1 / 14 NMR problems 1st part 1. The 1H NMR spectrum of the borohydride ion (BH4-) is given below. The element boron has two isotopes, both of which are NMR active: 10B (I=3, 20%) and 11B (I=3/2, 80%). Rationalize the appearance of the observed 1H spectrum. Answer The BH4- ion has a tetrahedral shape so all 4 hydrogens are equivalent. There will be only one resonance in the 1H NMR spectrum and this will be split by coupling to the boron nucleus. Boron has 2 isotopes 10B (I=3, 20%) and 11B (I=3/2, 80%). So for the 20% of molecules that contain 10B, the 1H resonance will be split into a 7-line multiplet by coupling to 10B (using the rule that multiplicity = 2nI +1 where n = 1and I=3). For the 80% of molecules that contain 11B, the 1H resonance will be split into a 4-line multiplet by coupling to 11B (using the rule that multiplicity = 2nI +1 where n = 1and I=3/2). The observed spectrum will be the superposition of these two contributing sub-spectra. 2 / 14 Note that if you could integrate the 7 lines from the spectrum arising from the molecules containing 10B and then the 4 lines the spectra arising from the molecules containing 11B, the relative intensities should be 1:4 (reflecting the ratio of the natural abundances of the 10B and 11B). If you measure the coupling constants from the spectra, you would find that 1JH-11B is approximately 80 Hz and that 1JH-10B is approximately 27 Hz. The ratio between these is 80/27=3 reflecting the fact that the magnetogyric ratio of 11B (11B) is approximately 3 times larger than the magnetogyric ratio and 10B ( 10B) and that coupling is proportional to the product of the magnetogyric ratios of the nuclei which are coupled (Jij i.j). 2.The 1H NMR spectrum of methane (CH4) consists of a single resonance and because the four protons have exactly the same chemical shift, no H-H couplings appear in the spectrum so it is not possible to measure 2JH-H from the spectrum. The 1H NMR spectrum of dideuteromethane (CD2H2) is given below. Rationalise the splitting pattern and explain how you could estimate 2JH-H using this spectrum. 3 / 14 Answer The CH2D2 molecule is tetrahedral and the two protons are equivalent. They have the same chemical shift and will be split by coupling to the two deuterium nuclei. Deuterium has a spin of 1. You would expect the proton signal to be split into 5 lines (using the rule that multiplicity = 2nI +1 where n = 2 and I=1) and 5 lines can be clearly seen in the spectrum with a splitting 2J1H-2H = 2.0 Hz. Given the fact that coupling is proportional to the product of the magnetogyric ratios of the nuclei which are coupled (Jij i.j), if we know 2J1H-2H (and this is proportional to 1H.2H) then we should be able to estimate 2J1H-1H in the same molecule. It should be fairly apparent that 2J1H-1H / 2J1H-2H = 1H / 2H and from the table in your hand-out material. 1H / 2H is approximately 6.5 so 2J1H-1H is approximately 6.5 x 2J1H-2H = 6.5 x 2.0 = 13.0 Hz. 3. The 1H NMR spectrum of propane consists of two signals ( 0.90 and 1.30) with areas in the ratio 3:1 respectively. Sketch the 1H spectrum of propane (to scale) for three different spectrometers with proton frequencies of 60 MHz, 300 MHz and 600 MHz. Answer The important thing to note is that the resonance frequency of a nucleus (expressed in Hz) depends on the strength of the magnetic field whereas the coupling constants (in Hz) are magnetic field independent. As the magnetic field is increased the signals get further apart (in Hz) but the widths of the multiplets remain the same. 4 / 14 4. The 1H NMR spectrum of dipropylether, (CH3CH2CH2)2O, is given below. The spectrum consists of 3 distinct resonances at 3.4 (2H, triplet), 1.6 (2H, multiplet) and 1.0 (3H, triplet). Assign the spectrum and clearly describe the spectra which would be obtained while applying a strong Rf field at: 1 3.4 ppm. 2 1.6 ppm. 3 1.0 ppm. Answer The signals in the spectrum can be readily assigned ( 3.4 is due to the CH2 group adjacent to the oxygen; 1.6 is due to the remaining CH2 group and 1.0 is due to the CH3 group). The assignment is based on (i) the relative intensities of the signals; (ii) the observed splitting pattern; (iii) the fact that the CH2 group adjacent to the oxygen must be the low field resonance. The resonance at 3.4 is a triplet because it is coupled to 2 protons on the adjacent carbon. The resonance at 1.6 is a triplet of quartets because it is coupled to the 2 protons at 3.4 (triplet splitting) and the 3 protons at 1.0 (quartet splitting). 5 / 14 The effect of irradiation at each of these frequencies is to decouple the nuclei which have signals at these frequencies. Decoupling effectively removes the nucleus from the spin system. 1 With irradiation at 3.4, the multiplicity of the CH2 resonance at 1.6 is simplified from a triplet of quartets to a quartet. The CH3 resonance at 1.0 remains unchanged. 2 With irradiation at 1.6, the multiplicity of the CH2 resonan