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Chapter 3: Problem 112

Statement 1 \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{2}-\mathrm{CH}_{3}(\) ethyl benzene \()\) on reaction with \(\mathrm{NBS}\) gives mainly \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CHBr}-\mathrm{CH}_{3} .\) and Statement 2 The reaction with NBS proceeds by free radical mechanism.

Short Answer

Expert verified
Answer: Yes, both statements are true. The reaction between ethyl benzene and NBS results in the formation of \(\mathrm{C}_{6}\mathrm{H}_{5}-\mathrm{CHBr}-\mathrm{CH}_{3}\), and the reaction proceeds via a free radical mechanism.

Step by step solution

01

Understanding the Reaction

Ethyl benzene reacts with NBS, which is a brominating agent, to form a substituted benzene compound \(\mathrm{C}_{6}\mathrm{H}_{5}-\mathrm{CHBr}-\mathrm{CH}_{3}\). The main reaction that takes place is bromination.
02

Importance of N-Bromosuccinimide (NBS)

The presence of NBS in the reaction mixture is important since it facilitates the selective replacement of an allylic or benzylic hydrogen with a bromine atom. This process is known as allylic or benzylic bromination.
03

Reaction Mechanism

The reaction between ethyl benzene and NBS proceeds via a free radical mechanism. This involves three main steps: 1. Initiation: Formation of bromine radicals from NBS 2. Propagation: Reaction between bromine radicals and ethyl benzene 3. Termination: Combination of free radicals to form the product
04

Checking the validity of both statements

Statement 1: \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\) (ethyl benzene) on reaction with \(\mathrm{NBS}\) gives mainly \(\mathrm{C}_{6}\mathrm{H}_{5}-\mathrm{CHBr}-\mathrm{CH}_{3}\). This statement is true, as the bromination reaction with NBS predominantly proceeds through a benzylic bromination, which results in the mentioned product. Statement 2: The reaction with NBS proceeds by a free radical mechanism. This statement is also true, as explained in Step 3. The reaction between ethyl benzene and NBS occurs via a free radical mechanism involving initiation, propagation, and termination steps.

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Most popular questions from this chapter

Mononitration of biphenyl by treating with nitrating mixture gives mainly

The carboxyl group is displaced when salicylic acid undergoes (a) bromination (b) nitration (c) either (a) or (b) (d) dehydration

$$ \begin{array}{l} \text { The organic product (X) formed in the following reaction is }\\\ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Cl}+\mathrm{C}_{6} \mathrm{H}_{5} \stackrel{\mathrm{AICl}_{3}}{\longrightarrow} \mathrm{X} \end{array} $$

Statement 1 Phenol reacts with \(\mathrm{PCl}_{5}\) to give only a small amount of chlorobenzene, the main product is triphenyl phosphate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}\right)_{3} \mathrm{P}=\mathrm{O}\) or \(\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \mathrm{PO}_{4}\) and Statement 2 \(\mathrm{C}-\mathrm{O}\) bond in phenol has partial double bond character.

o-Nitrophenol has lower \(\mathrm{pK}\) than (a) p-nitrophenol (b) o-cresol (c) o-fluorophenol (d) \(\mathrm{p}\) -methoxyphenol
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