Schoolwork

Synthesis and Determination of [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃

Abstract

The reaction of mesitylene with Mo(CO)₆ under reflux yields [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ in low percent yield (around 1%). ¹H NMR of [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ shows singlets at δ 2.25 and 5.23 with absorption ratios of 9:3, respectively. ¹H NMR of mesitylene shows singlets at δ 2.25 and 6.78, also with absorption ratios of 9:3, respectively. This suggests addition of the metal complex to mesitylene causes downfield shifting of the signal for protons attached directly to the ring as they are unshielded from the backbonding of carbonyl groups. The IR spectrum of [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ shows a strong antisymmetric C-O stretch at 1852 cm^-1 and a medium symmetric C-O stretch at 1942 cm^-1 with peak areas of 64.462 cm^-1 and 9.111 cm^-1 respectively. The calculated OC-Mo-CO bond angle is 108.32°.

Introduction

The reaction of mesitylene with Mo(CO)₆ under reflux yields [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃. The reaction specifically takes place in the following manner:

Scheme 1

This is compound of interest because it a metal-arene complex and can be considered to be an octahedral rather than tetrahedral complex. This is because the OC-Mo-CO bond angles are close to 90° instead of the expect 109.5° for tetrahedrals.¹ In order to determine the structure of said substance from its ¹H NMR, the peaks must be compared to the same spectrum of mesitylene for indication of identical methyl group peaks and downfield shifting a peak indicative of protons attached directly to the ring. The IR spectrum of [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ can be analyzed for peaks indicative of symmetrical and antisymmetrical carbonyl stretching, whose areas can be used to calculate the bond angle between the carbonyl groups attached the to metal.

Experimental

All syntheses were carried out in nitrogen and the reagents and solvents were purchased from commercial sources and used as received unless otherwise noted. The synthesis of [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ (1) was based on reports published previously.¹

[1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ (1). Mo(CO)₆ (2.083 g, 7.92 mmol) and mesitylene (10 mL, 72 mmol) were added subsequently to a 100 mL 3 neck round-bottom flask along with a small magnetic stir bar. A sand bath was constructed and set to 50% power. A greased sidearm, stopcock, and 30 cm cold water condenser were attached to the round-bottom flask. A greased gas inlet was then attached to the condenser and connected to a bubbler. The condenser was not connected to a cold water source; it was used only to allow air to circulate. The sidearm was connected to a nitrogen source, and the system was allowed to degas for 5 minutes. The system was then put on the sand bath and the stir bar was spun at a moderate speed via a magnetic stirring instrument.

After 5 minutes, the solution in the round-bottom flask was not boiling as outlined, so the sand bath was turned up to 70% power. The sand bath was turned up to 85% another 5 minutes later. A rigorous boil was achieved when the sand bath was set to 95% 5 minutes after that. It was then set to 85% power in efforts to obtain a less extreme boil. After a total of 0.33 h of reflux, the solution was taken off the sand bath and allowed to cool to room temperature. When the apparatus was removed from the sand bath, it was dropped and roughly more than 60% of the solution was lost. The remaining solution cooled to a blackish yellow color.

The following and final procedures took place in the presence of air. Once cool, the solution was washed with 15 mL of hexane via suction filtration in a 15 mL frit. The solution was then washed with another 5 mL of hexane. About 10 mL CH₂Cl₂ was added to the blackish yellow powder precipitate remaining in the frit. The powder was washed with 25 to 30 mL of hexane and vacuum dried. This powder was discarded and the collected yellowish washings were rotovapped for about 0.33 h to obtain the desired product. The product was vacuum filtered, as it would not completely dry under the rotovap. The resulting yellowish powder was determined to be 1 (0.028 g, 1.18% yield based on the amount of Mo(CO)₆ used). ¹H NMR (CH₂Cl₂): δ 2.25 (s, -CH₃), 5.23 (s, C-H). FTIR (ATR) ν_(C-O) 1852 cm^-1 (s, C-O linkage), ν_(C-O) 1942 cm^-1 (m, C-O linkage).

1,3,5-C₆H₃(CH₃)₃ (2). The ¹H NMR spectrum of 2 was extrapolated from the literature.^{1 1}H NMR (CHCl₃): δ 2.25 (s, -CH₃), 6.78 (s, C-H).

Results

The reaction of Mo(CO)₆ with mesitylene yielded 0.028 g of the product, [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃. This translated to 0.09328 mmol, and thus a 1.18% yield based on the amount of Mo(CO)₆ used, which was the limiting reagent in the reaction. Mo(CO)₆ reacted to form the product in a 1:1 ratio, and 7.92 mmol of Mo(CO)₆ was used to start, so that proportion was taken into account when calculating the percent yield. Proton NMR spectroscopy yielded a two peaks of interest. A peak found at δ 2.25 was indicative of methylhydrogens and a peak noted at δ 5.23 was suggestive of hydrogens attached directly to the aromatic ring.¹ These peaks were noted with relative intensities of 9 to 3, respectively. Peaks seen at δ 7.25 and 1.54 were attributed to solvent and hexane, respectively. The ¹H NMR spectrum of mesitylene in CHCl₃ showed absorptions at δ 2.25 and 6.78 with relative intensities of 9 to 3, respectively.¹ The peak at δ 2.25 hinted of methyl protons and the peak at δ 6.78 was suggestive of protons bonded directly the ring. The mass spectrum of [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ showed a peak at m/z = 302.0, which is nearly equal to the molar mass of said substance.¹ The IR spectrum showed a strong peak around 1852 cm^-1 of area 64.462 cm^-1 indicative of antisymmetrical C-O stretching and a medium peak around 2942 cm^-1 of area 9.111 cm^-1 indicative of symmetrical C-O stretching. These areas were used to calculate a OC-Mo-CO bond angle of 108.32°.

Discussion

The percent yield of [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ is very poor. Much of this quantitative shortcoming can be attributed to clumsiness, as much of the solution containing future product was lost when the reaction vessel was dropped. This not only resulted in a direct loss of solution, but also exposed the solution to air. The solution had been kept under nitrogen as to prevent decomposition of the products. This exposure to air undoubtedly had a contribution to the poor percent yield. The solution was also not heated as desired because it was difficult to control the sand bath. The solution was to be brought to a moderate boil, but could not be controlled to do so. It would not boil, then boiled rigorously a moment later. Attempts were made to subdue the boiling, but were unsuccessful. This overheating was probably not favorable for the reaction. More adept control of the sand bath would have resulted in a better yield of product.

Washing the product with excess amounts of hexane and CH₂Cl₂ also may have added to the loss of product. Excess washing would make it difficult to extract the product from the solution, as there would have been a relatively small amount of product compared to the amount of solution it was dissolved in. The solution could not be completely dried with the rotovap, which means there was an excess of hexane and/or mesitylene in the solution. Vacuum filtration then had to be used to collect the product, which was not ideal. Best case scenario, the rotovap would have completely dried the product and it would have been scraped out of the flask. Vacuum filtration gives a better chance for loss of product.

The product seemed pure as it produced clear ¹H NMR and IR spectra readings. The ¹H NMR spectrum shows a methyl peak at δ 2.25 and a C-H peak at δ 5.23, and the reagent in the reaction, mesitylene, also gives a peak at δ 2.25. This seems to confirm the structure and addition of the metal complex, as only the C-H peak was shifted downfield. The methyl protons are too shielded to be affected by the metal. The downfield shift is caused by backbonding of the carbonyls. IR spectroscopy revealed 3 C-O stretches, 2 of which were accounted for by a strong peak at 1852 cm^-1 and the 3^rd of which was accounted for by a medium peak at 1942 cm^-1. Two peaks were seen because of symmetrical and antisymmetrical stretching of the carbonyls.¹ The two antisymmetical modes have exactly identical absorption frequencies, and the symmetrical mode has a different absorption frequency than them, which means that a total of two peaks should be seen.¹ The areas of these peaks, 64.462 cm^-1 for antisymmetical stretch and 9.111 cm^-1 for symmetrical stretch, allowed for discovery of the bond angle between the CO ligands. The calculated angle was 108.32°. This seems to make sense as [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ is predicted to be a tetrahedral complex and the expected bond angle for tetrahedral complexes is 109.5°, but in reality [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ acts more like an octahedral complex and the bond angle should be slightly less than 90°.¹ This error could be due to excess solvent or impure an sample, which resulted in skewed IR spectrum readings and thus incorrect peak areas. The oxidation state and electron count of Mo in [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ are 0 and 6 electrons, so it is an 18 electron complex.

Conclusion

The main purpose of the experiment was to interpret the ¹H NMR and IR spectra of [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ to confirm the product and to decipher the bond angle between the carbonyls. The ¹H NMR spectrum of the product showed peaks at δ 2.25 and 5.23, while the ¹H NMR spectrum of mesitylene gave peaks at δ 2.25 and 6.78, both in ratios of 9:3, respectively. This ratio seems to confirm methyl groups and single protons attached to the ring. The downfield shifting of the second peak is attributed to the coordination of mesitylene to the metal complex. The protons attached directly to the ring are affected by backbonding of the carbonyl groups. The methyl hydrogens are shielded, and thus are not affected by the metal complex. The IR spectrum yielded two peaks near 2000 cm^-1. These two peaks account for 3 C-O stretches, 2 of which are accounted for by a strong peak at 1852 cm^-1 and the 3^rd of which are accounted for by a medium peak at 1942 cm^-1. The strong peak accounts for antisymmetrical stretching of the carbonyls and the medium peak accounts for symmetrical stretching of the carbonyls. The areas of these peaks, 62.462 cm^-1 and 9.111 cm^-1 respectively, provide for a theoretical angle between the carbonyls of 108.32°. In reality, this angle should be only nearly 90°. The percent yield for the reaction was poor and could have improved with a more steady hand, more precise heating of the reagents, less exposure of the solution to air, and less solvent used for washing.

References

(1) Angelici, R. J.; Girolami, G. S.; Rachufuss T. B. Synthesis and Technique in Inorganic Chemistry: A Laboratory Manual; University Science Books: Sausilito, CA, 1999; pp 161-170.

Synthesis and Determination of Polypyrazolylborates:

K[HB(3,5-C₅H₇N₂)₃] and HB(3,5-C₅H₇N₂)₃Cu(CO)

Abstract

The reaction of KBH₄ heated with 3,5-dimethylpyrazole produces potassium tris(3,5-dimethylpyrazolyl)hydroborate at an unknown percent yield. ¹H NMR spectroscopy of K[HB(3,5-C₅H₇N₂)₃] shows singlets at δ 1.79 and 2.06 indicative of –CH₃ whereas the same spectroscopy of 3,5-dimethylpyrazole shows only one singlet in the same area. IR spectroscopy of K[HB(3,5-C₅H₇N₂)₃] shows a B-H stretch at 2420 cm^-1, which is absent from the IR spectrum of 3,5-dimethylpyrazole. The reaction of K[HB(3,5-C₅H₇N₂)₃] with CuI and CO gives rise to HB(3,5-C₅H₇N₂)₃Cu(CO) at 265% yield. ¹³C NMR spectroscopy of HB(3,5-C₅H₇N₂)₃Cu(CO) produces a singlet at δ 172.4 indicative of C-O bonding, which does not appear on the ¹³C NMR spectrum of K[HB(3,5-C₅H₇N₂)₃]. IR spectroscopy of HB(3,5-C₅H₇N₂)₃Cu(CO) produces a C-O stretch at 2053 cm^-1, which is absent from the IR spectrum of K[HB(3,5-C₅H₇N₂)₃]. The analysis of these spectra seems to validate the supposed products from the reactions.

Introduction

The reaction of KBH₄ with 3,5-dimethylpyrazole yields tris(3,5-dimethylpyrazolyl)hydroborate. The reaction specifically takes place in the following manner:

This is compound of interest because it a polypyrazolylborate, or scorpionate, formed from a binary born hydride, which are difficult to handle.^1,2 In order to determine the structure of said substance from a ¹H NMR and IR spectrum, the peaks and stretches must be compared to the same spectra for 3,5-dimethylpyrazole to look for indications of boron in the structure and differentiation in methyl groups. The spectra for the two compounds should be similar save for those two main differentiations.

Metal complex HB(3,5-C₅H₇N₂)₃Cu(CO) can be synthesized from the following reaction:

The identity of this product can be confirmed by comparing its ¹³C NMR and IR spectra to the same spectra of the reagent K[HB(3,5-C₅H₇N₂)₃]. In this case, the spectra are compared to look for the presence of C-O bonding. The spectra should appear similar aside from peaks and stretches indicative that bond. The systematic addition of functional groups to these two compounds is what makes them comparable and identifiable to and from one another in their ¹H NMR, ¹³C NMR, and IR spectra.

Experimental

All syntheses were carried out in air and the reagents and solvents were purchased from commercial sources and used as received unless otherwise noted. The synthesis of K[HB(3,5-C₅H₇N₂)₃] (1) and HB(3,5-C₅H₇N₂)₃Cu(CO) (2) were based on reports published previously.¹

K[HB(3,5-C₅H₇N₂)₃] (1). KBH₄ (1.028 g, 19.1 mmol) and 3,5-dimethylpyrazole (7.002 g, 72.8 mmol) were added subsequently to a 100 mL round-bottom flask along with a small magnetic stir bar. A cold water condenser with greased joint was inserted into the round-bottom flask containing the solution. This connection was further secured with a keck clip. A silicone oil bath was constructed with a glass dish containing a paper clip as a stirring instrument. The bath was placed on a hot plate and the round bottom flask was placed in the oil bath. The cold water condenser was not connected to a cold water source; it was used to allow air to circulate.

Once secure, the hot plate was turned on to 230 °C and the stirring instruments were spun at a moderate speed. A thermometer was inserted into oil bath to monitor the temperature, which fluctuated between 230 °C and 250 °C during the experiment. The solution was allowed to heat for 1 h. After this time, the round-bottom flask was taken off the oil bath and the condenser was removed. A white solid precipitate submerged in liquid remained and was allowed to cool to 90 °C, again using the thermometer to measure temperature. 50 mL of toluene was added to the flask and the solution was vacuum filtered with a 60 mL frit. A total of about 100 mL more toluene was added to wash the resulting white solid precipitate. The precipitate was washed a final time with 50 mL of diethyl ether. The precipitate was then vacuum dried for 0.33 h. The precipitate was a powdery white substance 1. ¹H NMR (D₂O): δ 1.79 (s, -CH₃), 2.06 (s, -CH₃), 5.82 (s, C-H). ¹³C NMR (D₂O): δ 11.05 (s, -CH₃), 12.44 (s, -CH₃), 104.9 (s, C-H), 146.1 (s, C-CH₃), 148.9 (s, C-CH₃). FTIR (ATR) ν_(C=N) 1560 cm^-1 (s, pyrazolyl), ν_(B-H) 2420 cm^-1 (s, B-H linkage), ν_(C-H) 2950 cm^-1 (br, C-H linkage).

HB(3,5-C₅H₇N₂)₃Cu(CO) (2). CuI powder (0.394 g, 2.07 mmol) and acetone (35 mL) were subsequently added to a 100 mL round-bottom flask along with a small magnetic stir bar. A septum was attached to the flask. Previously synthesized 1 (0.191 g, 0.567 mmol) was dissolved in a minimal amount of acetone (3 to 5 mL). CO gas was bubbled into the round-bottom flask for about 5 min. At this time, the solution of 1 and acetone was injected into the round-bottom flask, and CO gas was allowed bubbled in for another few minutes. A yellowish liquid resulted and the flask was put on ice for 1 h to allow for recrystallization. The solution was then roto-vaporized for 5 to 10 minutes to make up for inadequate recrystallization.

A grayish, greenish powder remained in the round-bottom flask, which was scraped out using a spatula and determined to be 2 (0.584 g, 265% yield based on the amount of 1 used). ¹H NMR (CDCl₃): δ 2.30 (s, -CH₃), 2.50 (s, B-H), 5.68 (s, C-H). ¹³C NMR (CDCl₃): δ 12.50 (s, -CH₃), 13.92 (s, -CH₃), 104.37 (s, C=C), 143.60 (s, C=N), 147.42 (s, C-N), 172.4 (s, C-O). FTIR (ATR) ν_(C=N) 1543 cm^-1 (s, pyrazolyl), ν_(C-O) 2053 cm^-1 (s, carbonyl), ν_(B-H) 2499 cm^-1 (s, B-H linkage), ν_(C-H) 2921 cm^-1 (br, C-H linkage).

C₅H₈N₂ (3). The ¹H NMR and IR spectra of (3) were obtained from Sigma Aldrich.^{3,4 1}H NMR (CDCl₃): δ 2.25 (s, -CH₃), 5.8 (s, C-H). FTIR (ATR) ν_(C-N) 1030 cm^-1 (s, pyrazolyl), ν_(C-H) 2860 cm^-1 (br, C-H linkage), ν_(C-H) 2930 cm^-1 (br, C-H linkage).

Results

The reaction of KBH₄ and 3,5-dimethylpyrazole was not measured for yield of the product, K[HB(3,5-C₅H₇N₂)₃], but theoretical yield would be 19.1 mmol. Theoretical yield of H₂ gas, though not measured, was 57.3 mmol, based on the amount of KBH₄ used, which was the limiting reagent. KBH₄ reacts to form H₂ in a 1:3 ratio, and 19.1 mmol of KBH₄ was used to start, so that proportion was taken into account when calculating the theoretical yield. ¹H and ¹³C NMR spectroscopy of the product yielded several peaks. The ¹H NMR spectrum presented a two singlets found at δ 1.79 and 2.06, representative of methyl groups. A singlet found at δ 5.82 was indicative of the hydrogen attached directly to pyrazolyl ring. The ¹³C NMR spectrum yielded a pair of singlets found at δ 11.05 and 12.44, which was suggestive of methyls attached to the pyrazolyl ring. A singlet found at δ 104.9 was from the C-H bond on the ring, and two final singlets found at δ 146.1 and 148.9 were from the carbons on the ring attached to the methyl groups. The IR spectrum showed a sharp peak around 1560 cm^-1 indicative of a C=N bond forming the pyrazolyl ring, a sharp peak around 2420 cm^-1 indicative of B-H linkage, and finally a broad peak near 2950 cm^-1 suggestive of C-H bonding.

The reaction of CuI, K[HB(3,5-C₅H₇N₂)₃], CO, and acetone yielded 0.584 g of product, HB(3,5-C₅H₇N₂)₃Cu(CO). This translated to 1.502 mmol, and thus was a 265% yield. ¹H and ¹³C NMR spectroscopy of the product yielded several peaks. The ¹H NMR spectrum contained a singlet found at δ 2.30, representative of methyl groups. A singlet found at δ 2.50 was indicative of the hydrogen bonded to boron. A third singlet found at δ 5.68 was from protons bonded to the pyrazolyl ring. The ¹³C NMR spectrum produced a two singlets found at δ 12.50 and 13.92, which suggested methyls carbons. A singlet found at δ 104.37 was from double bonded carbons, a singlet found at δ 143.60 was from carbon double bonded to nitrogen, another singlet found at δ 147.42 was from carbon singly bonded to nitrogen, and one final singlet at δ 172.4 was representative of carbon bonded to oxygen. The IR spectrum yielded a sharp peak around 1543 cm^-1 indicative of a C=N bond forming the pyrazolyl ring, a sharp peak around 2053 cm^-1 indicative of C-O bonding, a sharp peak around 2499 cm^-1 indicative of B-H linkage, and finally a broad peak near 2921 cm^-1 suggestive of C-H bonding.

Discussion

The percent yield of K[HB(3,5-C₅H₇N₂)₃] was not able to be determined. The weight of this product was either never obtained or the figure was lost during the experiment. Percent error, though not measured, could possibly have been affected from heating the KBH₄ and 3,5-dimethylpyrazole solution at too high a temperature, as it went above the 230 °C limit specified by the experimental guidelines.¹ The solution was heated for only 1 h, when the suggested time was 1 to 1.5 h, which means the reagents may not have completely reacted. When the solution was taken off the oil bath and allowed to cool, it cooled more quickly than expected, and dropped to 90 °C or lower before adding the 50 mL of toluene when 100 °C was specified the addition temperature.¹ There was some confusion as far as the protocol at this point, so the solution with the toluene added was allowed to cool for a short while, when the guidelines asked for the residue to be filtered and washed hot. The solution was still warm when filtered and washed, but not nearly as hot as it could have been. These types of errors would have resulted in loss of potential product and negatively affected the percent yield, had it been measured.

The product did seem pure, as it was a clean white color, and its ¹H NMR, ¹³C NMR, and IR spectra yielded clear readings. The ¹H spectrum shows methyl peaks at δ 1.79 and 2.06 whereas the ¹H spectrum for the reagent in the reaction, 3,5-dimethylpyrazole, shows only one methyl peak at δ 2.25. This seems to validate that addition of the boron to the molecule, as it would cause make each methyl group slightly different from the other. Polypyrazolylborates produce sharp a B-H stretch in their IR spectra, and this is evident in the IR spectrum reading for K[HB(3,5-C₅H₇N₂)₃].² A sharp peak is noted at 2420 cm^-1, whereas the IR spectrum for 3,5-dimethylpyrazole does not contain said stretch, again supporting the claim for addition of boron to the molecule.

The percent yield for HB(3,5-C₅H₇N₂)₃Cu(CO) was not accurate. The product obtained from the roto-vaporization was not washed, so it is suspected that the other product of the reaction, KI, was mixed in with the desired product. It is believed that the greenish powder was HB(3,5-C₅H₇N₂)₃Cu(CO) while the greyish powder was KI. This is why the percent yield was above 100%. Aside from that inaccuracy, the only 0.567 mmol of K[HB(3,5-C₅H₇N₂)₃] was used, when the protocol called for 2 mmol to be used.¹ This would not affect the percent yield, as the amount of K[HB(3,5-C₅H₇N₂)₃] used would still be the limiting reagent, but could have affected the NMR and IR spectrums. However, it was intuitively noted that the greenish powder was the desired product, and an effort was made to extract only that powder from the product for the spectroscopy determinations.

The fact that recrystallization did not seem take place as detailed¹ and that a roto-vaporizer had to be used to dry the product most likely did not help the yield of product either. Product may have been lost during this process. If the powder had been washed with acetone, a more accurate percent yield would have been obtained because the KI would have been washed away, but this was not extremely necessary for the purposes of this experiment. The product obtained did give clear ¹H NMR, ¹³C NMR, and IR spectra, meaning it was fairly pure. The addition of the CO to the molecule from K[HB(3,5-C₅H₇N₂)₃] is evident in the ¹³C NMR and IR spectra. There is a distinct peak at δ 172.4 on the ¹³C NMR spectrum which is not noted on the same spectrum for K[HB(3,5-C₅H₇N₂)₃]. The IR spectrum of HB(3,5-C₅H₇N₂)₃Cu(CO) shows a tall sharp stretch at 2053 cm^-1 distinctive of C-O bonding; the IR spectrum of K[HB(3,5-C₅H₇N₂)₃] shows no such stretch. Peaks and stretches for the spectra were labeled with the help of colleagues. An acknowledgement is made that are more than likely downfield or upfield shifts of some of the peaks from one product to another because of changes in chemical structure, but these postulates were not explored. The oxidation state and electron count of Cu in HB(3,5-C₅H₇N₂)₃Cu(CO) are +1 and 10 electrons, so it is a 18 electron complex.

Conclusion

The main purpose of the experiment was to decipher the structural changes from 3,5-dimethylpyrazole to potassium tris(3,5-dimethylpyrazolyl)hydroborate to a copper complex of potassium tris(3,5-dimethylpyrazolyl)hydroborate through ¹H NMR, ¹³C NMR, and IR spectra. Addition of boron to 3,5-dimethylpyrazole was apparent in the ¹H NMR and IR spectra of the first product. The ¹H spectrum shows methyl peaks at δ 1.79 and 2.06 whereas the ¹H spectrum for the reagent in the reaction, 3,5-dimethylpyrazole, shows only one methyl peak at δ 2.25, seemingly confirming the addition of boron as this would make each methyl group differentiable. The IR spectrum of this product showed a sharp stretch around 2420 cm^-1, indicative of B-H bonding, which is absent in the IR spectrum for 3,5-dimethylpyrazole. All of these finding seem to validate K[HB(3,5-C₅H₇N₂)₃] as being the product of the reaction.

The ¹H NMR, ¹³C NMR, and IR spectra of the second product also seem to confirm its expected structure. The ¹³C NMR spectrum for the second product shows a peak at δ 172.4, which is an area suggestive of C-O bonding. The ¹³C NMR spectrum of K[HB(3,5-C₅H₇N₂)₃] contains no peak in this area. The IR spectrum of the second product shows a sharp stretch around 2053 cm^-1, which is also indicative of C-O bonding. The IR spectrum of K[HB(3,5-C₅H₇N₂)₃] contains no stretch in this area. These noted findings on the spectra all point towards to product being HB(3,5-C₅H₇N₂)₃Cu(CO).

The percent yield for the first reaction was not monitored, but would have been aversely affected by factors such as poor temperature control, short reaction time, and more prompt washing technique. The percent yield for the second reaction was poor, but could have been improved by washing the product with acetone and by allowing for a longer recrystallization period.

References

(1) Bochmann, M. Preparation and Complexation of Tris(3,5-dimethylpyrazoyl)hydroborate. pp 33-35.

(2) Trofimenko, S. Polypyrazolylborates: Scorpionates. Journal of Chemical Education. 2005, 82, 1715-1720.

(3) http://www.sigmaaldrich.com/spectra/fnmr/FNMR010068.PDF

(4) http://www.sigmaaldrich.com/spectra/ftir/FTIR007818.PDF

Purpose

To determine the mathematical relationship between current, potential difference, and resistance in a simple circuit, and also to compare the potential versus current behavior of a resistor to that that of a light bulb.

Hypothesis

The graph of potential difference versus current for a resistor should result in a linear plot, in accordance with the equation R = V / I. To extract even more from this equation, the slope of said linear plot will be equal to the resistance. The potential versus current behavior for that of a light bulb is predicted to not follow Ohm’s law, as light bulbs give off heat and light. It is not expected to have the same linear plot of current versus time that a resistor is characteristic of.

Labeled Diagrams

See attached sheet.

Data

Graphs

Resistor 10 Ω

Resistor 51 Ω

Light Bulb First 3 Points

Light Bulb Last 10 Points

Questions

1. As the potential across the resistor increased, the current through the resistor increased. If the change in current is proportional to voltage, the data should be in a straight line and it should go through zero. In these two examples how close is the y-intercept to zero? Is there a proportional relationship between voltage and current? If so, write the equation for each run in the form potential = constant x current. (Use a numerical value for the constant.)

The y-intercept is extremely close to zero for both the 10 Ω and 51 Ω resistor, being 0.0004256 V and -0.009978 V respectively. Yes, there is a proportional relationship between voltage and current. The equation for the first run (10 Ω resistor) is U = 9.655 I and the equation for the second run (51 Ω resistor) is U = 48.94 I.

2. Compare the constant in each of the above equations to the resistance of each resistor.

The constants are nearly equal to the resistance of their respective resistor.

3. The constant you determined in each equation should be similar to the resistance of each resistor. However, resistors are manufactured such that their actual value is within a tolerance. For most resistors used in this lab, the tolerance is 5% or 10%. Check with your instructor to determine the tolerance of the resistors you are using. Calculate the range of values for each resistor. Does the constant in each equation fit within the appropriate range of values for each resistor?

The range for the 10 Ω resistor is 9.5 Ω to 10.5 Ω. The determined constant of 9.655 fits into this range. The 51 Ω resistor has a range of 48.45 Ω and 53.55 Ω. The determined constant of 48.94 fits within that range.

4. Do your resistors follow Ohm’s law? Base your answer on your experimental data.

Yes, they follow Ohm’s law because the correlation for linear fit for each trial is equal to 1.000.

5. Describe what happened to the current through the light bulb as the potential increased. Was the change linear? Since the slope of the linear regression line is a measure of resistance, describe what happened to the resistance as the voltage increased. Since the bulb gets brighter as it gets hotter, how does the resistance vary with temperature?

The change was more curved at low voltage, but became more linear as the voltage was raised higher. Resistance increased as the voltage increased, as the slope was 3.261 V/A for the first 3 points and 37.66 V/A for the last 10 points. Since the bulb gets brighter as it gets hotter, that means resistance increases with temperature.

6. Does your light bulb follow Ohm’s law? Base your answer on your experimental data.

The light bulb seemed to follow Ohm’s law for the first few points, but the data began to curve into a steeper linear plot as voltage was increased. This is evident by the change in slop from 3.261 V/A for the first 3 points and 37.66 V/A for the last 10 points. The y-intercept for the last 10 points is also non-zero, which goes against Ohm’s law.

Purpose

To determine the spring constant of a spring by measuring its stretch versus applied force, to determine the spring constant of a spring by measuring the period of oscillation for different masses, and also to investigate the dependence of period of oscillation on the value of mass and amplitude of motion.

Hypothesis

If the applied force (mass) is to remain the same while the vertical displacement is increased, there period will remain the same. However, if the vertical displacement is held constant while the applied force in increased, the period will increase. This is in accordance with the derived equation T = 2π (M / k)^0.5.

Labeled Diagrams

See attached sheet.

Data

Part I

k = 3.53 N/m

Part II

Part III

m_s = 0.00940 kg

Slope = 10.5 s²/kg

Y-Intercept = 0.0687 N

k = 3.75 N/m

C = 0.694

% difference = 6.04%

Graphs

Part I

Part II

Part III

Questions

1. Do the data from Part 1 verify Hooke’s Law? State clearly the evidence for your answer.

The data correlate close to Hooke’s Law, but not quite. The law states that F = -ky, where F is in this case Mg and y equals the negative displacement. After graphing forces versus displacement, a value of 3.53 N/m was determined as the spring constant. However, when applying this value to the equation and using recorded displacement values, the calculated force come up less than the actual for used. For example, in the first trial y = -0.055 m. Multiplying that value by the extrapolated spring constant gives a theoretical force of 0.194 N, but the actual force used was 0.244 N. All other trials yield a similar lowball theoretical force.

2. How is the period T expected to depend upon the amplitude A? Do your data confirm this expectation?

Period is not expected to depend upon amplitude, as suggested by the equation T = 2π (M / k)^0.5, where amplitude is absent as a variable. The data confirms this expectation, as the period was nearly the same for each trial.

3. Consider the value you obtained for C. If you were to express C as a whole number fraction, which of the following would best fit your data (1/2, 1/3, 1/4, 1/5)?

The obtained value of C is 0.694, which is closest to 1/3.

4. Calculate T predicted by the equation T = 2π (M / k)^0.5 for M = 0.050 kg. Calculate T predicted by T = 2π ( [M + Cm_s] / k )^0.5 for M = 0.050 kg and your value of C. What is the percent difference between them? Repeat for a value for M of 1.000 kg. Is there a difference in the percent differences? If so, which is greater and why?

T predicted using the first equation and M = 0.050 kg is 0.726 s. T predicted using the second equation and M = 0.050 kg is 0.771 s. This is a percent difference of 6.01%.

T predicted using the first equation and M = 1.000 kg is 3.24 s. T predicted using the second equation and M = 1.000 kg is 3.26 s. This is a percent difference of only 0.615%. The greater percent difference occurs at the lower weight because the weight of the spring is almost insignificant at higher weight. The proportion of the mass to the spring is so great that it has almost no effect on the calculation.

Conclusion

During part one of the experiment, the vertical displacement of a spring was measured as a function of force applied to it. The starting position of the spring was recorded using a stretch indicator. Mass was added to the spring, and the displacement was recorded. This was repeated with various amounts of mass. From these data, a graph of force versus displacement was plotted, and a linear fit slope revealed the spring constant. In this endeavor, the spring constant was valued at 3.53 N/m.

However, when applying this spring constant to the recorded displacements in the Hooke’s Law equation, the calculated forces are lower than the recorded forces. In similar manner, when rearranging Hooke’s Law to solve for displacement, the calculated displacements are larger than the actual recorded displacements. This means there was human error, most likely in terms of not being precise with the displacement readings because the recordings for the masses used were accurate. Because such small masses were used, any error in displacement readings was augmented. The spring used may also not have been perfect.

During part two of the experiment, the period of the spring was measured as amplitude changed while mass remained constant. The period remained nearly the same throughout every trial, which was to be expected. Any differences in period may be accounted to inadequate stopwatch usage and inaccurate starting displacements throughout the trials. It should be noted that at amplitude of 0.100 m, the hook lost contact with the spring for a split second at the apex of oscillation, which accounts for its oddity in period. This was something that could not be avoided; at that amplitude the spring pulled the hook up too quickly which caused the loss of contact. The resulted graph of period versus amplitude yielded a linear fit slope of close to 0 (-0.247 s/m), which was predicted.

During part three of the experiment, the period of the spring was measured as mass was varied while amplitude remained constant. As the mass was increased, the period also increased. This was not surprising considering the given equations. The square of the period versus mass for each trial was plotted and a linear fit was taken. The slope and y-intercept of this line was then used to determine the spring constant and C, the fraction of the spring’s mass that should be taken into account for the equation T = 2π ( [M + Cm_s] / k )^0.5. The calculated value of k was 3.75 N/m, which is only 6.04% different from the value determined earlier of 3.53 N/m. The value of C was determined to be 0.694, which is closest to the whole number fraction of 1/3. Any error during parts two and three can be attributed to inaccurate stopwatch recordings and slight variance in displacement and release of the masses at each amplitude.

Equations

T = 2π (M / k)^0.5

T = 2π ( [M + Cm_s] / k )^0.5

slope = 4π² / k

y-intercept = 4π²Cm_s / k

Tinkering with Tin: Synthesis of SnCl(CH₂C₆H₅)₃ and SnCl₄[OS(CH₃)₂]₂

Abstract

The reaction of tin with benzyl chloride under reflux yields SnCl(CH₂C₆H₅)₃ in low percent yield (around 15%) due to various factors. ¹H NMR of SnCl(CH₂C₆H₅)₃ shows a singlet at δ 3.15 with satellites containing coupling constant between Sn-H of 78.1 Hz. The reaction of SnCl₄ with OS(CH₃)₂ yields SnCl₄[OS(CH₃)₂]₂ with poor percent yield (around 140%) due to various factors. The IR spectrum of SnCl₄[OS(CH₃)₂]₂ shows a large S-O stretch at 899.7 cm^-1, which is a lower frequency than the S-O stretch of 1042.5 cm^-1 from the IR spectrum of OS(CH₃)₂, suggesting coordination on the metal complex at oxygen.

Introduction

The reaction of tin with benzyl chloride yields the metal complex SnCl(CH₂C₆H₅)₃. The reaction specifically takes place in the following manner:

[Figure missing.]

This is compound of interest because of the different isotopes of tin that naturally occur. In order to determine the structure of said substance from a ¹H NMR spectrum, the peaks must be analyzed for satellites occurring from coupling of hydrogen to isotopes of tin. The observed spectrum is a combination of SnCl(CH₂C₆H₅)₃ molecules containing different isotopes of Sn, namely ¹¹⁷Sn and ¹¹⁹Sn, among others. The value for J_Sn-H between satellites can confirm the presence of SnCl(CH₂C₆H₅)₃. Mass spectrum can also be used to confirm the presence of SnCl(CH₂C₆H₅)₃ in the product by looking for a peak at m/z equal to the molecular mass of the substance (427 g/mol).

SnCl₄[OS(CH₃)₂]₂ can be synthesized from the following reaction:

[Figure missing.]

It is a reaction of interest because it can be used to determine where OS(CH₃)₂ coordinates to Sn. Coordination of S to Sn would result in a higher frequency of S-O stretch on an IR spectrum than the frequency of the S-O stretch on the IR spectrum of OS(CH₃)₂ due to a strengthening of the S-O bond. Coordination of O to Sn would result in a lower frequency of S-O stretch because of a weakened S-O bond from the resonance form needed to secure that coordination.¹

Experimental

All syntheses were carried out in air and the reagents and solvents were purchased from commercial sources and used as received unless otherwise noted. The synthesis of SnCl(CH₂C₆H₅)₃ (1) and SnCl₄[OS(CH₃)₂]₂ (2) were based on reports published previously.¹

SnCl(CH₂C₆H₅)₃ (1). 325 mash-Aldrich Sn powder (1.939 g, 16.34 mmol), 99% Aldrich benzyl chloride (6.0 mL, 52 mmol), and deionized H₂O (4.0 mL) were added subsequently to a 50 mL round-bottom flask. A small stir bar was added then added to this solution. A sand bath was placed over a stir plate and a cold water condenser with greased joint was inserted into the round-bottom flask containing the solution. This connection was further secured with a keck clip. The round bottom flask was placed in the sand bath and the condenser was connected to the cold water source.

Once secure, the sand bath was set to 50% power and the stir bar was spun a shade over moderate speed. The solution was allowed to reflux for 2.75 h. Once reflux was complete, the 50 mL round-bottom flask was removed from the condenser and allowed to cool in an ice bath until it was cold. The liquid was decanted from the white precipitate and discarded. The white precipitate was saved. Ethyl acetate (10 mL) was added to the precipitate and this solution was heated using a sand bath at 65% power and stirred at moderate speed until the white solid had completely dissolved. The round-bottom flask was then taken off the heat and submerged in an ice bath for approximately 0.5 h to allow for recrystallization.

The stir bar was removed and a high vacuum was used for about 10 minutes to extract the extraneous liquid and leave a white powder. Several mL of ether were added to the round-bottom flask and a glass stir rod was used to break up the chunks of powder and dissolve it in the ether. This solution was then suction filtered with a 30 mL glass frit and allowed the dry. The precipitate collected was 1 (0.471 g, 13.49% based on the amount of Sn powder used). ¹H NMR (CDCl₃): δ 7.25 (s, CDCl₃), 7.19 (s, Ph-H), 7.17 (s, Ph-H), 7.05 (s, Ph-H), 3.15 (s, J_Sn-H = 78.1 Hz, CH₂).

SnCl₄[OS(CH₃)₂]₂ (2). SnCl₄ (2.25 mL, 19 mmol), anhydrous diethyl ether (45 mL), DMSO (2.9 mL, 41 mmol), and ether (5 mL) were subsequently added to a 125 mL Erlenmeyer flask. The DMSO was added using a syringe for safety purposes while the other reagents were measured and added using graduated cylinders. This solution yielded a precipitate which was isolated by suction filtering the solution through a 50 mL glass frit. Several extra mL of ether were added to the Erlenmeyer flask to help aid in transfer of all the precipitate to the filter. Once dry, the white powder precipitate 2 was collected and weighed. 11.04 g (26.49 mmol) were recovered, giving a yield of 139.4%. FTIR (ATR) ν_(S-O) 899.7 cm^-1 (s S-O coordination to Sn).

OS(CH₃)₂ (3). The IR spectrum of (3) was taken by Dr. Graham. FTIR (ATR) ν_(S-O) 1042.5 cm^-1 (s, S-O linkage).

Results

The reaction of Sn, benzyl chloride, and H₂O yielded 0.471 g of the product, SnCl(CH₂C₆H₅)₃^. This translated to 1.102 mmol, and thus was a 13.49% yield based on the amount of Sn used, which was the limiting reagent in the reaction. Sn reacted to form the product in a 2:1 ratio, and 16.34 mmol of Sn was used to start, so that proportion was taken into account when calculating the percent yield. Proton NMR yielded several peaks. A set of peaks were found at δ 7.19, 7.17, and 7.05 were indicative of phenyl resonances derived from protons on the phenyl ring of the product.¹ A sharp peak located at δ 7.25 was due to the chloroform solvent ¹. Coupling between Sn and H produced a singlet at δ 3.15 with satellites ¹J_Sn-H equal to 78.1 Hz due to the presence of isotopes ¹¹⁷Sn and ¹¹⁹Sn.¹ The mass spectrum of SnCl(CH₂C₆H₅)₃ showed a peak at m/z = 427.0,¹ which is the molar mass of said substance.

The reaction of SnCl₄ and DMSO yielded 11.04 g of product, SnCl₄[OS(CH₃)₂]₂. This translated to 26.49 mmol, and thus was a 139.4% yield. IR spectrum of SnCl₄[OS(CH₃)₂]₂ showed its largest peak at 899.7 cm^-1 while the IR spectrum of OS(CH₃)₂ gave its largest peak at 1042.51 cm^-1.

Discussion

The percent yields for each product are less than stellar. During the synthesis of SnCl(CH₂C₆H₅)₃, letting the reagents reflux for a longer time period, closer to 3 hours, may have been beneficiary to resulting in more product. The reagents may not have all completely reacted. Some of the precipitate may have been accidentally removed during decanting after reflux, and much ethyl acetate may have been added to that precipitate. A minimal amount should have been used while trying to dissolve the white solid precipitate in the sand bath. The less ethyl acetate needed and used would have resulted in a better percent yield.

The fact that recrystallization did not seem take place as detailed¹ and that a vacuum had to be used to dry the product most likely did not help the yield of product either. Product may have been lost during the vacuuming process. The glass frit used for filtering was not of the utmost quality, and so product may have escaped during that process, also. The product was also not completely dry, and gave a misleading mass measurement, meaning the percent yield is even lower than recorded. The product obtained did seem to give a clear ¹H NMR spectra, meaning it was fairly pure. The coupling between Sn and H due to isotopes ¹¹⁷Sn and ¹¹⁹Sn ¹ in the product is distinctly visible around δ 3.15 with a J_Sn-H of 78.1 Hz. The NMR chemical shifts of the Sn-CH₂ protons and the C­₆H₅ protons of SnCl(CH₂C₆H₅)₃ are so different because of the 117 and 119 isotopes of Sn. They affect the coupling with H, producing satellites. The C₆H₅ protons are not affected by this coupling. The literature states that peaks for phenyl resonance appear around δ 7 and chloroform appears around δ 7.24, which seems to validate the experimental values obtained (δ 7.05 to 7.19 for phenyl resonances ad δ 7.25 for chloroform).¹

The percent yield of SnCl₄[OS(CH₃)₂]₂ is likely thrown off because the product was not dry when it was weighed. Several extra mL of ether were used for transport of the precipitate to the glass frit for filtering, which would most likely lead to loss of product. Not all of the precipitate was able to be transferred from the Erlenmeyer flask to the frit. Product was distinctly lost due to a seemingly defective frit, too. Attempts to refilter the filtrate were unsuccessful. The fact that the frit quality was poor meant that the product was not able to be dried well and thus carried extra weight. The product seemed to smoke away as it was allowed to dry further in exposure to air, and lost mass over time. This conundrum could not be adequately explained. IR spectrum of the product SnCl₄[OS(CH₃)₂]₂ shows its largest stretch at 899.7 cm^-1, hinting that the coordination to Sn occurs at oxygen, as this frequency is lower than that of the largest stretch from the OS(CH₃)₂ spectrum (1042.51 cm^-1). This is due to resonance and weakening of the S-O bond.¹ The two spectra are similar save for the shifting of that one peak. The stretch at 1042.51 cm^-1 is consistent with the value stated in the literature for S-O stretching (approximately 1100 cm^-1).¹

Conclusion

The main purposes of the experiments were to decipher the ¹H NMR for SnCl(CH₂C₆H₅)₃ in order to determine J_Sn-H and to interpret the IR spectra of OS(CH₃)₂ and SnCl₄[OS(CH₃)₂]₂ to tell whether O or S coordinates to Sn. The J_Sn-H was found to be 78.1 Hz. The observed spectrum is a combination of SnCl(CH₂C₆H₅)₃ molecules containing different isotopes of Sn, namely ¹¹⁷Sn and ¹¹⁹Sn, among others. This is what makes the ¹H NMR difficult to interpret, but the J_Sn-H value is validation of the presence of SnCl(CH₂C₆H₅)₃. The percent yield for that reaction was poor, due to varying factors. Longer reflux time, less ethyl acetate used during recrystallization, and better filtration techniques all would have contributed to an improved percent yield.

The IR spectrum of OS(CH₃)₂ showed its largest stretch around 1050 cm^-1 while the IR spectrum of SnCl₄[OS(CH₃)₂]₂ gave its largest stretch around 900 cm^-1. This lower frequency of stretch suggests that O and not S coordinates to the Sn. This is due in part to the resonance form needed to form the complex, which weakens the S-O bonding, and thus lowers the stretching frequency. The yield for the reaction to form SnCl₄[OS(CH₃)₂]₂ was also poor, which can be mostly attributed to poor filtering and drying techniques, along with the overuse of ether.

References

(1) “Synthesis and Techniques in Inorganic Chemistry,” Third Edition, G. S. Girolami, T.R. Rauchfuss, and R. Angelici, University Science Books, 1999.

These are all my handwritten notes from the class.

Psychology of the Self Part 1 B/W

Psychology of the Self Part 2 B/W 1

Psychology of the Self Part 2 B/W 2

Psychology of the Self Part 3 B/W

Anthrax is an illness caused by the bacterium bacillus anthracis. The bacteria are spread through spores and can infect the host cutaneously, inhalationally, or gastrointestinally. If infected, anthrax can often be fatal to the host. The bacteria employ a synergistic binary mechanism in order to infect eukaryotic cells and inflict the host with the illness. The symbiotic nature of anthrax is what makes it especially potent and of such concern.

The basic binary mechanism works with two components; component “A” and component “B.” Precursor component B must first be activated via proteolysis to form an oligomer. These activated B components will either then form a heptamer in solution which will then bind to a receptor on the cell surface, or bind as monomers to the receptor and form a homoheptamer. This component B-receptor complex then acts as a docking station for component A. Under acidic conditions, enzymatic component A is able to be translocated through the component B-receptor complex into the cytosol.

Once in the cytosol, component A is able to then disarm the cell through a number of different methods. First, component A could potentially force mono-ADP-ribosylation of G-actin, which incites cytoskeletal disarry and cell death. Second, it could induce proteolysis of mitogen-activated protein kinase kinases (MAPKK), which prevents cell signaling. Finally, component A could increase cellular levels of cyclic AMP, which results in immunosuppression and edema.

Anthrax consists of three synergistically acting proteins. The first protein of interest is the protective antigen (PA), which serves as the component B. It can be activated proteolytically from PA83 to PA63 via either trypsin, serum, of furin. It should be noted that furin is a special case, in which PA83 binds to the receptor before being activated by furin located on the cell surface. Once the heptamer of PA63 is formed, it is then able to send the A components into the cytosol under acidic conditions.

In the case of anthrax, component A actually consists of two separate proteins; the lethal factor (LF) and edema factor (EF). These two factors compete for docking rights on the PA63 heptamer in order to gain entrance into the cytosol. When the factors come into with the PA, they react to form lethal toxin and edema toxin. Upon entering the cytosol, each toxin does slightly different destruction on the cell.

Lethal toxin disarms the host immune system. It does this by targeting macrophages and dendritic cells, which eliminates any immunological response that the hosts would have. In essence, the host becomes deprived on pathogen killing cells. The edema toxin works in conjunction with the lethal factor by increasing cellular levels of cyclic AMP. This decreases the host immune response. The combination of these two toxins leads to a build up of bacteria and the host cannot attack the infection because its immune system is nearly non-existent thanks to anthrax.

Anthrax infection can be prevented by vaccine and treated with antibiotics. It was of national concern during the fall of 2001 when anthrax was found in the mail. This made people weary of anthrax as a possible biological weapon that could be used for terrorism. It can be made in vitro, which is part of what makes it such a threat. Anthrax spores can be destroyed with formaldehyde. The name anthrax comes from a Greek word for “coal,” which refers to the black ulcers that form from cutaneous infections. In all, anthrax can be a deadly disease and needs to be carefully dealt with.

Bibliography

Barth H, Aktories K, Popoff MR, Stiles BG. “Binary bacterial toxins: biochemistry, biology, and applications of common Clostridium and Bacillus proteins.” Microbiol Mol Biol Rev. 2004 Sep;68(3):373-402, table of contents. Review.

Guarner J, Zaki SR. “Histopathology and immunohistochemistry in the diagnosis of bioterrorism agents.” J Histochem Cytochem. 2006 Jan;54(1):3-11. Epub 2005 Sep 7.

Otherwise known as: “A Reusable Polymeric Asymmetric Hydrogenation Catalyst Made by Ring-Opening Olfein Metathesis Polymerization”

This was a PowerPoint presentation I did for class.

It is widely known among inorganic chemists that multiply bonded metal-ligand species take part in a diverse set of atom and group transfer reactions. It is common to witness CR₂ groups transferred to unsaturated organic substrates, but viewing the insertion of CH₂ into C-H bonds to yield saturated product is quite unusual. In the case of [TolC(NSiMe₃)₂]₂Ta(CH₂)CH₃, its electrophilic nature allows for an improbable double group transfer to occur when exposed to pyridine N-oxides. This reactions yields [TolC(NSiMe₃)₂]₂Ta(O)CH₃ due to simultaneous deoxygenation and regioselective methylation of the pyridine N-oxide.

The benzamidinate tantalum ethylidene complex is also able to react with nitrones, which are similar in structure for pyridine N-oxides. It is not however able to react with weak oxidants such as styrene oxide and triphenylphosphine oxide. Only one equivalent of the pyridine N-oxide was needed for the aforementioned reaction to take place. 2-Methylpryidine is produced as well, as confirmed by comparison using NMR integration versus a trimethoxybeneze internal standard. The trimethoxybenze reacts further with 2-methylpyridin N-oxide to ultimately yield 2,6-dimethylpyridine and oxo complex.

It should be noted that methylation occurs regioselectively at the unsubstituted ortho position in each pyridine N-oxide. Also, the substituted pyridine N-oxide species react much slower than the unsubstituted variant, comparatively in minutes versus microseconds.

Proton and carbon 13 NMR, IR spectroscopy, and X-ray crystallography were all used to verify the tantalum oxo complex product. The IR spectrum shows a strong stretch at 922 cm^-1, which is a feasible number to indicative of terminal Ta-O multiple bonds (typically 850-1000 cm^-1). X-ray crystallography reveals a distorted-octahedral coordination geometry surrounding the tantalum and thus confirmed the presence of a terminal oxo character. The measured bond length of the tantalum atom to oxygen bond is reported to be 1.76 Å, which is in line with previously reported figures for Ta-O multiple bonds. Thus, all the statistics seem to confirm that a double group transfer does indeed take place.

The mechanism of this reaction is thought to take place via two possible schemes involving a total of three mechanisms, but it is not known which scheme or mechanism is correct. There is an absence of intermediates in the reaction as evidenced by UV, IR, and ¹H NMR spectroscopy, so deuterium labeling is used to distinguish these potential routes of formation. GC-MS shows parent ion at m/z 95 and 111 corresponding to the methyl and dimethylpyridine products, respectively. At 2.40 ppm there is a 1:1:1 triplet indicative of the CH₂D group. This group also appears in both the proton and carbon 13 NMR spectras, which in all suggests that the mechanism of reaction takes place via scheme one and a mechanism label B.

Finally, nitrones which is similar in structure to pyridine N-oxides are also reacted with the benzamidinate tantalum ethylidene complex to see if they have a comparable interaction. Only after heating the complex with N-tert-butyl-α-phenylnitrone at 45 °C for 40 hours did styrene and another new organometallic product come to fruition. The new product is suspected to be [TolC(NSiMe₃)₂]­₂TA(O)(N^tBuMe) through ¹H, ¹³C{¹H} NMR, IR, and mass spectroscopic techniques.

In conclusion, it is the enhanced electrophilicity of the benzamidinate tantalum ethylidene which allows for the reaction pathway to occur. The atom transfer reactions allow for Ta-O double bonds and organic product with new C-C bonds to be formed. Further investigation into these matters is ongoing. I believe that following steps that could be taken would to delve into other metals complexes that could allow for double group transfers. Logically, I would think that the next metals to investigate would be other group 5 metals, possibly replacing Ta with Nb or Db. These metals should have the most similar properties in relationship to Ta. Reactants other than N-oxides and nitrones could also be analyzed to see if it is possible to replicate the double group transfer.

In a related study performed by ….

Purpose

To compare the moments of inertia calculated using two different methods, and to verify that angular momentum is conserved in an interaction between a rotating disk and a ring dropped onto the disk.

Hypothesis

If a weighted ring is added to the disk, the moment of inertia will be the same as the disk without the weighted ring. The angular momentum before the ring is dropped on the disk during part two will be greater than the angular momentum after the ring is dropped.

Labeled Diagrams

See attached sheet.

Data

Part 1

Mass of disk (M): 1.500 kg
Radius of disk (R): 0.114 m
Radius of shaft (r): 0.006 m
Mass of ring (m): 1.420 kg
Inner radius of ring (R₁): 0.054 m
Outer radius of ring (R₂): 0.064 m

Disk Alone

Disk plus ring

Part 2

Graphs

Part One (Disk Alone)

Part One (Disk Plus Ring)

Part Two

Questions

Part 1

1. In your data table in Part 1, you have two values for the moment of inertia. One is found from the theoretical equation for moment of inertia that is introduced in the Theory section and other is an experimental value obtained using Newton’s 2^nd law for rotational motion, τ = Iα, in conjunction with the definition of torque, τ =rF. How well do your two values agree with each other? What is the percent difference? Which do you think is likely a better way to calculate a value for moment of inertia?

The values are extremely close, as the percent difference for the disk alone is 9.5% and the disk plus the ring is 0% (they are of equal value). I think the better way to calculate the moment of inertia is to use I = ½ MR², as it is a more elegant equations that takes into account less variables. The other equation takes more variables into account, mainly for calculation torque, which I feel leads to increased error.

Part 2

1. How do your values for the angular momentum before and after the ring is dropped onto the disk compare? What is the percent difference?

The angular momentum before the ring is dropped onto the disk is greater than the angular momentum after the ring is dropped onto the disk. The percent difference is 11.7%.

2. Does there appear to be an inverse relationship between moment of inertia and angular velocity?

No, there appears to be a direct relationship between moment of inertia and angular velocity. As the angular velocity decreased, so did the moment of inertia.

3. How well do your results support the theory of conservation of momentum? What are the limitations of the experimental setup?

The results somewhat support the theory of conservation of momentum. The percent difference is 11.7%, which I suppose isn’t a huge discrepancy, but it could be better. The limitations of the experimental setup were that it is difficult to drop the ring on the spinning disk perfectly. We were able to drop the ring into the grooves of the disk, but there was still some wiggle room in those grooves. The ring would need to fit in the grooves like a puzzle piece in order to be positioned dead center to yield the least amount of error.

Conclusion

Lab Summarized

During the first part of the lab, the moments of inertia for a spinning disk with and without a weighted ring on top were calculated using two different methods. The force coercing the disk to spin was a 300 g weight attached to the shaft of the disk using a string a pulley system. The weight was allowed to free fall and the resulting graph of velocity versus time was used to find the final angular velocity by taking the mean of the segment after which the string had completely unraveled from the shaft. The angular acceleration was found from the slope of this graph up to that point.

These values, along with the force of kinetic friction, found by determining the minimum force needed to get the disk spinning, were used to find the moment of inertia. The moment of inertia was also calculated a second way, using the radii and masses of the disk and ring.

The second part of the experiment was performed much like part one of the experiment using the disk alone, only this time shortly after the string had unraveled, the ring was dropped onto the spinning disk. Using the angular velocities and moment of inertias, determined much like they were in part one, the angular moments before and after the ring were dropped were calculated and compared.

The percent differences between the two different calculations of the moments of inertia in part one were quite low. Using the disk alone, the percent difference was 9.5% and with the disk plus the ring, the percent difference as 0%. Under perfect conditions, the values should have been equal. The angular acceleration and final velocity for the disk along were greater than that of the measurement for the disk plus the ring, which could be expected, due to the extra mass. The percent difference between the angular momentums before and after the ring was dropped in part two was 11.7%. They should have been equal under ideal conditions.

As stated previously in the questions, some of this error is most likely due from the ring not being placed dead center around the spinning disk. If the disk was dropped at and angle and did not make complete contact with the disk at the same instant, this could have also caused error. If the shaft was not properly lubricated, this would have caused error throughout the experiment. Lastly, if the string ever caught a snag while unraveling, this would have also contributed to the error.

Equations

I =τ/α
I = ½ MR2
L = I_fω_f
I_iω_i = I_fω_f