Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Organic Spectroscopy => Topic started by: HB on February 26, 2013, 01:07:20 PM
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the protons of 4-chloropyridine appeared as two doublets upon conducting the NMR on 300 MHz apparatus
upon using 500 MHz apparatus, they appeared as two double doublets dd, can any one explain that to me
also, how can I calculate the ratio of Z and E isomers from 1HNMR, each proton show two signals but with different intensities
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The proton couplings in this case are dd, it is just that they are not seen in 300 MHz NMR, due to the fact that it operates at a lower frequency. The higher frequency of the 500 MHz NMR means that the magnetic field in the NMR is stronger and thus the signal to noise ratio is higher and the sensitivity of a 500 MHz NMR is thus higher. The stronger magnetic field spreads out the chemical shifts and thus gives a more accurate look of the FID of your compound.
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You calculate the E/Z ratio from the hight (integral) of the two different peaks, each for one isomer. If one integral for the E isomer is 2 and the integral for the other, Z isomer is, say, 0.4 [of course, you have to accurately assign the correct protonos of each isomer], this means that the ratio of E/Z = 2/0.4 which is 5/1
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4-chloropyridine should produce a spectrum of the type AA'XX' if I am not mistaken.