$^{1}\mathrm{H}$ NMR spectrum of an organic compound recorded on a 500 MHz spectrometer showed a quartet with line positions at 1759, 1753, 1747, 1741 Hz. Chemical shift ($\delta$) and coupling constant (Hz) of the quartet are
(a)3.5 ppm, 6 Hz
(b)3.5 ppm, 12 Hz
(c)$3.6 \mathrm{ppm}, 6 \mathrm{Hz}$
(d)$3.6 \mathrm{ppm}, 12 \mathrm{Hz}$
Answer
Answer (as printed): A
Explanation
Frequency in Hz = Operating frequency of instrument (in Hz) $\times$ chemical shift in ppmChemical shift = Frequency in Hz / Operating frequency of instrumentAverage frequency = (1759 + 1753 + 1747 + 1741) / 4 = 1750 Hz. Thus chemical shift = 1750/500 = 3.5 ppmCoupling constant J = difference between any two adjacent line frequencies = 1759 - 1753 Hz = 6 Hz