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Chapter 22: Problem 534

Synthesize m-chloroaniline from benzene. Any inorganic reagents may be used.

Short Answer

Expert verified
To synthesize m-chloroaniline from benzene, follow these steps: 1. Perform nitration of benzene using concentrated sulfuric acid (H2SO4) and concentrated nitric acid (HNO3) to form nitrobenzene. 2. Reduce the nitro group to an amine using tin (Sn) and concentrated hydrochloric acid (HCl), then neutralize with sodium hydroxide (NaOH) to obtain aniline. 3. Protect the amine group by acetylating with acetyl chloride (CH3COCl) to form N-acetylaniline. 4. Introduce a chlorine atom in the meta-position via electrophilic aromatic substitution using aluminum chloride (AlCl3) and chlorine gas (Cl2) to form m-chloro-N-acetylaniline. 5. Remove the N-acetyl group by hydrolyzing with concentrated hydrochloric acid (HCl) and heating, resulting in m-chloroaniline and acetic acid as a side product.

Step by step solution

01

Nitrate Benzene

To introduce a nitro group on the benzene ring, we need to carry out a nitration reaction. This involves the use of a nitrate electrophile, which can be prepared by reacting concentrated sulfuric acid (H2SO4) with concentrated nitric acid (HNO3). The benzene is then reacted with the nitrate electrophile, forming the nitrobenzene product. #Step 2: Reduction of Nitrobenzene#
02

Reduce Nitro Group to Amine

To convert the nitro group to the amine, we need to perform a reduction reaction. One common method for reducing nitro groups is using tin (Sn) metal with concentrated hydrochloric acid (HCl) as the reducing agent. This reaction will yield anilinium chloride as an intermediate, which can then be neutralized with a base, such as sodium hydroxide (NaOH), to obtain aniline. #Step 3: Activation of Aniline#
03

Protect the Amine Group

Before we can carry out the electrophilic aromatic substitution to introduce the chlorine atom, we need to protect the amine group on the aniline to prevent it from undergoing multiple substitution reactions. This can be achieved by acetylating the amine group using acetyl chloride (CH3COCl) to form an N-acetylaniline. #Step 4: Chlorination of N-acetylaniline#
04

Introduce Chlorine Atom on the Benzene Ring

Next, we will introduce a chlorine atom in the meta-position, using an electrophilic aromatic substitution reaction. We can use aluminum chloride (AlCl3) as the Lewis acid catalyst and react it with chlorine gas (Cl2). This generates the electrophile to form m-chloro-N-acetylaniline as our product. #Step 5: De-protection of N-acetyl group#
05

Remove the N-acetyl Group

Finally, to obtain m-chloroaniline as the final product, we need to remove the N-acetyl group from m-chloro-N-acetylaniline. This can be achieved using hydrolysis with concentrated hydrochloric acid (HCl) and heating. The resulting product will be m-chloroaniline and acetic acid as a side product.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electrophilic Aromatic Substitution
Electrophilic Aromatic Substitution (EAS) is a common reaction mechanism used to introduce substituents onto an aromatic ring, such as benzene. The process involves the aromatic ring acting as a nucleophile, reacting with an electrophile to form a substituted aromatic compound.
The aromatic ring's stable 蟺-electron system tends to resist addition reactions that break the ring's aromaticity. Instead, substitution preserves the aromatic character while replacing a hydrogen atom with a new substituent.
  • The mechanism typically starts with the generation of a powerful electrophile.
  • The electrophile attacks the 蟺-electron cloud of the benzene, forming a high-energy 蟽-complex (also known as an arenium ion).
  • The arenium ion quickly loses a proton to restore the aromaticity of the ring, completing the substitution.
Through EAS, chemists can introduce various functional groups to the benzene ring, altering its chemical properties and reactivity.
Nitration Reaction
The nitration reaction is a specific type of electrophilic aromatic substitution where a nitro group ( O_2 nO_2 ) is introduced onto the aromatic ring. This is accomplished using a mixture of concentrated nitric acid ( HNO_3 HNO_3 ) and sulfuric acid ( H_2SO_4 H_2SO_4 ).
The combination of these acids creates the nitronium ion ( NO_2^+ NO_2^+ ), a strong electrophile capable of reacting with benzene. The process follows the typical EAS mechanism:
  • Formation of the nitronium ion in situ by protonation of nitric acid by sulfuric acid.
  • The generated nitronium ion attacks the benzene ring to form an intermediate 蟽-complex.
  • A hydrogen ion is then lost from the 蟽-complex, resulting in the regeneration of aromaticity and the formation of nitrobenzene.
Nitration not only introduces useful functional groups but also influences the reactivity and orientation of subsequent reactions on the benzene ring.
Reduction Reaction
Reduction reactions are crucial for converting functional groups, like nitro groups, into amines. In the synthesis of m-chloroaniline, the nitrobenzene produced from the nitration step is reduced to aniline, an important aromatic amine.
An efficient method uses tin (Sn) and concentrated hydrochloric acid ( HCl HCl ) for this reduction. Here's how the process works:
  • The tin primarily serves as a source of electrons, facilitating the reduction of the nitro group ( NO_2 NO_2 ) to an amine group ( NH_2 NH_2 ).
  • A temporary intermediate, anilinium chloride, forms during the reaction.
  • Neutralization with a base, like sodium hydroxide ( NaOH NaOH ), converts the anilinium intermediate to free base aniline.
Reduction is a key technique that enhances molecular diversity, expanding the array of accessible chemical compounds.
Amine Protection
Amine protection strategies are vital in organic synthesis to shield the reactive amine groups from unwanted reactions. In this process, the amine group of aniline is protected to ensure the desired meta-chlorination.
The protection is achieved by acetylating the amine with acetyl chloride ( CH_3COCl CH_3COCl ), forming N-acetylaniline:
  • The acetyl group prevents the amine from participating in additional substitution reactions by decreasing its nucleophilicity.
  • Introduction of the chlorine atom follows via another EAS reaction, yielding m-chloro-N-acetylaniline.
The protected form is later deprotected after chlorination by hydrolysis, removing the acetyl group and revealing m-chloroaniline.
Using protective groups is an essential strategy in multi-step organic syntheses, ensuring regioselectivity and control by temporarily deactivating reactive sites on molecules.

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Most popular questions from this chapter

Explain the feature of amine basicity.

Choline, a constituent of phospholipids (fat-like phosphate esters of great physiological importance), has the formula \(\mathrm{C}_{5} \mathrm{H}_{15} \mathrm{O}_{2} \mathrm{~N}\). It dissolves readily in water to form a strongly basic solution. It can be prepared by the reaction of ethylene oxide with trimethylamine in the presence of water. (a) What is a likely structure for choline? (b) What is a likely structure for its acety1 derivative, acetylcholine, \(\mathrm{C}_{7} \mathrm{H}_{17} \mathrm{O}_{3} \mathrm{~N}\), important in nerve action?

Reaction of acid chlorides with sodium azide, \(\mathrm{NaN}_{3}\), yields acyl azides, RCON \(_{3}\). When heated, these undergo the Curtius rearrangement to amines, \(\mathrm{RNH}_{2}\), or, in a non-hydroxylie solvent, to isocyanates, RNCO. Using the structure for the azide, suggest a mechanism for the rearrangement.

Novocaine, a local anesthetic, is a compound of formula \(\mathrm{C}_{13} \mathrm{H}_{20} \mathrm{O}_{2} \mathrm{~N}_{2}\). It is insoluble in water and dilute \(\mathrm{NaOH}\), but soluble in dilute HC1. Upon treatment with \(\mathrm{NaNO}_{2}\) and \(\mathrm{HCl}\) and then with \(\beta\) -naphthol, a highly colored solid is formed. When Novocaine is boiled with aqueous \(\mathrm{NaOH}\), it slowly dissolves. The alkaline solution is shaken with ether and the layers are separated. Acidification of the aqueous layer causes the precipitation of a white solid A; continued addition of acid causes \(A\) to redissolve. Upon isolation \(A\) is found to have a melting point of \(185-6^{\circ}\) and the formula \(\mathrm{C}_{7} \mathrm{H}_{7} \mathrm{O}_{2} \mathrm{~N}\). Evaporation of the ether layer leaves a liquid \(\mathrm{B}\) of formula \(\mathrm{C}_{6} \mathrm{H}_{15} \mathrm{ON}\). B dissolves in water to give a solution that turns litmus blue. Treatment of \(B\) with acetic anhydride gives C, \(\mathrm{C}_{8} \mathrm{H}_{17} \mathrm{O}_{2} \mathrm{~N}\), which is insoluble in water and dilute base, but soluble in dilute HCl. \(\mathrm{B}\) is found to be identical with the compound formed by the action of diethylamine on ethylene oxide. (a) What is the structure of Novocaine? (b) Outline all steps in a complete synthesis of Novocaine from toluene and readily available aliphatic and inorganic reagents. $$ \begin{array}{|l|l|l|} \hline {\text { CARBOXYLIC ACIDS }} \\ \hline \text { Name } & \text { Formula } & \begin{array}{c} \text { M.p. } \\ { }^{\circ} \mathrm{C} \end{array} \\ \hline \text { o-Nitrobenzoic } & \circ-\mathrm{O}_{2} \mathrm{NC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 147 \\ \hline \text { m-Nitrobenzoic } & \mathrm{m}-\mathrm{O}_{2} \mathrm{NC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 141 \\ \hline \text { p-Nitrobenzoic } & \mathrm{p}-\mathrm{O}_{2} \mathrm{NC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 242 \\ \hline \text { Phthalic } & \mathrm{o}-\mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{COOH})_{2} & 231 \\ \hline \text { Isophthalic } & \mathrm{m}-\mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{COOH})_{2} & 348 \\ \hline \text { Terephthalic } & \mathrm{p}-\mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{COOH})_{2} & 300 \text { sub1. } \\ \hline \text { Salicylic } & \mathrm{o}-\mathrm{HOC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 159 \\ \hline \text { p-Hydroxybenzoic } & \mathrm{p}-\mathrm{HOC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 213 \\ \hline \text { Anthranilic } & \mathrm{o}-\mathrm{H}_{2} \mathrm{NC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 146 \\ \hline \text { m-Aminobenzoic } & \mathrm{m}-\mathrm{H}_{2} \mathrm{NC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 179 \\ \hline \text { p-Aminobenzoic } & \mathrm{p}-\mathrm{H}_{2} \mathrm{NC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 187 \\ \hline \text { o-Methoxybenzoic } & \mathrm{o}-\mathrm{CH}_{3} \mathrm{OC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 101 \\ \hline \text { m-Methoxybenzoic } & \mathrm{m}-\mathrm{CH}_{3} \mathrm{OC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 110 \\ \hline \text { p-Methoxybenzoic(Anisic) } & \mathrm{p}-\mathrm{CH}_{3} \mathrm{OC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 184 \\ \hline \end{array} $$

Describe exactly how you would go about separating a mixture of the three water-insoluble liquids, aniline (b. p. \(\left.184^{\circ}\right), \mathrm{n}\) -butylbenzene (b. p. \(183^{\circ}\) ), and n-valeric acid (b. p. \(187^{\circ}\) ), recovering each compound pure and in essentially quantitative yield. Do the same for a mixture of the three water-insoluble solids, \(p\) -toluidine, o-bromobenzoic acid, and \(\mathrm{p}\) -nitroanisole.
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