Chapter 13, Problem 13.63AP

Interpretation Introduction

Interpretation:

The proton decoupled ${}^{\text{13}}\text{C}\text{NMR}$ spectra of ethyl bromide is to be stated.

Concept introduction:

The ${}^{13}\text{C}$ NMR spectroscopy is similar kind of spectroscopy to NMR the only difference them is that it is used to detect the presence of ${}^{13}\text{C}$ nucleus. The natural abundance of ${}^{13}\text{C}$ nucleus is only $1.1\%$ while ${}^{12}\text{C}$ nucleus is not detectable by NMR spectroscopy. The difference between its resonance frequency and that of the reference standard is known as the chemical shift of a nucleus. The $\left(n+1\right)$ rule is applicable on the protons attached to the same carbon whose splitting pattern is to be determined.

Answer to Problem 13.63AP

The proton decoupled ${}^{\text{13}}\text{CNMR}$ spectra of ethyl bromide is shown below.

Explanation of Solution

The carbon atom exists in two isotopes ${}^{12}\text{C}$ and ${}^{13}\text{C}$ their relative intensity $0.989$ and $0.011$ respectively. The bromine atom exists in two isotopes ${}^{79}\text{Br}$ and ${}^{81}\text{Br}$ their relative intensity $\left(0.505\right)$ and $\left(0.495\right)$ respectively. In the given case consider ${}^{79}\text{Br}$ isotope only. The hydrogen atom exists basically in ${}^1\text{H}$ isotope with maximum intensity. The general formula for calculation of relative intensity is shown below.

$\text{Relative}\text{intensity}=\text{Multiplication}\text{of}\text{natural}\text{abundance}\text{of}\text{elements}$ …(1)

The relative intensity for the molecule in which no ${}^{13}\text{C}$ isotope is present is calculated by substituting the value of the element’s abundance in equation (1) as shown below.

$\begin{array}{rll}\text{Relative}\text{intensity}& =& \left(\begin{array}{c}{\left(\text{Natural}\text{abundance}\text{of}{}^1\text{H}\right)}^5\times {\left(\text{Natural}\text{abundance}\text{of}{}^{12}\text{C}\right)}^2\\ \times \left(\text{Natural}\text{abundance}\text{of}{}^{79}\text{Br}\right)\end{array}\right)\\ & =& {\left(0.999\right)}^5\times {\left(0.989\right)}^2\times \left(0.505\right)\\ & =& 0.995\times 0.9781\times 0.505\\ & =& 0.491\end{array}$

The coefficient represents the number of atoms present of each element. Here the possibility of two ${}^{12}\text{C}$ is present. The relative intensity for the molecule in which one ${}^{12}\text{C}$ and one ${}^{13}\text{C}$ isotope are present is calculated by substituting the value of elements abundance in equation (1) as shown below.

$\begin{array}{rll}\text{Relative}\text{intensity}& =& \left(\begin{array}{c}{\left(\text{Natural}\text{abundance}\text{of}{}^1\text{H}\right)}^5\times 2\left(\text{Natural}\text{abundance}\text{of}{}^{12}\text{C}\right)\\ \times \left(\text{Natural}\text{abundance}\text{of}{}^{13}\text{C}\right)\times \left(\text{Natural}\text{abundance}\text{of}{}^{79}\text{Br}\right)\end{array}\right)\\ & =& {\left(0.999\right)}^5\times \left(\left(0.989\right)\times \left(0.011\right)\times 2\right)\times \left(0.505\right)\\ & =& \left(0.995\right)\times \left(0.989\right)\times \left(0.011\right)\times 2\times \left(0.505\right)\\ & =& 0.0109\end{array}$

Here, ${}^{12}\text{C}$ and ${}^{13}\text{C}$ both can be present in vice-versa form that is why coefficient $2$ is multiplied in the calculation. Similarly, the relative intensity for the molecule in which two ${}^{\text{13}}\text{C}$ isotopes are present is calculated by substituting the value of elements abundance in equation (1) as shown below.

$\begin{array}{rll}\text{Relative}\text{intensity}& =& \left(\begin{array}{c}{\left(\text{Natural}\text{abundance}\text{of}{}^1\text{H}\right)}^5\times {\left(\text{Natural}\text{abundance}\text{of}{}^{13}\text{C}\right)}^2\\ \times \left(\text{Natural}\text{abundance}\text{of}{}^{79}\text{Br}\right)\end{array}\right)\\ & =& {\left(0.999\right)}^5\times {\left(0.011\right)}^2\times \left(0.505\right)\\ & =& \left(0.995\right)\times \left(0.000121\right)\times \left(0.505\right)\\ & =& 6.079\times {10}^{-5}\end{array}$

Here the possibility of two ${}^{13}\text{C}$ is present. Terefore, the relative ratio is calculated by comparing relative intensity as shown below.

$\begin{array}{rll}\text{no}{}^{13}\text{C}& :& \text{one}{}^{13}\text{C}:\text{two}{}^{13}\text{C}\\ 0.491& :& 0.0109:6.079\times {10}^{-5}\end{array}$

The compound ethyl bromide contains two types of proton containing carbon. It gives rise to a quartet signal of $-{\text{CH}}_{\text{3}}$ carbon and a triplet signal of $-{\text{CH}}_{\text{2}}-$ carbon. In the given case, proton decoupled spectra is reported which means a phenomena DEPT is used. DEPT is expanded as distortionless enhancement by polarization transfer which is used to differentiate $1°$, $2°$, $3°$ and $4°$ carbon presence. Three different types of DEPT is used DEPT $90°$, $45°$ and $135°$. DEPT $90°$ give a signal of $-\text{CH}$ carbon only. DEPT $45°$ give signals of all carbon present. DEPT $135°$ give signals of $-\text{CH}$ and $-{\text{CH}}_{\text{3}}$ in the upper direction and a signal of $-{\text{CH}}_{\text{2}}-$ carbon in the opposite direction. The proton decoupled ${}^{\text{13}}\text{CNMR}$ spectra of ethyl bromide is shown below.

Figure 1

Conclusion

The proton decoupled ${}^{\text{13}}\text{C}\text{NMR}$ of ethyl bromide is shown in Figure 1.