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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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