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1. |
This is a very straightforward question once you have correctly written down the 3 dimensional structure of each of the compounds in question. The 3 dimensional geometry and in particular the symmetry associated with each compound allows you to determine which atoms are (or are not) chemically equivalent. |
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i) |
(CH3)4C is a tetrahedral molecule with all 4 methyl groups equivalent. In the 1H NMR spectrum, there would be 1 NMR signal for the 4 equivalent CH3 groups. |
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ii) |
(CH3)4C is a tetrahedral molecule with all 4 methyl groups equivalent. In the 13C NMR spectrum, there would be 1 NMR signal for the 4 equivalent carbons of the methyl groups and one for the quaternary carbon at the centre of the molecule (2 signals in all). |
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iii) |
There is only one phosphorus in this molecule - there would be only one signal in the 31P NMR spectrum. |
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iv) |
The bonding in (CH3CH2)3PO is tetrahedral at phosphorus. Each ethyl group has 2 non equivalent C atoms, but the 3 ethyl groups are equivalent. In the 13C NMR spectrum, there would be 1 NMR signal for the CH3 groups and 1 for the CH2 groups (2 signals in all). |
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v) |
The 3 ethyl groups are equivalent. In the 1H NMR spectrum there would be 1 NMR signal (a multiplet) for the protons of the CH3 groups and 1 for the protons of the CH2 groups (2 signals in all). |
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vi) |
(CH3)3SiSi(CH3CH2)3 contains 2 Si atoms - they are non-equivalent since one has 3 methyl substituents and one has 3 ethyl substituents. In the 29Si NMR spectrum there would be 2 signals. |
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vii) |
The bonding in (CH3)3SiSi(CH3CH2)3 is tetrahedral at each of the silicon atoms. All 3 methyl substituents attached to one silicon are equivalent and would give rise to one signal in the 1H spectrum. All 3 ethyl substituents on the second silicon are equivalent and each would have 1 NMR signal for the CH3 groups and 1 for the CH2 groups. So overall there would be 3 signals in the 1H spectrum. |
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viii) |
The bonding in (CH3)3SiSi(CH3CH2)3 is tetrahedral at each of the silicon atoms. All 3 methyl substituents attached to one silicon are equivalent and would give rise to one signal in the 13C NMR spectrum. All 3 ethyl substituents on the second silicon are equivalent and each would have 1 NMR signal for the CH3 groups and 1 for the CH2 groups. So overall there would be 3 signals in the 13C NMR spectrum. |
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2. |
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i) |
Signals appear at 2.1 and 2.3 ppm in a spectrometer operating at 200 MHz. At 200 MHz, 1 ppm corresponds to 200 Hz. The chemical shift difference between the signals is 2.1 - 2.3 = 0.2 ppm. 0.2 ppm corresponds to a frequency difference of 0.2 x 200 = 40 Hz. |
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ii) |
At 600 MHz, 1 ppm corresponds to 600 Hz so 0.2 ppm corresponds to a frequency difference of 0.2 x 600 = 120 Hz. Note that tripling the spectrometer frequency also triples the frequency difference between signals in the NMR spectrum. |
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iii) |
Dispersion is directly proportional to magnetic field strength. As in the example above, if you triple the magnetic field strength of the spectrometer, you triple the resonance frequency of any specific nucleus and therefore triple the frequency difference between signals in the spectrum. |
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3. |
The resonance frequency of a nucleus (expressed in Hz) is directly proportional to the strength of the applied magnetic field and changes from spectrometer to spectrometer, depending on the strength of the magnet. It is more convenient to express the resonance frequency in dimensionless units (termed "parts per million" or "ppm") by dividing the actual resonance frequency of the nucleus by the Larmor frequency of the type of nucleus being observed. In this way, all chemical shifts are effectively "normalised" to take account of the fact that each different spectrometer may have a different magnetic field strength. Chemical shifts are typically measured relative to the frequency of some standard compound, taken by convention as a reference, and chemical shifts are usually expressed in units of ppm from the resonance of the reference compound. Chemical shifts expressed in ppm are independent of Bo and are tabulated as characteristic molecular properties. If the Larmor frequency for observing a nucleus in particular spectrometer is 400 MHz then 1 ppm corresponds to 400Hz. |
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4. |
The answer to this question requires an understanding of the Larmor equation. One of the things that is obvious from the Larmor equation is that the resonance frequency required to observe a certain type of nucleus is proportional to its magnetogyric ratio (g). The higher the magnetogyric ratio, the higher the frequency required to observe the NMR signal of the nucleus. The question asks for a comparison between the frequencies required for the observation of protons (1H) and phosphorus (31P) in the same magnet. The magnetogyric ratio of 31P is 0.405 times that of 1H so the frequency required to observe 31P will be scaled by the same factor: 31P frequency = 1H frequency x 0.405 = (0.405 x 60) MHz = 24.3 MHz. |
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5. |
The answer to this question again requires an understanding of the Larmor equation. From the Larmor equation, the resonance frequency required to observe a certain type of nucleus is proportional to the strength of the magnetic field (Bo). The higher the magnetic field, the higher the resonance frequency of a given type of nucleus. The question asks for a comparison between the frequencies required for the observation of 15N in two different magnets. The magnetic field strength is increased by a factor of (11.745/2.349) = 5 so the frequency required to observe 15N will be scaled by this factor: 15 N frequency at 11.745T = 15N frequency at 2.349T x 5= (10.13 x 5) MHz = 50.65 MHz. |
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6. |
The answer to this question is very similar to the answer to Question 4. The question asks for a comparison between the frequencies required for the observation of protons (1H) and deuterium (2H) in the same magnet. The magnetogyric ratio of 1H is 6.5 times that of 2H, so the frequency required to observe 2H will be 1/6.5 that required to observe 1H. 2 H frequency = 1H frequency / 6.5= (400 / 6.6) MHz = 61.54 MHz. |