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The two fine-structure components of a nuclear magnetic resonance transition are observed at chemical shifts of 2.142 and 2.208 ppm in a 300 MHz NMR spectrometer. Calculate the coupling constant (in Hz):
Correct answer is between '19.5,20.0'. Can you explain this answer?
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The two fine-structure components of a nuclear magnetic resonance tran...
Coupling constant (J) = distance between peaks (ppm) * Frequency of NMR Spectrometer (MHz)
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**Explanation:**
To calculate the coupling constant, we need to determine the difference in chemical shift (Δδ) between the two fine-structure components and convert it to frequency using the formula:
Coupling constant (J) = Δδ * spectrometer frequency
In this case, the chemical shift difference (Δδ) is given as:
Δδ = 2.208 ppm - 2.142 ppm = 0.066 ppm
The spectrometer frequency is given as 300 MHz.
**Calculating the coupling constant:**
Coupling constant (J) = Δδ * spectrometer frequency
= 0.066 ppm * 300 MHz
= 19.8 Hz
The calculated coupling constant is 19.8 Hz.
Since the correct answer is given as between 19.5 and 20.0 Hz, the calculated value falls within this range.
**Reasoning:**
The coupling constant represents the interaction between the two coupled nuclei, and it depends on the strength of the magnetic interaction and the relative orientation of the nuclei. In this case, the coupling constant is influenced by the chemical environment and the electronic structure of the molecules.
The chemical shift difference (Δδ) between the two fine-structure components is small, indicating a small coupling constant. This suggests a weak magnetic interaction between the two coupled nuclei.
The spectrometer frequency is directly proportional to the coupling constant. A higher spectrometer frequency results in a higher coupling constant. In this case, the 300 MHz spectrometer frequency contributes to the calculated value of 19.8 Hz.
The given range of 19.5 to 20.0 Hz for the correct answer allows for some experimental error and variations in the measurement. The calculated value of 19.8 Hz falls within this range, indicating that it is a reasonable and acceptable value for the coupling constant in this scenario.
To calculate the coupling constant, we need to determine the difference in chemical shift (Δδ) between the two fine-structure components and convert it to frequency using the formula:
Coupling constant (J) = Δδ * spectrometer frequency
In this case, the chemical shift difference (Δδ) is given as:
Δδ = 2.208 ppm - 2.142 ppm = 0.066 ppm
The spectrometer frequency is given as 300 MHz.
**Calculating the coupling constant:**
Coupling constant (J) = Δδ * spectrometer frequency
= 0.066 ppm * 300 MHz
= 19.8 Hz
The calculated coupling constant is 19.8 Hz.
Since the correct answer is given as between 19.5 and 20.0 Hz, the calculated value falls within this range.
**Reasoning:**
The coupling constant represents the interaction between the two coupled nuclei, and it depends on the strength of the magnetic interaction and the relative orientation of the nuclei. In this case, the coupling constant is influenced by the chemical environment and the electronic structure of the molecules.
The chemical shift difference (Δδ) between the two fine-structure components is small, indicating a small coupling constant. This suggests a weak magnetic interaction between the two coupled nuclei.
The spectrometer frequency is directly proportional to the coupling constant. A higher spectrometer frequency results in a higher coupling constant. In this case, the 300 MHz spectrometer frequency contributes to the calculated value of 19.8 Hz.
The given range of 19.5 to 20.0 Hz for the correct answer allows for some experimental error and variations in the measurement. The calculated value of 19.8 Hz falls within this range, indicating that it is a reasonable and acceptable value for the coupling constant in this scenario.
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The two fine-structure components of a nuclear magnetic resonance transition are observed at chemical shifts of 2.142 and 2.208 ppm in a 300 MHz NMR spectrometer. Calculate the coupling constant (in Hz):Correct answer is between '19.5,20.0'. Can you explain this answer?
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The two fine-structure components of a nuclear magnetic resonance transition are observed at chemical shifts of 2.142 and 2.208 ppm in a 300 MHz NMR spectrometer. Calculate the coupling constant (in Hz):Correct answer is between '19.5,20.0'. 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 magnetic resonance transition are observed at chemical shifts of 2.142 and 2.208 ppm in a 300 MHz NMR spectrometer. Calculate the coupling constant (in Hz):Correct answer is between '19.5,20.0'. 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 magnetic resonance transition are observed at chemical shifts of 2.142 and 2.208 ppm in a 300 MHz NMR spectrometer. Calculate the coupling constant (in Hz):Correct answer is between '19.5,20.0'. Can you explain this answer?.
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