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Chapter 9: Problem 10

The magnetic field strength is measured in units of tesla (T). Calculate the energy involved in an NMR transition at a field strength of \(4.7 \mathrm{~T}\). For \(\mathrm{H}, \gamma\) is \(2.7 \times 10^{8} \mathrm{~T}^{-1} \mathrm{~s}^{-1}\), and Planck's constant is \(9.5 \times 10^{-14}(\mathrm{kcal} / \mathrm{mol})(\mathrm{s})\).

Short Answer

Expert verified
The energy involved in the NMR transition is approximately \(1.20555 \times 10^{-4} \, \mathrm{kcal/mol} \).

Step by step solution

01

Identify the Given Values

List all given values from the problem: - Magnetic field strength: \( B = 4.7 \, \mathrm{T} \) - Gyromagnetic ratio for hydrogen (\( \gamma \)): \( \gamma = 2.7 \times 10^8 \, \mathrm{T}^{-1} \, \mathrm{s}^{-1} \) - Planck's constant (\( h \)): \( h = 9.5 \times 10^{-14} \, \mathrm{kcal/mol} \cdot \mathrm{s} \)
02

Calculate the Angular Frequency

Use the formula \( \omega = \gamma B \) to calculate the angular frequency. Substituting the values: \( \omega = 2.7 \times 10^8 \, \mathrm{T}^{-1} \, \mathrm{s}^{-1} \times 4.7 \, \mathrm{T} \) \( \omega = 1.269 \times 10^9 \, \mathrm{s}^{-1} \)
03

Calculate the Energy of the NMR Transition

Use the formula \( E = h \omega \) to calculate the energy. Substituting the values: \( E = 9.5 \times 10^{-14} \, \mathrm{kcal/mol} \cdot \mathrm{s} \times 1.269 \times 10^9 \, \mathrm{s}^{-1} \) \( E = 1.20555 \times 10^{-4} \, \mathrm{kcal/mol} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic field strength
Magnetic field strength is a measure of the intensity of a magnetic field. It is denoted by the symbol \( B \) and is measured in tesla (T). The magnetic field strength directly influences the energy levels and transition frequencies of nuclear spins in NMR spectroscopy. A stronger magnetic field increases the energy difference between nuclear spin states, which in turn affects the resonance frequency (\( \omega \)) during NMR transitions. To better understand these relationships:
Gyromagnetic ratio
The gyromagnetic ratio (\( \gamma \)) is a constant that relates the magnetic moment of a nucleus to its angular momentum. It varies between different types of nuclei and influences the resonance frequency in NMR. Mathematically, it is expressed in units of \( \text{T}^{-1} \text{s}^{-1} \). In the presence of a magnetic field, the angular frequency (\( \omega \)) of the nuclear spins can be calculated using the formula \( \omega = \gamma B \), where \( B \) is the magnetic field strength. For hydrogen nuclei (protons), \( \gamma \) is given as \( 2.7 \times 10^8 \text{T}^{-1} \text{s}^{-1} \).
Planck's constant
Planck's constant (\( h \)) is a fundamental constant in quantum mechanics that relates the energy of a photon to its frequency. For calculations in NMR, Planck's constant units are typically expressed in \( \text{kcal/mol} \cdot \text{s} \). It is used to calculate the energy (\( E \)) of a nuclear spin transition via the formula \( E = h \omega \), where \( \omega \) is the angular frequency.

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Most popular questions from this chapter

IR spectra can be obtained for samples in solution. The IR spectrum of an alcohol has a broad band around \(3300 \mathrm{~cm}^{-1}\) due to the \(\mathrm{O}-\mathrm{H}\) stretch, and this portion of the IR spectrum sharpens as the solution becomes more dilute. The \(\mathrm{O}-\mathrm{H}\) stretching band is also sharper for tert-butyl alcohol than it is for methyl alcohol at the same concentration. Explain why the \(\mathrm{O}-\mathrm{H}\) stretch sharpens upon dilution and for tert-butyl alcohol.

You have bottles containing three isomeric compounds, \(\mathbf{A}, \mathbf{B}\), and \(\mathbf{C}\), each of the formula \(\mathrm{C}_{10} \mathrm{H}_{14}\). The only significant bands in the \(\mathrm{IR}\) spectra of all three compounds are at about 3050,2950 , and \(1610 \mathrm{~cm}^{-1}\). The \({ }^{1} \mathrm{H}\) NMR spectrum of A shows signals at \(\delta 2.13(\mathrm{~s}, 6 \mathrm{H})\) and \(6.88(\mathrm{~s}, 1 \mathrm{H}) ;\) B shows signals at \(\delta 2.18(\mathrm{~s}, 3 \mathrm{H}), 2.25(\mathrm{~s}, 3 \mathrm{H})\), and \(6.89(\mathrm{~s}, 1 \mathrm{H})\); and \(\mathrm{C}\) shows signals at \(\delta 2.11(\mathrm{~s}, 3 \mathrm{H})\), \(2.19(\mathrm{~s}, 3 \mathrm{H}), 2.22(\mathrm{~s}, 6 \mathrm{H})\), and \(6.80(\mathrm{~s}, 2 \mathrm{H})\). The \({ }^{13} \mathrm{C}\) NMR spectrum of \(\mathrm{A}\) shows lines at \(\delta 131.02,133.52\), and \(19.04 \mathrm{ppm}\). Compound \(\mathbf{B}\) shows signals at \(\delta 133.76\), \(133.79,126.88,20.65\), and \(15.78\) ppm. Compound C shows signals at \(\delta 136.15\), \(134.43,131.69,128.29,20.74,20.20\), and \(14.86 \mathrm{ppm}\). What are the structures?

(a) A hydrogen attached directly to the carbon of a carbonyl group (aldehyde hydrogen) appears at \(\delta\) 9-10 ppm and must be strongly deshielded. Explain why. (b) Vinylic hydrogens that are on the \(\beta\) position of a conjugated carbonyl compound are also substantially downfield. Use resonance structures to explain why.

Explain carefully why the following two \({ }^{1} \mathrm{H}\) NMR spectra of 1-bromobutane look so different: (a) Spectrum taken at \(60 \mathrm{MHz}\) (b) Spectrum taken at 300 MHz

Isomeric compounds \(\mathbf{A}\) and \(\mathbf{B}\) have the composition \(\mathrm{C}_{11} \mathrm{H}_{12} \mathrm{O}_{4}\). Spectral data are summarized below. Deduce structures for \(\mathbf{A}\) and \(\mathbf{B}\) and explain your reasoning. Compound A IR (KBr): 1720 (s), 1240 (s) \(\mathrm{cm}^{-1}\) \({ }^{1} \mathrm{H} \mathrm{NMR}\left(\mathrm{CDCl}_{3}\right): \delta 2.46(\mathrm{~s}, 3 \mathrm{H})\) $$ \begin{aligned} &3.94(\mathrm{~s}, 6 \mathrm{H}) \\ &8.05(\mathrm{~d}, J=2 \mathrm{~Hz}, 2 \mathrm{H}) \\ &8.49(\mathrm{t}, J=2 \mathrm{~Hz}, 1 \mathrm{H}) \end{aligned} $$ Compound B IR (Nujol): 1720 (s), 1245 (s) \(\mathrm{cm}^{-1}\) \({ }^{1} \mathrm{H} \mathrm{NMR}\left(\mathrm{CDCl}_{3}\right): \delta 2.63(\mathrm{~s}, 3 \mathrm{H})\) $$ \begin{aligned} &3.91(\mathrm{~s}, 6 \mathrm{H}) \\ &7.28(\mathrm{~d}, J=8 \mathrm{~Hz}, 1 \mathrm{H}) \\ &8.00(\mathrm{dd}, J=8 \mathrm{~Hz}, 2 \mathrm{~Hz} ; 1 \mathrm{H}) \\ &\mathrm{dd}=\text { doublet of doublets } \\ &8.52(\mathrm{~d}, J=2 \mathrm{~Hz}, 1 \mathrm{H}) \end{aligned} $$
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