Chapter 24, Problem 24.84P

Interpretation Introduction

(a)

Interpretation:

Heating the given compound at ${\text{225 - 235}}^{\text{o}}\text{C}$ for $\text{8 h}$ produced a new compound with the formula ${\text{C}}_{\text{23}}{\text{H}}_{\text{16}}{\text{N}}_{\text{2}}\text{O}$. The mechanism for this reaction is believed to consist of a $\left[\text{4+2}\right]$ cycloaddition followed by a $\left[\text{4+2}\right]$ cycloelimination. The ${}^{\text{1}}\text{H NMR}$ spectrum contained the following signals: $\text{δ 8}\text{.33 - 8}\text{.25 and 7}\text{.58 - 6}\text{.97 (m, 14H), 5}\text{.46 (s, 2H)}$. The mechanism for this reaction is to be drawn.

Concept introduction:

The Diels-Alder reaction, $\left[\text{4+2}\right]$ cycloaddition is reversible at high temperature and this process is called a retro Diels-Alder reaction. It is also known as a $\left[\text{4+2}\right]$ cycloelimination 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 ${}^{\text{1}}\text{H NMR}$ spectrum.

Answer to Problem 24.84P

The mechanism for the given reaction is

Explanation of Solution

The given reaction is

Since it is given that the mechanism for the given reaction consists of a $\left[\text{4+2}\right]$ cycloaddition followed by a $\left[\text{4+2}\right]$ cycloelimination and the formula of the product is ${\text{C}}_{\text{23}}{\text{H}}_{\text{16}}{\text{N}}_{\text{2}}\text{O}$. From the formula, it is noticed that two N atoms are less in the product than the starting compound (${\text{C}}_{\text{23}}{\text{H}}_{\text{16}}{\text{N}}_4\text{O}$). So the overall reaction is a loss of ${\text{N}}_{\text{2}}$ molecule only. It is suggested that a $\left[\text{4+2}\right]$ cycloaddition takes place first, so there is one possibility for that reaction that is given below:

After $\left[\text{4+2}\right]$ cycloaddition, the $\left[\text{4+2}\right]$ cycloelimination and a ${\text{N}}_{\text{2}}$ molecule are losses from the molecule as shown below:

The formula of the product16 is matched with the given formula. Now checking for the given ${}^{\text{1}}\text{H NMR}$ data: there are $\text{14H}$ in an aromatic region and $\text{2H}$ of a methylene group is deshielded due to the adjacent oxygen atom and gives a signal at $\text{5}\text{.46 (s, 2H)}$. Which is also matched with the given spectra. Thus the suggested structure of the product is the correct one.

Thus the mechanism for the given reaction is

Conclusion

The mechanism for this reaction is drawn.

Interpretation Introduction

(b)

Interpretation:

Heating the given compound at ${\text{225 - 235}}^{\text{o}}\text{C}$ for $\text{8 h}$ produced a new compound with the formula ${\text{C}}_{\text{23}}{\text{H}}_{\text{16}}{\text{N}}_{\text{2}}\text{O}$. The mechanism for this reaction is believed to consist of a $\left[\text{4+2}\right]$ cycloaddition followed by a $\left[\text{4+2}\right]$ cycloelimination. The ${}^{\text{1}}\text{H NMR}$ spectrum contained the following signals: $\text{δ 8}\text{.33 - 8}\text{.25 and 7}\text{.58 - 6}\text{.97 (m, 14H), 5}\text{.46 (s, 2H)}$. The product for this reaction is to be drawn.

Concept introduction:

The Diels-Alder reaction, $\left[\text{4+2}\right]$ cycloaddition is reversible at high temperature and this process is called a retro Diels-Alder reaction it is also known as a $\left[\text{4+2}\right]$ cycloelimination 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.

Answer to Problem 24.84P

the product for the given reaction is

Explanation of Solution

The given reaction is

Since it is given that the mechanism for the given reaction is consists of a $\left[\text{4+2}\right]$ cycloaddition followed by a $\left[\text{4+2}\right]$ cycloelimination and the formula of the product is ${\text{C}}_{\text{23}}{\text{H}}_{\text{16}}{\text{N}}_{\text{2}}\text{O}$. From the formula, it is noticed that two N atoms are less in the product than the starting compound (${\text{C}}_{\text{23}}{\text{H}}_{\text{16}}{\text{N}}_4\text{O}$). So the overall reaction is a loss of ${\text{N}}_{\text{2}}$ molecule only. It is suggested that a $\left[\text{4+2}\right]$ cycloaddition takes place first, so there is one possibility for that reaction is given below:

After $\left[\text{4+2}\right]$ cycloaddition, the $\left[\text{4+2}\right]$ cycloelimination and a ${\text{N}}_{\text{2}}$ molecule are losses from the molecule as shown below:

The formula of the product ${\text{C}}_{\text{23}}{\text{H}}_{\text{16}}{\text{N}}_{\text{2}}\text{O}$ is matched with the given formula. Now checking for the given ${}^{\text{1}}\text{H NMR}$ data: there are $\text{14H}$ in an aromatic region and $\text{2H}$ of a methylene group is deshielded due to the adjacent nitrogen atom and gives a signal at $\text{5}\text{.46 (s, 2H)}$. Which is also matched with the given spectra. Thus the suggested structure of the product is the correct one.

Thus the product for the given reaction is

Conclusion

The product for the given reaction is drawn.