${}^{1}\mathrm{H}$ 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 and coupling constant (Hz) of the quartet are
(a)$3.5\,\mathrm{ppm}, 6\,\mathrm{Hz}$
(b)$3.5\,\mathrm{ppm}, 12\,\mathrm{Hz}$
(c)$3.6\,\mathrm{ppm}, 6\,\mathrm{Hz}$
(d)$3.6\,\mathrm{ppm}, 12\,\mathrm{Hz}$
Answer
Answer: A ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
Centre of the quartet = (1759 + 1741)/2 = 1750 Hz; at 500 MHz that is 1750/500 = 3.5 ppm. Line spacing = 6 Hz = J.