Calculate the frequency separation between two proton resonances having d=3.49 a

Question

Calculate the frequency separation between two proton resonances having d=3.49 and d=8.50 on an NMR spectrometer operating at 500 MHz.

A proton in a given molecule interacts with 1 neighboring protons. The expected proton resonance, therefore, will be split into 2 lines. Determine the relative peak intensity distribution for the resonance. Enter each peak intensity separated by a colon, for example, a resonance split into three lines having relative intensities 3 to 11 to 4 would be entered as 3:11:4.

Explanation / Answer

1. Chemical shift = (difference of frequencies with TMS)/ spectrometer frequency

For d = 8.5 = frequency of peak / 500

frequency of peak = 8.5 x 500 = 4250Hz

For d = 3.49, frequency of peak = 3.49 x 500 = 1745 Hz

Frequency separation = 4250-1745 = 2505Hz

2. Because the proton has only 1 neighbouring proton, no of peaks splitting = N+1 where N = no. of neighbouring proton.

So, splitting peaks = 1+1 = 2 peaks.

The intensities of the peaks are given by coefficients of Pascals's Triangle. So for a DOUBLET, the ratio is 1:1

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