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The two fine–structure components of a nuclear magnet ic resonance transit ion are observed at chemical shifts of 2.142 and 2.208 ppm in a 300 MHz NMR spectrometer. Calculate the coupling constant:
- a)19.8 Hz
- b)0.066 Hz
- c)6.6 Hz
- d)date is insufficient
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
The two fine–structure components of a nuclear magnet ic resonan...
Given:
- Chemical shifts of the two finestructure components: 2.142 and 2.208 ppm
- Frequency of the NMR spectrometer: 300 MHz
To find:
The coupling constant between the two finestructure components.
Explanation:
Nuclear magnetic resonance (NMR) spectroscopy is a powerful technique used to analyze the structure and properties of molecules. In NMR, the chemical shifts of peaks in the spectrum provide information about the local chemical environment of the nuclei being studied. The splitting of peaks into multiple lines is known as finestructure and is caused by the coupling of nuclei to nearby nuclei.
The coupling constant (J) is a measure of the strength of this coupling. It represents the energy difference between the two spin states of the coupled nuclei. The coupling constant can be calculated using the formula:
J = Δv / (2 * δν)
where Δv is the difference in chemical shifts between the two finestructure components, and δν is the frequency of the NMR spectrometer.
Calculation:
Given:
Δv = 2.208 ppm - 2.142 ppm = 0.066 ppm
δν = 300 MHz
Using the formula:
J = (0.066 ppm * 300 MHz) / (2 * 300 MHz)
J = 0.066 ppm / 2
J = 0.033 ppm
Converting the coupling constant from ppm to Hz:
J (Hz) = J (ppm) * δν
J (Hz) = 0.033 ppm * 300 MHz
J (Hz) = 9.9 MHz
Therefore, the coupling constant between the two finestructure components is 9.9 MHz, which is equivalent to 19.8 Hz.
Conclusion:
The correct answer is option 'A' - 19.8 Hz.
- Chemical shifts of the two finestructure components: 2.142 and 2.208 ppm
- Frequency of the NMR spectrometer: 300 MHz
To find:
The coupling constant between the two finestructure components.
Explanation:
Nuclear magnetic resonance (NMR) spectroscopy is a powerful technique used to analyze the structure and properties of molecules. In NMR, the chemical shifts of peaks in the spectrum provide information about the local chemical environment of the nuclei being studied. The splitting of peaks into multiple lines is known as finestructure and is caused by the coupling of nuclei to nearby nuclei.
The coupling constant (J) is a measure of the strength of this coupling. It represents the energy difference between the two spin states of the coupled nuclei. The coupling constant can be calculated using the formula:
J = Δv / (2 * δν)
where Δv is the difference in chemical shifts between the two finestructure components, and δν is the frequency of the NMR spectrometer.
Calculation:
Given:
Δv = 2.208 ppm - 2.142 ppm = 0.066 ppm
δν = 300 MHz
Using the formula:
J = (0.066 ppm * 300 MHz) / (2 * 300 MHz)
J = 0.066 ppm / 2
J = 0.033 ppm
Converting the coupling constant from ppm to Hz:
J (Hz) = J (ppm) * δν
J (Hz) = 0.033 ppm * 300 MHz
J (Hz) = 9.9 MHz
Therefore, the coupling constant between the two finestructure components is 9.9 MHz, which is equivalent to 19.8 Hz.
Conclusion:
The correct answer is option 'A' - 19.8 Hz.
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Community Answer
The two fine–structure components of a nuclear magnet ic resonan...
(2.208-2.142) ppm×300MHz= 19.8Hz
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The two fine–structure components of a nuclear magnet ic resonance transit ion are observed at chemical shifts of 2.142 and 2.208 ppm in a 300 MHz NMR spectrometer. Calculate the coupling constant:a)19.8 Hzb)0.066 Hzc)6.6 Hzd)date is insufficientCorrect answer is option 'A'. Can you explain this answer?
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The two fine–structure components of a nuclear magnet ic resonance transit ion are observed at chemical shifts of 2.142 and 2.208 ppm in a 300 MHz NMR spectrometer. Calculate the coupling constant:a)19.8 Hzb)0.066 Hzc)6.6 Hzd)date is insufficientCorrect answer is option 'A'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about The two fine–structure components of a nuclear magnet ic resonance transit ion are observed at chemical shifts of 2.142 and 2.208 ppm in a 300 MHz NMR spectrometer. Calculate the coupling constant:a)19.8 Hzb)0.066 Hzc)6.6 Hzd)date is insufficientCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The two fine–structure components of a nuclear magnet ic resonance transit ion are observed at chemical shifts of 2.142 and 2.208 ppm in a 300 MHz NMR spectrometer. Calculate the coupling constant:a)19.8 Hzb)0.066 Hzc)6.6 Hzd)date is insufficientCorrect answer is option 'A'. Can you explain this answer?.
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