Complete the following table. Which nucleus requires the least energy to flip its spin at this applied field? Which nucleus requires the most energy? \begin{tabular}{lccc} \hline Nucleus & Applied Field (tesla, T) & Radio Frequency \((\mathrm{MHz})\) & Energy \((\mathrm{J} / \mathbf{m o l})\) \\ \hline\({ }^{1} \mathrm{H}\) & \(7.05\) & 300 & \\ \({ }^{13} \mathrm{C}\) & \(7.05\) & \(75.5\) & \(-\) \\ \({ }^{19} \mathrm{~F}\) & \(7.05\) & 282 & \(-\) \\ \hline \end{tabular}

Short Answer

Expert verified
Answer: The nucleus that requires the least energy to flip its spin is \({ }^{13} \mathrm{C}\) (29.95 J/mol), and the nucleus that requires the most energy is \({ }^{1} \mathrm{H}\) (119.7 J/mol).

Step by step solution

01

Calculate the energy for each nucleus

To find the energy required to flip the spin for each nucleus, we need to first convert the radio frequency from MHz to Hz by multiplying with 10^6. After that, we can apply the formula E = h * f to find the energy. Since the given unit for energy is in J/mol, we also need to account for that by multiplying by Avogadro's number (6.022 x 10^23). For \({ }^{1} \mathrm{H}\): Frequency (Hz) = 300 MHz * 10^6 = 3.00 x 10^8 Hz Energy = (6.626 x 10^(-34) J*s) * (3.00 x 10^8 Hz) * (6.022 x 10^23) Energy = 119.7 J/mol For \({ }^{13} \mathrm{C}\): Frequency (Hz) = 75.5 MHz * 10^6 = 7.55 x 10^7 Hz Energy = (6.626 x 10^(-34) J*s) * (7.55 x 10^7 Hz) * (6.022 x 10^23) Energy = 29.95 J/mol For \({ }^{19} \mathrm{~F}\): Frequency (Hz) = 282 MHz * 10^6 = 2.82 x 10^8 Hz, Energy = (6.626 x 10^(-34) J*s) * (2.82 x 10^8 Hz) * (6.022 x 10^23) Energy = 112.71 J/mol
02

Determine the nucleus with the least and the most energy for spin flip

Now we can compare the energy for each nucleus to determine which one requires the least energy and which one requires the most energy. Least energy required: \({ }^{13} \mathrm{C}\) (29.95 J/mol) Most energy required: \({ }^{1} \mathrm{H}\) (119.7 J/mol) Updated table: \begin{tabular}{lccc} \hline Nucleus & Applied Field (tesla, T) & Radio Frequency \((\mathrm{MHz})\) & Energy \((\mathrm{J} / \mathbf{m o l})\) \\\ \hline\({ }^{1} \mathrm{H}\) & \(7.05\) & 300 & 119.7\\\ \({ }^{13} \mathrm{C}\) & \(7.05\) & \(75.5\) & 29.95 \\\ \({ }^{19} \mathrm{~F}\) & \(7.05\) & 282 & 112.71 \\\ \hline \end{tabular} The nucleus that requires the least energy to flip its spin is \({ }^{13} \mathrm{C}\), and the nucleus that requires the most energy is \({ }^{1} \mathrm{H}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Groups of atoms in which substitution of one atom by an isotope creates an achiral molecule are called homotopic. \- Homotopic groups always have identical chemical shifts. Those in which such substitution produces a chiral molecule are enantiotopic. Enantiotopic groups have identical chemical shifts, except in a chiral environment.

\({ }^{13} \mathbf{C}-\mathrm{NMR}\) is like \({ }^{1} \mathrm{H}-\mathrm{NMR}\), except the nuclear spins of \({ }^{13} \mathrm{C}\) nuclei are being analyzed. \- \({ }^{13}\) C-NMR spectra are commonly recorded in a hydrogen-decoupled instrumental mode. In this mode, all \({ }^{13} \mathrm{C}\) signals appear as singlets. \- The number of different signals in a \({ }^{13} \mathrm{C}-\mathrm{NMR}\) spectrum tell you how many nonequivalent carbon atoms are in a molecule. \- \({ }^{13}\) CNMR chemical shifts tell you what kind of carbon atoms are present.

Compound M, molecular formula \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}\), readily decolorizes \(\mathrm{Br}_{2}\) in \(\mathrm{CCl}_{4}\) and is converted by \(\mathrm{H}_{2} / \mathrm{Ni}\) into compound \(\mathrm{N}\), molecular formula \(\mathrm{C}_{3} \mathrm{H}_{12} \mathrm{O}\). Following is the \({ }^{1} \mathrm{H}-\mathrm{NMR}\) spectrum of compound \(\mathrm{M}\). The \({ }^{19} \mathrm{C}-\mathrm{NMR}\) spectrum of compound \(\mathrm{M}\) shows signals at \(\delta 146.12,110.75,71.05\), and \(29.88\). Deduce the structural formulas of compounds \(M\) and N.

The \({ }^{1} \mathrm{H}-\mathrm{NMR}\) of compound \(\mathrm{R}, \mathrm{C}_{6 i} \mathrm{H}_{14} \mathrm{O}\), consists of two signals: \(\delta 1.1\) (doublet) and \(\delta 3.6\) (septet) in the ratio \(6: 1\). Propose a structural formula for compound \(R\) consistent with this information.

The experimental conditions required to cause nuclei to resonate are affected by the local chemical and magnetic environments. \- Electrons around a hydrogen or carbon create local magnetic fields that shield the nuclei of these atoms from the applied field. \- Any factor that increases the exposure of nuclei to an applied field is said to deshield them and shifts their signal downfield to a larger \(\delta\) value. \- Conversely, any factor that decreases the exposure of nuclei to an applied field is said to shield them and shifts their signal upfield to a smaller \(\delta\) value.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free