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1. |
Clearly the methine hydrogen in this "weird" metacyclophane molecule is constrained to be exactly above the centre of the benzene ring i.e. within the shielding zone of the aromatic system. On a Johnson-Bovey plot, the methine hydrogen will have a r-coordinate of zero and be located somewhere on the z axis.
The methine proton in 4-propylheptane (d 1.2 ppm) is a reasonable model for the "natural" chemical shift of the methine proton in (1) if there were no ring current shift. The observed shift (d -4.03 ppm) is 5.23 ppm to higher field (more shielded) than that expected by comparison with the model compound and this upfield shift can be attributed to the effect of the aromatic ring. |
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On a Johnson-Bovey plot, shielding of 5.23 ppm for a proton constrained along the z-axis would be expected if the proton was held about 1.9 Å above the centre of the aromatic ring. |
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2. |
Protons constrained in the shielding or deshielding zones of an aromatic ring frequently have abnormal chemical shifts due to the ring-current effect. The "natural shift of the methyl protons in alanine is d 1.2 ppm. |
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i) |
If the group is constrained above the centre of a benzene ring its r coordinate will be zero. With a z-coordinate of 2.8 Å, the Johnson-Bovey plot indicates there should be a shielding of approximately 2.4 ppm so the methyl protons would be expected to appear at 2.4 ppm to higher field than normal i.e. at 1.2 - 2.4 = -1.2 ppm. |
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ii) |
If the group is constrained above the centre of a benzene ring its r coordinate will be zero. With a z-coordinate of 3.8 Å, the Johnson-Bovey plot indicates there should be a shielding of approximately 1.0 ppm so the methyl protons would be expected to appear at 1.0 ppm to higher field than normal i.e. at 1.2 - 1.0 = 0.2 ppm. |
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iii) |
If the group is constrained in the plane of the benzene ring, its z-coordinate will be zero. With a r-coordinate of 4.9 Å, the Johnson-Bovey plot indicates there should be a de-shielding of approximately 0.25 ppm so the methyl protons would be expected to appear at 0.25 ppm to lower field than normal i.e. at 1.2 - 0.25 = 1.45 ppm. |
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3. |
The 13C NMR spectrum of CH2Cl2 has a single resonance for the single carbon in the molecule. The carbon is coupled to 2 equivalent protons so the signal is split to a multiplet of (2nI + 1) lines where I = ½ (for 1H) n = 2 because there are 2 protons: (2nI + 1) = (2 x 2 x ½) + 1 = 3. The signal will appear as a triplet. You should ask why the 13C spectrum exhibits coupling to protons but the 1H spectrum does not show coupling to 13C. The answer lies in the low natural abundance of 13C. Only about 1% of all C is 13C so when you are observing the 13C you are only observing about 1% of the carbons in the sample. However, the natural abundance of 1H is (effectively) 100% so every 13C you observe will have two NMR-active protons coupled to it. The converse is not true - most 1H nuclei will be in molecules where the carbon is not 13C (since most carbon is 12C and this is an NMR-silent nucleus). |
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4. |
The geometry of the trimethyl phosphite molecule is tetrahedral at phosphorus so the 3 methoxy substituents are all identical. There will be only one resonance in the 1H NMR spectrum and this would encompass all 9 protons. The 1H NMR spectrum appears as a doublet and this indicates that there must be coupling to some other nucleus. This other nucleus must be phosphorus since 31P is 100% abundant and it is the only other abundant NMR-active nucleus in the molecule. The 1H spectrum appears as a doublet - using the (2nI + 1) rule because I for 31P is ½, n = 1 because the protons are coupled to one 31P: (2nI + 1) = (2 x 1 x ½) + 1 = 2 The splitting between the lines in the 1H NMR spectrum must be the 3-bond coupling (3JP-H) between each of the protons of the methoxy group and the 31P. The 31P spectrum appears as a 10-line multiplet - using the (2nI + 1) rule because I for 1H is ½, n = 9 because the phosphorus is coupled to nine protons: (2nI + 1) = (2 x 9 x ½) + 1 = 10 The splitting between the lines in the 31P NMR spectrum must again be the 3-bond coupling (3JP-H) between each of the protons of the methoxy group and the 31P. Note that the outer lines of the multiplet are very weak and difficult to see without some sort of vertical expansion, because the relative intensity of the lines (as given by the binomial coefficients of order 10) is 1:9:36:84:126:126:84:36:9:1. |
