Chemical Bonding and Structure · Chemical Bonding and Structure A. Chemical Bonding 1. Review of...

CHEM 6320 Fall 2005

Chemical Bonding and Structure A. Chemical Bonding

1. Review of Quantum Mechanics a. Any moving entity behave like a wave with an associated wavelength λ

given by:

mvh

ph==λ

b. The wave treatment of the matter is necessary for light particles like electrons, atoms and/or molecules.

c. The properties and behavior of a particle are specified by the wavefunction, which is obtained by solving the Schrödinger equation:

0)(82

2

2

2

2

2

2

2=Ψ−+

∂Ψ∂

+∂Ψ∂

+∂Ψ∂ VE

hM

zyxπ

where M is the mass of particle, E is the total energy, and V is the potential energy.

d. Wave equation can be solved exactly for the H atom (and a few other systems) but approximations are needed for multielectron systems and especially for polyatomic molecules.

e. For polyatomic molecules, Born-Oppenheimer approximation allows the separation of the total wavefunction ),(molecular RrΨ into an electronic wavefunction );(e Rrψ and a nuclear wavefunction )(N Rχ .

)();(),( Nemolecular RRrRr χψ=Ψ f. The electronic wavefunction is obtained by solving the Schrödinger

equation at a fixed nuclear arrangement:

0)(82

2

2

2

2

2

2

2=−+

∂∂

+∂∂

+∂∂ ψπψψψ VE

hm

zyx

where m is the mass of electron, E is the total (electronic) energy, and V is the potential energy.

– Rewrite using Laplacian operator: 0)(82

22 =−+ ψπψ VE

hm

∇

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CHEM 6320 Fall 2005

– Rearrange: ψψπ

EVmh

=

+∇

− 22

2

8

– Introduce Hamiltonian operator (the sum of the kinetic and potential energies in operator form): ψψ EH =ˆ

g. The solutions are equations that represent set of “orbitals” (or eigenfunctions) of increasing energy (or eigenvalues), orbitals that have an increasing number of nodes.

h. Eigenfunctions must be: – Orthogonal: jidji ≠=∫ ,0τψψ

– Normalized: jidji ==∫ ,1τψψ

i. An important property of the electronic wavefunction is that ),,(2 zyxψ gives the ______________________________________

_______________. 2. Valence Bond (VB) Theory

a. Consider wave equation for each various possible Lewis electronic configuration (canonical forms). Total electronic wave function is a linear combination of these, each with weighting factor:

...332211e +++= ψψψψ ccc

b. Example for H2: HH − , −+ H:H , +− H:H 3. Hybrid Orbitals and Hybridization

a. Hybrid orbitals are formed by mixing atomic orbitals; the wavefunction for the hybrid orbitals is obtained as a linear combination of (s and p) atomic wavefunctions:

...22222222orbital hybrid ++++=zzyyxx ppppppss cccc ψψψψψ

– The s character of the hybrid orbital is given by ( 22 )sc

– The p character of the hybrid orbital is given by 2

22

22

2 )()()(zyx ppp ccc ++

b. A hybrid orbital that is % s character and (x )100 x− % p character is said to be an hybrid orbital.

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CHEM 6320 Fall 2005

c. An sp hybrid orbital is y % s character and % p character.

d. Hybrid orbitals give directionality for bonding due to a more effective overlap.

sp

sp2

sp3

sp∞

sp

sp2

sp3

sp∞ e. A hybrid atomic orbital can be written as . As z increases: zsp

– the orbital become more directional – the bond length increases – the bond strength decreases – the s character decreases

f. The angle θ between any two hybrid orbitals λsp and δsp :

λδθ 1cos −=

g. Rule: Carbon (and other elements) uses more p character in hybrid orbitals bonding to more electronegative atoms and less p character in bonding to less electronegative atoms.

4. Hybridization Examples a. Carbon in methane

– Electronic configuration of C atom: 1 222 22 pss– One electron is easily promoted from s2 to 2 (promotion

energy is small). p

2s 2p sp3

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CHEM 6320 Fall 2005

– The carbon valence becomes four. – Four hybrid orbitals are formed, orbitals that are one part s

and three parts p (25% s character and 75% p character):

3sp

( ) pspsps 3412

44

43

414 +=+=+×

– The 41 in equation above represents the s character of the

hybrid orbital (and is not the coefficient). sc2

– The shape of hybrid orbitals: 3sp

+ =

– The orientation of hybrid orbitals: 3sp

x

y z

zy

x

109.5°

xps 223

221 ψψ + ;

yx pps 2223

2223

221 ψψψ ++

b. Nitrogen in ammonia – Electronic configuration of N: 1 322 22 pss

– The promotion energy of one electron from 2 2s to s3 (that would increase the valence to five) is much higher than the energy of the two additional bonds so the valence remains 3.

– Lewis structure: N HH

H – Nitrogen can use, in principle, the three p orbitals for bonding to

three H s1 orbitals but that would put the non-bonding electron pair in s2 orbital, situation that is not favored because it wastes s orbital character on non-bonding.

– Nitrogen can use, in principle, three orbitals for bonding to three H

2sps1 orbitals but that would make 90° angles between the

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CHEM 6320 Fall 2005

non-bonding electron pair in orbital and the three orbitals, also not desired situation.

p2 2sp

3sp

NHH

H – The preferred case is making orbitals, three used for bonding

to three H s1 orbitals and one for the lone pair.

HH

N

H c. Carbon in difluoromethane

F

F

HH

sp>3sp<3

– The angles are not exactly 109.5 degrees. – The carbon uses hybrid orbitals with less p character (than )

to bind to H and more p character to bind to F.

3sp

– A consequence: F–C–F angle < 109.5° < H–C–H angle – Experimental: F–C–F angle = 108.2°; H–C–H angle = 112.5°

d. Carbon in ethylene – Assume it comes from ethane from which two H were removed

( hybridization for C): 3sp

H

H

H

H109.5°

This structure is (possible buy) not energetically favored because contains two bad σ overlaps.

– In reality, each C is hybridized: 2sp

120°H

H

H

H

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CHEM 6320 Fall 2005

This structure contains one σ C–C bond (with strong or good overlap) and one π C–C bond (with weak or bad overlap).

e. Carbon in acetylene – Each C is ___ hybridized. sp

H H

– An hybridization would produce in an ineffective overlap. 3spf. Carbon in allene

H2C C CH2 C C C

H

HH

H

H

H120°

H

H

– The molecule is not planar. – The central C ____ hybridized while the other two are ____

hybridized. – This is a case of a cumulative double bond.

g. Carbon in cyclopropane – If C is sp hybridized there would be a large angle strain

(109 ).

3

o − oo 5.49605. =

– Assuming C uses pure orbitals to bond with the other C (the

angle strain would be only 30°), the C–H bonds would involve p

sp hybrids so the H–C–H angle would be 180°.

H

H

– Make a compromise and the result is carbon atoms being

hybridized to form C–C bonds (106° interorbital angle).

63.3sp

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CHEM 6320 Fall 2005

– What kind of hybrid orbitals are used to form the C–H bonds? – Each sp hybrid orbital (used in C–C bonding) has

21.6% s character.

63.3

– Each sp hybrid orbital (used in C–C bonding) has 78.4% p character.

63.3

– If two hybrid orbitals have 21.6% s character, the other two hybrid orbitals (used in C–H bonding) should have 28.4% s character.

– If two hybrid orbitals have 78.4% p character, the other two hybrid orbitals (used in C–H bonding) should have 71.6% p character.

– A hybrid orbital that has 28.4% s character and 71.6% p character is an hybrid orbital. 52.24.28/6.71 spsp =

5. Resonance a. The actual molecule is a resonance hybrid (weighed average) of

contributing resonance forms. Contribution of individual resonance forms increases:

– As number of covalent bond increases (minimizing unpaired electrons)

– With minimization of charge separation (unlike charges) – With placement of + or – on electropositive or electronegative

atoms, respectively Note: All resonance forms must have the same spin multiplicity

b. Resonance energy is the extra stability of resonance hybrid as a result of resonance possibility (difficult to determine experimentally).

c. Note that resonance involves only the distribution of electrons in a delocalized (π) system. There is no movement of atomic nuclei, implying a fixed geometry and fixed hybridization (if rehybridization is necessary to allow for resonance, this must be done before considering resonance forms):

– No resonance for methyl amine. (N is hybridized.) 3spCH3 NH2

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CHEM 6320 Fall 2005

– Acetamide:

CH3C

O

NH2 CH3C NH2

O

Resonancehybrid: CH3C NH2

Oδ+

δ−

d. Another way to think about resonance forms is to consider all possible

ways of distributing the given number of π electrons in the π system. – Example butadiene:

C C C C CCCC C C C C

_ + +_

CCCC

_ +

Resonance hybrid:

e. If resonance energy turns out to be more than predicted on the basis of simple resonance, the system is aromatic, e.g., a system like benzene:

24 +n

f. If resonance energy turns out to be less than predicted on the basis of

simple resonance, the system is antiaromatic, e.g., a system like cyclobutadiene:

n4

g. Other examples

– −2NO

– +2NO

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CHEM 6320 Fall 2005

– Pyridine:

– Pyrrole:

h. Steric inhibition of resonance – Example of uninhibited resonance:

N

O

OO +

_

_

_ +N

O

O O

_

– Example of inhibited resonance:

_

N

O

O O

C

H

HH

+_

– Nitro group should be in the (ring) plane to be able to write the

resonance structures. – Rule of six: The can be steric hindrance in between atoms

connected through 4 other atoms (for a total of 6 atoms). – Resonance is even more inhibited for:

_ +N

O

O O

C

H

HH

C

H

HH

_

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CHEM 6320 Fall 2005

B. Bond Energy, Polarity, and Polarizability 1. Bond Length

It is mostly a reflection of hybridization, not just the “normal” notion of single > double > triple:

CH2 CH CH3CH2

sp2 sp2 sp3 sp3

1.34 1.50 1.54 1.341.461.34

sp2sp2sp2sp2

CH CH2CHCH2

2. Bond Energy

a. Homolytic bond energy (BE) reflects stability of two radicals produced by bond scission:

3232 CHPhCHCHPhCH ⋅+⋅→− kcal/mol70=∆H

323323 CHCHCHCHCHCH ⋅+⋅→− kcal/mol85=∆H b. Heats of formation and heats of hydrogenation can be used to estimate

relative stability between related compounds through a thermodynamic cycle.

CH3CH3 CH3

CH3

CH3CH2CH2CH3

vs.

+ H2 + H2 ∆H = -27.6 kcal/mol

∆H = -28.6 kcal/mol

E

n-butane

cis-2-butenetrans-2-butene

3. Bond Polarity

a. Bond polarity in A–B is based on the relative electronegativities of the attached atoms/groups.

b. Molecular dipole is the vector sum of all the bond dipoles and electron-pair dipoles.

– Example: NH3 has dipole moment; CCl4 has no dipole moment. c. C is slightly electronegative of H in a normal bond. 3sp

d. The electronegativity increases sp , explaining the acidity of terminal acetylenes:

spsp << 23

R C C Hδ− δ+

HCCR ++−

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CHEM 6320 Fall 2005

4. Acid Dissociation (I) COOHCH3

+− + HCOOCH3

(II) CCOOHCl3+− + HCCOOCl3

(III) COOHCHCH 23+− + HCOOCHCH 23

a. (II) is more acidic than (I) on account of the fact that the inductive effect of the electron-withdrawing Cl selectively stabilizes the conjugate base.

b. One typically explains the lower solution acidity of (III) over (I) on the notion that the extra methyl group is inductively electron donating.

c. However, in gas phase, (III) is more acidic than (I). Actually, the intrinsic inductive effect of alkyl substitution is either electron-withdrawing or electron-donating, depending on the electron demand of the reaction center.

d. Why is (III) less acidic that (I) in solution? The reason is the steric inhibition of solvation:

O

OCH3

O

OH

HH

e. The same dichotomy between solution and gas-phase acidities applies

to the alcohol series MeOH, EtOH, i-PrOH, t-BuOH. – Example: t-BuOH is less acid in liquid phase and more acid in

gas phase than MeOH. (t-BuO– has a larger volume than MeO– so a better redistribution of the charge - a better stability - in gas phase.)

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CHEM 6320 Fall 2005

C. Molecular Symmetry 1. Symmetry Elements

a. Identity (E ) b. n-Fold axis of symmetry (C ) n

– The axis with the highest value of n is called principal axis. c. Plane of symmetry (σ )

– vσ : the plane of symmetry is parallel to a unique axis or to a principal axis

– hσ : the plane of symmetry is perpendicular to a unique axis or to a principal axis

– dσ : the plane of symmetry bisects the angle between C axes that are perpendicular to a principal axis (

2

dσ is a special type of a vσ plane)

d. Center of symmetry (i ) e. n-Fold rotation reflection axis of symmetry (improper rotation) ( ) nS

2. Symmetry Operations a. E : no change b. : Rotation about the axis by 360/n degrees nCc. σ : Reflection in a plane d. : Reflection through the center of symmetry ie. : Rotation about the axis by 360/n degrees followed by reflection

through a plane perpendicular to the axis nS

3. Point Groups a. A set of symmetry operations constitutes a point group. b. Each point group consists of a number of symmetry elements. c. A symmetry element may have more than one symmetry operation

associated with it. d. The total number of symmetry operations (which may be greater than

the total number of symmetry elements) is called the order of the point group.

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CHEM 6320 Fall 2005

e. Examples Group Symmetry Elements Examples

1C E ClH

IBr

sC (C ) h E , hσ Br

iC (S ) 2 E , i

H

H

ClBrCl Br

2C E , C 2

H

ClCl

H

v2C E , C , 22 vσ H

Cl

H

Cl F

F

HH

OH H

v3C E , C , 33 vσ H

HH

Cl

N

HHH

v4C E , C , 24 vσ , 2 dσ

F

HF

H

F

F

H

H

h2C E , C , 2 hσ , i Cl

H

H

Cl

h2D E , 3C , 32 σ , i H

H

H

H

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CHEM 6320 Fall 2005

e. Examples - continuation

h3D E , C , 3C , 3 2 hσ , 3 vσ , 3S B

H

H H

OS

O

O

h4D E , C , 4C , 4 2 hσ , 2 vσ , 2 dσ , , i 4S

H

HH

H

H

H

H

H

Xe

F

F F

F

h6D E , C , 3C , 6 2 hσ , 3 vσ , 3 dσ ,

, i 6S

H

H

H

H

H

H

d2D E , , 3C , 24S 2 dσ C C CH

HH

H

dT E , 4C , 3C , 3S , 63 2 4 dσ HH

HH

v∞C E , ∞C , ∞ vσ H

Cl h∞D E , C , ∞C , ∞ 2 ∞S , ∞ vσ , hσ , i H

H f. Some helpful rules

– The group contains an n-fold axis and n nvC vσ mirror planes. – The group contains an n-fold axis and a mirror plane

perpendicular to the n-fold axis. nhC

– The group contains an n-fold axis and n 2-fold axes perpendicular to n-fold axis.

nD

– The group contains an n-fold axis and n 2-fold axes perpendicular to n-fold axis plus a plane perpendicular to n-fold axis.

nhD

– The group contains same symmetry elements as plus n-dihedral mirror planes.

ndD nD

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4. Summary of the Shapes Corresponding to Different Point Groups

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5. Flow Diagram for Determining the Point Group of a Molecule

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CHEM 6320 Fall 2005

D. Molecular Orbital Theory 1. Introduction

a. Bonding arises from the overlap of atomic orbitals, combining to form molecular orbitals, whose “clouds” surround two or more atomic nuclei.

b. The number of molecular orbitals is equal to the number of contributing atomic orbitals.

c. In a molecular orbital (MO) diagram, the orbitals are filled up with available electrons (consistent with Pauli exclusion principle and Hund’s rules), filling orbitals with lowest energy first.

d. Types of molecular orbitals – σ : overlap where centers of electron density are on the axis

common to the two nuclei ( ∞C symmetry = infinite fold rotation axis)

– π : “sideways” overlap with C symmetry ( 2-fold rotation axis)

and mirror plane 2

2. Qualitative Molecular Orbital Approach

a. Total number of MO’s is equal to total number of contributing AO’s in the basis set (s, p, d). Usually neglect the low-lying s1 electrons for organic molecules. One could use hybrid AO’s, and would ultimately get the same solution. For delocalized approach, it is easier to use pure AO’s; for symmetry approach, it is easier to use hybrid AO’s.

b. Symmetry of MO’s must conform to the symmetry of the molecule (all MO’s must be either symmetric or antisymmetric to all symmetry elements).

c. Orthogonal AO’s do not interact.

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CHEM 6320 Fall 2005

d. Relative energy of AO’s (and thus the derived MO’s) goes down as the atom electronegativity increases.

e. Relative energies of MO’s go up with increasing number of nodes (but π nodes are less destabilizing than σ nodes).

– The lowest energy MO is always totally bonding between all nuclei.

– The highest energy MO is always totally antibonding between all nuclei.

f. Bonding MO’s assign larger coefficients to more electronegative atoms. Bonding MO’s assign smaller coefficients to less electronegative atoms. Antibonding MO’s have opposite behavior.

3. Two-Atom Molecular Orbital Diagram a. Used for diatomic molecules and for a localized bonding picture b. Two atomic orbitals (on different atoms) combine to form two (two-

atom) molecular orbitals (one bonding with positive overlap and lower energy and one anti-bonding with node and higher energy).

∆E

∆E ϕBϕA

ψMO = ϕA + ϕB

atom A atom B

ψ*MO = ϕA – ϕB

c. The “interaction energy” increases

– as extend of overlap increases (sigma > pi) – as energies of the contributing atomic orbitals become more

similar

strong interaction strong interaction weak interaction weak interaction

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CHEM 6320 Fall 2005

4. Molecular Orbital Diagram for Carbon Monoxide

πx* πy

*

πx πy

σ′*

σ′

σ*

σ

2p (O)

2s (O)

2p (C)

2s (C)

a. Atomic orbitals of O are lower in energy than those of C. b. Molecular orbitals have different contributions from C and O atomic

orbitals. 5. Delocalized Bonding

a. For delocalized bonding (MO’s represent more than two nuclei), general mathematical treatment of MO’s as linear combination of atomic orbitals (LCAO) with appropriate weighing factors:

...332211 +++= ϕϕϕψ ccc Note similarity to VB theory equation. In practice, both approaches give similar results, though VB calculations are more cumbersome.

b. Electron density at each atom for any given molecular orbital One needs the overall sum of the square of the atom coefficients for all MO’s containing electrons. The electron density at atom r is given by:

∑=j

jrjr cnq 2

where j is the number of MO’s, n is the number of electrons in that orbital, and is the coefficient of AO in jjrc th MO.

n = 2 if the molecular orbital is doubly occupied n = 1 if the molecular orbital is singly occupied n = 0 if the molecular orbital is empty

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CHEM 6320 Fall 2005

6. Molecular Orbital Diagram for Methane a. Calculated MO energy (in a.u.) diagram for methane

–11.2714

–0.9320 –0.5418

+0.6441 +0.6887

–11.2714

–0.9320 –0.5418

+0.6441 +0.6887

– Photoelectron spectroscopy shows indeed two different

ionization energies for methane. b. Consider methane in a cubic frame of reference (above) c. Atomic orbitals of carbon

d. Molecular orbitals of methane

e. Observations

– The MO involving 2s orbital of C has no node. – The MOs involving 2p orbitals of C have one node. – The MO involving 2s orbital of C is totally symmetric. – The MOs involving 2p orbitals of C show C2 axis of symmetry. – The antibonding MOs are similar but with opposite signs in the

overlap region.

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CHEM 6320 Fall 2005

7. Molecular Orbital Diagram for Ethylene

C CH

H

H

Hx

y

za. MO energy (in a.u.) diagram for ethylene

–1.0 –0.78 –0.64–0.56 –0.51–0.37+0.24+0.59 +0.62+0.63 +0.84+0.89

σ1

σ2

AE, σ3

Bππ*C GH

F, σ4

D

–1.0 –0.78 –0.64–0.56 –0.51–0.37+0.24+0.59 +0.62+0.63 +0.84+0.89

σ1

σ2

AE, σ3

Bππ*C GH

F, σ4

D

2pzC

π*

2pz C

π

2pzC

π*

2pz C

π

b. Ethylene molecular orbitals

1σ 2σ 3σ 4σ

E F G H

A B C D c. Observations

– σ1-σ4 involves the 2s orbitals, E-H involves the 2px orbitals, A-D involves the 2py orbitals, and π-π* involves the 2pz orbitals of C.

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CHEM 6320 Fall 2005

– σ3 and E orbitals are not orthogonal to each other. (They have the same symmetry.) A combination of σ3 and E is a better description of the real orbital. Same for σ4 and F orbitals.

σ3

E

σ3+E

σ3σ3

EE

σ3+Eσ3+E

– The energy difference between π bonding and antibonding MOs

is smaller than that for the σ orbitals due to a weaker π interaction (overlap) compared to σ.

– The orbitals and the nodes have some symmetry elements as the whole molecule.

8. Symmetry Approach to the Construction of Molecular Orbitals a. An equivalent approach to construction of delocalized (polyatomic)

molecular orbitals considers these as a linear combination of localized (2-atom) molecular orbitals using molecular symmetry.

b. Procedure – Create a bonding model of localized MO from a basis set of AO – Identify the symmetry elements of the molecule – Identify which MOs might already be symmetry adapted – For those which are not, group together orbitals which are

transformed into each other by symmetry operations, and form + and – combinations. The total number of orbitals at the end must equal that at the start.

c. Observations – All symmetry elements may not be needed, but conclusions can

be based only on those symmetry elements to which the orbitals are adapted.

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CHEM 6320 Fall 2005

– The situation becomes more complicated if there is a 3-fold or higher rotational axis.

d. The example of water

H H

O

σv'

σv

C2

H H

O

σv'

σv

C2– Lewis structure predicts sp3 hybridization, two of which are used for σ bonding and σ* antibonding with two H 1s and the other two contain lone pairs.

– The elements of symmetry for water: – There are two localized molecular orbitals:

H H

O

H H

O

H H

O

H H

O

1ψ 2ψ

– The localized molecular orbitals transform into themselves by reflection in σ′. Neither is symmetry adapted with respect to σ or C2. (For these symmetry operations, they are transformed into each other.)

– Therefore, take positive and negative combinations of these localized molecular orbitals to obtain:

H H

O

H H

O

H H

O

H H

O

21S ψψ +=Ψ 21A ψψ −=Ψ

– Both new orbitals are either symmetric or antisymmetric to all symmetry elements.

e. The example of methane

yx

z

yx

z

H

H

H

H

C

H

H

H

H

C

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CHEM 6320 Fall 2005

– The four localized molecular orbitals:

HH

HH

C

ψ1

HH

HH

C

ψ1

HH

HH

C

ψ2

HH

HH

C

ψ2

HH

HH

C

ψ3

HH

HH

C

ψ3

HH

HH

C

ψ4

HH

HH

C

ψ4 – Note that:

ψ1 ←→ (read transform into) ψ3 by C2(x); ψ1 ←→ ψ2 by C2(y); ψ1 ←→ ψ4 by C2(z)

– Delocalized molecular orbitals: 4321S ψψψψ +++=Ψ 4321Axz

ψψψψ −−+=Ψ

HH

HH

C

ΨS

HH

HH

C

ΨAxz

HH

HH

C

ΨAxz

HH

HH

C

ΨS

yzAΨ and Ψ are similar. xyA

– Final picture of the molecular orbital energy diagram is similar to the one before:

ΨS

ΨA

ΨS

ΨA

– Molecular orbitals as combination of atomic orbitals (from

before):

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CHEM 6320 Fall 2005

9. Approximate LCAO-MO Theory a. Writing a molecular orbital as a linear combination of atomic orbitals

(LCAO) leads to secular determinant. – Molecular orbitals:

∑=

=n

ssisi c

1ϕψ

for n basis set atomic orbitals ϕ. (cis is the coefficient of atomic orbital s in ith molecular orbital.)

– Introduce in Schrödinger equation:

∑∑==

===n

ssisiii

n

ssisi cEEHcH

11ˆˆ ϕψϕψ

– Recall orthogonality and normalization requirements, which can be combined into one by saying that eigenfunctions constitute an orthonormal set:

rssr Sd =∫ τϕϕ (Srs = 0 if r ≠ s, Srs = 1 if r = s) – Multiply Schrödinger equation above by ψi and integrate:

∑ ∑ ∫∑ ∑ ∫= == =

=n

r

n

ssrisjr

n

r

n

ssrisjr dccEdHcc

1 11 1ˆ τϕϕτϕϕ

– Abbreviate ∫= τϕϕ dHH srrs ˆ (called Coulomb integral) and ∫= τϕϕ dsS rrs (called overlap integral).

– Rewriting: ∑ ∑

∑ ∑

= =

= == n

r

n

srsisjr

n

r

n

srsisjr

Scc

HccE

1 1

1 1

– The set of orbitals and energies are obtained by putting the condition that energy is minimum with respect to the coefficients (Variational Principle), i.e., the derivative of the energy with respect to the coefficients is zero. These conditions led to a set of n equations with n unknowns, cis.

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CHEM 6320 Fall 2005

– These equations are called secular equations, and, written out, they are:

0)(...)()(

0)(...)()(0)(...)()(

222111

222222212121

112121211111

=−++−+−

=−++−+−=−++−+−

innninninininin

inniniiii

inniniiii

cSEHcSEHcSEH

cSEHcSEHcSEHcSEHcSEHcSEH

MMMM

– These secular equations have non-trivial solutions if and only if the determinant of coefficients (called the secular determinant) is equal to zero:

0

...............

...

...

2211

2222222121

1112121111

=

−−−

−−−−−−

nninnninnin

ninii

ninii

SEHSEHSEH

SEHSEHSEHSEHSEHSEH

When expanded, this represents an nth order polynomial in E, with n roots Ei.

b. Electron correlation – The determinant above cannot be solved exactly because of our

ability to calculate the repulsion (interaction) terms Hrs. This is due to an obstacle known as electron correlation.

– For example, consider adding an electron to a system of one electron. The motion of the electron is influenced by its location relative to the first electron. But the location of the first electron is now dependent on the location of the second electron, which is exactly the problem we are trying to solve.

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CHEM 6320 Fall 2005

E. Hückel Molecular Orbital Theory 1. Method Description

a. Only π electrons are treated. In the LCAO equation above, ϕs is a 2p orbital on atom s, so that our basis set consists on n such 2p orbitals in the π system.

b. The Hamiltonian integral Hrs, r ≠ s, (also called resonance integral) which represents the interaction energy of the electron in ϕr with the electron in ϕs, is assumed to be:

= 0 when r and s are not directly bonded = β (some average non-zero value) when r and s are directly

bonded Both of these assumptions are severe approximations.

c. The Hamiltonian integral Hrr, representing the energy of the electron in the basis set atomic orbital ϕr, is the same value α if all the atoms in the π system are carbons.

d. Altogether, remembering what restrictions our use of an orthogonal set of orbitals places on the overlap integral Srs:

1=rrS 0=rsS , if r ≠ s α=rrH , the Coulomb integral β=rsH , if atoms r and s are adjacent 0=rsH , if atoms r and s are neither identical nor adjacent

– Note #1: Both α and β are negative numbers. – Note #2: The Coulomb and resonance integrals, Hrr = α and Hrs

= β, assume different values for 2p orbitals on oxygen, nitrogen, etc.

e. With these approximations, the secular determinant can be reduced according to:

– diagonal elements are α – E. – off-diagonal elements are either β or 0, depending upon the

connectivity of the molecule.

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CHEM 6320 Fall 2005

2. Examples a. The case of ethylene

22 CHCH = – The two molecular orbitals are given by:

)(21)(211 BpBApA cc ϕϕψ +=

)(22)(222 BpBApA cc ϕϕψ +=

– The secular determinant will be:

0=−

−E

Eαββα

– Introduce a new variable β

α Ex −=

01

1=

xx

– There are two solutions (x = 1 and x = –1) that are associated

with two orbitals. E1 = α + β and E2 = α – β

– After getting the energy levels, plug each solution for x into the original set of secular equations and solve, using normalization

requirements that 11

=∑=

n

iijc .

– For x = –1 solution ⇒ E1 = α + β – The two secular equations for this solution:

0)()( 1111 =−+− BABABAAAAA cSEHcSEH 0)( 111 =⋅+− BA ccE βα

011 =⋅+⋅− BA cc ββ ccc BA == 11

where the two carbon atoms are labeled A and B.

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CHEM 6320 Fall 2005

– To obtain the values for c1A and c1B, one need to put the normality condition for the molecular orbital.

)(2)(2)(21)(211 BpApBpBApA cccc ϕϕϕϕψ +=+=

12

22

)(2)(221)(2)(211

)(2)(22111

==

++

+=

∫∫

∫∫

c

dcdcc

dcd

BpBpBBpApBA

ApApA

τϕϕτϕϕ

τϕϕτψψ

2/1=c

)(221

)(221

1 BpAp ϕϕψ +=

– For x = 1 solution ⇒ E2 = α – β – The two secular equations for this solution:

0)()( 2222 =−+− BABABAAAAA cSEHcSEH 0)( 222 =⋅+− BA ccE βα

022 =⋅+⋅ BA cc ββ ccc BA =−= 22

– To obtain the values for c2A and c2B, one need to put the normality condition for the molecular orbital.

)(2)(2)(22)(222 BpApBpBApA cccc ϕϕϕϕψ −=+=

12

22

)(2)(222)(2)(222

)(2)(22211

==

+−

=

∫∫

∫∫

c

dcdcc

dcd

BpBpBBpApBA

ApApA

τϕϕτϕϕ

τϕϕτψψ

2/1=c

)(221

)(221

2 BpAp ϕϕψ −=

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CHEM 6320 Fall 2005

– Molecular orbital energy diagram:

α

α – β

α + β

– The energy of an electron in a pure 2p orbital is α. – The total π energy of ethylene is 2(α + β) = 2α + 2β

Eπ = 2α + 2β (= the energy of a localized double bond) – More general, for ground state:

Eπ = Σ ni Ei – The values of α and β are not calculated but rather obtained

empirically by comparison with experiment. By doing so, β = –75 kJ/mol.

b. The case of butadiene 22 CHCHCHCH =−=

– The secular determinant is:

0

000

000

=

−−

−−

EE

EE

αββαβ

βαββα

– Introducing β

α Ex −= (⇔ xE βα −= )

0

100110011001

=

xx

xx

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CHEM 6320 Fall 2005

– Expand

13

)1()2(010

10011

110

1101

24

23

+−=

=−−−==−

xx

xxxxx

xx

xxx

– Solutions are:

618.1,618.0,618.0,618.12

51,2

51++−−=

±±−=x


βα 618.1+

βα 618.0+

βα 618.1−

βα 618.0−

βα 618.1+

βα 618.0+

βα 618.1−

βα 618.0−

– The π wavefunctions for butadiene:

– The energy of the π electrons (total π electronic energy): ( ) ( ) βαβαβαπ 472.44618.02618.12 +=+++=E

– Compared with two ethylenes, 2(2α + 2β ), there is an additional 0.472β of energy. This is the delocalization energy (Edeloc).

kJ/mol35472.0)HC(2)HC( 4284deloc −≈=−= βππ EEE

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CHEM 6320 Fall 2005

c. The case of benzene – The secular determinant is:

0

100011100001100001100001110001

=

xx

xx

xx

6

54

3

21

– There are six solutions: x = ±1, ±1, ±2 – Molecular orbital energy diagram:

βα 2+

βα +

βα 2−

βα −

βα 2+

βα +

βα 2−

βα −

– The energy of the π electrons:

( ) ( ) ( ) βαβαβαβαπ 862222 +=+++++=E – The delocalization energy:

kJ/mol1502)HC(3)HC( 4266deloc −≈=−= βππ EEE

– The π wavefunctions for benzene:

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CHEM 6320 Fall 2005

d. The case of naphthalene 1

2

3

456

7

89

10

– The secular determinant is:

0

100011000110000001011000000001100000000110000100010000100000100000000110000000011010000001

=

xx

xx

xx

xx

xx

e. The case of allyl – The secular determinant is:

010

1101=

xx

x CH2 CH CH2

023 =− xx – The three solutions are: x = 2− , 0, 2+ – Obtaining the molecular orbitals for solution x = 2− :

βα 21 +=E

=−+=+−+=+−

0)2(0)2(0)2(

32

321

21

ccccc

cc

– From first equation: c 21

22 2c=

– From first and third equations: c 31 c=

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CHEM 6320 Fall 2005

– Normalization condition: 12

322

21 =++ ccc ⇒ ⇒ 12 2

121

21 =++ ccc 14 2

1 =c

21

1 =c ⇒ 21

3 =c ⇒ 2

12 =c

)3(221

)2(221

)1(221

1 ppp ϕϕϕψ ++=

(no node) – Obtaining the molecular orbitals for solution x = 0:

α=2E

==+=

000

2

31

2

ccc

c

02 =c ⇒ 2

11 =c ⇒

21

3 −=c

)3(221

)1(221

2 pp ϕϕψ −=

(one node) – Obtaining the molecular orbitals for solution x = 2+ :

βα 23 −=E

Find as above: 21

1 =c ; 21

3 =c ; 2

12 −=c

)3(221

)2(221

)1(221

3 ppp ϕϕϕψ +−=

(two nodes)

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CHEM 6320 Fall 2005


α

α + β2

α – β2

α

α + β2α + β2

α – β2α – β2

– Energy of the π electrons:

βββ

βαβαβα

π π

0.820.820.82

2244anion2233radical2222cation

electrons#species allyl delocEE

+++

– The extra energy (compared with pure p electrons) and the delocalization energy are the same for the allyl cation, radical, and anion.

3. Limitations and Improvements a. Filling orbitals with electrons in this way ignores the fact that there will

be increased repulsion as we add successive electrons to the system. Such repulsion is considered only in some average way incorporated into the parameters α and β. Again, the solutions are numerically inaccurate, but relative comparisons consistently predict physical properties.

b. We assumed that the overlap integral Sij = 0 for i ≠ j. For non-neighboring 2p orbitals, this assumption is valid, but overlap for neighboring 2p orbitals is about 0.25. How would the calculation change if we took this into account?

– The case of ethylene

022222121

12121111 =−−−−ESHESHESHESH

0=−−−−EESESE

αββα

where S is the adjacent 2p – 2p overlap

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CHEM 6320 Fall 2005

– Introduce a new variable ESEx

−−

=βα (

SxxE

−−

=1

βα )

01

1=

xx

– There are two solutions (x = ±1) and the associated energies

are S

E++

=11

βα and S

E−−

=12

βα

– Define a new nonbonding level 21 SS

−−

=′βαα and a new

bond (resonance) integral 21 SS

−−

=′αββ .

– With these definitions, the energy levels are given by: βα ′±′=E (the same form as before)

c. Since the exact choices of α and β are numerically inaccurate (arbitrary) anyway, one can make believe that the used α and β incorporate an average correction for Sij when i and j are adjacent. The spacing between energy levels will be slightly different, but are reasonably good for relative comparisons.

4. Simplified Solutions of Energy Levels and Coefficients a. For linear polyene of n atoms:

– Secular determinant solutions

1cos2

+−=

njx jπ where nj ,...,3,2,1=

– The coefficient of atom i in jth molecular orbital wavefunction

1sin

12

++=

nij

ncij

π

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CHEM 6320 Fall 2005

b. For rings of n atoms: – Secular determinant solutions

njx jπ2cos2−=

where

±

±±±=

−

even for odd for

,...2,1,02

21

nn

j n

n

– Trigonometric solutions can be obtained through use of Frost’s circle “trick”: For a ring of n atoms, inscribe a regular polygon of n side in a circle of diameter 4β with one corner at the lowest point. (The lowest energy is always E = α + 2β). The points of the polygon will then correspond to the appropriate energy levels.

– Note: Frost’s circle works for complete π systems. It should not be used for cyclopentadiene that should be treated as butadiene.

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CHEM 6320 Fall 2005

5. Qualitative Pictures for Molecular Orbitals in Linear Polyene a. Procedure

– Draw a straight line connecting 2 points at one C–C bond length on each terminus.

– Draw vibrating strings with increasing number of nodes (but keep the same amplitude).

– Draw the atomic orbital at each C to touch the string. b. Example butadiene:

– The coefficients of atomic orbitals in molecular orbitals are given

by geometric condition above. – Observations:

– First molecular orbital has no nodes. – Second molecular orbital has one node at the middle of the

molecule. – The nth molecular orbital has (n – 1) nodes.

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CHEM 6320 Fall 2005

6. Molecular Orbital Analysis a. Electron densities and charge densities

– Total electron density at atom i:

∑=j

ijji cnq 2

where j is the number of MO, nj is the number of electrons in that orbital, and is the coefficient of an atomic orbital on atom i in j

jrcth MO.

n = 2 if the molecular orbital is doubly occupied n = 1 if the molecular orbital is singly occupied n = 0 if the molecular orbital is empty

– Charge density at atom i: iq−=1density Charge

– The case of ethylene 1)(2 2

21

21 === qq

0density Charge = – The case of allyl cation

There are only two electrons in ψ1.

∑∑ +==+

+

122/12/14/12/13

012/12/122/12/14/12/11

density charge2iii qcci

– The case of allyl radical and anion For allyl radical, one additional electron in ψ2 For allyl anion, two additional electrons in ψ2

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CHEM 6320 Fall 2005

∑∑∑ −=−

−

===⋅+⋅+

++=⋅+⋅+

12/1

02/1

density chargeanion

435.1)2/1(22/1)2/1(12/14/12/13

01010025.1)2/1(22/1)2/1(12/14/12/11

anion)(radical)(2iiii qqcci

– Note that charge densities match those predicted by valence

bond theory resonance forms:

+ + +1/2 +1/2≡

+ + +1/2 +1/2≡

– – –1/2 –1/2≡

– – –1/2 –1/2≡

b. Bond orders

– Partial π bond order in jth MO between atoms r and s:

jsj

jrjjrs ccnp ∑=

– The case of ethylene 1)2/1)(2/1(212 ==p (as expected)

– The case of allyl (cation, radical, anion): 707.02/1)2/1)(2/1(22312 ==== pp

– Note that the value is in contrast to our usual sense of a half π bond order from valence bond resonance forms. (The bond is stronger than given by resonance theory.)

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CHEM 6320 Fall 2005

7. Summary a. Delocalization energy is calculated relative to 2(α + β) which is the

energy of a double bond in ethylene (and α which is the energy of a pure p orbital not participating in a double bond = if the molecule is a radical)

αβαπ mnEE −+−= )22()system(deloc where n is the number of double bonds and m is the number of unpaired electrons (e.g. m = 1 in allyl radical)

b. Resonance energy is calculated relative to α which is the energy of a pure p orbital in an atom

c. Procedure for Hückel method – Write the molecule and label the atoms – Write the secular determinant – Solve the secular determinant to calculate the energy levels

( βα xE −= where highest value of –x corresponds to the lowest energy)

– Plug in the values of x into secular equations – Solve secular equations (including the normalization condition)

to determine the coefficient of atomic orbitals in each molecular orbital

– Represent (draw) each molecular orbital – Fill in the molecular orbital energy diagram with the appropriate

number of electrons – Calculate π electron energy (Eπ = Σ ni Ei ) – Calculate delocalization energy (and resonance energy) – Calculate electron densities, charge densities, and bond orders

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CHEM 6320 Fall 2005

F. Application of Molecular Orbital Theory to Reactivity and Stability 1. Hyperconjugation (Hyperconjugative Stabilization Effect)

a. It explains why ethyl cation is more stable than methyl cation. b. It is sometimes referred to as “no-bond” resonance:

CH2CH

H

H H

H

H C CH2

Ha

Hb

Hc

HH

Ha

Hb

Hc

HH

c. The CH3CH2+ ion has a plane of symmetry:

d. The three bonding C – H orbitals of the methyl group in CH3CH2+:

Ha

Hb

Hc

HH

Ha

Hb

Hc

HH

Ha

Hb

Hc

HH

Ha

Hb

Hc

HH

Ha

Hb

Hc

HH

Ha

Hb

Hc

HH

e. The symmetry adapted C – H orbitals ΨS and ΨA:

Ha

Hb

Hc

HH

Ha

Hb

Hc

HH

ΨS

Ha

Hb

Hc

HH

Ha

Hb

Hc

HH

ΨA

f. Interaction of the methyl group orbital ΨA with the p orbital of the

cationic carbon stabilizes a filled orbital (therefore the molecule):

C –– C

H

H

C –– C

H

H

C –– C

H

H

C –– C

H

H

C –– C

H

H

C –– C

H

H

C –– C

H

H

C –– C

H

H

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CHEM 6320 Fall 2005

2. Bimolecular Substitutions a. Reactivity can be explained in most cases by looking at the

HOMO-LUMO interaction (in the transition state) – HOMO stands for the highest occupied molecular orbital – LUMO stands for the lowest unoccupied molecular orbital

b. Molecular orbital diagram for bimolecular electrophilic substitution

C X

E

C X

E

C X

E

C X

E

C X

E

C X

E

LUMOLUMO

LUMOLUMO

HOMOHOMOHOMO

C X

E

C X

E

– The substitution can occur from the front because symmetry

permits HOMO-LUMO interaction and stabilization. – A LUMO-LUMO interaction is not possible due to symmetry

(and would not produce any stabilization anyway). c. Molecular orbital diagrams for bimolecular nucleophilic substitution

– Frontside attack:

C X

Nu

C X

Nu

C X

Nu

C X

NuC X

Nu

C X

NuLUMOLUMO

HOMOHOMOHOMO

HOMOHOMOHOMO

– Symmetry prevents HOMO-LUMO interaction so the only

interaction is between filled orbitals. The reaction will not take this path.

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CHEM 6320 Fall 2005

– Backside attack:

C XNu C XNu

C XNu C XNu

C XNu C XNuLUMOLUMO

C XNu C XNu

HOMOHOMOHOMO

HOMOHOMOHOMO

– Now the HOMO-LUMO interaction is possible and the

transition state is stabilized. 3. Substituent Effects or Heteroatom Replacements in Conjugated System

a. Example: ethylene vs. formaldehyde

α

α – β

α + β

π*

π

π*

π

C – O

– As electronegative increases, the energy of a 2p atomic orbital

goes down in energy. – There is a greater contribution of the atomic orbital of O than

that of C in the bonding molecular orbital and reverse for the antibonding one.

– As a consequence, the electron density over O will be higher than over C (⇒ O will have a partial negative charge and C will have a partial positive charge).

– An electrophile will react with filled π orbital and will attack O. – A nucleophile will react with empty π* orbital and will attack C. – CH2=O will be more reactive than CH2=CH2 toward nucleophile. – CH2=O will be less reactive than CH2=CH2 toward electrophile.

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CHEM 6320 Fall 2005

b. Example: ethylene vs. vinylamine

CH2 CH NH2

..

NH2CHCH2 – Vinylamine is isoelectronic with allyl anion.

α

α – β

α + β

α – β2α – β2

α + β2α + β2

C NC N

– The shape of vinylamine HOMO (ψ2):

– The theory predicts that vinyl amine will be more reactive than ethylene toward an electrophile, and that the attack will occur on terminal C (as predicted also by resonance).

– Both compounds are equally reactive toward a nucleophile. 4. Symmetry Approach to Interaction between Reactants Molecular Orbitals

a. Use perturbation MO analysis of reaction probability between two molecules (each defined in MO terms)

– Transition state (‡) for reaction involves partial bonding therefore partial interaction between orbitals.

– The most important interaction will be between HOMO of nucleophile and LUMO of electrophile (frontier orbital control).

– The strength of interaction increases as interacting orbitals are closer in the energy.

– Only molecular orbitals of matching symmetry can interact. To do that, define a symmetry element, which is preserved from the beginning to the end of the reaction (R → ‡ → P). One needs to assign reactant and product orbitals as symmetric or antisymmetric with respect to chosen symmetry element.

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CHEM 6320 Fall 2005

b. Correlation diagrams – Consider the following electrocyclic reactions:

++ +

vs.

++ +

– Assume concerted bond formation with geometry shown.

‡ ‡

– Observations

– The element of symmetry is a plane of symmetry (although a C2 axis can also be considered)

– In the representation above, we did not consider the correct amplitude of the atomic p orbitals in each molecular orbital (because it doesn’t change the conclusions).

– For allyl cation + ethylene reaction there is no net stabilization of the transition state compared to reactants.

– For allyl cation + butadiene reaction there is a net stabilization of an occupied orbital through an interaction with an empty orbital (HOMO-LUMO interaction).

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CHEM 6320 Fall 2005

G. Problems and Exercises 1. For the molecule CX2Y2, if carbon uses sp2 orbitals for bonding to X, what

hybrid orbitals does it use for bonding to Y? 2. The H–N–H angle in ammonia is 107.3°. Calculate the hybrid orbital N

uses to bond to H. Then calculate the hybridization of the lone-pair and the interorbital angle between the H and the lone pair.

3. a. The strain energy of spiropentane (62.5 kcal/mol) is considerably

greater than twice that of cyclopropene (27.5 kcal/mol). Suggest an explanation.

1

2

3 b. The fractional s character in bonds to carbon in organic molecules may

be estimated by its relation to 13C–13C coupling constants, as determined by NMR. Estimate the fractional s character of C-1 in its bond to C-3 of spiropentane, given the following information

)1(3

CC)3(1

1313

KsJ

s −=

where K is a constant equal to 550 Hz, the 13C–13C coupling constant J between C-1 and C-3 is observed to be 20.2 Hz, and s3(1) is the s character at C-3 in its bond to C-1.

4. Provide electronic configurations (Lewis dot structures) for the following

species, also indicating the appropriate hybridization of the “heavy” atoms (everything but H). Give resonance forms and the resonance hybrid when resonance is important (you may use line bonds to indicate electron pairs). a. CO3

2– b. HCO3

– c. NO2

+ d. NO2

– e. N3

– (azide)

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CHEM 6320 Fall 2005

f. SO42–

g. SO32–

h. S2O32– (thiosulfate)

i. NCO– (isocyanate) j. CNO– (cyanate) k. CH3NC (methyl isocyanate) l. ClO4

– 5. Write resonance forms for the following (when appropriate)

a. 2-cyanoimidazole b. 2- and 3- nitrofuran c. 2-, 3-, and 4-aminobenzaldehyde d. 2-, 3-, and 4-methoxypyridine

6. Suggest a possible explanation for the following observations:

a. The dipole moment of the hydrocarbon calicene has been estimated to be as large as 5.6 D.

calicene b. The dipole moment of furan is smaller than and in opposite direction

from that of pyrrole.

O

0.71 D

_

_

1.80 D

NH

7. Predict the energetically preferred site of protonation for each of the

following molecules and explain the basis of your prediction. a. N C6H5C6H5CH

b. C

O

NHCH3

CH3

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CHEM 6320 Fall 2005

c. NH

d. NNH2

8. Use thermochemical relationships to obtain the required information.

a. The heats of formation of cyclohexane, cyclohexene, and benzene are, respectively, –29.5, –1.1, and +19.8 kcal/mol. Estimate the resonance energy of benzene.

b. The heats of formation of 2-methyl-1-pentene and 2-methylpenthane are, respectively, –13.6, and +41.7 kcal/mol. Calculate the heat of hydrogenation of 2-methyl-1-pentene.

9. Identify the point group and the elements of symmetry for the following

molecules: a. ortho-dichlorobenzene b. meta-dichlorobenzene c. para-dichlorobenzene d. naphthalene e. 1,5-dichloronaphthalene f. anthracene

10. Consider construction of full MO depiction of acetylene (HC≡CH) the

same way ethylene was handled in the lecture notes. There are 10 total atomic orbitals. (The 1s orbitals of carbon are not included.) Draw all the molecular orbitals (considering nodal symmetry requirements). Describe how many nodes each molecular orbital has and show where these nodes are located.

11. Calculate the energy levels and coefficients for 1,3-butadiene using Hückel

MO theory.

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CHEM 6320 Fall 2005

12. Estimate from Hückel MO theory the delocalization energy, expressed in units of β, of cyclobutadienyl dication (C4H4

+). 13. Calculate Eπ for cation, radical and anion of and 14. Set up and solve the secular determinant for system below. Calculate Eπ

and delocalization energy.

15. a. Sketch the nodal properties of the highest occupied molecular orbital of

pentadienyl cation (CH2=CH–CH=CH2–CH2+).

b. Two of the π-MOs of pentadienyl are given below. Specify which one is of lower energy, and classify each as to whether it is bonding, nonbonding, or antibonding. Explain your reasoning.

1 2 3 4 5

5421 50.050.050.050.0 φφφφψ −−+=x

531 58.058.058.0 φφφψ +−=y

16. Construct a qualitative MO diagram showing how the π-molecular orbitals

in the following molecules are modified by the addition of the substituent: a. vinyl fluoride, compared to ethylene b. acrolein, compared to ethylene c. propene, compared to ethylene d. fluorobenzene, compared to benzene

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CHEM 6320 Fall 2005

Principles of Stereochemistry A. Definitions

1. Isomers – different compounds with the same molecular formula 2. Stereoisomers – isomers with the same connectivity but different

configuration (non-inconvertible by single bond rotation) – Note distinction between configuration and conformation

(different spatial orientation interconvertible by single bond rotation)

3. Atropisomers – stereoisomers that result from restricted single bond rotation

Cl

CH3

CH3

Cl

Cl

CH3

Cl

CH3

4. Enantiomers – stereoisomers that are mirror images of one another

a. Chirality – the property of non-superimposability on mirror image b. Homochiral – the property of being enantiomerically pure

5. Diastereomers – non-mirror image stereoisomers (stereoisomers in which the distance between atoms, i.e. nuclei, is different)

6. Optical Activity – property of being able to rotate the plane of plane-polarized light a. Optical rotation – a physical property but varies with wavelength of

light λ, conditions (solvent, temperature), and concentration. b. Specific rotation [α]D – the associated physical constant, quoted for a

given solvent, temperature, and concentration (to exclude complications of non-Beer’s law behavior) (D refers to the sodium D line at 589 nm).

c. Optical purity – defined for a mixture of enantiomers (the enantiomers have equal but opposite in sign [α]D values) as:

100senantiomer pure of ][

mixture of ][ (%)purity optical ×=α

α

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CHEM 6320 Fall 2005

d. Enantiomeric excess (e.e.) – an equivalent property defined as:

−×

enantiomerminor offraction mole

enantiomermajor offraction mole

100

7. Optical Rotatory Dispersion (ORD) – relates to how [α] changes with the wavelength (λ), and is a reflection of configuration

8. Circular Dichroism (CD) – relates to the ability to show differential absorption of circularly polarized light. The constant analogous to [α] is θ, called the molecular ellipticity.

9. Chiral Center – typically a tetrahedral (sp3) carbon with 4 non-identical ligands. Other examples are unsymmetrical sulfoxides and unsymmetrical tetraalkylammonium salts, but not unsymmetrical amines.

S

..

PhCH3

OS

O

CH3

Ph

..

slow

fast Ph EtNMe

..

Nt

a. Unsymmetrical amines

N

..

PhMe

Et

fast

NPhEtMe

..sp2

sp3

b. Unsymmetrical tetraalkylammon

NPh

MeEt

CH2Ph

Br

– The rehybridization is impec. Unsymmetrical sulfoxides

S

..

PhCH3

OS

Ph

..

slow

– The configuration is more sO and d orbitals of S.

52

bu

..Me

EtPh

NEtMe

Ph

..

sp3

ium salts

ded in this case.

O

CH3

S

O

RR ..

table due to interaction between the

CHEM 6320 Fall 2005

B. Conventions 1. Cahn-Ingold-Prelog Convention

a. Use R (rectus) and S (sinister) descriptors that are assigned using the sequence rule to assign a priority order to the substituents on the atom to which a configuration is being assigned.

– The R and S descriptors are not related to (+) and (–) (clockwise and counterclockwise rotation of polarized light, respectively).

b. Procedure – Assign decreasing priority in order of decreasing atomic number.

– When two or more of the substituent atoms are the same element, the priority is assign based on the next attached atom in those substituents.

– An atom that is multiply bonded is counted once for each formal bond.

– When two isotopes of the same elements are substituents, the isotope with larger atomic mass takes priority over the isotope with lower atomic mass.

– D has priority over H – 13C has priority over 12C

– Lone pair is equivalent with atomic number 0. – Look at the molecule with the substituent with the lowest priority

(#4) going away (going back) and the substituents with priorities 1, 2, and 3 toward you.

12

3

4

4

32

1

– Assign the configuration R if the substituents decrease in priority

(1→2→3) in a clockwise fashion or assign the configuration S if the substituents decrease in priority in a counterclockwise sense.

R4

3 2

1

1

2 3

4 S

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CHEM 6320 Fall 2005

c. Examples – Assign priorities for the following substituents

– What is the smallest hydrocarbon that can have a chiral carbon?

D

H

CH3

T

ethane:

– Assign R or S configuration for all chiral centers below:

CH3CH2

H

CH3

OH

S

..CH3

OPh

ClH

OH

H 2. Fisher Projections

a. A Fisher projection is a two-dimensional representation of a three-dimensional structure in which the horizontal bonds are directed toward the viewer (toward front) while the vertical bonds are directed away from the viewer (toward back):

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CHEM 6320 Fall 2005

b. It is used mainly for carbohydrates. c. Rotating a Fisher projection by 90° gives the enantiomer of the initial

molecule, and rotating a Fisher projection by 180° gives back the initial molecule.

CH3

H Cl

Br

ClH

CH3

Br H

Br CH3

ClCH3Br

H

Cl 3. Fisher (D-L) Convention

a. It was initially developed and is used mainly for carbohydrates. b. Procedure

– Align the molecule with the highest oxidation state C at top – Write the molecule in a Fisher projection – Look at the chiral center closest to the bottom – If the substituent (e.g. OH) projects to right ⇒ D isomer – If the substituent (e.g. OH) projects to left ⇒ L isomer – Note: The D and L descriptors are not related to (+) and (–)

(clockwise and counterclockwise rotation of polarized light, respectively).

HO CH2 CHCH O

OHb. Examples

– Glyceraldehydes CHO

H OH

CH2OH

D-glyceraldehyde(+)

D-(+)-glyceraldehyde

R

S

L-(-)-glyceraldehyde(-)

L-glyceraldehyde

CHO

HO H

CH2OH

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CHEM 6320 Fall 2005

– Erythrose and threose OH

OCH CHCH2HO CH

OH

– The presence of another chiral center increase the numbers

of stereoisomers (There are 2n possible stereoisomers for n chiral centers)

CHO

HO H

OHH

CH2OH

D-threoseD-erythrose

CHO

H OH

OHH

CH2OH

CHO

HO H

HHO

CH2OH

L-erythrose L-threose

CHO

H O

H

HHO

CH2OH

– D-erythrose and D-threose are diastereoisomers (so they have different names)

– D-erythrose and L-erythrose are enantiomers (mirror image of each other)

– IUPAC names: – D-erythrose ≡ – L-erythrose ≡ – D-threose ≡ – L-threose ≡

4. Erythro-Threo Terminology a. It can be used only for the case of two chiral centers. b. The erythro and threo are defined (based on erythrose and threose) as:

A

H X

XH

B

erythro

A

X H

XH

B

threo

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CHEM 6320 Fall 2005

c. The case of A ≠ B: – There are two chiral centers therefore 4 stereoisomers:

– an erythro enantiomeric pair (optically active) A

X H

HX

B

A

H X

XH

B – a threo enantiomeric pair (optically active)

A

H X

HX

B

A

X H

XH

B d. The case of A = B:

– There are still two chiral centers but because the substituents for the these centers are the same there are only 3 different stereoisomers:

– a threo enantiomeric pair (optically active) A

X H

XH

A

A

H X

HX

A – These are (R,R) and (S,S) isomers.

– an achiral molecule due to symmetry (the molecule has a plane of symmetry), called “meso”:

A

H X

XH

A – This is the (R,S) isomer and is an example of a molecule that has chiral center but not optical activity.

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CHEM 6320 Fall 2005

C. Stereochemistry of Selected Classes of Compounds 1. Amino Acids

a. Natural amino acids found in proteins have the L-configuration COOH

CH3

HH2NL-alanine: L-serine:

COOH

CH2OH

HH2N

2. Cycloalkanes - Introduction

a. Many cycloalkanes present different possible conformations that are (easily) interchangeable into one another. Example cyclohexane:

a

g

j

dh

ec

li

k b

fd

j

g

a

fb

ki

e

c l

h

b. If the conformers are easily interchanged, in investigating the stereochemistry of the molecule, it is more convenient to look at the carbon ring as being planar.

fed

lkj

ac b

gi h

c. In general, if a molecule has a plane of symmetry or a center of inversion, that molecule is optically inactive even though it has chiral centers (two or more).

3. Monosubstituted Cycloalkanes a. Compounds in this class do not have a chiral center therefore they are

optically inactive. b. These molecules have a plan of symmetry.

XX

XX

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CHEM 6320 Fall 2005

4. Disubstituted Cycloalkanes a. Disubstituted cyclopropane

– 1,2-dimethyl cyclopropane – A pair of enantiomers (trans)

CH3

CH3

CH3

CH3R,R S,S

– A meso, achiral stereoisomer (cis) CH3CH3

CH3 CH3R,SR,S

b. Disubstituted cyclobutane – 1,2-dimethyl cyclobutane

– A pair of enantiomers (trans)

S,S

CH3

CH3

CH3

CH3

CH3

CH3R,R S,S

– A meso, achiral stereoisomer (cis)

CH3

CH3

CH3

CH3

R,S R,S – 1,3-dimethyl cyclobutane

– Two achiral stereoisomers (not meso) (cis and trans)

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CHEM 6320 Fall 2005

c. Disubstituted cyclopentane – 1,2-dimethyl cyclopentane


CH3

CH3CH3

CH3

CH3CH3

S,SR,R S,S – A meso, achiral stereoisomer (cis)

R,S R,S

CH3

CH3

CH3

CH3

– 1,3-dimethyl cyclopentane

–

S,SR,R S,S

CH3CH3

CH3

CH3

CH3

CH3


CH3

CH3

CH3

CH3

R,SR,S

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CHEM 6320 Fall 2005

d. Disubstituted cyclohexane – 1,2-dimethyl cyclohexane

– A pair of enantiomers (trans) CH3

CH3

CH3

CH3

CH3

CH3S,SR,R S,S

– A meso, achiral stereoisomer (cis) CH3

CH3

CH3CH3

R,S R,S – 1,3-dimethyl cyclohexane


S,SR,R S,S

CH3

CH3CH3

CH3CH3

CH3


R,SR,S

CH3CH3

CH3CH3

– 1,4-dimethyl cyclohexane

–

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CHEM 6320 Fall 2005

5. Tetrasubstituted Cycloalkanes a. 1,3-Dibromo-2,4-dimethylcyclobutane

– The trans, cis, cis stereoisomer

BrCH3

Br

CH3

BrCH3

Br

CH3 – –

– The trans, trans, cis stereoisomer

CH3

Br

CH3Br

CH3

Br

CH3Br

– –

b. 1,4-Dibromo-2,5-dimethylcyclohexane – The molecule has four chiral centers for a total of 16 possible

stereoisomers but some of them are meso forms (optically inactive, superimposable on their mirror image):

– The S, R, R, S stereoisomer (or R, S, S, R stereoisomer)

Br

Br

CH3

CH3CH3

Br

CH3

Br

1 1

– The R, R, S, S stereoisomer (or S, S, R, R stereoisomer)

11CH3

CH3Br Br

CH3

CH3

BrBr

– –

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CHEM 6320 Fall 2005

D. Resolution of Racemic Mixtures 1. General Method

The overall idea is to make covalent or noncovalent (e.g., salt) derivative with an optically pure substance. This results in two diastereomeric compounds or complexes that can be separated on the basis of different physical properties (e.g., fractional crystallization). Once separated, the derivatizing agent can be removed to obtain the pure enantiomers:

(R)-A ~ (R)-B + (S)-A ~ (R)-B

(R/S)-A + (R)-B

separate(S)-A ~ (R)-B(R)-A ~ (R)-B

(R)-A (S)-A

remove (R)-B

2. Alternate Methods

a. Kinetic resolution – Relies on the fact that two enantiomers react at different rates

with optically pure reagent, so that one enantiomer will be depleted more rapidly. Thus, if the reaction is stopped at about 50% completion, the unreacted starting material will be enantiomerically enriched.

– Example:

O CO R*OH

R/S + (R) R*-COCl O CO R*

R.....R S.....R

+

1 : 1

O C R*

1 : 0.5

(R) R*-COCl+R/S

OH O-

Cl

transitionalintermediate

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CHEM 6320 Fall 2005

reactant

‡

intermediate

– The basis of the separation is that the two respective transition states are diastereomeric, and thus of different energies. Most common example is enzymatic resolution.

b. Chromatographic separation using a homochiral support – The basis is that two respective noncovalent interactions which

occur are diastereomeric: the enantiomer which ha higher affinity for the support is selectively retained and elutes last.

3. Related Methods a. Mosher’s reagent

– Mosher’s reagent is used to determine enantiomeric excess by 1H NMR (optical rotation will provide this information only if the [α]D is known).

COOH

C

CF3

OCH3

– Either enantiomer of Mosher’s reagent can be used. – Results in time-averaged diastereomeric interactions (complexes)

with the two enantiomeric constituents, giving rise to distinct 1H NMR signals, which can be integrated to give the % e.e.

– The special feature of Mosher’s reagent which makes it work better than other homochiral reagents are:

– the polar groups which can interact through hydrogen-bonding and dipolar interactions

– the phenyl group which produces large anisotropic shifts (good separation of the signal)

b. Chiral shift reagents – A metal ion, usually a lanthanide, in conjunction with a

homochiral ligand. – The metal ion binds to Lewis base functionalities. – Diastereomeric ternary complexes result that display distinct 1H

NMR signals: (R)-B:/La3+/(+)L vs. (S)-B:/La3+/(+)L

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CHEM 6320 Fall 2005

E. Stereochemistry of Alkenes 1. Cis-Trans Convention

a. Used mainly for molecules with two identical substituents on each carbon atom participating in the double bond.

b. The cis isomer (stereoisomer) has the identical substituents on the same side of the double bond; the trans stereoisomer has these substituents on opposite sides of the double bond.

c. Example 2-butene: H3C

CH3

H3C CH3

trans-2-butenecis-2-butene 2. Z-E Convention

a. More general, less confusing convention, to be used for all types of compounds, not only for those with identical substituents.

b. Procedure: – Assign priorities (similar to the R-S convention) for the two

ligands on each carbon participating in the double bond. – The Z stereoisomer has the substituents with the highest priority

on the same side of the double bond. (Z comes from zusammen = together)

– The E stereoisomer has the substituents with the highest priority on opposite sides of the double bond. (E comes from entgegen = opposite)

c. Example 3-methyl-2-pentene: H3C CH3

CH2CH3H

H3C

CH3H

CH2CH3

d. Note that these E-Z stereoisomers are diastereoisomers (isomers with

the same connectivity in which distances between atoms are different).

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CHEM 6320 Fall 2005

F. Stereochemistry of Reactions 1. Definitions

a. Stereospecific reactions – A stereospecific reaction is one in which stereoisomeric starting

materials afford stereoisomerically different products under the same reaction conditions.

– It is determined by the reaction mechanism. – Examples (see below) are concerted reactions (single-step

reactions) and stepwise reactions (with intermediates which retain stereochemistry).

b. Stereoselective reactions – A stereoselective reaction is one in which a single reactant has

the capacity of forming two or more stereoisomeric products in a particular reaction but one is formed preferentially.

– It depends on the mechanism and the structure (of reactant). – If a given reactant stereoisomer produces a non 50-50 mixture

(any percentage between 0 and 100%) of product stereoisomers, the reaction is stereoselective (and every single step in this reaction will be stereospecific).

2. Stereospecific Reactions a. One (of two) possible enantiomers of reactant give one (of two)

possible enantiomers of product – SN2:

C YXδ-δ-

Inversion at C

– Repulsion between electron pair on incoming nucleophile and C–Y bonding electron pair prevent frontside attack.

– SE2:

Retention at C δ+

δ+C

X

Y

– Electrophile is “directed” to frontside of C–Y bonding

electron pair.

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CHEM 6320 Fall 2005

b. One (of two) possible diastereomers of reactant give one (of two) possible diastereomers of product

– E2:

CH

CX

B:

Anti periplanar

CH

CXB:

Syn periplanar - less favorable – Electron pair generated upon base-abstraction of proton

“displaces” the leaving group from backside in analogy to SN2.

– Addition to C=C:

CA

CB

CA

CB

C CA−B

- A−B - A−B

A−B

Anti Syn

Reaction Stereochemistry Reason

Epoxidation by peracids Syn

Concerted formation of 2 bonds between :Ö: and

the two C=C carbons

Addition of Br2, Cl2 (or BrX) Anti

Two steps: “syn” addition forming

halonium ion followed by SN2 inversion opening

Addition of H2 (Wilkinson’s cat.) Syn

Multistep process occurring at one face of

olefin bound to Rh

Addition of OH,OH: KMnO4

Syn Both O–C bonds created

in one step via cyclic transition state

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CHEM 6320 Fall 2005

3. Relating Configuration

C

CH3

CH2OCH3

HBr

HO-

SN2C

CH3

CH2OCH3

HHO

C

CH3

CH2OCH3

HNCSN2CN-

4. Examples

a. Addition of hydrochloric acid to C=C H3C CH3

H H

H3C

CH3

H

H

CH3

CH

CH2

CH3

CH3

CH

CH2

CH3

Cl+ Cl-+ H+

R or S50% : 50%

samecarbocation

– b. Addition of hydrochloric acid to C=C in the presence of a chiral center

R+ H+

ClR,R

R,S

+ Cl-

– The reaction produces a non 50:50 mixture of two

diastereoisomers. – The reaction is stereoselective.

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CHEM 6320 Fall 2005

c. Bimolecular nucleophilic substitution

CH3CH2Cl CH3CH2OH+ HO-

SN2 – This reaction is stereospecific (because of the stereospecific

mechanism). One would find 100% inversion if use CH3C*HDCl.

d. Nucleophilic substitution at sp3 carbon

Br

OH

OH

OHaq. base

SN2

H2O, ∆SN1 OH

OH

Br

OHR

R R

R

– Are both reaction stereospecific? First reaction is stereospecific while the second one is not. The determination is (and should be) done only based on the mechanism.

– The chiral C is not involved in the reaction so its unchanged configuration is not relevant to reaction stereospecificity.

e. Bimolecular elimination

R

SI

E2I

E2

+

major minor –

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CHEM 6320 Fall 2005

G. Prochiral Relationships 1. Introduction

a. Prochirality: distinction between identical but topologically distinguishable ligands or faces (when sp2 bonding).

b. To test for a prochiral center, make an imaginary replacement of one of two identical ligands by a heavier isotope of the same ligand.

– If get same molecule, then the two ligands are homotopic. – If get different molecules, then the two ligands are heterotopic.

– If get enantiomers then the two ligands are enantiotopic – If get diastereoisomers, then the two ligands are

diastereotopic c. Enantiotopic ligands/faces are chemically identical (e.g., same NMR

signals) except toward reaction with a homochiral reagent. d. Diastereotopic ligand/faces are chemically and physically different

(e.g., different NMR signals due to different environments). e. Examples

HO OHHH

HH HH

– Pro-R atom is the one that would produce an R stereoisomer if

replaced by a heavier isotope. – Pro-S atom is the one that would produce an S stereoisomer if

replaced by a heavier isotope.

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CHEM 6320 Fall 2005

CH3H

Et

HO

HH

– Imaginary replacement of H will give (“R”,S)- and (“S”,S)-

Et

CH3H

Ha

HbOH

Et

CH3H

OH

Ha Hb

2. Prochiral Faces

a. For example, consider the addition of hydride ion (H–) to carbonyl:

C

O

CH3 CH3

OHH-

– Adding H– to top or to bottom face results in the same

compound. ⇒ The two faces are homotopic.

H-OH

C

O

CH3CH2 CH3 – The reaction will produce (R)- or (S)- depending on

whether the addition takes place to top or to bottom face. ⇒ The two faces are enantiotopic.

C

O

CH3

Ph

CH3

OH

CH3

Ph

H-

– Consider the configuration of the reactant chiral center is R. – The reaction will produce (R,R)- or (R,S)- depending on

whether the addition takes place to top or to bottom face. ⇒ The two faces are diastereotopic.

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CHEM 6320 Fall 2005

b. Nomenclature for Heterotopic Faces – Assign priority numbers to the three sp2 ligands, similar to the

R-S convention. – If the ligands decrease in priority (1→2→3) in a clockwise

fashion, the face is re (coming from rectus). – If the ligands decrease in priority (1→2→3) in a

counterclockwise fashion, the face is si (coming from sinister).

C

O

CH3CH2 CH3

– The top side is , the bottom side is . 3. Reactivity Toward Homochiral Reagents

a. Distinction of prochiral ligands/faces by homochiral reagents -O2C

CO2-

OHH

-O2C

CO2-

fumarase, H2O

– Fumarate ion is achiral. – Addition of water is performed to si face only.

esteraseR

H

CH2OHCH2OAcRC

CH2OAc

CH2OAcH

– Esterase carried out the hydrolysis of pro-R “arm”. – S stereoisomer is formed with 95% e.e.

H2 RCH2CNHAc

COOHH

R

COOH

NHCOCH3

H Rh catalyst

– Rh catalyst has homochiral phosphine ligand. – The reaction produces high % e.e., R or S stereoisomer

depending on (+) or (–) ligand.

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CHEM 6320 Fall 2005

aconitase H

CO2--O2C

CH2CO2-

cis-aconitate

CH2 CO2-

OH

CCO2

-

-O2C CH2

citrate – The citrate ion has 3 prochiral centers. – All 4 hydrogen atoms are distinct. – Dehydration involves strictly pro-R hydrogen on pro-R

“arm”. b. How can chiral reagents (e.g., enzymes) distinguish prochirality?

– Consider CH3CH2OH → CH3CH=O via H– abstraction mechanism:

CH3

HHO H CH3

H

OCH3

HHO H

oxoxpro-R

– –

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CHEM 6320 Fall 2005

H. Problems and Exercises 1. Indicate whether the relationship in each of the following pairs of

compounds is identical, enantiotopic, or diastereotopic:

a.

CHO

NH2H

H

CH2OH

HO

CH2OH

OHH

NH2

CHO

H

b.

CH3H HH3C

c. O

O

d. H

CH3 CH3

H

e. O

Cl

H

O

H

Cl

f.

H

PhClHH

CH3 H

H CH3Cl

PhH

g. SO

SO

2. The structure originally proposed for cordycepic acid, [α]D = +40.3°, has

been shown to be incorrect. Suggest a reason to be skeptical about the original structure given below.

OH

HO

HO

CO2H

OH

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CHEM 6320 Fall 2005

3. Each reaction in the sequence shown is reported to proceed with retention of configuration, yet the starting material has the R configuration and the product has the S configuration. Reconcile this apparent contradiction.

Ph

PO-CH3CH2CH2

CH3

PhSiH3Ph

PCH3CH2CH2CH3

PhSiH3Ph

PCH2PhCH3CH2CH2

CH3

Br

4. Using the sequence rule, specify the configuration at each stereogenic

center in the following molecules:

a.

H

(CH2)3CH3 b.

Ph

Ph3Si HO

C Br

O

c. OO

CH3

CH3H

OH

d.

(CH2)6CO2HO

PhHOH

e.

CH3

CO2HO

CH3H3C

CH3

OH

f.

NH2

CO2H

g.

S

H3C

O

CH3

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CHEM 6320 Fall 2005

5. The racemization of medium-ring trans-cycloalkenes depends upon ring size and substitution, as indicated in the data below. Discuss these relative reactivities in terms of the structures of the cycloalkenes and the mechanism of racemization.

C CR

R(CH2)n

Ring size n R t1/2 for racemization 8 6 H 105 years at 25°C 9 7 H 10 seconds at 25°C

10 8 CH3 3 days at 100°C 12 10 CH3 1 day at 25°C

6. Give the product(s) described for each reaction. Specify all aspects of

stereochemistry. a. stereospecific anti addition of bromine to Z- and E-cinnamic acid b. solvolysis of (S)-3-bromooctane in methanol with 6% racemization c. stereospecific syn elimination of acetic acid from (R,S)-1,2-

diphenylpropyl acetate d. stereoselective epoxidation of bicyclo[2.2.1]hept-2-ene proceeding

94% from the exo direction 7. The enzyme enolase catalyzes the following reaction:

CO2-

OPO32-H

CH2OH H H

OPO32--O2C

+ H2O

When (2R,3R)-2-phosphoglycerate-3-d was used as the substrate, the E-isomer of phosphoenolpyruvate-3-d was produced. Is the stereochemistry of elimination syn or anti?

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CHEM 6320 Fall 2005

8. Some of the compounds shown contain diastereotopic atoms or groups. Which posses this characteristic? For those that do, indicate the atoms or groups that are diastereotopic and indicate which atom or group is pro-R and which is pro-S.

a. O

CH3H3C

HH b. O

HH3C

3CHH

c.

CHCO2CH(CH3)2

OCH3

C6H5

d. NH

CHCO2H

2

(CH3)2CH

e.

CH

NH2

C6H5CH2 CNHCH2CO2H

O

f. BrCH2CH(OC2H5)2 9. An important sequence in valine biosynthesis in bacteria is

(CH3)2C CHCO2H

O NH2H

(CH3)2CH C CO2H

O

(CH3)2CH CHCO2H

NH2

valine The stereochemical aspects of this sequence have been examined, using a diol substrate in which one of the methyl groups has been replaced by CD3. Given the information that labeled starting diol of configuration 2R,3R produces labeled valine of configuration 2S,3S, deduce whether the C-3 hydroxyl group is replaced with overall retention or inversion of configuration.

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CHEM 6320 Fall 2005

10. Enzymatic oxidation of naphthalene by bacteria proceeds by way of the intermediate cis-diol shown. Which prochiral faces of C-1 and C-2 of naphthalene are hydroxylated in this process?

OH

OH

11. a. One of the diastereomers of 2,6-dimethylcyclohexyl benzyl ether

exhibits two doublets for the benzylic protons in its NMR spectrum. Deduce the stereochemistry of this isomer.

b. The NMR spectrum of the highly hindered molecule trimesitylmethane indicates that there are two enantiomeric species present in solution, the interconversion of which is separated by a barrier of 22 kcal/mol. Discuss the source of the observed chirality of this molecule.

CH3

CH3

CH3 CH

3 12. A synthesis of the important biosynthetic intermediate mevalonic acid

starts with the enzymatic hydrolysis of the diester below by pig liver esterase. The pro-R group is selectively hydrolyzed. Draw a three-dimensional structure of the product.

CCH2CO2CH3

CH3

OHH3CO2CCH2

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CHEM 6320 Fall 2005

Conformational, Steric, and Stereoelectronic Effects A. Conformational Analysis in Saturated Compounds

1. Strain Energy – excess energy caused by nonideal geometry )()()()(strain dEEErEE +++= φθ

a. Bond stretching: 2r )(5.0)( rkr ∆=E

b. Bond angle distortion: E 2θ )(5.0)( θθ ∆= k

c. Torsional strain: )3cos1(5.0)( 0 φφ += VE for tetrahedral carbon d. Non-bonded repulsions or van der Waals repulsions, )(dE , when two

non-bonded groups are closer than the sum of the van der Waals radii 2. Linear Alkanes

a. For ethane and propane, the strain energy inherent in the eclipsed as opposed to staggered conformation is a quantum mechanical “repulsion” between pairs of bonding electrons:

-1.00.01.02.03.0

0 60 120 180 240 300 360Torsion angle (degree)

E (k

cal/m

ol)

2.9 kcal/mol for ethane

2.9 kcal/mol for ethane

H

H (CH3)

HH

H

HH

H

H

H H (CH3)

H

b. For n-butane, an additional factor comes into play when one considers rotation about the central C–C:

-1.0

0.0

1.0

2.0

3.0

4.0

5.0

6.0

0 60 120 180 240 300 360Torsion angle (degree)

E (k

cal/m

ol)

3.4 kcalmol

0.8kcalmol

3.4 kcalmol3.4 kcalmolkcalmol

0.8kcalmol0.8kcalmolkcalmol

H

CH3

HC

H H

H3

H

CH3

H

H

H CH3

H

CH3

HH

H CH3

H

CH3

H

CH3

H H

H

CH3

HH

H3C H

H

CH3

H

H

CH3 H

H

CH3

HC

H H

H3

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CHEM 6320 Fall 2005

3. Conformational Equilibrium a. The relative distribution of a molecule among various conformations

can be calculated if the energy differences are known: KRTG ln−=∆ o

where K represents the conformational equilibrium constant b. For example, to determine the distribution of n-butane between the

gauche and anti conformations: anti gauche

)2ln(kcal/mol8.0 RTG −−−=∆ where RTln 2 term is a statistical correction due two gauche to one anti conformers. – At room temperature:

kcal/mol39.0)41.0(8.0 −=−−−=∆G gaucheantiK 34%,66%9.1 ⇒−=⇒

c. Composition-equilibrium-free-energy relationships More stable isomer

(%) Equilibrium constant (K)

Free energy (∆G°298) (kcal/mol)

50 1.00 0.0 55 1.22 –0.119 60 1.50 –0.240 65 1.86 –0.367 70 2.33 –0.502 75 3.00 –0.651 80 4.00 –0.821 85 5.67 –1.028 90 9.00 –1.302 95 19.00 –1.744 98 49.00 –2.306 99 99.00 –2.722

99.9 999.00 –4.092

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CHEM 6320 Fall 2005

4. Rotational Energy Barrier a. Rotational energy barriers of compounds of the type CH3–X

Compound Barrier height (kcal/mol) CH3–CH3 2.88

CH3–CH2CH3 3.4 CH3–CH(CH3)2 3.9 CH3–C(CH3)3 4.7

CH3–SiH3 1.7 CH3–NH2 2.0 CH3–OH 1.1

R3

R2R1

H

H

H

b. Observations and explanations – The interelectronic repulsion between C–H and C–C is larger

than between C–H and C–H and larger than between C–H and lone-pair.

– The big jump in rotational barrier height between isobutene and neopentane is due to a decreased flexibility in methyl substituents that further hinder the rotation.

– Barrier height is lower for CH3–SiH3 because C–Si bond is longer.

5. Another Example a. The case of 1-chloropropane

CH3 Cl

– One predicts the same conformer as in n-butane (anti favored over gauche).

– In gauche conformer there are actually attractions between methyl and chlorine (and not so large steric repulsions) that make gauche to be thermodynamically preferred over anti.

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CHEM 6320 Fall 2005

B. Conformational Analysis in Compounds with Double Bond 1. Propene and Acetaldehyde

a. There are two possible conformers:

XH

H

HH

XH

HHH

vs. X = O, CH2

H

H

XH

H H

H XH

Heclipsed bisected

b. The eclipsed conformer is more stable. c. The bisected conformer is less stable due to unfavorable interactions

(repulsion) between the π system and the two C–H σ bonds. d. Note that the rotational barriers in these cases are less than the case of

ethane. 2. More Examples

a. 1-butene have a methyl group connected to C-3 so there are two different eclipsed conformations:

CH2

H

H

H3CH

CH2

H

CH3

HH

is more stable than

due to more favorable van der Waals interactions.

b. If another methyl group is present at C-2, the two conformations are similar in energy:

CH2

CH3

CH3

HH

CH2

CH3

H

H3CH

vs.

– Note the gauche interaction (energetically less favored) between

methyl groups in the first conformer.

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CHEM 6320 Fall 2005

c. A similar case but for a ketone:

OR'

H

RH

OR'

R

HH

majorminor – In the major conformer, the repulsion between R and O is not as

great and the R and R′ are in anti position (preferred over gauche).

3. Compounds with Conjugated Double Bonds a. 1,3-butadiene has two conformers in equilibrium, with the s-trans

conformer being preferred:

b. Acrolein has also two conformers in equilibrium, (s-trans conformer is still being preferred but less than the case of 1,3-butadiene):

– The 1-6 interaction is less important in acrolein than in 1,3-butadiene.

c. 3-methyl acrolein has also two conformers in equilibrium, (s-trans conformer is still being preferred but less than the case of acrolein):

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CHEM 6320 Fall 2005

C. Conformational Analysis in Small Rings 1. Cyclopropane

a. The molecule is “planar” (i.e. the carbon atoms form a plane).

2. Cyclobutane

a. If the molecule would be planar, there would be a smaller angle strain. But all the (C–H) bonds are eclipsed so there is bigger torsion strain.

b. The molecule is not planar (i.e. one carbon atom is above or below the

plane made by the other three carbon atoms). In this way, the angle strain in increased (the C–C–C bond angle is decreased) but the torsion strain is decreased.

c. cis-1,3-dimethylcyclobutane is more stable than trans-1,3-

dimethylcyclobutane because cis has two pseudoequatorial bonds.

3. Cyclopentane a. Again, a planar molecule would have little angle strain (108 vs. 109.5

degrees) but it would have large torsion strain (due to eclipsed bonds). The most stable conformations have the carbon atoms not in the same plane, reducing therefore the torsion strain:

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CHEM 6320 Fall 2005

D. Conformational Analysis in 6-Membered Rings 1. Cyclohexane

Pote

ntia

l Ene

rgy

11 kcal

5.5 kcal

1.6 kcal1.6 kcal

chair half-chair boattwist-boat I twist-boat IIconformer conformer conformer

a. “Ring-flip” of chair conformation of cyclohexane interchanges all equatorial groups and axial groups.

AB

C

AC

B

2. Monosubstituted Cyclohexane

a. There is equilibrium between the two conformers axial and equatorial: CH3

H

H

CH3

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CHEM 6320 Fall 2005

b. The equatorial conformer is energetically preferred. – In the axial conformed of methylcyclohexane there are two

unfavorable “gauche” butane interactions (and unfavorable 1,3-diaxial interactions).

CH3

H

CH3

H

H

HCH3

H

C

C

– The Newman projection about C1-C2 bond: – Conformational restriction makes this “gauche” interaction a

little worse than normal (0.9 rather than 0.8 kcal/mol). – Equatorial methyl is favored over axial methyl by 1.8 kcal/mol.

c. The free energy difference between conformers is referred to as the conformational free energy (∆G°).

– Values of ∆G° (in kcal/mol) for axial-equatorial equilibrium. –CH3 1.8 –Ph 2.9 –CH2CH3 1.8 –OCH3 0.6 –iPr 2.1 –COOC2H5 1.2 –tBu 4.5 –C≡N 0.2

– Relative ∆G° values for Me, Et, iPr, tBu.

CH3HH

CHH

H CH3CH3 CH

H

H3C CH3CH3

– For isopropylcyclohexane, the van der Waals repulsion is

no worse than in the case of methyl but there is restricted rotation therefore an entropy loss.

– For t-butylcyclohexane, the van der Waals repulsion cannot be relived by rotation and ∆G° increases to 4.5 kcal/mol.

– Equatorial preference for t-Bu has comparable ∆G° to that between chair and twist-boat conformations of cyclohexane

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CHEM 6320 Fall 2005

d. Measurements of axial-equatorial equilibrium – Take NMR at low enough temperature where conformational

interconversion is “slow on the NMR time scale” then integrate the separated signals to get the equilibrium constant K = [eq]/[ax]

– Measure ∆G° for two diastereomers where the only difference is whether the substituent is equatorial or axial

X

locked

X

locked 3. Disubstituted Cyclohexane

a. trans-1,2-dimethylcyclohexane

– The axial-axial conformer has four gauche interactions:

– The equatorial-equatorial conformer has one gauche interaction:

– The axial-axial conformer is destabilized by 2.7 kcal/mol (approximately three time the energy of a gauche interaction).

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CHEM 6320 Fall 2005

b. cis-1,4-dimethylcyclohexane H

CH3H

CH3

H

CH3

CH3

H

A – The two conformers are the same.

c. trans-1,4-dimethylcyclohexane CH3

H

CH3

H

H

CH3CH3

HB C

– A is less stable than B by 1.9 kcal/mol (same as ∆G° for one axial -CH3).

– C is less stable than B by ~3.7 kcal/mol (close to twice ∆G° for axial -CH3 ≅ about 4 “gauche” interactions).

d. cis-1,3-dimethylcyclohexane

CH3

HH

CH3

CH3

CH3 H

H

– The axial-axial conformer is less stable than B by about 5.5 kcal/mol (the additional ~1.8 kcal/mol is due to particularly unfavorable 1,3-Me/Me diaxial interactions).

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CHEM 6320 Fall 2005

e. cis-1,4-di-t-butylcyclohexane – The presence of t-butyl group in an axial position is so

energetically unfavorable that the conformation of the ring changes to a twist-boat to accommodate both t-butyl groups in pseudoequatorial positions:

HHH

H

4. Decalin

H

H H

H

H

Hvs.

trans cis a. The trans isomer has the ring locked. b. The cis isomer has three additional gauche interactions. One would

predict 3 × (0.9 kcal/mol) = 2.7 kcal/mol less stable. c. The cis isomer can do “ring-flip” (as shown above).

5. Cyclohexene

H

H a. The two sp2 carbon atoms and the four atoms directly bonded to these

are in one plane, while the other two carbon atoms (C-4 and C-5) are one above the plane and one below the plane.

b. There are four pseudoaxial and four pseudoequatorial valences for the sp3 carbon atoms in the ring.

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CHEM 6320 Fall 2005

6. Cyclohexanone and Derivatives a. In cyclohexanone (and methylenecyclohexane) there is less torsional

strain associated sp2 (compared with sp3) therefore a more facile ring inversion.

X

X = O, CH2

b. Methylcyclohexanone – For 2-methylcyclohexanone, the conformer with methyl in

equatorial position is more stable that the conformer with methyl in axial position by about 1.8 kcal/mol (same as methylcyclohexane).

O

CH3HH

vs.

OCH3

H

– For 3-methylcyclohexanone, the conformer with methyl in

equatorial position is more stable that the conformer with methyl in axial position by only about 1.3-1.4 kcal/mol (because in this case there is only one unfavorable 1,3-diaxial interaction).

OCH3

H

vs.

OCH3

H

H

c. Chloromethylcyclohexanone – This is an example of the conformational preference is

influenced than other factors than steric.

OCl

Hvs.

O

Cl

H

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CHEM 6320 Fall 2005

– The conformer with methyl in equatorial position has a very large dipole moment.

– The conformer with methyl in axial position has the unfavorable axial interaction (between Cl and two H) but has a smaller dipole moment and is the favored form in nonpolar solvents.

7. 1,3-Dioxan and Derivatives a. The presence of oxygen makes the ring more flexible and the 1,3-

diaxial interactions less significant. b. The C–O–C angles are also different than the C–C–C angles in

cyclohexane. c. ilibrium between conformers is shifted. Example: The equ

8. More Comments a. When considering conformational preferences of multisubstituted

cyclohexanes, one has to consider dipole-dipole and/or stereoelectronic factors in addition to simple steric effects.

b. Axial-equatorial preferences can change in complicated molecules:

CH3

H

CH3

CH3

– The equatorial conformer has unfavorable (“bad”) van der Waals

repulsion. c. One cannot necessarily predict exactly which conformation is the most

stable in a molecule but one can make correlations and comparisons with similar compounds.

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CHEM 6320 Fall 2005

E. The Anomeric Effect 1. Observations

a. In cyclic compounds with heteroatom(s) in the ring and with electron-withdrawing groups at C-1, the axial conformer is (may be) favored over the equatorial conformer especially in less polar solvents.

vs.O

X

H

..

.. O

H

X

..

..

X = OR, Cl, F, NR 2. Explanations

a. Minimizing unfavorable dipole-dipole interactions – In the equatorial conformer the two dipoles are coplanar

therefore a larger dipole vector sum. – In the axial conformer there is 120° dihedral angle twist resulting

in a reduced dipole vector sum.

O X

.... .. ..

O

Xequatorial axial

a. Hyperconjugative overlap – Anti periplanar arrangement in the axial conformer allows for

possible hyperconjugative overlap of oxygen lone pair with empty C–X σ* orbital.

..

X

..

O OX X

..

O+

("no bond" resonance)

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CHEM 6320 Fall 2005

F. Conformational Effects on Reactivity 1. Cyclohexane Conformer Reactivity

a. The question is which conformation reacts faster and an example is which conformation reacts faster with hydride ion (H–) or Rδ––Mδ+?

H

CH=OH

CH=O

vs.

eq. ax. – This is not the same question as which conformer is more stable,

for which the answer is equatorial. b. The issue here is to assess ∆G between reactant state and transition

state (‡), i.e. ∆G‡. c. The question is: “Is the substituent getting more crowded or less

crowded in going from the reactant state to the transition state?” – If getting more crowded (e.g., sp2 → sp3), then equatorial

conformer reacts faster. – If getting less crowded (e.g., sp3 → sp2), then axial conformer

reacts faster.

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CHEM 6320 Fall 2005

2. Stereoelectronic Effects a. They are related to anomeric effects. b. In reactions that involve sp2 ↔ sp3, formation and collapse of sp3 state

is favored when the leaving/entering group is antiperiplanar to lone pairs on both of the other heteroatoms.

faster slower

O..

OCH3

+

O

H

OCH3

..

..

..O

H

O CH3

..

– For axial conformer, stabilization is coming from only one pair

of electrons. c. More examples

O

OCH3

OCH3O

OCH3

++ H+

- CH3OH ..

....O

O

OCH3

CH3

– D-labeling shows exclusive loss of axial methoxy. – The reason is maximization of π overlap at the reaction transition

states.

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CHEM 6320 Fall 2005

G. Reactivity of Cyclic Compounds 1. Effect of Angle Strain on Reactivity

a. Manifested especially for cyclopropane that reacts with ring-opening with electrophiles (X2, H+, H2-cat) that normally react only with double bond (C=C) and not other cycloalkanes.

2. Stability of trans-Cycloalkanes a. trans-cycloheptene has only been trapped as a transient intermediate b. trans-cyclooctene is known is known but is unstable and reacts rapidly

to relieve the strain. (It reacts with X2 faster than cyclohexene.) – Note that trans-cyclooctene can be resolved into (+) and (–)

enantiomers (atropisomerism): H

H

H

H 3. Brendt’s Rule

a. Brendt’s rule says that no C=C is permitted at the bridgehead position of bicyclic compounds.

possible not possible b. Six atoms cannot all be coplanar unless bridgehead “loops” are long

enough. 4. Stereoelectronic/Steric Inhibition of Resonance

a. First example

N

O..

N

O

+_

CR' NR2

O

.. CR' NR2

O _

+

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CHEM 6320 Fall 2005

– Nitrogen’s lone pair is perpendicular to carbonyl π system (resonance form is a Brendt’s rule violation).

b. Second example

O O O O- H+

_

_O O

OO

- H+

OO

_O

O_

5. Ring Closure Rates (Rate of Cyclization) Versus Ring Size

a. The general trend is 5 > 6 > 3 > 7 > 4 > 8-10 – Ring size of 3 has unfavorable ∆H‡ but favorable ∆S‡. – Ring sizes of 8-10 have normal ∆H‡ but large negative ∆S‡. – This rule applies to, for example:

Br(CH2)nNH2 (CH2)n NH

but varies with type of reaction and hybridization of reacting ends.

– Exceptions from general trend:

XR

O

_ R

O

often occur faster than

_ R

O

XR

O

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CHEM 6320 Fall 2005

b. Baldwin’s rules – Baldwin’s rules provide a systematic analysis on ring formation

by taking into account ring size, hybridization at reaction center, whether “reacting bond” is endocyclic or exocyclic to forming ring

C X

Nu:n-exo-tet

(sp3)

OK for any ring size

C X

Nu:

C X

Nu:n-exo-tet

(sp3)

OK for any ring size OK for any ring siz

n-exo-trig(sp2)

e

C

Y

Nu:X

n-exo-trig(sp2)

eOK for any ring siz

C

Y

Nu:XC

Y

Nu:X

n-exo-dig(sp)

OK for n ≥ 5

C C

Nu:

C-RO

n-exo-dig(sp)

OK for n ≥ 5

C C

Nu:

C-RO

C C

Nu:

C-RO

n-endo-trig(sp2)

OK for n ≥ 6

C

Nu:

ZC

n-endo-trig(sp2)

OK for n ≥ 6

C

Nu:

ZC C

Nu:

ZC

OK for any ring siz

n-endo-dig(sp)

e

C C

Nu:Z

n-endo-dig(sp)

eOK for any ring siz

C C

Nu:Z

C C

Nu:Z

– Example:

CH CHPh

O

HO: O

O

Ph

5-endo-trignot allowed

5-endo-dig allowed

O

O

PhHO:

C C-Ph

O

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CHEM 6320 Fall 2005

H. Torsional and Stereoelectronic Effects on Reactivity 1. Torsional Effects on Reactivity

a. For reactions of cyclic compounds – sp2 → sp3 transformation is more favorable for 6- than 5-

membered ring – sp3 → sp2 transformation is more favorable for 5- than 6-

membered ring b. Example

O OH

[H]

[O] vs.

[H]

[O]

O OH

– For reduction reaction (involving sp2 → sp3), the 6-membered

ring is faster than the 5-membered ring – For oxidation reaction (involving sp3 → sp2), the 5-membered

ring is faster than the 6-membered ring 2. Stereoelectronic Effects of Carbonyl Reactivity

a. Example of nucleophilic attack on cyclohexanone

O

H

H

HH

HH

:Nu (ax)

:Nu (eq)

b. Equatorial attack requires eclipsing of C–O bond with two C–Heq

bonds at transition state. O-

Nu

Nu

O- OH

H

axialatack

equatorialatack

H

H

H

H – Axial attack is favored for small nucleophiles. – Equatorial attack is preferred for large, bulky nucleophiles

(because of 1,3-diaxial problem).

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CHEM 6320 Fall 2005

3. Rules for Predicting the Major Product a. Cram’s Rule

– Predict the correct major and minor diastereoisomers.

CR C

OL

MS

Nu:

CC

OH

Nu

R ML

S

– When the three substituents at the vicinal C differ in size (called them L=large, M=medium, S=small), and the molecule is oriented such that the largest group is anti to the carbonyl oxygen, the major product arises from addition of the nucleophile syn to the smaller substituent.

R

O

L

MS

L

MS OH

RNu L

MS OH

NuR+

major minor – Transition-state theory-based

explanation: the molecule is oriented such that the largest group is oriented perpendicularly to the carbonyl group, the nucleophilic attack occurs from the opposite side:

c. Example CH3CH3

exo attack

endo attack

CH3CH3

O

endo attack

exo attack

– endo attack is preferred because of methyl groups from bridge

(steric effects) – exo attack is preferred when there are H in the bridge

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CHEM 6320 Fall 2005

I. Problems and Exercises 1. Estimate ∆H° for each of the following conformational equilibria:

a.

H CH3

CH3

CH3

H HH3C C 3

H

HCH3

H H

b.

CH3 CH3

H

CH3

H3C HH3C CH3

H

H

H3C CH3

c. O

CH3

CH3

O CH3

CH3

HOH

H

H

A

2. Draw a clear three-dimensional representation

showing the preferred conformation of cis,cis,trans-perhydro-9b-phenalenol (A):

3. The trans : cis ratio at equilibrium for 4-t-butylcyclohexanol has been

established for several solvents near 80°C: Solvent trans (%) cis (%) Cyclohexane 70.0 30.0 Tetrahydrofuran (THF) 72.5 27.5 i-Propyl Alcohol 79.0 21.0

From these data, calculate the conformational energy of the hydroxyl group in each solvent. Explain any correlation between the observed conformational preference and the properties of the solvent.

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CHEM 6320 Fall 2005

4. Explain the basis for the selective formation of the product shown over the alternative product.

a.

CH3H3C CH3H3C

O

CH3H3C

Oversus

RCO3H

(88 %) (12 %)

b.

N

CH3H

H3C+ BH3 N

CH3H

H3B-

H3Cpreferred

toN

CH3H

H3C

H3B-++

c. LiAlH4 rather

than

HO

CH3

HOHH

CH3

O

CH3

d.

N

N

S CH3

O

O O

HH

CO2CH3

CH3

RCO3H N

N

S CH3

O

O O

HH

CO2CH3

CH3

O-

+

rather than the stereoisomeric sulfoxide

e.

CH2CH2

O

SnCl4

CH2OH

rather than

OH

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CHEM 6320 Fall 2005

5. For the following pairs of reactions, indicate which you would expect to be more favorable and explain the basis of your prediction. a. Which isomer will solvolyze more rapidly in acetic acid?

OSO2Ph

OSO2PhorA B

b. Which isomer will be converted to a quaternary salt more rapidly?

or

N(CH3)2

(CH3)3CN(CH3)2

(CH3)3CDC

c. Which lactone will be formed more rapidly? O ODMSO-O2C(CH2)2CH2Br

or

-O2C(CH2)2CH2Br DMSOO O

F

E

d. Which compound will undergo hydrolysis more rapidly?

O

O N 2O

NO2

O

O

CH3 or

HG e. Which compound will aromatize more rapidly by loss of ethoxide ion?

NO2

NO2

OC2H5N

_ or _

NO2

NO2

OC2H5N

I J

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CHEM 6320 Fall 2005

6. Predict the most stable conformation for each of the following molecules and explain the basis of your prediction.

a.

CO2CH3

Cl b. C(

C(CH3)3

CH3)3

OH

c.

O

CH3

CH3 d. Br

O

OCOCH3

OCOCH3

OCOCH3

e. O

O

F

F

f. OC

O O

OCH3

H3

g.

O

O

h. FCH2CH2-

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CHEM 6320 Fall 2005

7. Predict the preferred conformation of the isomeric (cis- and trans-) 3-penten-2-ones, A, R=CH3. How would you expect the conformational picture to change as R becomes progressively larger?

RCCH CHCH3

OA

8. Estimate the energy difference between the stable and unstable chair

conformations of each of the following trimethylcyclohexanes:

CH3

CH3

CH3

CH3

CH3

CH3

CH3

CH3

CH3

CH3

CH3

CH3

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CHEM 6320 Fall 2005

Study and Description of Reaction Mechanisms A. Chemical Thermodynamics

1. Review a. Thermodynamics tells us the relative energy levels and thus the

feasibility of a particular process, but nothing about the reaction path (and thus no insight into the mechanism or the rate).

b. Relation between the Gibbs free energy, enthalpy, and entropy: STHG ∆−∆=∆

c. Spontaneous processes – In general, a system evolves spontaneously in a direction that

lowers its energy (or enthalpy) and that increases its disorder (i.e., entropy).

– For an isolated system (no matter or energy exchange with the surrounding):

=>

process reversible0process sspontaneou0

dSdS

– The Helmholtz energy (A = U – TS):

– The Gibbs energy (G = H – TS = A + PV):

2. Enthalpy of reaction (∆rH) a. Usually computed from data tables representing gas-phase reactions. b. For a general case aA + bB → cC + dD, can be calculated as:

∑∑ −=∆ of energiesformed bonds

of energiesbroken bondsr H

∑∑ ∆−∆=∆ react)(prod)( freactfprodrooo HnHnH

where ∆fH is the enthalpy of formation

∑ ∑ ∆−∆=∆ )prod(react)( cprodcreactrooo HnHnH

where ∆cH is the enthalpy of combustion c. ∆H predictions in solution require estimates of solvation effects.

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CHEM 6320 Fall 2005

3. Chemical Equilibrium a. Relation between the Gibbs free energy and the equilibrium constant:

ZYBA ZYBA νννν +→+

eqBA

ZYBA

ZY

= νν

νν

aaaa

K ; KRTG ln−=∆ o

KQRTQRTTGTG lnln)()( r =+∆=∆ o

where BA

ZY

BA

ZYνν

νν

aaaa

=Q is the reaction quotient

– If Q < K ⇒ ∆G < 0 ⇒ – If Q > K ⇒ ∆G > 0 ⇒

b. Le Châtelier’s Principle: If a chemical reaction at equilibrium is subjected to a change in conditions that displaces it from equilibrium, then the reaction adjust toward a new equilibrium state. The reaction proceeds in the direction that partially offsets the change in conditions.

c. Gibbs-Helmholtz equation:

2/

TH

TTG

P

oo ∆−=

∂∆∂

2)(ln)(ln

RTH

dTTKd

TTK

P

o∆==

∂∂

– –

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CHEM 6320 Fall 2005

B. Chemical Kinetics 1. Reaction Rate

a. Consider a general reaction ZYBA ZYBA νννν +→+ b. The number of moles of reactants and products:

)((0))( AAA tntn ξν−= ; )((0))( BBB tntn ξν−= )((0))( YYY tntn ξν+= ; )((0))( ZZZ tntn ξν+=

where ξ is the extent of reaction c. The change in time of the number of moles of reactants and products:

dttd

dttdn )()(

AA ξν−= ;

dttd

dttdn )()(

BB ξν−=

dttd

dttdn )()(

YY ξν+= ;

dttd

dttdn )()(

ZZ ξν+=

d. Dividing by the volume V, one obtain the change in time of the concentrations:

dttd

Vdtd

dttdn

V)(]A[)(1 AA ξν

−== ; dt

tdVdt

ddt

tdnV

)(]B[)(1 BB ξν−==

dttd

Vdtd

dttdn

V)(]Y[)(1 YY ξν

+== ; dt

tdVdt

ddt

tdnV

)(]Z[)(1 ZZ ξν+==

e. Define the rate of reaction v(t):

2. Rate Laws a. The relationship between v(t) and the concentrations is called the rate

law. The rate laws describe the experimentally determined dependence of the rate of the reaction with the reactant and products concentrations.

b. Rate law expression: KBA ]B[]A[)( mmktv =

where k is the rate constant of the reaction, and mA, mB, … are the orders in A, B, … (are constants)

c. In general, the rate law cannot be deduced from the chemical equation:

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CHEM 6320 Fall 2005

d. The overall order of the reaction in equal to the sum of the order of the reaction with respect to each species appearing in the rate law (Σmj).

e. The reaction rates can be followed by spectral changes, conductance changes (for reactions which create or consume ions), pH changes (for reactions which create or consume H+ or HO–), polarimetry, etc.

f. Examples of rate laws: )(gNO2(g)ONO(g)2 22 →+ ]O[]NO[ 2

2kv =)HI(g2(g)I(g)H 22 →+ ]I][H[ 22kv =

CO(g)(g)CHCHO(g)CH 43 +→ 2/33 ]CHOCH[kv =

)CO(gClCO(g)(g)Cl 22 →+ ]CO[]Cl[ 2/32kv =

)HBr(g2(g)Br(g)H 22 →+ 12

2/122

]Br][HBr[1]Br][H[

−′′+

′=

kk

v

g. Integrated rate law expression: – Give the time-dependence of the concentration(s) – First-order reaction (consider A + B → products reaction)

]A[]A[)( kdt

dtv =−= ⇒

– The half-life of the reaction (t1/2) is time required for half of

the reactant to disappear: kk

t 693.02ln2/1 ==

– Second-order reaction (consider A + B → products reaction) 2]A[]A[)( k

dtdtv =−= ⇒

– The half-life: 0

2/1 ]A[1

kt =

– n-order reaction (consider A → P reaction) nk

dtdx

dtd

dtdtv ]A[]P[]A[)( ===−=

⇒

−−−

=−− 1

01

0 ]A[1

)]A([1

11

nnxnkt where x = [P]

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CHEM 6320 Fall 2005

3. Temperature Dependence of the Rate Constants a. The rates of chemical reactions strongly depend on the temperature,

and almost always they increase with the temperature. The dependence of the rate with the temperature is due to the dependence of the rate constant with the temperature.

b. Commonly the dependence of the rate constant with the temperature can be described by the empirical equation called Arrhenius equation:

2aln

RTE

dTkd= ln k

1000/T

ln k

1000/T

RTE

Ak alnln −=

RTEAek /a−= where Ea is the activation energy and A is the pre-exponential factor

c. The activation energy can be determined from the slope of ln k versus 1/T representation or from a two-point fit (knowing k at two different temperatures)

)()(ln

2

1

21

21a Tk

TkTT

TTRE ⋅−

⋅=

d. A simplistic (but not completely true) interpretation of the activation energy is as the energy that the reactants should have in order for the reaction to proceed from the reactants to products along a reaction coordinate.

e. The ln k versus 1/T representation may not be linear so the dependence of k to T can be written as:

RTEmeaTk /′−= where m = 1, ½, -½, etc ⇒ mRTEE +′=a and mmeaTA =

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CHEM 6320 Fall 2005

4. Understanding the Phenomenological Rate Laws a. One should consider how reactants transform into products and how

this path determines the kinetics of the chemical reaction. b. Many reactions involve chemical intermediates and therefore can be

written as: Reactants → Intermediate(s) → Products

– Example: Nucleophilic substitution for tertiary alkyl halides:

OH-C)(CHC)(CHCl-C)(CH 33HO

33Cl33 → →−

−++

−

c. Intermediates: are bound chemical compounds, relative unstable and therefore reactive, located at a minimum energy along the reaction coordinate (not to be confused with transition state that is located at or in the close proximity of a maximum in energy along the reaction coordinate).

d. Chemical reactions can be divided into: – Elementary reactions = reactions that occur in a single step (and

that do not involve any reaction intermediate) – Complex reactions = reactions that does not occur in a single

step (and that do involve reaction intermediates) e. Chemical kinetics deals with determining the sequence of elementary

reactions by which a complex reaction occurs, i.e. with determining the mechanism of the complex reactions.

5. Kinetics of Elementary Reactions a. Elementary reactions or processes

– are reactions that occur in a single step and that do not involve any reaction intermediate

– are processes that involve a limited (a maximum of 3-4) number of chemical bonds being broken and/or formed

– are denoted (in some Physical Chemistry textbooks) by ⇒ for a forward reaction and ⇐ for a reverse reaction. (These notations distinguish the elementary reactions (or elementary steps) from the other reactions.)

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CHEM 6320 Fall 2005

b. The rate law of an elementary reaction (step or process) can be deduced from the balanced chemical equation itself.

c. The number of reactant molecules involved in an elementary chemical reaction is called molecularity. The elementary reactions are termed unimolecular, bimolecular, and termolecular (if they have one reactant, two reactants, or three reactants).

– Unimolecular reactions: A ⇒ Products

– Bimolecular reactions: A + B ⇒ Products

– Termolecular reactions: A + B + C ⇒ Products

6. Kinetics of Complex Reactions a. Kinetics of some special chemical reactions

– Reversible reactions: reactions that occur in both directions k1

k-1A B

– Assuming that both reactions are first-order (or are

elementary) with respect to A and assuming [B]0 = 0:

]B[]A[dt

]A[11 −−=− kkd ; [ ]A[]A[]B 0 −=

0111 ]A[]A)[(]A[−− −+=− kkk

dtd

⇒ 11

01)(

01 ]A[]A[]A

11[

−

−+−

++

=−

kkkek tkk

or [ eq)(

eq0 ]A[)]A[]A([]A 11 +−= −+− tkke

t

00 ]A[]B[;

]A[]A[

t

00 ]A[]B[;

]A[]A[

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CHEM 6320 Fall 2005

– Parallel reactions: reactions in which one reactant is transformed in two (or more) products

B

CA

k1

k2 – Assuming that both reactions are first-order (or are

elementary) with respect to A: tkke )(

021]A[]A[ +−= ;

2

1

2

1]C[]B[

kk

vv

==

– The half-life for A: 21

2/1693.0

kk +=t

– The activation energy (Ea) for the disappearance of A:

21

a,22a,11a kk

EkEkE

+

+=

where Ea,1 is the activation energy for the first reaction (A → B) and Ea,2 is the activation energy for the second reaction (A → C)

– Consecutive reactions: a series of reactions in which the product of one reaction is the reactant in another one

k2k1A CB – Assuming that both reactions are first-order (or are

elementary) and assuming [B]0 = [C]0 = 0:

]A[]A[1k

dtd

−= ⇒ [ tke 10]A[]A −=

]B[]A[]B[]A[]B[20121

1 kekkkdt

d tk −=−= −

⇒ 012

1 ]A)[(]B 21 tktk eekk

k −− −−

=[

]B[]C[2k

dtd

= ⇒ 012

21 ]A[1]C12

−−

+=−−

kkekek tktk

[

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CHEM 6320 Fall 2005

0.0

0.5

1.0

time

[C]/[A]0

[B]/[A]0

[A]/[A]0

[A]/[

A] 0,

[B]/[

A] 0,

[C]/[

A] 0

k1 = 3k2

0.0

0.5

1.0

time

[C]/[A]0

[B]/[A]0

[A]/[A]0

[A]/[

A] 0,

[B]/[

A] 0,

[C]/[

A] 0

k1 = 3k2

b. The rate-determining step

– If one step of the reaction mechanism is much slower than any other steps, that step controls the overall reaction rate and is called rate-determining step.

– The presence of a rate-determining step in the mechanism of a complex reaction is not required though.

– Working out the kinetics of a complex reaction, the final result is the rate law which is an algebraic expression containing one or more rate constants and concentrations of all reactant species involved in or prior to the rate determining step.

– Kinetic data provide information only about the rate determining step and steps preceding it.

– Example (g)CONO(g)CO(g)(g)NO 22obs +→+ k

– Two-step mechanism:

(g)NONO(g)(g)NO(g)NO 3221

+⇒+k

(g)CO(g)NOCO(g)(g)NO 2232

+⇒+k

– The first step is much slower than the second step (v1 << v2) – The rate of the overall reaction is given by the rate of the

rate-determining step: v1 = k1[NO2]2

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CHEM 6320 Fall 2005

– Example: A + B C D E + F

k1

k-1

k2 k3

=dt

d ]C[

=dt

d ]D[

– If k2 is slow relative to k3, then D will go on to E + F as soon as it is formed, so that the rate becomes:

]C[]FE[2k

dtd

=+

=rate

– If the first step is a rapid but unfavorable equilibrium (k–1 >> k1) then one can write:

]B][A[]C[

1

1 ==−kkK

and the rate becomes:

]B][A[]B][A[ obs1

12 k

kkk ==−

rate

c. The steady-state approximation – If the concentration of a reaction intermediate is approximately

constant over a period of time, one can approximate that d[I]/dt = 0 which means that the rate of formation is equal with the rate of consumption for that intermediate. This happens when the elementary steps in which the intermediate is consumed are faster, and very little intermediate can build up.

– This approximation (d[I]/dt = 0) is called the steady-state approximation and is used often to simplify the mathematics of a kinetic model.

– The steady-state assumption corresponds to a case in which the intermediate is so reactive that [I] ≅ 0.

– The steady-state approximation is a poor approximation when the rate constants of the two steps are comparable.

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CHEM 6320 Fall 2005

– Example:

k1

k-1CA + B k2 ProductsC + D

– Assume C in reactive intermediate and apply stead-state

approximation for its concentration:

=dt

d ]C[

]D[]B][A[]C[

21

1kk

k+

=⇒−

]D[]D][B][A[]D][C[rate

21

212 kk

kkk+

==⇒−

– If the first step is rate-limiting (⇔ k2[D] >> k–1): =⇒ rate

– If the second step is rate-limiting (⇔ k2[D] << k–1):

]D][C[]D][B][A[rate 21

21 kk

kk==⇒

−

– The first step is a pre-equilibrium. d. The principle of microscopic reversibility

– The same pathway that is traveled in the forward direction of a reaction will be traveled in the reverse direction, since it affords the lowest energy barrier for either process. (Each reaction coordinate defines two reactions: the forward one and the reverse one.)

e. The principle of detailed balance – It states that when a complex reaction is at equilibrium, the rate

of the forward process is equal to the rate of the reverse process for each and every step of the reaction mechanism.

f. More comments – If a proposed mechanism supports an experimentally determine

rate law, it doesn’t mean that the reaction follow that mechanism.

– Kinetics itself cannot prove a mechanism but can rule out possible mechanisms that are inconsistent with kinetics.

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CHEM 6320 Fall 2005

– Kinetics indistinguishability refers to the case of two possible mechanisms providing the same rate law. “Extrakinetic evidence” (like spectroscopic evidence of one intermediate that appear in one mechanism but not in the other) is needed to distinguish between the two mechanisms.

g. Guidelines for deducing mechanisms from rate laws The following guidelines are helpful in deducing mechanistic information from a known rate law. The more complex the rate law, the more information can be deduced.

– All species in the numerator of a rate law must appear in the mechanism as reactants. They will appear as many times in the mechanism (before or during the rate-determining step) as the order to which they appear in the rate law.

– Species missing from the numerator of the rate law occur in the mechanism after the rate-determining step.

– Fractional orders (1/2, 3/2, etc) indicate the dissociation of the molecule in question.

– A simple rate law with a species having a negative order (–1, –2, etc.) implies a rapid equilibrium step involving that species before the rate-determining step.

– A complex rate law (having two or more terms in the denominator) means that there is at least one intermediate chemical species, and suggests the use of the steady state approximation.

– The terms in the denominator of a complex rate law give specific information about what the intermediates react with (how they disappear).

– The species in the numerator of a complex rate law give information about how an intermediate is formed.

– If the numerator of the rate law has a high power of any species, that species must enter into the mechanism several times sequentially.

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7. Catalysis a. Catalysis is another way to increase the reaction rate (other than

increasing the temperature) by making the reaction occur through a different mechanism that has lower activation energy (lower rate constant).

b. A catalyst is a substance that participates in a chemical reaction but is not consumed in the process. The catalyst provides a new mechanism by which reaction can occur.

– When a catalyst is present, the heat of reaction [exothermicity (or endothermicity) of the reaction] does not change. ⇒ The position of the chemical equilibrium does not change with the presence of the catalyst.

– A catalyst lowers the activation energy for both the forward and reverse reaction ⇔ it increases the reaction rate for both the forward and reverse reaction.

– The reaction coordinate for an uncatalyzed chemical reaction is different than the reaction coordinate of a catalyzed chemical reaction.

c. Depending on the phase in which the catalyst is compared to the reactant (and product), one finds homogeneous catalysis (same phase) and heterogeneous catalysis (different phases).

d. Enzymatic reactions are catalyzed biological processes that involve enzymes.

– Enzymes are protein molecules that catalyze specific biochemical reactions.

– The reactant molecule that the enzyme acts upon is called substrate.

– The region of the enzyme where the substrate reacts is called active site.

– The specificity of an enzyme depends on the geometry of the active site and the spatial constrains imposed by the rest of the enzyme.

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8. Transition State Theory a. Transition state theory also called activated-complex theory is a theory

of the rate of elementary reactions. b. The theory focuses on transient species (called activated complex or

transition state) locate din the vicinity of the top of the barrier height (activation energy) of a reaction.

c. Deducing the rate law: – Consider an elementary reaction: A + B ⇒ Products

– The rate law is given by: ]B][A[]P[ kdt

dv ==

– According to the activated complex theory, the model of the elementary reaction is a two-step process (process here does not mean reaction but rather half-process):

PABBA ‡ →→←+

where AB‡ is the activated complex. – The transition state (activated complex) quantities are denoted by

a double dagger sign, ‡ (not ≠). – Another important assumption is that the reactants and the

activated complex are in equilibrium with each other, and the equilibrium constant for this equilibrium is:

]B][A[]AB[

/]B[/]A[/]AB[ ‡‡

‡o

oo

o ccc

cKc ==

where c° is the standard-state concentration (this is often 1.00 mol/dm3).

– The activated complexes are assumed to be stable within a small region of width δ center at the barrier top. According to the transition-state theory, the rate of the reaction is given by the rate of a unimolecular process, with a rate constant k‡, that transform the activated complex into products:

][AB][ ‡‡kdtPd

=

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CHEM 6320 Fall 2005

– The rate constant k‡ is proportional to the frequency with which the activated complexes cross over the barrier top (νc), where the proportionality constant (κ) is called transmission coefficient (and is assumed to be 1 if no other information is available):

cc‡ νκν ≅=k

– Because [ ocKc /]B][A[]AB ‡‡ =

]B][A[/]B][A[][ ‡c‡

c oo

cKcK

dtPd c

cνν ==⇒

– Compare to ]B][A[]P[ kdt

d==v

ocKk c

‡cν=⇒ (k has units of conc–1s–1)

– The equilibrium constant written in terms of partition functions:

)/)(/()/(

]B][A[]AB[

BA

‡‡‡

VqVqcVqcKc

oo

==

where qA, qB, and q‡ are the partition functions of A, B and AB‡. – Assume that the motion of the reacting system over the barrier

top is a one-dimensional vibrational motion. The vibrational

partition function, qvib, for this degree of freedom is c

Bvib νh

Tkq =

– Write the partition function of the activated complex as a product of the partition function of the motion of passing over the energy barrier and a partition function for the rest of degrees of freedom of the activated complex:

‡intvib

‡ qqq = ; )/)(/(

)/(

BA

‡int

c

B‡

VqVqcVq

hTkKc

o

ν=

‡B

BA

‡intB

‡c

)/)(/()/(

Khc

TkVqVq

cVqhc

TkcKk c

o

o

oo===

ν⇒

where K‡ is the equilibrium constant for the formation of the transition state from the reactants but with one degree of freedom

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CHEM 6320 Fall 2005

(the vibrational mode describing the motion along the reaction coordinate) excluded from the activated complex partition function.

– One can define a standard Gibbs energy of activation, ∆‡G°, as the change in Gibbs energy when going from the reactants at a concentration c° to transition state at concentration c°.

‡‡ ln KRTG −=∆ o RTGehc

Tkk /B ‡ o

o∆−=⇒

– But ∆‡G° = ∆‡H° – T∆‡S° where ∆‡H° is the standard enthalpy of activation and ∆‡S° is the standard entropy of activation:

RTHRS eehc

Tkk //B ‡‡ oo

o∆−∆= (compare with RTEAek /a−= )

d. The relationship between the Arrhenius activation energy Ea and ∆‡H° and the relationship between the Arrhenius preexponential factor A and ∆‡S°:

2a

2

‡

2

‡‡ 2ln1lnRTE

RTRTH

RTRTU

dTKd

TdTkd

=+∆

=+∆

=+=oo

RTHE 2‡a +∆=⇒ o RTEH 2a

‡ −=∆⇔ o

RTERS eehc

Tkek //B

2a

‡ −∆=o

oRSe

hcTkeA /B

2 ‡ o

o∆=⇒

( for reactions in solution) RTHE +∆= o‡a

e. The value of ∆‡S° gives information about the relative structure of the activated complex and the reactants:

– a positive value ⇒ the structure of the activated complex is less ordered than the reactants

– a negative value ⇒ the structure of the activated complex is more ordered than the reactants

– Examples: bimolecular reactions with one product (∆‡S° ≅ –25 to –35 cal/mol⋅K); unimolecular reorganization (∆‡S° ≅ –5 cal/mol⋅K); unimolecular decomposition (∆‡S° ≅ +10 to +20 cal/mol⋅K).

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CHEM 6320 Fall 2005

C. Substituent Effects and Linear Free-Energy Relationships 1. Linear Free-Energy Relationships

a. Consider the following set of reactions:

COH

O

XCO- + H+O

X

K

– Ionization will be aided by electron withdrawing groups and

discouraged by electron donating groups. One can think that this is the effect of X on ∆G of ionization.

b. Consider the following set of reactions:

k C

OH

O-

OCH3X

COCH3 + HO-O

X‡

– Rate will be increased by electron withdrawing groups and

decreased by electron donating groups. One can think of the effect of X on the ∆G‡ for saponification.

c. Consider the effect of changing from substituent X to substituent Y. In such case, if the change for the second reaction, ∆∆G‡, is found to be m times the change for the first reaction, ∆∆G, then we would expect the same proportionality constant m to relate ∆∆G‡ to ∆∆G for some other substituent change.

d. This expectation (linear free-energy relationship = LFER) would be true if the reaction being compared are affected by substituents according to the same mechanism of transmission of the electronic effect of the substituent.

– An alternate way to express LFER is ∆∆G‡ = m·∆∆G (or ∆∆G1 = m·∆∆G2 if comparing two equilibriums or ∆∆G1

‡ = m·∆∆G2‡ if comparing two reaction rates).

log = m·log KK0

kk0

acidity constant forsubstituent case

rate forsubstituent case

rate forreference case

acidity constant forreference case

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CHEM 6320 Fall 2005

2. Hammett Equation a. Given by:

where σ is the substituent constant and ρ is reaction constant (give the sensitivity of reaction to substituent changes)

– Parameters σ and ρ can have different signs. b. The reference for the substituent is usually H. c. The standard reference reaction is the ionization of X–C6H4–COOH for

which ρ = 1.

– For this reaction: 0

logKK

=σ

– Considering that RTGeK /∆−=

RTGG

303.20∆+∆−

=⇒σ

)(73.0 0C25 GG ∆−∆=oσ

– For para-nitro group (p-NO2), σp = +0.81 ⇒ ∆∆G = 1.1 kcal/mol (⇔ p-NO2 strengthen acidity by 1.1 kcal/mol)

d. Substituent constant σ – For electron withdrawing groups (e ← gps), σ > 0 (⇔ acidity

increases). – For electron donating groups (e → gps), σ < 0 (⇔ acidity

decreases). – There are large tables for meta and para substituents (σm and σp)

but not for orto substituents (σo) because there can be large steric effects and the reliability of the predictions is questionable.

– Example: Me–O–C6H4–COOH: para and meta substituents will have opposite signs of σ.

– The σ values are averaged for a large number of reactions. Because of that, they may not be very accurate for a particular reaction.

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CHEM 6320 Fall 2005

e. Reaction constant ρ – For reactions facilitated by electron withdrawing groups

(e ← gps), ρ > 0. – For reactions facilitated by electron donating groups (e → gps), ρ < 0.

3. Electronic Effects a. Polar effect – Reflects mainly a difference in electronegativity

– Field effect – transmitted through space – Inductive effect – transmitted through bonds

– We will not discuss the field effect – Polar effects are going down about 35% (1/2.8) per carbon

atom b. Resonance effect

– Example: π delocalization – Cannot extend across saturated centers.

4. Modifications of the Hammett Equation a. Clearly, some reactions will be more sensitive to resonance effects than

others. b. In addition to σm and σp, one can use two new parameters introduced

by Hammett: σ + when there is direct interaction of electron donating substituent

with a cationic reaction center

CH3O CH2+..

CH3O C 2H+

σ – when there is direct interaction of electron withdrawing substituent with an anionic reaction center

N OO

O

+N OO

O

+

– These new parameters cannot be used if an extra methylene is

present. In this case, one should use the initial σ parameter. c. More advance equations were proposed by Yakawa & Tsuno, Swain &

Lupton, and Taft (see textbook).

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CHEM 6320 Fall 2005

5. Applications to Reaction Mechanisms a. Magnitude of ρ (as well as the sign) tells sensitivity of reaction to

electronic effects of substituents, in turn indicating the extend of charge development at the reaction center at the transition state.

b. Hammett plots may show breaks. log K

σ

log K

σ – This is most likely seen for multistep reaction where there is a

change in rate limiting step as substituents vary to make reaction faster or slower.

– Each step will have its own value of ρ. c. Swain–Lupton or Taft analysis lead to assessment of polar vs.

resonance effects not evident from simple Hammett analysis. One can classify substituents as: e → resonance (–M)

e → polar (–I) e → resonance (–M)

e ← polar (+I) e ← resonance (+M)

e ← polar (+I) M > I I > M

all alkyls AcNH– Br– CH3CO– O2N– AcO– Cl– ROCO– CF3– R2N– (H2N–) F– N≡C– Me3N– RO– (HO–) Ph–

– M stand for mesomeric (other word for resonance) – Comments:

– I value is more important for meta substituents. – M value is more important for para substituents.

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CHEM 6320 Fall 2005

D. Mechanistic Concepts 1. Kinetic versus Thermodynamic Control

a. General free-energy diagram:

R

A B

∆GB‡∆GA

‡

thermodynamic product

kinetic product

b. Thermodynamically, R → A transformation is more favorable than R → B transformation (A is the thermodynamic product).

c. Kinetically, R → B transformation is more favorable than R → A transformation (B is the kinetic product).

d. Commonly, the thermodynamic and kinetic products are the same. e. Example:

O

O Odeprotonation

OO+ CH3I

- I-+ CH3I

- I-

– For thermodynamic product:

– weak base = t-BuOK/t-BuOH (to make deprotonation process close to equilibrium)

– higher T – For kinetic product:

– strong base = RLi (can attack higher pKa) – lower T

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CHEM 6320 Fall 2005

2. Hammond’s Postulate a. Two states which occur consecutively during a reaction process and

which have similar energies, will have only small differences in molecular structure.

b. Implications: – For a highly exothermic reaction, the transition state resembles

reactants (“early” transition state). – For a highly endothermic reaction, the transition state resembles

products (“late” transition state).

A–X+Y

A–Y+X

X···A···Y

A–X+Y

A–Y+X

X···A···Y

A–X+Y

A–Y+X

X··A····Y

A–X+Y

A–Y+X

X··A····Y

A–X+Y

A–Y+X

X····A··Y

A–X+Y

A–Y+X

X····A··Y

c. Example: Selectivity of R–H + ⋅Cl vs. R–H + ⋅Br reactions

– The R–H + ⋅Br → R⋅ + H–Br reaction – Weak H–Br bond ⇒ – Late transition states ⇒ – Reaction is

– The R–H + ⋅Cl → R⋅ + H–Cl reaction – Strong H–Cl bond ⇒ – Early transition states ⇒ – Reaction is

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CHEM 6320 Fall 2005

d. High-energy intermediate – Electrophilic aromatic substitution (EAS)

H

+ E+ ‡

HE

+-H+

E

σ complex – Esther hydrolysis (saponification)

+ HO- ‡ C

O-

OEtCH3

OH

- EtO-CH3 C

O

OEt

CH3 C

O

OHtetrahedral

intermediate – The energy profile along the reaction coordinate for both reaction

above is as below where the high-energy intermediate I is either the σ complex or the tetrahedral intermediate.

Energy

Reaction coordinate

PR

‡‡

I

Energy

Reaction coordinate

PR

‡‡

I

– One should realize that the high-energy intermediate is alike the

two transition states because they have close energy and similar structures. (High energy intermediate can be seen as an “approximation” to the transition states.)

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CHEM 6320 Fall 2005

3. Curtin-Hammett Principle a. Assume one the two situations below:

or∆GAB

PA

∆GB‡

∆GA‡

PBA B

∆GAB

PA

∆GB‡

∆GA‡

PBA B

∆GABA BPA

∆GB‡

∆GA‡

PB

∆GABA BPA

∆GB‡

∆GA‡

PB

b. If ∆GAB for two equilibrating reactant states (i.e., different conformations) is small with respect to both ∆GA

‡ and ∆GB‡, then the

product ration is not determined by the A/B conformational distribution, but only be the relative energies of the transition states leading to A and B.

– Usually A B is a conformational equilibrium. c. Position of equilibrium between conformations (which conformation is

favored) cannot control the product ratio. d. A reaction may proceed through a minor configuration if it is the one

that leads to the lowest-energy transition state. e. Example:

Cl

ClSN2 E

HO-‡1‡2

HO-

2 – The analysis should be made between the transition state

energies of the two reactions (‡1 and ‡2) not between the energies of the conformers.

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CHEM 6320 Fall 2005

E. Catalysis by Acids and Bases 1. Definitions

a. Acid-base catalysis occurs when conjugate acid or base of the reactant reacts faster.

b. Specific acid catalysis (SAC) – The rate depends on equilibrium protonation of the reactant. – Example:

B + H+ BH+Ka

BH+ + C B-C + H+r.d.s.k2

– Direct B + C reaction is slower. – The rate of reaction is given by:

]B][C[][H]][CB][H[]C][BH[rate obsa

22

+++ === kKkk

– The rate is a direct function of pH. c. Specific base catalysis (SBC)

– The rate depends on equilibrium participation of HO–. – Example:

X + C Products r.d.s.

k2

B + HO- XK

– X is the conjugate base of B or HO– adduct. – The rate of reaction is given by:

]B][C[][HO]][CB][HO[]C][X[rate obs2

2−− === k

Kkk

– The rate is a direct function of pH.

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d. General acid catalysis (GAC) – The rate is influenced by proton donors other than just H+ itself.

...]X][Y[][HA]X][Y[][HA]X][Y[][Hrate 22110 +++= + kkk – HA1 and HA2 are “buffer” species. – The term k0[H+][X][Y] is the SAC term.

– The general acid catalysis can be distinguished by observing change in rate with [HA] at constant pH. If there is no change then only SAC operates.

rate

SAC[buffer]

OH

Nu

..

+ +

Nu

..

OCH3COO H

– Example:

e. General base catalysis (GBC) – The rate is influenced by proton acceptors other than just HO–

itself. ...]X][Y[][B]X][Y[][B]X][Y[][HOrate 22110 +++= − kkk

– B1 and B2 are “buffer” species. – The term k0[HO–][X][Y] is the SBC term.

2. Kinetic Indistinguishability of General Acid/Base Catalysis a. Acid catalysis vs. base catalysis

– Consider general acid catalysis:

R + HA RH+ + A- r.d.s.

Productsfast

– The rate is: rate = kobs[R][HA] – Note that there is no free H+.

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CHEM 6320 Fall 2005

– Consider general base catalysis: R + HA RH+ + A- (preequilibrium)

RH+ + A- product + HAr.d.s.

– The rate is: rate = k′obs[RH+][A–]

– But ]RH[]H][R[RH

a +

+=

+K ( +

++ =⇔

RHa

]H][R[]RH[K

) and

]HA[]H][A[HA

a

+−=K (

]H[]HA[]A[

HAa

+− =⇔

K )

– Substituting in rate law expression:

]R][HA[rateRHa

HAa

obs +′=

K

Kk

– This expression is the same as the one for GAC with

+′=

RHa

HAa

obsobsK

Kkk

b. One cannot distinguish kinetically between GAC by HA and GBC by the conjugate base of HA, A–.

c. Likewise, one cannot distinguish kinetically between GBC by B and GAC by the conjugate acid of B, BH+.

d. One should therefore just called general catalysis (GC) or general acid-base catalysis (GABC).

3. Other Aspects of Acid-Base Catalysis a. Key distinguishing features of specific versus general catalysis:

– For SAC/SBC, H+ transfer occurs in a rapid equilibrium. – For GAC/GBC, H+ transfer occurs at the rate-limiting step.

b. How test the relative effectiveness of various GAC’s as proton donors? – One would expect that the best catalyst is the hydronium ion (or

the “proton”). – One would expect correlation of catalytic activity with pKa.

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CHEM 6320 Fall 2005

– Bronsted catalysis law: bKk += acat loglog α – The parameter α gives the sensitivity of catalytic step to

acidity differences. – Usually, as α → 1, the greater is the degree of proton

transfer at the transition state. – Example:

– Exothermic reaction ⇒ early transition state ⇒ the strength of the acid is not that important.

– Endothermic reaction ⇒ late transition state ⇒ the strength of the acid is more important.

c. Example >C=O + HONH2 → >C=N–OH (oxime) – Reaction is catalyzed by AcOH. – Possible mechanism explaining the experiment rate law:

rate = kobs[HONH2][>C=0][HOAc]

HONH2

..C O H OAc.... r.d.s.

– The real mechanism:

+ H OAcC Ofast

C O H+ + AcO-

AcO- NH

OH

H+C O H r.d.s.

– The rate constant of the second step of this mechanism

should be larger than the rate constant of the rate-determining step of the mechanism above.

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4. Acidity Functions a. In any solvent the strongest acid possible is the solvent conjugate acid

and the strongest base possible is the solvent conjugate base. b. In water, one is limited to the range between H3O+ (pKa = –1.7) and

HO– (pKa = 15.7). c. In non-aqueous solvent or mixed solvent media, how do one express

acids stronger than H3O+ and bases weaker than HO–? d. Can define an acidity scale on basis of the fractional ability of some

reference set of weak bases to be protonated: B + H+ BH+

BH

BH

BH

BH]B[

]BH[γ

γ

+

+

+

++

==aaa

aK (γ are activity coefficients)

B

BH0H ]B[

]BH[γ

γ

Kha

++

+

=≡⇒

– Choose B as “dye” such that B BH+ is determined colorimetrically.

– A new scale defined below can be used like a pH scale. 00 log hH −=

e. Procedure to how to determine H0 and K: – Choose a B/BH+ pair where K is determinable at the extreme in

measurable pH. – Example: At pH = 14 (pOH = 0) , if pKb = 0 (⇔ pKa = 14) ⇒ B:BH+ = 50:50

– Use K to determine H0 just beyond the pH scale. – If B:BH+ = 91:9.1 ⇒ pH = 15 (H0 = 15) – If B:BH+ = 99:0.99 ⇒ H0 = 16

– Using the determined H0, now determine K for a second B/BH+ pair.

– Continue extending H0 scale little by little.

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F. Kinetic Isotope Effects 1. Molecular Vibrations

a. Consider a system (i.e., molecule) with N nuclei and n electrons. – This will be described by a wave function Ψ(r,R) depending on

the coordinates of all the nuclei and all the electrons. – According to Born-Oppenheimer approximation, the wave

function can be separated: )();(),( nuel RRrRr ψψ=Ψ

– Solve for ψel considering a fixed position of the nuclei (at one nuclear configuration).

– Electrons create a potential in which nuclei move. This is called a potential energy surface for polyatomic and potential energy curve for a diatomic. The nuclear movement of this potential is also quantized.

b. For a diatomic molecule, the potential for nuclear movement is a Morse curve. A first-order approximation to this curve is a harmonic curve.

c. Vibrational energy levels based on harmonic-oscillator model ( ) νhvEv 2

1+=

v = 0, 1, 2, … is the vibrational quantum number

µπν k

21

= is called fundamental (vibrational) frequency

BA

BAmm

mm+

=µ is the reduced mass

k is the force constant d. The minimum energy (the ground-state energy) is not zero even for

ν = 0. This energy is called the zero point energy (ZPE): νh2

1ZPE =

e. The number of molecular vibrations (or normal modes or frequencies) is (3n – 6) for a non-linear molecule and (3n – 7) for a linear molecule where n is the number of atoms in the molecule.

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2. Primary Kinetic Isotope Effect (1º KIE) a. 1º KIE is observed in a reaction where a bond to the isotopically

substituted atom is broken in the rate-limiting step. If the bond is broken after the rate-determining step, then no KIE is observed.

b. 1º KIE arises because the zero-point vibrational energy of the bond to the heavier nucleus is lower than that of the bond to the lighter nucleus.

– R3C–D breaks slower than R3C–H – 13C–H breaks slower than 12C–H

c. At the transition state, the bond vibrational degree of freedom is being converted to a translational degree of freedom, such that the difference between C–H and C–D gets much smaller.

+A H BBH

A H B

A +

BHA

A H B

early

late

C–DC–H

C–DC–H

C–DC–H

C–DC–H

C–DC–H

C–DC–H

d. For a classical mechanism (no tunneling involved), the 1º KIE will thus

be maximal (kH/kD ≈ 7-10 at 25ºC) when there is maximal bond breaking at the transition state, which will be the case for isothermal reactions (when the transition state occurs near the 50% point along the reaction coordinate).

e. For example, the reaction A–H + ⋅B → A⋅ + H–B will have a symmetrical transition state (A⋅⋅⋅H⋅⋅⋅B) if the A–H and B–H bond energies are similar.

f. However, if the reaction is very exothermic (early transition state = A⋅⋅H⋅⋅⋅⋅B) or the reaction is very endothermic (late transition state = A⋅⋅⋅⋅H⋅⋅B), there are significant differences in the zero-point vibrational energies between C–H and C–D at the transition states, and thus the 1º KIE will be small.

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3. Secondary Kinetic Isotope Effect (2º KIE) a. Occurs in reactions where the bond to the isotopically substituted atom

is perturbed (but not broken) in the rate-determining step. (kH/kD ≈ 0.7-1.5 in such cases – most common are hybridization changes)

b. 2º KIE can be either – normal (kH/kD > 1)

– Example: sp3 → sp2 at transition state; C–H bending becomes freer (e.g., SN1 reaction)

– inverse (kH/kD < 1) – Example: sp2 → sp3 at transition state; C–H bending

becomes more restricted (e.g., nucleophilic addition to carbonyls)

c. 1º KIE vs. 2º KIE – Consider the finding that a D-for-H substitution results in an

observed KIE of 1.5. – This could indicate either a full normal 2º KIE or a small 1º KIE

(such as if C–H bond breaking is only partly rate-limiting). – However, on observed KIE ≥ 2 provides strong evidence for C–

H bond breaking in the rate-limiting step. 4. Isotope Labeling

a. It can be used to follow stereochemistry at a prochiral center – Example: CH3CH2Br + HO– → SN2

b. It can be used to check for the existence of intermediates that allow for exchange.

– Example: E1cb mechanism results in incorporation of D into recovered substrate when reaction is run in deuterated solvent (not true for E2 or E1).

Ph-CH2-CH2-Br Ph-CH--CH2-Br Ph-CH=CH2-Br-

c. One can use radioisotopes (3H, 14C) or NMR-active nuclei (1H, 13C,

15N).

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G. Solvent Effects 1. General Aspects of Solvation

a. Nature of solvent effects – Bulk property effects of the solvent, reflecting its dipolar nature

(dielectric constant) and/or polarizability – Local solvation factors (e.g., hydrogen bonding and ion-dipole

interactions) b. Solvents can be classified as protic or aprotic, polar or nonpolar. c. Factors determining solvent polarity

– Dielectric constant (ignores local solvation) – The speed of a reaction in various solvents is a measure of the

solvent polarity – Example: t-BuCl → t-Bu+ + Cl– – Can use relative rate as a measure of “ionizing power” of

the solvent. – Winstein formula: Y )/log( EtOH aq. 80%solvent kk=

– Spectrophotometric measurements of dye molecules with charge separated excited state

– Example: A – B → δ+A – Bδ– 2. Solvent Effects on Reaction Rates

a. To predict the solvent effect on the rate of a reaction, one should consider the charge properties of the transition state compared to that of the reactant state.

– Charges at the transition state can be estimated by taking the weighted average of the reactant state charges and the product state charges.

b. In going from reactants to the transition state, if there is: – Charge creation ⇒ rate increases as solvent polarity increases – Charge localization (same overall charge) ⇒ rate increases

slightly as solvent polarity increases – Charge destruction ⇒ rate increases as solvent polarity decreases – Charge dispersal (same overall charge) ⇒ rate increases slightly

as solvent polarity decreases

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CHEM 6320 Fall 2005

c. Solvent effects for the four charge types of SN1/SN2 reactions – Reaction of type Nu– + S+ (negatively charge nucleophile +

positively charged substrate) Example: HO– + Me3S+ → MeOH + Me2S

SN1: Me3S+ → [Meδ+ ⋅⋅⋅ δ+SMe2]‡ ⇔ charge dispersal SN2: HO– + Me3S+ → [HOδ– ⋅⋅⋅ Me ⋅⋅⋅ δ+SMe2]‡ ⇔ charge

destruction ⇒ Increasing solvent polarity will slow down both SN1 and SN2.

– Reaction of type Nu + S+ (neutral nucleophile + positively charged substrate) Example: Et2S + Me4N+ → Et2S+Me + Me3N

SN1: Me4N+ → [Meδ+ ⋅⋅⋅ δ+NMe3]‡ ⇔ charge dispersal SN2: Et2S + Me4N+ → [Et2Sδ+ ⋅⋅⋅ Me ⋅⋅⋅ δ+NMe3]‡ ⇔ charge

dispersal ⇒ Increasing solvent polarity will slow down both SN1 and SN2.

– Reaction of type Nu + S (neutral nucleophile + neutral substrate) Example: Me3N + EtI → Me3N+Et + I–

SN1: EtI → [Etδ+ ⋅⋅⋅ δ–I]‡ ⇔ charge creation SN2: Me3N + EtI → [Me3Nδ+ ⋅⋅⋅ Et ⋅⋅⋅ δ–I]‡ ⇔ charge creation

⇒ Increasing solvent polarity will speed up both SN1 and SN2. – Reaction of type Nu– + S (negatively charge nucleophile +

neutral substrate) Example: NC– + CH3Br → CH3CN + Br–

SN1: CH3Br → [CH3δ+ ⋅⋅⋅ δ–Br]‡ ⇔ charge creation

SN2: NC– + CH3Br → [NCδ– ⋅⋅⋅ CH3 ⋅⋅⋅ δ–Br]‡ ⇔ charge dispersal

⇒ Increasing solvent polarity will speed up the SN1 reaction but will slow down the SN2 reaction.

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CHEM 6320 Fall 2005

3. Choosing a Solvent a. The problem:

– The most common SN2 charge type (RX +Y– → RY +X–) predicts that the maximal rate will be achieved in less polar solvents.

– But the Y– reactant is part of some salt M+Y–, which, if M+ = Na+, K+, etc., will not be soluble in nonpolar solvents.

b. Solutions to the problem:

– A solution is to use alcohols that are less polar than water, but Y– reactivity will still be diminished by H-bonding and ion-dipole interactions.

– A better compromise is often a polar aprotic solvent (e.g., DMF, DMSO) that will dissolve M+Y– without unduly solvating Y–.

– Another alternative is to use M+ = R4N+, e.g., n-Bu4N+, so sufficient solubility can be achieved in less polar organic solvents.

– The ideal situation would be a nonpolar aprotic solvent like benzene. In this case, one can use 18-crown-6 to solubilize K+ so that K+Y– can be used as the reactant nucleophile. As a result, Y– will be a powerful (unsolvated, “bare”) nucleophile.

c. The concept of phase-transfer catalysis takes advantage of the ability to “dissolve” a strong nucleophile in the “organic layer” due to the inclusion of organic cations in the system.

– Examples of cation are tetraalkylammonium and tetraalkylphosphonium ions, which has good solubility in organic solvents.

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CHEM 6320 Fall 2005

4. Solvent Isotope Effects a. Obtained when replacing H2O with D2O as the solvent. b. D3O+ is a stronger acid than H3O+ ⇒ species that can be protonated

will exist more as conjugate acids (i.e., protonated) in D2O. – Thus, if rate is proportional with [RH(D)+], kH2O/kD2O < 1.

c. D–O bond breaks slower than H–O bond due to a 1º KIE. – Thus, if the reaction involves H+ transfer in the rate determining

step (like the case of the general acid-base catalysis), kH2O/kD2O > 1.

d. If one observes kD2O ≈ kH2O, then: – Neither (b) or (c) apply, or – Both (b) and (c) are operating, but offsetting each other.

5. Gas Phase vs. Solution Phase a. Sensitivity to electronic effects is much higher in gas phase.

CH2

Br

X

SN1CH2

X

++ Br-

– The same sign (negative) of ρ.

b. Acidity rank orders t-BuOH i-PrOH EtOH MeOH

increase in solution

increase in gas phase – In gas-phase there is a better charge delocalization for t-BuOH. – The trend in gas phase is a result of intrinsic electronic effect.

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H. More Concepts 1. Intermediate Detection and Reaction Mechanism

a. One would try to “catch” an intermediate in order to verify a proposed mechanism.

R → I → P b. The presence of I can be verified by detecting it spectroscopically or by

trapping it. c. Its mere presence does not necessarily prove the mechanism though.

– For example the mechanism can be of form: R P

I 2. Lewis Acid/Base Catalysis

a. A Lewis acid is an electron acceptor, a Lewis base in an electron donor.

– A somewhat unusual Lewis acid is a metallic ion: Mn+. b. Hard-soft acid-base theory

– Soft Lewis acid/base – Large size, low charge density, high polarizability – They react between themselves.

– Hard Lewis acid/base – Small size, high charge density, low polarizability – They react between themselves.

– Example: F– is hard, I– is soft

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CHEM 6320 Fall 2005

I. Problems and Exercises 1. Calculate the enthalpy and entropy of activation (∆H‡ and ∆S‡) at 40ºC for

the acetolysis of m-chlorobenzyl p-toluenesulfonate from the data given: Temp (ºC) 25.0 40.0 50.1 58.8

k × 105 (s–1) 0.0136 0.085 0.272 0.726 2. 2-Vinylmethylenecyclopropane rearranges in the gas phase to

3-methylenecyclopentene. Two possible reaction mechanisms are:

or

a. Sketch a reaction energy diagram for each process. b. How might an isotopic labeling experiment distinguish between these

mechanisms? 3. The phenyl radical has assigned both σ + and σ – values, and these are:

σ + = –0.18 and σ – = +0.08. Notice that the signs of σ + and σ – are different. Discuss the reasons that the phenyl group has both a σ + and σ – value and explain why they are of different signs.

4. Match the ρ values with the appropriate reactions. Explain your reasoning.

Reaction constants: +2.45, +0.75, –2.39, –7.29. a. Nitration of substituted benzenes b. Ionization of substituted benzenethiols c. Ionization of substituted benzenephosphonic acids d. Reaction of substituted N,N-dimethylanilines with methyl iodide

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CHEM 6320 Fall 2005

5. The pseudo-first-order rate constants for the acid-catalyzed hydration of substituted styrenes in 3.5 M HClO4 at 25ºC are given. Plot the data against σ and σ + and determine ρ and ρ +. Interpret the significance of the results.

Substituent k × 108 (s–1) σ σ + p-CH3O 488,000 –0.12 –0.78 p-CH3 16,400 –0.14 –0.31

H 811 0 0 p-Cl 318 +0.24 +0.11

p-NO2 1.44 +0.81 - 6. Write the rate law that would describe the rate of product formation for the

following system:

C CH

BrBr

H+ R3N

k1

k-1

+ R3NH+C CBrBr

H

C CBrBr

H k2 C C BrH + Br

C C BrHk3+ R3N C C NR3H

+ + Br if the second step is rate-controlling and the first step is a preequilibrium.

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CHEM 6320 Fall 2005

7. Predict whether normal or inverse isotope effects will be observed for each reaction below. Explain. Indicate any reactions in which you would expect kH/kD > 2. The isotopically substituted hydrogens are marked with asterisk.

a. CH3CH3CH3CH3

CH3CH3CH3CH3

+

CH2

C

CH2

OEt

CH2

C

CH2

OH+

CH2

C

CH2EtOHH2O

CH2

C

CH2

OSO2Ar HHH H

b.

CH3CHCH3

NO2

+ OH + H2ONO2

CH3CCH3+

c.

C OPh2C PhCH CH2+Ph2C

C CH2

C

H

O

Ph

Ph2C

C CH2

C

H

O

Ph

d.

CH3CH2C

OC2H5

OC2H5

HH+

fastCH3CH2C

OC2H5

OC2H5

H

H+

slow

+CH3CH2C

OC2H5

H

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CHEM 6320 Fall 2005

8. A mechanism for alkene arylation by palladium(II) is given below. The isotope effect kH/kD was found to be 5 when benzene-d6 was used. When styrene-β,β-d2 was used, no isotope effect was observed. Which step is rate-determining?

k2

k3

k1

PhCH CH2+

Pd(O2CCH3)2

Ph Pd(II)O2CCH3 Ph Pd CH CHPh + CH3COOH

Pd(O)+Ph Pd CH CHPh PhCH CHPh

CH3COOH++ Ph Pd(II)O2CCH3

9. Consider the decomposition reaction of : )g(ON 52

(g)O(g)4NO(g)O2N 2252obs +→k

A proposed mechanism for this reaction is:

(g)NO(g)NO(g)ON 32

1

1

52 +⇒⇐−

k

k

(g)O(g)NONO(g)(g)NO(g)NO 222

32 ++⇒+k

(g)2NONO(g)(g)NO 23

3

k⇒+

Assume that the steady-state approximation applies to both the and reaction intermediates to show that this mechanism is

consistent with the experimentally observed rate law

(g)NO3NO(g)

]]O[5

2d

ON[ 2obskt

d= .

Express in terms of the rate constants for the individual steps of the reaction mechanism.

obsk

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CHEM 6320 Fall 2005

10. The rate law for the reaction between and to form phosgene

: is

CO(g)

(COCl2

(g)Cl2

)COCl( 2 )g)g(CO)g(Cl obs2 →+

k

][CO][Cl 2/32obskCO]Cl[ 2

dtd

= .

Show that the following mechanism is consistent with this rate law:

)g(M)g(2Cl )g(M)g(Cl1

1

2 +⇒⇐+

−

k

k (fast equilibrium)

)g(M)g(ClCO )g(M)g(CO)g(Cl2

2

+⇒⇐++−

k

k (fast equilibrium)

)g(Cl)g(COCl )g(Cl)g(ClCO 23

2 +⇒+k

(slow) where M is any gas molecule present in the reaction container. Express

in terms of the rate constants for the individual steps of the reaction mechanism.

obsk

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CHEM 6320 Fall 2005

Concerted Pericyclic Reactions A. Electrocyclic Reactions

1. Introduction a. Concerted reactions are single-step processes, and they can be both

unimolecular and bimolecular reactions. b. Pericyclic reactions are cyclic unimolecular electronic rearrangements

(occurring through cyclic transition states). c. Another name is valence tautomerizations (≡ electronic

isomerizations). d. Woodward and Hoffmann proposed that the pathways of the concerted

pericyclic reactions were determined by the symmetry properties of the orbitals directly participating in the reaction.

e. Electrocyclic reactions are a class of pericyclic reactions in which formation of a single bond occurs between the ends of a linear conjugated system of π electrons (as well as the reverse process).

f. Electrocyclic reactions involve redistributing π and σ bonds but the total number of bonds remains constant.

2. Common Electrocyclic Reactions a. The case of two π bonds ↔ π bond + σ bond

– This is a four-electron system. A

D

B

C

2π π + σ

AB

CD

– The question is what diastereoisomer will be formed. – According to microscopic reversibility principle both forward

and reverse reaction will follow the same mechanism. – If the mechanism were concerted, one would expect a single

stereochemical result.

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CHEM 6320 Fall 2005

– Example: 2,4-hexadiene ↔ 3,4-dimethylcyclobutene reaction Me

H

H

Me

∆

Me

HMe

Hcis

E, Z

Me

HH

Me∆

∆

trans

E, E

Z, Z

– Each reaction is highly stereospecific (no other stereoisomer formed).

– Additional example H

H H

H

EZ

– The E,Z isomer cannot be obtained (sterically impossible). – The Z,Z isomer cannot be obtained (sterically possible but

electronically “forbidden”).

ZZ

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CHEM 6320 Fall 2005

b. The case of three π bonds ↔ two π bonds + σ bond – This is a six-electron system. – This is a different system from the ones above because there are

two extra π electrons. The rules for this system will be different. – Example: 2,4,6-octatriene → 5,6-dimethyl-1,3-cyclohexadiene

Me

MeZ

E

EMeH

H

Me

cis

trans

HH

Me

Me

Z

E

Z

Me

HMe

3. Rules for Predicting the Products

a. The question is how the molecule ends should twist to make the product?

– The two options are disrotatory and conrotatory motion. b. Electronic rules do not address the equilibrium constant (i.e., the

preference of the forward or the reverse reaction). c. Analysis of the symmetry of reactant HOMO

– HOMO for butadiene has one node, and one predicts conrotatory motion (both clockwise and counterclockwise).

– For hexatriene ↔ cyclohexadiene reaction, looking at ψ3 (two

nodes), one predicts disrotatory motion.

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CHEM 6320 Fall 2005

d. Analysis of the topology of the transition state – Like drawing resonance for cyclic conjugated systems – Disrotatory motion for 4n + 2 systems:

– Transition state has Hückel topology. – A Hückel system has zero (or any even number) of phase

changes around an orbital array. – A Hückel system is aromatic for 4n + 2 electrons,

antiaromatic for 4n electrons. – Conrotatory motion for 4n systems:

– Transition state has Möbius topology. – A Möbius system has one (or any odd number) of phase

changes around an orbital array. – A Möbius system is aromatic for 4n electrons, antiaromatic

for 4n + 2 electrons. e. Using correlation diagrams

– It is based on analyzing orbital symmetry of reactants and products.

– The procedure: – Find a symmetry element preserved R → ‡ → P. – Analyze all reacting orbitals (orbitals changing during the

reaction) as symmetric (S) or antisymmetric (A) with respect to the symmetry element.

– Fill the orbitals with electrons. – Identify correlations between the reactant and product

orbitals.

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CHEM 6320 Fall 2005

f. Correlation diagram for cyclobutene ↔ butadiene

disrotatory motionplane of symmetry

HH H H

HH HH

H

H

H

H

conrotatory motionaxis of symmetry

HH H H

HH HH

H

H

H

H

– Symmetry properties of cyclobutene and butadiene orbitals

+ +

+ +

__ _ _

++

+ _

classificationwith respect to

the plane

the axis

S S A A

S A S A

σ π π∗ σ∗

__

++ __

++

+_

+ __ +

+ _

__

++ ++

__

_ +

+ __+

+_

classificationwith respect to

the plane

the axis

S A S A

A S A S

ψ1 ψ2 ψ3 ψ4

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CHEM 6320 Fall 2005

– Correlation diagram for the conrotatory motion – The element of symmetry is the axis of symmetry.

cyclobutene butadiene

σ*

π*

π

σ

ψ4

ψ3

ψ2

ψ1 – This is the symmetry-allowed conrotatory reaction. (The

symmetry is conserved.) – Correlation diagram for the disrotatory motion

– The element of symmetry is the plane of symmetry. cyclobutene butadiene

σ*

π*

π

σ

ψ4

ψ3

ψ2

ψ1 – This is the symmetry-forbidden disrotatory reaction. – The HOMO in butadiene (antisymmetric) corresponds to

LUMO in cyclobutene and reverse. g. Correlation diagram for hexatriene ↔ cyclohexadiene

ψ1, ..., ψ6 ψ1, ..., ψ4, σ, σ* – This is an example of six-electron system that is allowed for

disrotatory motion (maintaining a plane of symmetry).

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CHEM 6320 Fall 2005

– Correlation diagram for the disrotatory motion hexatriene cyclohexadiene

ψ3

ψ2

ψ1

σ

ψ4

σ*

ψ4

ψ3

ψ2

ψ1

ψ5

ψ6

4. More examples

a. Dewar benzene H

H

H

H

H

H

H

conrotatorymotion

disrotatory motion

electronically forbidden sterically allowed

electronically allowedsterically forbidden

4 e-

Z

Z

Z

E

– Dewar benzene is unexpectedly stable.

b. Cycloheptatriene – Disrotatory motion (allowed for

six electrons). – Sterically allowed and

electronically allowed. – Could not make the trans isomer.

H

H

H

H

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c. Charged systems – They works in the same way. – Examples of two-electron systems

+

+

– Two-electron electrocyclic reaction is allowed for

disrotatory motion (same as for 6 electrons). CH3

CH3H

H

orCH3

CH3

H

H

+H

H

CH3

CH3

+disrotatory 2 e-

sterically forbidden

2 e-disrotatory CH3

H

H

CH3

+

CH3

HH

CH3

– For a four-electron system, reaction is allowed for conrotatory motion.

O

OH

H H

OH

OH

+H+

5p - π

3p - πσ

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B. Sigmatropic Rearrangements 1. Introduction

a. Sigmatropic rearrangements are reactions that involve a concerted reorganization of electrons during which a group attached to a σ bond migrates to the terminus of an adjacent π-electron system with a simultaneous shift of the π electrons.

b. Sigmatropic rearrangements are labeled [i, j] where i represents the number of atoms in the migrating fragment and j represents the number of atoms in the π system directly involved in the bonding changes.

c. Topology of sigmatropic migration. – Suprafacial (or supra) migration involves a process in which the

migrating group remains associated with the same face of the conjugated π system throughout the process.

– Antarafacial (or antara) migration involves a process in which the migrating group moves to the opposite face of the π system during the migration.

d. Examples

b d

cHa

db

aHc

db

ac

H

b d

cHa

R RR R

OO

OSR OR2N

CH2 H3C CH2R CH2 CH2 RCH2CH3 CH2

S OR + _R2N O

_+

1,3-suprafacial shift of hydrogen 1,3-antarafacial shift of hydrogen

1,5-shift of alkyl group1,3-shift of alkyl group

3,3-sigmatropic rearrangement of allyl vinyl ether (oxi-Cope rearrangement)

3,3-sigmatropic rearrangement of 1,5-hexadiene(Cope rearrangement)

1,7-sigmatropic shift of alkyl group1,7-sigmatropic shift of hydrogen

2,3-sigmatropic rearrangement of amine oxide2,3-sigmatropic rearrangement of allyl sulfoxide

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2. Classification of Sigmatropic Hydrogen/Alkyl Shifts

suprafacial 1,3 1,3 suprafacial retention 1,3 suprafacial inversion Hückel system Hückel system Möbius system 4 electrons 4 electrons 4 electrons

suprafacial 1,5 1,5 suprafacial retention 1,5 suprafacial inversion Hückel system Hückel system Möbius system 6 electrons 6 electrons 6 electrons

antarafacial 1,7 suprafacial 1,7 Möbius system Hückel system 8 electrons 8electrons

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3. Selection Rules for Sigmatropic Shifts of Order [i, j] Order [1, j] 1+ j supra/ret* supra/inv* antara/ret* antara/inv*

4n forbidden allowed allowed forbidden 4n + 2 allowed forbidden forbidden allowed

Order [i, j] i + j supra/supra supra/antara antara/antara 4n forbidden allowed forbidden

4n + 2 allowed forbidden allowed (*ret = retention, inv = inversion)

4. Examples a. 1,3-alkyl shift occurs thermally with inversion of configuration of

migrating carbon

H

OAc

H

D

HH

H

OAc

D

H

HH

H

OAc

D

H

HH

product if retention(not observed)

H

OAc

D

H

HH

product if inversion(observed)

OAc

H

D

H

∆

– This reaction is not possible for hydrogen.

b. 1,5-alkyl shift occurs thermally with retention of configuration of migrating carbon

H

H

Me

Me

H

H

Me

Me

H

H

Me

Me Me

Me

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c. Cope rearrangement (3,3-sigmatropy on 1,5-diene)

CN

COOEt

CN

COOEt

∆

– The product is stabilized because it allows conjugation. – Reaction proceeds usually through chair-like transition state.

(Boat-like transition state is less stable by about 6 kcal/mol.)

Ph

MeMe

MePh

Me

Ph

Me

Me

Ph

MeMe

E (major)

Z (minor) – Both transformations are electronically allowed. – Chirality in reactant is translated into the product. (This is a

general feature of sigmatropic shifts.) d. Degenerate rearrangement

H

H

H

H

fast at high T

– The molecule (homotropilidene) transforms into itself. – At low temperature, the transformation is slow so the 13C NMR

gives five signals (five different carbons). – At high temperature, the transformation is fast so the 13C NMR

gives three signals (three different carbons). – Another example is bullvalene:

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C. Cycloaddition Reactions 1. Diels-Alder Reaction

a. Cycloaddition reactions involve two molecules as reactants. b. Diels-Alder reaction, also called the [4π + 2π] cycloaddition (or 4 + 2

cycloaddition), is the addition of an alkene (or dienophile) to a diene. c. Diels-Alder reaction involves 6 electrons so the transition state has a

Hückel topology. The reaction is stereospecific syn (suprafacial) with respect to both diene and alkene.

d. Correlation diagram for [4π + 2π] cycloaddition – Define a symmetry element preserved

R → ‡ → P. In this case, this is a plane of symmetry.

HH

HH

H

H

H

H

– Assign all reacting orbitals as symmetric (S) or antisymmetric (A) with respect to the symmetry element.

– Draw the correlation diagram and identify correlations between the reactant and product orbitals

– Fill the orbitals with electrons and decide if the reaction is allowed.

– Same arguments show that 2 + 2 cycloaddition is forbidden.

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2. Selection Rules for m + n Cycloadditions m + n supra/supra supra/antara antara/antara

4n forbidden allowed forbidden 4n + 2 allowed forbidden allowed

3. Stereochemistry of Diels-Alder Reaction a. Alder Rule

– If two isomeric adducts are possible, the one that has an unsaturated substituent(s) on the alkene oriented toward the newly formed cyclohexene double bond is the preferred product.

– In other words, endo stereoisomer is preferred over exo stereoisomer.

R

R

X

R

R

X+

X

R

R

XH

Hendo addition

exo addition

R

R

H

H

X+

R

R

R

R

X

R

R

X

R

R

X

– Example:

O

O

OH

H

exoproduct

endoproduct

+ O

O

O

O

H

H

O

O

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b. Regioselectivity of the Diels-Alder reaction – It gives the preferred (major) product of a Diels –Alder reaction

when both diene and dienophile are not symmetric (i.e., are substituted).

– The reactivity is enlarged when the substituent on one reactant is electron-withdrawing group (EWG) while the substituent on the other reactant is electron-releasing group (ERG).

– The four possibilities and the major products are given below:

EWG

ERG ERG

EWG

EWG

ERG

EWG

ERG

ERG

EWG

ERG

EWG

EWG

ERG

EWG

ERG

type A:

type B:

type C:

type D:

– The preferred orientation is governed by matching largest

coefficients on interacting frontier orbitals (HOMO of one molecule and LUMO of the other molecule).

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D. Problems and Exercises 1. Show, by constructing a correlation diagram, whether each of the following

disrotatory cyclizations is symmetry allowed: a. pentadienyl cation to cyclopentenyl cation b. pentadienyl anion to cyclopentenyl anion

2. Which of the following reactions are allowed according to the orbital

symmetry conservation rules? Explain.

a.

H

H

COOCH3

COOCH3 COOCH3

H

H

COOCH3

∆

b. .

.∆

c.

O

O

O

O

H

H

d.

CH2OSi(CH3)3

CH3

H3C

∆CH3CH2

CH3

OSi(CH3)3

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3. Suggest a mechanism by which each transformation could occur. (More than one step can be involved.)

a.

CH3HO

100°C CH3CH C

CH3

CH O

b.

370°C

4. Predict the regiochemistry and stereochemistry of the following

cycloadditions reactions and indicate the basis of your prediction.

a. CHH

CH3 H

CH2

O

O

O

+

b.

+CH2CHH

CH3O2CNH HH2C CHCOOCH3

c.

H

CH3

CN

H

H

H+ N

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5. Give the structure, including stereochemistry, of the products expected for the following reaction.

a. CN +

b. + Ph2C C O

c.

∆

d.

OSiR3

CHO

heat

e. + C C COOCH3CH3OOC

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Chemical Bonding and Structure · Chemical Bonding and Structure A. Chemical Bonding 1. Review of...

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