Question 2. (10 points) When obtaining NMR spectra of compounds containing a hyd

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Question 2. (10 points) When obtaining NMR spectra of compounds containing a hydrogen attached to a heteroatom (such as oxygen or nitrogen), why does the peak for the corresponding hydrogen appear broad and typically integrates as less than the expected number of protons? Question 2. Answers A. A single hydrogen will always integrate for one proton and give a sharp peak no matter what atom it is bound to. B. Broad peak is due to the slow (in terms of the NMR time scale) interchange for the acidic oir ionizable proton of the compound for a deuterium atom (an isotope of hydrogen) present in the NMR solvent, additionally the lower integration is due to the fact that the NMR cannot identify the deuterium atom, only the hydrogen atom. C. The strength of the hydrogen bond and its various vibrations lead to this broad peak. D. The shielding effect of the heteroatom leads to the peak broadening and lessened intensity. Question 3. (10 points) True or False running NMR samples in various deuterated solvents will provide identical spectra, with the only difference being the peak associated with the NMR solvent. A. Tru

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Question 2. (10 points) When obtaining NMR spectra of compounds containing a hyd

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