Chapter 18, Problem 18.18P

Propose a structural formula for each compound consistent with its ¹H-NMR and ¹³C-NMR spectra.

1. (a) C₅H₁₀O₂

1. (b) C₇H₁₄O₂

1. (c) C₆ H₁₂O₂

1. (d) C₇H₁₂O₄

1. (e) C₄H₇ClO₂

1. (f) C₄H₆O₂

Interpretation Introduction

Interpretation:

The structural formula for compound having molecular formula ${\text{C}}_{\text{5}}{\text{H}}_{\text{10}}{\text{O}}_{\text{2}}$ has to be identified using the given spectral data.

Concept Introduction:

${}^{\text{1}}\text{H}$ NMR spectroscopy:

It is the study of the interaction between electromagnetic radiation and the nuclei of hydrogen atoms and the hydrogen atoms.

Chemical shift: The frequency of the proton signal in the spectrum with reference to the standard compound  which may be TMS(Tetramethylsilane) shows signal at 0 ppm(parts per million).

Integration value (I): The integration value at the bottom of the ${}^{\text{1}}\text{H}\text{NMR}$ spectrum represents the number of protons giving rise to the signal.

Chemical shift values for protons in different electronic environment

$\begin{array}{cc}Typesofprotons& Chemicalshift\\ Methyl& \text{~0}\text{.85}\\ Methylene& \text{~1}\text{.2}\\ Methine& \text{~1}\text{.55}\\ Allylic& \text{~2}\\ \begin{array}{l}Alkynyl\hfill \end{array}& \text{~2}\text{.5}\\ R-O-C{\boxed{H}}_3& \text{~}3.3\\ Br-C{H}_2-\boxed{H}& \text{~2}\text{.5}-\text{4}\\ Cl-C{H}_2-\boxed{H}& ~3-4\\ Vinyl& ~4.5-6.5\\ Aryl& ~6.5-8\\ Aldehyde& ~10\\ carboxylic& ~12\end{array}$

FUNCTIONAL GROUP	EFFECT ON THE ALPHA PROTONS	EFFECT ON THE BETA PROTONS
Oxygen of an alcohol or ether	$\text{+2}\text{.5}$	$\text{+0}\text{.5}$
Oxygen of ester	$\text{+3}\text{.0}$	$\text{+0}\text{.6}$
Carbonyl groups	$\text{+1}\text{.0}$	$\text{+0}\text{.2}$

${}^{\text{13}}\text{C}$ NMR spectroscopy:

It is the study of the interaction between electromagnetic radiation and the nuclei of the carbon atoms.

The signals in the ${}^{\text{13}}\text{C NMR}$ spectrum of a compound is proportional to the number of carbons that are present in the different environment within the molecule. The carbon that is present in the electron-rich environment shows a signal at a lower frequency and vice-versa. Therefore, the carbon that is present nearest to the electron-withdrawing groups produces a high-frequency signal.

Explanation of Solution

Given compound has the molecular formula ${\text{C}}_{\text{5}}{\text{H}}_{\text{10}}{\text{O}}_{\text{2}}$.

Spectral data is given below:

  $\begin{array}{cc}{}^{\text{1}}\text{H-NMR}& {}^{\text{13}}\text{C-NMR}\\ \text{0}\text{.96}\left(\text{d,6H}\right)& \text{161}\text{.11}\\ \text{1}\text{.96}\left(\text{m,1H}\right)& \text{70}\text{.01}\\ \text{3}\text{.95}\left(\text{d,2H}\right)& \text{27}\text{.71}\\ \text{8}\text{.08}\left(\text{s,1H}\right)& \text{19}\text{.00}\end{array}$

By analyzing the spectral data, the structure can be predicted.

The value of 8.08 suggests an aldehyde proton. 0.96 will indicate the presence of methyl groups and 1.96 indicates the methylene proton $\text{β}$ to carbonyl group.3.95 suggest the presence of methylene proton $\text{α}$

The structure proposed according to the data is given below:

Interpretation Introduction

Interpretation:

The structural formula for compound having molecular formula ${\text{C}}_7{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}$ has to be identified using the given spectral data.

Concept Introduction:

${}^{\text{1}}\text{H}$ NMR spectroscopy:

It is the study of the interaction between electromagnetic radiation and the nuclei of hydrogen atoms and the hydrogen atoms.

Chemical shift: The frequency of the proton signal in the spectrum with reference to the standard compound  which may be TMS(Tetramethylsilane) shows signal at 0 ppm(parts per million).

Integration value (I): The integration value at the bottom of the ${}^{\text{1}}\text{H}\text{NMR}$ spectrum represents the number of protons giving rise to the signal.

Chemical shift values for protons in different electronic environment

$\begin{array}{cc}Typesofprotons& Chemicalshift\\ Methyl& \text{~0}\text{.85}\\ Methylene& \text{~1}\text{.2}\\ Methine& \text{~1}\text{.55}\\ Allylic& \text{~2}\\ \begin{array}{l}Alkynyl\hfill \end{array}& \text{~2}\text{.5}\\ R-O-C{\boxed{H}}_3& \text{~}3.3\\ Br-C{H}_2-\boxed{H}& \text{~2}\text{.5}-\text{4}\\ Cl-C{H}_2-\boxed{H}& ~3-4\\ Vinyl& ~4.5-6.5\\ Aryl& ~6.5-8\\ Aldehyde& ~10\\ carboxylic& ~12\end{array}$

FUNCTIONAL GROUP	EFFECT ON THE ALPHA PROTONS	EFFECT ON THE BETA PROTONS
Oxygen of an alcohol or ether	$\text{+2}\text{.5}$	$\text{+0}\text{.5}$
Oxygen of ester	$\text{+3}\text{.0}$	$\text{+0}\text{.6}$
Carbonyl groups	$\text{+1}\text{.0}$	$\text{+0}\text{.2}$

${}^{\text{13}}\text{C}$ NMR spectroscopy:

It is the study of the interaction between electromagnetic radiation and the nuclei of the carbon atoms.

The signals in the ${}^{\text{13}}\text{C NMR}$ spectrum of a compound is proportional to the number of carbons that are present in the different environment within the molecule. The carbon that is present in the electron-rich environment shows a signal at a lower frequency and vice-versa. Therefore, the carbon that is present nearest to the electron-withdrawing groups produces a high-frequency signal.

Explanation of Solution

Given compound has the molecular formula ${\text{C}}_7{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}$.

Spectral data is given below:

  $\begin{array}{l}\begin{array}{cc}{}^{\text{1}}\text{H-NMR}& {}^{\text{13}}\text{C-NMR}\\ \text{0}\text{.92}\left(\text{d,6H}\right)& 171.15\\ 1.52\left(\text{m,6H}\right)& 63.12\\ 1.70\left(\text{m,1H}\right)& 37.31\\ 2.09\left(\text{s,3H}\right)& 25.05\\ 4.10\left(\text{t,2H}\right)& 22.45\\ & 21.06\end{array}\\ \end{array}$

By analyzing the spectral data, the structure can be predicted.

The value of 2.09 indicates a methyl group $\text{α}$ to carbonyl group. 4.09 indicate the presence of methylene group directly attached to oxygen of ether and 1.70 suggests the presence of methane group.1.52 indicates the presence of methylene group $\text{β}$ to ether.

The structure proposed according to the data is given below:

Interpretation Introduction

Interpretation:

The structural formula for compound having molecular formula ${\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_{\text{2}}$ has to be identified using the given spectral data.

Concept Introduction:

${}^{\text{1}}\text{H}$ NMR spectroscopy:

It is the study of the interaction between electromagnetic radiation and the nuclei of hydrogen atoms and the hydrogen atoms.

Chemical shift: The frequency of the proton signal in the spectrum with reference to the standard compound  which may be TMS(Tetramethylsilane) shows signal at 0 ppm(parts per million).

Integration value (I): The integration value at the bottom of the ${}^{\text{1}}\text{H}\text{NMR}$ spectrum represents the number of protons giving rise to the signal.

Chemical shift values for protons in different electronic environment

$\begin{array}{cc}Typesofprotons& Chemicalshift\\ Methyl& \text{~0}\text{.85}\\ Methylene& \text{~1}\text{.2}\\ Methine& \text{~1}\text{.55}\\ Allylic& \text{~2}\\ \begin{array}{l}Alkynyl\hfill \end{array}& \text{~2}\text{.5}\\ R-O-C{\boxed{H}}_3& \text{~}3.3\\ Br-C{H}_2-\boxed{H}& \text{~2}\text{.5}-\text{4}\\ Cl-C{H}_2-\boxed{H}& ~3-4\\ Vinyl& ~4.5-6.5\\ Aryl& ~6.5-8\\ Aldehyde& ~10\\ carboxylic& ~12\end{array}$

FUNCTIONAL GROUP	EFFECT ON THE ALPHA PROTONS	EFFECT ON THE BETA PROTONS
Oxygen of an alcohol or ether	$\text{+2}\text{.5}$	$\text{+0}\text{.5}$
Oxygen of ester	$\text{+3}\text{.0}$	$\text{+0}\text{.6}$
Carbonyl groups	$\text{+1}\text{.0}$	$\text{+0}\text{.2}$

${}^{\text{13}}\text{C}$ NMR spectroscopy:

It is the study of the interaction between electromagnetic radiation and the nuclei of the carbon atoms.

The signals in the ${}^{\text{13}}\text{C NMR}$ spectrum of a compound is proportional to the number of carbons that are present in the different environment within the molecule. The carbon that is present in the electron-rich environment shows a signal at a lower frequency and vice-versa. Therefore, the carbon that is present nearest to the electron-withdrawing groups produces a high-frequency signal.

Explanation of Solution

Given compound has the molecular formula ${\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_{\text{2}}$.

Spectral data is given below:

  $\begin{array}{l}\\ \begin{array}{cc}{}^{\text{1}}\text{H-NMR}& {}^{\text{13}}\text{C-NMR}\\ 1.18\left(\text{d,6H}\right)& 177.16\\ 1.26\left(\text{t,3H}\right)& 60.17\\ 2.51\left(\text{m,1H}\right)& 34.04\\ 4.13\left(\text{q,2H}\right)& 19.01\\ & 14.25\end{array}\end{array}$

By analyzing the spectral data, the structure can be predicted.

1.18 and 1.26 will indicate the presence of methyl groups and 4.13 indicates the methylene proton $\text{α}$ to oxygen. 2.51 indicate the presence of methine proton attached to a carbonyl group.

The structure proposed according to the data is given below:

Interpretation Introduction

Interpretation:

The structural formula for compound having molecular formula ${\text{C}}_7{\text{H}}_{\text{12}}{\text{O}}_4$ has to be identified using the given spectral data.

Concept Introduction:

${}^{\text{1}}\text{H}$ NMR spectroscopy:

It is the study of the interaction between electromagnetic radiation and the nuclei of hydrogen atoms and the hydrogen atoms.

Chemical shift: The frequency of the proton signal in the spectrum with reference to the standard compound  which may be TMS(Tetramethylsilane) shows signal at 0 ppm(parts per million).

Integration value (I): The integration value at the bottom of the ${}^{\text{1}}\text{H}\text{NMR}$ spectrum represents the number of protons giving rise to the signal.

Chemical shift values for protons in different electronic environment

$\begin{array}{cc}Typesofprotons& Chemicalshift\\ Methyl& \text{~0}\text{.85}\\ Methylene& \text{~1}\text{.2}\\ Methine& \text{~1}\text{.55}\\ Allylic& \text{~2}\\ \begin{array}{l}Alkynyl\hfill \end{array}& \text{~2}\text{.5}\\ R-O-C{\boxed{H}}_3& \text{~}3.3\\ Br-C{H}_2-\boxed{H}& \text{~2}\text{.5}-\text{4}\\ Cl-C{H}_2-\boxed{H}& ~3-4\\ Vinyl& ~4.5-6.5\\ Aryl& ~6.5-8\\ Aldehyde& ~10\\ carboxylic& ~12\end{array}$

FUNCTIONAL GROUP	EFFECT ON THE ALPHA PROTONS	EFFECT ON THE BETA PROTONS
Oxygen of an alcohol or ether	$\text{+2}\text{.5}$	$\text{+0}\text{.5}$
Oxygen of ester	$\text{+3}\text{.0}$	$\text{+0}\text{.6}$
Carbonyl groups	$\text{+1}\text{.0}$	$\text{+0}\text{.2}$

${}^{\text{13}}\text{C}$ NMR spectroscopy:

It is the study of the interaction between electromagnetic radiation and the nuclei of the carbon atoms.

The signals in the ${}^{\text{13}}\text{C NMR}$ spectrum of a compound is proportional to the number of carbons that are present in the different environment within the molecule. The carbon that is present in the electron-rich environment shows a signal at a lower frequency and vice-versa. Therefore, the carbon that is present nearest to the electron-withdrawing groups produces a high-frequency signal.

Explanation of Solution

Given compound has the molecular formula ${\text{C}}_7{\text{H}}_{\text{12}}{\text{O}}_4$.

Spectral data is given below:

  $\begin{array}{l}\\ \begin{array}{cc}{}^{\text{1}}\text{H-NMR}& {}^{\text{13}}\text{C-NMR}\\ 1.28\left(\text{t,6H}\right)& 166.52\\ \text{3}\text{.36}\left(\text{s,2H}\right)& 61.43\\ 4.21\left(\text{q,4H}\right)& 41.69\\ & 14.07\end{array}\end{array}$

By analyzing the spectral data, the structure can be predicted.

The value of 3.36 suggests a methylene directly attach to a carbonyl group. 1.28 will indicate the presence of methyl groups and 4.21 indicates the methylene proton $\text{β}$ to carbonyl group $\text{α}$ to oxygen.

The structure proposed according to the data is given below:

Interpretation Introduction

Interpretation:

The structural formula for compound having molecular formula ${\text{C}}_4{\text{H}}_7{\text{ClO}}_{\text{2}}$ has to be identified using the given spectral data.

Concept Introduction:

${}^{\text{1}}\text{H}$ NMR spectroscopy:

It is the study of the interaction between electromagnetic radiation and the nuclei of hydrogen atoms and the hydrogen atoms.

Chemical shift: The frequency of the proton signal in the spectrum with reference to the standard compound  which may be TMS(Tetramethylsilane) shows signal at 0 ppm(parts per million).

Integration value (I): The integration value at the bottom of the ${}^{\text{1}}\text{H}\text{NMR}$ spectrum represents the number of protons giving rise to the signal.

Chemical shift values for protons in different electronic environment

$\begin{array}{cc}Typesofprotons& Chemicalshift\\ Methyl& \text{~0}\text{.85}\\ Methylene& \text{~1}\text{.2}\\ Methine& \text{~1}\text{.55}\\ Allylic& \text{~2}\\ \begin{array}{l}Alkynyl\hfill \end{array}& \text{~2}\text{.5}\\ R-O-C{\boxed{H}}_3& \text{~}3.3\\ Br-C{H}_2-\boxed{H}& \text{~2}\text{.5}-\text{4}\\ Cl-C{H}_2-\boxed{H}& ~3-4\\ Vinyl& ~4.5-6.5\\ Aryl& ~6.5-8\\ Aldehyde& ~10\\ carboxylic& ~12\end{array}$

FUNCTIONAL GROUP	EFFECT ON THE ALPHA PROTONS	EFFECT ON THE BETA PROTONS
Oxygen of an alcohol or ether	$\text{+2}\text{.5}$	$\text{+0}\text{.5}$
Oxygen of ester	$\text{+3}\text{.0}$	$\text{+0}\text{.6}$
Carbonyl groups	$\text{+1}\text{.0}$	$\text{+0}\text{.2}$

${}^{\text{13}}\text{C}$ NMR spectroscopy:

It is the study of the interaction between electromagnetic radiation and the nuclei of the carbon atoms.

The signals in the ${}^{\text{13}}\text{C NMR}$ spectrum of a compound is proportional to the number of carbons that are present in the different environment within the molecule. The carbon that is present in the electron-rich environment shows a signal at a lower frequency and vice-versa. Therefore, the carbon that is present nearest to the electron-withdrawing groups produces a high-frequency signal.

Explanation of Solution

Given compound has the molecular formula ${\text{C}}_4{\text{H}}_7{\text{ClO}}_{\text{2}}$.

Spectral data is given below:

  $\begin{array}{l}\\ \begin{array}{cc}{}^{\text{1}}\text{H-NMR}& {}^{\text{13}}\text{C-NMR}\\ 1.68\left(\text{d,3H}\right)& 170.51\\ \text{3}\text{.80}\left(\text{s,3H}\right)& 52.92\\ 4.42\left(\text{q,4H}\right)& 52.32\\ & 21.52\end{array}\end{array}$

By analyzing the spectral data, the structure can be predicted.

The value of 4.42 suggests a proton group directly attach to a chlorine group. 3.80 will indicate the presence of methyl group which is $\text{α}$ to oxygen and $\text{β}$ to carbonyl group. 1.68represent a methyl group $\text{β}$ to chlorine.

The structure proposed according to the data is given below:

Interpretation Introduction

Interpretation:

The structural formula for compound having molecular formula ${\text{C}}_4{\text{H}}_6{\text{O}}_{\text{2}}$ has to be identified using the given spectral data.

Concept Introduction:

${}^{\text{1}}\text{H}$ NMR spectroscopy:

It is the study of the interaction between electromagnetic radiation and the nuclei of hydrogen atoms and the hydrogen atoms.

Chemical shift: The frequency of the proton signal in the spectrum with reference to the standard compound  which may be TMS(Tetramethylsilane) shows signal at 0 ppm(parts per million).

Integration value (I): The integration value at the bottom of the ${}^{\text{1}}\text{H}\text{NMR}$ spectrum represents the number of protons giving rise to the signal.

Chemical shift values for protons in different electronic environment

$\begin{array}{cc}Typesofprotons& Chemicalshift\\ Methyl& \text{~0}\text{.85}\\ Methylene& \text{~1}\text{.2}\\ Methine& \text{~1}\text{.55}\\ Allylic& \text{~2}\\ \begin{array}{l}Alkynyl\hfill \end{array}& \text{~2}\text{.5}\\ R-O-C{\boxed{H}}_3& \text{~}3.3\\ Br-C{H}_2-\boxed{H}& \text{~2}\text{.5}-\text{4}\\ Cl-C{H}_2-\boxed{H}& ~3-4\\ Vinyl& ~4.5-6.5\\ Aryl& ~6.5-8\\ Aldehyde& ~10\\ carboxylic& ~12\end{array}$

FUNCTIONAL GROUP	EFFECT ON THE ALPHA PROTONS	EFFECT ON THE BETA PROTONS
Oxygen of an alcohol or ether	$\text{+2}\text{.5}$	$\text{+0}\text{.5}$
Oxygen of ester	$\text{+3}\text{.0}$	$\text{+0}\text{.6}$
Carbonyl groups	$\text{+1}\text{.0}$	$\text{+0}\text{.2}$

${}^{\text{13}}\text{C}$ NMR spectroscopy:

It is the study of the interaction between electromagnetic radiation and the nuclei of the carbon atoms.

The signals in the ${}^{\text{13}}\text{C NMR}$ spectrum of a compound is proportional to the number of carbons that are present in the different environment within the molecule. The carbon that is present in the electron-rich environment shows a signal at a lower frequency and vice-versa. Therefore, the carbon that is present nearest to the electron-withdrawing groups produces a high-frequency signal.

Explanation of Solution

Given compound has the molecular formula ${\text{C}}_{\text{5}}{\text{H}}_{\text{10}}{\text{O}}_{\text{2}}$.

Spectral data is given below:

  $\begin{array}{l}\\ \begin{array}{cc}{}^{\text{1}}\text{H-NMR}& {}^{\text{13}}\text{C-NMR}\\ 2.29\left(\text{m,2H}\right)& 177.81\\ \text{2}\text{.50}\left(\text{t,2H}\right)& 68.58\\ 4.36\left(\text{t,2H}\right)& 27.79\\ & 22.17\end{array}\end{array}$

By analyzing the spectral data, the structure can be predicted.

The value of 4.36 suggests a methylene directly attach to a oxygen group. 2.50 will indicate the presence of methylene proton $\text{α}$ to carbonyl and 2.29 indicates methylene proton $\text{β}$ to carbonyl group.

The structure proposed according to the data is given below: