Question

te 8.24 1.46 2.00 1.54 200 824mwweemamn 14 12 10 PPM USER:DATE: 1106 18 (11:28) PTS1d: 8192 89.983 OFI: 540.0 SWI: 1497 PW:24.4 us PD: 3.0 sec

tabulate the nmr data and provide a concise explanation for the assignment of the nmr data. give its chemical shift, multiplicity, coupling constant, assignment of peaks. molecule is [(C5H5N)2I]Br

Answer 1

The NMR data given in the question is proton NMR. Generally ¹H-NMR are performed in solvents like D₂O, CDCl₃ etc. However the question does not specify the solvent. This is required because the deutariated solvents have a signal in the NMR data.

However we can still solve the NMR data assuming the solvent. First of all the leak at 0 ppm is due to Tetramethylsilane which is the standard used in NMR spectroscopy. From the molecular formula of the compound given I have drawn the predicted image of the compound and assigned each of the proton by a, b or c which types in NMR. As it can be seen that there are 4 a type, 4 b type and 2 c type protons present. All these protons are aromatic in nature. The ortho-protons i.e. the “a” type protons will be most deshielded by the effect of aromatic ring, proximity to N and positive I atoms. So they will show most downfield shift and we can assign the peak around 8.2 ppm to “a” protons. The signal is a doublet as there are 1 “b” type proton in the next carbon atom. The para-protons or the “c” type protons should be more deshielded than the “b” type protons as the electronegative effect of N-atom will be more in the para position. The “b” type protons will be slightly upfield shifted than the “c” type protons. The “c” protons have 2 “b” protons in the next carbon and therefore they show a multiplet/ triplet of triplet and the “b” protons have one “a” and one “c” type protons in the next carbon atom. SO they show multiple doublets. All the peaks due to “b” and “c” are in the region of 7.2.to 7.8 ppm. So these are all the signals from the compound. The coupling constants are difficult to guess from the complicated feature of the spectra.

This leaves us with the two peaks around ~3.7 and ~ 1.5 ppm. My best guess is that one of these peaks originated from the deuterium of the solvent and another from the residual water present in the solvent and the sample. May be D6-DMSO was used as the solvent to record this NMR spectra.