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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
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
In a 200 MHz NMR spectrometer, a molecule shows two doublets separated...
Correct answer is 'D' bcos J is constat is not changing so it remains as 10 Hz
& chemical shift is 2ppm it is also same in 600mhz
hence ,
2ppm =oscillating frequency in hz/operating freq
2×10^-6 = X/600×10^-6
.. 2 =X/600
X=1200 Hz
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Community Answer
In a 200 MHz NMR spectrometer, a molecule shows two doublets separated...
Explanation:
Nuclear Magnetic Resonance (NMR) is a spectroscopic technique used to observe local magnetic fields around atomic nuclei. It is used to determine the structure of organic molecules by analyzing the chemical shift and coupling constants of the nuclei.

Given:
Frequency of NMR spectrometer (v1) = 200 MHz
Chemical shift difference between the two doublets = 2 ppm
Coupling constant (J) = 10 Hz

To find:
Chemical shift difference and coupling constant in a 600 MHz NMR spectrometer.

Solution:

Chemical shift difference:
Frequency of NMR spectrometer is directly proportional to the chemical shift. Therefore, we can use the equation:

v1/v2 = δ2/δ1

where v1 and v2 are the frequencies of the two NMR spectrometers, and δ1 and δ2 are the chemical shifts of the molecule in the two spectrometers.

Substituting the given values:

200/600 = δ2/δ1

δ2/δ1 = 1/3

The chemical shift difference is given as 2 ppm in the 200 MHz spectrometer. Therefore, in the 600 MHz spectrometer, the chemical shift difference will be:

2 ppm x (1/3) = 0.67 ppm

Coupling constant:
The coupling constant is independent of the frequency of the NMR spectrometer. Therefore, the coupling constant in the 600 MHz spectrometer will be the same as in the 200 MHz spectrometer, which is 10 Hz.

Therefore, the separation between the two signals and the coupling constant in a 600 MHz spectrometer will be 0.67 ppm and 10 Hz, respectively. Hence, the correct option is (C).
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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 Hzb)1200 Hz and 30 Hzc)600 Hz and 10 Hzd)1200 Hz and 10 HzCorrect answer is option 'C'. Can you explain this answer?
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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 Hzb)1200 Hz and 30 Hzc)600 Hz and 10 Hzd)1200 Hz and 10 HzCorrect answer is option 'C'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about 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 Hzb)1200 Hz and 30 Hzc)600 Hz and 10 Hzd)1200 Hz and 10 HzCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 Hzb)1200 Hz and 30 Hzc)600 Hz and 10 Hzd)1200 Hz and 10 HzCorrect answer is option 'C'. Can you explain this answer?.
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