Chapter 14, Problem 14.1P

The ${}^{\text{1}}\text{H}$ NMR spectrum of ${\text{CH}}_{\text{3}}\text{OH}$ recorded on a $\text{500 MHz}$ NMR spectrometer consists of

two signals, one due to the ${\text{CH}}_{\text{3}}$ protons at $17\text{15 Hz}$ and one due to the $\text{OH}$ proton at

$1830\text{ Hz}$, both measured downfield from TMS. (a) Calculate the chemical shift of each

absorption. (b) Do the ${\text{CH}}_{\text{3}}$ protons absorb upfield or downfield from the $\text{OH}$ proton?

Interpretation Introduction

(a)

Interpretation: The chemical shifts of the given absorptions are to be calculated.

Concept introduction: In NMR spectrum, peaks are known as resonances, lines or absorptions. On the horizontal axis, the position of absorption is generally referred to as chemical shift. The chemical shift of any absorption is calculated by the formula,

$\text{Chemical shift}=\frac{\text{Observed chemical shift}\left(\text{in Hz}\right)\text{downfield from TMS}}{\upsilon \text{ of the NMR spectrometer}\left(\text{in MHz}\right)}$

Answer to Problem 14.1P

The chemical shifts of the given absorptions are $3.\text{43 ppm}$ and $3.66\text{ ppm}$.

Explanation of Solution

The observed chemical shift due to the ${\text{CH}}_{\text{3}}$ protons is $17\text{15 Hz}$.

The observed chemical shift due to the $\text{OH}$ protons is $1830\text{ Hz}$.

The operating frequency is $50\text{0 MHz}$.

The conversion of $\text{MHz}$ to $\text{Hz}$ is done as,

$\text{1 MHz}=1{\text{0}}^{\text{6}}\text{ Hz}$

Therefore, the conversion of $\text{500 MHz}$ to $\text{Hz}$ is done as,

$\text{500 MHz}=500\times 1{\text{0}}^{\text{6}}\text{ Hz}$

Chemical shift $\left(\text{in ppm}\right)$ for first absorption:

Chemical shift due to ${\text{CH}}_{\text{3}}$ protons:

The chemical shift of absorption is calculated by the formula,

$\text{Chemical shift}\left(\text{ppm}\right)=\frac{\text{Observed chemical shift}\left(\text{in Hz}\right)\text{downfield from TMS}}{\upsilon \text{ of the NMR spectrometer}\left(\text{in MHz}\right)}$

Where,

$•$ $\upsilon$ is the proportionality constant in $\text{MHz}$ and is known as operating frequency.

Substitute the values of observed chemical shift and operating frequency in the above formula to calculate the chemical shift due to ${\text{CH}}_{\text{3}}$ protons.

$\begin{array}{rll}\text{Chemical shift}& =& \frac{17\text{15 Hz}}{500\times 1{\text{0}}^{\text{6}}\text{ Hz}}\\ & =& 3.\text{43}\times 1{\text{0}}^{-6}\text{ Hz}\end{array}$

Therefore, the chemical shift is $3.43\times 1{\text{0}}^{-6}\text{ Hz}$.

The conversion of $\text{Hz}$ to $\text{ppm}$ is done as,

$\text{1 Hz}=1{\text{0}}^{\text{6}}\text{ ppm}$

Therefore, the conversion of $3.43\times 1{\text{0}}^{-6}\text{ Hz}$ to $\text{ppm}$ is done as,

$\begin{array}{rll}3.43\times 1{\text{0}}^{-6}\text{ Hz}& =& 3.43\times 1{\text{0}}^{-6}\times 1{\text{0}}^{\text{6}}\text{ ppm}\\ & =& 3.43\text{ ppm}\end{array}$

Hence, the chemical shift is $3.\text{43 ppm}$.

Chemical shift due to $\text{OH}$ protons:

Substitute the values of observed chemical shift and operating frequency in the above formula to calculate the chemical shift due to $\text{OH}$ protons.

$\begin{array}{rll}\text{Chemical shift}& =& \frac{1830\text{ Hz}}{50\text{0}\times 1{\text{0}}^{\text{6}}\text{ Hz}}\\ & =& 3.66\times 1{\text{0}}^{-6}\text{ Hz}\end{array}$

Therefore, the chemical shift is $3.66\times 1{\text{0}}^{-6}\text{ Hz}$.

The conversion of $\text{Hz}$ to $\text{ppm}$ is done as,

$\text{1 Hz}=1{\text{0}}^{\text{6}}\text{ ppm}$

Therefore, the conversion of $3.66\times 1{\text{0}}^{-6}\text{ Hz}$ to $\text{ppm}$ is done as,

$\begin{array}{rll}3.66\times 1{\text{0}}^{-6}\text{ Hz}& =& 3.66\times 1{\text{0}}^{-6}\times 1{\text{0}}^{\text{6}}\text{ ppm}\\ & =& 3.66\text{ ppm}\end{array}$

Hence, the chemical shift is $3.66\text{ ppm}$.

Conclusion

The chemical shifts of the given absorptions are $3.\text{43 ppm}$ and $3.66\text{ ppm}$.

Interpretation Introduction

(b)

Interpretation: Whether the ${\text{CH}}_{\text{3}}$ protons absorb upfield or downfield from the $\text{OH}$ proton is to be predicted.

Concept introduction: In NMR spectrum, peaks are known as resonances, lines or absorptions. On the horizontal axis, the position of absorption is generally referred to as chemical shift. The increasing order of chemical shift is plotted from right to left in NMR spectrum.

Answer to Problem 14.1P

The peak of ${\text{CH}}_{\text{3}}$ protons is upfield from the $\text{OH}$ proton.

Explanation of Solution

The terms, upfield and downfield expresses the relative location of signals. The meaning of upfield is to the right and of downfield is to the left. The ${\text{CH}}_{\text{3}}$ protons absorb upfield from the $\text{OH}$ proton as shown below.

Figure 1

Hence, the peak of ${\text{CH}}_{\text{3}}$ protons of ${\text{CH}}_{\text{3}}\text{OH}$ is upfield from the peak of $\text{OH}$ proton.

Conclusion

The peak of ${\text{CH}}_{\text{3}}$ protons is upfield.