The proton decoupled 13 C NMR spectra of ethyl bromide is to be stated. Concept introduction: The 13 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 C nucleus. The natural abundance of 13 C nucleus is only 1.1 % while 12 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 ( n + 1 ) rule is applicable on the protons attached to the same carbon whose splitting pattern is to be determined.

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Chapter 13, Problem 13.63AP

Interpretation Introduction

Interpretation:

The proton decoupled $13C NMR$ spectra of ethyl bromide is to be stated.

Concept introduction:

The $13C$ NMR spectroscopy is similar kind of spectroscopy to NMR the only difference them is that it is used to detect the presence of $13C$ nucleus. The natural abundance of $13C$ nucleus is only $1.1 %$ while $12C$ 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 $(n+1)$ rule is applicable on the protons attached to the same carbon whose splitting pattern is to be determined.

Expert Solution & Answer

Answer to Problem 13.63AP

The proton decoupled $13CNMR$ spectra of ethyl bromide is shown below.

Organic Chemistry, Ebook And Single-course Homework Access, Chapter 13, Problem 13.63AP , additional homework tip 1

Explanation of Solution

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

$Relative intensity=Multiplication of natural abundance of elements$ …(1)

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

$Relative intensity=((Natural abundance of 1H)5×(Natural abundance of 12C)2×(Natural abundance of 79Br))=(0.999)5×(0.989)2×(0.505)=0.995×0.9781×0.505=0.491$

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

$Relative intensity=((Natural abundance of 1H)5×2(Natural abundance of 12C)×(Natural abundance of 13C)×(Natural abundance of 79Br))=(0.999)5×((0.989)×(0.011)×2)×(0.505)=(0.995)×(0.989)×(0.011)×2×(0.505)=0.0109$

Here, $12C$ and $13C$ 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 $13C$ isotopes are present is calculated by substituting the value of elements abundance in equation (1) as shown below.

$Relative intensity=((Natural abundance of 1H)5×(Natural abundance of 13C)2×(Natural abundance of 79Br))=(0.999)5×(0.011)2×(0.505)=(0.995)×(0.000121)×(0.505)=6.079×10−5$

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

$no 13C:one 13C:two 13C0.491:0.0109:6.079×10−5$

The compound ethyl bromide contains two types of proton containing carbon. It gives rise to a quartet signal of $−CH3$ carbon and a triplet signal of $−CH2−$ 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 $−CH$ carbon only. DEPT $45°$ give signals of all carbon present. DEPT $135°$ give signals of $−CH$ and $−CH3$ in the upper direction and a signal of $−CH2−$ carbon in the opposite direction. The proton decoupled $13CNMR$ spectra of ethyl bromide is shown below.

Organic Chemistry, Ebook And Single-course Homework Access, Chapter 13, Problem 13.63AP , additional homework tip 2

Figure 1

Conclusion

The proton decoupled $13C NMR$ of ethyl bromide is shown in Figure 1.

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