Rep:Mod:TSML16

Year 3 Computational labs: Transition States and Reactivity

Introduction

Minimum: The minimum is the lowest point on a potential energy surface. The gradient at this point is zero and the curvature is negative. There should not be any imaginary frequencies for molecules at the minimum.

Transition state: The highest energy point on a potential energy surface along the reaction coordinate. At this point the gradient is zero and the curvature is positive. It can be confirmed that it is a transition state if there is an imaginary frequency.

Nf710 (talk) 09:47, 4 November 2016 (UTC) Well defined, could have put some info about the methods in here.

Exercise 1: Reaction of Butadiene with Ethylene

Figure 1: MO diagram of reaction between Butadiene and Ethene

The Molecular Orbital shows a normal electron demand Diels-Alder reaction. The gerade orbitals are closer in energy therefore have a greater interaction hence greater splitting than the ungerade orbitals. This results in the ungerade HOMO and LUMOs. ^1,2

Figure 2: The HOMO of Butadiene	Figure 3: The LUMO of Butadiene
This MO correlates to the HOMO of Butadiene in Figure 1	This MO correlates to the LUMO of Butadiene in Figure 1
Figure 4: The HOMO of Ethene	Figure 5: The LUMO of Ethene
This MO correlates to the HOMO of Ethene in Figure 1	This MO correlates to the LUMO of Ethene in Figure 1

These Molecular Orbitals resemble the Molecular orbitals drawn on Figure 1 hence shows a correlation between the theoretical and computed models.

Figure 6: The HOMO of the transition state of the reaction between Ethene and Butadiene	Figure 7: The LUMO of the transition state of the reaction between Ethene and Butadiene

Both the HOMO and the LUMO of the transition state shows molecular orbitals of ungerade symmetry. This correlates to the HOMO and LUMO drawn in figure 1. These transition state molecular orbitals are formed by the combination of the LUMO of Butadiene and the HOMO of Ethene. These are both of ungerade symmetry. This is evidence that two orbitals must be of the same symmetry in order to interact. The orbital overlap of ungerade-ungerade and gerade-gerade orbitals are non-zero. The orbital overlap of ungerade-gerade orbitals are zero. ³

Changing bond lengths

Figure 8: Bond labels that correspond to figure 9

C-C bond lengths of reactants

Carbon	Reactants	Transition state	Product
C1	1.32731	1.38607	1.54076
C2	1.33341	1.38307	1.50034
C3	1.47080	1.40721	1.33766
C4	1.33344	1.38307	1.50034
C5	n/a	2.27216	1.50034
C6	n/a	2.27215	1.50034

Figure 9: Tabulated changes in C-C bond lengths throughout the reaction.

	Typical values of C-C bonds/Å
Sp3-Sp3	1.54
Sp3-Sp2	1.50
Sp2-Sp2	1.46

Figure 10: Tabulated typical C-C bonds⁴

C3 decreases throughout the reaction and the double bond is formed. At the end of the reaction C3 forms a Sp2-Sp2 bond. The bond length is similar to the typical values of a Sp2-Sp2 bond. The discrepancy is due to the inaccuracy of the bond measurement tool on Gaussian. C1,C2 and C4 increases throughout the reaction. C5 and C6 are the Sp3-Sp3 bonds that are formed in the Diels-Alder reaction. The bond lengths formed are similar to typical Sp3-Sp3 bond length values. It can be seen that if the C-C bonds is closer to the double bond it has more Sp2 character therefore the bond lengths are slightly shorter than the typical Sp3-Sp3 bond.⁴

Nf710 (talk) 10:04, 4 November 2016 (UTC)Your analysis of bond lengths is excellent however you havent shown any frequency analysis and therefore havent showning if the reaction is synchronous or a synchronous.

Exercise 2: Reaction of Benzoquinone with Cyclopentadiene

Figure 11: MO diagram of reaction between Benzoquinone and Cyclopentadiene

This MO diagram in figure 11 is very similar to that of the MO between Butadiene and Ethene. However the HOMO and LUMO of Benzoquinone is lower in energy compared to that of Ethene. This is due to the electron withdrawing oxygens on the Benzoquinone which make the double bond more electron deficient. This reaction should still be a normal electron demand reaction as the dienophile is electron-poor and the diene is electron-rich. ¹

Nf710 (talk) 10:07, 4 November 2016 (UTC) Good understanding of the theory

Endothermic Diels-Alder reaction

(Not "endothermic" or "exothermic"! Tam10 (talk) 15:31, 27 October 2016 (BST))

Figure 12: The HOMO and LUMO of the Endothermic Diels-Alder reaction transition state.

	Sum of electronic and thermal free energies (Hartree/particle)
Reactants	-575.431613
Transition state	-575.383855
Product	-575.426694
Reaction Barrier	0.047758
Reaction Energy	0.004919

Figure 13: Tabulated energies throughout the reaction

Nf710 (talk) 10:12, 4 November 2016 (UTC) Make sure you specify the level of theory!

Exothermic Diels-Alder reaction

Figure 14: The HOMO and LUMO of the Exothermic Diels-Alder reaction transition state.

	Sum of electronic and thermal free energies (Hartree/particle)
Reactants	-575.431613
Transition state	-575.381307
Product	-575.427052
Reaction Barrier	0.050306
Reaction Energy	0.004561

Figure 15:Tabulated energies throughout the reaction

Figure 12 and 14 show that the HOMO and LUMO of both the transition states of the exothermic and endothermic Diels-Alder reaction have a gerade symmetry. This suggests that the reaction has occured by inverse electron demand. This is where the gerade MOs formed by the HOMO of Cyclopentadiene and the LUMO of Benqzoquinone are further apart in energy than the ungerade MOs formed by LUMO of Cyclopentadiene and the HOMO of Benzoquinone. Due to the larger energy difference of the gerade orbitals they will split less causing them to become the HOMO and LUMO. This discrepancy between the theory and the computationally obtained models can be due to the interactions between the HOMO and LUMO and the molecular orbitals of lower energies. This causes the computationally calculated molecular orbitals to be inaccurate. ³

Figure 13 and 15 show tabulated energy values of molecules throughout the reaction in Hartree/Particles. The reaction energy of the exothermic reaction is greater than that of the endothermic reaction. This implies that the exothermic reaction is more thermodynamically favorable. ³

The energy barrier of the endothermic reaction is slightly lowered due to the secondary orbital interactions. Therefore the Endothermic reaction has a lower reaction barrier than the exothermic reaction and hence is more kinetically favorable.³

Nf710 (talk) 10:45, 4 November 2016 (UTC) (Some confusion here. Your Transition state energies are corrrect but your calculated reactants and products are incorrect so your energy barriers and reaction energies are incorrect. You have correctly said the endo is the kenetic product and this is because of the SOO interaction in the TS. )

Exercise 3: Diels-Alder vs Cheletropic

Exothermic Diels-Alder reaction

Figure 16: Shows the Exothermic Diels-Alder reaction Intrinsic reaction coordinate. It can be seen in figure 15 that the IRC path was computed from the product to the reactants which is why the path starts from high energy and ends in a lower energy.

	Sum of electronic and thermal free energies (Hartree/particle)
Reactants	0.083511
Transition state	0.092077
Product	0.021455
Reaction Barrier	0.008566
Reaction Energy	0.062056

Figure 17: Tabulated energies throughout the reaction

Endothermic Diels-Alder reaction

Figure 18: Shows the intrinsic reaction coordinate of the Endothermic Diels-Alder reaction. It can be seen in figure 17 that the IRC path was computed from the reactants to products as the path starts at higher energy and ends at a lower energy.

	Sum of electronic and thermal free energies (Hartree/particle)
Reactants	0.087746
Transition state	0.090559
Product	0.021704
Reaction Barrier	0.002813
Reaction Energy	0.066042

Figure 19: Tabulated energies throughout the reaction

(The data would be easier to see if it was stacked together (DA endo/exo and Cheletropic)

Cheletropic reaction

Figure 20: Shows the intrinsic reaction coordinate of the Cheletropic reaction. In Figure 19 it can be seen that the IRC path was computed from the products to the reactants as the path starts from lower energy and goes to higher energy.

	Sum of electronic and thermal free energies (Hartree/particle)
Reactants	0.071004
Transition state	0.099063
Product	0.000002
Reaction Barrier	0.028059
Reaction Energy	0.071002

Figure 21: Tabulated energies throughout the reaction

Figure 22: Reaction profile obtained from the computed reaction energies

For all reactions the transition state has the highest energy and the product has the lowest energy. All three reaction profiles look similar to that of Figure 22.

(But how do they look when placed together? After all you are comparing them. Transition states will always have the highest energy along a reactant-product reaction coordinate! Tam10 (talk) 15:36, 27 October 2016 (BST))

By looking at figures 17, 19 and 21 it is evident that the Cheletropic reaction has the highest activation energy and the endothermic Diels-Alder reaction has the lowest activation energy. This suggests that the endothermic reaction is the most kinetically preferred. The Cheletropic reaction also has the greatest Reaction energy therefore is also the most thermodynamically preferred.

By looking at Figures 16, 18 and 20 it can be seen that during the course of the reaction in all three reactions an aromatic 6 membered ring is formed from a diene. This is one of the causes for the stability of the product.

(Last two paragraphs are good Tam10 (talk) 15:36, 27 October 2016 (BST))