Question

Bromination of benzene must be carried out in the presence of reagent ' \(X\) ' . ' \(X\) ' is (A) \(\mathrm{CCl}_{4}\) (B) \(\mathrm{NaBr}\) (C) \(\mathrm{FeBr}_{3}\) (D) \(\mathrm{HBr}\)

Short answer

The correct reagent 'X' for bromination of benzene is (C) \(\mathrm{FeBr}_{3}\), as it acts as a strong Lewis acid and generates a suitable electrophile for the reaction.

Step-by-step solution

Understanding the bromination of benzene

The bromination of benzene is an electrophilic aromatic substitution reaction where a hydrogen atom on the benzene ring is replaced by a bromine atom. This process requires a strong Lewis acid as a catalyst to generate a suitable electrophile (the bromine-BR2 molecule) for the reaction to proceed. Now let's move to identifying the correct reagent 'X'.

Analyzing the given answer choices

We have the following four options for reagent 'X': (A) \(\mathrm{CCl}_{4}\) (Carbon tetrachloride) (B) \(\mathrm{NaBr}\) (Sodium bromide) (C) \(\mathrm{FeBr}_{3}\) (Iron(III) bromide) (D) \(\mathrm{HBr}\) (Hydrogen bromide)

Determine the correct reagent 'X' for bromination of benzene

Among these four options, the correct reagent 'X' should act as a strong Lewis acid, and generate an electrophile for the reaction. Let's evaluate each option: (A) \(\mathrm{CCl}_{4}\): Carbon tetrachloride is a nonpolar solvent and doesn't act as a strong Lewis acid. It can not generate an electrophile required for the bromination of benzene. Therefore, this option is not correct. (B) \(\mathrm{NaBr}\): Sodium bromide is a salt and it doesn't act as a strong Lewis acid. It fails to generate an electrophile required for the reaction. Hence, this option is not correct. (C) \(\mathrm{FeBr}_{3}\): Iron(III) bromide is a strong Lewis acid and forms a complex with bromine molecules, generating a suitable electrophile for the bromination of benzene. It effectively catalyzes the reaction. This option is correct. (D) \(\mathrm{HBr}\): Hydrogen bromide is a weak acid and not a strong Lewis acid, it is incapable of generating the necessary electrophile for the reaction. So, this option is not correct. The correct answer is (C) \(\mathrm{FeBr}_{3}\), as it is the strong Lewis acid required for the bromination of benzene.

Key concepts

Electrophilic Aromatic Substitution

In electrophilic aromatic substitution (EAS), a hydrogen atom on an aromatic ring, like benzene, is replaced by an electrophile. This type of reaction preserves the aromaticity of the ring, a key feature of benzene, through a two-step mechanism.

The first step involves the formation of a positively charged intermediate known as the arenium ion. This is temporary and occurs when the electrophile attacks the aromatic ring. In the second step, this intermediate quickly stabilizes by losing a hydrogen, thereby restoring the aromatic structure of the benzene.

The benzene ring is highly stable due to its conjugated π-electron system. During EAS, this stability is slightly disrupted but then restored, allowing the reaction to proceed under specific conditions. This makes benzene less reactive towards addition reactions compared to alkenes. Therefore, the use of catalysis and strong electrophiles is crucial in these reactions.

Lewis Acid Catalysis

Lewis acid catalysis is essential in the bromination of benzene. A Lewis acid, like \(\mathrm{FeBr}_{3}\), accepts electron pairs, which helps in the formation of more reactive species.

During the process, \(\mathrm{FeBr}_{3}\) combines with bromine (\(\mathrm{Br}_{2}\)) to create a powerful bromine cation, \([\mathrm{Br}^{+}]\), which is the true electrophilic species in the reaction. This complexation helps in breaking the Br-Br bond more efficiently and facilitates its attack on the electron-rich benzene ring.

Without the presence of a Lewis acid, bromine alone isn't a strong enough electrophile to react with benzene. The Lewis acid catalyst thus speeds up the reaction by making the electrophile more accessible and reactive.

Electrophile Generation

Electrophile generation is a critical step in the electrophilic aromatic substitution of benzene. This involves creating an active electrophile that can successfully attack the benzene ring.

In the bromination reaction, the electrophile is generated when \(\mathrm{Br}_{2}\) interacts with the Lewis acid \(\mathrm{FeBr}_{3}\). This interaction polarizes the bromine molecule, making one bromine atom more positive and thus a better electrophile.

The formation of this electrophilic bromine cation is essential for the reaction to occur, as it is reactive enough to attack the stable benzene ring, breaking the aromatic π-system momentarily. This step is central to the success of the substitution process, ensuring that the bromine can replace a hydrogen atom on the ring.