有机化合物波谱解析课件 核磁共振部分习题及答案_1.pdfVIP

st NMR problems 1 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 hydrogen’s 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. 1 / 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). 1 If you measure the coupling constants from the spectra, you would find that JH-11B is 1 approximately 80 Hz and that JH-10B is approximately 27 Hz. The ratio between these is 80/27=3 reflecting the fact that the magnetogyric ratio of 11B (γ11B) is approximately 3 times larger than the