Archives for November 2009

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

Purpose

To utilize two different methods of determining the initial velocity of a fired ball, namely a ballistic pendulum and treating the ball as a projectile, and then compare these two calculated values. The loss of kinetic energy from firing the ball into the pendulum is also an area of interest.

Hypothesis

The initial velocity determined by firing the ball into the ballistic pendulum should theoretically be equal to the initial velocity determined by firing the ball as a projectile.

Labeled Diagrams

See attached sheet.

Data

Part One

v= 6.23 m/s α_v=0.0187 m/s

Part Two

Y=1.003 m v= 8.75 m/s α_v=0.120 m/s

Questions

1. Compare the two different values of v average. Calculate the percent difference between them. State whether the two measurements agree within the combined standard errors of the two values of v average.

The average initial velocity for the ballistic pendulum was 6.23 m/s while the average initial velocity for the projectile determination was 8.75 m/s. This is a percent difference of 33.6%. It should have been expected that these two values would be equal. The two measurements also do not agree within the combined standard errors of the two values for v average, as the standard errors only total 0.1387 m/s, and the average velocities fall out of that range.

2. Calculate the loss in kinetic energy when the ball collides with the pendulum as the difference between ½ mv² (the kinetic energy before) and ½ (m + M)V² (the kinetic energy immediately after the collision). What is the fractional loss in kinetic energy? Calculate by dividing the loss by the original kinetic energy.

The average kinetic energy before the collision is 1.35 J and the average kinetic energy immediately after the collision is 0.272 J, so the loss of kinetic energy is 1.08 J. The fractional loss in kinetic energy is 0.8.

3. Calculate the ratio M / (m + M) for the values of m and M in Part 1. Compare this ratio with the ratio calculated in the previous question. Express the fractional loss of kinetic energy in symbol form and use equations from the Theory section to show it should equal M / (m + M).

The ratio M / (m + M) is equal to 0.8. This ratio is exactly the same as the fractional loss of kinetic energy.

The fractional loss of kinetic energy equals ( ½ mv² – ½ (m + M)V² ) / ( ½ mv² ).

Conclusion

During part one of the experiment, a ball was fired into a ballistic pendulum to ultimately determine its initial velocity. This process was repeated five times in order to obtain average values to work with in order to eliminate error. By massing the ball and the pendulum, recording the initial and final heights, the values for V and finally v could be calculated. It was found that the average initial velocity of the ball was 6.23 m/s.

During part two of the experiment, the same ball was fired as a projectile instead of into a ballistic pendulum. The ball was fired from a table horizontally to the ground. A piece of carbon paper was used to capture the spot where the ball first struck the ground. Height and horizontal distance the ball traveled were then measured in order to determine the initial velocity of the ball. The average initial velocity of the ball was 8.75 m/s.

As far as the accuracy of the results from the lab, the percent difference between the average velocities calculated is 33.6%. This is a fairly significant difference, which suggests that there sources of error during the procedure. The notched part of the ballistic setup could have had finer groves to yield more accurate measurements. The major contributor of error, however, was most likely from the distance measurements from the projectile part of the lab. One positive to come from the results was that the fractional loss in kinetic energy was identical to the mass ratios from the ballistic pendulum setup, which is theoretically expected.

Equations

∆KE = ½ mv²

½ (m + M)V² = (m + M)gh

mv = (m + M)V

V = (2gh)^0.5 v = (m + M) (2gh)^0.5 / m v = ∆x / (2∆y / g)^0.5

This is a PowerPoint presentation I did for class.