Chapter 22, Problem 49P

Interpretation Introduction

Interpretation:

The synthesis of each of the given compound from the indicated starting material is to be stated.

Concept introduction:

The hydration of alkenes in the presence of an acid catalyst such as sulfuric acid results in the formation of alcohols. Aldehydes and Ketones contain carbonyl $\left(\text{C=O}\right)$ functional group in their parent chain. They are obtained by the oxidation of secondary alcohols in the presence of oxidizing agents such as ${\text{KMnO}}_{\text{4}}$ or ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$.

The reagent ${\text{HNO}}_{\text{2}}$ in the presence of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is used in nitration reaction. The reagent $\text{Sn}$ in the presence of $\text{HCl}$ converts the nitro group into amine group. The reagent ${\text{NaNO}}_{\text{2}}$ in the presence of $\text{HCl}$ forms diazonium compound.

Lithium aluminum hydride and sodium borohydride are strong reducing agents. They are inorganic compounds and are used as the reducing agents in organic synthesis. They are used for the conversion of carboxylic acids, aldehydes and ketones into primary and secondary alcohols.

Answer to Problem 49P

Solution:

a) The synthesis of $p\text{-aminobenzoic acid}$ from $p\text{-methylaniline}$ is shown below.

b) The synthesis of $p{\text{-FC}}_{\text{6}}{\text{H}}_{\text{4}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{3}}$ from benzene is shown below.

c) The synthesis of $1\text{-bromo-2-fluoro-3,5-dimethylbenzene}$ from $m\text{-xylene}$ is shown below.

d) The synthesis of $\text{1-bromo-4-fluoro-2-methylnaphthalene}$ from its starting material is shown below.

e) The synthesis of $o{\text{-BrC}}_{\text{6}}{\text{H}}_{\text{4}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$ from $p{\text{-O}}_{\text{2}}{\text{NC}}_{\text{6}}{\text{H}}_{\text{4}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$ is shown below.

f) The synthesis of $m{\text{-ClC}}_{\text{6}}{\text{H}}_{\text{4}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$ from $p{\text{-O}}_{\text{2}}{\text{NC}}_{\text{6}}{\text{H}}_{\text{4}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$ is shown below.

g) The synthesis of $1\text{-bromo-3,5-diethylbenzene}$ from $m\text{-diethylbenzene}$ is shown below.

h) The synthesis of $\text{1-bromo-2-iodo-3-}\left(\text{trifluoromethyl}\right)\text{benzene}$ from $\text{N-}\left(\text{4-amino-2-bromo-6-}\left(\text{trifluoromethyl}\right)\text{phenyl}\right)\text{acetamide}$ is shown below.

i) The synthesis of the desired compound from its starting material is shown below.

Explanation of Solution

a) $p\text{-Aminobenzoic acid}$ from $p\text{-methylaniline}$

The preparation of aromatic carboxylic acid takes place by the oxidation of alkylbenzene in the presence of an oxidising agent such as potassium dichromate or potassium permanganate. The synthesis of $p\text{-aminobenzoic acid}$ from $p\text{-methylaniline}$ is shown below.

b) $p{\text{-FC}}_{\text{6}}{\text{H}}_{\text{4}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{3}}$ from benzene

In the preparation of $p{\text{-FC}}_{\text{6}}{\text{H}}_{\text{4}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{3}}$ from benzene, the first step is the nitration of benzene to form nitrobenzene. In the next step, nitrobenzene reacts with $\text{Sn}$ in the presence of $\text{HCl}$ to form aniline. The resulting aniline reacts with acetic anhydride, ${\text{AC}}_{\text{2}}\text{O}$ to give $N\text{-phenylacetamide}$ that further reacts with propanoyl chloride followed by nitration and diazonium compound is formed as a result. In the final step, the diazonium compound reacts with ${\text{HBF}}_{\text{4}}$ to form the desired product $p{\text{-FC}}_{\text{6}}{\text{H}}_{\text{4}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{3}}$. The synthesis of $p{\text{-FC}}_{\text{6}}{\text{H}}_{\text{4}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{3}}$ from benzene is shown below.

c) $1\text{-Bromo-2-fluoro-3,5-dimethylbenzene}$ from $m\text{-xylene}$

In the preparation of $1\text{-bromo-2-fluoro-3,5-dimethylbenzene}$ from $m\text{-xylene}$, the first step is the nitration of $m\text{-xylene}$ to form $\text{2,4-dimethyl-1-nitrobenzene}$. In the second step, the resulting compound reacts with $\text{Sn}$ in the presence of $\text{HCl}$ to convert the nitro group of the compound into amine group.

In the next step, further nitration takes place followed by the addition of $\text{Sn}$ in the presence of $\text{HCl}$ to form $\text{2-fluoro-3,5-dimethylaniline}$. This is then followed by the reaction of $\text{2-fluoro-3,5-dimethylaniline}$ with ${\text{NaNO}}_{\text{2}}$ in the presence of $\text{HCl}$ to form diazonium compound. In the final step, the diazonium compound reacts with $\text{CuBr}$ in the presence of $\text{HBr}$. This results in the formation of the desired product $1\text{-bromo-2-fluoro-3,5-dimethylbenzene}$. The synthesis of $1\text{-bromo-2-fluoro-3,5-dimethylbenzene}$ from $m\text{-xylene}$ is shown below.

d)

In the first step of the reaction, the starting material reacts with ${\text{NaNO}}_{\text{2}}$ in the presence of $\text{HCl}$ to form the diazonium compound. In the second step, the diazonium compound reacts with $\text{CuBr}$ in the presence of $\text{HBr}$ to form $\text{1-bromo-2-methyl-4-nitronaphthalene}$ that further reacts with $\text{Sn}$ in the presence of $\text{HCl}$ to convert the nitro group of the compound into amine group.

In the next step, the resulting compound reacts with ${\text{NaNO}}_{\text{2}}$ in the presence of $\text{HCl}$ to form diazonium compound that finally yields the desired compound on heating with ${\text{HBF}}_{\text{4}}$. The complete synthesis is shown below.

e) $o{\text{-BrC}}_{\text{6}}{\text{H}}_{\text{4}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$ from $p{\text{-O}}_{\text{2}}{\text{NC}}_{\text{6}}{\text{H}}_{\text{4}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$

The preparation of $o{\text{-BrC}}_{\text{6}}{\text{H}}_{\text{4}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$ can be done by treating $p{\text{-O}}_{\text{2}}{\text{NC}}_{\text{6}}{\text{H}}_{\text{4}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$ with $\text{Sn}$ in the presence of $\text{HCl}$ to convert the nitro group into amine group. In the second step, the resulting compound reacts with ${\text{NaNO}}_{\text{2}}$ in the presence of $\text{HCl}$ to form a diazonium compound. The diazonium compound reacts with ethanol to give $tert\text{-butyl}$ benzene that undergoes nitration to form $\text{1-}tert\text{-butyl-2-nitrobenzene}$. The resulting compound again reacts with $\text{Sn}$ in the presence of $\text{HCl}$ to convert the nitro group into amine group.

In the next step, the resulting compound reacts with ${\text{NaNO}}_{\text{2}}$ in the presence of $\text{HCl}$ to form diazonium compound. Finally, the diazonium compound reacts with $\text{CuBr}$ in the presence of $\text{HBr}$ to form the desired product. The synthesis of $o{\text{-BrC}}_{\text{6}}{\text{H}}_{\text{4}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$ from $p{\text{-O}}_{\text{2}}{\text{NC}}_{\text{6}}{\text{H}}_{\text{4}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$ is shown below.

f) $m{\text{-ClC}}_{\text{6}}{\text{H}}_{\text{4}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$ from $p{\text{-O}}_{\text{2}}{\text{NC}}_{\text{6}}{\text{H}}_{\text{4}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$

The preparation of $m{\text{-ClC}}_{\text{6}}{\text{H}}_{\text{4}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$ from $p{\text{-O}}_{\text{2}}{\text{NC}}_{\text{6}}{\text{H}}_{\text{4}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$ occurs in many steps such as the treatment of $p{\text{-O}}_{\text{2}}{\text{NC}}_{\text{6}}{\text{H}}_{\text{4}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$ with $\text{Sn}$ in the presence of $\text{HCl}$ to convert the nitro group into amine group. Hydrolysis, chlorination and nitration finally yield $m{\text{-ClC}}_{\text{6}}{\text{H}}_{\text{4}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$ as the final product. The synthesis of $m{\text{-ClC}}_{\text{6}}{\text{H}}_{\text{4}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$ from $p{\text{-O}}_{\text{2}}{\text{NC}}_{\text{6}}{\text{H}}_{\text{4}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$ is shown below.

g) $1\text{-Bromo-3,5-diethylbenzene}$ from $m\text{-diethylbenzene}$

In the synthesis of $1\text{-bromo-3,5-diethylbenzene}$, $m\text{-diethylbenzene}$ reacts with ${\text{SeO}}_2$ to form $\text{1,1-}\left(\text{1,3-phenylene}\right)\text{diethanone}$ that undergoes bromination in the presence of ${\text{AlCl}}_{\text{3}}$ to form bromo substituted compound.

In the final step, reduction of ketone occurs to yield the final product $1\text{-bromo-3,5-diethylbenzene}$. The synthesis of $1\text{-bromo-3,5-diethylbenzene}$ from $m\text{-diethylbenzene}$ is shown below.

h)

The preparation of $\text{1-bromo-2-iodo-3-}\left(\text{trifluoromethyl}\right)\text{benzene}$ from $\text{N-}\left(\text{4-amino-2-bromo-6-}\left(\text{trifluoromethyl}\right)\text{phenyl}\right)\text{acetamide}$ occurs in many steps such as reaction with ${\text{NaNO}}_{\text{2}}$ in the presence of $\text{HCl}$ and hydrolysis.

The synthesis of $\text{1-bromo-2-iodo-3-}\left(\text{trifluoromethyl}\right)\text{benzene}$ from $\text{N-}\left(\text{4-amino-2-bromo-6-}\left(\text{trifluoromethyl}\right)\text{phenyl}\right)\text{acetamide}$ is shown below.

i)

In this conversion, the formation of seven-membered ring present between two benzene rings occurs. Lithium aluminum hydride and sodium borohydride are strong reducing agents. The synthesis of the desired compound from its starting material is shown below.