Notes Lecture 1 Conformational Analysis
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Chem 531D. A. Evans, F. Michael Acyclic Conformational Analysis-1
The following discussion is intended to provide a generaloverview of acyclic conformational analysis
Carey & Sundberg, Advanced Organic Chemistry, Part A. Chapter 3Conformational, Steric and Stereoelectronic Effects
R. W. Hoffmann, Chem. Rev . 1989, 89, 1841-1860 Allylic 1-3-Strain as a Controlling Element in Stereoselective Transformations
Ethane & Propane
∆ E = +3.4 kcal mol-1 (R = Me)
staggeredconformation
eclipsedconformation
∆ E = +3.0 kcal mol-1 (R = H)
1 x 1.4
2 x 1.0
Incremental contributions to the barrier:
3 x 1.0
1 (H↔Me)
2 (H↔H)
3 (H↔H)
propane
ethane
∆E (kcal/mol)Eclipsed atomsStructure
For purposes of analysis, each eclipsed conformer may be broken up into its
component destabilizing interactions.
Ethane Rotational Barrier: The FMO View
One can see from space-filling models that the van der Waals radii of thehydrogens do not overlap in the eclipsed ethane conformation. This makes thesteric argument for the barrier untenable.
An alternative explanation for the rotational barrier in ethane is that betterhyperconjugative overlap is possible in the staggered conformation than in theeclipsed conformation as shown below.
σ* C–H
LUMO
σ C–H
HOMO
In the staggered conformation there are 3 anti-periplanar C–H Bonds
σ C–H
HOMO
σ* C–H
LUMO
σ C–H
σ∗ C–H
In the eclipsed conformation there are 3 syn-periplanar C–H Bonds
σ∗ C–H
σ C–H
Following this argument one might conclude that:
The staggered conformer has a better orbital match between bonding andantibonding states.
The staggered conformer can form more delocalized molecular orbitals.
R
C
H
HC
H
HHH
RH
H
H
H
C C
C CC
H
C
H
C C
HH
Eliel & Wilen, Stereochemistry of Organic Compounds, Wiley, 1994
Me
Me
Me Calculate the the rotational barrier about the C1-C2bond in isobutane
H H
H
H
3.4
3.0
Conventional Explanation: Steric interaction between eclipsed groups.
Big debate: See:Pophristic, V. Nature, 2001, 411, 565. (It's not sterics)Bickelhaupt, F. M. Angew. Chem., Int. Ed. 2003, 42 , 4183. (Yes, it is)Weinhold, F. Angew. Chem., Int. Ed. 2003, 42 , 4188. (No, it's not)
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Chem 531D. A. Evans, F. Michael Acyclic Conformational Analysis: Butane
The 1,2-Dihaloethanes
X = Cl; ∆H° = + 0.9–1.3 kcal/molX = Br; ∆H° = + 1.4–1.8 kcal/mol
X = F; ∆H° = – 0.6-0.9 kcal/mol
Eliel. page 609
Observation: While the anti conformers are favored for X = Cl, Br, the gaucheconformation is prefered for 1,2-difluroethane. Explain.
X
C
X
H
HH
H
H
C
X
H
HH
X
Discuss with class the origin of the gauche stabiliation of the difluoro anaolg.
pKeq
0
-1
-2
0
–1.4
1.0
10
100
∆GºKeq
∆ Gº298 = 1.4 x pKeq
which is = 1.4 x ∆pKa
pK = – log [H+]
∆ Gº298 = –1.4 log10Keq
∆ Gº = –2.3RTlog10K
At 298 K: 2.3RT = 1.4 (∆G in kcal/mol)
∆ G° = –RT Ln K
Relationship between ∆G and Keq and pKa
–2.8 kcal/mol
Recall that:or
Since
Hence, pK is proportional to the free energy change
+3.6
+5.1
+0.88Ref = 0
G
E1
E2
n-Butane Torsional Energy Profile
∆ E = ?
Eclipsed atoms ∆E (kcal/mol)
+1.0 kcal/mol1 (H↔H)
+2.8 kcal/mol2 (H↔Me)
∆E est = 3.8 kcal/mol
The estimated value of +3.8 agrees quite well with the value of +3.6 reportedby Allinger (J. Comp. Chem. 1980, 1, 181-184)
eclipsedconformation
staggeredconformation
Using the eclipsing interactions extracted from propane & ethane we should beable to estimate all but one of the eclipsed butane conformations
Butane
H
C
Me
HHH
Me
C
Me
H H
H
Me
H
Me
C
Me
C
H
HHH H H
HH
Me
Me
Me
C
Me
H
C
H
H
HH
HH
H
Me
Me
e n e
r g y
A
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From the energy profiles of ethane, propane, and n-butane, one may extractthe useful eclipsing interactions summarized below:
Hierarchy of Eclipsing Interactions
+1.0
+1.4
+3.1
eclipsedconformation
staggeredconformation
+3.1
Incremental contributions to the energy.
+2.0
1 (Me↔Me)
2 (H↔H)
∆E (kcal/mol)Eclipsed atoms
From the torsional energy profile established by Allinger, we should be able to
extract the contribution of the Me↔
Me eclipsing interaction to the barrier:
Butane continued
Acyclic Conformational Analysis: ButaneD. A. Evans, F. Michael Chem 531
Me
C
H
C
H
H MeH Me H
H
Me
H
H
C C
X Y
H
H
H
H
X Y
H H
H Me
Me Me
General nomenclature for diastereomers resulting from rotation about a
single bond R
C
R
R
C
R
R
CR
sp
sc
(Klyne, Prelog, Experientia 1960
, 16 , 521.)
sc
acac
ap
C
R
R
C
R
R
C
R
R
0°
+60°
+120°
180°
-60°
-120°
Torsion angle Designation Symbol
0 ± 30°
+60 ± 30°
+120 ± 30°
180 ± 30°
-120 ± 30°
-60 ± 30°
± syn periplanar
+ syn-clinal
+ anti-clinal
antiperiplanar
- anti-clinal
- syn-clinal
± sp
+ sc (g+)
+ ac
ap (anti or t)
- ac
- sc (g-)
Energy Maxima
Energy Minima
E2
G
E1
A
E1
G
n-ButaneConformer
Nomenclature forstaggered conformers:
C
H
H H
H
Me
Me
C
H
H Me
H
H
Me
C
H
Me H
H
H
Me
trans or tor (anti)
gauche(+)
or g+gauche(-)
or g-
Conformer populationat 298 K: 70% 15% 15%
Let's extract out the magnitude of the Me–Me interaction:
∆E (kcal/mol)
+5.1 kcal/mol
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~ 3.1
It may be concluded that in-plane 1,3(Me↔Me) interactions are Ca +4
kcal/mol while 1,2(Me↔Me) interactions are destabliizing by Ca 3.1 kcal/mol.
~ 3.7 ~3.9 ~ 7.6
Estimates of In-Plane 1,2 &1,3-Dimethyl Eclipsing Interactions
1,3(Me↔Me) = + 3.7 kcal mol -1
∆ G° = +5.5 kcal/mol
∆ G° = X + 2Y where:
X = 1,3(Me↔Me)
Y = 1,3(Me↔H) = gauche butane
Estimate of 1,3-Dimethyl Eclipsing Interaction
1,3(Me↔Me) = ∆ G° – 2Y = 5.5 –1.8 = +3.7 kcal/mol
1,3(Me↔H) = gauche butane = 0.9 kcal/mol
The double-gauche pentane
conformation
n-Pentane
Acyclic Conformational Analysis: PentaneD. A. Evans, F. Michael Chem 531
Me Me MeMe Me MeMeMe
Me Me
H H
H H
Me H
H H
H Me Me H
H Me
H H
Me Me
Me Me
Me MeMe Me
Me
Me Me
Me
Rotation about both the C2-C3 and C3-C4 bonds in either direction (+ or -):
tg+g-g+
g-t
g-g-
tg-
g+g-
g+t
g+g+t,t
From prior discussion, you should be able to estimate energies of 2 & 3 (relative to 1).On the other hand, the least stable conformer 4 requires additional data before isrelative energy can be evaluated.
Anti(2,3)-Anti(3,4)
1
1
1
1
3 3
3
3
5 5
5
5
Gauche(2,3)-Anti(3,4)
Gauche(2,3)-Gauche(3,4)Gauche(2,3)-Gauche'(3,4)double gauche pentane
syn-pentane
1 (t,t)
4 (g+g –) 3 (g+g+)
2 (g+t)
The new high-energy conformation: (g+g –)
H H
Me Me
g+g-
Me Me
t,t
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Conformational Analysis: Cyclic Systems-1Evans, Kim, Breit, Michael Chem 531
Three Types of Strain:
Prelog Strain: van der Waals interactions
Baeyer Strain: bond angle distortion away from the ideal
Pitzer Strain: torsional rotation about a σ bond
Baeyer Strain for selected ring sizes
size of ring Heat of Combustion(kcal/mol)
Total Strain(kcal/mol)
Strain per CH2(kcal/mol)
"angle strain"deviation from 109°28'
3456789
101112131415
499.8656.1793.5944.8
1108.31269.21429.61586.81743.11893.42051.92206.12363.5
27.526.36.20.16.29.7
12.612.411.34.15.21.91.9
9.176.581.240.020.891.211.401.241.020.340.400.140.13
24°44'9°44'0°44'
-5°16'
Eliel, E. L., Wilen, S. H. Stereochemistry of Organic Compounds Chapter 11, John Wiley & Sons, 1994.
Baeyer "angle strain" is calculated from the deviation of the
planar bond angles from the ideal tetrahedral bond angle.
Discrepancies between calculated strain/CH2 and the "angle
strain" results from puckering to minimize van der Waals or
eclipsing torsional strain between vicinal hydrogens.
Why is there an increase in strain for medium sized rings even
though they also can access puckered conformations free of
angle strain? The answer is transannular strain- van der Waals
interactions between hydrogens across the ring.
Introductory Concepts
H
H
Nonbonding
Cyclopropane
Necessarily planar . Substituents are therefore eclipsed.
Disubstitution prefers to be trans.
υ = 3080 cm-1
ϕ = 120 °
Almost sp2, not sp3
Walsh Model for Strained Rings:
Rather than σ and σ* c-c bonds, cyclopropane has sp2 and p-type
orbitals instead.
H
H
side view
σ –1 (bonding)
σ (antibonding) σ (antibonding)
π (antibonding)
π (bonding) π (bonding)
3
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Evans, Kim, Breit, Michael Chem 531
eq
ax ax
eq
ax
eq
eqax
Cyclobutane
ϕ = 28 °
Eclipsing torsional strain overridesincreased bond angle strain by puckering.
Ring barrier to inversion is 1.45 kcal/mol.
∆G = 1 kcal/mol favoring R = Me equatorial
1,3 Disubstitution prefers cis diequatorial totrans by 0.58 kcal/mol for di-bromo cmpd.
1,2 Disubstitution prefers trans diequatorial tocis by 1.3 kcal/mol for diacid (roughly equivalent tothe cyclohexyl analogue.)
145-155°
A single substituent prefers the equatorial position of the flap of the envelope
(barrier ca. 3.4 kcal/mol, R = CH3).
X
X
Cyclopentane
C2 Half-Chair Cs Envelope
Two lowest energy conformations (10 envelope and 10 half chair conformations
Cs favored by only 0.5 kcal/mol) in rapid conformational flux (pseudorotation)
which causes the molecule to appear to have a single out-of-plane atom "bulge"
which rotates about the ring.
Since there is no "natural" conformation of cyclopentane, the ring conforms to
minimize interactions of any substituents present.
1,2 Disubstitution preferstrans for steric/torsionalreasons (alkyl groups) and
dipole reasons (polar groups).
Cs Envelope
Conformational Analysis: Cyclic Systems-2
Cs Envelope
X
A carbonyl or methylene prefers the planar position ofthe half-chair (barrier 1.15 kcal/mol for cyclopentanone).
Me
Me 1,3 Disubstitution: Cis-1,3-dimethyl cyclopentane0.5 kcal/mol more stable than trans.
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D. A. Evans, F. Michael Chem 531Conformational Preferences About sp3 –sp2 Bonds
How does one account for this observation?
Torsional Strain: the resistance to rotation about a bond
Torsional energy: the energy required to obtain rotation about a bond
also known as dihedral angle
Torsional steering: Stereoselectivity originating from transition state torsionalenergy considerations
Torsional Angle:
∆G = +3 kcal mol-1
Torsional Strain (Pitzer Strain): Ethane
staggeredeclipsed
Relevant Orbital Interactions:
Dorigo, A. E.; Pratt, D. W.; Houk, K. N. JACS 1987, 109, 6591-6600.
Propene Eliel, pg 615+
HC H
H
H
C HH
H
H
C C
Confomation A is more stable than B and represents the most stable
conformation of propene
Wiberg K. B.; Martin, E. J. Amer. Chem. Soc. 1985, 107 , 5035-5041.
H
H
H
H
HC H
H
H
C HH
H
H
CC
H
H
H
H
C H
H
H
C
H
H
σ C–H's properly aligned for π∗ overlap
hence better delocalization
C HH
H
H
C
H
H
σ C–H & π electrons are
destabilizing
H
H
H
H
H
H
H
H
O O
Conformational Preferences: Acetaldehyde
The eclipsed conformation (conformation A) is preferred.Polarization of the carbonyl decreases the 4 electron destabilizing
Rotational barrier: 1.14 kcal/mol
Houk, JACS 1983, 105 , 5980-5988.
MeH
H
H
HMe
H
H
X X
Conformational Preferences
1-Butene (X = CH2); Propanal (X = O)
Conformation A is preferred. There is little steric repulsion between themethyl and the X-group in conformation A.
B
A B
AB
A B
A
H
H
H
H H
H
H
H H
H
H
H
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Φ = 180Φ = 0
Φ = 0
Φ = 60
Φ = 120
Φ = 180
+1.18 kcal
+0.37 kcal
+2.00kcal
+1.33kcal
+1.32 kcal
+0.49 kcal
Φ = 180
Φ = 120
Φ = 50
Φ = 0
Φ = 180Φ = 0
E ( k
c a
l / m o
l )
The Torsional Energy ProfileThe Torsional Energy Profile
Chem 531Evans, Duffy, Ripin, Michael Conformational Barriers to Rotation: Olefin A1,2 Interactions
Φ (Deg)
2-propen-1-ol 1-butene
Conforms to ab initio (3-21G) values:Wiberg, K. B.; Martin, E. J. Am. Chem. Soc. 1985, 107, 5035.
Φ (Deg)
Me
H H
C HCH
H
HC
HC H
HH
OH
HC
HC H
H
HH
CH
C H
H
H
HC
HC H
H
H
Me Me
H
HC HC
OH
HO
HC
HC H
H
H
OH
HO
H
H
C HCH
H
H
H
C HCH
H
H
H
H
H
MeMe
C HCH
H
ΦΦ
Me OH
1,2 Eclipsing is worse than 1,3 eclipsing because thehydrogens are closer in 1,3 eclipsing (2.4 vs. 2.5 A)
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Chem 531Evans, Duffy, Ripin, Michael Conformational Barriers to Rotation: Olefin A1,2 Interactions-2
Φ (Deg)
2-methyl-1-butene
E (
k c a l / m o
l )
+2.68kcal
+1.39 kcal
+0.06 kcal
Φ = 180
Φ = 110
Φ = 50
Φ = 0
Φ = 180Φ = 0
The Torsional Energy Profile
Φ (Deg)
2-methyl-2-propen-1-ol
E (
k c a l /
m o
l )
The Torsional Energy Profile
Φ = 0 Φ = 180
Φ = 0
Φ = 60
Φ = 120
Φ = 180
+0.21 kcal
+1.16 kcal
+2.01kcal
HC
HC Me
H
H
H C
H
C Me
H
H
HC
H
Me
H H
C MeCH
H
C Me
H
H
Me
Me
H
HC MeC
H
H
Me
Me
HC
HC Me
HH
OH
OH
HO
HC
H
H
C MeH
OH
HO
H
H
C MeCH
H
H
H
C MeCH
H
H
H
C MeCH
H
Φ Φ
Me
Me OH
Me
1,2 Me-Me eclipsing is strongly disfavored (2.7 kcal/mol)
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Values calculated using MM2 (molecular mechanics) force fieldsvia the Macromodel multiconformation search. Review: Hoffman, R. W. Chem. Rev. 1989, 89, 1841.
(Z)-2-buten-1-ol (Z)-2-pentene
Φ (Deg) Φ (Deg)
Chem 531Evans, Duffy, Ripin, Michael Conformational Barriers to Rotation: Olefin A1,3 Interactions
+0.86kcal
+1.44 kcal
Φ = 180
Φ = 120
Φ = 0
Φ = 180Φ = 0
The Torsional Energy ProfileThe Torsional Energy Profile
Φ = 0 Φ = 180
Φ = 0
Φ = 90
Φ = 180+3.88 kcal
+0.52kcal
H C
Me
C H
H
H
HC
Me
Me
H H
C HCMe
HH
CMe
C H
HH
OH
C H
H
H
HO
OH
H
HC HC
Me
H
OH
Me
HC
MeC H
H H
Me
Me
H
H
C HCMe
H
H
H
C HCMe
H
Φ Φ
Me
Me
Me
OH
Now, 1,3 Me-Me eclipsing interaction is most disfavored (3.9 kcal/mol)
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+0.66
+4.68
+0.40 kcal+0.34
Φ = -80
Φ = 0
Φ = 80
+2.72
Φ = 150
Φ = 110
Φ = -140
Φ = 180Φ = 0
The Torsional Energy Profile
Chem 531Evans, Duffy, Ripin, Michael Conformational Barriers to Rotation: Olefin A1,3 Interactions-2
(Z)-2-hydroxy-3-penteneRotate clockwise
Φ (Deg)
HC
MeC H
HOH
Me
HC
MeC H
H
HO
H
CMe
H
CMe
C H
HHO
Me
C H
H
HO Me
Me
OH
HC HC
Me
H
Me
HMe
OH
C HCMe
H
HC
MeC H
HHO
Me
Φ
H
H
R RL
RS
C HCH
R
H
RL
RS
C HCH
R
RL
RS
H
C HCR
H
H
RL
RS
R
H
H RL
RS
H
R
H RL
RS
H
MeCH
C
RS
RL
H
put largest group hereto avoid botheclipsing interactions
put H here to havesmallest eclipsinginteraction
no real preference
A1,2 strain
Neither
A1,3 strain
C RCR
H
H
RL
RS
R
R
H RL
RS
A1,2 strain and A1,3 strain
A1,3 strain is more important than A1,2 strain
(4.0 vs. 2.7 kcal/mol)
put H here to havesmallest eclipsinginteraction
Summary of A1,2 and A1,3 strain
Me
Me
OH
Best conformations place the smallest substituent (H) next to the (Z )-methyl group to minimize eclipsing interactions