Q14 · CSIR NET Chemistry, December 2016
Paper: CSIR NET December 2016 · Subject: Organic Chemistry · Chapter: Organic Spectroscopy · Topic: NMR 1H · Marks: 2 · Difficulty: Medium
${}^1$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 ppm, 6 Hz
(d)3.6 ppm, 12 Hz
Answer
Answer (as printed): A
Explanation
Frequency in Hz = (Operating frequency of instrument (in Hz) ´ chemical shift in ppm) Chemical shift = Frequency in Hz / Operating frequency of instrument Average frequency = (1759 + 1753 + 1747 + 1741) / 4 = 1750 Hz. Thus Chemical shift = 1750/ 500 = 3.5 ppm Coupling constant J = difference between any two adjacent line frequencies = 1759-1753 Hz = 6 Hz
Explanation as extracted from the printed page; notation may be imperfect.
Study loop for Organic Spectroscopy