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1. |
This molecule is clearly undergoing an exchange process and this is due to restricted or hindered rotation about the bond between the aromatic ring and the amine nitrogen. If you envisage the molecule "frozen" with the -NH(CH3) group locked in the plane of the aromatic ring, the aromatic protons are clearly non-equivalent with one of the protons on the same side of the molecule as the CH3 group on the amine (syn) and one on the opposite side (anti). At low temperature (slow exchange on the NMR timescale) there are two different aromatic protons and these would couple to each other giving the observed two sets of doublets (one doublet for each aromatic proton at nA and nB). At the high temperature (fast exchange on the NMR timescale) the rotation of the -NH(CH3) group is rapid and the aromatic resonances are averaged to a singlet. Note that the singlet appears at the midpoint of nA and nB. At intermediate temperatures the spectra are broadened due to exchange and even though they are featureless, the rate of exchange (at various temperatures) can be obtained by analysis of the exchange broadened spectra. A simple estimate of the rate of exchange can be obtained at the coalescence point (the temperature at which the two separate signals merge to a single broad resonance). In this series of spectra this occurs at -32.7oC. At this temperature the rate of exchange (k) is give by: k = p.Dn / Ö2 where Dn is the chemical shift difference (nA - nB ) at slow exchange. From the slow exchange spectrum, Dn is approximately 17 Hz so k = 3.14 x 17 / 1.1414 = 38 sec-1 at 32.7 oC. Once the rate constant has been determined, it is then possible to calculate a range of thermodynamic and kinetic values for the exchange process (e.g. barrier heights or enthalpies and entropies for the exchange from the temperature dependence of the rate constant). |
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2. |
1,4-dioxan is a 6-membered ring and adopts a chair conformation (like other 6-membered rings such as cyclohexane). In the chair conformation, there are two proton environments and protons can be either axial or equatorial. At room temperature, rapid ring flipping causes protons to exchange between axial and equatorial sites so the proton spectrum is a singlet. In 1,4-dioxan-d7, seven out of the 8 protons have been substituted deuterium, leaving only 1 proton. Again this proton would be expected to be in either the axial or equatorial position and at low temperature (slow exchange) the protons in axial positions will have a different chemical shift to protons in the equatorial sites. At low temperature where the ring flip is slow, there will be molecules with the single proton axial some with the proton equatorial. There would be two signals one corresponding to axial protons and one corresponding to equatorial protons. The fact that these signals are of approximately equal in intensity indicates that there is no significant thermodynamic preference to have the proton to be equatorial or axial. At high temperature, the ring flip is rapid and the proton is rapidly exchanged between axial and equatorial sites and the shift is averaged to a single resonance at a shift half way between the signals between the signals in the slow exchange spectrum. |
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3. |
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i) |
The energy difference between the a and b states is proportional to the strength of the magnetic field (the Larmor equation). The greater the strength of the magnetic field, the larger the energy difference. As the energy difference increases, the population difference between the aand b states also increases (the Boltzmann equation) and the greater the population difference, the more intense is the NMR signal that can be observed. So signal intensity will be larger in larger magnetic fields. |
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ii) |
The Boltzmann equation also has a temperature dependence. As temperature is increased, the population difference decreases - at absolute zero, all the population is in the lowest energy state so the population difference is a maximum. At infinitely high temperature, the population difference is zero so there would be no signal. So, in general, the higher the temperature, the less intense the NMR signal. |
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iii) |
2H has a lower magnetogyric ratio than 1H so according to the Larmor equation its resonance frequency will be lower reflecting the lower energy gap between the states. As in (I) above, the NMR signal intensity would be less if 2H was observed instead of 1H. |
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iv) |
The NMR signal intensity basically reflects the amount of sample in the NMR probe so, assuming the concentration of the sample remains constant, the NMR signal intensity will be greater if more sample can be put into the spectrometer. |
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v) |
As for (iv), if the concentration of NMR-active spins in the sample, is increased, the intensity of the NMR signal increases. |
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4. |
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i) |
Fe(III) is paramagnetic so a soluble paramagnetic salt in an NMR sample will efficiently relax the nuclei in the sample. Nuclei which relax rapidly give rise to broad signals in the NMR spectrum. |
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ii) |
Oxygen is paramagnetic and even small amounts of oxygen dissolved in the sample contributes to relaxation. So if the sample was rigorously degassed, this would remove oxygen and the lines would get sharper because the relaxation would be less efficient. |
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iii) |
When a solution is cooled to near its freezing point, it becomes viscous and less mobile. Solutes dissolved in the solution are less mobile and tumble more slowly as the viscosity is increased. Relaxation is more efficient when molecular motion is slowed so lines become broader as the solution becomes more viscous. |