Chapter 19, Problem 19.66AP

Interpretation Introduction

(a)

Interpretation:

A suitable structure for ${\text{C}}_{\text{10}}{\text{H}}_{\text{10}}{\text{O}}_{\text{2}}$ that is in accordance with the given spectroscopic data is to be stated.

Concept introduction:

Spectroscopy method is used to identify the structure of the molecule. It is based on the interactions between matter and electromagnetic radiations. IR spectroscopy is used for the detection of functional group present in the compound. Proton NMR spectroscopy identifies the number of hydrogen atoms present in a molecule and the nature of the functional group. The value of chemical peaks depends upon the chemical environment around the hydrogen atom.

Answer to Problem 19.66AP

A suitable structure for ${\text{C}}_{\text{10}}{\text{H}}_{\text{10}}{\text{O}}_{\text{2}}$ that is in accordance with the given spectroscopic data is shown below.

Explanation of Solution

The degree of unsaturation or double bond equivalent (DBE) is calculated by the formula stated below.

$\text{DBE}=\text{C}+1-\frac{\text{H}}{2}-\frac{\text{X}}{2}+\frac{\text{N}}{2}$ …(1)

Where,

• $\text{C}$ is the number of carbon atoms.
• $\text{H}$ is the number of hydrogen atoms.
• $\text{X}$ is the number of halogen atoms.
• $\text{N}$ is the number of nitrogen atoms.

For ${\text{C}}_{\text{10}}{\text{H}}_{\text{10}}{\text{O}}_{\text{2}}$, substitute $\text{C}$ as $10$, $\text{H}$ as $10$ and $\text{N,}\text{X}$ as zero in equation (1).

$\begin{array}{rll}\text{DBE}& =& 10+1-\frac{\text{10}}{2}-\frac{\text{0}}{2}+\frac{0}{2}\\ & =& 11-5\\ & =& 6\end{array}$

The value of double bond equivalence may indicate the presence of five double bonds and a ring.

The NMR data of the compound ${\text{C}}_{\text{10}}{\text{H}}_{\text{10}}{\text{O}}_{\text{2}}$ shows a singlet at $\text{δ}=\text{2}\text{.82}$ for six protons indicating the presence of two methyl groups. The singlet peak at $\text{δ}=8.13$ for four hydrogens confirms the presence of disubstituted benzene ring. The substituents must be present at ortho and para position as the ortho and meta protons do not give a singlet for four aromatic protons.

The IR peak at $1681{\text{cm}}^{-\text{1}}$ indicates the presence of a ketone group.

So, the structure of the compound ${\text{C}}_{\text{10}}{\text{H}}_{\text{10}}{\text{O}}_{\text{2}}$ is shown below.

Figure 1

Conclusion

The structure of the compound ${\text{C}}_{\text{10}}{\text{H}}_{\text{10}}{\text{O}}_{\text{2}}$ is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

A suitable structure for ${\text{C}}_5{\text{H}}_{\text{10}}\text{O}$ that is in accordance with the given spectroscopic data is to be stated.

Concept introduction:

Spectroscopy method is used to identify the structure of the molecule. It is based on the interactions between matter and electromagnetic radiations. IR spectroscopy is used for the detection of functional group present in the compound. Proton NMR spectroscopy identifies the number of hydrogen atoms present in a molecule and the nature of the functional group. The value of chemical peaks depends upon the chemical environment around the hydrogen atom.

Answer to Problem 19.66AP

A suitable structure for ${\text{C}}_5{\text{H}}_{\text{10}}\text{O}$ that is in accordance with the given spectroscopic data is shown below.

Explanation of Solution

The degree of unsaturation or double bond equivalent (DBE) is calculated by the formula stated below.

$\text{DBE}=\text{C}+1-\frac{\text{H}}{2}-\frac{\text{X}}{2}+\frac{\text{N}}{2}$ …(1)

Where,

• $\text{C}$ is the number of carbon atoms.
• $\text{H}$ is the number of hydrogen atoms.
• $\text{X}$ is the number of halogen atoms.
• $\text{N}$ is the number of nitrogen atoms.

For ${\text{C}}_5{\text{H}}_{\text{10}}\text{O}$, substitute $\text{C}$ as $5$, $\text{H}$ as $10$ and $\text{N,}\text{X}$ as zero in equation (1).

$\begin{array}{rll}\text{DBE}& =& 5+1-\frac{\text{10}}{2}-\frac{\text{0}}{2}+\frac{0}{2}\\ & =& 6-5\\ & =& 1\end{array}$

The value of DBE indicates that either there is a double bond or a ring. The NMR spectrum data shows the singlet for 9 protons at $\delta 1.1$ which indicates the presence of three methyl groups of tertiary butyl. The singlet at $\delta 9.8$ for one proton states the presence of aldehyde group. So, the structure of the compound ${\text{C}}_5{\text{H}}_{\text{10}}\text{O}$ is shown below.

Figure 2

Conclusion

A suitable structure for ${\text{C}}_5{\text{H}}_{\text{10}}\text{O}$ that is in accordance with the given spectroscopic data is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

A suitable structure for ${\text{C}}_5{\text{H}}_{\text{10}}\text{O}$ that is in accordance with the given spectroscopic data is to be stated.

Concept introduction:

Spectroscopy method is used to identify the structure of the molecule. It is based on the interactions between matter and electromagnetic radiations. IR spectroscopy is used for the detection of functional group present in the compound. Proton NMR spectroscopy identifies the number of hydrogen atoms present in a molecule and the nature of the functional group. The value of chemical peaks depends upon the chemical environment around the hydrogen atom.

Answer to Problem 19.66AP

A suitable structure for ${\text{C}}_6{\text{H}}_{\text{10}}\text{O}$ that is in accordance with the given spectroscopic data is shown below.

Explanation of Solution

The degree of unsaturation or double bond equivalent (DBE) is calculated by the formula stated below.

$\text{DBE}=\text{C}+1-\frac{\text{H}}{2}-\frac{\text{X}}{2}+\frac{\text{N}}{2}$ …(1)

Where,

• $\text{C}$ is the number of carbon atoms.
• $\text{H}$ is the number of hydrogen atoms.
• $\text{X}$ is the number of halogen atoms.
• $\text{N}$ is the number of nitrogen atoms.

For ${\text{C}}_6{\text{H}}_{\text{10}}\text{O}$, substitute $\text{C}$ as $6$, $\text{H}$ as $10$ and $\text{N,}\text{X}$ as zero in equation (1).

$\begin{array}{rll}\text{DBE}& =& 6+1-\frac{\text{10}}{2}-\frac{\text{0}}{2}+\frac{0}{2}\\ & =& 7-5\\ & =& 2\end{array}$

The degree of unsaturation indicates the presence of either two double bonds or one ring and one double bond or two rings.

The IR spectrum shows the frequency at $1701{\text{cm}}^{-1},\text{970}{\text{cm}}^{-1}$ indicating the presence of carbonyl group and olefinic carbon respectively.

The NMR data shows the doublet at $\delta 9.51$ for single hydrogen indicating the presence of an aldehydic proton. The peak at around $\text{δ}\text{5}\text{.5}\text{and}\text{δ}\text{6}\text{.5}$ indicates the presence of olefinic hydrogens. The multiplet observed at around $\delta 2.2$ is for the methylene proton.

The UV-visible spectroscopy data shows the ${\text{λ}}_{\text{max}}=215\left(\in =17,400\right),329\left(\in =26\right)$ for $\text{n}\to \text{π*}$ and $\text{π}\to \text{π*}$ transitions respectively in $\text{α,β}-\text{unsaturated}$ aldehydes. So, the structure for ${\text{C}}_6{\text{H}}_{\text{10}}\text{O}$ is shown below.

Figure 3

Conclusion

A suitable structure for ${\text{C}}_6{\text{H}}_{\text{10}}\text{O}$ that is in accordance with the given spectroscopic data is shown in Figure 3.