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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 'C'. Can you explain this answer?
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Most Upvoted Answer
In a 200 MHz NMR spectrometer, a molecule shows two doublets separated...
Explanation:
Given:
- Frequency of NMR spectrometer = 200 MHz
- Two doublets are separated by 2 ppm
- Observed coupling constant = 10 Hz
To find:
- Separation and coupling constant in a 600 MHz NMR spectrometer
Solution:
Chemical shift:
The separation between the two doublets is given by the chemical shift difference between them.
Δδ = 2 ppm
The chemical shift is independent of the magnetic field strength, so it remains constant across different NMR spectrometers.
Coupling constant:
The coupling constant is proportional to the magnetic field strength.
J1 / J2 = ν1 / ν2
where J1 and J2 are the coupling constants in two different magnetic field strengths, and ν1 and ν2 are the corresponding frequencies.
We can use this equation to find the new coupling constant in a 600 MHz spectrometer.
J2 = J1 (ν2 / ν1)
J2 = 10 Hz (600 MHz / 200 MHz)
J2 = 30 Hz
Separation:
The chemical shift remains constant, so the separation between the two doublets remains the same in a 600 MHz spectrometer.
Δδ = 2 ppm
Final answer:
The separation between the two signals and the coupling constant in a 600 MHz spectrometer will be 600 Hz and 10 Hz respectively.
Hence, option (C) is the correct answer.
Given:
- Frequency of NMR spectrometer = 200 MHz
- Two doublets are separated by 2 ppm
- Observed coupling constant = 10 Hz
To find:
- Separation and coupling constant in a 600 MHz NMR spectrometer
Solution:
Chemical shift:
The separation between the two doublets is given by the chemical shift difference between them.
Δδ = 2 ppm
The chemical shift is independent of the magnetic field strength, so it remains constant across different NMR spectrometers.
Coupling constant:
The coupling constant is proportional to the magnetic field strength.
J1 / J2 = ν1 / ν2
where J1 and J2 are the coupling constants in two different magnetic field strengths, and ν1 and ν2 are the corresponding frequencies.
We can use this equation to find the new coupling constant in a 600 MHz spectrometer.
J2 = J1 (ν2 / ν1)
J2 = 10 Hz (600 MHz / 200 MHz)
J2 = 30 Hz
Separation:
The chemical shift remains constant, so the separation between the two doublets remains the same in a 600 MHz spectrometer.
Δδ = 2 ppm
Final answer:
The separation between the two signals and the coupling constant in a 600 MHz spectrometer will be 600 Hz and 10 Hz respectively.
Hence, option (C) is the correct answer.
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Community Answer
In a 200 MHz NMR spectrometer, a molecule shows two doublets separated...
Two doublets separated by 2 ppm so for one doublet 1ppm there for the separation in 600MHz is ( dv×10^6 / 600×10^6 )=1ppm dv 10^6=1×600×10^6 V=600
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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 '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 '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 '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 'C'. Can you explain this answer?.
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 '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 '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 'C'. Can you explain this answer?.
Solutions 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 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for IIT JAM.
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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 'C'. Can you explain this answer?, a detailed solution 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 'C'. Can you explain this answer? has been provided alongside types of 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 'C'. Can you explain this answer? theory, EduRev gives you an
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