Jn307Module2:testbed
MINI PROJECT 2
In this investigation, i shall be investigating the bonds lengths and vibrations of the Tantalum complex in a Schrock Alkylidenes. in a Schrock Alkylidenes, the double bond is formed by the backbonding from the p orbital into an empty d orbital orbital of the metal. as well as this, one electron will be donated to the metal in a sigma fashion to a p orbital on the metal. this would form a strong M=C bond. i shall investigate changing the R group on the molecule below and to investigates how the bond lengths, angle and the frequency of the IR of the CO bond changes as the R groups increases. The spin of all molecules will thus be a triplet spin as there are two spins of unpaired electrons.
Schrock Allidenes have some conditions in order for it to not be a fisher carbene. The metal must have a high oxidation number, prefable a early transition metal though there are some exceptions and pi-donor ligands which are usually alkyl. Electron withdrawing groups would distrupt the strong M=C bond and thus would get either a single bond or a fischer carbene.
Preliminary work.
Initially the decision was to have extremes R groups in order to better quantify how the molecules will changed with increasing sterics affects. the groups initially chosen was to have R=H, 6 carbons straight and branched and then investigate a pentadiene, a benzene and a substituted benzene ring. the conditions set in gaussview as:
Method: DFT Basis Set: B3LYP Additional Keywords: opt=loose pseudo-potential = LANL2MB Maxcycle =50
However, after waiting for the molecules to optimised, it was discovered that not all the values was converged except for when R=H. I then had to rethink, and believed that i have overextended myself too much by drawing bigger molecules that gaussview could not handle.
Imaged of an unsuccessful optimision when R=benzene: DOI:10042/to-3450
i then changed my method in making R=H,Ch3, Ch2ChH3, CH2CH2CH3, CH(CH3)CH2CH3, CH2CH(CH3)CH3. this would allow me to investigate the affect of increasing the R groups in the sChrock complexes. These molecules became optimised using the above method shown. It is now possible to investigate how by either changing the length or making a branch on an alpha or beta carbon will affect the molecular complex.
After the rough optimition has occured, i then created another optimising gfl file with the following conditions and ran on the Scan.
Method: B3LYP Additional Keywords: int=ultrafine scf=conver=9 pseudo-potential: LANL2DZ Maxcycle =50
As there are 3 different scans for 6 different molecules, the time and resources that we have are not enough in order to be able to measure the energies of the singlet and triplet spin of two identical molecules. However, i was able to scan the first optimision when R=H of which the energy of the triplet scan was slightly lower. Therefore, i have decided to use the triplet spin throughout my investigation. After Each optimization, the data was checked in word pad to check that all the values has converged before moving on to the next part. For R=H, the MO and NBO was also investigated because as the smallest molecule, it would be the quickest to scan.
Bond Lengths
After the second optimisation, the bond lengths were checked to see whether the bond remains a double bond compared with the methyl groups. the data is shown in the table below:
| R group | bond length of Ta=C | Bond length Ta-C | Difference | angle C-M=C |
|---|---|---|---|---|
| H | 2.14773 | 2.23860 | 83.147 | |
| CH3 | 2.15029 | 2.23914 | 86.521 | |
| CH2CH3 | 2.15172 | 2.23926 | 86.711 | |
| CH2CH2CH3 | 2.15155 | 2.23965 | 86.521 | |
| CH(CH3)CH2CH3 | 2.15457 | 2.23858 | 86.711 | |
| CH2CH(CH3)CH3 | 2.15225 | 2.23931 | 86.554 |
it can be seen from the table above that as the chain is increased in a linear fashion, the bond length of Ta=C gradually increases, this may be because as the chain increases, the bond length needs to increases in order to minimise sterics effects. the backbonding to the carbon may have also increases with an increasing linear chain which would also destablised the Ta=C slightly. As well as the Ta=C bond increasing, it can be seen that the Ta-C also increases as the linear chain increases, this may be because of the increasing backbonding from the metal to the carbon single bond which would result in a higher bond length. The angles also increases as the linear chain increases, this is also due to the increases steric bulk the molecule felt but the molecule needs to be careful not to be too close to the two cyclopentadiene rings cis to them.
On average, the bond difference between both bonds are between 0.15 a.u. i believe that this is an measurable difference between the double bond and the single bond bonded to the Ta but an NBO analysis of when R=H would help distinguish between the two different Schrock Alkylidenes molecules. Therefore a NBO with the conditions of:
NBO=FULL
was performed after the second optimisation.
NBO when R=H
The NBO when R=HDOI:10042/to-3448 and ch2ch2ch3DOI:10042/to-3447 was investigated and compared with each other. This would allow me to see the nuclephilicity of the Ta=C carbon atom.
 |
 |
The NBO when R=H and when R=CH2CH2CH£ shows a drastic change in the polarity of the nucleophilic carbon. When R=H, the carbon double bonded to the metal Ta is -0.758 however when R=CH2CH2CH3, the carbon has a decreased polarity of -0.451. This is a large change and represents the idea that the carbon electron density is becoming more diffused and hence a poorer nucleophile against electrophiles. As the R groups increases, the reaction with the nucleophilic carbon will become slower, this may also be in part to the increased stability a bulky group would have on the R group of the carbon because of the spread of the electron density.
 |
 |
The Summary text file associated with the log file shows an interesting development. The methyl group itself appears to be a sp2 orbital while, the double bonded Ta-C appears to be more like a sp orbital. nevertheless, a difference can clearly be seen in the hybridised orbitals of the molecule. This may also show that the double bonded Ta=C may exhibit some triple bond characteristics and the methyl may form some kind of single bond/ double bond overlap, which also supports the argument that the carbon would be more nuclephilic that due to the orbital overlap shown above. Comparison of the two groups shows is given in the table below:
| R group | Atom | s contriubution/% | p contribution /% |
| H | C= | 46.11 | 53.01 |
| C- | 32.38 | 67.62 | |
| CH2CH2CH3 | C= | 43.06 | 56.02 |
| C- | 32.52 | 67.48 |
The data above shows the hybridized orbitals on the double bonded C= and the methyl group. As it can be seen, as we increased the R group substitutes on the double bonded carbon, the s character contributes less while the contribution of the p orbital increases. Earlier, it was found that the bond lengths increased as the R groups increased, the electron density in the nucleophilic carbon also decreases as it becomes more diffused along the chain carbons. Therefore, it would be higher up in a molecular orbital diagram as the bond between the Ta and the carbon becomes weaker, which would result in less contribution from the s orbital. The methyl group has no noticeable change in contribution to the Ta metal complex.
Summary of data optimized and frequency analysis
| R Group | H | CH3 | CH2CH3 | CH2CH2CH3 | CH(CH3)CH2CH3 | CH2CH(CH3)CH3 |
| 1st Optimization | DOI:10042/to-3360 | DOI:10042/to-3363 | DOI:10042/to-3365 | DOI:10042/to-3368 | DOI:10042/to-3370 | DOI:10042/to-3372 |
| 2nd Optimization | DOI:10042/to-3361 | DOI:10042/to-3364 | DOI:10042/to-3366 | DOI:10042/to-3369 | DOI:10042/to-3371 | DOI:10042/to-3373 |
| Frequency analysis | DOI:10042/to-3362 | DOI:10042/to-3367 |
MO when R=H
In the Optimisation of all the molecules, i was unable to procure molecular orbitals due to using the triplet spin state instead of the singlet. If more time was given the molecular orbitals between the chain's R group are of significant interest in order to see how a secondary or tertiary carbon would affect the molecular orbital overlap between the Ta=C, this would also allow us to view the energies of the HOMO and LUMO and to see how the stabilization will be affected by changing the position of the methyl group.
Vibrational differences when R= H, Ch3 and CH2CH2CH3
| R=H | R=Ch3 | R=CH2CH2CH3 | |
 |
 |
 | |
| Frequency/cm-1 | 200.15 | 193.65 | 177.16 |
| Intensity/cm-1 | 1.1435 | 0.9380 | 1.1593 |
 |
 |
 | |
| Frequency/cm-1 | 474.20 | 478.35 | 476.77 |
| Intensity/cm-1 | 12.1928 | 17.0089 | 13.1083 |
 |
 |
 | |
| Frequency/cm-1 | 641.48 | 613.02 | 705.02 |
| Intensity/cm-1 | 34.9699 | 10.2854 | 6.4633 |
The graphs tries to show a graphical representation of the methyl and the R groups vibrational movement as we increases the R group. in order to chart the stability of the molecule, a constant was sought which was the methyl group cis to the R group. When we changed the R group from H to Ch3, a drastic change can be seen in the frequency. The middle one is associated with the movement of the C-H in the methyl groups, as we changed from a H to a CH3, the intensity increases by around 5 a.u. and the frequency increases indicating a longer bond and hence a more destablised bond. However when we add two carbons to the CH3 groups, the C-H becomes lower in frequency and the intensity slightly increases, in some ways, the bond becomes stronger and more rigid, this may be because due to the -CH3 slightly further out as well as the cyclopendaidene on the other side, the methyl must conform closer to the Ta metal and hence cannot be as free as it wants to be due to sterics. The other frequencies does not tell much about the different stability of the molecules, although, from H to CH£ there is a much lower intensity in the frequency which may be because of the increased sterics around the molecule which prohibits the movement of the atoms.
In Conclusion, it can be seen that as the R groups increases, the bond length of C=O generally increases to limit steric interaction, the NBO also shows a more diffused carbon which suggests that the carbon has become less nuclephilic and therefore cannot an electrophile as well as it can. However, with a diffused electron density, the bond should be more stable and thus cannot degrade as easily allowing for more complicated reactions to occur. If time would allow me, i shall uses the singlet and see how benzene MO diagram would affect the Schrock Alkylidenes.