Chapter 21, Problem 21.16P

Which of the molecules and ions given in Problem 21.15 are aromatic according to the Hückel criteria? Which, if planar, would be antiaromatic?

21.15 State the number of 2p orbital electrons in each molecule or ion.

Interpretation Introduction

Interpretation:

Which of the given molecules and ions are aromatic according to the $\text{Huckel}$ criteria and which, if planar, would be anti-aromatic has to be determined.

Concept Introduction:

The term aromaticity means “extreme stability”. So, aromatic compounds are highly stable compounds whereas anti-aromatic compounds are highly unstable compounds.

The aromatic compounds and anti-aromatic compounds can be distinguished based on Huckel’s rule of aromaticity.

Huckel’s rule of aromaticity is$\text{4n}\text{+}\text{2}\text{=}{\text{π}}_{\text{e}}$

If $n=0,1,2,\mathrm{....}$, then the given compound is an aromatic compound.

If $n=fractionalnumbers$, then the given compound is an non-aromatic compound.

If ${\pi }_e=4n$, then the given compound is an anti-aromatic compound.

Explanation of Solution

The given compound is shown here:

There are five $\text{π-}$ bonds. Each $\text{π-}$ bond has two electrons. Hence, there are a total of $\text{5}\text{×}\text{2}\text{=}\text{10}\text{-π-electrons}\left({\text{π}}_{\text{e}}\right)$.

Huckel’s rule of aromaticity $4n+2={\pi }_e$

$\begin{array}{c}\text{4n}\text{+}\text{2}\text{=}10\\ \text{n}=\frac{10-2}{4}\\ =\frac{8}{4}\\ =2\end{array}$

$\because \text{n=}\text{Whole number, }\therefore \text{the given compound is aromatic}\text{.}$

Interpretation Introduction

Interpretation:

Which of the given molecules and ions are aromatic according to the $\text{Huckel}$ criteria and which, if planar, would be anti-aromatic has to be determined.

Concept Introduction:

The term aromaticity means “extreme stability”. So, aromatic compounds are highly stable compounds whereas anti-aromatic compounds are highly unstable compounds.

The aromatic compounds and anti-aromatic compounds can be distinguished based on Huckel’s rule of aromaticity.

Huckel’s rule of aromaticity is$\text{4n}\text{+}\text{2}\text{=}{\text{π}}_{\text{e}}$

If $n=0,1,2,\mathrm{....}$, then the given compound is an aromatic compound.

If $n=fractionalnumbers$, then the given compound is an non-aromatic compound.

If ${\pi }_e=4n$, then the given compound is an anti-aromatic compound.

Explanation of Solution

The given compound is shown here:

There are six $\text{π-}$ bonds. Each $\text{π-}$ bond has two electrons. Hence, there are a total of $\text{6}\text{×}\text{2}\text{=}\text{12}\text{-π-electrons}\left({\text{π}}_{\text{e}}\right)$ .

Huckel’s rule of aromaticity $4n+2={\pi }_e$

$\begin{array}{c}4n+2={\pi }_e\\ {\pi }_e=12\\ \Rightarrow 4\times 3\\ \Rightarrow 4n\end{array}$

$\because {\pi }_e\text{=}4n\text{, }\therefore \text{the given compound is anti-aromatic}\text{.}$

The given compound is planar. So, it is expected to possess aromaticity due to the expected effective delocalization of $\text{π-electrons}$ on the planar structure. But, even though the given compound is planar, yet is being anti-aromatic which has been proved by the $\text{Huckel's}\text{rule}$.

Interpretation Introduction

Interpretation:

Which of the given molecules and ions are aromatic according to the $\text{Huckel}$ criteria and which, if planar, would be anti-aromatic has to be determined.

Concept Introduction:

The term aromaticity means “extreme stability”. So, aromatic compounds are highly stable compounds whereas anti-aromatic compounds are highly unstable compounds.

The aromatic compounds and anti-aromatic compounds can be distinguished based on Huckel’s rule of aromaticity.

Huckel’s rule of aromaticity is$\text{4n}\text{+}\text{2}\text{=}{\text{π}}_{\text{e}}$

If $n=0,1,2,\mathrm{....}$, then the given compound is an aromatic compound.

If $n=fractionalnumbers$, then the given compound is an non-aromatic compound.

If ${\pi }_e=4n$, then the given compound is an anti-aromatic compound.

Explanation of Solution

The given compound is shown here:

There are two $\text{π-}$ bonds. Each $\text{π-}$ bond has two electrons. Hence, there are a total of $\text{2}\text{×}\text{2}\text{=}\text{4}\text{-π-electrons}\left({\text{π}}_{\text{e}}\right)$ .

Huckel’s rule of aromaticity $4n+2={\pi }_e$

$\begin{array}{c}4n+2={\pi }_e\\ {\pi }_e=4\\ \Rightarrow 4\times 1\\ \Rightarrow 4n\end{array}$

$\because {\pi }_e\text{=}4n\text{, the given compound is anti-aromatic}\text{.}$

The given compound is planar. So, it is expected to possess aromaticity due to the expected effective delocalization of $\text{π-electrons}$ on the planar structure. But, even though the given compound is planar, yet is being anti-aromatic which has been proved by the $\text{Huckel's}\text{rule}$.

Interpretation Introduction

Interpretation:

Which of the given molecules and ions are aromatic according to the $\text{Huckel}$ criteria and which, if planar, would be anti-aromatic has to be determined.

Concept Introduction:

The term aromaticity means “extreme stability”. So, aromatic compounds are highly stable compounds whereas anti-aromatic compounds are highly unstable compounds.

The aromatic compounds and anti-aromatic compounds can be distinguished based on Huckel’s rule of aromaticity.

Huckel’s rule of aromaticity is$\text{4n}\text{+}\text{2}\text{=}{\text{π}}_{\text{e}}$

If $n=0,1,2,\mathrm{....}$, then the given compound is an aromatic compound.

If $n=fractionalnumbers$, then the given compound is an non-aromatic compound.

If ${\pi }_e=4n$, then the given compound is an anti-aromatic compound.

Explanation of Solution

The given compound is shown here:

There are two $\text{π-}$ bonds and also there is an unpaired electron. Each $\text{π-}$ bond has two electrons. Hence, there are a total of $\text{2}\text{×}\text{2}\text{=}\text{4}+1=5\text{-π-electrons}\left({\text{π}}_{\text{e}}\right)$ .

Huckel’s rule of aromaticity $4n+2={\pi }_e$

$\begin{array}{c}\text{4n}\text{+}\text{2}\text{=}11\\ \text{n}=\frac{5-2}{4}\\ =\frac{3}{4}\\ =0.75\end{array}$

$\because \text{n=}\text{fractional number, }\therefore \text{the given compound is non-aromatic}\text{.}$

Interpretation Introduction

Interpretation:

Which of the given molecules and ions are aromatic according to the $\text{Huckel}$ criteria and which, if planar, would be anti-aromatic has to be determined.

Concept Introduction:

The term aromaticity means “extreme stability”. So, aromatic compounds are highly stable compounds whereas anti-aromatic compounds are highly unstable compounds.

The aromatic compounds and anti-aromatic compounds can be distinguished based on Huckel’s rule of aromaticity.

Huckel’s rule of aromaticity is$\text{4n}\text{+}\text{2}\text{=}{\text{π}}_{\text{e}}$

If $n=0,1,2,\mathrm{....}$, then the given compound is an aromatic compound.

If $n=fractionalnumbers$, then the given compound is an non-aromatic compound.

If ${\pi }_e=4n$, then the given compound is an anti-aromatic compound.

Explanation of Solution

The given compound is shown here:

There are two $\text{π-}$ bonds and also there is a pair of electrons. Each $\text{π-}$ bond has two electrons. Hence, there are a total of $\text{2}\text{×}\text{2}\text{=}\text{4}+2=6\text{-π-electrons}\left({\text{π}}_{\text{e}}\right)$.

Huckel’s rule of aromaticity $4n+2={\pi }_e$

$\begin{array}{c}\text{4n}\text{+}\text{2}\text{=}6\\ \text{n}=\frac{6-2}{4}\\ =\frac{4}{4}\\ =1\end{array}$

$\because \text{n=}\text{Whole number, }\therefore \text{the given compound is aromatic}\text{.}$

Interpretation Introduction

Interpretation:

Which of the given molecules and ions are aromatic according to the $\text{Huckel}$ criteria and which, if planar, would be anti-aromatic has to be determined.

Concept Introduction:

The term aromaticity means “extreme stability”. So, aromatic compounds are highly stable compounds whereas anti-aromatic compounds are highly unstable compounds.

The aromatic compounds and anti-aromatic compounds can be distinguished based on Huckel’s rule of aromaticity.

Huckel’s rule of aromaticity is$\text{4n}\text{+}\text{2}\text{=}{\text{π}}_{\text{e}}$

If $n=0,1,2,\mathrm{....}$, then the given compound is an aromatic compound.

If $n=fractionalnumbers$, then the given compound is an non-aromatic compound.

If ${\pi }_e=4n$, then the given compound is an anti-aromatic compound.

Explanation of Solution

The given compound is shown here:

There are two $\text{π-}$ bonds. Each $\text{π-}$ bond has two electrons. Hence, there are a total of $\text{2}\text{×}\text{2}\text{=}\text{4}\text{-π-electrons}\left({\text{π}}_{\text{e}}\right)$ .

Huckel’s rule of aromaticity $4n+2={\pi }_e$

$\begin{array}{c}4n+2={\pi }_e\\ {\pi }_e=4\\ \Rightarrow 4\times 1\\ \Rightarrow 4n\end{array}$

$\because {\pi }_e\text{=}4n\text{, the given compound is anti-aromatic}\text{.}$

The given compound is planar. So, it is expected to possess aromaticity due to the expected effective delocalization of $\text{π-electrons}$ on the planar structure. But, even though the given compound is planar, yet is being anti-aromatic which has been proved by the $\text{Huckel's}\text{rule}$.

Interpretation Introduction

Interpretation:

Which of the given molecules and ions are aromatic according to the $\text{Huckel}$ criteria and which, if planar, would be anti-aromatic has to be determined.

Concept Introduction:

The term aromaticity means “extreme stability”. So, aromatic compounds are highly stable compounds whereas anti-aromatic compounds are highly unstable compounds.

The aromatic compounds and anti-aromatic compounds can be distinguished based on Huckel’s rule of aromaticity.

Huckel’s rule of aromaticity is$\text{4n}\text{+}\text{2}\text{=}{\text{π}}_{\text{e}}$

If $n=0,1,2,\mathrm{....}$, then the given compound is an aromatic compound.

If $n=fractionalnumbers$, then the given compound is an non-aromatic compound.

If ${\pi }_e=4n$, then the given compound is an anti-aromatic compound.

Explanation of Solution

The given compound is shown here:

There are three $\text{π-}$ bonds. Each $\text{π-}$ bond has two electrons. Hence, there are a total of $\text{3}\text{×}\text{2}\text{=}6\text{-π-electrons}\left({\text{π}}_{\text{e}}\right)$.

Huckel’s rule of aromaticity $4n+2={\pi }_e$

$\begin{array}{c}\text{4n}\text{+}\text{2}\text{=}6\\ \text{n}=\frac{6-2}{4}\\ =\frac{4}{4}\\ =1\end{array}$

$\because \text{n=}\text{Whole number, }\therefore \text{the given compound is aromatic}\text{.}$

Interpretation Introduction

Interpretation:

Which of the given molecules and ions are aromatic according to the $\text{Huckel}$ criteria and which, if planar, would be anti-aromatic has to be determined.

Concept Introduction:

The term aromaticity means “extreme stability”. So, aromatic compounds are highly stable compounds whereas anti-aromatic compounds are highly unstable compounds.

The aromatic compounds and anti-aromatic compounds can be distinguished based on Huckel’s rule of aromaticity.

Huckel’s rule of aromaticity is$\text{4n}\text{+}\text{2}\text{=}{\text{π}}_{\text{e}}$

If $n=0,1,2,\mathrm{....}$, then the given compound is an aromatic compound.

If $n=fractionalnumbers$, then the given compound is an non-aromatic compound.

If ${\pi }_e=4n$, then the given compound is an anti-aromatic compound.

Explanation of Solution

The given compound is shown here:

There are three $\text{π-}$ bonds. Each $\text{π-}$ bond has two electrons. Hence, there are a total of $\text{3}\text{×}\text{2}\text{=}6\text{-π-electrons}\left({\text{π}}_{\text{e}}\right)$.

Huckel’s rule of aromaticity $4n+2={\pi }_e$

$\begin{array}{c}\text{4n}\text{+}\text{2}\text{=}6\\ \text{n}=\frac{6-2}{4}\\ =\frac{4}{4}\\ =1\end{array}$

$\because \text{n=}\text{Whole number, }\therefore \text{the given compound is aromatic}\text{.}$

Interpretation Introduction

Interpretation:

Which of the given molecules and ions are aromatic according to the $\text{Huckel}$ criteria and which, if planar, would be anti-aromatic has to be determined.

Concept Introduction:

The term aromaticity means “extreme stability”. So, aromatic compounds are highly stable compounds whereas anti-aromatic compounds are highly unstable compounds.

The aromatic compounds and anti-aromatic compounds can be distinguished based on Huckel’s rule of aromaticity.

Huckel’s rule of aromaticity is$\text{4n}\text{+}\text{2}\text{=}{\text{π}}_{\text{e}}$

If $n=0,1,2,\mathrm{....}$, then the given compound is an aromatic compound.

If $n=fractionalnumbers$, then the given compound is an non-aromatic compound.

If ${\pi }_e=4n$, then the given compound is an anti-aromatic compound.

Explanation of Solution

The given compound is shown here:

There are four $\text{π-}$ bonds. Each $\text{π-}$ bond has two electrons. Hence, there are a total of $\text{4}\text{×}\text{2}\text{=}8\text{-π-electrons}\left({\text{π}}_{\text{e}}\right)$ .

Huckel’s rule of aromaticity $4n+2={\pi }_e$

$\begin{array}{c}\text{4n}\text{+}\text{2}\text{=}8\\ \text{n}=\frac{8-2}{4}\\ =\frac{6}{4}\\ =1.5\end{array}$

$\because \text{n=}\text{fractional number, }\therefore \text{the given compound is non-aromatic}\text{.}$

Interpretation Introduction

Interpretation:

Which of the given molecules and ions are aromatic according to the $\text{Huckel}$ criteria and which, if planar, would be anti-aromatic has to be determined.

Concept Introduction:

The term aromaticity means “extreme stability”. So, aromatic compounds are highly stable compounds whereas anti-aromatic compounds are highly unstable compounds.

The aromatic compounds and anti-aromatic compounds can be distinguished based on Huckel’s rule of aromaticity.

Huckel’s rule of aromaticity is$\text{4n}\text{+}\text{2}\text{=}{\text{π}}_{\text{e}}$

If $n=0,1,2,\mathrm{....}$, then the given compound is an aromatic compound.

If $n=fractionalnumbers$, then the given compound is an non-aromatic compound.

If ${\pi }_e=4n$, then the given compound is an anti-aromatic compound.

Explanation of Solution

The given compound is shown here:

There are three $\text{π-}$ bonds and also there are two pairs of electrons. Each $\text{π-}$ bond has two electrons. Hence, there are a total of $\text{3}\text{×}\text{2}\text{=}6+4=10\text{-π-electrons}\left({\text{π}}_{\text{e}}\right)$.

Huckel’s rule of aromaticity $4n+2={\pi }_e$

$\begin{array}{c}\text{4n}\text{+}\text{2}\text{=}10\\ \text{n}=\frac{10-2}{4}\\ =\frac{8}{4}\\ =2\end{array}$

$\because \text{n=}\text{Whole number, }\therefore \text{the given compound is aromatic}\text{.}$