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Chapter 3: Problem 98

The major product formed by the reaction of \(\mathrm{N}\) -phenylbenzamide with \(\mathrm{Br}_{2} / \mathrm{Fe}\) is

Short Answer

Expert verified
Answer: The major product formed in the reaction is 3-bromo-N-phenylbenzamide.

Step by step solution

01

Identify the reactants

We are given the reactants N-phenylbenzamide (an amide groups attached to a benzene ring) and Bromine with Iron catalyst (Br2/Fe). The general reaction we are looking into is an electrophilic aromatic substitution (EAS) involving benzene rings.
02

Understand the mechanism with Fe catalyst

The catalyst here is Fe, which reacts with bromine to form iron (III) bromide (FeBr3). This is an important step as the presence of FeBr3 makes the bromine a better electrophile. The reaction can be written as: Fe + Br2 → FeBr3
03

Electrophilic Attack on N-phenylbenzamide

Now, the electrophilic aromatic substitution takes place, with the formation of a sigma complex or arenium ion between the electrophile (bromine) and the benzene ring. The bromine electrophile is generated as a result of interacting with FeBr3 (bromine gains a positive charge).
04

Locate the preferred position for substitution

N-Phenylbenzamide has an amide group (\(\ce{-CONHPh}\)) attached to the benzene ring which acts as an electron donor. Due to resonance effect of amide group, it acts as a meta-directing group. This means that the bromine substituent prefers to attach at the meta position on the benzene ring, maintaining the resonance structure.
05

Formation of the major product

Once the bromine is added to the meta position, deprotonation by FeBr4- anion, which acts as a base, occurs, and the intermediate rearranges to form the final product. The resonance is now restored in the benzene ring, forming the para-brominated N-phenylbenzamide as the major product. The final product is: 3-bromo-N-phenylbenzamide.

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