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Chapter 4: Problem 142

\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3} \frac{\mathrm{Cl}_{2}}{\mathrm{Fe}}{ }^{\mathrm{C}}(\mathrm{A}) \stackrel{\mathrm{NBS} / \mathrm{CCl}_{4}}{\longrightarrow}(\mathrm{B})\)

Short Answer

Expert verified
Answer: The final product formed is benzyl bromide (C6H5CH2Br).

Step by step solution

01

Reaction of Toluene with Cl2

For the first step, toluene (C6H5CH3) reacts with Cl2 in the presence of a Fe catalyst. The Fe catalyst helps to polarize Cl2, making it easier for one chlorine atom to react with the toluene's CH3 group. The H atom in the methyl group of toluene is replaced by Cl.
02

Formation of Intermediate Compound A

At the end of the first step, we obtain the compound A. The structure of compound A is chloromethylbenzene (C6H5CH2Cl).
03

Reaction of Chloromethylbenzene with NBS

In this step, N-bromosuccinimide (NBS) acts as a selective brominating agent, which is used to replace the H atom on the benzylic position of the chloromethylbenzene. CCl4 acts as a solvent in this reaction.
04

Formation of Final Product B

At the end of step 3, we obtain the final product B. The structure of compound B is benzyl bromide (C6H5CH2Br). So, the overall reaction can be described as: Toluene (C6H5CH3) + Cl2/Fe → Chloromethylbenzene (C6H5CH2Cl) + NBS/CCl4 → Benzyl Bromide (C6H5CH2Br)

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Most popular questions from this chapter

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