Mod:Lab Toolbox

Basis Sets

In this lab, a functional has been provided for you (B3LYP) and you much select which basis set to use (3-21+G, 6-31++G(d,p), 6-311++G(d,p)).

The B3LYP functional has been selected since due to its hybrid nature and therefore potential for more accurate results. This functional is, however, not good alone for modelling dispersion interactions and therefore either a dispersion term must be added to the end e.g. B3LYP-D3 or a new functional must be selected. For this work, we are not looking at the effects of dispersion however they are important in many systems.

There are various factors which need to be considered when selecting a basis set for calculations. Ideally, you will read various literature surrounding the subject, evaluating the suitability of the basis set for your computational investigations. Several different basis sets exist and you must determine what differentiates each basis set from another and make a judgement on which is best to use for the particular task. Bond lengths, angles etc are a potential method for this, but there are drawbacks - some of which you should discuss in your report. The same logic applies to selecting a functional.

In this case, in addition, you will need to to consider the time it takes to run calculations using the different basis sets and whether they can be successfully completed in the time allocated on your desktop computer or in good time on the HPC. You are also advised to research the size of the basis sets and report and discuss any relevant findings.

It is best to watch the video on DFT at this point to get a better understanding of what is going on in a chemical simulation with DFT, and what a basis set is. Does the functional or basis set determine the Hamiltonian used for calculations?

Use the keyword below in all calculations:

integral=grid=ultrafine

Fructose

Fructose is a sugar which is found naturally in foods such as honey and various fruits. It is one of the sweetest sugars and is produced from starch or sucrose, amongst other methods. The Shi catalyst is also is derived from fructose.

It is well established that Fructose can exist in five different forms in solution: the open ring, α-D-fructopyranse, β-D-fructopyranose, α-D-fructofuranose and β-D-fructofuranose. Experiments determined that two of these forms dominate the equilibrium between the five forms in water. These are the β-D-fructopyranose (~68%) and β-D-fructofuranose (~32%). Dissolved in other solvents, different ratios of each of the forms dominates. . Assuming, however, that each of these forms exists in equilibrium in a vacuum, it is possible to determine which forms of fructose are the most stable at 25 °C. It can be seen from the figure below that the most stable are the pyranose forms, then the furanose and then the open chain. This allows us to determine the relative stability of the structures. It is also important to note at this point that energies can be compared because there are same number and type of atoms on each isomer, and the same level of theory (i.e. basis set and functional) has been used for all calculations. Running frequency calculations allows one to find out the energy of the molecule at 25 °C.

Fig. 1 Equilibrium of fructose. Adapted from 1 and 2

Insert energy chemdraw here of fructose.

Fructose can exist as a pair of enantiomers, L and D-Fructose. It is often more common to see these as (+) and (-)-Fructose, which is indicative of the rotation of the plane of polarised light. Whilst there are several other methods (which can also be calculated using G09) to determining whether two molecules are enantiomers, such as ECD and VCD, optical rotation is the one which can easily be carried out in the lab. Here, we have conducted an optical rotation on L-Fructose.

In order to determine the optical rotation of fructose, it is necessary to add some additional keywords to the Gaussian .gif file after it has been generated and saved, but not run.

1) The output from any prior optimisation and frequency analysis of the open chain of fructose was copied into a new Gaussview window. A calculation was set up with the following:

1a) An Energy calculation was conducted

1b) The solvation of the system was set to CPCM, and water as a solvent.

1c) The additional keywords polar(optrot) cphf=rdfreq were added to the input line.

1d) The job was saved but not run.

2) The corresponding .gif file for that molecule was opened using Notepad++

For example, the input line should be:

%chk=\\icnas4.cc.ic.ac.uk\...
# b3lyp/6-31++g(d,p) scrf=(cpcm,solvent=water) geom=connectivity
integral=grid=ultrafine polar(optrot) CPHF=RdFreq

Title Card Required

0 1
x,y,z co-ordinates and atoms
...

3) The wavelengths at which the optical rotation was taken was added at the end of the document.

end x,y,z co-ordinates
...blank line...
Other information
...blank line...
589nm

4) This job was then saved and then run on either Gaussian09/Gaussview or on the Imperial HPC.

We find our calculated result gives [α]_{589 nm} = + 131° and the literature value reported is + 95° . What could be the differences for these values, both computational and experimental?

Enantiomeric excess

Enantiomeric excess (ee) allows chemists to determine the percentage of which enantiomer is favourably produced in a chemical reaction. Methods to develop chiral catalysts which synthesise one enantiomer over another are therefore highly valued.

The ee of a reaction can also be calculated from the Gibbs Free Energy of the transition state of the reaction.

1) This can be done by calculating the difference in the transition state energies for the two diasteromeric transition states

2) Then calculating the rate constant k, and relating the result above to the rate constant.

3) Then calculating the enantiomeric excess.

Conduct research on this topic in order to find a paper which yields additional information and equations from the above. Refer to this page for additional information.