notes lecture 1 conformational analysis

8/9/2019 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)



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



+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


15/18

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)


16/18

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 )


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)


17/18

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


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

notes lecture 1 conformational analysis

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