2018-03-13T13:08:39Z

Mg5715: /* Potential energy surface */

==Introduction==

===Potential energy surface===
[[File:MG5715_TS_INTRO_DIATOMIC.PNG|thumb|500px|x500px|center|Fig.1 The potential energy curve of a diatomic molecule<ref>L., D. J. (1957). Model of a potential energy surface. J. Chem. Educ., 34(5), 215.</ref>]]
The potential energy surface of a diatomic molecule is anharmonic oscillation (as shown in Fig.1). The lowest point in this energy potential is a stationary point, which means it has a zero first derivative:
<center><math>\frac{dE(R)}{dR}=0</math></center>
<small><center>Equation 1. First derivative of 1-D potential energy</center></small>

R stands for the bond length and the physical meaning of the first derivative of potential energy is the force and the second derivative is the force constant (k). The bond will vibrate; so the vibration wave number could be calculated by Equation 2.
<center><math>\tilde{v}=\frac{1}{2c\pi}\sqrt{\frac{k}{\mu}}</math></center>
<small><center>Equation 2. Vibrational wavenumber of a diatomic molecule</center></small>
where <math>\mu</math> is the reduced mass and it could be calculated by Equation 3.
<center><math>\mu=\frac{M_AM_B}{M_A+M_B}</math></center>
<small><center>Equation 3. Reduced mass of a diatomic molecule</center></small>

[[File:MG5715_TS_INTRO_TRIATOMIC.PNG|thumb|500px|x500px|center|Fig.3 Potential energy surface of triatomic molecule<ref>Ot, W. J. (1990). Computational quantum chemistry. Journal of Molecular Structure: THEOCHEM (Vol. 207). http://doi.org/10.1016/0166-1280(90)85035-L</ref>]]
For a triatomic molecule (e.g.H<sub>2</sub>O), the potential energy surface has two coordinates including bond length R and bond angle <math>\theta</math> and the potential energy surface is shown in Fig.3. For a non-linear molecule including N atoms, it will have '''3N-6''' independent geometric variables. For each atom, it will have three variables (bond length, bond angle and torsional angles) and three global rotations and three global translations should be subtracted. If it is a linear molecule, there only have two rotation axes, so it has '''3N-5''' independent geometric variables.
Hence, a stationary point in this potential energy surface could be defined by equation 4.

<center><math>\frac{dE(\mathbf{R})}{dR_i}=0 i=1,2,3,...3N-6</math></center>
<small><center>Equation 4. General equation of a stationary point with 3N-6 variables</center></small>
where '''R''' is the set of all nuclear coordiantes.<ref>Ot, W. J. (1990). Computational quantum chemistry. Journal of Molecular Structure: THEOCHEM (Vol. 207). http://doi.org/10.1016/0166-1280(90)85035-L</ref>

[[File:MG5715_TS_INTRO_PES.PNG|thumb|500px|x500px|center|Fig.4 The 1-D potential energy surface of a reaction<ref>L., D. J. (1957). Model of a potential energy surface. J. Chem. Educ., 34(5), 215.</ref>]]
Fig.4 is a 1-D potential energy surface of a reaction. Products and reactants are the minimum points in the lowest energy pathway. For transition state, it is the maximum point on the lowest energy pathway. All of reactants, products and transition state are the stationary points, which means the first derivative of potential energy surface at that point is zero. (<math>\frac{dE(\mathbf{R})}{dR}=0</math>). The second derivative of potential energy surface, the force constant, is used to distinguish the products and reactants with the transition state. Reactants and products are {{fontcolor1|blue|'''minima'''}}, so the second derivative is positive.(<math>\frac{d^2E(\mathbf{R})}{dR^2}>0</math>) Transition state is the {{fontcolor1|blue|'''saddle point'''}} of the potential energy surface, which means its second derivative is negative.(<math>\frac{d^2E(\mathbf{R})}{dR^2}<0</math>)

===Basis set===

===Computational Methods===
Although Gauss View contains lots of different calculation method, only two of them are used in this lab: (1) Semi-empirical method PM6 and (2) Density Functional Theory(DFT) method B3LYP.

===Methods to find Transition State===
* '''Method 1'''
(1) predict and draw a guess transition state structure

(2) optimise the guess transition state structure to TS(Berny) and set ''Calculate force constants'' to '''once''' and the output is the optimised transition state (Only have '''one''' imaginary frequency)

This method is quite easy and it is the fastest method.However, this method is not very reliable and it only works for small systems because it will fail easily if the predicted transition state is not close to the real transition state.

* '''Method 2'''
(1) predict and draw a guess transition state structure

(2) freeze the distance between atoms where the bond will be formed in the reaction

(3) optimise the frozen-bond structure to a minimum

(4) repeat method 1(2)

This method is similar to method 1 but this one is more reliable than method 1 because it freezes the atoms to prevent them from moving. This makes sure the system close to the transition state before calculation. However, this also will fail easily.

* '''Method 3'''
(1) draw the reactant(s) or product(s) and choose the one which has fewer molecules.

(2) optimise the reactant(s) or product(s) to a minimum

(3) break or form the bond and freeze the distance between atoms

(4) repeat method 2 (2)-(4)

This method is much more reliable than the previous two methods and it does not need to predict the possible transition state. However, this method is more complicated and it requires more steps.

==Excercise 1: Reaction of Butadiene with Ethylene==
[[file:MG5715_TS_EX1_REACTION SCHEME.JPG|thumb|550px|center|Fig. The reaction scheme of cycloaddition of butadiene and ethylene]]
The reaction of butadiene with ethylene is a traditional '''[4+2] cycloaddition''', which also known as '''Diels-Alder reaction'''.In Diels-Alder reaction, a conjugated diene (butadiene) will react with a dienophile (ethylene) to form a cyclohexene and the reaction scheme is shown in Fig. Although the s-''trans'' conformation is more energetically favourable, the conformation of diene should be s-''cis'' because of the interaction between the frontier molecular orbitals(FMO). Moreover, the energy barrier between s-''trans'' and s-''cis'' is not extremely high. In this exercise, all reactants, products and transition state were optimised by {{fontcolor1|blue|'''semi-empirical method PM6 '''}} in Gauss View and {{fontcolor1|blue|'''Method 2'''}} is used to locate the transition state.
===MO analysis===

[[file:MG5715_TS_EX1_MO_1.JPG|thumb|550px|center|Fig. The molecular orbital of cycloaddition of butadiene with ethylene]]

The MO diagram is shown in Fig. and this graph is drawn according to the energy calculated by PM6. The energy of product ({{fontcolor1|black|'''black'''}} line) should be lower than that of the reactants, and the energy of transition state ({{fontcolor1|#fe7af9|'''pink'''}} line) is higher than that of reactants. The HOMO and LUMO are shown in Talbe.

{| class="wikitable" style="margin: auto;"
|
! center; style="background: #0D4F8B; color: white; text-align: center;"| Butadiene
! center; style="background: #0D4F8B; color: white; text-align: center;" | Ethylene
! center; style="background: #0D4F8B; color: white; text-align: center;" colspan="2"| MO of Transition State
|-
| center;| LUMO

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| <jmol>
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|-
| center;| HOMO
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For a [p+q]-cycloaddtion reaction, it could be driven either thermally or photochemically. For photochemical reaction, the electrons on HOMO will be excited to LUMO and become two SOMOs.Therefore, the photochemical cycloaddition is a little bit different with the thermal cycloaddition. A general rule called '''''Woodward Hoffmann Rule''''' is used to determine whether the cycloaddition is allowed or forbidden.

*Woodward-Hoffmann Rule for cycloaddition reactions
For a [p+q]-cylcoaddtion reaction, only two components are involved, where one contains p π-electrons and the other one has q π-electrons. s stands for suprafacial and a means antarafacial

Table .Woodward-Hoffmann Rule <ref>Semis, K. L., & Rules, B. W. (1965). Woodward-Hoff mann Rules : Electrocyclic Reactions.</ref>
{| class="wikitable" style="margin: auto;"
! center; style="background: #0D4F8B; color: white; text-align: center;"| p+q
! center; style="background: #0D4F8B; color: white; text-align: center;" | Thermally allowed
! center; style="background: #0D4F8B; color: white; text-align: center;" | Photochemically allowed

|-
| center; | 4n
| p<sub>s</sub>+q<sub>a</sub> or p<sub>a</sub>+q<sub>s</sub>
| p<sub>s</sub>+q<sub>s</sub> or p<sub>a</sub>+q<sub>a</sub>

|-
| center;| 4n+2
| p<sub>s</sub>+q<sub>s</sub> or p<sub>a</sub>+q<sub>a</sub>
| p<sub>s</sub>+q<sub>a</sub> or p<sub>a</sub>+q<sub>s</sub>

|}

In this reaction, butadiene has 4 π-electrons and ethylene contains 2 π-electrons. The sum of p and q is 6 and they are all suprafacial. Hence, this [4+2] cycloaddition is {{fontcolor1|blue|'''thermally allowed'''}}.

*overlap integral
<center><math>S=\int\psi\psi^*\,d\tau</math></center>
<center><small>Equation. Calculation of overlap integral</small></center>

The overlap integral is calculated by equation and it is telling how well two orbitals are overlapped. The value of overlap integral is between 0 and 1. 0 means that the two orbitals do not have any overlap and 1 means that they perfectly overlap with each other. Table is used to determine whether the wavefunction is symmetric or antisymmetric.

{| class="wikitable" style="margin: auto;"
! center; style="background: #0D4F8B; color: white; text-align: center;"| Symmetric
! center; style="background: #0D4F8B; color: white; text-align: center;" | Antisymmetric
|-
| center;|ψ(r<sub>1</sub>,r<sub>2</sub>)={{fontcolor1|red|'''+'''}}ψ(r<sub>2</sub>,r<sub>1</sub>)
| center;|ψ(r<sub>1</sub>,r<sub>2</sub>)={{fontcolor1|red|'''-'''}}ψ(r<sub>2</sub>,r<sub>1</sub>)
|}

For two wavefunctions, the overlap integral will be 0 if the product of two wavefunctions is antisymmetric. If the product of two wavefunctions is symmetric, the overlap integral will be 1. Only when '''both of the wavefunctions are antisymmetric''' or '''both are antisymmetric''', the overlap integral will be 1. As shown in MO diagram, the antisymmetric orbital will overlap with the antisymmetric orbital.

===Analysis of bond lengths===
{| class="wikitable" style="margin: auto;"
|center; | [[file:MG5715_TS_E1_IRC_DISTANCE.jpg|thumb|none|500px|x500px|center|Fig. Disatance VS reaction coordination]]
|center; | [[file:MG5715_TS_E1_LABEL.PNG|thumb|none|500px|x500px|center|Fig. Labeled molecule]]
|+ Table. C
|}

In Fig., it shows that the distances between C atoms change with reaction coordination. and Fig. label the molecule. Initially, the distance between C<sub>14</sub> and C<sub>1</sub> and distance between C<sub>7</sub> and C<sub>11</sub> are around 3.4 Å, which indicates there does not have any bond between those two atoms. The Van der Waal radius of C atom is 1.77Å <ref>Batsanov, S. S. (2001). Van der Waals Radii of Elements. Inorganic Materials Translated from Neorganicheskie Materialy Original Russian Text, 37(9), 871–885. http://doi.org/10.1023/A:1011625728803</ref> 3.4 Å is smaller than twice of Van der Waals radius of C; therefore, C<sub>14</sub> is interacting with C<sub>1</sub> and a bond will be formed between them.

{| class="wikitable" style="margin: auto;"
|
! center; style="background: #0D4F8B; color: white;"| C(sp<sup>3</sup>)-C(sp<sup>3</sup>)
! center; style="background: #0D4F8B; color: white;"| C(sp<sup>3</sup>)-C(sp<sup>2</sup>)
! center; style="background: #0D4F8B; color: white;"| C(sp<sup>2</sup>)-C(sp<sup>2</sup>)
|-
| center;| Bond length/Å
| 1.542
| 1.488
| 1.459
|+ Table. Bond length of C-C bond<ref>Bernstein, H. J. (1961). BOND DISTANCES IN HYDROCARBONS * t. Retrieved from http://pubs.rsc.org/-/content/articlepdf/1961/tf/tf9615701649</ref>
|}

Table . shows the literature value of C-C bond length with different hybridisation and comparing those with the Fig. It suggests C<sub>1</sub>, C<sub>7</sub>,C<sub>11</sub> and C<sub>14</sub> change from sp<sup>2</sup> to sp<sup>3</sup>; C<sub>4</sub> and C<sub>6</sub> remain sp<sup>2</sup>. Besides that, this table also confirms the product is hexene. The distance between C<sub>4</sub> and C<sub>6</sub> is 1.338 Å, which is similar to the bond lenght of C(sp<sup>2</sup>)-C(sp<sup>2</sup>). The rest of bond lengths are around 1.5 Å; so the rest of bonds are C(sp<sup>3</sup>)-C(sp<sup>3</sup>).

===Vibration===
{| class="wikitable" style="margin: auto;"
! center; style="background: #0D4F8B; color: white;" | Transition State vibration
! center; style="background: #0D4F8B; color: white;" | Inrinsic Reaction Coordinate(IRC)

|-
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| [[file:MG5715_TS_E1_IRC.gif‎]]

|-
| center;| [[file:MG5715_TS_E1_VF.PNG|thumb|none|500px|x500px|center|Fig. Vibration Frequencies of Transition State]]
| [[file:MG5715_TS_E1_IRC_PATH.PNG|thumb|none|500px|x500px|center|Fig. sumarry of IRC plot ]]

|}

As shown in Fig., there is only one imaginary frequency (at -948.8 cm<sup>-1</sup>); so it justifies the transition state is right. However, it is not enough for determining the transition state. It is critical to do the IRC calculation. RMS Gradient should be zero in reactants, products and also transition state because they are local minima. However, for transition state, it is saddle point; hence, the second derivative should be negative.

According to the gif shown above, it shows that bonds are formed '''synchronously'''.

===LOG file===

==Excercise 2:Reaction of Cyclohexadiene and 1,3-Dioxole==

[[File:MG5715_TS_E2_REACTION.jpg|thumb|550px|center|Fig. The reaction scheme of Cyclohexadiene and 1,3-Dioxole]]

The reaction of cyclohexadiene and 1,3-Dioxole is also a Diels-Alder reaction, but this reaction will give two products (endo and exo) because of two possible transition states. The reaction scheme is shown in Fig. In this exercise, the transition state is opitimised by {{fontcolor1|blue|'''PM6'''}} firstly and then optimised again by {{fontcolor1|blue|'''B3LYP'''}}. Moreover, the transition state is determined by {{fontcolor1|blue|'''method 3'''}} in this exercise.

=== MO analysis of exercise 2===

[[File:MG5715_TS_E2_MO.jpg|thumb|550px|center|Fig. MO diagram of Cyclohexadiene and 1,3-Dioxole]]

The MO diagram of this reaction is displayed in Fig.The energies of product's MO is shown in '''black''' lines and the {{fontcolor1|#fe7af9|'''pink'''}} lines represent the energies of transition states' MO, and all of them are drawn based on the relative energies calculated by '''B3LYP'''. Furthermore, the HOMO and LUMO of reactants are shown in Table. and the MOs of transition states are displayed in Table.

{| class="wikitable" style="margin: auto;"
|
! style="background: #0D4F8B; color: white; text-align: center;"| HOMO
! style="background: #0D4F8B; color: white; text-align: center;" | LUMO

|-
| Clycohexadiene

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|-
| 1,3-dioxole
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|+ Table . LUMO and HOMO of reactants
|}

{| class="wikitable" style="margin: auto;"
|
! style="background: #0D4F8B; color: white; text-align: center;"| ENDO TS
! style="background: #0D4F8B; color: white; text-align: center;" | EXO TS

|-
| LUMO+1
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|-
| LUMO
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|-
| HOMO
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|-
| HOMO-1
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|+ Table . MO of Transition states
|}

====Secondary Orbital Interactions ====
{| class="wikitable"
|
! style="background: #0D4F8B; color: white;" | Endo-TS
! style="background: #0D4F8B; color: white;" | Exo-TS
|-
| Jmol of computed HOMO

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|-
| LCAO diagram of HOMO

| [[File:MG5715_TS_E2_ENDO_TS.jpg|thumb|550px|center]]

| [[File:MG5715_TS_E2_EXO_TS.jpg|thumb|550px|center]]
|+ Table. HOMO of Endo-TS and Exo- TS
|}

The computed HOMO and LACO diagram of HOMO are shown in Table. Endo-product is more stable than Exo-product due to lack of steric clash with the methyl group in cyclohexadiene. Furthermore, the Endo-TS is lower in energy because of '''Secondary Orbital Interaction'''. The lone pair of oxygen could donate the electron density to the LUMO of cyclohexadiene; hence, the Endo-TS will be stabilised. Both the Endo-TS and Endo-product are lower in energy; so this reaction is endo-selective.

==== Inverse VS Normal Demand Diels-Alder Reaction ====

[[File:MG5715_TS_E2_DA.jpg|thumb|550px|center|Fig. two tyes of Diels-Alder reaction]]

For a Diels-Alder reaction, the reactivity is controlled by relative energies of '''Frontier Molecular Orbitals (FMOs)'''. There are two different types as shown in Fig.:(1) Normal demand and (2) Inverse demand. Usually, the reaction of an all carbon diene with a hetero-dienophile will give a ''normal electron demand'' because the heteroatom in dienophile may withdraw the electron density from dienophile and lower the energy of MOs. The best way to justify whether the reaction is normal demand or inverse demand is to sequence the energy of reactants as shown in Table. If the HOMO of dienophile is lower than that of diene, it is normal demand, vice versa. The HOMO of cyclohexadiene is the lowest; hence, this reaction is {{fontcolor1|blue|'''inverse demand'''}} Diels-Alder reaction. The reason is that the two oxygen atom in 1,3-dioxole donate the electron density toward the π-bonding and raise the energy of MOs.

{| class="wikitable" style="margin: auto;"
! style="background: #0D4F8B; color: white;" | Endo-reaction
! style="background: #0D4F8B; color: white;" | Exo-reaction

|-

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|+ Table. Single point energy of reactants
|}

=== Activation Barrier and Energy changed ===
{| class="wikitable" style="margin: auto;"
|
! style="background: #0D4F8B; color: white;" | Energy calculated by PM6/kJmol<sup>-1</sup>
! style="background: #0D4F8B; color: white;" | Energy calculated by B3LYP/kJmol<sup>-1</sup>
|-
| Cylcohexadiene
| 306.9
| -612593

|-
| 1,3-dioxole
| -137.3
| -701189

|-
| Endo-product
| 99.2
| -1313849

|-
| Endo-TS
| 362.2
| -1313622

|-
| Exo-product
| 99.7
| -1313845

|-
| Exo-TS
| 364.7
| -1313614
|+ Table. Energies of reactants, products and transition states
|}

According to the energies of reactants, products and transition states (shown in Table), the reaction energy and activation energy could be caluclated by following equation:
<pre>
Activation Energy = Transition state energy - Sum of energies of reactants
Reaction Energy = product energy - Sum of energies of reactants
</pre>

{| class="wikitable" style="margin: auto;"
! rowspan="2"|
! style="background: #0D4F8B; color: white;" colspan="2"| Activation Energy/kJmol<sup>-1</sup>
! style="background: #0D4F8B; color: white;" colspan="2"| Reaction Energy/kJmol<sup>-1</sup>
|-
! style="background: #0D4F8B; color: white;"| PM6
! style="background: #0D4F8B; color: white;"| B3LYP
! style="background: #0D4F8B; color: white;"| PM6
! style="background: #0D4F8B; color: white;"| B3LYP
|-
| Endo-reaction
| +192.6
| +159.8
| -70.3
| -67.4
|-
| Exo-reaction
| +195.1
| +169.7
| -69.9
| -63.8
|+ Table. Reaction energy and activation energy for both Endo and Exo product.
|}

All the energy calculated is shown in Table. Both the activation energy and reaction energy for Endo-product are lower than those for Exo-product. A thermodynamically favourable product is more stable, which means it has more negative reaction energy. The kinetically favourable product has lower reaction barrier (activation energy). Hence, the endo-product of this exercise is both '''thermodynamically and kinetically favourable'''.

==Excercise 3 Diels-Alder vs Cheletropic==

[[File:MG5715_TS_E3_REACTION.jpg|thumb|550px|center|Fig. Reaction Scheme of o-Xylylene with SO<sub>2</sub>]]

The reaction of o-xylylene with SO<sub>2</sub> could be either Diels-Alder or cheletropic reaction. For Diels-Alder reaction, o-Xylylene acts as diene and SO<sub>2</sub> acts as dienophile to form a six-membered ring and there are two possible products (endo-product and exo-product). As discussed in exercise 2, the Diels-Alder is endo-selective. For the cheletropic reaction, a five-membered ring will be formed and two new bonds are formed with the same atom. The reaction scheme of those reactions is displayed in Fig. In this exercise, the reactants, products and transition states were optimised by {{fontcolor1|blue|'''PM6'''}} and {{fontcolor1|blue|'''B3LYP'''}} and {{fontcolor1|blue|'''Method 3'''}} was used to find the transition state. The reaction energy and activation energy were calculated to find out the favourable reaction.

=== IRC Analysis===
{| class="wikitable" style="margin: auto;"
|
! style="background: #0D4F8B; color: white;" | IRC
|-
| Cheletropic reaction
| [[File:MG5715_TS_E3_Cheletropic_IRC.gif]]
|-
| Endo Diels-Alder reaction
| [[File:MG5715_TS_E2_ENDO_irc.gif]]
|-
| Exo Diels-Alder reaction
| [[File:MG5715_TS_E3_EXO_irc.gif]]
|+ Table. The IRC of three possible reactions
|}

There are two different mechanisms for Diels-Alder reaction:(1)synchronous and symmetrical (concerted) mechanism and (2) multistage (non-concerted) and asynchronous mechanism.

* synchronous and symmetrical (concerted) mechanism
A very typical example of this mechanism is Exercise 1. The two new bonds are formed simultaneously and the two new bonds have the same bond length in the transition state.

* multistage (non-concerted) and asynchronous mechanism
The two bonds could not be formed at the same time because the bond length of them are different at transition state. Therefore, usually, a di-radical will be formed.

For Diels-Alder reaction, the distance between O and C in o-Xylylene is different from the distance between S and C in o-Xylylene at transition state because of distinct Van der Waal radii.
Therefore, the bond does not form synchronously, which could be observed in the IRC in Table. However, for cheletropic reaction, the two now bonds are both formed between S atom and C in o-Xylylene; so the two bond are formed synchronously as shown in IRC.

=== Activation Energy and Reaction Energy===
[[File:MG5715_TS_E3_ENERGY.jpg|thumb|550px|center|Fig. Energy diagram of o-Xylylene with SO<sub>2</sub>]]

{| class="wikitable" style="margin: auto;"
|
! style="background: #0D4F8B; color: white;" | Energy calculated by PM6/kJmol<sup>-1</sup>
! style="background: #0D4F8B; color: white;" | Energy calculated by B3LYP/kJmol<sup>-1</sup>
|-
| O-Xylylene
| 469.5
| -812604

|-
| SO<sub>2</sub>
| -311.4
| -1440362

|-
| Endo-product
| 57.0
| -2253040

|-
| Endo-TS
| 237.8
| -2252919.7

|-
| Exo-product
| 56.3
| -2253041

|-
| Exo-TS
| 241.7
| -2252919.8

|-
| Cheletropic product
| -0.005251
| -2253024

|-
| Cheletropic TS
| 260.1
| -2252902
|+ Table. Energies of reactants, products and transition states
|}

{| class="wikitable" style="margin: auto;"
! rowspan="2"|
! style="background: #0D4F8B; color: white;" colspan="2"| Activation Energy/kJmol<sup>-1</sup>
! style="background: #0D4F8B; color: white;" colspan="2"| Reaction Energy/kJmol<sup>-1</sup>
|-
! style="background: #0D4F8B; color: white;"| PM6
! style="background: #0D4F8B; color: white;"| B3LYP
! style="background: #0D4F8B; color: white;"| PM6
! style="background: #0D4F8B; color: white;"| B3LYP
|-
| Endo-reaction
| +79.7
| +46.2
| -101.1
| -73.9
|-
| Exo-reaction
| +83.7
| +46.1
| -101.8
| -73.2
|-
| Cheletropic
| +102.0
| +63.0
| -158.1
| -58.8
|+ Table. Reaction energy and activation energy for both Endo and Exo product.
|}

The energies of reactants, products and transition states are shown in Table and the calculated reaction energy and activation energy are displayed in Table. The energy diagram (Fig.) is drawn according to the energy calculated by PM6. The product of the cheletropic reaction is the most stable because of aromaticity. However, the activation energy of cheletropic product is quite high. Therefore, at low temperature, a sultine (kinetic product) will be formed but this reaction in reversible. At high temperature, a sulfolene will be formed and this reaction is irreversible.

=== Extension===
[[File:MG5715_TS_E3_EXTENSION_SCHEME.jpg|thumb|550px|center|Fig. Reaction Scheme of o-Xylylene with SO<sub>2</sub>]]
There are two dienes in o-Xylylene and therefore, it has another possible Diels-Alder reaction as shown in Fig. The energy calculated is displayed in Table and they are calculated by PM6.

{| class="wikitable" style="margin: auto;"
|
! style="background: #0D4F8B; color: white;" | Energy calculated by PM6/kJmol<sup>-1</sup>
|-
| O-Xylylene
| 469.5

|-
| SO<sub>2</sub>
| -311.4

|-
| Endo-product
| 172.3

|-
| Endo-TS
| 268.0

|-
| Exo-product
| 176.7

|-
| Exo-TS
| 275.8
|+
|}

{| class="wikitable" style="margin: auto;"
|
! style="background: #0D4F8B; color: white;" | Activation Energy/kJmol<sup>-1</sup>
! style="background: #0D4F8B; color: white;" | Reaction Energy/kJmol<sup>-1</sup>
|-
| Endo-reaction
| +109.9
| +14.2

|-
| Exo-reaction
| +117.7
| +18.6

|+ Table. Reaction energy and activation energy for both Endo and Exo product.
|}

Compare the energy in Table. with the data in Table, this reaction is quite unlikely to happen because the activation barrier is higher and the reaction energy is endothermic.

==Conclusion==

==Reference==

Rep:Mod:mg5715TS

2018-03-13T12:46:43Z

Mg5715: /* Potential energy surface */

==Introduction==

===Potential energy surface===
[[File:MG5715_TS_INTRO_DIATOMIC.PNG|thumb|500px|x500px|center|Fig.1 The potential energy curve of a diatomic molecule<ref>L., D. J. (1957). Model of a potential energy surface. J. Chem. Educ., 34(5), 215.</ref>]]
The potential energy surface of a diatomic molecule is anharmonic oscillation (as shown in Fig.1). The lowest point in this energy potential is a stationary point, which means it has a zero first derivative:
<center><math>\frac{dE(R)}{dR}=0</math></center>
<small><center>Equation 1. First derivative of 1-D potential energy</center></small>

R stands for the bond length and the physical meaning of the first derivative of potential energy is the force and the second derivative is the force constant (k). The bond will vibrate; so the vibration wave number could be calculated by Equation 2.
<center><math>\tilde{v}=\frac{1}{2c\pi}\sqrt{\frac{k}{\mu}}</math></center>
<small><center>Equation 2. Vibrational wavenumber of a diatomic molecule</center></small>
where <math>\mu</math> is the reduced mass and it could be calculated by Equation 3.
<center><math>\mu=\frac{M_AM_B}{M_A+M_B}</math></center>
<small><center>Equation 3. Reduced mass of a diatomic molecule</center></small>

[[File:MG5715_TS_INTRO_TRIATOMIC.PNG|thumb|500px|x500px|center|Fig.3 Potential energy surface of triatomic molecule<ref>Ot, W. J. (1990). Computational quantum chemistry. Journal of Molecular Structure: THEOCHEM (Vol. 207). http://doi.org/10.1016/0166-1280(90)85035-L</ref>]]
For a triatomic molecule (e.g.H<sub>2</sub>O), the potential energy surface has two coordinates including bond length R and bond angle <math>\theta</math> and the potential energy surface is shown in Fig.3. For a non-linear molecule including N atoms, it will have '''3N-6''' independent geometric variables. For each atom, it will have three variables (bond length, bond angle and torsional angles) and three global rotations and three global translations should be subtracted. If it is a linear molecule, there only have two rotation axes, so it has '''3N-5''' independent geometric variables.
Hence, a stationary point in this potential energy surface could be defined by equation 4.

<center><math>\frac{dE(\mathbf{R})}{dR_i}=0 i=1,2,3,...3N-6</math></center>
<small><center>Equation 4. General equation of a stationary point with 3N-6 variables</center></small>
where '''R''' is the set of all nuclear coordiantes.<ref>Ot, W. J. (1990). Computational quantum chemistry. Journal of Molecular Structure: THEOCHEM (Vol. 207). http://doi.org/10.1016/0166-1280(90)85035-L</ref>

[[File:MG5715_TS_INTRO_PES.PNG|thumb|500px|x500px|center|Fig.4 The 1-D potential energy surface of a reaction<ref>L., D. J. (1957). Model of a potential energy surface. J. Chem. Educ., 34(5), 215.</ref>]]
Fig.4 is a 1-D potential energy surface of a reaction. Products and reactants are the minimum points in the lowest energy pathway. For transition state, it is the maximum point on the lowest energy pathway. All of reactants, products and transition state are the stationary points, which means the first derivative of potential energy surface at that point is zero. (<math>\frac{dE(\mathbf{R})}{dR}=0</math>). The second derivative of potential energy surface, the force constant, is used to distinguish the products and reactants with the transition state.

the transition state is the '''saddle point''' of the potential energy surface. The second derivative of potential energy surface, the force constant, is used to distinguish the stationary point with saddle point.

===Basis set===

===Computational Methods===
Although Gauss View contains lots of different calculation method, only two of them are used in this lab: (1) Semi-empirical method PM6 and (2) Density Functional Theory(DFT) method B3LYP.

===Methods to find Transition State===
* '''Method 1'''
(1) predict and draw a guess transition state structure

(2) optimise the guess transition state structure to TS(Berny) and set ''Calculate force constants'' to '''once''' and the output is the optimised transition state (Only have '''one''' imaginary frequency)

This method is quite easy and it is the fastest method.However, this method is not very reliable and it only works for small systems because it will fail easily if the predicted transition state is not close to the real transition state.

* '''Method 2'''
(1) predict and draw a guess transition state structure

(2) freeze the distance between atoms where the bond will be formed in the reaction

(3) optimise the frozen-bond structure to a minimum

(4) repeat method 1(2)

This method is similar to method 1 but this one is more reliable than method 1 because it freezes the atoms to prevent them from moving. This makes sure the system close to the transition state before calculation. However, this also will fail easily.

* '''Method 3'''
(1) draw the reactant(s) or product(s) and choose the one which has fewer molecules.

(2) optimise the reactant(s) or product(s) to a minimum

(3) break or form the bond and freeze the distance between atoms

(4) repeat method 2 (2)-(4)

This method is much more reliable than the previous two methods and it does not need to predict the possible transition state. However, this method is more complicated and it requires more steps.

==Excercise 1: Reaction of Butadiene with Ethylene==
[[file:MG5715_TS_EX1_REACTION SCHEME.JPG|thumb|550px|center|Fig. The reaction scheme of cycloaddition of butadiene and ethylene]]
The reaction of butadiene with ethylene is a traditional '''[4+2] cycloaddition''', which also known as '''Diels-Alder reaction'''.In Diels-Alder reaction, a conjugated diene (butadiene) will react with a dienophile (ethylene) to form a cyclohexene and the reaction scheme is shown in Fig. Although the s-''trans'' conformation is more energetically favourable, the conformation of diene should be s-''cis'' because of the interaction between the frontier molecular orbitals(FMO). Moreover, the energy barrier between s-''trans'' and s-''cis'' is not extremely high. In this exercise, all reactants, products and transition state were optimised by {{fontcolor1|blue|'''semi-empirical method PM6 '''}} in Gauss View and {{fontcolor1|blue|'''Method 2'''}} is used to locate the transition state.
===MO analysis===

[[file:MG5715_TS_EX1_MO_1.JPG|thumb|550px|center|Fig. The molecular orbital of cycloaddition of butadiene with ethylene]]

The MO diagram is shown in Fig. and this graph is drawn according to the energy calculated by PM6. The energy of product ({{fontcolor1|black|'''black'''}} line) should be lower than that of the reactants, and the energy of transition state ({{fontcolor1|#fe7af9|'''pink'''}} line) is higher than that of reactants. The HOMO and LUMO are shown in Talbe.

{| class="wikitable" style="margin: auto;"
|
! center; style="background: #0D4F8B; color: white; text-align: center;"| Butadiene
! center; style="background: #0D4F8B; color: white; text-align: center;" | Ethylene
! center; style="background: #0D4F8B; color: white; text-align: center;" colspan="2"| MO of Transition State
|-
| center;| LUMO

| <jmol>
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|<jmol>
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| <jmol>
<jmolApplet>
<title>LUMO of Transition State</title>
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|-
| center;| HOMO
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For a [p+q]-cycloaddtion reaction, it could be driven either thermally or photochemically. For photochemical reaction, the electrons on HOMO will be excited to LUMO and become two SOMOs.Therefore, the photochemical cycloaddition is a little bit different with the thermal cycloaddition. A general rule called '''''Woodward Hoffmann Rule''''' is used to determine whether the cycloaddition is allowed or forbidden.

*Woodward-Hoffmann Rule for cycloaddition reactions
For a [p+q]-cylcoaddtion reaction, only two components are involved, where one contains p π-electrons and the other one has q π-electrons. s stands for suprafacial and a means antarafacial

Table .Woodward-Hoffmann Rule <ref>Semis, K. L., & Rules, B. W. (1965). Woodward-Hoff mann Rules : Electrocyclic Reactions.</ref>
{| class="wikitable" style="margin: auto;"
! center; style="background: #0D4F8B; color: white; text-align: center;"| p+q
! center; style="background: #0D4F8B; color: white; text-align: center;" | Thermally allowed
! center; style="background: #0D4F8B; color: white; text-align: center;" | Photochemically allowed

|-
| center; | 4n
| p<sub>s</sub>+q<sub>a</sub> or p<sub>a</sub>+q<sub>s</sub>
| p<sub>s</sub>+q<sub>s</sub> or p<sub>a</sub>+q<sub>a</sub>

|-
| center;| 4n+2
| p<sub>s</sub>+q<sub>s</sub> or p<sub>a</sub>+q<sub>a</sub>
| p<sub>s</sub>+q<sub>a</sub> or p<sub>a</sub>+q<sub>s</sub>

|}

In this reaction, butadiene has 4 π-electrons and ethylene contains 2 π-electrons. The sum of p and q is 6 and they are all suprafacial. Hence, this [4+2] cycloaddition is {{fontcolor1|blue|'''thermally allowed'''}}.

*overlap integral
<center><math>S=\int\psi\psi^*\,d\tau</math></center>
<center><small>Equation. Calculation of overlap integral</small></center>

The overlap integral is calculated by equation and it is telling how well two orbitals are overlapped. The value of overlap integral is between 0 and 1. 0 means that the two orbitals do not have any overlap and 1 means that they perfectly overlap with each other. Table is used to determine whether the wavefunction is symmetric or antisymmetric.

{| class="wikitable" style="margin: auto;"
! center; style="background: #0D4F8B; color: white; text-align: center;"| Symmetric
! center; style="background: #0D4F8B; color: white; text-align: center;" | Antisymmetric
|-
| center;|ψ(r<sub>1</sub>,r<sub>2</sub>)={{fontcolor1|red|'''+'''}}ψ(r<sub>2</sub>,r<sub>1</sub>)
| center;|ψ(r<sub>1</sub>,r<sub>2</sub>)={{fontcolor1|red|'''-'''}}ψ(r<sub>2</sub>,r<sub>1</sub>)
|}

For two wavefunctions, the overlap integral will be 0 if the product of two wavefunctions is antisymmetric. If the product of two wavefunctions is symmetric, the overlap integral will be 1. Only when '''both of the wavefunctions are antisymmetric''' or '''both are antisymmetric''', the overlap integral will be 1. As shown in MO diagram, the antisymmetric orbital will overlap with the antisymmetric orbital.

===Analysis of bond lengths===
{| class="wikitable" style="margin: auto;"
|center; | [[file:MG5715_TS_E1_IRC_DISTANCE.jpg|thumb|none|500px|x500px|center|Fig. Disatance VS reaction coordination]]
|center; | [[file:MG5715_TS_E1_LABEL.PNG|thumb|none|500px|x500px|center|Fig. Labeled molecule]]
|+ Table. C
|}

In Fig., it shows that the distances between C atoms change with reaction coordination. and Fig. label the molecule. Initially, the distance between C<sub>14</sub> and C<sub>1</sub> and distance between C<sub>7</sub> and C<sub>11</sub> are around 3.4 Å, which indicates there does not have any bond between those two atoms. The Van der Waal radius of C atom is 1.77Å <ref>Batsanov, S. S. (2001). Van der Waals Radii of Elements. Inorganic Materials Translated from Neorganicheskie Materialy Original Russian Text, 37(9), 871–885. http://doi.org/10.1023/A:1011625728803</ref> 3.4 Å is smaller than twice of Van der Waals radius of C; therefore, C<sub>14</sub> is interacting with C<sub>1</sub> and a bond will be formed between them.

{| class="wikitable" style="margin: auto;"
|
! center; style="background: #0D4F8B; color: white;"| C(sp<sup>3</sup>)-C(sp<sup>3</sup>)
! center; style="background: #0D4F8B; color: white;"| C(sp<sup>3</sup>)-C(sp<sup>2</sup>)
! center; style="background: #0D4F8B; color: white;"| C(sp<sup>2</sup>)-C(sp<sup>2</sup>)
|-
| center;| Bond length/Å
| 1.542
| 1.488
| 1.459
|+ Table. Bond length of C-C bond<ref>Bernstein, H. J. (1961). BOND DISTANCES IN HYDROCARBONS * t. Retrieved from http://pubs.rsc.org/-/content/articlepdf/1961/tf/tf9615701649</ref>
|}

Table . shows the literature value of C-C bond length with different hybridisation and comparing those with the Fig. It suggests C<sub>1</sub>, C<sub>7</sub>,C<sub>11</sub> and C<sub>14</sub> change from sp<sup>2</sup> to sp<sup>3</sup>; C<sub>4</sub> and C<sub>6</sub> remain sp<sup>2</sup>. Besides that, this table also confirms the product is hexene. The distance between C<sub>4</sub> and C<sub>6</sub> is 1.338 Å, which is similar to the bond lenght of C(sp<sup>2</sup>)-C(sp<sup>2</sup>). The rest of bond lengths are around 1.5 Å; so the rest of bonds are C(sp<sup>3</sup>)-C(sp<sup>3</sup>).

===Vibration===
{| class="wikitable" style="margin: auto;"
! center; style="background: #0D4F8B; color: white;" | Transition State vibration
! center; style="background: #0D4F8B; color: white;" | Inrinsic Reaction Coordinate(IRC)

|-
| center;| <jmol>
<jmolApplet>
<color>white</color>
<size>300</size>
<script>frame 23; vibration 1;rotate x -20; </script>
<uploadedFileContents>MG5715_TS_E1_TS_VIB.LOG</uploadedFileContents>
</jmolApplet>
</jmol>
| [[file:MG5715_TS_E1_IRC.gif‎]]

|-
| center;| [[file:MG5715_TS_E1_VF.PNG|thumb|none|500px|x500px|center|Fig. Vibration Frequencies of Transition State]]
| [[file:MG5715_TS_E1_IRC_PATH.PNG|thumb|none|500px|x500px|center|Fig. sumarry of IRC plot ]]

|}

As shown in Fig., there is only one imaginary frequency (at -948.8 cm<sup>-1</sup>); so it justifies the transition state is right. However, it is not enough for determining the transition state. It is critical to do the IRC calculation. RMS Gradient should be zero in reactants, products and also transition state because they are local minima. However, for transition state, it is saddle point; hence, the second derivative should be negative.

According to the gif shown above, it shows that bonds are formed '''synchronously'''.

===LOG file===

==Excercise 2:Reaction of Cyclohexadiene and 1,3-Dioxole==

[[File:MG5715_TS_E2_REACTION.jpg|thumb|550px|center|Fig. The reaction scheme of Cyclohexadiene and 1,3-Dioxole]]

The reaction of cyclohexadiene and 1,3-Dioxole is also a Diels-Alder reaction, but this reaction will give two products (endo and exo) because of two possible transition states. The reaction scheme is shown in Fig. In this exercise, the transition state is opitimised by {{fontcolor1|blue|'''PM6'''}} firstly and then optimised again by {{fontcolor1|blue|'''B3LYP'''}}. Moreover, the transition state is determined by {{fontcolor1|blue|'''method 3'''}} in this exercise.

=== MO analysis of exercise 2===

[[File:MG5715_TS_E2_MO.jpg|thumb|550px|center|Fig. MO diagram of Cyclohexadiene and 1,3-Dioxole]]

The MO diagram of this reaction is displayed in Fig.The energies of product's MO is shown in '''black''' lines and the {{fontcolor1|#fe7af9|'''pink'''}} lines represent the energies of transition states' MO, and all of them are drawn based on the relative energies calculated by '''B3LYP'''. Furthermore, the HOMO and LUMO of reactants are shown in Table. and the MOs of transition states are displayed in Table.

{| class="wikitable" style="margin: auto;"
|
! style="background: #0D4F8B; color: white; text-align: center;"| HOMO
! style="background: #0D4F8B; color: white; text-align: center;" | LUMO

|-
| Clycohexadiene

| <jmol>
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</jmol>
| <jmol>
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<uploadedFileContents>MG5715_TS_E2_REACTANTS_1.LOG</uploadedFileContents>
</jmolApplet>
</jmol>

|-
| 1,3-dioxole
| <jmol>
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|<jmol>
<jmolApplet>
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<uploadedFileContents>MG5715_TS_E2_REACTANTS_2.LOG</uploadedFileContents>
</jmolApplet>
</jmol>
|+ Table . LUMO and HOMO of reactants
|}

{| class="wikitable" style="margin: auto;"
|
! style="background: #0D4F8B; color: white; text-align: center;"| ENDO TS
! style="background: #0D4F8B; color: white; text-align: center;" | EXO TS

|-
| LUMO+1
| <jmol>
<jmolApplet>
<color>white</color>
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</jmol>

| <jmol>
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<uploadedFileContents>MG5715_TS_E2_EXO.LOG</uploadedFileContents>
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</jmol>

|-
| LUMO
| <jmol>
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</jmol>

|<jmol>
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<uploadedFileContents>MG5715_TS_E2_EXO.LOG</uploadedFileContents>
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</jmol>

|-
| HOMO
| <jmol>
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</jmol>

|<jmol>
<jmolApplet>
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<script>frame 20; mo 41; mo nodots nomesh fill translucent; mo titleformat ""; set antialiasdisplay on</script>
<uploadedFileContents>MG5715_TS_E2_EXO.LOG</uploadedFileContents>
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</jmol>

|-
| HOMO-1
| <jmol>
<jmolApplet>
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<script>frame 48; mo 40; mo nodots nomesh fill translucent; mo titleformat ""; set antialiasdisplay on</script>
<uploadedFileContents>MG5715_TS_E2_ENDO.LOG</uploadedFileContents>
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</jmol>

|<jmol>
<jmolApplet>
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</jmol>

|+ Table . MO of Transition states
|}

====Secondary Orbital Interactions ====
{| class="wikitable"
|
! style="background: #0D4F8B; color: white;" | Endo-TS
! style="background: #0D4F8B; color: white;" | Exo-TS
|-
| Jmol of computed HOMO

|<jmol>
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|<jmol>
<jmolApplet>
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|-
| LCAO diagram of HOMO

| [[File:MG5715_TS_E2_ENDO_TS.jpg|thumb|550px|center]]

| [[File:MG5715_TS_E2_EXO_TS.jpg|thumb|550px|center]]
|+ Table. HOMO of Endo-TS and Exo- TS
|}

The computed HOMO and LACO diagram of HOMO are shown in Table. Endo-product is more stable than Exo-product due to lack of steric clash with the methyl group in cyclohexadiene. Furthermore, the Endo-TS is lower in energy because of '''Secondary Orbital Interaction'''. The lone pair of oxygen could donate the electron density to the LUMO of cyclohexadiene; hence, the Endo-TS will be stabilised. Both the Endo-TS and Endo-product are lower in energy; so this reaction is endo-selective.

==== Inverse VS Normal Demand Diels-Alder Reaction ====

[[File:MG5715_TS_E2_DA.jpg|thumb|550px|center|Fig. two tyes of Diels-Alder reaction]]

For a Diels-Alder reaction, the reactivity is controlled by relative energies of '''Frontier Molecular Orbitals (FMOs)'''. There are two different types as shown in Fig.:(1) Normal demand and (2) Inverse demand. Usually, the reaction of an all carbon diene with a hetero-dienophile will give a ''normal electron demand'' because the heteroatom in dienophile may withdraw the electron density from dienophile and lower the energy of MOs. The best way to justify whether the reaction is normal demand or inverse demand is to sequence the energy of reactants as shown in Table. If the HOMO of dienophile is lower than that of diene, it is normal demand, vice versa. The HOMO of cyclohexadiene is the lowest; hence, this reaction is {{fontcolor1|blue|'''inverse demand'''}} Diels-Alder reaction. The reason is that the two oxygen atom in 1,3-dioxole donate the electron density toward the π-bonding and raise the energy of MOs.

{| class="wikitable" style="margin: auto;"
! style="background: #0D4F8B; color: white;" | Endo-reaction
! style="background: #0D4F8B; color: white;" | Exo-reaction

|-

|<jmol><jmolApplet>
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|<jmol><jmolApplet>
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|-

|<jmol><jmolApplet>
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<uploadedFileContents>MG5715_TS_E2_EXO_SINGLEPOINT.LOG</uploadedFileContents>
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|-

|<jmol><jmolApplet>
<color>white</color>
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|<jmol><jmolApplet>
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|-

|<jmol><jmolApplet>
<color>white</color>
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|<jmol><jmolApplet>
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|+ Table. Single point energy of reactants
|}

=== Activation Barrier and Energy changed ===
{| class="wikitable" style="margin: auto;"
|
! style="background: #0D4F8B; color: white;" | Energy calculated by PM6/kJmol<sup>-1</sup>
! style="background: #0D4F8B; color: white;" | Energy calculated by B3LYP/kJmol<sup>-1</sup>
|-
| Cylcohexadiene
| 306.9
| -612593

|-
| 1,3-dioxole
| -137.3
| -701189

|-
| Endo-product
| 99.2
| -1313849

|-
| Endo-TS
| 362.2
| -1313622

|-
| Exo-product
| 99.7
| -1313845

|-
| Exo-TS
| 364.7
| -1313614
|+ Table. Energies of reactants, products and transition states
|}

According to the energies of reactants, products and transition states (shown in Table), the reaction energy and activation energy could be caluclated by following equation:
<pre>
Activation Energy = Transition state energy - Sum of energies of reactants
Reaction Energy = product energy - Sum of energies of reactants
</pre>

{| class="wikitable" style="margin: auto;"
! rowspan="2"|
! style="background: #0D4F8B; color: white;" colspan="2"| Activation Energy/kJmol<sup>-1</sup>
! style="background: #0D4F8B; color: white;" colspan="2"| Reaction Energy/kJmol<sup>-1</sup>
|-
! style="background: #0D4F8B; color: white;"| PM6
! style="background: #0D4F8B; color: white;"| B3LYP
! style="background: #0D4F8B; color: white;"| PM6
! style="background: #0D4F8B; color: white;"| B3LYP
|-
| Endo-reaction
| +192.6
| +159.8
| -70.3
| -67.4
|-
| Exo-reaction
| +195.1
| +169.7
| -69.9
| -63.8
|+ Table. Reaction energy and activation energy for both Endo and Exo product.
|}

All the energy calculated is shown in Table. Both the activation energy and reaction energy for Endo-product are lower than those for Exo-product. A thermodynamically favourable product is more stable, which means it has more negative reaction energy. The kinetically favourable product has lower reaction barrier (activation energy). Hence, the endo-product of this exercise is both '''thermodynamically and kinetically favourable'''.

==Excercise 3 Diels-Alder vs Cheletropic==

[[File:MG5715_TS_E3_REACTION.jpg|thumb|550px|center|Fig. Reaction Scheme of o-Xylylene with SO<sub>2</sub>]]

The reaction of o-xylylene with SO<sub>2</sub> could be either Diels-Alder or cheletropic reaction. For Diels-Alder reaction, o-Xylylene acts as diene and SO<sub>2</sub> acts as dienophile to form a six-membered ring and there are two possible products (endo-product and exo-product). As discussed in exercise 2, the Diels-Alder is endo-selective. For the cheletropic reaction, a five-membered ring will be formed and two new bonds are formed with the same atom. The reaction scheme of those reactions is displayed in Fig. In this exercise, the reactants, products and transition states were optimised by {{fontcolor1|blue|'''PM6'''}} and {{fontcolor1|blue|'''B3LYP'''}} and {{fontcolor1|blue|'''Method 3'''}} was used to find the transition state. The reaction energy and activation energy were calculated to find out the favourable reaction.

=== IRC Analysis===
{| class="wikitable" style="margin: auto;"
|
! style="background: #0D4F8B; color: white;" | IRC
|-
| Cheletropic reaction
| [[File:MG5715_TS_E3_Cheletropic_IRC.gif]]
|-
| Endo Diels-Alder reaction
| [[File:MG5715_TS_E2_ENDO_irc.gif]]
|-
| Exo Diels-Alder reaction
| [[File:MG5715_TS_E3_EXO_irc.gif]]
|+ Table. The IRC of three possible reactions
|}

There are two different mechanisms for Diels-Alder reaction:(1)synchronous and symmetrical (concerted) mechanism and (2) multistage (non-concerted) and asynchronous mechanism.

* synchronous and symmetrical (concerted) mechanism
A very typical example of this mechanism is Exercise 1. The two new bonds are formed simultaneously and the two new bonds have the same bond length in the transition state.

* multistage (non-concerted) and asynchronous mechanism
The two bonds could not be formed at the same time because the bond length of them are different at transition state. Therefore, usually, a di-radical will be formed.

For Diels-Alder reaction, the distance between O and C in o-Xylylene is different from the distance between S and C in o-Xylylene at transition state because of distinct Van der Waal radii.
Therefore, the bond does not form synchronously, which could be observed in the IRC in Table. However, for cheletropic reaction, the two now bonds are both formed between S atom and C in o-Xylylene; so the two bond are formed synchronously as shown in IRC.

=== Activation Energy and Reaction Energy===
[[File:MG5715_TS_E3_ENERGY.jpg|thumb|550px|center|Fig. Energy diagram of o-Xylylene with SO<sub>2</sub>]]

{| class="wikitable" style="margin: auto;"
|
! style="background: #0D4F8B; color: white;" | Energy calculated by PM6/kJmol<sup>-1</sup>
! style="background: #0D4F8B; color: white;" | Energy calculated by B3LYP/kJmol<sup>-1</sup>
|-
| O-Xylylene
| 469.5
| -812604

|-
| SO<sub>2</sub>
| -311.4
| -1440362

|-
| Endo-product
| 57.0
| -2253040

|-
| Endo-TS
| 237.8
| -2252919.7

|-
| Exo-product
| 56.3
| -2253041

|-
| Exo-TS
| 241.7
| -2252919.8

|-
| Cheletropic product
| -0.005251
| -2253024

|-
| Cheletropic TS
| 260.1
| -2252902
|+ Table. Energies of reactants, products and transition states
|}

{| class="wikitable" style="margin: auto;"
! rowspan="2"|
! style="background: #0D4F8B; color: white;" colspan="2"| Activation Energy/kJmol<sup>-1</sup>
! style="background: #0D4F8B; color: white;" colspan="2"| Reaction Energy/kJmol<sup>-1</sup>
|-
! style="background: #0D4F8B; color: white;"| PM6
! style="background: #0D4F8B; color: white;"| B3LYP
! style="background: #0D4F8B; color: white;"| PM6
! style="background: #0D4F8B; color: white;"| B3LYP
|-
| Endo-reaction
| +79.7
| +46.2
| -101.1
| -73.9
|-
| Exo-reaction
| +83.7
| +46.1
| -101.8
| -73.2
|-
| Cheletropic
| +102.0
| +63.0
| -158.1
| -58.8
|+ Table. Reaction energy and activation energy for both Endo and Exo product.
|}

The energies of reactants, products and transition states are shown in Table and the calculated reaction energy and activation energy are displayed in Table. The energy diagram (Fig.) is drawn according to the energy calculated by PM6. The product of the cheletropic reaction is the most stable because of aromaticity. However, the activation energy of cheletropic product is quite high. Therefore, at low temperature, a sultine (kinetic product) will be formed but this reaction in reversible. At high temperature, a sulfolene will be formed and this reaction is irreversible.

=== Extension===
[[File:MG5715_TS_E3_EXTENSION_SCHEME.jpg|thumb|550px|center|Fig. Reaction Scheme of o-Xylylene with SO<sub>2</sub>]]
There are two dienes in o-Xylylene and therefore, it has another possible Diels-Alder reaction as shown in Fig. The energy calculated is displayed in Table and they are calculated by PM6.

{| class="wikitable" style="margin: auto;"
|
! style="background: #0D4F8B; color: white;" | Energy calculated by PM6/kJmol<sup>-1</sup>
|-
| O-Xylylene
| 469.5

|-
| SO<sub>2</sub>
| -311.4

|-
| Endo-product
| 172.3

|-
| Endo-TS
| 268.0

|-
| Exo-product
| 176.7

|-
| Exo-TS
| 275.8
|+
|}

{| class="wikitable" style="margin: auto;"
|
! style="background: #0D4F8B; color: white;" | Activation Energy/kJmol<sup>-1</sup>
! style="background: #0D4F8B; color: white;" | Reaction Energy/kJmol<sup>-1</sup>
|-
| Endo-reaction
| +109.9
| +14.2

|-
| Exo-reaction
| +117.7
| +18.6

|+ Table. Reaction energy and activation energy for both Endo and Exo product.
|}

Compare the energy in Table. with the data in Table, this reaction is quite unlikely to happen because the activation barrier is higher and the reaction energy is endothermic.

==Conclusion==

==Reference==