Chapter 20, Problem 20.44P

(a)

Interpretation Introduction

Interpretation: The structural formula for the Diels-Alder adduct formed by cyclopentadiene has to be drawn.

Concept Introduction:

Diels-Alder reaction:

It is the reaction of conjugated dienes with double or triple bonded compounds which are known as “dienophiles”. The reaction is a $4+2$ cycloaddition reaction that results in the formation of a six membered cyclic product which is known as “adduct”.

Example:

This mechanism shown that three $\text{π}$-bonds are broken to form two new $\text{σ}$ -bonds and a new $\text{π}$-bond.

Diels-Alder reaction to form bicyclic system:

The Diels-Alder adduct formed in the Diels-Alder reaction can also be a bicyclic system which will be obtained when cylopentadiene is used as the diene as shown here:

In this reaction, the cylopentadiene acts as both diene and dienophile and formed a bicyclic system. When it is heated to ${\text{170}}^{\text{o}}\text{C}$, a reverse Diels-Alder reaction takes place in which the bicyclic system will be decomposed and gives back the cylopentadiene.

Explanation of Solution

The given reaction is:

The structural formula for the Diels-Alder adduct can be found using the following mechanism of the given Diels-Alder reaction:

In this Diels-Alder reaction, one molecule of cyclopentadienone acts as a Diene whereas the other molecule of cyclopentadienone acts as a Dienophile. The resulted Diels-Alder adduct is a tricyclic product

(b)

Interpretation Introduction

Interpretation: The difference in the stability of the given ketones has to be accounted.

Concept Introduction:

Stability based on aromaticity:

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

Answer to Problem 20.44P

The difference in the stability of the given ketones has been accounted as:

Cycloheptatrienone is more stable than cyclopentadienone.

Explanation of Solution

The given ketones are:

The stability of these two ketones can be distinguished based on aromaticity as discussed below:

There are two $\text{π-}$ bonds inside the cyclopentane ring. Each $\text{π-}$ bond has two electrons. Hence, there are a total of $\text{2}\text{×}\text{2}\text{=}\text{4}\text{-π-electrons}\left({\text{π}}_{\text{e}}\right)$ which is equal to $\text{4n}$.

Huckel’s rule of aromaticity:

$\begin{array}{c}4n+2={\pi }_e\\ {\text{π}}_{\text{e}}\text{=}\text{4}\Rightarrow \text{4n}\Rightarrow \text{Anti}-\text{Aromatic}\text{.}\end{array}$

Therefore, the given molecule has $\text{4n}$ number of $\text{π-electrons}$. So, it is an anti-aromatic compound which means it is a highly unstable compound.

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

Huckel’s rule of aromaticity:

  $\begin{array}{c}4n+2={\pi }_e\\ {\text{π}}_{\text{e}}\text{=}\text{6}\end{array}$

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

  $\therefore \text{n}\text{=}1\Rightarrow \text{Aromatic}$

Therefore, the given molecule satisfies the Huckel’s rule of aromaticity. So, it is an anti-aromatic compound which means it is a highly stable compound.

Hence, the overall discussion makes it clear that cycloheptatrienone is more stable than cyclopentadienone.