Chapter 24, Problem 24.82P

Interpretation Introduction

(a)

Interpretation:

When the given compound is heated, ethene gas is evolved and a product with the formula ${\text{C}}_{\text{14}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$ is formed. The structure of ${\text{C}}_{\text{14}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$ is to be drawn from the given data.

Concept introduction:

The Diels-Alder reaction is reversible at high temperature, and this process is called a retro Diels-Alder reaction. The mechanism for this reaction is as the reverse of a Diels-Alder mechanism i.e. the entire six-membered ring is break-down into the corresponding diene and the dienophile. The aromatic proton given signal range from $\text{7 to 9}$ ppm in ${}^{\text{1}}\text{H NMR}$ spectrum and that is for ${}^{\text{13}}\text{C NMR}$ spectrum the signal range is $\text{120 to 140}$ ppm.

Answer to Problem 24.82P

The structure of ${\text{C}}_{\text{14}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$ is

Explanation of Solution

The given compound is

So when this compound is heated, ethene gas is evolved and a product with formula ${\text{C}}_{\text{14}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$ is formed. Let’s focus on the formula of the product. First, calculate the degree of unsaturation for the product.

$\text{DBE =}\frac{\text{2C - H - X + N + 2}}{\text{2}}$

$\text{= }\frac{\text{2 × 14 - 8 + 2}}{\text{2}}$

$\text{= 11}$ (so the compound is aromatic and contains more than one benzene ring)

The signal in ${}^{\text{1}}\text{H NMR}$ the spectrum between $\text{7 and 9}$ ppm suggests that the compound must be aromatic and more deshielding is due to the presence of carbonyl groups on the benzene ring. The ${}^{\text{13}}\text{C NMR}$ spectrum also gives the signal in the aromatic region and only four signals in ${}^{\text{13}}\text{C NMR}$ spectrum suggest that the product is highly symmetric. Thus the product must be aromatic. The signal at $\text{185 ppm}$ corresponds to the carbonyl carbons. Now focusing on the given compound, which on heating, ethene gas is evolved and the aromatic product with molecular formula ${\text{C}}_{\text{14}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$ is formed. Thus the retro Diels-Alder reaction may take place as shown below:

The product of the retro Diels-Alder reaction is also aromatic and matched with the given spectra and also ethene gas is evolved. so the above product is the correct one.

Conclusion

When the given compound is heated, ethene gas is evolved and a product with the formula ${\text{C}}_{\text{14}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$ is formed. The structure of ${\text{C}}_{\text{14}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$ is drawn from the given data.

Interpretation Introduction

(b)

Interpretation:

When the given compound is heated, ethene gas is evolved and a product with the formula ${\text{C}}_{\text{14}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$ is formed. The mechanism is to be drawn that accounts for its formation.

Concept introduction:

The Diels-Alder reaction is reversible at high temperature and this process is called a retro Diels-Alder reaction. The mechanism for this reaction is as the reverse of a Diels-Alder mechanism i.e. the entire six-membered ring is broken down into the corresponding diene and the dienophile. The aromatic proton given signal range from $\text{7 to 9}$ ppm in ${}^{\text{1}}\text{H NMR}$ the spectrum and that is for ${}^{\text{13}}\text{C NMR}$ spectrum the signal range is $\text{120 to 140}$ ppm.

Answer to Problem 24.82P

The mechanism for the given reaction is

Explanation of Solution

The given compound is

So when this compound is heated, ethene gas is evolved and product with formula ${\text{C}}_{\text{14}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$ is formed. Let focus on the formula of the product. First, calculate the degree of unsaturation for the product.

$\text{DBE =}\frac{\text{2C - H - X + N + 2}}{\text{2}}$

$\text{= }\frac{\text{2 × 14 - 8 + 2}}{\text{2}}$

$\text{= 11}$ (so the compound is aromatic and contains more than one benzene ring)

The signal in ${}^{\text{1}}\text{H NMR}$ spectrum between $\text{7 and 9}$ ppm suggests that the compound must be aromatic and more deshielding due to the presence of carbonyl groups on the benzene ring. The ${}^{\text{13}}\text{C NMR}$ spectrum also gives the signal in the aromatic region, and the signal at $\text{185 ppm}$ corresponds to the carbonyl carbons, and only four signals in ${}^{\text{13}}\text{C NMR}$ spectrum suggest that the product is highly symmetric. Thus the product must be aromatic. Now focusing on the given compound, which on heating, ethene gas is evolved and the aromatic product with molecular formula ${\text{C}}_{\text{14}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$ is formed. Thus the retro Diels-Alder reaction may take place as shown below:

The product of the retro Diels-Alder reaction is also aromatic and matched with the given spectra and also ethene gas is evolved. So the above product and the mechanism (retro Diels-Alder mechanism) is the correct one.

Conclusion

When the given compound is heated, ethene gas is evolved and a product with the formula ${\text{C}}_{\text{14}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$ is formed. The mechanism is drawn that accounts for its formation.

Interpretation Introduction

(c)

Interpretation:

When the given compound is heated, ethene gas is evolved and a product with the formula ${\text{C}}_{\text{14}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$ is formed. The main driving force that favours the product in the given reaction is to be stated.

Concept introduction:

The Diels-Alder reaction is reversible at high temperature and this process is called a retro Diels-Alder reaction. The mechanism for this reaction is as the reverse of a Diels-Alder mechanism i.e. the entire six-membered ring is break-down into the corresponding diene and the dienophile. The aromatic proton given signal range from $\text{7 to 9}$ ppm in ${}^{\text{1}}\text{H NMR}$ spectrum and that is for ${}^{\text{13}}\text{C NMR}$ spectrum the signal range is $\text{120 to 140}$ ppm. Aromatic compounds are more stable than the nonaromatic compound.

Answer to Problem 24.82P

The main driving force that favours the product in the given reaction is the formation of a more stable aromatic compound.

Explanation of Solution

The given reaction is

It is noticed that the starting material in the given reaction is overall non-aromatic and also possesses strain due to middle ethylene groups on both sides, but the product is aromatic and strain-free one. Since aromatic compounds are more stable than the non-aromatic compounds this is a key state and the main driving force for the given reaction.

Conclusion

When the given compound is heated, ethene gas is evolved and a product with the formula ${\text{C}}_{\text{14}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$ is formed. The main driving force that favours the product in the given reaction is stated.