Q13-1P - Organic Chemistry (Mcmurry)

Question

The amount of energy required to spin-flip a nucleus depends both on the strength of the external magnetic field and on the nucleus. At a field strength of 4.7 T, rf energy of 200 MHz is required to bring a ${}^{\mathbf{1}}\mathbf{H}$ nucleus into resonance, but energy of only 187 MHz will bring a ${}^{\mathbf{19}}\mathbf{F}$ nucleus into resonance. Calculate the amount of energy required to spin-flip a ${}^{\mathbf{19}}\mathbf{F}$ nucleus. Is this amount greater or less than that required to spin-flip a ${}^{\mathbf{1}}\mathbf{H}$ nucleus?

Step-by-Step Solution

Answer

The energy needed to spin flip a ${}^{\text{19}}\text{F}$ nucleus is $1.24\times {10}^{-25}\mathrm{J}$. This is lower in energy than the energy needed to spin flip a hydrogen nucleus. 

Step 1: Spin-flip a nucleus

The application of a very strong magnetic field makes the energy difference between the two spin states to be larger and high energy is required for spin-flip. The application of a weaker magnetic field leads to a lesser energy between the spin states.

Step 2: Amount of energy needed to spin flip the fluorine-19 nucleus

To determine the energy needed to spin flip of ${}^{\text{1}}\text{H}$ and ${}^{\text{19}}\text{F}$ nucleus, we will need to multiply the frequency of electromagnetic energy required to initiate the transition by the Planck’s constant and Avogadro number. This gives the energy in kilojoules per mole.

 

For ${}^{\text{1}}\text{H}$

 $$\begin{gathered} \mathrm{\Delta E}=\left(6.02\times {10}^{23}\right)\left(6.63\times {10}^{-34}\right)\left(200\times {10}^6\right) \\ =\frac{8.0\times {10}^5\mathrm{kJ}}{\mathrm{mol}} \end{gathered}$$

 

For ${}^{\text{19}}\text{F}$

$$\begin{gathered} \Delta E=\left(6.02\times {10}^{23}\right)\left(6.63\times {10}^{-34}\right)\left(187\times {10}^6\right) \\ =\frac{7.5\times {10}^{5}kJ}{mol} \end{gathered}$$

The electromagnetic energy needed to spin flip the ${}^{\text{1}}\text{H}$ nucleus is $\text{200 MHz}$, which is higher in frequency than the radiation needed for ${}^{\text{19}}\text{F}$ nucleus. Because the energy is directly proportional to the frequency of the radiation, the hydrogen nucleus will require higher energy.

The energy needed to spin flip a ${}^{\text{19}}\text{F}$ nucleus is $1.24\times {10}^{-25}\mathrm{J}$. This is lower in energy than the energy needed to spin flip a hydrogen nucleus.