Question

h. A shift to spectrum of a phosphane bef equal com- e with Self-study exercise The 100 MHz 'H NMR spectrum of butanone is shown in Fig. 4.19 and consists of a quartet, a The coupling constants J for the triplet and quartet are equal. Account for the observed spectrum. singlet and a triplet s. ublets called z). In o two Heteronuclear spin-spin coupling: 1c-H Each of the nuclei 'H and 13C has a magnetic spin quantum number 1 =, and when C and 'H nuclei are in close proximity, they can couple. However, in molecules contain- ing a natural isotopic distribution of carbon atoms, only 1% 13C nuclei. From a statistical consideration, it follows that in a H NMR spectrum of, for example acetone, 1C-H coupling is not observed, although it is observed in the 1C NMR spectrum of the same sample The 1C NMR spectrum of acetone exhibits a singlet due to the C-O carbon atom, and a quartet due to the two equivalent methyl 1C nuclei. otons of the ution, are moH e bai to bno ot g Self-study exercise Why do you not observe 13C-13C coupling in the 1'C NMR signal being spectrum of acetone? antum The following case studies illustrate the use of NM spectroscopy to study comnounds containing spin-active

Answer 1

Fig. 4.19 is not given hence I am answering 2nd part

C-13 has a natural abundance of 1.1% i.e out of 100 carbon nucleii only 1 carbon is C-13 which is NMR active and almost 99% C are C-12 which are NMR inactive so intensity of this C-13 is usually low

However probability of Two adjacent Carbon in acetone to be C-13 is actually 1% * 1% = which is 0.01% which is very very negligible hence coupling between C13 - C13 in acetone is not observed

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