30802341 harvard s advanced organic chemistry

D. A. Evans

An Introduction to Frontier Molecular Orbital Theory-1 Problems of the Day

Chem 206

http://www.courses.fas.harvard.edu/~chem206/ http://evans.harvard.edu/problems/

Chemistry 206 Advanced Organic ChemistryLecture Number 1

The molecule illustrated below can react through either Path A or Path B to form salt 1 or salt 2. In both instances the carbonyl oxygen functions as the nucleophile in an intramolecular alkylation. What is the preferred reaction path for the transformation in question? O O Br N H O BrPath B Path A 1

Br

+

N O H Br O

2

+

N O H Br

Br

Introduction to FMO Theory General Bonding Considerations The H2 Molecule Revisited (Again!) Donor & Acceptor Properties of Bonding & Antibonding States Hyperconjugation and "Negative" Hyperconjugation Anomeric and Related Effects

This is a "thought" question posed to me by Prof. Duilo Arigoni at the ETH in Zuerich some years ago

(First hr exam, 1999)The three phosphites illustrated below exhibit a 750fold span in reactivity with a test electrophile (eq 1) (Gorenstein, JACS 1984, 106, 7831).(RO)3P OMe O P O A O

Reading Assignment for week:Kirby, Stereoelectronic Effects Carey & Sundberg: Part A; Chapter 1 Fleming, Chapter 1 & 2 Fukui,Acc. Chem. Res. 1971, 4, 57. (pdf) Curnow, J. Chem. Ed. 1998, 75, 910 (pdf) Alabugin & Zeidan, JACS 2002, 124, 3175 (pdf) D. A. Evans1-01-Cover Page 9/15/03 8:56 AM

+

El(+)O P O B

+ (RO)3PEl

(1)

O P OMe O C

Rank the phosphites from the least to the most nucleophilic and provide a concise explanation for your predicted reactivity order.

Monday, September 15, 2003

D. A. Evans

An Introduction to Frontier Molecular Orbital Theory-1 Stereoelectronic Effects

Chem 206

Universal Effects Governing Chemical Reactions There are three: Steric EffectsNonbonding interactions (Van der Waals repulsion) between substituents within a molecule or between reacting moleculesMe S N2 Me Nu C R RO H H Me Me2CuLi RO Me H H R

Geometrical constraints placed upon ground and transition states by orbital overlap considerations. Fukui Postulate for reactions:"During the course of chemical reactions, the interaction of the highest filled (HOMO) and lowest unfilled (antibonding) molecular orbital (LUMO) in reacting species is very important to the stabilization of the transition structure."

Nu:

R

C R

Br

Br:

RO H

O

O

major

General Reaction TypesO

p. p.

minor

Radical Reactions (~10%): Polar Reactions (~90%):

A + B A(:) + B(+)Lewis Acid

A A

B B

Electronic Effects (Inductive Effects):The effect of bond and through-space polarization by heteroatom substituents on reaction rates and selectivities Inductive Effects: Through-bond polarization Field Effects: Through-space polarization

Lewis Base

FMO concepts extend the donor-acceptor paradigm to non-obvious families of reactions Examples to consider

Me R C R Br

S N1

+R R C Me + Br:

H2

+

2 Li(0) Mg(0)

2 LiH CH3MgBr

CH3I +

rate decreases as R becomes more electronegative "Organic chemists are generally unaware of the impact of electronic effects on the stereochemical outcome of reactions." "The distinction between electronic and stereoelectronic effects is not clear-cut."

1-02-Introduction-1

9/12/03 4:44 PM

D. A. Evans

Steric Versus Electronic Effects; A time to be careful!!O R3SiO EtO O R3Si OSiR3 O TiCl4 Nu OSiR3

Chem 206

Steric Versus electronic Effects: Some Case StudiesWhen steric and electronic (stereoelectronic) effects lead to differing stereochemical consequencesWoerpel etal. JACS 1999, 121, 12208.

diastereoselection >94:6 OSiR3 H O diastereoselection 93:7 H OSiR3

O

OAc

SnBr4 Me SiMe3

O Me

Me

AlCl3

stereoselection 99:1 stereoselection >95:5 OSiR3

O

OAc

SnBr4 BnO

O

O Danishefsky et al JOC 1991, 56, 387 BnO

p. p.

BnO

O EtO2C O EtO2C (R)2CuLi Bu OTBS O OTBS diastereoselection 8:1 O Ph N O only diastereomer Bu OTBS OAc OAc O H Ph N O H N N N N O OAc OAc H H N R R O TBS Al R O Yakura et al Tetrahedron 2000, 56, 7715 O Ph 60-94% AcO AcO H H N O only diastereomer Ph

Bu3AlR3 O Al O

EtO2C

EtO

Yakura's rationalization:

Mehta et al, Acc Chem. Res. 2000, 33, 278-286

1-03-Introduction-1a

9/15/03 8:14 AM

D. A. Evans

The H2 Molecular Orbitals & Antibonds The H2 Molecule (again!!)

Chem 206

Linear Combination of Atomic Orbitals (LCAO): Orbital Coefficients Rule Two: Each MO is constructed by taking a linear combination of the individual atomic orbitals (AO): Bonding MO Antibonding MO = C11 + C22 = C*11 C*22

Let's combine two hydrogen atoms to form the hydrogen molecule. Mathematically, linear combinations of the 2 atomic 1s states create two new orbitals, one is bonding, and one antibonding: Rule one: A linear combination of n atomic states will create n MOs. (antibonding) E Energy H 1 p. p. E (bonding) 1s 1s H 2

The coefficients, C1 and C2, represent the contribution of each AO. Rule Three: (C1)2 + (C2)2 = 1

The squares of the C-values are a measure of the electron population in neighborhood of atoms in question Rule Four: bonding(C1)2 + antibonding(C*1)2= 1 In LCAO method, both wave functions must each contribute one net orbital Consider the pibond of a C=O function: In the ground state pi-CO is polarized toward Oxygen. Note (Rule 4) that the antibonding MO is polarized in the opposite direction.

Let's now add the two electrons to the new MO, one from each H atom: (antibonding) E1

Energy

C

O

(antibonding)

H 1

1s E2

1s

H 2 Energy

(bonding)

C

O

Note that E1 is greater than E2. Why?C O

(bonding)

1-04-Introduction-2 9/15/03 8:38 AM

D. A. Evans

Bonding Generalizations

Chem 206

Bond strengths (Bond dissociation energies) are composed of a covalent contribution ( Ecov) and an ionic contribution ( Eionic). Bond Energy (BDE) = Ecovalent + Eionic (Fleming, page 27)

Orbital orientation strongly affects the strength of the resulting bond.For Bonds: A B Better than A B

When one compares bond strengths between CC and CX, where X is some other element such as O, N, F, Si, or S, keep in mind that covalent and ionic contributions vary independently. Hence, the mapping of trends is not a trivial exercise.

For Bonds:

A

B

Better than

A

B

Useful generalizations on covalent bonding Overlap between orbitals of comparable energy is more effective than overlap between orbitals of differing energy.p. p.

This is a simple notion with very important consequences. It surfaces in the delocalized bonding which occurs in the competing anti (favored) syn (disfavored) E2 elimination reactions. Review this situation. An anti orientation of filled and unfilled orbitals leads to better overlap. This is a corrollary to the preceding generalization. There are two common situations.

For example, consider elements in Group IV, Carbon and Silicon. We know that C-C bonds are considerably stronger by Ca. 20 kcal mol-1 than C-Si bonds.C C C CC CSi C

better than

C

Si

C

Si

Case-1: Anti Nonbonding electron pair & CX bondXSi-SP3

X Alone pair HOMO

C-SP3 CC

C-SP3

A

C

* CX LUMO

X Better thanlone pair HOMO

* CX LUMO

C-SP3 CSi

C

A

C

H3CCH3 BDE = 88 kcal/mol Bond length = 1.534

H3CSiH3 BDE ~ 70 kcal/mol Bond length = 1.87

Case-2: Two anti sigma bondsX C CY HOMO

This trend is even more dramatic with pi-bonds: CC = 65 kcal/mol CSi = 36 kcal/mol SiSi = 23 kcal/mol Weak bonds will have corresponding low-lying antibonds.Formation of a weak bond will lead to a corresponding low-lying antibonding orbital. Such structures are reactive as both nucleophiles & electrophiles A Y

Y* CX LUMO

X C

C

Better than

X C

CY HOMO

* CX LUMO

C

Y

1-05-Introduction-3 9/12/03 4:36 PM

D. A. Evans

Donor-Acceptor Properties of Bonding and Antibonding States

Chem 206

Donor Acceptor Properties of C-C & C-O Bonds Consider the energy level diagrams for both bonding & antibonding orbitals for CC and CO bonds.* C-C * C-O

Hierarchy of Donor & Acceptor States Following trends are made on the basis of comparing the bonding and antibonding states for the molecule CH3X where X = C, N, O, F, & H.-bonding States: (CX)CH3CH3 CH3H CH3NH2 CH3OH

C-SP

3

C-SP3 O-SP3

very close!!

decreasing -donor capacity C-C C-O

CH3F

poorest donor

p. p.

The greater electronegativity of oxygen lowers both the bonding & antibonding C-O states. Hence: CC is a better donor orbital than CO CO is a better acceptor orbital than CC

-anti-bonding States: (CX)CH3H

For the latest views, please read Alabugin & Zeidan, JACS 2002, 124, 3175 (pdf)CH3CH3 CH3NH2 CH3OH CH3F

Donor Acceptor Properties of CSP3-CSP3 & CSP3-CSP2 Bonds* CC * CC better acceptor Increasing -acceptor capacity

best acceptor

C-SP3

C-SP3 C-SP2

The following are trends for the energy levels of nonbonding states of several common molecules. Trend was established by photoelectron spectroscopy.Nonbonding States

CC

better donor

CC

H3P: H2S: H3N: H2O: HCl: decreasing donor capacitypoorest donor

The greater electronegativity of CSP2 lowers both the bonding & antibonding CC states. Hence: CSP3-CSP3 is a better donor orbital than CSP3-CSP2 CSP3-CSP2 is a better acceptor orbital than CSP3-CSP31-06-donor/acceptor states 9/12/03 5:16 PM

D. A. Evans

Hybridization vs Electronegativity

Chem 206

Electrons in 2S states "see" a greater effective nuclear charge than electrons in 2P states. This becomes apparent when the radial probability functions for S and P-states are examined: The radial probability functions for the hydrogen atom S & P states are shown below.100 % 100 %

There is a linear relationship between %S character & Pauling electronegativity5

N4.5

SP

Radial Probability

Pauling Electronegativity

Radial Probability

1 S Orbital

4

N N3.5SP3

SP2

C 3

SP

2 S Orbital

2 S Orbital 2 P Orbital

C

p. p.3 S Orbital 3 P Orbital

2.5

SP2

C

SP3

2

20

25

30

35

40

45

50

55

% S-Character

There is a direct relationship between %S character & hydrocarbon acidity60

CH (56)

S-states have greater radial penetration due to the nodal properties of the wave function. Electrons in S-states "see" a higher nuclear charge. Above observation correctly implies that the stability of nonbonding electron pairs is directly proportional to the % of S-character in the doubly occupied orbitalLeast stable Most stablePka of Carbon Acid

55

4

50

45

C H (44)6 6

40

CSP3

CSP2

CSP

35

PhCC-H (29)30

The above trend indicates that the greater the % of S-character at a given atom, the greater the electronegativity of that atom.1-07-electroneg/hybrization 9/12/03 4:49 PM

25 20 25 30 35 40 45 50 55

% S-Character

D. A. Evans

Hyperconjugation: Carbocation Stabilization

Chem 206

The interaction of a vicinal bonding orbital with a p-orbital is referred to as hyperconjugation. This is a traditional vehicle for using valence bond to denote charge delocalization.R H H C

Physical Evidence for Hyperconjugation Bonds participating in the hyperconjugative interaction, e.g. CR, will be lengthened while the C(+)C bond will be shortened. First X-ray Structure of an Aliphatic Carbocation

+C

R+ H H H H C C H H

The graphic illustrates the fact that the C-R bonding electrons can "delocalize" to stabilize the electron deficient carbocationic center. Note that the general rules of drawing resonance structures still hold: the positions of all atoms must not be changed.p. p.

+

1.431

[F5SbFSbF5]

+C

Stereoelectronic Requirement for Hyperconjugation: Syn-planar orientation between interacting orbitals

100.6

1.608

Me

Me Me

The Molecular Orbital Description CR CRT. Laube, Angew. Chem. Int. Ed. 1986, 25, 349

+C

H H

+C

H H

The Adamantane Reference (MM-2)H

1.528

CR

CR

110 Me Me Me

1.530

Take a linear combination of CR and CSP2 p-orbital: "The new occupied bonding orbital is lower in energy. When you stabilize the electrons is a system you stabilize the system itself."1-08-Hyperconj (+)-1 9/12/03 4:53 PM

D. A. Evans

"Negative" HyperconjugationSyn OrientationR

Chem 206antibonding CR R: H H C X+ H H antibonding CR R: R X+ H H C filled hybrid orbital X

Delocalization of nonbonding electron pairs into vicinal antibonding orbitals is also possible R H H C

R C X

R H H H H

H X H H H

C

X

filled hybrid orbital

X

C

This decloalization is referred to as "Negative" hyperconjugation

Anti OrientationR

Since nonbonding electrons prefer hybrid orbitals rather that P orbitals, this orbital can adopt either a syn or anti relationship to the vicinal CR bond.

H H

C

X

H H

C

The Molecular Orbital Description CR Overlap between two orbitals is better in the anti orientation as stated in "Bonding Generalizations" handout.

X

Nonbonding e pair

The Expected Structural PerturbationsChange in Structure Shorter CX bond Longer CR bond

Spectroscopic ProbeX-ray crystallography X-ray crystallography Infrared Spectroscopy Infrared Spectroscopy NMR Spectroscopy NMR Spectroscopy

CR

As the antibonding CR orbital decreases in energy, the magnitude of this interaction will increase Note that CR is slightly destabilized

Stronger CX bond Weaker CR bond Greater e-density at R Less e-density at X

1-09-Neg-Hyperconj 9/12/03 4:53 PM

D. A. Evans

Lone Pair Delocalization: N2F2The trans Isomer

Chem 206Now carry out the same analysis with the same 2 orbitals present in the trans isomer.

The interaction of filled orbitals with adjacent antibonding orbitals can have an ordering effect on the structure which will stabilize a particular geometry. Here are several examples: Case 1: N2F2 F N N This molecule can exist as either cis or trans isomers F F N N F There are two logical reasons why the trans isomer should be more stable than the cis isomer. The nonbonding lone pair orbitals in the cis isomer will be destabilizing due to electron-electron repulsion. The individual CF dipoles are mutually repulsive (pointing in same direction) in the cis isomer. In fact the cis isomer is favored by 3 kcal/ mol at 25 C. Let's look at the interaction with the lone pairs with the adjacent CF antibonding orbitals. The cis Isomer Ffilled N-SP2 F antibonding NF NF (LUMO) filled N-SP2 (HOMO)

filled N-SP2

N F

N

F antibonding NF filled N-SP2 (HOMO)

NF (LUMO)

In this geometry the "small lobe" of the filled N-SP2 is required to overlap with the large lobe of the antibonding CF orbital. Hence, when the new MO's are generated the new bonding orbital is not as stabilizing as for the cis isomer.

Conclusions Lone pair delocalization appears to override electron-electron and dipole-dipole repulsion in the stabilization of the cis isomer. This HOMO-LUMO delocalization is stronger in the cis isomer due to better orbital overlap. Important Take-home Lesson Orbital orientation is important for optimal orbital overlap.

N

N

A

B

forms stronger pi-bond than

A

B

Note that by taking a linear combination of the nonbonding and antibonding orbitals you generate a more stable bonding situation. Note that two such interactions occur in the molecule even though only one has been illustrated.1-10- N2F2 9/12/03 4:59 PM

A

B

forms stronger sigma-bond than

A

B

This is a simple notion with very important consequences. It surfaces in the delocalized bonding which occurs in the competing anti (favored) syn (disfavored) E2 elimination reactions. Review this situation.

D. A. Evans

Lone Pair Delocalization: The Gauche Effect

Chem 206

The interaction of filled orbitals with adjacent antibonding orbitals can have an ordering effect on the structure which will stabilize a particular conformation. Here are several examples of such a phenomon called the gauche effect: HydrazineH N H N H Hanti

The closer in energy the HOMO and LUMO the better the resulting stabilization through delocalization. Hence, N-lone pair NH delocalization better than NH NH delocalization. Hence, hydrazine will adopt the gauche conformation where both N-lone pairs will be anti to an antibonding acceptor orbital. The trend observed for hydrazine holds for oxygen derivatives as well Hydrogen peroxideH O O H anti

Hydrazine can exist in either gauche or anti conformations (relative to lone pairs).

H N H

H H

H N H H

gauche

observed HNNH dihedral angle Ca 90

H

There is a logical reason why the anti isomer should be more stable than the gauche isomer. The nonbonding lone pair orbitals in the gauche isomer should be destabilizing due to electron-electron repulsion. In fact, the gauche conformation is favored. Hence we have neglected an important stabilization feature in the structure. HOMO-LUMO Interactions Orbital overlap between filled (bonding) and antibonding states is best in the anti orientation. HOMO-LUMO delocalization is possible between: (a) N-lone pair NH; (b) NH NHH filled N-SP3 (HOMO) N N NH (LUMO) NH (HOMO) H NH (LUMO) N N H NH (LUMO)

H2O2 can exist in either gauche or anti conformations (relative to hydrogens). The gauche conformer is prefered.

H O

O H

gauche H

observed HOOH dihedral angle Ca 90

H

Major stabilizing interaction is the delocalization of O-lone pairs into the CH antibonding orbitals (Figure A). Note that there are no such stabilizing interactions in the anti conformation while there are 2 in the gauche conformation. Figure A Figure B OH (LUMO) (HOMO) filled O-SP3

filled N-SP3 (HOMO)

(HOMO) filled O-SP3

Note that you achieve no net stabilization of the system by generating molecular orbitals from two filled states (Figure B). better stabilization NH (HOMO)

Problem: Consider the structures XCH2OH where X = OCH3 and F. What is the most favorable conformation of each molecule? Illustrate the dihedral angle relationship along the CO bond.

1-11 Gauche Effect 9/11/01 11:27 PM

D. A. Evans

The Anomeric Effect: Negative HyperconjugationUseful LIterature Reviews

Chem 206

http://www.courses.fas.harvard.edu/~chem206/

Kirby, A. J. (1982). The Anomeric Effect and Related Stereoelectronic Effects at Oxygen. New York, Springer Verlag. Box, V. G. S. (1990). The role of lone pair interactions in the chemistry of the monosaccharides. The anomeric effect. Heterocycles 31: 1157. Box, V. G. S. (1998). The anomeric effect of monosaccharides and their derivatives. Insights from the new QVBMM molecular mechanics force field. Heterocycles 48(11): 2389-2417. Graczyk, P. P. and M. Mikolajczyk (1994). Anomeric effect: origin and consequences. Top. Stereochem. 21: 159-349. Juaristi, E. and G. Cuevas (1992). Recent studies on the anomeric effect. Tetrahedron 48: 5019. Plavec, J., C. Thibaudeau, et al. (1996). How do the Energetics of the Stereoelectronic Gauche and Anomeric Effects Modulate the Conformation of Nucleos(t)ides? Pure Appl. Chem. 68: 2137-44. Thatcher, G. R. J., Ed. (1993). The Anomeric Effect and Associated Stereoelectronic Effects. Washington DC, American Chemical Society.


Stereoelectronic Effects-2 Anomeric and Related Effects Electrophilic & Nucleophilic Substitution Reactions The SN2 Reaction: Stereoelectronic Effects Olefin Epoxidation: Stereoelectronic Effects Baeyer-Villiger Reaction: Stereoelectronic Effects Hard & Soft Acid and Bases (Not to be covered in class)

Problem 121

http://evans.harvard.edu/problems/

Sulfonium ions A and B exhibit remarkable differences in both reactivity and product distribution when treated with nucleophiles such as cyanide ion (eq 1, 2). Please answer the questions posed in the spaces provided below. KCN

Reading Assignment: Kirby, Chapters 1-3S BF4 Et

A

rel. rate = 8000

S

Et

+ PhCH2CN

(1)

D. A. Evans

Wednesday, September 17, 2003

KCN S Et B rel. rate = 1 S

+

MeCH2CN

(2)

2-00-Cover Page 9/17/03 8:35 AM

D. A. EvansThe Anomeric Effect

The Anomeric Effect: Negative Hyperconjugation

Chem 206

It is not unexpected that the methoxyl substituent on a cyclohexane ring prefers to adopt the equatorial conformation. H OMe H

OMe Gc = +0.6 kcal/mol What is unexpected is that the closely related 2-methoxytetrahydropyran prefers the axial conformation: H O OMe Gp = 0.6 kcal/mol O H OMe

Since the antibonding CO orbital is a better acceptor orbital than the antibonding CH bond, the axial OMe conformer is better stabilized by this interaction which is worth ca. 1.2 kcal/mol. Other electronegative substituents such as Cl, SR etc also participate in anomeric stabilization. H 1.781 H O Cl H O O Cl 1.819 ClThis conformer preferred by 1.8 kcal/mol Why is axial CCl bond longer ?

axial O lone pair CCl CCl

That effect which provides the stabilization of the axial OR conformer which overrides the inherent steric bias of the substituent is referred to as the anomeric effect. Let anomeric effect = A Gp = A = Gc + A Gp Gc

O Cl

H O HOMO

The Exo-Anomeric Effect

CCl

A = 0.6 kcal/mol 0.6 kcal/mol = 1.2 kcal/mol Principal HOMO-LUMO interaction from each conformation is illustrated below: H

There is also a rotational bias that is imposed on the exocyclic COR bond where one of the oxygen lone pairs prevers to be anti to the ring sigma CO bond H O O R O R favored R O O

O

O

OMe

O

H OMe

axial O lone pair CH

axial O lone pair CO

A. J. Kirby, The Anomeric and Related Stereoelectronic Effects at Oxygen, Springer-Verlag, 1983 E. Jurasti, G. Cuevas, The Anomeric Effect, CRC Press, 1995

2-01-Anomeric Effect-1 9/16/03 2:40 PM

D. A. Evans

The Anomeric Effect: Carbonyl GroupsAldehyde CH Infrared Stretching Frequencies

Chem 206

Do the following valence bond resonance structures have meaning?R C X O X R C O

Prediction: The IR CH stretching frequency for aldehydes is lower than the closely related olefin CH stretching frequency. For years this observation has gone unexplained. R R C H CH = 2730 cm CF3-1

R C C R

Prediction: As X becomes more electronegative, the IR frequency should increase O Me CH3 Me O CBr3 Me O

O H

CH = 3050 cm -1

C=O (cm-1)

Sigma conjugation of the lone pair anti to the H will weaken the bond. This will result in a low frequency shift.

1720

1750

1780

Infrared evidence for lone pair delocalization into vicinal antibonding orbitals.The NH stretching frequency of cis-methyl diazene is 200 cm-1 lower than the trans isomer. Me H N N Mefilled N-SP2 H antibonding NH

Prediction: As the indicated pi-bonding increases, the XCO bond angle should decrease. This distortion improves overlap.

R C X * CX O lone pair O

R C X O

N

N

NH = 2188 cm -1 Me N N H NH = 2317 cm -1

Me Nfilled N-SP2

N

antibonding NH H

Evidence for this distortion has been obtained by X-ray crystallography Corey, Tetrahedron Lett. 1992, 33, 7103-7106

..

The low-frequency shift of the cis isomer is a result of NH bond weakening due to the anti lone pair on the adjacent (vicinal) nitrogen which is interacting with the NH antibonding orbital. Note that the orbital overlap is not nearly as good from the trans isomer. N. C. Craig & co-workers JACS 1979, 101, 2480.

2-02-Anomeric Effect-2 9/16/03 2:41 PM

D. A. Evans

The Anomeric Effect: Nitrogen-Based SystemsCMe3 Me3C CH N N N CMe3 Me3C N N N Me3C

Chem 206

Observation: CH bonds anti-periplanar to nitrogen lone pairs are spectroscopically distinct from their equatorial CH bond counterparts

H H H H

N H

Me3C G = 0.35kcal/mol

N HOMO CH

A. R. Katritzky et. al., J. Chemm. Soc. B 1970 135

Favored Solution Structure (NMR)Me MeN NMe NMe Me N Me J. E. Anderson, J. D. Roberts, JACS 1967 96 4186 N N N Me MeN

Spectroscopic Evidence for Conjugation Infrared Bohlmann Bands Characteristic bands in the IR between 2700 and 2800 cm-1 for C-H4, C-H6 , & C-H10 stretch Bohlmann, Ber. 1958 91 2157 Reviews: McKean, Chem Soc. Rev. 1978 7 399 L. J. Bellamy, D. W. Mayo, J. Phys. Chem. 1976 80 1271 NMR : Shielding of H antiperiplanar to N lone pair H10 (axial): shifted furthest upfield H6, H4: = Haxial - H equatorial = -0.93 ppm Protonation on nitrogen reduces to -0.5ppm H. P. Hamlow et. al., Tet. Lett. 1964 2553 J. B. Lambert et. al., JACS 1967 89 37612-03-Anomeric Effect-3 9/16/03 2:43 PM

Favored Solid State Structure (X-ray crystallography)1.484 1.453

Bn N Me1.457

Me1.453

N N Bn1.459

N

A. R. Katrizky et. al., J. C. S. Perkin II 1980 1733

D. A. Evans

Anomeric Effects in DNA Phosphodiesters

Chem 206

Calculated Structure of ACGTGC Duplex

The Phospho-Diesters Excised from Crystal Structure

Guanine Cytosine Cytosine

1B

2B 1A Phosphate-1A p. p.Thymine Adenine

Phosphate-1B

The Anomeric EffectAcceptor orbital hierarchy: * POR * > * PO

R O

R O

P O

O O R

P

O O R

O R P O O RPhosphate-2A Phosphate-2B

Gauche-Gauche conformation O

R P O

O O R

O

O

Anti-Anti conformation Gauche-Gauche conformation affords a better donor-acceptor relationship

Oxygen lone pairs may establish a simultaneous hyperconjugative relationship with both acceptor orbitals only in the illustrated conformation.Plavec, et al. (1996). How do the Energetics of the Stereoelectronic Gauche & Anomeric Effects Modulate the Conformation of Nucleos(t)ides? Pure Appl. Chem. 68: 2137-44.

2-04-DNA Duplex/Anomeric 9/17/03 9:25 AM

D. A. Evans

Carboxylic Acids (& Esters): Anomeric Effects Again? Hyperconjugation: (Z) ConformerO O R' O Me H O Me

Chem 206

Conformations: There are 2 planar conformations.O O O O H R' R

Let us now focus on the oxygen lone pair in the hybrid orbital lying in the sigma framework of the C=O plane. * COO R R O

(Z) Conformer Specific Case: Methyl Formate

R

(E) Conformer

R

G = +4.8 kcal/mol

O

C R

O

In the (Z) conformation this lone pair is aligned to overlap with * CO.

The (E) conformation of both acids and esters is less stable by 3-5 kcal/mol. If this equilibrium were governed only by steric effects one would predict that the (E) conformation of formic acid would be more stable (H smaller than =O). Since this is not the case, there are electronic effects which must also be considered. These effects will be introduced shortly. Rotational Barriers: There is hindered rotation about the =COR bond. These resonance structures suggest hindered rotation about =COR bond. This is indeed observed:O R O R' R O O R'Energy

(E) ConformerR R O C O

In the (E) conformation this lone pair is aligned to overlap with * CR.

* CRO

R O Cbarrier ~ 10-12 kcal/mol

O R O R O RG ~ 2-3 kcal/mol

Since * CO is a better acceptor than * CR (where R is a carbon substituent) it follows that the (Z) conformation is stabilized by this interaction.

R

O R

O R R O

Esters versus Lactones: Questions to Ponder.Esters strongly prefer to adopt the (Z) conformation while small-ring lactones such as 2 are constrained to exist in the (Z) conformation. From the preceding discussion explain the following: 1) Lactone 2 is significantly more susceptible to nucleophilic attack at the carbonyl carbon than 1? Explain.

Rotational barriers are ~ 10-12 kcal/mol. This is a measure of the strength of the pi bond.

O 1 Et CH3CH2 O O O 2

Lone Pair Conjugation: The oxygen lone pairs conjugate with the C=O.

versus

R O

C R

O

The filled oxygen p-orbital interacts with pi (and pi*) C=O to form a 3-centered 4-electron bonding system.

2) Lactone 2 is significantly more prone to enolization than 1? In fact the pKa of 2 is ~25 while ester 1 is ~30 (DMSO). Explain. 3) In 1985 Burgi, on carefully studying O O O the X-ray structures of a number of lactones, noted that the O-C-C () & O O O O-C-O () bond angles were not equal. Explain the indicated trend in bond = 12.3 = 6.9 = 4.5 angle changes.

SP2 Hybridization

Oxygen Hybridization: Note that the alkyl oxygen is Sp2. Rehybridization is driven by system to optimize pi-bonding.

2-05 RCO2R Bonding 9/16/03 2:50 PM

D. A. Evans

Three-Center BondsCase 3: 2 p-Orbitals; 1 s-orbital

Chem 206

Consider the linear combination of three atomic orbitals. The resulting molecular orbitals (MOs) usually consist of one bonding, one nonbonding and one antibonding MO. Case 1: 3 p-Orbitals pi-orientationantibonding

antibonding

2

+

nonbonding

Energy

3

nonbonding

bonding

Case 4: 2 s-Orbitals; 1 p-orbitalbondingNote that the more nodes there are in the wave function, the higher its energy. H 2C H 2C CH CH + CH2 Allyl carbonium ion: both pi-electrons in bonding state

Do this as an exercise

Examples of three-center bonds in organic chemistry A. H-bonds: (3center, 4electron)O CH3 O H O H O CH3

CH2

Allyl Radical: 2 electrons in bonding obital plus one in nonbonding MO.

The acetic acid dimer is stabilized by ca 15 kcal/mol

H 2C

CH

Allyl Carbanion: 2 electrons in bonding obital plus 2 in CH2 nonbonding MO.

B. H-B-H bonds: (3-center, 2 electron)H B H B H H H H H B H H B H H

Case 2: 3 p-Orbitals sigma-orientation 3nonbonding antibonding

H

diborane stabilized by 35 kcal/molEnergy

C. The SN2 Transition state: (3center, 4electron)H Nu C H H BrThe SN2 transition state approximates a case 2 situation with a central carbon p-orbital The three orbitals in reactant molecules used are: 1 nonbonding MO from Nucleophile (2 electrons) 1 bonding MO CBr (2 electrons) 1 antibonding MO * CBr

bonding

2-06 3-center bonds/review 10/28/03 12:00 PM

D. A. Evans

Substitution Reactions: General Considerations

Chem 206

Why do SN2 Reactions proceed with backside displacement?R Nu: H H C X R Nu X

Electrophilic substitution at saturated carbon may occur with either inversion of retention Inversion + M

R Nu C H H X:

C H H

Ra El(+) H Rb C M

+ Nu

Ra C H

Ra Nu C Rb H M+

Given the fact that the LUMO on the electrophile is the CX antibonding orblital, Nucleophilic attack could occur with either inversion or retention.

Rb

InversionR Nu

RetentionRa El(+) Ra C M H Rb C M+

RetentionR X H H C

Ra H Rb C El M+

C H HLUMO

X

H Rb

El +

HOMO

Constructive overlap between Nu & *CX

Nu Overlap from this geometry results in no net bonding interaction El(+)LUMO

Ra Ra C H Rb

C M H Rb

M

Expanded view of *CX

HOMO

InversionLUMO

El(+)

Cantibonding HOMO

Xbonding Br2 H Br

Retention Examples

Li H

CO2 CO2Li H

NuFleming, page 75-76

predominant inversion

predominant retention

Stereochemistry frequently determined by electrophile structureSee A. Basu, Angew. Chem. Int. Ed. 2002, 41, 717-738

2-07-SN2-1 9/18/03 12:38 PM

D. A. EvansThe reaction under discussion:R R C H X

SN2 Reaction: Stereoelectronic EffectsThe use of isotope labels to probe mechanism. C H H X R Nu C H H

Chem 206

Nu:

H

Nu

X:

1 and 2 containing deuterium labels either on the aromatic ring or on the methyl group were prepared. A 1:1-mixture of 1 and 2 were allowed to react. If the rxn was exclusively intramolecular, the products would only contain only three deuterium atoms:O S O O CH3 SO3 D3C Nu

The NuCX bonding interaction is that of a 3-center, 4-electron bond. The frontier orbitals which are involved are the nonbonding orbital from Nu as well as CX and CX: CXD 3C

Nu:

exclusively intramolecular

(CD3ArNuCH3)

CH3

1

O S

O O CD3

SO3 H 3C Nu

(CH3ArNuCD3

energy

Nu: C X

Me

Nu:

exclusively intramolecular

CD3

CX RCH2X

Nu

Experiments have been designed to probe inherent requirement for achieving a 180 NuCX bond angle: Here both Nu and leaving group are constrained to be part of the same ring.R R

2 If the reaction was exclusively intermolecular, products would only contain differing amounts of D-label depending on which two partners underwent reaction. The deuterium content might be analyzed by mass spectrometry. Here are the possibilities: 1 + 1 D3-product 2 CD3ArNuCH3 D'3-product 2 + 2 2 CH3ArNuCD3 1 CD3ArNuCD3 1 CH3ArNuCH3 Hence, for the strictly intermolecular situation one should see the following ratios D0 : D3 : D'3 : D6 = 1 : 2 : 2 : 1. 1 + 2 The product isotope distribution in the Eschenmoser expt was found to be exclusively that derived from the intermolecular pathway! Other Cases: exclusively intermolecular(CH3)2N SO3CH3 (CH3)3N

Nu:

C X

C H H X

H H

Nu

D6-product D0-product

"tethered reactants"

"constrained transition state"

The Eschenmoser Experiment (1970): Helv. Chim Acta 1970, 53, 2059 The reaction illustrated below proceeds exclusively through bimolecular pathway in contrast to the apparent availability of the intramolecular path.O S O O CH3 SO3

+

SO3

CH3

16% intramolecular 84% intermolecular

SO3CH3 N(CH3)2

SO3

Nu:

N(CH3)3

+

Nu

2-08-The SN2 RXN-FMO 9/16/03 2:56 PM

Hence, the NuCX 180 transition state bond angle must be rigidly maintained for the reaction to take place.

D. A. EvansSO3CH3 N(CH3)2

Intramolecular methyl transfer: Speculation on the transition structures(CH3)2N

Chem 206+SO3

SO3

SO3CH3

N(CH3)3

+

(CH3)3N

16% intramolecular; 84% intermolecular9- membered cyclic transition state

exclusively intermolecular8- membered cyclic transition state

174 174

2-09-Intra alk TS's 9/16/03 2:56 PM

00000 00000 00000 00000 00000est CO bond length 2.1

000000 000000 000000 000000 000000est CO bond length 2.1

00000 00000000 00000 00000000 00000 00000 00 00 00 00 0000 00 00 0000 00000 0000 00000

est CN bond length 2.1

est CN bond length 2.1

Approximate representation of the transition states of the intramolecular alkylation reactions. Transition state CO and CN bond lengths were estimated to be 1.5x(CX) bond length of 1.4

D. A. Evans The General Reaction:R R O

Olefin Epoxidation via Peracids: An Introduction Per-arachidonic acid EpoxidationO

Chem 206

R O OH R

R

+R R R

R

+R OH

Me O

O H

HOMO CC

LUMO *OO

note labeled oxygen is transferfed

O-O bond energy: ~35 kcal/mol

Reaction rates are governed by olefin nucleophilicity. The rates of epoxidation of the indicated olefin relative to cyclohexene are provided below: OHOH OAc

1.0

0.6

0.05

0.4

The indicated olefin in each of the diolefinic substrates may be oxidized selectively.Me Me Me Me Me Me Me

H Me Me

The transition state:

Me

O H

O

E. J. Corey, JACS 101, 1586 (1979) For a more detailed study see P. Beak, JACS 113, 6281 (1991) View from below olefin2-10 Epoxidation-1 9/16/03 2:58 PM

For theoretical studies of TS see R. D. Bach, JACS 1991, 113, 2338 R. D. Bach, J. Org. Chem 2000, 65, 6715

D. A. Evans The General Reaction:R R

Olefin Epoxidation with DioxiranesAsymmetric Epoxidation with Chiral KetonesR R O R R

Chem 206

+R R

R R

O

Review: Frohn & Shi, Syn Lett 2000, 1979-2000Me O Me O O O Me

O

+R R note labeled oxygen is transferfed

HOMO CC

LUMO *OO

chiral catalystO Me

O

O-O bond energy: ~35 kcal/molR2 H R1 R R O R2

Synthesis of the Dioxirane OxidantK+ O R O O S O O H R R O O SO3

O R2 R1 R2

oxone, CH3CN-H2O pH 7-8Me Me Ph

O R

O

(Oxone)

Ph

Ph

Ph

Ph

>95% ee

84% ee

92% ee

Synthetically Useful Dioxirane SynthesisO Me O F 3C CF3 Me

Question: First hour Exam 2000 (Database Problem 34)Question 4. (15 points). The useful epoxidation reagent dimethyldioxirane (1) may be prepared from "oxone" (KO3SOOH) and acetone (eq 1). In an extension of this epoxidation concept, Shi has described a family of chiral fructose-derived ketones such as 2 that, in the presence of "oxone", mediate the asymmetric epoxidation of di- and tri-substituted olefins with excellent enantioselectivities (>90% ee) (JACS 1997, 119, 11224). Me O Me R2 R1 R2KO3SOOH CH3CN-H2O pH 10.5 1 equiv 2 oxone, CH3CN-H2O pH 10.5

oxone

O Me

O Me O CF3

co-distill to give ~0.1 M soln of dioxirane in acetone co-distill to give ~0.6 M soln of dioxirane in hexafluoroacetone

oxone

O F 3C

Me

Me Me

O O1 (1) 2

Me O

Curci, JOC, 1980, 4758 & 1988, 3890; JACS 1991, 7654. Transition State for the Dioxirane Mediated Olefin EpoxidationO R R O O R R

O

O

O R2(2)

O Me O Me

O

R1

R2

planar

>90% ee

O

spiro

rotate 90

Part A (8 points). Provide a mechanism for the epoxidation of ethylene with dimethyldioxirane (1). Use three-dimensional representations, where relevant, to illustrate the relative stereochemical aspects of the oxygen transfer step. Clearly identify the frontier orbitals involved in the epoxidation. Part B (7 points). Now superimpose chiral ketone 2 on to your mechanism proposed above and rationalize the sense of asymmetric induction of the epoxidation of trisubstituted olefins (eq 2). Use three-dimensional representations, where relevant, to illustrate the absolute stereochemical aspects of the oxygen transfer step.

stabilizing Olp * C=C cis olefins react ~10 times faster than trans Houk, JACS, 1997, 12982.

2-11 Epoxidation-2 9/16/03 3:01 PM

D. A. EvansO C RL RS + RCO3H RCO2H

The Baeyer-Villiger Reaction: Stereoelectronic EffectsO RL C O major RS O Me O minor RS Me O + RCO3H CMe3 Me3C O H O O O- MeCO2H

Chem 206CMe3 Me O H O R O O

Let RL and RS be Sterically large and small substituents.+C RL

R

The major product is that wherein oxygen has been inserted into theRLCarbonyl bond. O kR kR / KMe RR O O C R Me + CF3CO3H O C MeH

FavoredCMe3 O Me O OH O R

Migrating group

major

CH3CH2 CH3(CH2)2 (CH3)3C

72 150 830 >2000 Conformer AH

O Me O CMe3

kMe

C R O

Me

minor

PhCH2RS OH C RL O O

DisfavoredMe O

Migrating group

The Intermediate

R

O OH Me O CMe3

O

Me3C O

The important stereoelectronic components to this rearrangement:R

O

1. The RLCOO dihedral angle must be180 due to the HOMO LUMO interaction -RLCOO. 2. The COOC' dihedral angle will be ca. 60 due to the gauche effect (O-lone pairsCO). This gauche geometry is probably reinforced by intramolecular hydrogen bonding as illustrated on the opposite page:2-12- Baeyer Villiger Rxn 9/16/03 5:33 PM

Conformer B

The destabilizing gauche interaction

Steric effects destabilize Conformer B relative to Conformer A; hence, the reaction is thought to proceed via a transition state similar to A. For relevant papers see: Crudden, Angew. Chem. Int. Ed 2000, 39, 2852-2855 (pdf) Kishi, JACS 1998, 120, 9392 (pdf)

D. A. Evans

The Baeyer-Villiger Reaction: Stereoelectronic EffectsCMe3 Me

Chem 206

Conformer A in three dimensions

O Me

+ RCO3H CMe3

Me3C O H

O

Me O- MeCO2H

O O H O R O

O

R

FavoredH

Migrating group CMe3 O O OH O R O Me O CMe3

Me

Conformer AH

DisfavoredMe O Me3C O R OH O

Migrating group

O Me1

O

CMe32 3

Conformer B

The destabilizing gauche interaction

4

Steric effects destabilize Conformer B relative to Conformer A; hence, the reaction is thought to proceed via a transition state similar to A. For relevant papers see: Crudden, Angew. Chem. Int. Ed 2000, 39, 2852-2855 (pdf) Kishi, JACS 1998, 120, 9392 (pdf)2-13- Baeyer Villiger Rxn-2 9/16/03 5:41 PM

23 dihedral angle ~ 178 from Chem 3D

B. Breit

FMO-Theory/HSAB Principle 1Hard and Soft Acids and Bases (HSAB-Principle)Reading Assignment: Fleming, Chapter 3, p33-46 Pearson, JACS 1963, 85, 3533.

Chem 206

FMO-Theory and Klopman-Salem equation provide an understanding of this empirical principle: Hard Acids have usually a positive charge, small ion radii (high charge density), energy rich (high lying) LUMO. Soft Acids are usually uncharged and large (low charge density), they have an energy poor (low lying ) LUMO (usually with large MO coefficient). Hard Bases usually have a negative charge, small ion radii (high charge density), energy poor (low lying) HOMO. Soft Bases are usually uncharged and large (low charge density), energy rich (high lying) HOMO (usually with large MO coefficient).

Hard Acids prefer to interact with hard bases Soft acids prefer to interact with soft bases.Softness: Polarizability; soft nucleophiles have electron clouds, which can be polarized (deformed) easily. Charged species with small ion radii, high charge density.

Hardness:

Qualitative scaling possible: Molecular Orbital Energies of an

idealized Soft SpeciesE E small HOMO/LUMO gap

idealized Hard Species

large HOMO/LUMO gap

FMO-Theory for interaction:

Soft-SoftE E

Hard-Hard

Acid

Base Acid Base

Significant Energy gain through HOMO/LUMO interaction

Only neglectable energy gain through orbital interaction.

2-14-FMO HSAB 1 9/20/00 8:30 AM

B. Breit

FMO-Theory/HSAB Principle 2

Chem 206

Klopman-Salem Equation for the interaction of a Nucleophile N (Lewis-Base) and an Electrophile E (Lewis-Acid).

E =

QNQE RNE Coulomb Term

2(cNcE)2 EHOMO(N) - ELUMO(E) Frontier Orbital Term

Q: Charge density : Dielectricity constant R: distance (N-E) c: coefficient of MO : Resonance Integral E: Energy of MO

Soft-Soft Interactions: Coulomb term small (low charge density). Dominant interaction is the frontier orbital interaction because of a small E(HOMON/LUMOE). formation of covalent bonds Hard-Hard Interactions: Frontier orbital term small because of large E(HOMON/LUMOE). Dominant interaction is described by the Coulomb term (Q is large for hard species), i.e. electrostatic interaction. formation of ionic bonds Hard-Soft Interactions: Neither energy term provides significant energy gain through interaction. Hence, Hard-Soft interactions are unfavorable.

2-15-FMO HSAB 2 9/20/00 8:27 AM

B. Breit

FMO-Theory/HSAB Principle 3HSAB principle - Application to Chemoselectivity Issues(c) SN2 vs E2

Chem 206

(a) Enolate AlkylationO soft

CO2R

hard O soft C C

soft MeI

C-AlkylationMe OTMS

HC(COOR)2 H Br hard OC2H5

CO2R

S N2

hard TMSCl

E2

O-Alkylation

(d) Ambident Nucleophiles (b) 1,2- vs. 1,4-addition to ,-unsaturated carbonyl compounds + 0.29O H H

- 0.48O

soft Ag S C N hard Na

soft MeI

H3 C

S

C O

N

S-Alkylation

+ 0.01 Charge density

+ 0.62 LUMO-coefficients

hard RCOX

S

C

N

R

N-Acylation

hard softO

hard MeLi

OH Me O Ag soft MeI H3C NO2

N-Alkylation

1,2-Addition soft Me2CuLi Conjugate AdditionMe O

O hard

N Na soft hard t-BuCl ONO

O-Alkylation

2-16-FMO HSAB 3 9/20/00 8:27 AM

D. A. Evans

Rules for Ring Closure: Introduction

Chem 206


The Primary Literature Baldwin, J. Chem. Soc., Chem. Comm. 1976, 734, 736. Baldwin, J. Chem. Soc., Chem. Comm. 1977 233. Baldwin, J. Org. Chem. 1977, 42, 3846. Baldwin, Tetrahedron 1982, 38, 2939.

Chemistry 206 Advanced Organic Chemistry

Lecture Number 3

Stereoelectronic Effects-3 "Rules for Ring Closure: Baldwin's Rules"Kirby, "Stereoelectronic Effects" Chapters 4, 5

Useful LIterature Reviews Johnson, C. D. (1993). Stereoelectronic effects in the formation of 5and 6-membered rings: the role of Baldwin's rules. Acc. Chem. Res. 26: 476-82. (Handout) Beak, P. (1992). Determinations of transition-state geometries by the endocyclic restriction test: mechanisms of substitution at nonstereogenic atoms. Acc. Chem. Res. 25: 215. (Handout)

Problems of the DayPropose mechanisms for the following reactionsR O R HO H+ R R O O

+HO O

D. A. Evans3-00-Cover Page 9/19/03 8:36 AM

Friday, September 19, 2003Me

NH2NH2 OMe Me

HN NH O

D. A. Evans, J. Johnson

Rules for Ring Closure: Introduction

Chem 206

Ring Closure and Stereoelectronic Connsiderations An Examination of Baldwin's Rules"Baldwin's Rules" provides a qualitative set of generalizations on the probability of a given ring closure. There are circumstances where the "rules" don't apply. They do not apply to non-first-row elements participating in the cyclization event. The longer bond lengths and larger atomic radii of 2nd row elements result in relaxed geometrical constraints. For example, a change in a heteroatom from O to S could result in relaxation of a given geometric constraint. X = O vs S X Yendo

C. Nucleophilic ring closures sub-classified according to hybridization state of electrophilic component: (tetrahedral = tet; trigonal = trig; digonal = dig) D. Nucleophilic ring closures further subclassified according to size of the fomed ring. For example:5-exo-tet

X Y

X

Y

5-exo-trig

X Y5-exo-dig

X

Y

X

Y

X Y

X

Y

The "rules" do not apply to electrocyclic processes.

NomenclatureClasses of Ring Closing Processes A. Exo-cyclization modes identified by the breaking bond being positioned exocyclic to the forming cycle.exo

Required trajectories (Baldwin):X Y

= 180

X Y

X X

X

Y

Y Y B. Endo-cyclization modes identified by the breaking bond being positioned endocyclic to the forming cycle.endo

= 109

X

YWill come back to this case later

X Y X = first-row element N, O

X

* 120 *

X

Y

X

Y

Y

Baldwin, J. Chem. Soc., Chem. Commun., 1976, 734.

3-01-Baldwin Rules-1 9/18/03 3:38 PM


Rules for Ring Closure: SP3 Carbon & Related SystemsFRST-PLATTNER RULE

Chem 206

Tetrahedral Carbon All exo cyclization modes are allowed: (n-exo-tet, n = 3)exo

In this simple model, the transition-state leading to 1 involves the diaxial orientation of nucleophile and leaving group. This orientation affords the best overlap of the anti-bonding CY orbital and the nonbonding electron pairs on the nucleophile O. In the formation of the diastereomeric epoxide 2, the proper alignment of orbitals may only be achieved by cyclization through the less-favored boat conformer. Accordingly, while both cyclizations are "allowed", there are large rate differences the the rates of ring closure. While the FRST-PLATTNER RULE deals wilth the microscopic reverse, in the opening of epoxides by nucleophiles, the stereoelectronic arguments are the same.

X

C Y

X

C Y

There are stereoelectronic issues to consider for n-exo-tet cyclizations Formation of 3-Membered Rings (3-exo-tet)H H H

Y

X C H2

C

Y

X

C

X

CH2 C H2

C H2

H

+ Y

Stereoelectronic Effects in Epoxide Ring CleavageNu Me3C NuO H O Me3C H HO HO NuH Me Me3C H H Nu Me H H

Conformational Effects in Epoxide Ring Formation/cleavageThose stereoelectronic effects that operate in ring cleavage also influence ring formation. Consider a rigid cyclohexene oxide system:Y H H O H Y H O H O

H

Y

fasterH H H Me3C

OY

1

O

slowerH O

Nu-

Nu HO

O H H

chair

boat

2

"The diaxial nucleophilic ring cleavage of epoxides" For more information on epoxide cleavage see Handout 03A.

3-02-Baldwin Rules-2 9/18/03 3:39 PM


Rules for Ring Closure: SP3 Carbon & Related SystemsCase 2: King, J.C.S. Chem. Comm., 1979, 1140.O S O Me NMe2 O O

Chem 206

Tetrahedral CarbonEndo cyclization modes that are disallowed (n-endo-tet, n = 39)

8-endo-tet disfavored Rxn exclusively intermolecular 8-endo-tet disfavored Rxn exclusively intermolecular 9-endo-tet borderline 84% intermolecular, 16% intramolecular

O O

S

_

NMe3+

endo

X

Y

C

Y C(SP3)

X

SO2OMe NMe2

SO3 NMe3+

The stereoelectronic requirement for a 180 XCY bond angle is only met when the endo cyclization ring size reaches 9 or 10 members. Case 1: Eschenmoser, Helvetica Chim. Acta 1970, 53, 2059.O S O O CX3 O S O O S O OCX3 O S O SO2OMe NMe2

SO3 NMe3+

NaH 6-endo-tet disfavored Rxn exclusively intermolecular (lecture 2)

ConclusionsAllowed endo cyclization modes will require transition state ring sizes of at least nine members.

Intramolecular epoxidation has also been evaluatedCY3

CY3

O Cl OOH n-

Beak, JACS 1991, 113, 6281. 8-endo-tet disfavoredCl CO2H

Cyclization exclusively intermolecular. However the exocyclic analog is exclusively intramolecularO S O O CX2I O S O O S O O

n

O

NaH 6-exo-tet favored Rxn exclusively intramolecularO S

CX2 O

n = 1: rxn exclusively intermolecular n = 9: rxn is intramolecular

Beak states that the conclusions made with carbon substitution also hold for oxygen atom transfer.Beak, P. (1992). Determinations of transition-state geometries by the endocyclic restriction test: mechanisms of substitution at nonstereogenic atoms. Acc. Chem. Res. 25: 215.

CY3

CY3

3-03-Baldwin Rules-3 9/18/03 4:07 PM


Rules for Ring Closure: SP2 Carbon & Related Systems

Chem 206

Trigonal CarbonEndo cyclization modes that are disallowed (3 to 5-endo-trig)n-endo-trig X Y X C Y MeO2C NH2 CO2Me

C

X = first-row element The 5-endo-trig cyclization is a watershed case distance from reacting centers: 2.77 Case 1: Baldwin, J. Chem. Soc., Chem. Commun., 1976, 734.CO2Me OH baseX

CO2Me O

howeverCO2Me SH

5-endo-trig Disfavored base

It is possible that a "nonvertical" trajectory is operational like that suspected in C=O additionCO2Me

S

Second row atom relaxes the cyclization geometrical requirement Case 2: Baldwin, J. Chem. Soc., Chem. Commun., 1976, 736.MeO2C NH2 CO2MeX

MeO2C HN

CO2Me

5-endo-trig 0%MeO2C HN O

5-exo-trig 100%

3-04-Baldwin Rules-4 9/18/03 4:07 PM

D. A. Evans, J. JohnsonCase 2: continued...MeO2C NH2 CO2MeX


Chem 206

Apparent exceptions to disallowed 5-endo-trig cyclization processMeO2C HN CO2MeO

5-endo-trig 0%MeO2C HN O

N

CH3CO2H

+ N

OH

N

O

Filer, J. Am. Chem. Soc. 1979, 44, 285.

5-exo-trig 100%R HC N1

CO2Me KO Bu R2 CO2Met

CO2Me R1 HN R CO2Me2

CO2Me R 3:11

Control experiment: Intermolecular reaction favors conjugate addtion.Me CO2Me PhCH2NH2 H N Me CO2Me 100% Me H N O 0% Ph

HN

CO2Me R2

Ph

R1 = aryl, R2 = aryl, alkyl

Grigg, J. Chem. Soc., Chem. Commun. 1980, 648.

Case 3:O Ph OMe NH2NH2 65 oC Ph HN NH O

Does the illustrated ketalization process necessarily violate "the rules"?

R O R

(CH2OH)2 H+

R R

O O

O Ph MeI OK

1) EtO2CCl, pyridine 2) NH2NH2

Ph NH

O

R O R

(CH2OH)2

R HO

R O

OH

H+ H2O

R R O

OH

( )2

H 2N 200 oC O Ph OMe NH2NH2 65 oC HN X

+

( )25-endo-trig

5-endo-trig

H+R O

disfavored ?R R O O

R Ph CO2Me NH2 Ph O HN NH HO

+

OH 5-exo-tet

( )2

favored ?

5-exo-trig

Johnson, C. D. (1993). Stereoelectronic effects in the formation of 5- and 6-membered rings: the role of Baldwin's rules. Acc. Chem. Res. 26: 476-82.

3-05-Baldwin Rules-5 9/18/03 4:08 PM

D. A. Evans, J. JohnsonMore Exceptions

Rules for Ring Closure: SP2 Carbon & Related SystemsBu

Chem 206Bu

Zard, Org. Lett. 2002, 4, 1135MeO O MeO N S S OEt N O

X Y

NaHHO

DMF, 60 C Y F H Cl Br Cond

Y O

ROOR heat 80% OO

X F F

Yield 80 17 15

DMF, 60 C, 2 h DMF, 80 C, 43 h DMF, 60 C, 8 h DMF, 60 C, 5 h

O MeO

O

O MeO N

MeO O MeO N

Cl Br

5-endo-trig

Ichikawa, et al Synthesis 2002, 1917-1936, PDF on Course Website Numerous other cases are provided in this review.

O

O

O

O

Br

H O

Revisiting Case 2 with FluorinesO O MeO2C

Bu3SnH AIBN 82%O H

O

MeO2C N O

OMe

MeO2C N Ts

CO2Me

Ts

5-exo-trig

TsHN

5-endo-trig

FavoredH O

Not Observed

5-endo-trigO O

O MeO2C N Ts O CF2 MeO2C TsHN

O OMe CF2 MeO2C N Ts F CO2Me

5-exo-trig

5-endo-trig

Chem. Comm 2088, 28 Review: "5-Endo-Trig Radical Cyclizatons" Ishibashi, et al Synthesis 2002, 695-713, PDF on Course Website

Not Observed

Favored

3-06-Baldwin Rules-6 9/18/03 5:10 PM



Chem 206

Trigonal Carbon: Exocyclic Enolate Alkylationexo

distance between reacting centers: 3.37C O C C Y-

O

C C

C Y

BrX

By definition, an exo-tet cyclization, but stereoelectronically behaves as an endo trig.Me Me O Me Me MO BrX

MO

O

Me Me O

(1)

The overlap for C-alkylation is poor due to geometrical constraints of 5-membered ring

only observed product

However:Me Me O Me BrKOt-Bu or LDA

distance between reacting centers: 3.04Me Me O > 95% by NMR

The relaxed geometrical constraint provided by the added CH2 group now renders the 6-membered cyclization possible

Baldwin, J. Chem. Soc., Chem. Commun. 1977, 233.

Given the failure of the enolate alkylation shown above (eq 1), explain why these two cyclizations are successful.Br NHAr O R O Ar NH R OMs base O N Ar base N O R Ar R

MO

Br

O

Favorskii Rearrangement (Carey, Pt B, pp 609-610) Your thoughts on the mechanismO Cl MeO HCl O MeO

CO2Me

3-07-Baldwin Rules-7 9/18/03 4:09 PM


Rules for Ring Closure: SP2 & SP Carbon & Related Systems

Chem 206

Trigonal Carbon: Intramolecular Aldol CondensationsBaldwin, Tetrahedron 1982, 38, 2939

Digonal Carbon: Cyclizations on to AcetylenesDIGONAL: Angle of approach for attack on triple bonds

MO R

X Y

(Enolendo)-Exo-trigO R

X YM

Nu120 120 E+

Baldwin: - 3 and 4-Exo-dig are disfavored - 5 to 7-Exo-dig are favored - 3 to 7-Endo-dig are favored

Favored: 6-7-(enolendo)-exo-trig Disfavored: 3-5-(enolendo)-exo-trig

X MO R Y

(Enolexo)-exo-trigO R

X YMH _

Ab initio SCF 4-31G calculations for the interaction of hydride with acetylene:H2.13 127 o 148o H

Favored: 3-7-(enolexo)-exo-trigH O Me OI

4-31G basis set Houk, J.ACS.1979, 101, 1340.

H

5-(Enolendo)-Exo-trigMe Me O

6-(Enolendo)-Exo-trigMe O H

H 156o 1.22

C

C

1.5-2.0

110o -120o

STO-3G minimal basis set H Dunitz, Helv Chim. Acta 1978, 61, 2538.

favoredMe O Me O MeIII

H O

C

C

Me Me OII

Crystal Structures do not support BaldwinO

Statistical Distribution, (I + II)/III = 2:1 Experimental Distribution, = 0:100N

N2.92 104o

(KOH, MeOH, r.t., 5 min, 77% y.)

ON+

N

+N

O93

2.44o

No

Caution: Baldwin's conclusions assume that the RDS is ring closure; however, it is well known (by some!) that the rate determining step is dehydration in a base-catalyzed aldol condensation.

86

J. Dunitz and J. Wallis J. C. S. Chem. Comm. 1984, 671.

3-08-Baldwin Rules-8 9/18/03 5:49 PM


Rules for Ring Closure: SP Carbon & Related Systems Indole synthesis:CH32 equiv. LDA 2 equiv. RX -78 oC R = Me, Bu, CO2Me LiTMP+

Chem 206

Endo Digonal versus Endo Trigonal Cyclizations5-endo-trigY X:

CH2R N

+

N

C

-

C-

In-plane approach; nucleophile lone pair is orthogonal to *

Out-of-plane approach; nucleophile lone pair can't achieve Brgi-Dunitz angle

RSaegusa, J. Am. Chem. Soc. 1977, 99, 3532. _

N

5-endo-dig:X

Spiro dihydrofuranones:O Li HO MeOn n = 1,2

Li+

Y

Allowed due to in-plane pi orbitals

OMe

KOtBu

O

OMe

XFor an opposing viewpoint to Baldwin's view of nucleophile trajectories, see Menger's article on directionality in solution organic chemistry: Tetrahedron 1983, 39, 1013.O Me Me HO Ph NaOMe MeOH Me Me O Ph O

n

n

Developing negative charge on the central allenic carbon is in the same plane as the OMe groupMagnus, J. Am. Chem. Soc. 1978, 100, 7746.

5-endo-digO

5-exo-digO R OHR = H, OMe

NaOMeX

5-endo-trig

R O Ph

Li

Ph

Li

Ph

however, the acid catalyzed version does cyclizeBaldwin, J. Chem. Soc., Chem. Commun., 1976, 736. Johnson, Can. J. Chem. 1990, 68, 1780 J. Am. Chem. Soc. 1983, 105, 5090 J. Chem. Soc., Chem. Commun. 1982, 36.

4-endo-digX

Li

Li

Ph

3-09-Baldwin Rules-9 9/19/03 8:38 AM


Rules for Ring Closure: SP Carbon & Related Systems

Chem 206

Digonal Cyclizations: Interesting ExamplesMeO2C CN Et3N, Toluene, reflux 12 h, 65-70% y. CN

Trost, J. Am. Chem. Soc., 1979, 101, 1284. Proposes E-olefin geometry, E/Z > 95:5

5-exo-digO O CO2Me:

R R'

30-40 kcal/mol ?

R R':

H

OH

HO2C HHirsutic Acid C

O

Conclusions and CaveatsOTBS

O Me

LiCH2NC; TBS-Cl 71%+

Baldwin's Rules are an effective first line of analysis in evaluating the stereoelectronics of a given ring closure Baldwin's Rules have provided an important foundation for the study of reaction mechanism Competition studies between different modes of cyclization only give information about relative rates, and are not an absolute indicator of whether a process is "favored" or "disfavored" Structural modifications can dramatically affect the cyclization mode; beware of imines and epoxidesEXO Tet Trig Dig X X Tet ENDO Trig X X X X X X Dig

OTBS 1) RCOCl2) AgBF4 86%

Me C-

N+ C O

Me R

N

Works for varying ring sizes and R groups; acylnitrilium ion can also work as an electophile in a Friedel-Crafts type of reaction 5-endo-dig Livinghouse, Tetrahedron 1992, 48, 2209.O H N O Me R

3 4 5 6 7

3-10-Baldwin Rules-10 9/18/03 5:21 PM

D. A. Evans

Acyclic Conformational Analysis-1Professor Frank Weinhold Univ. of Wisconsin, Dept of Chemistry B.A. 1962, University of Colorado, Boulder

Chem 206



A.M. 1964, Harvard University Ph.D. 1967, Harvard University Physical and Theoretical Chemistry.

Useful LIterature ReviewsEliel, E. L., S. H. Wilen, et al. (1994). Stereochemistry of Organic Compounds. New York, Wiley. Juaristi, E. (1991). Introduction to Stereochemistry and Conformational Analysis. New York, Wiley. Juaristi, E., Ed. (1995). Conformational Behavior of Six-Membered Rings: Analysis, Dynamics and Stereochemical Effects. (Series: Methods in Stereochemical Analysis). Weinheim, Germany, VCH. Schweizer, W. B. (1994). Conformational Analysis. Structure Correlation, Vol 1 and 2. H. B. Burgi and J. D. Dunitz. Weinheim, Germany, V C H Verlagsgesellschaft: 369-404. Kleinpeter, E. (1997). Conformational Analysis of Saturated Six-Membered Oxygen-Containing Heterocyclic Rings. Adv. Heterocycl. Chem. 69: 217-69. Glass, R. R., Ed. (1988). Conformational Analysis of Medium-Sized Ring Heterocycles. Weinheim, VCH. Bucourt, R. (1973). The Torsion Angle Concept in Conformational Analysis. Top. Stereochem. 8: 159.

Acyclic Conformational Analysis-1 Ethane, Propane, Butane & Pentane Conformations Simple Alkene Conformations

Reading Assignment for week A. Carey & Sundberg: Part A; Chapters 2 & 3R. W. Hoffmann, Angew. Chem. Int. Ed. Engl. 2000, 39, 2054-2070 Conformation Design of Open-Chain Compounds (handout)

The Ethane Barrtier ProblemF. Weinhold, Nature 2001, 411, 539-541 "A New Twist on Molecular Shape" (handout) F. M. Bickemhaupt & E. J. Baerends, Angew. Chem. Int. Ed. 2003, 42, 4183-4188,"The Case for Steric Repulsion Causing the Staggered Conformation in Ethane" (handout) F. Weinhold,, Angew. Chem. Int. Ed. 2003, 42, 4188-4194,"Rebuttal of the BikelhauptBaerends Case for Steric Repulsion Causing the staggered Connformation of Ethane" (handout)

Problems of the DayO Predict the most stable conformation of the indicated dioxospiran? O

D. A. Evans4-00-Cover Page 9/22/03 9:01 AM

Monday, September 22, 2003

D. A. Evans

Acyclic Conformational Analysis-1

Chem 206

The following discussion is intended to provide a general overview of acyclic conformational analysis

Ethane Rotational Barrier: The FMO ViewF. Weinhold, Angew. nature 2001, 411, 539-541"A New Twist on Molecular Shape"

Ethane & PropaneThe conformational isomerism in these 2 structures reveals a gratifying level of internal consistency.H H RH C H H

One can see from the space-filling models that the Van der Waals radii of the hydrogens do not overlap in the eclipsed ethane conformation. This makes the steric argument for the barrier untenable. One explanation for the rotational barrier in ethane is that better overlap is possible in the staggered conformation than in the eclipsed conformation as shown below. In the staggered conformation there are 3 anti-periplanar CH Bonds

eclipsed conformationH E = +3.0 kcal mol-1 (R = H) Van derWaals radii of vicinal hydrogens do not overlap in ethane E = +3.4 kcal mol-1 (R = Me) C H C CH HOMO

HC C

* CH LUMO

CH

p. p.

H

CH

R H C H HIn propane there is a discernable interaction

H H

In the eclipsed conformation there are 3 syn-periplanar CH Bonds staggered conformationH C H C CH HOMO CH

HC

HC

* CH LUMO CH

For purposes of analysis, each eclipsed conformer may be broken up into its component destabilizing interactions. Incremental Contributions to the Barrier. Structure ethane propane Eclipsed atoms E (kcal mol -1) 3 (HH) 2 (HH) 1 (HMe) +1.0 kcal mol -1 +1.0 kcal mol -1 +1.4 kcal mol -1

Following this argument one might conclude that: The staggered conformer has a better orbital match between bonding and antibonding states. The staggered conformer can form more delocalized molecular orbitals. J. P. Lowe was the first to propose this explanation"A Simple Molecuar Orbital Explanation for the Barrier to Internal Rotation in Ethane and Other Molecules" J. P. Lowe, JACS 1970, 92, 3799 Me Me Me

Calculate the the rotational barrier about the C1-C2 bond in isobutane

4-01-introduction 9/22/03 8:28 AM

D. A. Evans

Acyclic Conformational Analysis: Butane The 1,2-Dihaloethanes Butane

Chem 206

XH C H H H H H

H H C

X X

X

X = Cl; H = + 0.91.3 kcal/mol X = Br; H = + 1.41.8 kcal/mol X = F; H = 0.6-0.9 kcal/mol

Using the eclipsing interactions extracted from propane & ethane we should be able to estimate all but one of the eclipsed butane conformationsMe C H Me H

Observation: While the anti conformers are favored for X = Cl, Br, the gauche conformation is prefered for 1,2-difluroethane. Explain. Discuss with class the origin of the gauche stabiliation of the difluoro anaolg. Recent Article: Chem. Commun 2002, 1226-1227 (handout)

staggered conformation

H

H

H H

Me H C MeH

eclipsed conformationE=?

Eclipsed atoms 1 (HH) 2 (HMe)

E (kcal mol -1) +1.0 kcal mol -1 +2.8 kcal mol -1

Relationship between G and Keq and pKap. p. Recall that: or

G = RT Ln K G = 2.3RT Log10K2.3RT = 1.4 (G in kcal Mol1 )

E est = 3.8 kcal mol -1 The estimated value of +3.8 agrees quite well with the value of +3.6 reported by Allinger (J. Comp. Chem. 1980, 1, 181-184)

At 298 K:

G298 = 1.4 Log10KeqSince

n-Butane Torsional Energy ProfileH H HH C Me Me H C H

pKeq = Log10KeqE1

E2

G298 = 1.4 pKeqenergy

H H

Me H C MeH H H

Hence, pK is proportional to the free energy change

H H

Me C

H

Keq 1.0 10 100

pKeq 0 1 2

G 0 1.4 2.8 kcal /mol

H Me ARef = 0

+3.6

Me Me G

+5.1

+0.88

4-02-introduction-2 9/22/03 8:33 AM

D. A. Evans Butane continued

Acyclic Conformational Analysis: ButaneNomenclature for staggered conformers:H H Me Me C H H H H H Me C H H Me Me

Chem 206Me C H H H

From the torsional energy profile established by Allinger, we should be able to extract the contribution of the MeMe eclipsing interaction to the barrier:H H C H H Me H H Me Me C HH

staggered conformation Me

eclipsed conformation

trans or t or (anti) Conformer population at 298 K: 70%

gauche(+) or g+ 15%

gauche(-) or g15%

E = +5.1 kcal mol-1

Let's extract out the magnitide of the MeMe interaction 2 (HH) + 1 (MeMe) = +5.1 1 (MeMe) = +5.1 2 (HH) p. p. 1 (MeMe) = +3.1

General nomenclature for diastereomers resulting from rotation about a single bond (Klyne, Prelog, Experientia 1960, 16, 521.)RR C 0 R R C -60

R C

R

sp sc sc

+60

Incremental Contributions to the Barrier. Eclipsed atoms 2 (HH) 1 (MeMe) E (kcal mol -1) +2.0 +3.1R Eclipsed Butane conformation

acR C -120

ac+120 R C R 180 R C R

ap

From the energy profiles of ethane, propane, and n-butane, one may extract the useful eclipsing interactions summarized below: Hierarchy of Eclipsing Interactions

XX Y

Y H

E kcal mol -1 +1.0 +1.4 +3.1Energy Maxima Energy Minima Torsion angle 0 30 +60 30 +120 30 180 30 -120 30 -60 30

HH

Designation syn periplanar + syn-clinal + anti-clinal antiperiplanar - anti-clinal - syn-clinal

Symbol sp + sc (g+) + ac

H H

C

C H

H Me Me Me

n-Butane Conformer E2 G E1 A E1 G

ap (anti or t) - ac - sc (g-)

4-03-butane 9/26/03 1:51 PM

D. A. Evans n-Pentane

Acyclic Conformational Analysis: Pentane The double-gauche pentane conformationThe new high-energy conformation: (g+g)Me Hg-g-

Chem 206

Rotation about both the C2-C3 and C3-C4 bonds in either direction (+ or -):Me Me H Me Hg+g+ g+t

Me H Me Metg-

H Meg+g-

Me H

H Me

Me Me H

t,t

Me Me Me H H

H Me

Estimate of 1,3-Dimethyl Eclipsing Interaction Y

Metg+

Me H

Xg-t 1 g-g+

p. p.

1

3

5

3

5

1 (t,t) Anti(2,3)-Anti(3,4)

2 (g+t) Gauche(2,3)-Anti(3,4)

G = +5.5 kcal mol -1 G = X + 2Y where: X = 1,3(MeMe) & Y = 1,3(MeH) 1,3(MeH) = Skew-butane = 0.88 kcal mol -1 1,3(MeMe) = G 2Y = 5.5 1.76 = + 3.7 kcal mol -1

1 1 5 3

1,3(MeMe) = + 3.7 kcal mol -1

Estimates of In-Plane 1,2 &1,3-Dimethyl Eclipsing Interactions3

4 (g+g) Gauche(2,3)-Gauche'(3,4) double gauche pentane

3 (g+g+)

5

Me

Me

Me

Me

Me

Me

Me

Me

Gauche(2,3)-Gauche(3,4) 3.1 ~ 3.7 ~3.9 ~ 7.6

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 is relative energy can be evaluated.

It may be concluded that in-plane 1,3(MeMe) interactions are Ca +4 kcal/mol while 1,2(MeMe) interactions are destabliizing by Ca 2.2 kcal/mol.

4-04-pentane 9/26/03 11:23 AM

D. A. Evans

Acyclic Conformational Analysis: Natural Products Lactol & Ketol Polyether Antibioitics

Chem 206

The syn-Pentane Interaction - ConsequencesR. W. Hoffmann, Angew. Chem. Int. Ed. Engl. 2000, 39, 2054-2070 Conformation Design of Open-Chain Compounds (handout)

The conformation of these structures are strongly influenced by the acyclic stereocentersMe O HO R O OH H Me Me OH O H O O Me OH Et Et OH Me

R Me R Me Me Me

R'

R

R'or

Me H RR'g -g -

Me H R'

Me H H Mett

Me

Me

Et

R'

R

Meor

Me

Me H H R'tg

H H R Hgt

Ferensimycin B, R = Me Lysocellin, R = H

Consequences for the preferred conformation of polyketide natural products p. p.Analyze the conformation found in the crystal state of a bourgeanic acid derivative!

The conformation of these structures are strongly influenced by the acyclic stereocenters and internal H-bondingAlborixin R = Me; X-206 R = H

Internal H-BondingOH Me Me Me Me O O OR Bourgeanic acid Me Me C O OH R Me OH H Me O OH O OH OH O Me Me Me H O Me Me OH O Et Me OH

Metal ion ligation sites (M = Ag, K)Me Me O C O O R Me OH H Me O OH O OH OH O Me Me Me H O Me Me OH O Et Me OH

M

4-05-Natural Products 9/22/03 8:42 AM

D. A. Evans

Conformational Analysis: Ionophore X-206/X-rays X-ray of Ionophore X-206 H2O

Chem 206

X-ray of Ionophore X-206 - Ag+ - ComplexMetal ion ligation sites (M = Ag, K)H Me Me OH Me O C O O R Me OH H Me O OH O OH OH O Me Me Me H O Me Me OH O Et Me OH

Internal H-BondingMe Me O C O OH Me OH H Me O OH O OH OH O Me Me Me O Me Me OH O Et

M

p. p.

"The Total Synthesis of the Polyether Antibiotic X-206". Evans, D. A.; Bender, S. L.; Morris, J. J. Am. Chem. Soc. 1988, 110, 2506-2526.

4-06-X-206 conformation 9/22/03 8:42 AM

D. A. Evans

Conformational Analysis: Ionophore X-206/X-ray overlay

Chem 206

p. p.

4-07-X-206 overlay 9/19/01 11:57 AM

D. A. Evans

Stabilized Eclipsed Conformations in Simple Olefins Butane versus 1-ButeneMe H C H H H Me H H Me Me

Chem 206

Simple olefins exhibit unusal conformational properties relative to their saturated counterpartsPropane versus Propene109 H H H H Me H H H H H 120 CH2


C H H


G = +4 kcal mol-1

Me

Hybridilzation change opens up the CCC bond angle The Propylene Barrierp. p.H H C H CH2 H H CH2 C H


CH2 C H H H

H H

Me CH2 C H


= 50 staggered conformation

G = 0.83 kcal mol-1

=0

The Torsional Energy Profile = 50Me H C H H C H H

+2.0 kcal/mol

= 180H H C H H C Me H


H

=0 New destabilizing effectH C H Me C

H H H +1.32 kcal H C H H C H Me H +1.33 kcal

H X H C H H

repulsive interaction between CX & CH

+0.49 kcal

= 120 = 180

H X C H H

=0

Conforms to ab initio (3-21G) values: Wiberg, K. B.; Martin, E. J. Am. Chem. Soc. 1985, 107, 5035.

Acetaldehyde exhibits a similar conformational biasO H H H H Me H H O H H H H O Me Me H H O Me

K. Wiberg, JACS 1985, 107, 5035-5041 K. Houk, JACS 1987, 109, 6591-6600

The low-energy conformation in each of above cases is eclisped

4-08-simple alkenes 9/22/03 8:54 AM

Useful Destabilizing Interactions to Remember

Hierarchy of Vicinal Eclipsing InteractionsXX Y

Y H Me Me

E kcal mol -1 +1.0 +1.4 +3.1

HH

H H

C

C H

H Me

Estimates of In-Plane 1,2 &1,3-Dimethyl Eclipsing InteractionsMe Me Me Me Me Me Me Me

~ 3.1

~ 3.7

~3.9

~ 7.6

It may be concluded that in-plane 1,3(MeMe) interactions are Ca +4 kcal/mol while 1,2(MeMe) interactions are destabliizing by Ca +3 kcal/mol.

0000 0000 000 0000 0000 0000 000 0000 0000 0000 000 0000 00 0 00 0 0 00 0 0 00 0 0000 00000000 0000 0000 00000000 0000 0000 00000000 0000

minimized structure

4-09-Destabilizing Effects 9/20/01 5:33 PM

D. A. Evans

Acyclic Conformational Analysis-2 Problems of the Day:

Chem 206


Can you predict the stereochemical outcome of this reaction? O OTs MeLiNR2

OLi EtO n-C4H9 H Me

OTs 1


EtO n-C4H9 H

+ 98:2

2

D. Kim & Co-workers, Tetrahedron Lett. 1986, 27, 943.

Acyclic Conformational Analysis-2 Conformations of Simple Olefinic Substrates Introduction to Allylic Strain Introduction to Allylic Strain-2: Amides and EnolatesO Me CH2OBn MeB2H6 H2O2

OH O Me

CH2OBn

Me

diastereoselection 8:1

Reading Assignment for week A. Carey & Sundberg: Part A; Chapters 2 & 3R. W. Hoffmann, Angew. Chem. Int. Ed. Engl. 2000, 39, 2054-2070 Conformation Design of Open-Chain Compounds (handout) R. W. Hoffmann, Chem. Rev. 1989, 89, 1841-1860 Allylic 1-3-Strain as a Controlling Element in Stereoselective Transformations (handout) F. Weinhold, Nature 2001, 411, 539-541 "A New Twist on Molecular Shape" (handout)

Y. Kishi & Co-workers, J. Am. Chem. Soc. 1979, 101, 259.

Me

MePhNCO Et3N

Me

Me

O N only one isomer

NO2

A. Kozikowski & Co-workers, Tetrahedron Lett. 1982, 23, 2081.

D. A. Evans

Wednesday, September 24, 2001

5-00-Cover Page 9/24/03 8:46 AM

D. A. Evans

Stabilized Eclipsed Conformations in Simple Olefins Butane versus 1-ButeneMe H C H H H Me H H Me Me

Chem 206

Simple olefins exhibit unusal conformational properties relative to their saturated counterpartsPropane versus Propene109 H H H H Me H H H H H 120 CH2


C H H


G = +4 kcal mol-1

Me

Hybridilzation change opens up the CCC bond angle The Propylene Barrierp. p.H H C H CH2 H H CH2 C H


CH2 C H H H

H H

Me CH2 C H


= 50 staggered conformation

G = 0.83 kcal mol-1

=0

The Torsional Energy Profile = 50Me H C H H C H H

+2.0 kcal/mol

= 180H H C H H C Me H


H

=0 New destabilizing effectH C H Me C

H H H +1.32 kcal H C H H C H Me H +1.33 kcal

H X H C H H

repulsive interaction between CX & CH

+0.49 kcal

= 120 = 180

H X C H H

=0


Acetaldehyde exhibits a similar conformational biasO H H H H Me H H O H H H H O Me Me H H O Me

K. Wiberg, JACS 1985, 107, 5035-5041 K. Houk, JACS 1987, 109, 6591-6600

The low-energy conformation in each of above cases is eclisped

5-01-simple alkenes 9/22/03 8:54 AM

Evans, Duffy, & Ripin5

Conformational Barriers to Rotation: Olefin A-1,2 Interactions5

Chem 206

1-butene4

2-propen-1-olH H H C C OH H H

H H H C C Me

H H

4

E (kcal/mol)

3

E (kcal/mol)

3

2

2

1

1

0 -180

-90

0

90

180

0 -180

-90

0

90

180

(Deg)The Torsional Energy Profile = 50Me H H H C H C H H H C H

(Deg)The Torsional Energy ProfileH H C H C OH H

= 180H C Me H H H

H

= 60HO C H C H H

= 180

=0C H Me H C

H H H

= 120H C H C

OH H H

+1.32 kcalH H C H C

Me H H

+1.33 kcalH

+2.00 kcal

=0C H HO C

H H H

H

+1.18 kcal

+0.49 kcal=0

= 120 = 180

+0.37 kcal = 180

=0


5-02-1-butene & 2-propen-1-ol 9/23/03 2:59 PM


Conformational Barriers to Rotation: Olefin A-1,2 Interactions-25

Chem 206

2-methyl-1-butene4

2-methyl-2-propen-1-ol4

H H H C C Me

H Me

H H H C C OH

H Me

E (kcal/mol)

E (kcal/mol)

3

3

2

2

1

1

0 -180

-90

0

90

180

0 -180

-90

0

90

180

(Deg)The Torsional Energy Profile = 180 = 50Me H H H C H C Me H H C H H H C H C H Me Me

(Deg)The Torsional Energy Profile = 180 = 60HO C Me H H H H C H C OH Me

= 120OH

=0C H Me =0 H C

= 110H Me H

+1.39 kcalH H C H C

Me Me H

+2.68 kcal

=0C H HO = 180 =0 H C

H Me H

+1.16 kcal

H H

+2.01 kcalMe

C H

C H

+0.06 kcal

+0.21 kcal = 180

5-03-methylbutene etal 9/23/03 3:00 PM


Conformational Barriers to Rotation: Olefin A-1,3 Interactions5

Chem 206

(Z)-2-pentene4 4

(Z)-2-buten-1-olH Me H C C OH H H

H Me

H

E (kcal/mol)

3

E (kcal/mol)

C

C Me

H

H

3

2

2

1

1

0 -180

-90

0

90

180

0 -180

-90

0

90

180

(Deg)H Me C H Me C H H

(Deg)The Torsional Energy Profile =0Me C H HO C H H H Me

The Torsional Energy Profile =0

= 180H C H C H OH H

+3.88 kcal

= 180 = 90Me Me H C H C H H Me H H C H C Me H

+1.44 kcalMe

= 120OH C H C H H

+0.86 kcal

+0.52 kcal = 180 =0

H

=0

= 180

Values calculated using MM2 (molecular mechanics) force fields via the Macromodel multiconformation search. 5-04-2-pentene/ z-2-buten-1-ol 9/23/03 3:00 PM

Review: Hoffman, R. W. Chem. Rev. 1989, 89, 1841.


Conformational Barriers to Rotation: Olefin A-1,3 Interactions-2OH Me Rotate clockwise C Me C H H

Chem 206

(Z)-2-hydroxy-3-pentene4

Me OH Me

H

4.6 kcal/mol

HO

H C Me H

E (kcal/mol)

Me3

C

H

Me

100 H OH

2

Me H

C H

C

0.3-0.4 kcal/mol

1

H Me H C C OH-90 0 90 180

Me H

30

0 -180

Lowest energy conformerMe Me Me OH H H C H

(Deg)The Torsional Energy ProfileOH

60 Me

=0H Me C H HO C H Me

Me C H Me

2.7 kcal/mol

C OH

C

H H

= -140

Needs to be redone = 110Me Me C H H H OH C H OH

+2.72

????? = -80HO Me C H C Me H H

+4.68

A(1,3) interaction 4.0 kcal/molR23

HO H H C C Me

H Me

30

= 80 MeMe C C H H

Y X2

R1

= 150H C C Me H

R3

*1

R large

+0.66H

Me

R small

Lowest energy conformer

+0.34=0

+0.40 kcal

HO = 180

A(1,2) interaction 2.7 kcal/mol (MM2)

5-05-z-2-hydroxy-3-pentene 9/23/03 6:22 PM

D. A. Evans

Acyclic Conformational Analysis: Allylic Strain The Definition of Allylic Strain

Chem 206

Can you predict the stereochemical outcome of this reaction?D. Kim & Co-workers, Tetrahedron Lett. 1986, 27, 943. O EtO R large 1 R small n-C4H9 H Me OTsLiNR2

F. Johnson, Chem. Rev. 1968, 68, 375; Allylic Strain in Six-Membered Rings R. W. Hoffmann, Chem. Rev. 1989, 89, 1841-1860 (handout) Allylic 1-3-Strain as a Controlling Element in Stereoselective Transformations Houk, Hoffmann JACS 1991, 113, 5006

Consider the illustrated general structure where X & Y are permutations of C, N, and O:

R2

3 Y X 2

R1

OLi EtO n-C4H9 O H Me Me

OTs 1

+ 98:2

2

R3

Typical examples:R2 R3 R1

R2 R large N

R1

R2 R large R + N

R1

R2 R large

R1 N

Relevant enolateconformationsR large 1

EtO n-C4H9 H

major

O +

p. p.

R small Olefin

R small Imine

R small Imonium ion

R small Nitrone TsO(H2C)4 Me C Bu H C OR OLi Me Bu C C H (CH2)4OTs OR OLi Bu Me (CH2)4OTs OR C C OLi

In the above examples, the resident allylic stereocenter () and its associated substituents frequently impart a pronounced bias towards reactions occuring at the pi-bond. A(1,3) interaction R2 3 R1 Nonbonding interactions between the allylic Y substituents (Rlarge, Rsmall) & substituents at X R large the 2- & 3-positions play a critical role in R3 2 1 defining the stereochemical course of such R small reactions A(1,2) interaction Representative Reactions controlled by Allylic Strain InteractionsHO H Me HO O R OBnHg(OAc)2 NaBH4

A1

B1critical conformations

C1 H

H OR Me TsO(H2C)4 C C Bu OLi Me

H C

Bu C

Bu OR OLi Me H C C OR OLi (CH2)4OTs

(CH2)4OTs

A2

B2

C2

H

O

R EtO2C OBn 2 n-C4H9

Me

HO Me

minorH

diastereoselection 10:1 M. Isobe & Co-workers, Tetrahedron Lett. 1985, 26, 5199.

5-06-Allylic Strain-1 9/23/03 6:24 PM

D. A. EvansO EtO n-C4H9 H O EtO n-C4H9 HLiNR2 MeI

Allylic Strain & Enolate Diastereoface SelectionOTs MeLiNR2

Chem 206PhNH4Cl

O EtO n-C4H9 O EtO n-C4H9

Me diastereoselection 98:2 H Me3Si

Ph

OM OMe RR-substituent R = Me diastereoselection 87:13 80:20 40:60

O OMe R

Me3Si

Me diastereoselection 89:11 H Ph Me3Si O

R = Et R = CHMe2

major diastereomer opposite to that shown

I. Fleming & Co-workers, Chem. Commun. 1985, 318. Y. Yamamoto & Co-workers, Chem. Commun. 1984, 904.

D. Kim & Co-workers, Tetrahedron Lett. 1986, 27, 943.

PhLiNR2

H

O OBn diastereoselection 90:10 at C3one isomer at C2

OBn

MeCHO Me3Si 71% yield

MeO

p. p.

RO2C H CO2Me

O

MeOLiNR2

RO2C H H CO2Me

O "one isomer" Me Bn N Boc O N S SSn(OTf)2 RCHO 91-95%

Me

OH

I. Fleming & Co-workers, Chem. Commun. 1986, 1198.

Br95% yield

Me Bn N Boc R

H

O N OH

S S diastereoselection >95%

G. Stork & Co-workers, Tetrahedron Lett. 1987, 28, 2088.

TBSOCH2

Me CH2LiNR2 MeI

Me CH2 TBSOCH2 Me H H CO2Me "one isomer" Me OPh(MeS)2CLi

T. Mukaiyama & Co-workers, Chem. Letters 1986, 637

H CO2Me

OMe

MeI 86%

MeS MeS MeS

Me

O OMe diastereoselection 99:1

T. Money & Co-workers, Chem. Commun. 1986, 288.

Me

K. Koga & Co-workers, Tetrahedron Letters 1985, 26, 3031.

R PhMe2Si

OM OEt

RMeI

O OEt MeR = Me: diastereoselection 99:1 R = Ph: diastereoselection 97:3

PhMe2Si

I CO2Et R

OLi O-t-Bu

H

CO2Et

R = H: one isomer

I. Fleming & Co-workers, Chem. Commun. 1984, 28.

KOt-Bu THF -78 C

CO2-t-Bu R = Me: > 15 :1 H R

Y. Yamaguchi & Co-workers, Tetrahedron Letters 1985, 26,1723.

5-07-A-strain enolates 9/20/01 4:45 PM

D. A. Evanss The basic processS H HR

Allylic Strain & Olefin HydroborationHydroborations dominated by A(1,3) StrainS HR

Chem 206

HR

OH

B

H HR

B

H2 BR

H C RR

O Me Me

CH2 OBn

B2 H6 H2 O2

O Me Me

CH2 OBn

C

C R

C R R

C

C R

R

diastereoselection 8:1 OMe Me O B2 H6 H2 O2 OH O Me Me Me diastereoselection 12:1 OH OMe OH

s Response to steric effects: Here is a good calibration system:ACH2 Me3 C H

Me

Me

OxidantMCPBA BH3, H2 O2 E

Ratio, A:E69:31 34:66

ReferenceJOC, 1967, 32, 1363 JOC, 1970, 35, 2654 BnO Me

Y. Kishi & Co-workers, J. Am. Chem. Soc. 1979, 101, 259.

OH OH Me Me B2 H6 H2 O2 BnO Me Me Me OH

s Acyclic hydroboration can be controlled by A(1,3) interactions:OH RL RM Me OH R2 BH H2 O2 RL RM Me R R B C RL OH RL RM Me R2 BH H2 O2 OH RL RM Me R R RM B C RL H H C Me TrO CH2 OR OH H H C CH2 OR TrO

Diastereoselection = 3:1 C. H. Heathcock et. al. Tetrahedron Lett 1984 25 243. OH

major diastereomer

Me Me

control elements

A(1,3) allylic strain Steric effects; RL vs RM Staggered transition states

RM H

Me ThexylBH2 , OH then BH3 OTr TrO OH

Me

Me OTr OH OH 5:1

Diastereoselection;

major diastereomer

Me

Me

Me

Me

ThexylBH2 , then BH3 TrO

Me

Me

Me

Me OTr

OH

OH

OH

OH OTr

OH

OH

OH 4: 1

OH

Diastereoselection;

See Houk, Tetrahedron 1984, 40, 2257

H

Still, W.C.; Barrish, J. C. J. Am. Chem. Soc. 1983, 105, 2487.

5-07a-A-strain hydroboration 9/24/03 9:45 AM

D. A. EvansConsider the resonance structures of an amide:O R3 C N1

Allylic Strain & Amide ConformationR2 R33

Chem 206

Y X2

R1 R large1

The selection of amide protecting group may be done with the knowledge that altered conformational preferences may result:O O H H

R1 R R

O R3

C N +

R1 R1

R small

Favored for R = H, alkylH O H

R

N R

H

N R O

Favored for R = COR

A(1,3) interactions between the "allylic substituent" and the R1 moiety will strongly influence the torsion angle between N & C1. FavoredO Me C N Me

N H R H O N O H C R H H R

N

H

Disfavored

Me Me

O

Disfavored

N

H C O R

Favored

p. p.

conformations of cyclic amidesO R C N + R R C O H Me N R C Me O H R C O H N N R A(1,3)Chow Can. J. Chem. 1968, 46, 2821

H R C O N

R

H strongly favored Me Me

R O 1 A(1,3) interaction between the C2 & amide C R substituents will strongly influence the torsion N angle between C1 & C2. R R2

R O1 2

C

R

N R + R

H strongly favored

As a result, amides afford (Z) enolates under all conditionsH H

O Me Ph N O

O published X-ray structure of this amide shows chair Me diaxial conformationQuick, J. Org. Chem. 1978, 43, 2705

O Me

C

N H

L L

OM base favored O Me C N H L L H

Me

N L

L

Problem: Predict the stereochemical outcome of this cyclization.OH D. Hart, JACS 1980, 102, 397 Ph N O HOCOHCO2H

(Z)-EnolateH

H O H C L N Me L base disfavored O H

H L

OM C N Me L H Me N L L

N Ph O diastereoselection >95%

identify HOMO-LUMO pair

5-08-A-strain Amides-1 9/23/03 6:09 PM

(E)-Enolate

D. A. Evans

Allylic Strain & Amide Conformation

Chem 206

A(1,3) Strain and Chiral Enolate DesignO Me N Bn H L L El O Me N Bn O LDA O or NaNTMS2 Me O M O O

Polypropionate Biosynthesis: The Acylation EventO O SR O SR Me O SR Me OH O SR Me

N Bn

enolization selectivity >100:1HO

R O

Acylation CO2

R

Reduction

R

30802341 harvard s advanced organic chemistry

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