In a 200 MHz NMR spectrometer, a molecule shows two doublets separated by 2 ppm. The observed coupling constant is 10 Hz. The separation between these two signals and the coupling constant in a 600 MHz spectrometer will be, respectively
(a)600 Hz and 30 Hz
(b)1200 Hz and 30 Hz
(c)600 Hz and 10 Hz
(d)1200 Hz and 10 Hz
Answer
Answer: D ✓ checked by 4AB · confidence high
Explanation
2 ppm at 600 MHz = 1200 Hz; J is field-independent, 10 Hz (d).