ChemWiki - User contributions [en]

Rep:Mod:gkuytuyg

2016-10-21T09:46:32Z

Bw1114:

'''Introduction'''

When Gaussian calculates a minimum, it varies each nuclear position as a function of the total energy of the molecule and finds the geometry such that the energy is lowest. The transition state is the maximum energy geometry on the reaction pathway between the reactants and products.

At a minimum the gradient is zero since it is a stationary point and consequently the curvature is upwards on either side. The first derivative of the energy with respect to each nuclear coordinate is equal to zero at a minimum and second derivative is positive.

Frequency calculations are useful since at a minimum there will be no imaginary frequencies. This is because the frequencies are calculated using a harmonic oscillator model where the force constant is equal to the second derivative of the energy with respect to nuclear positions and the frequency is proportional to the square root of the force constant. If it is a minimum the second derivative will be positive giving all positive frequencies. However for a transition state there will be an imaginary frequency since the curvature is downwards so the second derivative will be negative, hence giving an imaginary frequency.

'''Exercise 1'''

[[File:Drawn MO Diagram Corrected.PNG]]

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 15; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>BUTADIENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The HOMO of Butadiene calculated with B3LYP</title>
</jmolApplet>
</jmol>

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 16; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>BUTADIENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The LUMO of Butadiene calculated with B3LYP</title>
</jmolApplet>
</jmol>

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 8; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>ETHENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The HOMO of Ethene calculated with B3LYP</title>
</jmolApplet>
</jmol>

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 9; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>ETHENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The LUMO of Ethene calculated with B3LYP</title>
</jmolApplet>
</jmol>

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 17; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>TS PM6.LOG</uploadedFileContents>
<title>The HOMO of the transition state calculated with PM6</title>
</jmolApplet>
</jmol>

It can be seen that the drawn MO diagram HOMOs and LUMOs correspond roughly to those that Gaussian produced. The dotted red lines show the possible MO interactions in the Diels Alder reaction. For any reaction only MOs of the same symmetry may overlap. This is shown by the parity labels g and u, otherwise the orbitals interfere destructively and become antibonding making new bond formation impossible and forbidden. When the symmetry is the same these reactions are allowed. Woodward-Hoffman rules are a way of working out whether a reaction is thermally allowed or not based on this principle. for u-u and g-g interactions the overlap integral will be non-zero. For u-g interactions it will be zero.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" | Bonds in Reactants /Å
! style="background: #0D4F8B; color: white;" | Bonds in Transition State /Å
! style="background: #0D4F8B; color: white;" | Bonds in Product /Å
|-
| Diene C=C 1.33918
| Diene C~C 1.37978
| Diene C-C 1.50085
|-
| Diene C=C 1.33918
| Diene C~C 1.37975
| Diene C-C 1.50083
|-
| Diene C-C 1.47013
| Diene C~C 1.41111
| Diene C=C 1.33700
|-
| Ethene C=C 1.33087
| Ethene C~C 1.38176
| Ethene C-C 1.53464
|-
| -
| -
| New C-C 1.53700
|-
| -
| -
| New C-C 1.53722
|}

The average C-C sp3 bond is 1.54Å and for C=C sp2 it is 1.47Å. For the diene the single bond is shorted than in a normal single C-C bond due to the conjugation of the double bond, and the carbons involved being sp2 so the hybrid orbital forming the single bond has more s character than would a regular sp3 C-C bond. These bond length calculations demonstrate the breaking and making of bonds over the course of the reaction through bond lengthening going from a double to single bond and bond contraction with going from a single to a double bond.

The Van der Waals radius of carbon is is 1.70Å. In the transition state the distance between the centres of the carbon atoms which form bonds is 2.11435Å. Given the radius of carbon atoms, at this distance apart their radii would overlap by 0.64Å on each

the animation of the vibration corresponding to the reaction path at the transition state shows the bond formation as being synchranous, as is expected of a pericyclic reaction:

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 18; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>TS PM6.LOG</uploadedFileContents>
<title>The LUMO of the transition state calculated with PM6</title>
</jmolApplet>
</jmol>

The lowest possible frequency is an asymmetric stretch which is asynchronous

'''Exercise 2'''

[[File:Endo HOMO.png]]

[[File:Endo LUMO.png]]

[[File:Exo HOMO.png]]

[[File:Exo LUMO.png]]

A normal demand Diels-Alder reaction involves the diene acting as the nucleophile since its HOMO is closer to the LUMO of the dieneophile. This is facilitated by electron withdrawing groups on the dieneophile which the lower the MO energies. In inverse demand Diels-Alder reactions the dieneophile acts as the nucleophile, since its HOMO has a better overlap with the LUMO of the diene.

In this reaction of of benzoquinone with cyclopentadiene it is normal demand since the HOMO of the diene is -0.21154, which is higher in energy than the HOMO of the dieneophile which is -0.27061. The energy gap between the HOMO of the diene and the LUMO of the dieneophile is 0.0815 which is smaller than the gap between the HOMO of the dieneophile and the LUMO of the diene, hence it is a normal demand Diels-Alder.

The sum of electronic and thermal free energies are as follows for the endo and exo reactants, products and transition states. From which the reaction barrier (TS-reactants) and reaction energy (products-reactants) were calculated:

{| class="wikitable"
! style="background: #0D4F8B; color: white;" |
! style="background: #0D4F8B; color: white;" | Endo
! style="background: #0D4F8B; color: white;" | Exo
|-
| Reactants Energy /Ha
| -575.418836
| -575.418836
|-
| Transition State Energy /Ha
| -575.383855
| -575.381305
|-
| Product Energy /Ha
| -575.427051
| -575.428695
|-
| Reaction Barrier /Ha
| 0.034981
| 0.037531
|-
| Reaction Energy /Ha
| -0.00821499999
| -0.00985899999
|}

From this data it is clear that the endo reaction is kinetically favourable since the transition state is lower in energy and the reaction barrier is smaller. However the energy of the exo product is the lowest in energy and the overall reaction is shown to be more exothermic for the exo product. This shows that it is the thermodynamic product due to that product having less steric clash.

In the endo transition state secondary orbital interactions can be seen. These have the effect of stabilising the endo transition state lowering the activation energy. This compensates for the higher steric clash in the endo transition state.

'''Exercise 3'''

For the cheletropic reaction the the product was used as a starting point then the reactants derived through finding the transition state and running an IRC backwards to get the reactants. The images below show the procucts, transition state and reactants respectively:

[[File:CH Product.PNG]]

[[File:CH TS.PNG]]

[[File:CH Reactants.PNG]]

As it goes from reactants to the transition state the terminal C=C bonds lengthen and the ones in side the ring increase in double bond character. This can be seen in Gaussian by one of the double bonds changing to a dotted bond in the transition state, indication the bonding character is somewhere between a single and double bond.

The reaction path of the IRC going from transition state to reactants is below:

[[File:CH IRC Graph.PNG]]

The Diels-Alder reaction kept failing to properly optimise when tried from the products or reactants as the starting point and I ran out of time, so no comparison was possible.

Rep:Mod:gkuytuyg

2016-10-21T09:38:35Z

Bw1114:

Rep:Mod:gkuytuyg

2016-10-21T09:33:15Z

Bw1114:

'''Introduction'''

When Gaussian calculates a minimum, it varies each nuclear position as a function of the total energy of the molecule and finds the geometry such that the energy is lowest. The transition state is the maximum energy geometry on the reaction pathway between the reactants and products.

At a minimum the gradient is zero since it is a stationary point and consequently the curvature is upwards on either side. The first derivative of the energy with respect to each nuclear coordinate is equal to zero at a minimum and second derivative is positive.

Frequency calculations are useful since at a minimum there will be no imaginary frequencies. This is because the frequencies are calculated using a harmonic oscillator model where the force constant is equal to the second derivative of the energy with respect to nuclear positions and the frequency is proportional to the square root of the force constant. If it is a minimum the second derivative will be positive giving all positive frequencies. However for a transition state there will be an imaginary frequency since the curvature is downwards so the second derivative will be negative, hence giving an imaginary frequency.

'''Exercise 1'''

[[File:Drawn MO Diagram Corrected.PNG]]

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 15; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>BUTADIENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The HOMO of Butadiene calculated with B3LYP</title>
</jmolApplet>
</jmol>

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 16; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>BUTADIENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The LUMO of Butadiene calculated with B3LYP</title>
</jmolApplet>
</jmol>

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 8; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>ETHENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The HOMO of Ethene calculated with B3LYP</title>
</jmolApplet>
</jmol>

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 9; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>ETHENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The LUMO of Ethene calculated with B3LYP</title>
</jmolApplet>
</jmol>

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 17; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>TS PM6.LOG</uploadedFileContents>
<title>The HOMO of the transition state calculated with PM6</title>
</jmolApplet>
</jmol>

It can be seen that the drawn MO diagram HOMOs and LUMOs correspond roughly to those that Gaussian produced. The dotted red lines show the possible MO interactions in the Diels Alder reaction. For any reaction only MOs of the same symmetry may overlap. This is shown by the parity labels g and u, otherwise the orbitals interfere destructively and become antibonding making new bond formation impossible and forbidden. When the symmetry is the same these reactions are allowed. Woodward-Hoffman rules are a way of working out whether a reaction is thermally allowed or not based on this principle. for u-u and g-g interactions the overlap integral will be non-zero. For u-g interactions it will be zero.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" | Bonds in Reactants /Å
! style="background: #0D4F8B; color: white;" | Bonds in Transition State /Å
! style="background: #0D4F8B; color: white;" | Bonds in Product /Å
|-
| Diene C=C 1.33918
| Diene C~C 1.37978
| Diene C-C 1.50085
|-
| Diene C=C 1.33918
| Diene C~C 1.37975
| Diene C-C 1.50083
|-
| Diene C-C 1.47013
| Diene C~C 1.41111
| Diene C=C 1.33700
|-
| Ethene C=C 1.33087
| Ethene C~C 1.38176
| Ethene C-C 1.53464
|-
| -
| -
| New C-C 1.53700
|-
| -
| -
| New C-C 1.53722
|}

The average C-C sp3 bond is 1.54Å and for C=C sp2 it is 1.47Å. For the diene the single bond is shorted than in a normal single C-C bond due to the conjugation of the double bond, and the carbons involved being sp2 so the hybrid orbital forming the single bond has more s character than would a regular sp3 C-C bond. These bond length calculations demonstrate the breaking and making of bonds over the course of the reaction through bond lengthening going from a double to single bond and bond contraction with going from a single to a double bond.

The Van der Waals radius of carbon is is 1.70Å. In the transition state the distance between the centres of the carbon atoms which form bonds is 2.11435Å. Given the radius of carbon atoms, at this distance apart their radii would overlap by 0.64Å on each

the animation of the vibration corresponding to the reaction path at the transition state shows the bond formation as being synchranous, as is expected of a pericyclic reaction:

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 18; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>TS PM6.LOG</uploadedFileContents>
<title>The LUMO of the transition state calculated with PM6</title>
</jmolApplet>
</jmol>

The lowest possible frequency is an asymmetric stretch which is asynchronous

'''Exercise 2'''

[[File:Endo HOMO.png]]

[[File:Endo LUMO.png]]

[[File:Exo HOMO.png]]

[[File:Exo LUMO.png]]

A normal demand Diels-Alder reaction involves the diene acting as the nucleophile since its HOMO is closer to the LUMO of the dieneophile. This is facilitated by electron withdrawing groups on the dieneophile which the lower the MO energies. In inverse demand Diels-Alder reactions the dieneophile acts as the nucleophile, since its HOMO has a better overlap with the LUMO of the diene.

In this reaction of of benzoquinone with cyclopentadiene it is normal demand since the HOMO of the diene is -0.21154, which is higher in energy than the HOMO of the dieneophile which is -0.27061. The energy gap between the HOMO of the diene and the LUMO of the dieneophile is 0.0815 which is smaller than the gap between the HOMO of the dieneophile and the LUMO of the diene, hence it is a normal demand Diels-Alder.

The sum of electronic and thermal free energies are as follows for the endo and exo reactants, products and transition states. From which the reaction barrier (TS-reactants) and reaction energy (products-reactants) were calculated:

{| class="wikitable"
! style="background: #0D4F8B; color: white;" |
! style="background: #0D4F8B; color: white;" | Endo
! style="background: #0D4F8B; color: white;" | Exo
|-
| Reactants Energy /Ha
| -575.418836
| -575.418836
|-
| Transition State Energy /Ha
| -575.383855
| -575.381305
|-
| Product Energy /Ha
| -575.427051
| -575.428695
|-
| Reaction Barrier /Ha
| 0.034981
| 0.037531
|-
| Reaction Energy /Ha
| -0.00821499999
| -0.00985899999
|}

In the endo transition state secondary orbital interactions can be seen. These have the effect of stabilising the endo transition state lowering the activation energy and hence making the endo product the kinetic one. This compensates for the higher steric clash in the endo transition state.

'''Exercise 3'''

For the cheletropic reaction the the product was used as a starting point then the reactants derived through finding the transition state and running an IRC backwards to get the reactants. The images below show the procucts, transition state and reactants respectively:

[[File:CH Product.PNG]]

[[File:CH TS.PNG]]

[[File:CH Reactants.PNG]]

As it goes from reactants to the transition state the terminal C=C bonds lengthen and the ones in side the ring increase in double bond character. This can be seen in Gaussian by one of the double bonds changing to a dotted bond in the transition state, indication the bonding character is somewhere between a single and double bond.

The reaction path of the IRC going from transition state to reactants is below:

[[File:CH IRC Graph.PNG]]

Rep:Mod:gkuytuyg

2016-10-21T08:42:08Z

Bw1114:

'''Introduction'''

When Gaussian calculates a minimum, it varies each nuclear position as a function of the total energy of the molecule and finds the geometry such that the energy is lowest. The transition state is the maximum energy geometry on the reaction pathway between the reactants and products.

At a minimum the gradient is zero since it is a stationary point and consequently the curvature is upwards on either side. The first derivative of the energy with respect to each nuclear coordinate is equal to zero at a minimum and second derivative is positive.

Frequency calculations are useful since at a minimum there will be no imaginary frequencies. This is because the frequencies are calculated using a harmonic oscillator model where the force constant is equal to the second derivative of the energy with respect to nuclear positions and the frequency is proportional to the square root of the force constant. If it is a minimum the second derivative will be positive giving all positive frequencies. However for a transition state there will be an imaginary frequency since the curvature is downwards so second derivative will be negative, hence giving an imaginary frequency.

'''Exercise 1'''

[[File:Drawn MO Diagram Corrected.PNG]]

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 15; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>BUTADIENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The HOMO of Butadiene calculated with B3LYP</title>
</jmolApplet>
</jmol>

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 16; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>BUTADIENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The LUMO of Butadiene calculated with B3LYP</title>
</jmolApplet>
</jmol>

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 8; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>ETHENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The HOMO of Ethene calculated with B3LYP</title>
</jmolApplet>
</jmol>

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 9; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>ETHENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The LUMO of Ethene calculated with B3LYP</title>
</jmolApplet>
</jmol>

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 17; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>TS PM6.LOG</uploadedFileContents>
<title>The HOMO of the transition state calculated with PM6</title>
</jmolApplet>
</jmol>

It can be seen that the drawn MO diagram HOMOs and LUMOs correspond to those that Gaussian produced. The dotted red lines show the possible MO interactions in the Diels Alder reaction. For any reaction only MOs of the same symmetry may overlap. This is shown by the parity labels g and u, otherwise the orbitals interfere destructively and become antibonding making new bond formation impossible and forbidden. When the symmetry is the same these reactions are allowed. Woodward-Hoffman rules are a way of working out whether a reaction is thermally allowed or not based on this principle. for u-u and g-g interactions the overlap integral will be non-zero. For u-g interactions it will be zero.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" | Bonds in Reactants /Å
! style="background: #0D4F8B; color: white;" | Bonds in Transition State /Å
! style="background: #0D4F8B; color: white;" | Bonds in Product /Å
|-
| Diene C=C 1.33918
| Diene C~C 1.37978
| Diene C-C 1.50085
|-
| Diene C=C 1.33918
| Diene C~C 1.37975
| Diene C-C 1.50083
|-
| Diene C-C 1.47013
| Diene C~C 1.41111
| Diene C=C 1.33700
|-
| Ethene C=C 1.33087
| Ethene C~C 1.38176
| Ethene C-C 1.53464
|-
| -
| -
| New C-C 1.53700
|-
| -
| -
| New C-C 1.53722
|}

The average C-C sp3 bond is 1.54Å and for C=C sp2 it is 1.47Å. For the diene the single bond is shorted than in a normal single C-C bond due to the conjugation of the double bond, and the carbons involved being sp2 so the hybrid orbital forming the single bond has more s character than would a regular sp3 C-C bond. These bond length calculations demonstrate the breaking and making of bonds over the course of the reaction through bond lengthening going from a double to single bond and bond contraction with going from a single to a double bond.

The Van der Waals radius of carbon is is 1.70Å. In the transition state the distance between the centres of the carbon atoms which form bonds is 2.11435Å. Given the radius of carbon atoms, at this distance apart their radii would overlap by 0.64Å on each

the animation of the vibration corresponding to the reaction path at the transition state shows the bond formation as being synchranous, as is expected of a pericyclic reaction:

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 18; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>TS PM6.LOG</uploadedFileContents>
<title>The LUMO of the transition state calculated with PM6</title>
</jmolApplet>
</jmol>

The lowest possible frequency is an asymmetric stretch which is asynchronous

'''Exercise 2'''

[[File:Endo HOMO.png]]

[[File:Endo LUMO.png]]

[[File:Exo HOMO.png]]

[[File:Exo LUMO.png]]

A normal demand Diels-Alder reaction involves the diene acting as the nucleophile since its HOMO is closer to the LUMO of the dieneophile. This is facilitated by electron withdrawing groups on the dieneophile which the lower the MO energies. In inverse demand Diels-alder reactions the dieneophile acts as the nucleophile, since its HOMO has a better overlap with the LUMO of the diene.

In this reaction of of benzoquinone with cyclopentadiene it is normal demans since the HOMO of the diene is -0.21154, which is higher in energy than the HOMO of the dieneophile which is -0.27061. The energy gap between the HOMO of the diene and the LUMO of the dieneophile is 0.0815 which is smaller than the gap between the HOMO of the dieneophile and the LUMO of the diene, hence it is a normal demand Diels Alder.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" |
! style="background: #0D4F8B; color: white;" | Endo
! style="background: #0D4F8B; color: white;" | Exo
|-
| Reactants Energy /Ha
| -575.55483099
| -575.418836
|-
| Transition State Energy /Ha
| -575.383855
| -575.381305
|-
| Product Energy /Ha
| -575.427051
| -575.428695
|-
| Reaction Barrier /Ha
|
|
|-
| Reaction Energy /Ha
|
|
|}

In the endo transition state secondary orbital interactions can be seen. These have the effect of stabilising the endo transition state lowering the activation energy and hence making the endo product the kinetic one. This compensates for the higher steric clash in the endo transition state.

'''Exercise 3'''

For the cheletropic reaction the the product was used as a starting point then the reactants derived through finding the transition state and running an IRC backwards to get the reactants. The images below show the procucts, transition state and reactants respectively:

[[File:CH Product.PNG]]

[[File:CH TS.PNG]]

[[File:CH Reactants.PNG]]

The reaction path of the IRC going from transition state to reactants is below:

[[File:CH IRC Graph.PNG]]

As it goes from reactants to the transition state the terminal C=C bonds lengthen and the ones in side the ring increase in double bond character. This can be seen in Gaussian by one of the double bonds changing to a dotted bond in the transition state, indication the bonding character is somewhere between a single and double bond.

Rep:Mod:gkuytuyg

2016-10-21T08:34:00Z

Bw1114:

'''Introduction'''

When Gaussian calculates a minimum, it varies each nuclear position as a function of the total energy of the molecule and finds the geometry such that the energy is lowest. The transition state is the maximum energy geometry on the reaction pathway between the reactants and products.

At a minimum the gradient is zero since it is a stationary point and consequently the curvature is upwards on either side. The first derivative of the energy with respect to each nuclear coordinate is equal to zero at a minimum and second derivative is positive.

Frequency calculations are useful since at a minimum there will be no imaginary frequencies. This is because the frequencies are calculated using a harmonic oscillator model where the force constant is equal to the second derivative of the energy with respect to nuclear positions and the frequency is proportional to the square root of the force constant. If it is a minimum the second derivative will be positive giving all positive frequencies. However for a transition state there will be an imaginary frequency since the curvature is downwards so second derivative will be negative, hence giving an imaginary frequency.

'''Exercise 1'''

[[File:Drawn MO Diagram Corrected.PNG]]

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 15; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>BUTADIENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The HOMO of Butadiene calculated with B3LYP</title>
</jmolApplet>
</jmol>

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 16; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>BUTADIENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The LUMO of Butadiene calculated with B3LYP</title>
</jmolApplet>
</jmol>

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 8; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>ETHENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The HOMO of Ethene calculated with B3LYP</title>
</jmolApplet>
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It can be seen that the drawn MO diagram HOMOs and LUMOs correspond to those that Gaussian produced. The dotted red lines show the possible MO interactions in the Diels Alder reaction. For any reaction only MOs of the same symmetry may overlap. This is shown by the parity labels g and u, otherwise the orbitals interfere destructively and become antibonding making new bond formation impossible and forbidden. When the symmetry is the same these reactions are allowed. Woodward-Hoffman rules are a way of working out whether a reaction is thermally allowed or not based on this principle. for u-u and g-g interactions the overlap integral will be non-zero. For u-g interactions it will be zero.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" | Bonds in Reactants /Å
! style="background: #0D4F8B; color: white;" | Bonds in Transition State /Å
! style="background: #0D4F8B; color: white;" | Bonds in Product /Å
|-
| Diene C=C 1.33918
| Diene C~C 1.37978
| Diene C-C 1.50085
|-
| Diene C=C 1.33918
| Diene C~C 1.37975
| Diene C-C 1.50083
|-
| Diene C-C 1.47013
| Diene C~C 1.41111
| Diene C=C 1.33700
|-
| Ethene C=C 1.33087
| Ethene C~C 1.38176
| Ethene C-C 1.53464
|-
| -
| -
| New C-C 1.53700
|-
| -
| -
| New C-C 1.53722
|}

The average C-C sp3 bond is 1.54Å and for C=C sp2 it is 1.47Å. For the diene the single bond is shorted than in a normal single C-C bond due to the conjugation of the double bond, and the carbons involved being sp2 so the hybrid orbital forming the single bond has more s character than would a regular sp3 C-C bond. These bond length calculations demonstrate the breaking and making of bonds over the course of the reaction through bond lengthening going from a double to single bond and bond contraction with going from a single to a double bond.

The Van der Waals radius of carbon is is 1.70Å. In the transition state the distance between the centres of the carbon atoms which form bonds is 2.11435Å. Given the radius of carbon atoms, at this distance apart their radii would overlap by 0.64Å on each

the animation of the vibration corresponding to the reaction path at the transition state shows the bond formation as being synchranous, as is expected of a pericyclic reaction:

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 18; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>TS PM6.LOG</uploadedFileContents>
<title>The LUMO of the transition state calculated with PM6</title>
</jmolApplet>
</jmol>

The lowest possible frequency is an asymmetric stretch which is asynchronous

'''Exercise 2'''

[[File:Endo HOMO.png]]

[[File:Endo LUMO.png]]

[[File:Exo HOMO.png]]

[[File:Exo LUMO.png]]

A normal demand Diels-Alder reaction involves the diene acting as the nucleophile since its HOMO is closer to the LUMO of the dieneophile. This is facilitated by electron withdrawing groups on the dieneophile which the lower the MO energies. In inverse demand Diels-alder reactions the dieneophile acts as the nucleophile, since its HOMO has a better overlap with the LUMO of the diene.

In this reaction of of benzoquinone with cyclopentadiene it is normal demans since the HOMO of the diene is -0.21154, which is higher in energy than the HOMO of the dieneophile which is -0.27061. The energy gap between the HOMO of the diene and the LUMO of the dieneophile is 0.0815 which is smaller than the gap between the HOMO of the dieneophile and the LUMO of the diene, hence it is a normal demand Diels Alder.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" |
! style="background: #0D4F8B; color: white;" | Endo
! style="background: #0D4F8B; color: white;" | Exo
|-
| Reactants Energy /Ha
| -575.55483099
| -575.418836
|-
| Transition State Energy /Ha
| -575.383855
| -575.381305
|-
| Product Energy /Ha
| -575.427051
| -575.428695
|-
| Reaction Barrier /Ha
|
|
|-
| Reaction Energy /Ha
|
|
|}

In the endo transition state secondary orbital interactions can be seen. These have the effect of stabilising the endo transition state lowering the activation energy and hence making the endo product the kinetic one. This compensates for the higher steric clash in the endo transition state.

'''Exercise 3'''

For the cheletropic reaction the the product was used as a starting point then the reactants derived through finding the transition state and running an IRC backwards to get the reactants. The images below show the procucts, transition state and reactants respectively:

[[File:CH Product.PNG]]

[[File:CH TS.PNG]]

[[File:CH Reactants.PNG]]

The reaction path of the IRC going from transition state to reactants is below:

[[File:CH IRC Graph.PNG]]

File:CH IRC Graph.PNG

2016-10-21T08:33:05Z

Bw1114:

Rep:Mod:gkuytuyg

2016-10-21T08:27:24Z

Bw1114:

'''Introduction'''

When Gaussian calculates a minimum, it varies each nuclear position as a function of the total energy of the molecule and finds the geometry such that the energy is lowest. The transition state is the maximum energy geometry on the reaction pathway between the reactants and products.

At a minimum the gradient is zero since it is a stationary point and consequently the curvature is upwards on either side. The first derivative of the energy with respect to each nuclear coordinate is equal to zero at a minimum and second derivative is positive.

Frequency calculations are useful since at a minimum there will be no imaginary frequencies. This is because the frequencies are calculated using a harmonic oscillator model where the force constant is equal to the second derivative of the energy with respect to nuclear positions and the frequency is proportional to the square root of the force constant. If it is a minimum the second derivative will be positive giving all positive frequencies. However for a transition state there will be an imaginary frequency since the curvature is downwards so second derivative will be negative, hence giving an imaginary frequency.

'''Exercise 1'''

[[File:Drawn MO Diagram Corrected.PNG]]

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It can be seen that the drawn MO diagram HOMOs and LUMOs correspond to those that Gaussian produced. The dotted red lines show the possible MO interactions in the Diels Alder reaction. For any reaction only MOs of the same symmetry may overlap. This is shown by the parity labels g and u, otherwise the orbitals interfere destructively and become antibonding making new bond formation impossible and forbidden. When the symmetry is the same these reactions are allowed. Woodward-Hoffman rules are a way of working out whether a reaction is thermally allowed or not based on this principle. for u-u and g-g interactions the overlap integral will be non-zero. For u-g interactions it will be zero.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" | Bonds in Reactants /Å
! style="background: #0D4F8B; color: white;" | Bonds in Transition State /Å
! style="background: #0D4F8B; color: white;" | Bonds in Product /Å
|-
| Diene C=C 1.33918
| Diene C~C 1.37978
| Diene C-C 1.50085
|-
| Diene C=C 1.33918
| Diene C~C 1.37975
| Diene C-C 1.50083
|-
| Diene C-C 1.47013
| Diene C~C 1.41111
| Diene C=C 1.33700
|-
| Ethene C=C 1.33087
| Ethene C~C 1.38176
| Ethene C-C 1.53464
|-
| -
| -
| New C-C 1.53700
|-
| -
| -
| New C-C 1.53722
|}

The average C-C sp3 bond is 1.54Å and for C=C sp2 it is 1.47Å. For the diene the single bond is shorted than in a normal single C-C bond due to the conjugation of the double bond, and the carbons involved being sp2 so the hybrid orbital forming the single bond has more s character than would a regular sp3 C-C bond. These bond length calculations demonstrate the breaking and making of bonds over the course of the reaction through bond lengthening going from a double to single bond and bond contraction with going from a single to a double bond.

The Van der Waals radius of carbon is is 1.70Å. In the transition state the distance between the centres of the carbon atoms which form bonds is 2.11435Å. Given the radius of carbon atoms, at this distance apart their radii would overlap by 0.64Å on each

the animation of the vibration corresponding to the reaction path at the transition state shows the bond formation as being synchranous, as is expected of a pericyclic reaction:

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 18; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>TS PM6.LOG</uploadedFileContents>
<title>The LUMO of the transition state calculated with PM6</title>
</jmolApplet>
</jmol>

The lowest possible frequency is an asymmetric stretch which is asynchronous

'''Exercise 2'''

[[File:Endo HOMO.png]]

[[File:Endo LUMO.png]]

[[File:Exo HOMO.png]]

[[File:Exo LUMO.png]]

A normal demand Diels-Alder reaction involves the diene acting as the nucleophile since its HOMO is closer to the LUMO of the dieneophile. This is facilitated by electron withdrawing groups on the dieneophile which the lower the MO energies. In inverse demand Diels-alder reactions the dieneophile acts as the nucleophile, since its HOMO has a better overlap with the LUMO of the diene.

In this reaction of of benzoquinone with cyclopentadiene it is normal demans since the HOMO of the diene is -0.21154, which is higher in energy than the HOMO of the dieneophile which is -0.27061. The energy gap between the HOMO of the diene and the LUMO of the dieneophile is 0.0815 which is smaller than the gap between the HOMO of the dieneophile and the LUMO of the diene, hence it is a normal demand Diels Alder.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" |
! style="background: #0D4F8B; color: white;" | Endo
! style="background: #0D4F8B; color: white;" | Exo
|-
| Reactants Energy /Ha
| -575.55483099
| -575.418836
|-
| Transition State Energy /Ha
| -575.383855
| -575.381305
|-
| Product Energy /Ha
| -575.427051
| -575.428695
|-
| Reaction Barrier /Ha
|
|
|-
| Reaction Energy /Ha
|
|
|}

In the endo transition state secondary orbital interactions can be seen. These have the effect of stabilising the endo transition state lowering the activation energy and hence making the endo product the kinetic one. This compensates for the higher steric clash in the endo transition state.

'''Exercise 3'''

For the cheletropic reaction the the product was used as a starting point then the reactants derived through finding the transition state and running an IRC backwards to get the reactants. The images below show the procucts, transition state and reactants respectively:

[[File:CH Product.PNG]]

[[File:CH TS.PNG]]

[[File:CH Reactants.PNG]]

Rep:Mod:gkuytuyg

2016-10-21T08:14:23Z

Bw1114:

'''Introduction'''

When Gaussian calculates a minimum, it varies each nuclear position as a function of the total energy of the molecule and finds the geometry such that the energy is lowest. The transition state is the maximum energy geometry on the reaction pathway between the reactants and products.

At a minimum the gradient is zero since it is a stationary point and consequently the curvature is upwards on either side. The first derivative of the energy with respect to each nuclear coordinate is equal to zero at a minimum and second derivative is positive.

Frequency calculations are useful since at a minimum there will be no imaginary frequencies. This is because the frequencies are calculated using a harmonic oscillator model where the force constant is equal to the second derivative of the energy with respect to nuclear positions and the frequency is proportional to the square root of the force constant. If it is a minimum the second derivative will be positive giving all positive frequencies. However for a transition state there will be an imaginary frequency since the curvature is downwards so second derivative will be negative, hence giving an imaginary frequency.

'''Exercise 1'''

[[File:Drawn MO Diagram Corrected.PNG]]

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It can be seen that the drawn MO diagram HOMOs and LUMOs correspond to those that Gaussian produced. The dotted red lines show the possible MO interactions in the Diels Alder reaction. For any reaction only MOs of the same symmetry may overlap. This is shown by the parity labels g and u, otherwise the orbitals interfere destructively and become antibonding making new bond formation impossible and forbidden. When the symmetry is the same these reactions are allowed. Woodward-Hoffman rules are a way of working out whether a reaction is thermally allowed or not based on this principle. for u-u and g-g interactions the overlap integral will be non-zero. For u-g interactions it will be zero.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" | Bonds in Reactants /Å
! style="background: #0D4F8B; color: white;" | Bonds in Transition State /Å
! style="background: #0D4F8B; color: white;" | Bonds in Product /Å
|-
| Diene C=C 1.33918
| Diene C~C 1.37978
| Diene C-C 1.50085
|-
| Diene C=C 1.33918
| Diene C~C 1.37975
| Diene C-C 1.50083
|-
| Diene C-C 1.47013
| Diene C~C 1.41111
| Diene C=C 1.33700
|-
| Ethene C=C 1.33087
| Ethene C~C 1.38176
| Ethene C-C 1.53464
|-
| -
| -
| New C-C 1.53700
|-
| -
| -
| New C-C 1.53722
|}

The average C-C sp3 bond is 1.54Å and for C=C sp2 it is 1.47Å. For the diene the single bond is shorted than in a normal single C-C bond due to the conjugation of the double bond, and the carbons involved being sp2 so the hybrid orbital forming the single bond has more s character than would a regular sp3 C-C bond. These bond length calculations demonstrate the breaking and making of bonds over the course of the reaction through bond lengthening going from a double to single bond and bond contraction with going from a single to a double bond.

The Van der Waals radius of carbon is is 1.70Å. In the transition state the distance between the centres of the carbon atoms which form bonds is 2.11435Å. Given the radius of carbon atoms, at this distance apart their radii would overlap by 0.64Å on each

the animation of the vibration corresponding to the reaction path at the transition state shows the bond formation as being synchranous, as is expected of a pericyclic reaction:

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 18; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>TS PM6.LOG</uploadedFileContents>
<title>The LUMO of the transition state calculated with PM6</title>
</jmolApplet>
</jmol>

The lowest possible frequency is an asymmetric stretch which is asynchronous

'''Exercise 2'''

[[File:Endo HOMO.png]]

[[File:Endo LUMO.png]]

[[File:Exo HOMO.png]]

[[File:Exo LUMO.png]]

A normal demand Diels-Alder reaction involves the diene acting as the nucleophile since its HOMO is closer to the LUMO of the dieneophile. This is facilitated by electron withdrawing groups on the dieneophile which the lower the MO energies. In inverse demand Diels-alder reactions the dieneophile acts as the nucleophile, since its HOMO has a better overlap with the LUMO of the diene.

In this reaction of of benzoquinone with cyclopentadiene it is normal demans since the HOMO of the diene is -0.21154, which is higher in energy than the HOMO of the dieneophile which is -0.27061. The energy gap between the HOMO of the diene and the LUMO of the dieneophile is 0.0815 which is smaller than the gap between the HOMO of the dieneophile and the LUMO of the diene, hence it is a normal demand Diels Alder.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" |
! style="background: #0D4F8B; color: white;" | Endo
! style="background: #0D4F8B; color: white;" | Exo
|-
| Reactants Energy /Ha
| -575.55483099
| -575.422767
|-
| Transition State Energy /Ha
| -575.383855
| -575.381305
|-
| Product Energy /Ha
| -575.427051
| -575.428695
|-
| Reaction Barrier /Ha
|
|
|-
| Reaction Energy /Ha
|
|
|}

In the endo transition state secondary orbital interactions can be seen. These have the effect of stabilising the endo transition state lowering the activation energy and hence making the endo product the kinetic one. This compensates for the higher steric clash in the endo transition state.

'''Exercise 3'''

For the cheletropic reaction the the product was used as a starting point then the reactants derived through finding the transition state and running an IRC backwards to get the reactants. The images below show the procucts, transition state and reactants respectively:

[[File:CH Product.PNG]]

[[File:CH TS.PNG]]

[[File:CH Reactants.PNG]]

Rep:Mod:gkuytuyg

2016-10-20T16:18:19Z

Bw1114:

'''Introduction'''

When Gaussian calculates a minimum, it varies each nuclear position as a function of the total energy of the molecule and finds the geometry such that the energy is lowest. The transition state is the maximum energy geometry on the reaction pathway between the reactants and products.

At a minimum the gradient is zero since it is a stationary point and consequently the curvature is upwards on either side. The first derivative of the energy with respect to each nuclear coordinate is equal to zero at a minimum and second derivative is positive.

Frequency calculations are useful since at a minimum there will be no imaginary frequencies. This is because the frequencies are calculated using a harmonic oscillator model where the force constant is equal to the second derivative of the energy with respect to nuclear positions and the frequency is proportional to the square root of the force constant. If it is a minimum the second derivative will be positive giving all positive frequencies. However for a transition state there will be an imaginary frequency since the curvature is downwards so second derivative will be negative, hence giving an imaginary frequency.

'''Exercise 1'''

[[File:Drawn MO Diagram Corrected.PNG]]

<jmol>
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<jmol>
<jmolApplet>
<color>white</color>
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<script>frame 2; mo 17; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>TS PM6.LOG</uploadedFileContents>
<title>The HOMO of the transition state calculated with PM6</title>
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It can be seen that the drawn MO diagram HOMOs and LUMOs correspond to those that Gaussian produced. The dotted red lines show the possible MO interactions in the Diels Alder reaction. For any reaction only MOs of the same symmetry may overlap. This is shown by the parity labels g and u, otherwise the orbitals interfere destructively and become antibonding making new bond formation impossible and forbidden. When the symmetry is the same these reactions are allowed. Woodward-Hoffman rules are a way of working out whether a reaction is thermally allowed or not based on this principle. for u-u and g-g interactions the overlap integral will be non-zero. For u-g interactions it will be zero.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" | Bonds in Reactants /Å
! style="background: #0D4F8B; color: white;" | Bonds in Transition State /Å
! style="background: #0D4F8B; color: white;" | Bonds in Product /Å
|-
| Diene C=C 1.33918
| Diene C~C 1.37978
| Diene C-C 1.50085
|-
| Diene C=C 1.33918
| Diene C~C 1.37975
| Diene C-C 1.50083
|-
| Diene C-C 1.47013
| Diene C~C 1.41111
| Diene C=C 1.33700
|-
| Ethene C=C 1.33087
| Ethene C~C 1.38176
| Ethene C-C 1.53464
|-
| -
| -
| New C-C 1.53700
|-
| -
| -
| New C-C 1.53722
|}

The average C-C sp3 bond is 1.54Å and for C=C sp2 it is 1.47Å. For the diene the single bond is shorted than in a normal single C-C bond due to the conjugation of the double bond, and the carbons involved being sp2 so the hybrid orbital forming the single bond has more s character than would a regular sp3 C-C bond. These bond length calculations demonstrate the breaking and making of bonds over the course of the reaction through bond lengthening going from a double to single bond and bond contraction with going from a single to a double bond.

The Van der Waals radius of carbon is is 1.70Å. In the transition state the distance between the centres of the carbon atoms which form bonds is 2.11435Å. Given the radius of carbon atoms, at this distance apart their radii would overlap by 0.64Å on each

the animation of the vibration corresponding to the reaction path at the transition state shows the bond formation as being synchranous, as is expected of a pericyclic reaction:

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 18; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>TS PM6.LOG</uploadedFileContents>
<title>The LUMO of the transition state calculated with PM6</title>
</jmolApplet>
</jmol>

The lowest possible frequency is an asymmetric stretch which is asynchronous

'''Exercise 2'''

[[File:Endo HOMO.png]]

[[File:Endo LUMO.png]]

[[File:Exo HOMO.png]]

[[File:Exo LUMO.png]]

A normal demand Diels-Alder reaction involves the diene acting as the nucleophile since its HOMO is closer to the LUMO of the dieneophile. This is facilitated by electron withdrawing groups on the dieneophile which the lower the MO energies. In inverse demand Diels-alder reactions the dieneophile acts as the nucleophile, since its HOMO has a better overlap with the LUMO of the diene.

In this reaction of of benzoquinone with cyclopentadiene it is normal demans since the HOMO of the diene is -0.21154, which is higher in energy than the HOMO of the dieneophile which is -0.27061. The energy gap between the HOMO of the diene and the LUMO of the dieneophile is 0.0815 which is smaller than the gap between the HOMO of the dieneophile and the LUMO of the diene, hence it is a normal demand Diels Alder.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" |
! style="background: #0D4F8B; color: white;" | Endo
! style="background: #0D4F8B; color: white;" | Exo
|-
| Reactants Energy /Ha
| Column 2, Row 2
| -575.422767
|-
| Transition State Energy /Ha
| -575.383855
| -575.381305
|-
| Product Energy /Ha
| Column 2, Row 4
| Column 3, Row 4
|-
| Reaction Barrier /Ha
| Column 2, Row 5
| Column 3, Row 5
|-
| Reaction Energy /Ha
| Column 2, Row 6
| Column 3, Row 6
|}

In the endo transition state secondary orbital interactions can be seen. These have the effect of stabilising the endo transition state lowering the activation energy and hence making the endo product the kinetic one. This compensates for the higher steric clash in the endo transition state.

'''Exercise 3'''

For the cheletropic reaction the the product was used as a starting point then the reactants derived through finding the transition state and running an IRC backwards to get the reactants. The images below show the procucts, transition state and reactants respectively:

[[File:CH Product.PNG]]

[[File:CH TS.PNG]]

[[File:CH Reactants.PNG]]

File:CH Reactants.PNG

2016-10-20T16:17:28Z

Bw1114:

File:CH TS.PNG

2016-10-20T16:17:13Z

Bw1114:

File:CH Product.PNG

2016-10-20T16:16:53Z

Bw1114:

Rep:Mod:gkuytuyg

2016-10-20T14:42:53Z

Bw1114:

'''Introduction'''

When Gaussian calculates a minimum, it varies each nuclear position as a function of the total energy of the molecule and finds the geometry such that the energy is lowest. The transition state is the maximum energy geometry on the reaction pathway between the reactants and products.

At a minimum the gradient is zero since it is a stationary point and consequently the curvature is upwards on either side. The first derivative of the energy with respect to each nuclear coordinate is equal to zero at a minimum and second derivative is positive.

Frequency calculations are useful since at a minimum there will be no imaginary frequencies. This is because the frequencies are calculated using a harmonic oscillator model where the force constant is equal to the second derivative of the energy with respect to nuclear positions and the frequency is proportional to the square root of the force constant. If it is a minimum the second derivative will be positive giving all positive frequencies. However for a transition state there will be an imaginary frequency since the curvature is downwards so second derivative will be negative, hence giving an imaginary frequency.

'''Exercise 1'''

[[File:Drawn MO Diagram Corrected.PNG]]

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It can be seen that the drawn MO diagram HOMOs and LUMOs correspond to those that Gaussian produced. The dotted red lines show the possible MO interactions in the Diels Alder reaction. For any reaction only MOs of the same symmetry may overlap. This is shown by the parity labels g and u, otherwise the orbitals interfere destructively and become antibonding making new bond formation impossible and forbidden. When the symmetry is the same these reactions are allowed. Woodward-Hoffman rules are a way of working out whether a reaction is thermally allowed or not based on this principle. for u-u and g-g interactions the overlap integral will be non-zero. For u-g interactions it will be zero.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" | Bonds in Reactants /Å
! style="background: #0D4F8B; color: white;" | Bonds in Transition State /Å
! style="background: #0D4F8B; color: white;" | Bonds in Product /Å
|-
| Diene C=C 1.33918
| Diene C~C 1.37978
| Diene C-C 1.50085
|-
| Diene C=C 1.33918
| Diene C~C 1.37975
| Diene C-C 1.50083
|-
| Diene C-C 1.47013
| Diene C~C 1.41111
| Diene C=C 1.33700
|-
| Ethene C=C 1.33087
| Ethene C~C 1.38176
| Ethene C-C 1.53464
|-
| -
| -
| New C-C 1.53700
|-
| -
| -
| New C-C 1.53722
|}

The average C-C sp3 bond is 1.54Å and for C=C sp2 it is 1.47Å. For the diene the single bond is shorted than in a normal single C-C bond due to the conjugation of the double bond, and the carbons involved being sp2 so the hybrid orbital forming the single bond has more s character than would a regular sp3 C-C bond. These bond length calculations demonstrate the breaking and making of bonds over the course of the reaction through bond lengthening going from a double to single bond and bond contraction with going from a single to a double bond.

The Van der Waals radius of carbon is is 1.70Å. In the transition state the distance between the centres of the carbon atoms which form bonds is 2.11435Å. Given the radius of carbon atoms, at this distance apart their radii would overlap by 0.64Å on each

the animation of the vibration corresponding to the reaction path at the transition state shows the bond formation as being synchranous, as is expected of a pericyclic reaction:

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The lowest possible frequency is an asymmetric stretch which is asynchronous

'''Exercise 2'''

[[File:Endo HOMO.png]]

[[File:Endo LUMO.png]]

[[File:Exo HOMO.png]]

[[File:Exo LUMO.png]]

A normal demand Diels-Alder reaction involves the diene acting as the nucleophile since its HOMO is closer to the LUMO of the dieneophile. This is facilitated by electron withdrawing groups on the dieneophile which the lower the MO energies. In inverse demand Diels-alder reactions the dieneophile acts as the nucleophile, since its HOMO has a better overlap with the LUMO of the diene.

In this reaction of of benzoquinone with cyclopentadiene it is normal demans since the HOMO of the diene is -0.21154, which is higher in energy than the HOMO of the dieneophile which is -0.27061. The energy gap between the HOMO of the diene and the LUMO of the dieneophile is 0.0815 which is smaller than the gap between the HOMO of the dieneophile and the LUMO of the diene, hence it is a normal demand Diels Alder.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" |
! style="background: #0D4F8B; color: white;" | Endo
! style="background: #0D4F8B; color: white;" | Exo
|-
| Reactants Energy /Ha
| Column 2, Row 2
| -575.422767
|-
| Transition State Energy /Ha
| -575.383855
| -575.381305
|-
| Product Energy /Ha
| Column 2, Row 4
| Column 3, Row 4
|-
| Reaction Barrier /Ha
| Column 2, Row 5
| Column 3, Row 5
|-
| Reaction Energy /Ha
| Column 2, Row 6
| Column 3, Row 6
|}

In the endo transition state secondary orbital interactions can be seen. These have the effect of stabilising the endo transition state lowering the activation energy and hence making the endo product the kinetic one. This compensates for the higher steric clash in the endo transition state.

'''Exercise 3'''

Rep:Mod:gkuytuyg

2016-10-20T12:57:55Z

Bw1114:

'''Introduction'''

When Gaussian calculates a minimum, it varies each nuclear position as a function of the total energy of the molecule and finds the geometry such that the energy is lowest. The transition state is the maximum energy geometry on the reaction pathway between the reactants and products.

At a minimum the gradient is zero since it is a stationary point and consequently the curvature is upwards on either side. The first derivative of the energy with respect to each nuclear coordinate is equal to zero at a minimum and second derivative is positive.

Frequency calculations are useful since at a minimum there will be no imaginary frequencies. This is because the frequencies are calculated using a harmonic oscillator model where the force constant is equal to the second derivative of the energy with respect to nuclear positions and the frequency is proportional to the square root of the force constant. If it is a minimum the second derivative will be positive giving all positive frequencies. However for a transition state there will be an imaginary frequency since the curvature is downwards so second derivative will be negative, hence giving an imaginary frequency.

'''Exercise 1'''

[[File:Drawn MO Diagram Corrected.PNG]]

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It can be seen that the drawn MO diagram HOMOs and LUMOs correspond to those that Gaussian produced. The dotted red lines show the possible MO interactions in the Diels Alder reaction. For any reaction only MOs of the same symmetry may overlap. This is shown by the parity labels g and u, otherwise the orbitals interfere destructively and become antibonding making new bond formation impossible and forbidden. When the symmetry is the same these reactions are allowed. Woodward-Hoffman rules are a way of working out whether a reaction is thermally allowed or not based on this principle. for u-u and g-g interactions the overlap integral will be non-zero. For u-g interactions it will be zero.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" | Bonds in Reactants /Å
! style="background: #0D4F8B; color: white;" | Bonds in Transition State /Å
! style="background: #0D4F8B; color: white;" | Bonds in Product /Å
|-
| Diene C=C 1.33918
| Diene C~C 1.37978
| Diene C-C 1.50085
|-
| Diene C=C 1.33918
| Diene C~C 1.37975
| Diene C-C 1.50083
|-
| Diene C-C 1.47013
| Diene C~C 1.41111
| Diene C=C 1.33700
|-
| Ethene C=C 1.33087
| Ethene C~C 1.38176
| Ethene C-C 1.53464
|-
| -
| -
| New C-C 1.53700
|-
| -
| -
| New C-C 1.53722
|}

The average C-C sp3 bond is 1.54Å and for C=C sp2 it is 1.47Å. For the diene the single bond is shorted than in a normal single C-C bond due to the conjugation of the double bond, and the carbons involved being sp2 so the hybrid orbital forming the single bond has more s character than would a regular sp3 C-C bond. These bond length calculations demonstrate the breaking and making of bonds over the course of the reaction through bond lengthening going from a double to single bond and bond contraction with going from a single to a double bond.

The Van der Waals radius of carbon is is 1.70Å. In the transition state the distance between the centres of the carbon atoms which form bonds is 2.11435Å. Given the radius of carbon atoms, at this distance apart their radii would overlap by 0.64Å on each

the animation of the vibration corresponding to the reaction path at the transition state shows the bond formation as being synchranous, as is expected of a pericyclic reaction:

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The lowest possible frequency is an asymmetric stretch which is asynchronous

'''Exercise 2'''

[[File:Endo HOMO.png]]

[[File:Endo LUMO.png]]

[[File:Exo HOMO.png]]

[[File:Exo LUMO.png]]

A normal demand Diels-Alder reaction involves the diene acting as the nucleophile since its HOMO is closer to the LUMO of the dieneophile. This is facilitated by electron withdrawing groups on the dieneophile which the lower the MO energies. In inverse demand Diels-alder reactions the dieneophile acts as the nucleophile, since its HOMO has a better overlap with the LUMO of the diene.

In this reaction of of benzoquinone with cyclopentadiene it is normal demans since the HOMO of the diene is -0.21154, which is higher in energy than the HOMO of the dieneophile which is -0.27061. The energy gap between the HOMO of the diene and the LUMO of the dieneophile is 0.0815 which is smaller than the gap between the HOMO of the dieneophile and the LUMO of the diene, hence it is a normal demand Diels Alder.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" |
! style="background: #0D4F8B; color: white;" | Endo
! style="background: #0D4F8B; color: white;" | Exo
|-
| Reactants Energy /Ha
| Column 2, Row 2
| Column 3, Row 2
|-
| Transition State Energy /Ha
| -575.383855
| -575.381305
|-
| Product Energy /Ha
| Column 2, Row 4
| Column 3, Row 4
|-
| Reaction Barrier /Ha
| Column 2, Row 5
| Column 3, Row 5
|-
| Reaction Energy /Ha
| Column 2, Row 6
| Column 3, Row 6
|}

In the endo transition state secondary orbital interactions can be seen. These have the effect of stabilising the endo transition state lowering the activation energy and hence making the endo product the kinetic one. This compensates for the higher steric clash in the endo transition state.

'''Exercise 3'''

Rep:Mod:gkuytuyg

2016-10-20T12:06:13Z

Bw1114:

'''Introduction'''

When Gaussian calculates a minimum, it varies each nuclear position as a function of the total energy of the molecule and finds the geometry such that the energy is lowest. The transition state is the maximum energy geometry on the reaction pathway between the reactants and products.

At a minimum the gradient is zero since it is a stationary point and consequently the curvature is upwards on either side. The first derivative of the energy with respect to each nuclear coordinate is equal to zero at a minimum and second derivative is positive.

Frequency calculations are useful since at a minimum there will be no imaginary frequencies. This is because the frequencies are calculated using a harmonic oscillator model where the force constant is equal to the second derivative of the energy with respect to nuclear positions and the frequency is proportional to the square root of the force constant. If it is a minimum the second derivative will be positive giving all positive frequencies. However for a transition state there will be an imaginary frequency since the curvature is downwards so second derivative will be negative, hence giving an imaginary frequency.

'''Exercise 1'''

[[File:Drawn MO Diagram Corrected.PNG]]

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It can be seen that the drawn MO diagram HOMOs and LUMOs correspond to those that Gaussian produced. The dotted red lines show the possible MO interactions in the Diels Alder reaction. For any reaction only MOs of the same symmetry may overlap. This is shown by the parity labels g and u, otherwise the orbitals interfere destructively and become antibonding making new bond formation impossible and forbidden. When the symmetry is the same these reactions are allowed. Woodward-Hoffman rules are a way of working out whether a reaction is thermally allowed or not based on this principle. for u-u and g-g interactions the overlap integral will be non-zero. For u-g interactions it will be zero.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" | Bonds in Reactants /Å
! style="background: #0D4F8B; color: white;" | Bonds in Transition State /Å
! style="background: #0D4F8B; color: white;" | Bonds in Product /Å
|-
| Diene C=C 1.33918
| Diene C~C 1.37978
| Diene C-C 1.50085
|-
| Diene C=C 1.33918
| Diene C~C 1.37975
| Diene C-C 1.50083
|-
| Diene C-C 1.47013
| Diene C~C 1.41111
| Diene C=C 1.33700
|-
| Ethene C=C 1.33087
| Ethene C~C 1.38176
| Ethene C-C 1.53464
|-
| -
| -
| New C-C 1.53700
|-
| -
| -
| New C-C 1.53722
|}

The average C-C sp3 bond is 1.54Å and for C=C sp2 it is 1.47Å. For the diene the single bond is shorted than in a normal single C-C bond due to the conjugation of the double bond, and the carbons involved being sp2 so the hybrid orbital forming the single bond has more s character than would a regular sp3 C-C bond. These bond length calculations demonstrate the breaking and making of bonds over the course of the reaction through bond lengthening going from a double to single bond and bond contraction with going from a single to a double bond.

The Van der Waals radius of carbon is is 1.70Å. In the transition state the distance between the centres of the carbon atoms which form bonds is 2.11435Å. Given the radius of carbon atoms, at this distance apart their radii would overlap by 0.64Å on each

the animation of the vibration corresponding to the reaction path at the transition state shows the bond formation as being synchranous, as is expected of a pericyclic reaction:

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The lowest possible frequency is an asymmetric stretch which is asynchronous

'''Exercise 2'''

[[File:Endo HOMO.png]]

[[File:Endo LUMO.png]]

[[File:Exo HOMO.png]]

[[File:Exo LUMO.png]]

A normal demand Diels-Alder reaction involves the diene acting as the nucleophile since its HOMO is closer to the LUMO of the dieneophile. This is facilitated by electron withdrawing groups on the dieneophile which the lower the MO energies. In inverse demand Diels-alder reactions the dieneophile acts as the nucleophile, since its HOMO has a better overlap with the LUMO of the diene.

In this reaction of of benzoquinone with cyclopentadiene it is normal demans since the HOMO of the diene is -0.21154, which is higher in energy than the HOMO of the dieneophile which is -0.27061. The energy gap between the HOMO of the diene and the LUMO of the dieneophile is 0.0815 which is smaller than the gap between the HOMO of the dieneophile and the LUMO of the diene, hence it is a normal demand Diels Alder.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" |
! style="background: #0D4F8B; color: white;" | Endo
! style="background: #0D4F8B; color: white;" | Exo
|-
| Reactants Energy /Ha
| Column 2, Row 2
| Column 3, Row 2
|-
| Transition State Energy /Ha
| -575.383855
| -575.381305
|-
| Product Energy /Ha
| Column 2, Row 4
| Column 3, Row 4
|-
| Reaction Barrier /Ha
| Column 2, Row 5
| Column 3, Row 5
|-
| Reaction Energy /Ha
| Column 2, Row 6
| Column 3, Row 6
|}

In the endo transition state secondary orbital interactions can be seen. These have the eddect of stabilising the endo transition state lowering the activation energy and hence making the endo product the kinetic one. This compensates for the higher steric clash in the endo transition state.

'''Exercise 3'''

Rep:Mod:gkuytuyg

2016-10-20T11:47:51Z

Bw1114:

'''Introduction'''

When Gaussian calculates a minimum, it varies each nuclear position as a function of the total energy of the molecule and finds the geometry such that the energy is lowest. The transition state is the maximum energy geometry on the reaction pathway between the reactants and products.

At a minimum the gradient is zero since it is a stationary point and consequently the curvature is upwards on either side. The first derivative of the energy with respect to each nuclear coordinate is equal to zero at a minimum and second derivative is positive.

Frequency calculations are useful since at a minimum there will be no imaginary frequencies. This is because the frequencies are calculated using a harmonic oscillator model where the force constant is equal to the second derivative of the energy with respect to nuclear positions and the frequency is proportional to the square root of the force constant. If it is a minimum the second derivative will be positive giving all positive frequencies. However for a transition state there will be an imaginary frequency since the curvature is downwards so second derivative will be negative, hence giving an imaginary frequency.

'''Exercise 1'''

[[File:Drawn MO Diagram Corrected.PNG]]

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<script>frame 2; mo 17; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>TS PM6.LOG</uploadedFileContents>
<title>The HOMO of the transition state calculated with PM6</title>
</jmolApplet>
</jmol>

It can be seen that the drawn MO diagram HOMOs and LUMOs correspond to those that Gaussian produced. The dotted red lines show the possible MO interactions in the Diels Alder reaction. For any reaction only MOs of the same symmetry may overlap. This is shown by the parity labels g and u, otherwise the orbitals interfere destructively and become antibonding making new bond formation impossible and forbidden. When the symmetry is the same these reactions are allowed. Woodward-Hoffman rules are a way of working out whether a reaction is thermally allowed or not based on this principle. for u-u and g-g interactions the overlap integral will be non-zero. For u-g interactions it will be zero.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" | Bonds in Reactants /Å
! style="background: #0D4F8B; color: white;" | Bonds in Transition State /Å
! style="background: #0D4F8B; color: white;" | Bonds in Product /Å
|-
| Diene C=C 1.33918
| Diene C~C 1.37978
| Diene C-C 1.50085
|-
| Diene C=C 1.33918
| Diene C~C 1.37975
| Diene C-C 1.50083
|-
| Diene C-C 1.47013
| Diene C~C 1.41111
| Diene C=C 1.33700
|-
| Ethene C=C 1.33087
| Ethene C~C 1.38176
| Ethene C-C 1.53464
|-
| -
| -
| New C-C 1.53700
|-
| -
| -
| New C-C 1.53722
|}

The average C-C sp3 bond is 1.54Å and for C=C sp2 it is 1.47Å. For the diene the single bond is shorted than in a normal single C-C bond due to the conjugation of the double bond, and the carbons involved being sp2 so the hybrid orbital forming the single bond has more s character than would a regular sp3 C-C bond. These bond length calculations demonstrate the breaking and making of bonds over the course of the reaction through bond lengthening going from a double to single bond and bond contraction with going from a single to a double bond.

The Van der Waals radius of carbon is is 1.70Å. In the transition state the distance between the centres of the carbon atoms which form bonds is 2.11435Å. Given the radius of carbon atoms, at this distance apart their radii would overlap by 0.64Å on each

the animation of the vibration corresponding to the reaction path at the transition state shows the bond formation as being synchranous, as is expected of a pericyclic reaction:

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 18; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>TS PM6.LOG</uploadedFileContents>
<title>The LUMO of the transition state calculated with PM6</title>
</jmolApplet>
</jmol>

The lowest possible frequency is an asymmetric stretch which is asynchronous

'''Exercise 2'''

[[File:Endo HOMO.png]]

[[File:Endo LUMO.png]]

[[File:Exo HOMO.png]]

[[File:Exo LUMO.png]]

A normal demand Diels-Alder reaction involves the diene acting as the nucleophile since its HOMO is closer to the LUMO of the dieneophile. This is facilitated by electron withdrawing groups on the dieneophile which the lower the MO energies. In inverse demand Diels-alder reactions the dieneophile acts as the nucleophile, since its HOMO has a better overlap with the LUMO of the diene.

In this reaction of of benzoquinone with cyclopentadiene it is normal demans since the HOMO of the diene is -0.21154, which is higher in energy than the HOMO of the dieneophile which is -0.27061. The energy gap between the HOMO of the diene and the LUMO of the dieneophile is 0.0815 which is smaller than the gap between the HOMO of the dieneophile and the LUMO of the diene, hence it is a normal demand Diels Alder.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" |
! style="background: #0D4F8B; color: white;" | Endo
! style="background: #0D4F8B; color: white;" | Exo
|-
| Reactants Energy /Ha
| Column 2, Row 2
| Column 3, Row 2
|-
| Transition State Energy /Ha
| Column 2, Row 3
| Column 3, Row 3
|-
| Product Energy /Ha
| Column 2, Row 4
| Column 3, Row 4
|-
| Reaction Barrier /Ha
| Column 2, Row 5
| Column 3, Row 5
|-
| Reaction Energy /Ha
| Column 2, Row 6
| Column 3, Row 6
|}

In the endo transition state secondary orbital interactions can be seen. These have the eddect of stabilising the endo transition state lowering the activation energy and hence making the endo product the kinetic one. This compensates for the higher steric clash in the endo transition state.

'''Exercise 3'''

Rep:Mod:gkuytuyg

2016-10-18T16:48:58Z

Bw1114:

'''Introduction'''

When Gaussian calculates a minimum, it varies each nuclear position as a function of the total energy of the molecule and finds the geometry such that the energy is lowest. The transition state is the maximum energy geometry on the reaction pathway between the reactants and products.

At a minimum the gradient is zero since it is a stationary point and consequently the curvature is upwards on either side. The first derivative of the energy with respect to each nuclear coordinate is equal to zero at a minimum and second derivative is positive.

Frequency calculations are useful since at a minimum there will be no imaginary frequencies. This is because the frequencies are calculated using a harmonic oscillator model where the force constant is equal to the second derivative of the energy with respect to nuclear positions and the frequency is proportional to the square root of the force constant. If it is a minimum the second derivative will be positive giving all positive frequencies. However for a transition state there will be an imaginary frequency since the curvature is downwards so second derivative will be negative, hence giving an imaginary frequency.

'''Exercise 1'''

[[File:Drawn MO Diagram Corrected.PNG]]

<jmol>
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<jmol>
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<jmol>
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<jmol>
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<uploadedFileContents>TS PM6.LOG</uploadedFileContents>
<title>The HOMO of the transition state calculated with PM6</title>
</jmolApplet>
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It can be seen that the drawn MO diagram HOMOs and LUMOs correspond to those that Gaussian produced. The dotted red lines show the possible MO interactions in the Diels Alder reaction. For any reaction only MOs of the same symmetry may overlap. This is shown by the parity labels g and u, otherwise the orbitals interfere destructively and become antibonding making new bond formation impossible and forbidden. When the symmetry is the same these reactions are allowed. Woodward-Hoffman rules are a way of working out whether a reaction is thermally allowed or not based on this principle. for u-u and g-g interactions the overlap integral will be non-zero. For u-g interactions it will be zero.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" | Bonds in Reactants /Å
! style="background: #0D4F8B; color: white;" | Bonds in Transition State /Å
! style="background: #0D4F8B; color: white;" | Bonds in Product /Å
|-
| Diene C=C 1.33918
| Diene C~C 1.37978
| Diene C-C 1.50085
|-
| Diene C=C 1.33918
| Diene C~C 1.37975
| Diene C-C 1.50083
|-
| Diene C-C 1.47013
| Diene C~C 1.41111
| Diene C=C 1.33700
|-
| Ethene C=C 1.33087
| Ethene C~C 1.38176
| Ethene C-C 1.53464
|-
| -
| -
| New C-C 1.53700
|-
| -
| -
| New C-C 1.53722
|}

The average C-C sp3 bond is 1.54Å and for C=C sp2 it is 1.47Å. For the diene the single bond is shorted than in a normal single C-C bond due to the conjugation of the double bond, and the carbons involved being sp2 so the hybrid orbital forming the single bond has more s character than would a regular sp3 C-C bond. These bond length calculations demonstrate the breaking and making of bonds over the course of the reaction through bond lengthening going from a double to single bond and bond contraction with going from a single to a double bond.

The Van der Waals radius of carbon is is 1.70Å. In the transition state the distance between the centres of the carbon atoms which form bonds is 2.11435Å. Given the radius of carbon atoms, at this distance apart their radii would overlap by 0.64Å on each

the animation of the vibration corresponding to the reaction path at the transition state shows the bond formation as being synchranous, as is expected of a pericyclic reaction:

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 18; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>TS PM6.LOG</uploadedFileContents>
<title>The LUMO of the transition state calculated with PM6</title>
</jmolApplet>
</jmol>

The lowest possible frequency is an asymmetric stretch which is asynchronous

'''Exercise 2'''

[[File:Endo HOMO.png]]

[[File:Endo LUMO.png]]

[[File:Exo HOMO.png]]

[[File:Exo LUMO.png]]

A normal demand Diels-Alder reaction involves the diene acting as the nucleophile since its HOMO is closer to the LUMO of the dieneophile. This is facilitated by electron withdrawing groups on the dieneophile which the lower the MO energies. In inverse demand Diels-alder reactions the dieneophile acts as the nucleophile, since its HOMO has a better overlap with the LUMO of the diene.

In this reaction of of benzoquinone with cyclopentadiene it is normal demans since the HOMO of the diene is -0.21154, which is higher in energy than the HOMO of the dieneophile which is -0.27061. The energy gap between the HOMO of the diene and the LUMO of the dieneophile is 0.0815 which is smaller than the gap between the HOMO of the dieneophile and the LUMO of the diene, hence it is a normal demand Diels Alder.

In the endo transition state secondary orbital interactions can be seen. These have the eddect of stabilising the endo transition state lowering the activation energy and hence making the endo product the kinetic one. This compensates for the higher steric clash in the endo transition state.

'''Exercise 3'''

Rep:Mod:gkuytuyg

2016-10-18T16:44:59Z

Bw1114:

'''Introduction'''

When Gaussian calculates a minimum, it varies each nuclear position as a function of the total energy of the molecule and finds the geometry such that the energy is lowest. The transition state is the maximum energy geometry on the reaction pathway between the reactants and products.

At a minimum the gradient is zero since it is a stationary point and consequently the curvature is upwards on either side. The first derivative of the energy with respect to each nuclear coordinate is equal to zero at a minimum and second derivative is positive.

Frequency calculations are useful since at a minimum there will be no imaginary frequencies. This is because the frequencies are calculated using a harmonic oscillator model where the force constant is equal to the second derivative of the energy with respect to nuclear positions and the frequency is proportional to the square root of the force constant. If it is a minimum the second derivative will be positive giving all positive frequencies. However for a transition state there will be an imaginary frequency since the curvature is downwards so second derivative will be negative, hence giving an imaginary frequency.

'''Exercise 1'''

[[File:Drawn MO Diagram Corrected.PNG]]

<jmol>
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<size>500</size>
<script>frame 2; mo 15; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
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<jmol>
<jmolApplet>
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<script>frame 2; mo 16; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
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<title>The LUMO of Butadiene calculated with B3LYP</title>
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<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 8; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>ETHENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The HOMO of Ethene calculated with B3LYP</title>
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<jmol>
<jmolApplet>
<color>white</color>
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<script>frame 2; mo 9; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>ETHENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The LUMO of Ethene calculated with B3LYP</title>
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<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 17; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>TS PM6.LOG</uploadedFileContents>
<title>The HOMO of the transition state calculated with PM6</title>
</jmolApplet>
</jmol>

It can be seen that the drawn MO diagram HOMOs and LUMOs correspond to those that Gaussian produced. The dotted red lines show the possible MO interactions in the Diels Alder reaction. For any reaction only MOs of the same symmetry may overlap. This is shown by the parity labels g and u, otherwise the orbitals interfere destructively and become antibonding making new bond formation impossible and forbidden. When the symmetry is the same these reactions are allowed. Woodward-Hoffman rules are a way of working out whether a reaction is thermally allowed or not based on this principle. for u-u and g-g interactions the overlap integral will be non-zero. For u-g interactions it will be zero.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" | Bonds in Reactants /Å
! style="background: #0D4F8B; color: white;" | Bonds in Transition State /Å
! style="background: #0D4F8B; color: white;" | Bonds in Product /Å
|-
| Diene C=C 1.33918
| Diene C~C 1.37978
| Diene C-C 1.50085
|-
| Diene C=C 1.33918
| Diene C~C 1.37975
| Diene C-C 1.50083
|-
| Diene C-C 1.47013
| Diene C~C 1.41111
| Diene C=C 1.33700
|-
| Ethene C=C 1.33087
| Ethene C~C 1.38176
| Ethene C-C 1.53464
|-
| -
| -
| New C-C 1.53700
|-
| -
| -
| New C-C 1.53722
|}

The average C-C sp3 bond is 1.54Å and for C=C sp2 it is 1.47Å. For the diene the single bond is shorted than in a normal single C-C bond due to the conjugation of the double bond, and the carbons involved being sp2 so the hybrid orbital forming the single bond has more s character than would a regular sp3 C-C bond. These bond length calculations demonstrate the breaking and making of bonds over the course of the reaction through bond lengthening going from a double to single bond and bond contraction with going from a single to a double bond.

The Van der Waals radius of carbon is is 1.70Å. In the transition state the distance between the centres of the carbon atoms which form bonds is 2.11435Å. Given the radius of carbon atoms, at this distance apart their radii would overlap by 0.64Å on each

the animation of the vibration corresponding to the reaction path at the transition state shows the bond formation as being synchranous, as is expected of a pericyclic reaction:

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 18; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>TS PM6.LOG</uploadedFileContents>
<title>The LUMO of the transition state calculated with PM6</title>
</jmolApplet>
</jmol>

The lowest possible frequency is an asymmetric stretch which is asynchronous

'''Exercise 2'''

[[File:Endo HOMO.png]]

[[File:Endo LUMO.png]]

[[File:Exo HOMO.png]]

[[File:Exo LUMO.png]]

A normal demand Diels-Alder reaction involves the diene acting as the nucleophile since its HOMO is closer to the LUMO of the dieneophile. This is facilitated by electron withdrawing groups on the dieneophile which the lower the MO energies. In inverse demand Diels-alder reactions the dieneophile acts as the nucleophile, since its HOMO has a better overlap with the LUMO of the diene.

In this reaction of of benzoquinone with cyclopentadiene it is normal demans since the HOMO of the diene is -0.21154, which is higher in energy than the HOMO of the dieneophile which is -0.27061. The energy gap between the HOMO of the diene and the LUMO of the dieneophile is 0.0815 which is smaller than the gap between the HOMO of the dieneophile and the LUMO of the diene, hence it is a normal demand Diels Alder.

In the endo transition state secondary orbital interactions can be seen. These have the eddect of stabilising the endo transition state lowering the activation energy and hence making the endo product the kinetic one. This compensates for the higher steric clash in the endo transition state.

Rep:Mod:gkuytuyg

2016-10-18T15:58:05Z

Bw1114:

'''Introduction'''

When Gaussian calculates a minimum, it varies each nuclear position as a function of the total energy of the molecule and finds the geometry such that the energy is lowest. The transition state is the maximum energy geometry on the reaction pathway between the reactants and products.

At a minimum the gradient is zero since it is a stationary point and consequently the curvature is upwards on either side. The first derivative of the energy with respect to each nuclear coordinate is equal to zero at a minimum and second derivative is positive.

Frequency calculations are useful since at a minimum there will be no imaginary frequencies. This is because the frequencies are calculated using a harmonic oscillator model where the force constant is equal to the second derivative of the energy with respect to nuclear positions and the frequency is proportional to the square root of the force constant. If it is a minimum the second derivative will be positive giving all positive frequencies. However for a transition state there will be an imaginary frequency since the curvature is downwards so second derivative will be negative, hence giving an imaginary frequency.

'''Exercise 1'''

[[File:Drawn MO Diagram Corrected.PNG]]

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 15; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>BUTADIENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The HOMO of Butadiene calculated with B3LYP</title>
</jmolApplet>
</jmol>

<jmol>
<jmolApplet>
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<script>frame 2; mo 16; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>BUTADIENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The LUMO of Butadiene calculated with B3LYP</title>
</jmolApplet>
</jmol>

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 8; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>ETHENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The HOMO of Ethene calculated with B3LYP</title>
</jmolApplet>
</jmol>

<jmol>
<jmolApplet>
<color>white</color>
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<script>frame 2; mo 9; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>ETHENE MINIMUM B3LYP.LOG</uploadedFileContents>
<title>The LUMO of Ethene calculated with B3LYP</title>
</jmolApplet>
</jmol>

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 17; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>TS PM6.LOG</uploadedFileContents>
<title>The HOMO of the transition state calculated with PM6</title>
</jmolApplet>
</jmol>

It can be seen that the drawn MO diagram HOMOs and LUMOs correspond to those that Gaussian produced. The dotted red lines show the possible MO interactions in the Diels Alder reaction. For any reaction only MOs of the same symmetry may overlap. This is shown by the parity labels g and u, otherwise the orbitals interfere destructively and become antibonding making new bond formation impossible and forbidden. When the symmetry is the same these reactions are allowed. Woodward-Hoffman rules are a way of working out whether a reaction is thermally allowed or not based on this principle. for u-u and g-g interactions the overlap integral will be non-zero. For u-g interactions it will be zero.

{| class="wikitable"
! style="background: #0D4F8B; color: white;" | Bonds in Reactants /Å
! style="background: #0D4F8B; color: white;" | Bonds in Transition State /Å
! style="background: #0D4F8B; color: white;" | Bonds in Product /Å
|-
| Diene C=C 1.33918
| Diene C~C 1.37978
| Diene C-C 1.50085
|-
| Diene C=C 1.33918
| Diene C~C 1.37975
| Diene C-C 1.50083
|-
| Diene C-C 1.47013
| Diene C~C 1.41111
| Diene C=C 1.33700
|-
| Ethene C=C 1.33087
| Ethene C~C 1.38176
| Ethene C-C 1.53464
|-
| -
| -
| New C-C 1.53700
|-
| -
| -
| New C-C 1.53722
|}

The average C-C sp3 bond is 1.54Å and for C=C sp2 it is 1.47Å. For the diene the single bond is shorted than in a normal single C-C bond due to the conjugation of the double bond, and the carbons involved being sp2 so the hybrid orbital forming the single bond has more s character than would a regular sp3 C-C bond. These bond length calculations demonstrate the breaking and making of bonds over the course of the reaction through bond lengthening going from a double to single bond and bond contraction with going from a single to a double bond.

The Van der Waals radius of carbon is is 1.70Å. In the transition state the distance between the centres of the carbon atoms which form bonds is 2.11435Å. Given the radius of carbon atoms, at this distance apart their radii would overlap by 0.64Å on each

the animation of the vibration corresponding to the reaction path at the transition state shows the bond formation as being synchranous, as is expected of a pericyclic reaction:

<jmol>
<jmolApplet>
<color>white</color>
<size>500</size>
<script>frame 2; mo 18; mo nodots nomesh fill translucent; set antialiasdisplay on</script>
<uploadedFileContents>TS PM6.LOG</uploadedFileContents>
<title>The LUMO of the transition state calculated with PM6</title>
</jmolApplet>
</jmol>

The lowest possible frequency is an asymmetric stretch which is asynchronous

'''Exercise 2'''

[[File:Endo HOMO.png]]

[[File:Endo LUMO.png]]

[[File:Exo HOMO.png]]

[[File:Exo LUMO.png]]

A normal demand Diels-Alder reaction involves the diene acting as the nucleophile since its HOMO is closer to the LUMO of the dieneophile. This is facilitated by electron withdrawing groups on the dieneophile which the lower the MO energies. In inverse demand Diels-alder reactions the dieneophile acts as the nucleophile, since its HOMO has a better overlap with the LUMO of the diene.

In this reaction of of benzoquinone with cyclopentadiene it is normal demans since the HOMO of the diene is -0.21154, which is higher in energy than the HOMO of the dieneophile which is -0.27061. The energy gap between the HOMO of the diene and the LUMO of the dieneophile is 0.0815 which is smaller than the gap between the HOMO of the dieneophile and the LUMO of the diene, hence it is a normal demand Diels Alder.

Rep:Mod:gkuytuyg

2016-10-18T15:38:22Z

Bw1114:

File:Exo LUMO.png

2016-10-18T15:37:16Z

Bw1114: Bw1114 uploaded a new version of File:Exo LUMO.png

File:Exo HOMO.png

2016-10-18T15:36:55Z

Bw1114: Bw1114 uploaded a new version of File:Exo HOMO.png

File:Endo LUMO.png

2016-10-18T15:36:35Z

Bw1114:

File:Endo HOMO.png

2016-10-18T15:36:14Z

Bw1114: Bw1114 uploaded a new version of File:Endo HOMO.png

Rep:Mod:gkuytuyg

2016-10-18T13:18:22Z

Bw1114:

Rep:Mod:gkuytuyg

2016-10-18T13:14:28Z

Bw1114:

Rep:Mod:gkuytuyg

2016-10-18T12:09:45Z

Bw1114:

Rep:Mod:gkuytuyg

2016-10-17T16:39:39Z

Bw1114:

Rep:Mod:gkuytuyg

2016-10-17T16:25:07Z

Bw1114:

Rep:Mod:gkuytuyg

2016-10-17T16:01:45Z

Bw1114:

Rep:Mod:gkuytuyg

2016-10-17T15:52:38Z

Bw1114:

Rep:Mod:gkuytuyg

2016-10-17T15:50:00Z

Bw1114:

Rep:Mod:gkuytuyg

2016-10-17T15:28:37Z

Bw1114:

Rep:Mod:gkuytuyg

2016-10-17T14:49:26Z

Bw1114:

File:Drawn MO Diagram Corrected.PNG

2016-10-17T14:49:07Z

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Rep:Mod:gkuytuyg

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