D. A. EvansAn Introduction to Frontier Molecular Orbital
Theory-1 Problems of the DayChem
206http://www.courses.fas.harvard.edu/~chem206/
http://evans.harvard.edu/problems/Chemistry 206 Advanced Organic
ChemistryLecture Number 1The 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 1Br+N
O H Br O2+N O H Br BrIntroduction 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 EffectsThis 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 CRank the
phosphites from the least to the most nucleophilic and provide a
concise explanation for your predicted reactivity order.Monday,
September 15, 2003D. A. EvansAn Introduction to Frontier Molecular
Orbital Theory-1 Stereoelectronic EffectsChem 206Universal 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 RGeometrical 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:RC RBrBr:RO HOOmajor General Reaction
TypesOp. p.minorRadical Reactions (~10%): Polar Reactions (~90%):A
+ B A(:) + B(+)Lewis AcidA AB 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 polarizationLewis BaseFMO concepts extend the
donor-acceptor paradigm to non-obvious families of reactions
Examples to considerMe R C R BrS N1+R R C Me + Br:H2+2 Li(0) Mg(0)2
LiH CH3MgBrCH3I +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-19/12/03 4:44 PMD. A. EvansSteric
Versus Electronic Effects; A time to be careful!!O R3SiO EtO O R3Si
OSiR3 O TiCl4 Nu OSiR3Chem 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 OSiR3OOAcSnBr4 Me SiMe3O
MeMeAlCl3stereoselection 99:1 stereoselection >95:5
OSiR3OOAcSnBr4 BnOOO Danishefsky et al JOC 1991, 56, 387 BnOp.
p.BnOO 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 PhBu3AlR3 O Al
OEtO2CEtOYakura's rationalization:Mehta et al, Acc Chem. Res. 2000,
33, 278-2861-03-Introduction-1a9/15/03 8:14 AMD. A. EvansThe H2
Molecular Orbitals & Antibonds The H2 Molecule (again!!)Chem
206Linear 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*22Let'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 2The coefficients, C1 and C2, represent the
contribution of each AO. Rule Three: (C1)2 + (C2)2 = 1The 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) E1EnergyCO (antibonding)H 11s E21sH 2
Energy (bonding)CONote that E1 is greater than E2. Why?C O
(bonding)1-04-Introduction-2 9/15/03 8:38 AMD. A. EvansBonding
GeneralizationsChem 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 BWhen
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:ABBetter thanABUseful
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 Cbetter thanCSiCSiCase-1:
Anti Nonbonding electron pair & CX bondXSi-SP3X Alone pair
HOMOC-SP3 CCC-SP3A C* CX LUMOX Better thanlone pair HOMO* CX
LUMOC-SP3 CSiCACH3CCH3 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 HOMOThis 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
YY* CX LUMOX CCBetter thanX C CY HOMO* CX LUMOCY1-05-Introduction-3
9/12/03 4:36 PMD. A. EvansDonor-Acceptor Properties of Bonding and
Antibonding StatesChem 206Donor 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-OHierarchy 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 CH3OHC-SP3C-SP3 O-SP3very close!!decreasing -donor capacity
C-C C-OCH3Fpoorest donorp. 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)CH3HFor the latest views,
please read Alabugin & Zeidan, JACS 2002, 124, 3175 (pdf)CH3CH3
CH3NH2 CH3OH CH3FDonor Acceptor Properties of CSP3-CSP3 &
CSP3-CSP2 Bonds* CC * CC better acceptor Increasing -acceptor
capacitybest acceptorC-SP3C-SP3 C-SP2The following are trends for
the energy levels of nonbonding states of several common molecules.
Trend was established by photoelectron spectroscopy.Nonbonding
States CCbetter donor CCH3P: 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 PMD.
A. EvansHybridization vs ElectronegativityChem 206Electrons 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 electronegativity5N4.5SPRadial
ProbabilityPauling ElectronegativityRadial Probability1 S Orbital4N
N3.5SP3SP2C 3SP2 S Orbital2 S Orbital 2 P OrbitalCp. p.3 S Orbital
3 P Orbital2.5SP2CSP322025303540455055% S-CharacterThere is a
direct relationship between %S character & hydrocarbon
acidity60CH (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 Acid5545045C H (44)6
640CSP3CSP2CSP35PhCC-H (29)30The 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 PM25 20 25 30 35 40 45 50 55% S-CharacterD. A.
EvansHyperconjugation: Carbocation StabilizationChem 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 CPhysical
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+CR+ H H H H C C H HThe 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]+CStereoelectronic Requirement for Hyperconjugation:
Syn-planar orientation between interacting orbitals100.6 1.608 MeMe
MeThe Molecular Orbital Description CR CRT. Laube, Angew. Chem.
Int. Ed. 1986, 25, 349+CH H+CH HThe Adamantane Reference
(MM-2)H1.528 CR CR110 Me Me Me1.530 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 PMD. A.
Evans"Negative" HyperconjugationSyn OrientationRChem 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 CR C XR H H H HH X H H
HCXfilled hybrid orbitalXCThis decloalization is referred to as
"Negative" hyperconjugationAnti OrientationRSince 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 HCXH HCThe Molecular Orbital Description CR Overlap
between two orbitals is better in the anti orientation as stated in
"Bonding Generalizations" handout.XNonbonding e pairThe Expected
Structural PerturbationsChange in Structure Shorter CX bond Longer
CR bondSpectroscopic ProbeX-ray crystallography X-ray
crystallography Infrared Spectroscopy Infrared Spectroscopy NMR
Spectroscopy NMR Spectroscopy CRAs 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
X1-09-Neg-Hyperconj 9/12/03 4:53 PMD. A. EvansLone Pair
Delocalization: N2F2The trans IsomerChem 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-SP2N FNF 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.NNABforms
stronger pi-bond thanAB 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 PMABforms stronger sigma-bond thanABThis 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. EvansLone Pair Delocalization: The Gauche EffectChem 206The
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 HantiThe
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 HH HH N H Hgaucheobserved HNNH
dihedral angle Ca 90HThere 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 OO Hgauche Hobserved HOOH dihedral angle Ca 90H
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-SP3filled 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 PMD. A. EvansThe Anomeric
Effect: Negative HyperconjugationUseful LIterature ReviewsChem
206http://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.Chemistry 206 Advanced Organic
ChemistryLecture Number 2Stereoelectronic 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
121http://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.
KCNReading Assignment: Kirby, Chapters 1-3S BF4 EtArel. rate =
8000SEt+ PhCH2CN(1)D. A. EvansWednesday, September 17, 2003KCN S Et
B rel. rate = 1 S+MeCH2CN(2)2-00-Cover Page 9/17/03 8:35 AMD. A.
EvansThe Anomeric EffectThe Anomeric Effect: Negative
HyperconjugationChem 206It is not unexpected that the methoxyl
substituent on a cyclohexane ring prefers to adopt the equatorial
conformation. H OMe HOMe 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 CClThat 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 GcO ClH
O HOMOThe Exo-Anomeric Effect CClA = 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 OOOOMeOH OMeaxial O lone pair CHaxial O lone pair COA.
J. Kirby, The Anomeric and Related Stereoelectronic Effects at
Oxygen, Springer-Verlag, 1983 E. Jurasti, G. Cuevas, The Anomeric
Effect, CRC Press, 19952-01-Anomeric Effect-1 9/16/03 2:40 PMD. A.
EvansThe Anomeric Effect: Carbonyl GroupsAldehyde CH Infrared
Stretching FrequenciesChem 206Do the following valence bond
resonance structures have meaning?R C X O X R C OPrediction: 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 CF3-1R C C RPrediction:
As X becomes more electronegative, the IR frequency should increase
O Me CH3 Me O CBr3 Me OO H CH = 3050 cm -1C=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.172017501780Infrared
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 NHPrediction: As the indicated pi-bonding increases,
the XCO bond angle should decrease. This distortion improves
overlap.R C X * CX O lone pair OR C X ONN NH = 2188 cm -1 Me N N H
NH = 2317 cm -1Me Nfilled N-SP2Nantibonding NH HEvidence 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
PMD. A. EvansThe Anomeric Effect: Nitrogen-Based SystemsCMe3 Me3C
CH N N N CMe3 Me3C N N N Me3CChem 206Observation: CH bonds
anti-periplanar to nitrogen lone pairs are spectroscopically
distinct from their equatorial CH bond counterpartsH H H HN HMe3C G
= 0.35kcal/molN HOMO CHA. R. Katritzky et. al., J. Chemm. Soc. B
1970 135Favored Solution Structure (NMR)Me MeN NMe NMe Me N Me J.
E. Anderson, J. D. Roberts, JACS 1967 96 4186 N N N Me
MeNSpectroscopic 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 PMFavored
Solid State Structure (X-ray crystallography)1.484 1.453Bn N
Me1.457Me1.453N N Bn1.459NA. R. Katrizky et. al., J. C. S. Perkin
II 1980 1733D. A. EvansAnomeric Effects in DNA PhosphodiestersChem
206Calculated Structure of ACGTGC DuplexThe Phospho-Diesters
Excised from Crystal StructureGuanine Cytosine Cytosine1B2B 1A
Phosphate-1A p. p.Thymine AdeninePhosphate-1BThe Anomeric
EffectAcceptor orbital hierarchy: * POR * > * POR OR OP OO O RPO
O RO R P O O RPhosphate-2A Phosphate-2BGauche-Gauche conformation
OR P OO O R O OAnti-Anti conformation Gauche-Gauche conformation
affords a better donor-acceptor relationshipOxygen 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
AMD. A. EvansCarboxylic Acids (& Esters): Anomeric Effects
Again? Hyperconjugation: (Z) ConformerO O R' O Me H O MeChem 206
Conformations: There are 2 planar conformations.O O O O H R' RLet
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 FormateR(E) ConformerRG = +4.8 kcal/molOC
ROIn 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 OIn the
(E) conformation this lone pair is aligned to overlap with * CR.*
CROR O Cbarrier ~ 10-12 kcal/molO R O R O RG ~ 2-3 kcal/molSince *
CO is a better acceptor than * CR (where R is a carbon substituent)
it follows that the (Z) conformation is stabilized by this
interaction.RO RO R R OEsters 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 OC ROThe 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 PMD. A.
EvansThree-Center BondsCase 3: 2 p-Orbitals; 1 s-orbitalChem
206Consider 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-orientationantibondingantibonding2+nonbondingEnergy3nonbondingbondingCase
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 stateDo this
as an exerciseExamples of three-center bonds in organic chemistry
A. H-bonds: (3center, 4electron)O CH3 O H O H O CH3CH2Allyl
Radical: 2 electrons in bonding obital plus one in nonbonding
MO.The acetic acid dimer is stabilized by ca 15 kcal/molH 2CCH
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 HCase 2: 3 p-Orbitals sigma-orientation 3nonbonding
antibondingHdiborane stabilized by 35 kcal/molEnergyC. 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 * CBrbonding2-06 3-center bonds/review
10/28/03 12:00 PMD. A. EvansSubstitution Reactions: General
ConsiderationsChem 206Why do SN2 Reactions proceed with backside
displacement?R Nu: H H C X R Nu XElectrophilic substitution at
saturated carbon may occur with either inversion of retention
Inversion + MR Nu C H H X:C H HRa El(+) H Rb C M+ NuRa C HRa 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.RbInversionR NuRetentionRa El(+) Ra C M H Rb
C M+ RetentionR X H H CRa H Rb C El M+C H HLUMOXH RbEl
+HOMOConstructive overlap between Nu & *CXNu Overlap from this
geometry results in no net bonding interaction El(+)LUMORa Ra C H
RbC M H RbMExpanded view of *CXHOMOInversionLUMOEl(+)Cantibonding
HOMOXbonding Br2 H BrRetention ExamplesLi HCO2 CO2Li HNuFleming,
page 75-76predominant inversionpredominant retentionStereochemistry
frequently determined by electrophile structureSee A. Basu, Angew.
Chem. Int. Ed. 2002, 41, 717-7382-07-SN2-1 9/18/03 12:38 PMD. A.
EvansThe reaction under discussion:R R C H XSN2 Reaction:
Stereoelectronic EffectsThe use of isotope labels to probe
mechanism. C H H X R Nu C H HChem 206Nu: HNuX: 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 3CNu:exclusively
intramolecular(CD3ArNuCH3)CH31O SO O CD3SO3 H 3C Nu
(CH3ArNuCD3energyNu: C XMeNu:exclusively intramolecularCD3CX
RCH2XNu 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 R2 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)3NNu:C XC H H XH HNuD6-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+SO3CH316% intramolecular 84% intermolecularSO3CH3
N(CH3)2SO3Nu:N(CH3)3+Nu2-08-The SN2 RXN-FMO 9/16/03 2:56 PMHence,
the NuCX 180 transition state bond angle must be rigidly maintained
for the reaction to take place.D. A. EvansSO3CH3
N(CH3)2Intramolecular methyl transfer: Speculation on the
transition structures(CH3)2NChem 206+SO3SO3SO3CH3N(CH3)3+(CH3)3N16%
intramolecular; 84% intermolecular9- membered cyclic transition
stateexclusively intermolecular8- membered cyclic transition
state174 1742-09-Intra alk TS's 9/16/03 2:56 PM00000 00000 00000
00000 00000est CO bond length 2.1 000000 000000 000000 000000
000000est CO bond length 2.1 000000 00000 000000 00000 00 00000 00
00000 00 00 00 00 0000 00 00 0000 00000 0000 00000est 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
OOlefin Epoxidation via Peracids: An Introduction Per-arachidonic
acid EpoxidationOChem 206R O OH RR+R R RR+R OHMe OO HHOMO CCLUMO
*OOnote labeled oxygen is transferfedO-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 OAc1.00.60.050.4 The indicated olefin in each
of the diolefinic substrates may be oxidized selectively.Me Me Me
Me Me Me MeH Me Me The transition state:MeO HOE. 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
PMFor theoretical studies of TS see R. D. Bach, JACS 1991, 113,
2338 R. D. Bach, J. Org. Chem 2000, 65, 6715D. A. Evans The General
Reaction:R ROlefin Epoxidation with DioxiranesAsymmetric
Epoxidation with Chiral KetonesR R O R RChem 206+R RR ROReview:
Frohn & Shi, Syn Lett 2000, 1979-2000Me O Me O O O Me O +R R
note labeled oxygen is transferfedHOMO CCLUMO *OOchiral catalystO
MeOO-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 SO3O R2 R1 R2oxone,
CH3CN-H2O pH 7-8Me Me PhO R O(Oxone)PhPhPhPh>95% ee84% ee92%
eeSynthetically Useful Dioxirane SynthesisO Me O F 3C CF3
MeQuestion: 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 10.5oxoneO MeO Me O
CF3co-distill to give ~0.1 M soln of dioxirane in acetone
co-distill to give ~0.6 M soln of dioxirane in
hexafluoroacetoneoxoneO F 3CMeMe MeO O1 (1) 2Me OCurci, JOC, 1980,
4758 & 1988, 3890; JACS 1991, 7654. Transition State for the
Dioxirane Mediated Olefin EpoxidationO R R O O R ROOO R2(2)O Me O
MeOR1R2planar>90% eeOspirorotate 90Part 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 PMD. A. EvansO C RL RS + RCO3H RCO2HThe
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- MeCO2HChem
206CMe3 Me O H O R O OLet RL and RS be Sterically large and small
substituents.+C RLRThe 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 MeHFavoredCMe3 O Me O OH O RMigrating
groupmajorCH3CH2 CH3(CH2)2 (CH3)3C72 150 830 >2000 Conformer AHO
Me O CMe3kMeC R OMeminorPhCH2RS OH C RL O ODisfavoredMe OMigrating
groupThe IntermediateRO OH Me O CMe3OMe3C OThe important
stereoelectronic components to this rearrangement:RO1. 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 PMConformer BThe
destabilizing gauche interactionSteric 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. EvansThe Baeyer-Villiger Reaction:
Stereoelectronic EffectsCMe3 MeChem 206Conformer A in three
dimensionsO Me+ RCO3H CMe3Me3C O HOMe O- MeCO2HO O H O R
OORFavoredHMigrating group CMe3 O O OH O R O Me O CMe3MeConformer
AHDisfavoredMe O Me3C O R OH OMigrating groupO Me1OCMe32 3Conformer
BThe destabilizing gauche interaction4Steric 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
PM23 dihedral angle ~ 178 from Chem 3DB. BreitFMO-Theory/HSAB
Principle 1Hard and Soft Acids and Bases (HSAB-Principle)Reading
Assignment: Fleming, Chapter 3, p33-46 Pearson, JACS 1963, 85,
3533.Chem 206FMO-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 anidealized Soft SpeciesE E small HOMO/LUMO
gapidealized Hard Specieslarge HOMO/LUMO gapFMO-Theory for
interaction:Soft-SoftE EHard-HardAcidBase Acid BaseSignificant
Energy gain through HOMO/LUMO interactionOnly neglectable energy
gain through orbital interaction.2-14-FMO HSAB 1 9/20/00 8:30 AMB.
BreitFMO-Theory/HSAB Principle 2Chem 206Klopman-Salem Equation for
the interaction of a Nucleophile N (Lewis-Base) and an Electrophile
E (Lewis-Acid).E =QNQE RNE Coulomb Term2(cNcE)2 EHOMO(N) - ELUMO(E)
Frontier Orbital TermQ: Charge density : Dielectricity constant R:
distance (N-E) c: coefficient of MO : Resonance Integral E: Energy
of MOSoft-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 AMB. BreitFMO-Theory/HSAB Principle 3HSAB
principle - Application to Chemoselectivity Issues(c) SN2 vs E2Chem
206(a) Enolate AlkylationO softCO2Rhard O soft C Csoft
MeIC-AlkylationMe OTMSHC(COOR)2 H Br hard OC2H5CO2RS N2hard
TMSClE2O-Alkylation(d) Ambident Nucleophiles (b) 1,2- vs.
1,4-addition to ,-unsaturated carbonyl compounds + 0.29O H H-
0.48Osoft Ag S C N hard Nasoft MeIH3 CSC ONS-Alkylation+ 0.01
Charge density+ 0.62 LUMO-coefficientshard RCOXSCNRN-Acylationhard
softOhard MeLiOH Me O Ag soft MeI H3C NO2N-Alkylation1,2-Addition
soft Me2CuLi Conjugate AdditionMe OO hardN Na soft hard t-BuCl
ONOO-Alkylation2-16-FMO HSAB 3 9/20/00 8:27 AMD. A. EvansRules for
Ring Closure: IntroductionChem
206http://www.courses.fas.harvard.edu/~chem206/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 ChemistryLecture Number 3Stereoelectronic
Effects-3 "Rules for Ring Closure: Baldwin's Rules"Kirby,
"Stereoelectronic Effects" Chapters 4, 5Useful 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 OD. A. Evans3-00-Cover
Page 9/19/03 8:36 AMFriday, September 19, 2003MeNH2NH2 OMe MeHN NH
OD. A. Evans, J. JohnsonRules for Ring Closure: IntroductionChem
206Ring 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 YendoC. 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-tetX
YXY5-exo-trigX Y5-exo-digXYXYX YXY 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.exoRequired
trajectories (Baldwin):X Y = 180 X YX XXYY Y B. Endo-cyclization
modes identified by the breaking bond being positioned endocyclic
to the forming cycle.endo = 109 XYWill come back to this case
laterX Y X = first-row element N, OX * 120 *XYXY YBaldwin, J. Chem.
Soc., Chem. Commun., 1976, 734.3-01-Baldwin Rules-1 9/18/03 3:38
PMD. A. Evans, J. JohnsonRules for Ring Closure: SP3 Carbon &
Related SystemsFRST-PLATTNER RULEChem 206Tetrahedral CarbonAll exo
cyclization modes are allowed: (n-exo-tet, n = 3)exoIn 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.XC YXC YThere are stereoelectronic issues to
consider for n-exo-tet cyclizations Formation of 3-Membered Rings
(3-exo-tet)H H HYX C H2CYXCXCH2 C H2C H2H+ YStereoelectronic
Effects in Epoxide Ring CleavageNu Me3C NuO H O Me3C H HO HO NuH Me
Me3C H H Nu Me H HConformational 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 HY fasterH H H Me3C OY
1OslowerH ONu-Nu HOO H Hchairboat2"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 PMD. A. Evans, J.
JohnsonRules for Ring Closure: SP3 Carbon & Related SystemsCase
2: King, J.C.S. Chem. Comm., 1979, 1140.O S O Me NMe2 O OChem
206Tetrahedral 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% intramolecularO
OS_NMe3+endoXYCY C(SP3)XSO2OMe NMe2SO3 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 NMe2SO3 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 evaluatedCY3CY3O
Cl OOH n-Beak, JACS 1991, 113, 6281. 8-endo-tet disfavoredCl
CO2HCyclization exclusively intermolecular. However the exocyclic
analog is exclusively intramolecularO S O O CX2I O S O O S O OnONaH
6-exo-tet favored Rxn exclusively intramolecularO SCX2 On = 1: rxn
exclusively intermolecular n = 9: rxn is intramolecularBeak 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.CY3CY33-03-Baldwin Rules-3 9/18/03 4:07 PMD. A. Evans, J.
JohnsonRules for Ring Closure: SP2 Carbon & Related SystemsChem
206Trigonal CarbonEndo cyclization modes that are disallowed (3 to
5-endo-trig)n-endo-trig X Y X C Y MeO2C NH2 CO2MeCX = 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 baseXCO2Me OhoweverCO2Me SH5-endo-trig
Disfavored baseIt is possible that a "nonvertical" trajectory is
operational like that suspected in C=O additionCO2MeSSecond row
atom relaxes the cyclization geometrical requirement Case 2:
Baldwin, J. Chem. Soc., Chem. Commun., 1976, 736.MeO2C NH2
CO2MeXMeO2C HNCO2Me5-endo-trig 0%MeO2C HN O5-exo-trig
100%3-04-Baldwin Rules-4 9/18/03 4:07 PMD. A. Evans, J. JohnsonCase
2: continued...MeO2C NH2 CO2MeXRules for Ring Closure: SP2 Carbon
& Related SystemsChem 206Apparent exceptions to disallowed
5-endo-trig cyclization processMeO2C HN CO2MeO5-endo-trig 0%MeO2C
HN ONCH3CO2H+ NOHNOFiler, J. Am. Chem. Soc. 1979, 44,
285.5-exo-trig 100%R HC N1CO2Me KO Bu R2 CO2MetCO2Me R1 HN R
CO2Me2CO2Me R 3:11Control experiment: Intermolecular reaction
favors conjugate addtion.Me CO2Me PhCH2NH2 H N Me CO2Me 100% Me H N
O 0% PhHNCO2Me R2PhR1 = aryl, R2 = aryl, alkylGrigg, J. Chem. Soc.,
Chem. Commun. 1980, 648.Case 3:O Ph OMe NH2NH2 65 oC Ph HN NH ODoes
the illustrated ketalization process necessarily violate "the
rules"?R O R(CH2OH)2 H+R RO OO Ph MeI OK1) EtO2CCl, pyridine 2)
NH2NH2Ph NHOR O R(CH2OH)2R HOR OOHH+ H2OR R OOH( )2H 2N 200 oC O Ph
OMe NH2NH2 65 oC HN X+( )25-endo-trig5-endo-trigH+R Odisfavored ?R
R O OR Ph CO2Me NH2 Ph O HN NH HO+OH 5-exo-tet( )2favored
?5-exo-trigJohnson, 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 PMD.
A. Evans, J. JohnsonMore ExceptionsRules for Ring Closure: SP2
Carbon & Related SystemsBuChem 206BuZard, Org. Lett. 2002, 4,
1135MeO O MeO N S S OEt N OX YNaHHODMF, 60 C Y F H Cl Br CondY
OROOR heat 80% OOX F FYield 80 17 15DMF, 60 C, 2 h DMF, 80 C, 43 h
DMF, 60 C, 8 h DMF, 60 C, 5 hO MeOOO MeO NMeO O MeO NCl
Br5-endo-trigIchikawa, et al Synthesis 2002, 1917-1936, PDF on
Course Website Numerous other cases are provided in this
review.OOOOBrH ORevisiting Case 2 with FluorinesO O MeO2CBu3SnH
AIBN 82%O HOMeO2C N OOMeMeO2C N
TsCO2MeTs5-exo-trigTsHN5-endo-trigFavoredH ONot
Observed5-endo-trigO OO MeO2C N Ts O CF2 MeO2C TsHNO OMe CF2 MeO2C
N Ts F CO2Me5-exo-trig5-endo-trigChem. Comm 2088, 28 Review:
"5-Endo-Trig Radical Cyclizatons" Ishibashi, et al Synthesis 2002,
695-713, PDF on Course WebsiteNot ObservedFavored3-06-Baldwin
Rules-6 9/18/03 5:10 PMD. A. Evans, J. JohnsonRules for Ring
Closure: SP2 Carbon & Related SystemsChem 206Trigonal Carbon:
Exocyclic Enolate Alkylationexodistance between reacting centers:
3.37C O C C Y-OC CC YBrX By definition, an exo-tet cyclization, but
stereoelectronically behaves as an endo trig.Me Me O Me Me MO
BrXMOOMe Me O(1)The overlap for C-alkylation is poor due to
geometrical constraints of 5-membered ringonly observed
productHowever:Me Me O Me BrKOt-Bu or LDAdistance between reacting
centers: 3.04Me Me O > 95% by NMRThe relaxed geometrical
constraint provided by the added CH2 group now renders the
6-membered cyclization possibleBaldwin, 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
RMOBrOFavorskii Rearrangement (Carey, Pt B, pp 609-610) Your
thoughts on the mechanismO Cl MeO HCl O MeOCO2Me3-07-Baldwin
Rules-7 9/18/03 4:09 PMD. A. Evans, J. JohnsonRules for Ring
Closure: SP2 & SP Carbon & Related SystemsChem 206Trigonal
Carbon: Intramolecular Aldol CondensationsBaldwin, Tetrahedron
1982, 38, 2939Digonal Carbon: Cyclizations on to AcetylenesDIGONAL:
Angle of approach for attack on triple bondsMO RX
Y(Enolendo)-Exo-trigO RX YMNu120 120 E+Baldwin: - 3 and 4-Exo-dig
are disfavored - 5 to 7-Exo-dig are favored - 3 to 7-Endo-dig are
favoredFavored: 6-7-(enolendo)-exo-trig Disfavored:
3-5-(enolendo)-exo-trigX MO R Y(Enolexo)-exo-trigO RX YMH _Ab
initio SCF 4-31G calculations for the interaction of hydride with
acetylene:H2.13 127 o 148o HFavored: 3-7-(enolexo)-exo-trigH O Me
OI4-31G basis set Houk, J.ACS.1979, 101,
1340.H5-(Enolendo)-Exo-trigMe Me O6-(Enolendo)-Exo-trigMe O HH 156o
1.22CC1.5-2.0110o -120oSTO-3G minimal basis set H Dunitz, Helv
Chim. Acta 1978, 61, 2538.favoredMe O Me O MeIIIH OCCMe Me
OIICrystal Structures do not support BaldwinOStatistical
Distribution, (I + II)/III = 2:1 Experimental Distribution, =
0:100NN2.92 104o(KOH, MeOH, r.t., 5 min, 77%
y.)ON+N+NO932.44oNoCaution: 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.86J. Dunitz and J. Wallis J. C. S. Chem. Comm. 1984,
671.3-08-Baldwin Rules-8 9/18/03 5:49 PMD. A. Evans, J.
JohnsonRules for Ring Closure: SP Carbon & Related Systems
Indole synthesis:CH32 equiv. LDA 2 equiv. RX -78 oC R = Me, Bu,
CO2Me LiTMP+Chem 206Endo Digonal versus Endo Trigonal
Cyclizations5-endo-trigY X:CH2R N+NC-C-In-plane approach;
nucleophile lone pair is orthogonal to *Out-of-plane approach;
nucleophile lone pair can't achieve Brgi-Dunitz angleRSaegusa, J.
Am. Chem. Soc. 1977, 99, 3532. _N5-endo-dig:X Spiro
dihydrofuranones:O Li HO MeOn n = 1,2Li+YAllowed due to in-plane pi
orbitalsOMeKOtBuOOMeXFor 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 OnnDeveloping 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-digO5-exo-digO R OHR = H,
OMeNaOMeX5-endo-trigR O PhLiPhLiPhhowever, 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-digXLiLiPh3-09-Baldwin Rules-9 9/19/03 8:38 AMD. A.
Evans, J. JohnsonRules for Ring Closure: SP Carbon & Related
SystemsChem 206Digonal 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 >
95:55-exo-digO O CO2Me:R R'30-40 kcal/mol ?R R':HOHHO2C HHirsutic
Acid COConclusions and CaveatsOTBSO MeLiCH2NC; 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 DigOTBS 1) RCOCl2) AgBF4 86%Me C-N+ C OMe
RNWorks 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 R3 4
5 6 7 3-10-Baldwin Rules-10 9/18/03 5:21 PMD. A. EvansAcyclic
Conformational Analysis-1Professor Frank Weinhold Univ. of
Wisconsin, Dept of Chemistry B.A. 1962, University of Colorado,
BoulderChem
206http://www.courses.fas.harvard.edu/~chem206/Chemistry 206
Advanced Organic ChemistryLecture Number 4A.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? OD. A. Evans4-00-Cover Page 9/22/03 9:01
AMMonday, September 22, 2003D. A. EvansAcyclic Conformational
Analysis-1Chem 206The following discussion is intended to provide a
general overview of acyclic conformational analysisEthane
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 HOne 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 Bondseclipsed 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 HOMOHC C* CH LUMO CHp.
p.H CHR H C H HIn propane there is a discernable interactionH HIn
the eclipsed conformation there are 3 syn-periplanar CH Bonds
staggered conformationH C H C CH HOMO CHHCHC* CH LUMO CHFor
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 -1Following 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 MeCalculate the the rotational
barrier about the C1-C2 bond in isobutane4-01-introduction 9/22/03
8:28 AMD. A. EvansAcyclic Conformational Analysis: Butane The
1,2-Dihaloethanes ButaneChem 206XH C H H H H HH H CX XXX = Cl; H =
+ 0.91.3 kcal/mol X = Br; H = + 1.41.8 kcal/mol X = F; H = 0.6-0.9
kcal/molUsing the eclipsing interactions extracted from propane
& ethane we should be able to estimate all but one of the
eclipsed butane conformationsMe C H Me HObservation: 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
conformationHHH HMe H C MeHeclipsed conformationE=?Eclipsed atoms 1
(HH) 2 (HMe) E (kcal mol -1) +1.0 kcal mol -1 +2.8 kcal mol
-1Relationship 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 Log10KeqSincen-Butane Torsional Energy
ProfileH H HH C Me Me H C HpKeq = Log10KeqE1E2 G298 = 1.4
pKeqenergyH HMe H C MeH H HHence, pK is proportional to the free
energy changeH HMe CHKeq 1.0 10 100pKeq 0 1 2G 0 1.4 2.8 kcal /molH
Me ARef = 0+3.6Me Me G+5.1+0.884-02-introduction-2 9/22/03 8:33
AMD. A. Evans Butane continuedAcyclic Conformational Analysis:
ButaneNomenclature for staggered conformers:H H Me Me C H H H H H
Me C H H Me MeChem 206Me C H H HFrom 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 HHstaggered conformation Meeclipsed
conformationtrans or t or (anti) Conformer population at 298 K:
70%gauche(+) or g+ 15%gauche(-) or g15% E = +5.1 kcal mol-1Let'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) = +3.1General
nomenclature for diastereomers resulting from rotation about a
single bond (Klyne, Prelog, Experientia 1960, 16, 521.)RR C 0 R R C
-60R CRsp sc sc+60Incremental Contributions to the Barrier.
Eclipsed atoms 2 (HH) 1 (MeMe) E (kcal mol -1) +2.0 +3.1R Eclipsed
Butane conformationacR C -120ac+120 R C R 180 R C RapFrom the
energy profiles of ethane, propane, and n-butane, one may extract
the useful eclipsing interactions summarized below: Hierarchy of
Eclipsing InteractionsXX YY 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 30HHDesignation syn periplanar + syn-clinal + anti-clinal
antiperiplanar - anti-clinal - syn-clinalSymbol sp + sc (g+) + acH
HCC HH Me Me Men-Butane Conformer E2 G E1 A E1 Gap (anti or t) - ac
- sc (g-)4-03-butane 9/26/03 1:51 PMD. A. Evans n-PentaneAcyclic
Conformational Analysis: Pentane The double-gauche pentane
conformationThe new high-energy conformation: (g+g)Me Hg-g-Chem
206Rotation about both the C2-C3 and C3-C4 bonds in either
direction (+ or -):Me Me H Me Hg+g+ g+tMe H Me Metg-H Meg+g-Me HH
MeMe Me Ht,tMe Me Me H HH MeEstimate of 1,3-Dimethyl Eclipsing
Interaction YMetg+Me HXg-t 1 g-g+p. p.135351 (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 -11 1 5 31,3(MeMe) = + 3.7 kcal mol -1Estimates of
In-Plane 1,2 &1,3-Dimethyl Eclipsing Interactions34 (g+g)
Gauche(2,3)-Gauche'(3,4) double gauche pentane3
(g+g+)5MeMeMeMeMeMeMeMeGauche(2,3)-Gauche(3,4) 3.1 ~ 3.7 ~3.9 ~
7.6From 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 AMD. A.
EvansAcyclic Conformational Analysis: Natural Products Lactol &
Ketol Polyether AntibioiticsChem 206The 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 MeR Me R Me Me MeR'RR'orMe H RR'g -g -Me H R'Me H H
MettMeMeEtR'R MeorMeMe H H R'tgH H R HgtFerensimycin B, R = Me
Lysocellin, R = HConsequences 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 =
HInternal 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
OHMetal 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 OHM4-05-Natural Products
9/22/03 8:42 AMD. A. EvansConformational Analysis: Ionophore
X-206/X-rays X-ray of Ionophore X-206 H2OChem 206X-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 OHInternal 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 EtMp. 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 AMD. A. EvansConformational Analysis: Ionophore X-206/X-ray
overlayChem 206p. p.4-07-X-206 overlay 9/19/01 11:57 AMD. A.
EvansStabilized Eclipsed Conformations in Simple Olefins Butane
versus 1-ButeneMe H C H H H Me H H Me MeChem 206Simple olefins
exhibit unusal conformational properties relative to their
saturated counterpartsPropane versus Propene109 H H H H Me H H H H
H 120 CH2staggered conformationC H Heclipsed conformation G = +4
kcal mol-1MeHybridilzation change opens up the CCC bond angle The
Propylene Barrierp. p.H H C H CH2 H H CH2 C Hstaggered
conformationCH2 C H H HH HMe CH2 C Heclipsed conformation = 50
staggered conformation G = 0.83 kcal mol-1=0The Torsional Energy
Profile = 50Me H C H H C H H+2.0 kcal/mol = 180H H C H H C Me
Heclipsed conformationH=0 New destabilizing effectH C H Me CH H H
+1.32 kcal H C H H C H Me H +1.33 kcalH X H C H Hrepulsive
interaction between CX & CH+0.49 kcal = 120 = 180H X C H
H=0Conforms 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 MeK.
Wiberg, JACS 1985, 107, 5035-5041 K. Houk, JACS 1987, 109,
6591-6600The low-energy conformation in each of above cases is
eclisped4-08-simple alkenes 9/22/03 8:54 AMUseful Destabilizing
Interactions to RememberHierarchy of Vicinal Eclipsing
InteractionsXX YY H Me Me E kcal mol -1 +1.0 +1.4 +3.1HHH HCC HH
MeEstimates of In-Plane 1,2 &1,3-Dimethyl Eclipsing
InteractionsMe Me Me Me Me Me Me Me~ 3.1~ 3.7~3.9~ 7.6It 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 000 0000 0000 0000 0000 000 0000
00 0 00 0 0 00 0 0 00 0 0000 0000 0000 0000 0000 0000 0000 0000
00000000 0000 0000minimized structure4-09-Destabilizing Effects
9/20/01 5:33 PMD. A. EvansAcyclic Conformational Analysis-2
Problems of the Day:Chem
206http://www.courses.fas.harvard.edu/~chem206/Can you predict the
stereochemical outcome of this reaction? O OTs MeLiNR2OLi EtO
n-C4H9 H MeOTs 1Chemistry 206 Advanced Organic ChemistryLecture
Number 5EtO n-C4H9 H+ 98:22D. 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 H2O2OH O Me CH2OBnMediastereoselection 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.MeMePhNCO Et3NMeMeO N only one isomerNO2A. Kozikowski
& Co-workers, Tetrahedron Lett. 1982, 23, 2081.D. A.
EvansWednesday, September 24, 20015-00-Cover Page 9/24/03 8:46 AMD.
A. EvansStabilized Eclipsed Conformations in Simple Olefins Butane
versus 1-ButeneMe H C H H H Me H H Me MeChem 206Simple olefins
exhibit unusal conformational properties relative to their
saturated counterpartsPropane versus Propene109 H H H H Me H H H H
H 120 CH2staggered conformationC H Heclipsed conformation G = +4
kcal mol-1MeHybridilzation change opens up the CCC bond angle The
Propylene Barrierp. p.H H C H CH2 H H CH2 C Hstaggered
conformationCH2 C H H HH HMe CH2 C Heclipsed conformation = 50
staggered conformation G = 0.83 kcal mol-1=0The Torsional Energy
Profile = 50Me H C H H C H H+2.0 kcal/mol = 180H H C H H C Me
Heclipsed conformationH=0 New destabilizing effectH C H Me CH H H
+1.32 kcal H C H H C H Me H +1.33 kcalH X H C H Hrepulsive
interaction between CX & CH+0.49 kcal = 120 = 180H X C H
H=0Conforms 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 MeK.
Wiberg, JACS 1985, 107, 5035-5041 K. Houk, JACS 1987, 109,
6591-6600The low-energy conformation in each of above cases is
eclisped5-01-simple alkenes 9/22/03 8:54 AMEvans, Duffy, &
Ripin5Conformational Barriers to Rotation: Olefin A-1,2
Interactions5Chem 2061-butene42-propen-1-olH H H C C OH H HH H H C
C MeH H4E (kcal/mol)3E (kcal/mol)322110 -180-900901800
-180-90090180 (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 HH = 60HO C H C H H = 180=0C H Me H CH H H = 120H C H COH H
H+1.32 kcalH H C H CMe H H+1.33 kcalH+2.00 kcal=0C H HO CH H
HH+1.18 kcal+0.49 kcal=0 = 120 = 180+0.37 kcal = 180=0Conforms to
ab initio (3-21G) values: Wiberg, K. B.; Martin, E. J. Am. Chem.
Soc. 1985, 107, 5035.5-02-1-butene & 2-propen-1-ol 9/23/03 2:59
PMEvans, Duffy, & Ripin5Conformational Barriers to Rotation:
Olefin A-1,2 Interactions-25Chem
2062-methyl-1-butene42-methyl-2-propen-1-ol4H H H C C MeH MeH H H C
C OHH MeE (kcal/mol)E (kcal/mol)3322110 -180-900901800
-180-90090180 (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 CMe Me H+2.68 kcal=0C H HO = 180 =0
H CH Me H+1.16 kcalH H+2.01 kcalMeC HC H+0.06 kcal+0.21 kcal =
1805-03-methylbutene etal 9/23/03 3:00 PMEvans, Duffy, &
Ripin5Conformational Barriers to Rotation: Olefin A-1,3
Interactions5Chem 206(Z)-2-pentene4 4(Z)-2-buten-1-olH Me H C C OH
H HH MeHE (kcal/mol)3E (kcal/mol)CC MeHH322110 -180-900901800
-180-90090180 (Deg)H Me C H Me C H H (Deg)The Torsional Energy
Profile =0Me C H HO C H H H MeThe 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
=0H=0 = 180Values 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 PMReview: Hoffman, R. W. Chem. Rev.
1989, 89, 1841.Evans, Duffy, & Ripin5Conformational Barriers to
Rotation: Olefin A-1,3 Interactions-2OH Me Rotate clockwise C Me C
H HChem 206(Z)-2-hydroxy-3-pentene4Me OH MeH4.6 kcal/molHOH C Me HE
(kcal/mol)Me3CHMe100 H OH2Me HC HC0.3-0.4 kcal/mol1H Me H C C OH-90
0 90 180Me H30 0 -180Lowest energy conformerMe Me Me OH H H C H
(Deg)The Torsional Energy ProfileOH60 Me=0H Me C H HO C H MeMe C H
Me2.7 kcal/molC OHCH H = -140Needs to be redone = 110Me Me C H H H
OH C H OH+2.72????? = -80HO Me C H C Me H H+4.68A(1,3) interaction
4.0 kcal/molR23HO H H C C MeH Me30 = 80 MeMe C C H HY X2R1 = 150H C
C Me HR3*1R large+0.66HMeR smallLowest energy conformer+0.34=0+0.40
kcalHO = 180A(1,2) interaction 2.7 kcal/mol
(MM2)5-05-z-2-hydroxy-3-pentene 9/23/03 6:22 PMD. A. EvansAcyclic
Conformational Analysis: Allylic Strain The Definition of Allylic
StrainChem 206Can 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 OTsLiNR2F. 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, 5006Consider the illustrated general
structure where X & Y are permutations of C, N, and O:R23 Y X
2R1OLi EtO n-C4H9 O H Me MeOTs 1+ 98:22R3Typical examples:R2 R3
R1R2 R large NR1R2 R large R + NR1R2 R largeR1 N Relevant
enolateconformationsR large 1EtO n-C4H9 HmajorO +p. p.R small
OlefinR small ImineR small Imonium ionR 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
OLiIn 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 NaBH4A1B1critical
conformationsC1 HH OR Me TsO(H2C)4 C C Bu OLi MeH CBu CBu OR OLi Me
H C C OR OLi (CH2)4OTs(CH2)4OTsA2B2C2HOR EtO2C OBn 2 n-C4H9MeHO
MeminorHdiastereoselection 10:1 M. Isobe & Co-workers,
Tetrahedron Lett. 1985, 26, 5199.5-06-Allylic Strain-1 9/23/03 6:24
PMD. A. EvansO EtO n-C4H9 H O EtO n-C4H9 HLiNR2 MeIAllylic Strain
& Enolate Diastereoface SelectionOTs MeLiNR2Chem 206PhNH4ClO
EtO n-C4H9 O EtO n-C4H9Me diastereoselection 98:2 H Me3SiPhOM OMe
RR-substituent R = Me diastereoselection 87:13 80:20 40:60O OMe
RMe3SiMe diastereoselection 89:11 H Ph Me3Si OR = Et R = CHMe2major
diastereomer opposite to that shownI. Fleming & Co-workers,
Chem. Commun. 1985, 318. Y. Yamamoto & Co-workers, Chem.
Commun. 1984, 904.D. Kim & Co-workers, Tetrahedron Lett. 1986,
27, 943.PhLiNR2HO OBn diastereoselection 90:10 at C3one isomer at
C2OBnMeCHO Me3Si 71% yieldMeOp. p.RO2C H CO2MeOMeOLiNR2RO2C H H
CO2MeO "one isomer" Me Bn N Boc O N S SSn(OTf)2 RCHO 91-95%MeOHI.
Fleming & Co-workers, Chem. Commun. 1986, 1198.Br95% yieldMe Bn
N Boc RHO N OHS S diastereoselection >95%G. Stork &
Co-workers, Tetrahedron Lett. 1987, 28, 2088.TBSOCH2Me CH2LiNR2
MeIMe CH2 TBSOCH2 Me H H CO2Me "one isomer" Me OPh(MeS)2CLiT.
Mukaiyama & Co-workers, Chem. Letters 1986, 637H CO2MeOMeMeI
86%MeS MeS MeSMeO OMe diastereoselection 99:1T. Money &
Co-workers, Chem. Commun. 1986, 288.MeK. Koga & Co-workers,
Tetrahedron Letters 1985, 26, 3031.R PhMe2SiOM OEtRMeIO OEt MeR =
Me: diastereoselection 99:1 R = Ph: diastereoselection 97:3PhMe2SiI
CO2Et ROLi O-t-BuHCO2EtR = H: one isomerI. Fleming &
Co-workers, Chem. Commun. 1984, 28.KOt-Bu THF -78 CCO2-t-Bu R = Me:
> 15 :1 H RY. Yamaguchi & Co-workers, Tetrahedron Letters
1985, 26,1723.5-07-A-strain enolates 9/20/01 4:45 PMD. A. Evanss
The basic processS H HRAllylic Strain & Olefin
HydroborationHydroborations dominated by A(1,3) StrainS HRChem
206HROHBH HRBH2 BRH C RRO Me MeCH2 OBnB2 H6 H2 O2O Me MeCH2 OBnCC
RC R RCC RRdiastereoselection 8:1 OMe Me O B2 H6 H2 O2 OH O Me Me
Me diastereoselection 12:1 OH OMe OHs Response to steric effects:
Here is a good calibration system:ACH2 Me3 C HMeMeOxidantMCPBA BH3,
H2 O2 ERatio, A:E69:31 34:66ReferenceJOC, 1967, 32, 1363 JOC, 1970,
35, 2654 BnO MeY. Kishi & Co-workers, J. Am. Chem. Soc. 1979,
101, 259.OH OH Me Me B2 H6 H2 O2 BnO Me Me Me OHs 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
TrODiastereoselection = 3:1 C. H. Heathcock et. al. Tetrahedron
Lett 1984 25 243. OHmajor diastereomerMe Mecontrol elementsA(1,3)
allylic strain Steric effects; RL vs RM Staggered transition
statesRM HMe ThexylBH2 , OH then BH3 OTr TrO OHMeMe OTr OH OH
5:1Diastereoselection;major diastereomerMeMeMeMeThexylBH2 , then
BH3 TrOMeMeMeMe OTrOHOHOHOH OTrOHOHOH 4: 1OHDiastereoselection;See
Houk, Tetrahedron 1984, 40, 2257HStill, W.C.; Barrish, J. C. J. Am.
Chem. Soc. 1983, 105, 2487.5-07a-A-strain hydroboration 9/24/03
9:45 AMD. A. EvansConsider the resonance structures of an amide:O
R3 C N1Allylic Strain & Amide ConformationR2 R33Chem 206Y X2R1
R large1The selection of amide protecting group may be done with
the knowledge that altered conformational preferences may result:O
O H HR1 R RO R3C N +R1 R1R smallFavored for R = H, alkylH O HRN RHN
R OFavored for R = CORA(1,3) interactions between the "allylic
substituent" and the R1 moiety will strongly influence the torsion
angle between N & C1. FavoredO Me C N MeN H R H O N O H C R H H
RNHDisfavoredMe MeODisfavoredNH C O RFavoredp. 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, 2821H R C O NRH strongly favored
Me MeR O 1 A(1,3) interaction between the C2 & amide C R
substituents will strongly influence the torsion N angle between C1
& C2. R R2R O1 2CRN R + RH strongly favoredAs a result, amides
afford (Z) enolates under all conditionsH HO Me Ph N OO published
X-ray structure of this amide shows chair Me diaxial
conformationQuick, J. Org. Chem. 1978, 43, 2705O MeCN HL LOM base
favored O Me C N H L L HMeN LL Problem: Predict the stereochemical
outcome of this cyclization.OH D. Hart, JACS 1980, 102, 397 Ph N O
HOCOHCO2H(Z)-EnolateHH O H C L N Me L base disfavored O HH LOM C N
Me L H Me N L LN Ph O diastereoselection >95%identify HOMO-LUMO
pair5-08-A-strain Amides-1 9/23/03 6:09 PM(E)-EnolateD. A.
EvansAllylic Strain & Amide ConformationChem 206A(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 OPolypropionate Biosynthesis: The Acylation EventO
O SR O SR Me O SR Me OH O SR MeN Bnenolization selectivity
>100:1HOR OAcylation CO2RReductionREl(+) JACS. 1982,104, 1737.O
OO MeCN HFirst laboratory analogue of the acylation eventO Et Cl O
Me Li O N R O Me Me R O N O O Ofavored enolization geometry p. p.
In the enolate alkylation process product epimerization is a
serious problem. Allylic strain suppresses product enolization
through the intervention of allylic strainH O El C N Me L L O Me C
N H El L L Me O C H El N L Lwith M. Ennis JACS 1984, 106,
1154.Diastereoselection ~ 97 : 3Why does'nt the acylation product
rapidy epimerize at the exocyclic stereocenter?? R O Me
CfavoredBACH N H R R O R C N Me R RWhile conformers B and C meet
the stereoelectronic requirement for enolization, they are much
higher in energy than conformer A. Further, as deprotonation is
initiated, A(1,3) destabilization contributes significantly to
reducing the kinetic acidity of the system These allylic strain
attributes are an integral part of the design criteria of chiral
amide and imide-based enolate systemsO O Me N Bn CH2OH Me N O O O
Me N Me OH MeEvans Tetr Lett. 1977, 29, 2495Evans JACS 1982,104,
1737.Myers JACS 1997, 119, 6496X-ray structure5-09-A-strain
Amides-2 9/20/01 5:22 PMD. A. EvansDiscodermolideChem 206hinge Me
HO O O H16Me17MeMe Me Me OH MeOHO ONH2Me OH- immunosuppressive
activity - potent microtubule-stabilizing agent (antitumor activity
similar to that of taxol)The epimers at C16 and C17 have no or
almost no biological activity.The conformation about C16 and C17 is
critical to discodermolide's biological activity.S. L. Schreiber et
al. JACS 1996, 118, 11061.5-10-Discodermolide 9/19/01 12:14 AMD. A.
EvansConformational Analysis - Discodermolide X-ray 1Me HO Me Me Me
OH Me OH O O NH2 Me MeChem 206OO HMe OH5-11-Discodermolide X-ray1
9/21/01 8:34 AMD. A. EvansConformational Analysis - Discodermolide
X-ray 2Me Me MeChem 206HO O O H16 Me Me Me OH Me OH O O NH2Me
OH165-12-Discodermolide X-ray2 9/21/01 8:40 AMD. A.
EvansConformational Analysis: Part3Chem
206http://www.courses.fas.harvard.edu/~chem206/Conformational
Analysis of Cyclic SystemsThree Types of Strain:Chemistry 206
Advanced Organic ChemistryPrelog Strain: van der Waals interactions
Baeyer Strain: bond angle distortion away from the ideal Pitzer
Strain: torsional rotation about a sigma bondLecture Number 6Baeyer
Strain for selected ring sizes"angle strain" size of ring Ht of
Combustion Total StrainStrain per CH2 (kcal/mol) (kcal.mol)
deviation from 10928' (kcal/mol)Conformational Analysis-3
Conformational Analysis of C4 C6 Rings Reading Assignment for week
A. Carey & Sundberg: Part A; Chapter 3Eliel & Wilen,
"Stereochemistry of Organic Compounds, "Chapter 11, Configuration
and Conformation of Cyclic Molecules, Wiley, 19943 4 5 6 7 8 9 10
11 12 13 14 15499.8 656.1 793.5 944.8 1108.3 1269.2 1429.6 1586.8
1743.1 1893.4 2051.9 2206.1 2363.527.5 26.3 6.2 0.1 6.2 9.7 12.6
12.4 11.3 4.1 5.2 1.9 1.99.17 6.58 1.24 0.02 0.89 1.21 1.40 1.24
1.02 0.34 0.40 0.14 0.132444' 944' 044' -516'Eliel, E. L., Wilen,
S. H. Stereochemistry of Organic Compounds Chapter 11, John Wiley
& Sons, 1994.Ribeiro & Rittner, "The Role of
Hyperconjugation in the Conformational Analysis of
Methylcyclohexane and Methylheterocyclohexanes" J. Org. Chem., 2003
, 68 , 6780-6787 (handout)de Meijere, "Bonding Properties of
Cyclopropane & their Chemical Characteristics" Angew Chem. Int.
Ed. 1979, 18, 809-826 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.D. A. Evans6-00-Cover Page 9/26/03 8:44
AMFriday, September 26, 2003Evans, Kim, BreitCyclopropane: Bonding,
Conformation, Carbonium Ion Stabilization CyclopropaneHChem
206Carbocation Stabilization via CyclopropylgroupsH H H Necessarily
planar. Subtituents are therefore eclipsed. Disubstitution prefers
to be trans.C = 120 Almost sp2, not sp3 = 3080 cm-1MeWalsh Model
for Strained Rings: Rather than and * c-c bonds, cyclopropane has
sp2 and p-type orbitals instead.A rotational barrier of about 13.7
kcal/mol is observed in following example:HMeNMR in super acids
(CH3) = 2.6 and 3.2 ppm OX-ray Structures support this orientation
1.302 1.517 1.478 1.222 R1.474 (antibonding) H (antibonding)
Nonbondingside view (antibonding)1.464 1.541 1.409 H3H1.444 1.534
OPh (bonding) (bonding)1 (bonding) de Meijere, "Bonding Properties
of Cyclopropane & their Chemical Characteristics" Angew Chem.
Int. Ed. 1979, 18, 809-826 (handout)de Meijere, A.; Wessjohann, L.
"Tailoring the Reactivity of Small Ring Building Blocks for Organic
Synthesis." Synlett 1990 , 20.R. F. Childs, JACS 1986, 108,
16926-01-Cyclopropane 9/25/03 7:55 PMEvans, Kim, Breit145-155ax eq
eq eq eq ax ax axConformational Analysis: Cyclic Systems-2
CyclobutaneH H HChem 206H H H H H H H H H H H H H H H H
HCyclopentaneH H H H H H Eclipsing torsional strain overrides
increased bond angle strain by puckering. Ring barrier to inversion
is 1.45 kcal/mol.H H H H = 28 CsEnvelopeC2 Half-ChairCsEnvelope 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.(MM2) Since there is no "natural" conformation of
cyclopentane, the ring conforms to minimize interactions of any
substituents present.HCsEnvelope (MM2)H HHH HHH H G = 1 kcal/mol
favoring R = Me equatorial 1,3 Disubstitution prefers cis
diequatorial to trans by 0.58 kcal/mol for di-bromo cmpd. A single
substituent prefers the equatorial position of the flap of the
envelope (barrier ca. 3.4 kcal/mol, R = CH3). 1,2 Disubstitution
prefers trans for steric/torsional reasons (alkyl groups) and
dipole reasons (polar groups). Me Me XX 1,3 Disubstitution:
Cis-1,3-dimethyl cyclopentane 0.5 kcal/mol more stable than trans.
1,2 Disubstitution prefers trans diequatorial to cis by 1.3
kcal/mol for diacid (roughly equivalent to the cyclohexyl
analogue.)H H A carbonyl or methylene prefers the planar position
of the half-chair (barrier 1.15 kcal/mol for cyclopentanone).
X6-02-Conform/cyclic-2 9/25/03 7:55 PMEvans, Kim,
BreitConformational Analysis: Cyclic Systems-3Chem
206Methylenecyclopentane and CyclopenteneStrain trends: > >
Decrease in eclipsing strain more than compensates for the increase
in angle strain.Cyclohexane Energy Profile (kcal/mol)Half-ChairBoat
+1.01.5 10.711.5Relative to cyclohexane derivatives, those of
cyclopentane prefer an sp2 center in the ring to minimize eclipsing
interactions. "Reactions will proceed in such a manner as to favor
the formation or retention of an exo double bond in the 5-ring and
to avoid the formation or retention of the exo double bond in the
6-ring systems." Brown, H. C., Brewster, J. H.; Shechter, H. J. Am.
Chem. Soc. 1954, 76, 467.+5.5 +5.5 Chair Twist Boat Inverted
ChairExamples:H H H H H H H H O O NaBH4 HH OHk6H H HHNaBH4H H OH
Hk6 = 23 k5k5Brown, H. C.; Ichikawa, K. Tetrahedron 1957, 1, 221.O
OE = +6.5-7.0 E = +5.5OOhydrolyzes 13 times faster thanConan, J-Y.;
Natat, A.; Priolet, D. Bull. Soc. Chim., Fr. 1976, 1935.O O OEt OH
O OEtThe barrier: +10.7-11.5 E=095.5:4.5 keto:enol76:24
enol:ketoBrown, H. C., Brewster, J. H.; Shechter, H. JACS 1954, 76,
467.6-03-Conform/cyclic-3 9/25/03 7:57 PMEvans, BreitConformational
Analysis: Cyclic Systems-4Chem 206Monosubstituted Cyclohexanes: A
ValuesR HA Values depend on the relative size of the particular
substituent.H H H H Me H H H Me Me H Me Me Me HKeqHRG = RTlnKeq
Meaxial has 2 gauche butane interactions more than Meequatorial.
Expected destabilization: 2(0.88) kcal/mol = ~1.8 kcal/mol;
Observed: 1.74 kcal/molMe H C H H H H Me H H C Me H C
HHAValue1.741.802.155.0The "relative size" of a substituent and the
associated A-value may not correlate. For example, consider the
CMe3 and SiMe3 substituents. While the SiMe3 substituent has a
larger covalent radius, it has a smaller A-value:Me C Me H Me Si Me
H Me Sn Me H The A Value, or -G, is the preference of the
substituent for the equatorial
position.MeMeMeAValue4.5-5.02.51.1Can you explain these
observations? The impact of double bonds on A-values:Lambert,
Accts. Chem. Res. 1987, 20, 454R H H Rsubstituent R = Me R = OMe R
= OAcG 0.8 0.8 0.6A-value (cyclohexane) 1.74 0.6 0.71The Me
substituent appears to respond strictly to the decrease in
nonbonding interactions in axial conformer. With the more polar
substituents, electrostatic effects due to the trigonal ring carbon
offset the decreased steric environment.6-04-Conform/cyclic-4
9/25/03 7:59 PMRigberio & Rittner, The Role of Hyperconjugation
in the Conformational Analysis of Methylcyclohexane and
Methylheterocyclohexanes JOC 2003, 68, 6780 Commentary by Ken Houk
University of California, Los Angeles Department of Chemistry Dear
David, The calculations in the Ribeiro article look fine, but I am
not convinced by the interpretation. It does seem to work pretty
well for many systems, but not obviously for the isomeric
1,3-dioxane cases they note early on. There seems no explanation of
why CH hyperconjugates better than CC. Further, the results with
alkyls larger than methyl still require traditional steric
arguments. I would say that the equatorial methyl preference has
been attributed in part to hyperconjugative effects that occur when
the CH bonds are anti-periplanar. But I would not yet go much
beyond that! Best regards, (b. 1943) A.B. 1964, Ph.D. 1968, Harvard
University; Assistant-Full Professor, Louisiana State University,
1968-1980; Alfred P. Sloan Fellow, 1975-1977; Camille and Henry
Dreyfus Teacher-Scholar, 1972-1977; LSU Distinguished Research
Master, 1978; Professor, University of Pittsburgh, 1980-1985;
Alexander von Humboldt Senior U.S. Scientist Award, 1982; Akron
Section, American Chemical Society Award, 1983; Arthur C. Cope
Scholar Award, 1988; Director, Chemistry Division, National Science
Foundation, 1988-1990; James Flack Norris Award in Physical Organic
Chemistry, 1991; Schrdinger Medal, World Association of
Theoretically Oriented Chemists, 1998; Tolman Medal, Southern
California Section, American Chemical Society, 1999; Fellow of the
American Academy of Arts and Sciences, 2002; American Chemical
Society Award for Computers in Chemical and Pharmaceutical
Research, 2003; International Academy of Quantum Molecular Science,
2003.D. A. EvansBond Lengths and A-Values of Methyl HalidesChem
206CF: 1.39 CCl: 1.79 CBr: 1.95 CI: 2.16 F A-value: 0.250.42Cl
A-value: 0.5364Br Avalue: 0.48-0.67I A-value: 0.470.61p. p.Chem 3D
Pro (Verson 5.0)6-04a methyl halides 9/26/01 8:26 AMEvans,
BreitConformational Analysis: Cyclic Systems-5Chem 206The impact of
trigonal Carbon Let's now compare look at the carbonyl analog in
the 3-positionMe H O O H MePolysubstituted Cyclohexane A Values As
long as the substituents on the ring do not interact in either
conformation, their A-values are roughly additive 1,4
Disubstitution: A Values are roughly additive.Me Me Me Me Me Me Me
MeG = 1.36 kcal/mol versus 1.74 for cyclohexane Let's now compare
look at the carbonyl analog in the 2-positionMe Me3C O H base
epimerization Me3C O H MeG = 0 kcal/molG = 2(1.74) = 3.48 kcal/molG
= 1.56 kcal/mol versus 1.74 for cyclohexane However, the larger
alkyl groups do not follow the expected trend. Can you explain?
(see Eliel, page 732)CHMe2 Me3C O H base epimerization Me3C O H
CHMe21,3 Disubstitution: A Values are only additive in the trans
diastereomerH X H Me H Me H XG = A(Me) A(X)HH X H Me X HG = 0.59
kcal/mol versus 2.15 for cyclohexaneCMe3 Me3C O H base
epimerization Me3C O HMeThe new interaction! For X = MeH CMe3 Me H
Me H H H H Me MeG = 1.62 kcal/mol versus 5.0 for
cyclohexane6-05-Conform/cyclic-5 9/25/03 8:01 PM+ 0.88+ 0.88G =
2(.9) + 1(+3.7)= 5.5 kcal/mol + 3.7Evans, BreitMe PhConformational
Analysis: Cyclic Systems-6 Let's now consider vicinal
substitutionMe H PhChem 206Let's now consider geminal
substitutionMeCase 1:MeHH Me HMeThe prediction:G = A(Ph) A(Me) G =
+2.8 1.7 = +1.1 kcal/molObserved:G = 0.32 kcal/molThe prediction:G
= 1 gauche butane 2A(Me) G = +0.88 2(1.74) = +2.6 kcal/molHence,
when the two substituents are mutually interacting you can predict
neither the magnitude or the direction of the equilibrium. Let's
analyze this case. Allinger, Tet. Lett. 1971, 3259Observed:G =
+2.74 kcal/molIf the added gauche butane destabilization in the
di-equatorial conformer had not been added, the estimate would have
been off.6-06-Conform/cyclic-6 9/25/03 8:13 PM00 000000 000000 0 0
0 000000 00 000A G = +2.80000 00 00000 00 0 0 00 000 00 000B G =
0.32Case 2:H Me Me OH H OH H H Me MeHHThe conformer which places
the isopropyl group equatorial is much more strongly preferred than
would be suggested by A- Values. This is due to a syn pentane OH/Me
interaction. Problem: Can you rationalize the stereochemical
outcome of this reaction?00 0000000 0 00 0 000000 00000000 0 000000
0 000 0O EtO LiNR2 MeI H EtOOMeCDn-C4H9n-C4H9Hdiastereoselection
89:11Note the difference in the Ph substituent in A & B.D. Kim
& Co-workers, Tetrahedron Lett. 1986, 27, 943.Evans,
BreitConformational Analysis: Cyclic Systems-7Chem
206Heteroatom-Substituted 6-Membered Rings A-values at the
2-position in both the O & N heterocycles are larger than
expected. This is due to the shorter CO (1.43 ), and CN (1.47 )
bond lengths relative to carbon (CC; 1.53 ). The combination of
bond length and bond angle change increases the indicated
1,3-diaxial interaction (see eq 1, 4).A-Values