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1H 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 (s) and coupling constant (Hz) of the quartet are:
- a)3.5 ppm, 6Hz
- b)3.5 ppm, 12Hz
- c)3.6 ppm, 6Hz
- d)3.6 ppm, 12Hz
Correct answer is option 'A'. Can you explain this answer?
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1H NMR spectrum of an organic compound recorded on a 500 MHz spectrome...
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Given information:
- 1H NMR spectrum of an organic compound was recorded on a 500 MHz spectrometer.
- The spectrum showed a quartet with line positions at 1759, 1753, 1747, and 1741 Hz.
To determine the chemical shift and coupling constant of the quartet:
1. Determine the chemical shift:
- The chemical shift is the location of a signal in the NMR spectrum, expressed in parts per million (ppm) relative to a reference compound.
- To calculate the chemical shift, we need to convert the line positions from Hz to ppm.
- The formula to convert from Hz to ppm is:
chemical shift (ppm) = line position (Hz) / spectrometer frequency (MHz)
- In this case, the spectrometer frequency is 500 MHz, so we can use the formula to calculate the chemical shift for each line position.
For the first line position:
chemical shift = 1759 Hz / 500 MHz = 3.518 ppm
For the second line position:
chemical shift = 1753 Hz / 500 MHz = 3.506 ppm
For the third line position:
chemical shift = 1747 Hz / 500 MHz = 3.494 ppm
For the fourth line position:
chemical shift = 1741 Hz / 500 MHz = 3.482 ppm
- The chemical shift values are approximately 3.518 ppm, 3.506 ppm, 3.494 ppm, and 3.482 ppm.
2. Determine the coupling constant:
- The coupling constant is a measure of the splitting pattern observed in the NMR spectrum.
- In this case, the spectrum shows a quartet, which indicates that the proton is coupled to three neighboring protons.
- The coupling constant is the distance between the peaks in the quartet, expressed in Hz.
- To determine the coupling constant, we need to calculate the difference between adjacent line positions.
For the first and second line positions:
coupling constant = 1759 Hz - 1753 Hz = 6 Hz
For the second and third line positions:
coupling constant = 1753 Hz - 1747 Hz = 6 Hz
For the third and fourth line positions:
coupling constant = 1747 Hz - 1741 Hz = 6 Hz
- The coupling constant is 6 Hz for all three pairs of adjacent line positions.
Conclusion:
- Based on the calculations, the chemical shift of the quartet is approximately 3.5 ppm. The coupling constant of the quartet is 6 Hz.
- Therefore, the correct answer is option 'A': 3.5 ppm, 6 Hz.
- 1H NMR spectrum of an organic compound was recorded on a 500 MHz spectrometer.
- The spectrum showed a quartet with line positions at 1759, 1753, 1747, and 1741 Hz.
To determine the chemical shift and coupling constant of the quartet:
1. Determine the chemical shift:
- The chemical shift is the location of a signal in the NMR spectrum, expressed in parts per million (ppm) relative to a reference compound.
- To calculate the chemical shift, we need to convert the line positions from Hz to ppm.
- The formula to convert from Hz to ppm is:
chemical shift (ppm) = line position (Hz) / spectrometer frequency (MHz)
- In this case, the spectrometer frequency is 500 MHz, so we can use the formula to calculate the chemical shift for each line position.
For the first line position:
chemical shift = 1759 Hz / 500 MHz = 3.518 ppm
For the second line position:
chemical shift = 1753 Hz / 500 MHz = 3.506 ppm
For the third line position:
chemical shift = 1747 Hz / 500 MHz = 3.494 ppm
For the fourth line position:
chemical shift = 1741 Hz / 500 MHz = 3.482 ppm
- The chemical shift values are approximately 3.518 ppm, 3.506 ppm, 3.494 ppm, and 3.482 ppm.
2. Determine the coupling constant:
- The coupling constant is a measure of the splitting pattern observed in the NMR spectrum.
- In this case, the spectrum shows a quartet, which indicates that the proton is coupled to three neighboring protons.
- The coupling constant is the distance between the peaks in the quartet, expressed in Hz.
- To determine the coupling constant, we need to calculate the difference between adjacent line positions.
For the first and second line positions:
coupling constant = 1759 Hz - 1753 Hz = 6 Hz
For the second and third line positions:
coupling constant = 1753 Hz - 1747 Hz = 6 Hz
For the third and fourth line positions:
coupling constant = 1747 Hz - 1741 Hz = 6 Hz
- The coupling constant is 6 Hz for all three pairs of adjacent line positions.
Conclusion:
- Based on the calculations, the chemical shift of the quartet is approximately 3.5 ppm. The coupling constant of the quartet is 6 Hz.
- Therefore, the correct answer is option 'A': 3.5 ppm, 6 Hz.
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1H 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 (s) and coupling constant (Hz) of the quartet are:a)3.5 ppm, 6Hzb)3.5 ppm, 12Hzc)3.6 ppm, 6Hzd)3.6 ppm, 12HzCorrect answer is option 'A'. Can you explain this answer?
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1H 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 (s) and coupling constant (Hz) of the quartet are:a)3.5 ppm, 6Hzb)3.5 ppm, 12Hzc)3.6 ppm, 6Hzd)3.6 ppm, 12HzCorrect 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 1H 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 (s) and coupling constant (Hz) of the quartet are:a)3.5 ppm, 6Hzb)3.5 ppm, 12Hzc)3.6 ppm, 6Hzd)3.6 ppm, 12HzCorrect 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 1H 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 (s) and coupling constant (Hz) of the quartet are:a)3.5 ppm, 6Hzb)3.5 ppm, 12Hzc)3.6 ppm, 6Hzd)3.6 ppm, 12HzCorrect answer is option 'A'. Can you explain this answer?.
1H 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 (s) and coupling constant (Hz) of the quartet are:a)3.5 ppm, 6Hzb)3.5 ppm, 12Hzc)3.6 ppm, 6Hzd)3.6 ppm, 12HzCorrect 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 1H 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 (s) and coupling constant (Hz) of the quartet are:a)3.5 ppm, 6Hzb)3.5 ppm, 12Hzc)3.6 ppm, 6Hzd)3.6 ppm, 12HzCorrect 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 1H 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 (s) and coupling constant (Hz) of the quartet are:a)3.5 ppm, 6Hzb)3.5 ppm, 12Hzc)3.6 ppm, 6Hzd)3.6 ppm, 12HzCorrect answer is option 'A'. Can you explain this answer?.
Solutions for 1H 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 (s) and coupling constant (Hz) of the quartet are:a)3.5 ppm, 6Hzb)3.5 ppm, 12Hzc)3.6 ppm, 6Hzd)3.6 ppm, 12HzCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
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