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1. |
The 13C spectrum of the compound (A) contains 5 signals and from the multiplicity these can be assigned to the two CH2 carbons, one CH carbon, one CH3 signal and one quaternary carbon. |
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i) |
The signal-to-noise ratio (S/N) of the spectrum increases as the square root of the number of acquisitions that are added together. If 128 acquisitions are added, this 4 times the number of scans as the 32 used to accumulate the basic spectrum. The S/N increases by a factor of Ö 4 = 2. |
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ii) |
With only a 1 sec delay between acquisitions, it very likely that all of the carbons in the sample are not fully relaxed between scans and hence the relative intensities of the signals are not correct. A long delay between acquisitions would ensure that all the nuclei are fully relaxed and therefore all 5 signals should all be of the same intensity. |
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iii) |
A paramagnetic species added to the sample will cause more efficient relaxation of all nuclei. Any differences in intensity resulting from differences in relaxation times would be reduced. All of the lines in the spectrum would be broadened. |
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iv) |
Irradiation of the proton spectrum before (but not during) the acquisition would provide an NOE enhancement to the carbon spectrum (without decoupling carbon from protons). So the intensity of the signals from protonated carbons in the sample would be increased. The spectrum would not be proton decoupled so the multiplicities of signals would remain the same. |
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v) |
There are 8 protons in this molecule and the spin system can be assigned as an AA'GG'RX3 spin system. Because the spin system contains nuclei which are chemically equivalent but magnetically non equivalent it cannot be analysed by first order rules. |
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2. |
The FID and the spectrum are derived from a sample containing 2 species. The spectrum clearly contains one sharp resonance and one broad resonance and you would predict that the broad signal arises from a species with a short relaxation time and the narrow resonance arises from a species with a long relaxation time. |
|
i) |
The pulse angle is 45o so increasing this by a factor of 2 will mean that a 90o pulse is delivered to the sample. The response of the sample is a sinusoidal function of the pulse angle. Sin(45) = 1/Ö
2 and Sin(90) = 1 so the intensity of the signal in the FID and in the spectrum will be increased by a factor of Ö
2 (1.414). |
|
ii) |
The pulse angle is 45o so increasing this by a factor of 4 will mean that a 180o pulse is delivered to the sample. The response of the sample is a sinusoidal function of the pulse angle. Sin(180) = 0 so the intensity of the signal in the FID and in the spectrum will be zero. Both the FID and the spectrum would contain only noise. |
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iii) |
The pulse angle is 45o so increasing this by a factor of 3 will mean that a 135o pulse is delivered to the sample. The response of the sample is a sinusoidal function of the pulse angle. Sin(135) = Sin(45) = 1/Ö
2 so the intensity of the signal in the FID and in the spectrum will be unchanged. |
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iv) |
The signal-to-noise ratio (S/N) of the spectrum increases as the square root of the number of acquisitions that are added together. If 64 acquisitions are added the S/N increases by a factor of Ö
64 = 8. |