?
Is a rhinoceros more unicorn or more dragon?
 contributor “stability” =  resemblance to reality =  contribution to hybrid
Therefore we need contributor preference (“stability”) rules:
As the number and/or magnitude of rules violations ,
= importance of individual contributor
= contribution to resonance hybrid
Resonance Contributor Preference Rules
For practice: see Klein Sec 2.8
Rule #1: The most important contributor has the maximum number of atoms with
full valence shells.
Example:
H2C
OH
H2C
OH
Open valence shell on carbon
Less important contributor
In some cases it may not be possible for all atoms to have full valence shells.
Rule #1 is more influential than all the other preference rules.
Rules #2-6 have no particular order of preference.
Resonance Contributor Preference Rules
Rule #2: The most significant contributor has the maximum number of covalent
bonds.
H
H
Example:
C
H
C
O
O
H
Three covalent bonds
Less important contributor
Resonance Contributor Preference Rules
Rule #3: The most significant contributor has the least number of formal charges.
H
H
Example:
C
C
O
H
No formal charges
More important contributor
O
H
Two formal charges
Less important contributor
Resonance Contributor Preference Rules
Rule #4: If a contributor must have formal charge(s), the most important contributors
has these charges on the atom(s) that can best accommodate them.
•Negative formal charges best on atoms of high electronegativity
O- better than C-
•Positive formal charges best on atoms of low electronegativity
C+ better than O+
•Minimize formal charge magnitude
+1 better than +2
Carbon EN = 2.5
Less important contributor
Oxygen EN = 3.5
More important contributor
Resonance Contributor Preference Rules
Rule #5: Resonance interaction (i.e., pi bond) is strongest between atoms in the
same row of the periodic table.
•Usually CNOF
•Usually outweighs electronegativity considerations (rule #4)
Example:
F
C
Cl
F
C
Cl
H
H
F, C both 2nd row
More important contributor
Even though EN F > EN Cl
C 2nd row; Cl 3rd row
Less important contributor
Resonance Contributor Preference Rules
Rule #6: Other factors (such as aromaticity) that we will encounter later.
Violations to the resonance contributor preference rules exist, but are uncommon.
Conjugated Molecules - Part 1
Lecture Supplement page 28
What is Conjugation?
For any molecule best structure = lowest energy
•Familiar examples: Which structure is lowest energy?
Methane
H
H
C
H
H
H
H
H
Square planar or tetrahedral?
H
H
H
C
C
H
Ethane
H
H
H
H
C
H
H
Staggered
C
H
or
C
H
H
eclipsed?
Lowest energy from minimizing electron repulsion and maximizing p orbital overlap
•Maximum p orbital overlap allows...
Resonance
Conjugation
Aromaticity
} What are these?
What is Conjugation?
Case #1: Relative stability of C4H6 isomers
•Isomers: Same molecular formula, different structure
More stable isomer
4 C (graphite) + 3 H2 (g)
DHof = 29.9 kcal mol-1
4 C (graphite) + 3 H2 (g)
DHof = 47.7 kcal mol-1
Isomers
•Heat of formation (enthalpy of formation; DHof): Hypothetical enthalpy change when a
substance is synthesized from elements in their standard states
• Importance: Lower DHof = more stable isomer
• Restriction: DHof comparisons only valid among isomers
What is Conjugation?
Case #1: Relative stability of C4H6 isomers
H
91.8
H2C
C
C
1,2-butadiene (an allene)
CH3
No ring strain
Two pi bonds
67.7
64.6
Why this order?
•Ring strain?
•Number of pi bonds?
•Position of pi bonds?
Ring strain
One pi bond
47.7
45.8
Alternating pi-sigma-pi bonds
H
C
C
CH2CH3
E
n
th
a
lp
y
o
f
fo
rm
a
ti
o
n
(D
H
f)o
,k
ca
l
m
o
l1
(c
al
cu
la
te
d
)
37.5
32.0
H3C
C
C
CH3
30.7
s-cis-1,3-butadiene
29.9
s-trans-1,3-butadiene
1,3-dienes
0.0 4 C (graphite) + 3 H2 (g)
No ring strain
Two pi bonds
What is Conjugation?
Case #2: Catalytic Hydrogenation of 1,3-Dienes versus 1,4-Dienes
Catalytic hydrogenation: Addition of H2 to a pi bond with a catalyst
H
Example:
CH3CH2
H2
C
C
H
CH3CH2
H
H
C
C
H
H
H
Pt
H
Thermodynamics
•Lose: H-H sigma bond, C-C pi bond
•Gain: 2 x C-H sigma bond
•Sigma bonds usually stronger than pi bonds
•Therefore catalytic hydrogenation is exothermic (DH < 0)
DH = -30 kcal mol-1
What is Conjugation?
Case #2: Catalytic hydrogenation of 1,3-dienes versus 1,4-dienes
How can we use catalytic hydrogenation to probe C4H6 isomer stability?
•Fact: 1,3-butadiene more stable than 2-butyne
H3C
C
C
CH3 + 2 H2
1,3-butadiene more stable than 2-butyne
+ 2 H2
E
n
th
a
lp
y
(H
;
k
ca
l
m
o
l)1
DH = -65.1 kcal mol-1
DH = -56.5 kcal mol-1
Same molecule = same enthalpy (H)
•Observation: DH (1,3-butadiene  butane) < DH (2-butyne  butane) by 8.6 kcal/mol
•Conclusion: Lowest DH (for catalytic hydrogenation) belongs to most stable isomer
What is Conjugation?
Case #2: Catalytic hydrogenation of 1,3-dienes versus 1,4-dienes
The experiment
•Use DH (cat H2) to compare pi-sigma-pi (1,3-diene) versus pi-sigma-sigma-pi (1,4-diene)
1-Pentene (alkene energy benchmark)
H2
DH = -30 kcal mol-1
Pt
1,4-Pentadiene (a 1,4-diene)
2 H2
3
1
2
5
4
Pt
1,3-Butadiene (a 1,3-diene)
2 H2
2
1
3
4
Pt
Predict: DH = 2 x (-30) = -60 kcal mol-1
Observe: DH = -60 kcal mol-1
Conclusion: No special stability for 1,4-diene
Predict: DH = -60 kcal mol -1 if stability = 1,4-diene
Observe: DH = -56.5 kcal mol-1 3.5 kcal mol-1 less than expected
Conclusion: 1,3-diene more stable than 1,4-diene.
General observation: 1,3-dienes more stable than similar 1,4-dienes
Lecture from Wed, 4/11/12
-Conformations of 1,3-butadiene (s-cis vs. s-trans)
-Discussion of how dihedral angle relates to extent of p-orbital overlap
-Resonance picture of 1,3-butadiene correlates well to physical properties
-Conjugated molecules and color
1,3-Butadiene: A Closer Look
What is origin of special 1,3-diene stability?
H
H
H
H
2.21 Å
H
C
C
H
C
H
C
C
C
K eq > 1
H
s-cis*
5%
H
H
C
C
H
H
s-trans
95%
Major conformation? Torsional strain: s-trans < s-cis
Stability: s-trans > s-cis
Planarity: Nearly always planar
* s-cis = two pi bonds, separated by sigma bond, in cis arrangement
2.50 Å
1,3-Butadiene: A Closer Look
Rotation around Csp2 - Csp2 bond:
32
31.5
31
Energy difference
between most and
least stable
conformations
Barrier to rotation
E
ne
rg
y
(k
ca
l
m
o
l)1
Pi bonds
perpendicular
Highest energy point
30.5
30
29.5
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
C=C-C=C Dihedral angle (degrees)
Angle between planes formed by three atoms each
Conclusion: More than just torsional strain is at work
1,3-Butadiene: Resonance Model
What is the origin of this extra stability, planarity, and barrier to rotation?
Resonance is often a strong influence on molecular structure, so start there
H
H
C
C
H
H
H
C
H
H
C
C
H
C
H
H
H
H
H
C
C
H
C
C
H
H
C
C
H
H
H
H
C
C
H
H
C
C
H
H
Resonance hybrid
Resonance hybrid observations
•Partial C2-C3 pi bond  explains barrier to rotation
•C2-C3 pz orbital overlap  explains planarity
1,3-Butadiene: Resonance Model
The resonance model looks useful, but simplistic. Is it accurate?
The resonance model predicts...
Model molecules
1,3-Butadiene
C2-C3 bond length (Å)
1.54
C2-C3 barrier to rotation
(kcal mol-1)
4.5
1.33
~60
Conclusion: Resonance model is accurate despite its simplicity
1.48
7.5
1,3-Butadiene: Resonance Model
How does resonance explain why a 1,3-diene is more stable than a 1,4-diene?
1,4-diene
H
e-

H
H
C
C
C
C
H
H
H
1,3-diene
H
H
H
e- 
H
C
C
H
C
C
C
C
H
H
H
H
•No significant resonance
•Pi electrons confined between two carbons
•Pi electrons have shorter wavelength
H
•Has some resonance
•Pi electrons roam over four carbons
•Pi electrons have longer wavelength
E = hc/ so  wavelength =  energy
Molecule is less stable
Molecule is more stable
Discussion of Handout:
“p orbital overlap: Resonance, Conjugation, and
Aromaticity”
Conjugated Molecules - Part 2
Lecture Supplement page 37
chlorophyll a
lycopene
-carotene
Part 1 Summary
Adjacent, overlapping p orbitals allows for...
•...more resonance
•...more electron delocalization
•...lower electron energy
Example:
Consequences of p orbital overlap
•Atoms with p orbitals must be planar
•Partial pi bond(s)
•Barrier to rotation
Greater stability
more stable than
Extra Stability Limited to 1,3-Dienes?
•1,3-diene has four adjacent p orbitals
•Three adjacent p orbitals is enough to provide extra stability
Example: An amide
Amide resonance contributors:
O
H
Amide resonance hybrid:
O d-
O
NH2
H
Pi electron delocalization
provides increased stability
NH2
H
d+
NH2
Extra Stability Limited to 1,3-Dienes?
Another special stabilization example: An amide
O d-
H
O
d+
NH2
Resonance hybrid
H
NH2
Nitrogen hybridization?
•Predict: Four attachments = sp3
•sp3 lacks p orbital needed for resonance
•Therefore sp2 to accommodate resonance
•Therefore sp2 to increase stability
I thought hybridization is controlled only by the number of attachments?!
•Energy causes geometry; geometry causes hybridization
Influenced by electron repulsion, resonance, etc.
Extra Stability Limited to 1,3-Dienes?
More adjacent p orbitals = larger electron “playground”
An amide has three adjacent, parallel p orbitals:
p orbital overlap gives delocalized pi bonds
O
p orbital overlap forms
pi bonds
Sigma
bonds
H
Build your own model
C
N
H
H
Compare with 1,3-butadiene: Four adjacent, parallel p orbitals:
H
H
H
C
C
C
H
H
C
H
Build your own model
In general: Adjacent, parallel p orbitals improve molecular stability
Conjugation: A Definition
Finally!
Decrease in electron energy
Conjugation: Special stability provided by electron delocalization in three
or more adjacent, parallel, overlapping p orbitals.
Not limited to pi-sigma-pi
(four carbon pz orbitals)
Consequences of Conjugation
Conjugation influences widespread
•Chemical reactivity: Is molecule reactive or inert?
•Molecular structure: Is a molecule (or portion of molecule) flat?
Is bond rotation hindered?
•Physical properties: Color
•Etc.
Consequences of Conjugation
Consequence #1: More extensive conjugation = greater stability
•Influences distribution of products in chemical reaction
•Reaction products:  stability =  amount produced
•Example: Determine major product of this reaction:
O
O
CH3
H
- H2O
O
CH3
CH3
CH3
or
CH3
HO
CH3
H
CH3
Which product isomer
is more stable?
CH3
CH3
Six conjugated p orbitals
Ten conjugated p orbitals
More stable
Less stable
Produced in greatest amount
Consequences of Conjugation
Consequence #2: Partial pi bond character
•Resistance to conformational change (barrier to rotation)
•Causes planarity of atoms conjugated p orbitals
•Example: Amide linkage between two amino acids in a protein
O
O
Barrier to
rotation
H
N
H
Planar
O d-
N
R
Less torsional strain
More stable
R
More torsional strain
Less stable
N d+
H
R
Resonance hybrid
Barrier to rotation and planarity critical to protein function.
Consequences of Conjugation
Consequence #3: Highly conjugated molecules may be colored
Examples:
Chlorophyll
Lycopene
-Carotene
Origin of color: •Some portion of visible (white) light spectrum is absorbed
•Brain perceives remaining light as color
So how does molecular structure control energy of photons absorbed?
Consequences of Conjugation
How does molecular structure control energy of photons absorbed?
Controlled by electron energies
•Molecule absorbs photon (hn)
•Electron is excited to higher energy
molecular orbital
DE
Absorb photon (DE = hn)
E
ecl
rt
o
n
e
ern
g
y
•Energy of photon absorbed must equal
HOMO/LUMO orbital energy difference (DE)
Lowest Unoccupied
Molecular Orbital
(LUMO)
Highest Occupied
Molecular Orbital
(HOMO)
Ground state
Excited state
Consequences of Conjugation
How does molecular structure control energy of photons absorbed?
• number of conjugated p orbitals  DE
•When DE low enough, photons of visible light absorbed
•Unabsorbed portion of visible light spectrum perceived as color
Photon absorbed
Observed color
Ultraviolet (UV)
Colorless
P
h
o
to
n
en
er
g
y
Visible light
Violet
Indigo
Blue
Green
Yellow
Orange
Red
Yellow
Orange
Red
Violet
Indigo
Blue
Green
Violet
Indigo
Blue
How does molecular structure control energy of photons absorbed? Green
Yellow
Orange
Red
Consequences of Conjugation
Yellow
Orange
Red
Violet
Indigo
Blue
Green
Example Molecules
1,3-Butadiene
Four conjugated p orbitals
DE = ultraviolet
Perceived color = colorless
1,3,5-Hexatriene
Six conjugated p orbitals
DE = ultraviolet
Perceived color = colorless
Styrene
Eight conjugated p orbitals
DE = ultraviolet
Perceived color = colorless
Violet
Indigo
Blue
How does molecular structure control energy of photons absorbed? Green
Yellow
Orange
Red
Example Molecules
Consequences of Conjugation
OH
Retinol (vitamin A)
Ten conjugated p orbitals
Photons of hn = DE are violet
Perceived color = yellow
Yellow
Orange
Red
Violet
Indigo
Blue
Green
O
Retinal
Twelve conjugated p orbitals
Photons of hn = DE are indigo
Perceived color = orange
Violet
Indigo
Blue
How does molecular structure control energy of photons absorbed? Green
Yellow
Orange
Red
Yellow
Orange
Red
Violet
Indigo
Blue
Green
Consequences of Conjugation
Example Molecules
CH3
H3C
Not conjugated
CH3
CH3
CH3
CH3
Not conjugated
H3C
CH3
Lycopene
22 conjugated p orbitals
Photons of hn = DE are blue
Perceived color = red
CH3
CH3
Friday’s topic: Aromaticity
(pre-read!!!)
1,3-Butadiene: A Closer Look
Rotation around the C2-C3 bond
•The s-cis conformation is planar
Indicates movie to play
•Filename: s-cis change perspective.mov
1,3-Butadiene: A Closer Look
Rotation around the C2-C3 bond
•Perpendicular conformation has lowest torsional strain
•Filename: s-cis to perpendicular.mov
1,3-Butadiene: A Closer Look
Rotation around the C2-C3 bond
•The s-trans conformation is planar
•Filename: perpendicular to s-trans_prof.mov