Chapter 20, Problem 37P

Interpretation Introduction

Interpretation:

Thegiven compounds are to be prepared by using the toluene, sodium cyanide, and carbon dioxide as the sources of the carbon atoms, along with any necessary inorganic reagents.

Concept introduction:

The oxidation of alkyl benzene using strong oxidizing agent ${\text{KMnO}}_{\text{4}}$ or chromic acid forms benzoic acid.

The hydroxyl group of carboxylic acid can be substituted with chlorine atom by reaction with thionyl chloride to form acyl chloride.

The condensation of alcohol with carboxylic acid forms an ester.

The primary amide can be prepared by the acylation of ammonia.

Electron donating groups activates the arenes and gives electrophilic substitution at ortho-para position.

Electron withdrawing groups deactivates the arenes and gives electrophilic substitution at meta position.

The treatment of Grignard’s reagent on carbon dioxide forms a carboxylic acid.

The alkyl cyanide (nitrile) on hydrolysis yields carboxylic acid.

Alkyl cyanide (nitrile) can be prepared by the nucleophilic substitution of alkyl halide with cyanide ion.

Aryl cyanide can be synthesized by using a Sandmeyer reaction where, the aryl diazonium salt is treated with $\text{copper(I) cyanide}$.

The aryl diazonium salt is a key intermediate in synthesis of aromatic compounds, which is prepared by nitrosation of amine (primary amide) bysodium nitrite in acidic condition.

The aryl chlorides can be converted to acid anhydride by nucleophilic acyl substitution using carboxylic acid.

The nitration is the electrophilic substitution of aromatic compound using reagent nitric acid in concentrated sulfuric acid ${\text{HNO}}_{\text{3}}{\text{/H}}_{\text{2}}{\text{SO}}_{\text{4}}$.

The nitrobenzene on reduction with $\text{Sn/HCl}$ forms aniline.

The benzylic bromide can be synthesized from toluene using $\text{N-Bromosuccinimide }\left(\text{NBS}\right)$.

The benzoic acid on heating with calcium oxide reduced to benzene.

Answer to Problem 37P

Solution:

a)

b)

c)

d)

e)

f)

g)

h)

i)

Explanation of Solution

a) Benzoyl chloride

The benzoyl chloride can be synthesized starting with toluene.

The reaction sequence is shown below:

The toluene is oxidized to benzoic acid using a strong oxidizing agent ${\text{KMnO}}_{\text{4}}$. The benzoic acid is then treated with thionyl chloride in pyridine to form benzoyl chloride.

b) Benzoic anhydride

The reaction sequence for the synthesis of Benzoic anhydride is shown below:

The toluene is first converted to benzoic acid using strong oxidizing agent ${\text{KMnO}}_{\text{4}}$, and then benzoic acid is reacted with thionyl chloride in pyridine to form benzoyl chloride. The benzoate ion formed by benzoic acid in basic condition treated with benzoyl chloride which produced benzoic anhydride.

c) Benzyl benzoate

The reaction sequence for the synthesis of benzyl benzoate is shown below:

The benzyl benzoate is an ester prepared by reacting the benzyl alcohol and benzoic acid. The benzoic acid is prepared by oxidizing toluene by using strong oxidizing agent ${\text{KMnO}}_{\text{4}}$. The benzyl alcohol is prepared by reducing the benzoic acid with the help of reducing agent lithium aluminum hydride $\left({\text{LiAlH}}_{\text{4}}\right)$.

d) Benzamide

The benzamide is synthesized by the reaction of benzoyl chloride with ammonia. The benzoyl chloride is prepared by starting with toluene. The toluene is first converted to benzoic acid using strong oxidizing agent ${\text{KMnO}}_{\text{4}}$, and then benzoic acid is reacted with thionyl chloride in pyridine to form benzoyl chloride.

The reaction sequence for the synthesis of benzamide is shown below:

e) Benzonitrile

In the first step, the toluene is converted to benzoic acid using a strong oxidizing agent ${\text{KMnO}}_{\text{4}}$, and then benzoic acid is heated with calcium oxide to produced benzene. The benzene is subjected to nitration with nitric acid and sulfuric acid to form nitrobenzene. The nitro group in nitrobenzene is reduced to amino group using the reagent $\text{Sn, HCl}$ and formed aniline. The aniline is then reacted with sodium nitrite in acidic condition which produced aryl diazonium salt. This aryl diazonium salt treated with $\text{copper(I) cyanide}$ to form a desired product of benzonitrile.

The reaction sequence for the synthesis of benzonitrile is shown below:

f) Benzyl cyanide

The reaction sequence for the synthesis of benzyl cyanide is shown below:

In the first step, the toluene is converted to benzyl bromide using a reagent $\text{N-Bromosuccinimide }\left(\text{NBS}\right)$. The bromine atom in benzyl bromide is substituted by cyanide to form benzyl cyanide. The reagent used is $\text{NaCN}$ in solvent $\text{DMSO}$.

g) Phenylacetic acid

The reaction sequence for the synthesis of phenylacetic acid is shown below:

In the first step, the toluene is converted to benzyl bromide using a reagent $\text{N-Bromosuccinimide }\left(\text{NBS}\right)$. The benzyl bromide is converted to Grignard reagent by treating it with magnesium metal in diethyl eher. The Grignard’s reagent benzyl magnesium bromide ${\text{PhCH}}_{\text{2}}\text{MgBr}$ is reacted with carbon dioxide to form phenylacetic acid.

h) $\text{p-Nitrobenzoyl chloride}$

The reaction sequence for the synthesis of $\text{p-Nitrobenzoyl chloride}$ is shown below:

The toluene on nitration gave the mixture of ortho-nitro toluene and para-nitro toluene. The para-nitro toluene is then oxidized using a strong oxidizing agent ${\text{KMnO}}_{\text{4}}$, and formed para-nitro benzoic acid, which is then on reaction with thionyl chloride produced $\text{p-Nitrobenzoyl chloride}$.

i) $\text{m-Nitrobenzoyl chloride}$

The reaction sequence for the synthesis of $\text{m-Nitrobenzoyl chloride}$ is shown below:

In first step, the toluene is oxidized to benzoic acid using a strong oxidizing agent ${\text{KMnO}}_{\text{4}}$. The benzoic acid treated with thionyl chloride in pyridine to form benzoyl chloride. The benzoyl chloride on nitration using nitric acid in concentrated sulfuric acid ${\text{HNO}}_{\text{3}}{\text{/H}}_{\text{2}}{\text{SO}}_{\text{4}}$ and produced $\text{m-Nitrobenzoyl chloride}$.