Welcome to one of the most important classes of reaction in aromatic chemistry! In this topic, we will look at how the stable benzene ring reacts.

What you'll learn:

• Why benzene undergoes substitution reactions rather than addition reactions.
• How to draw the mechanism for electrophilic aromatic substitution.
• The reagents, conditions, and mechanisms for nitration and Friedel–Crafts acylation.
• How to write the equations for generating the electrophiles in these reactions.

---

Why does benzene undergo substitution?

Benzene has a delocalised ring of $\pi$ (pi) electrons above and below the plane of its carbon atoms. This ring is extremely stable.

If benzene were to undergo an addition reaction (like an alkene does), the delocalised ring would be permanently broken, and that special thermodynamic stability would be lost. Instead, benzene prefers to undergo substitution.

Because the delocalised ring is an area of high electron density, it naturally attracts electrophiles. When the electrophile attacks, the delocalised ring temporarily breaks to form an intermediate. However, the ring is quickly restored when a hydrogen atom is kicked out as an $H^{+}$ ion. At A-Level, we restrict our focus to monosubstitutions (replacing just one hydrogen atom at a time).

---

Reaction 1: Nitration

Nitration replaces a hydrogen atom on the benzene ring with a nitro group (${-NO}_2$).

Reagents and conditions

To nitrate benzene, you must heat it with a mixture of two concentrated acids:

• Reagents: Concentrated nitric acid (${HNO}_3$) and concentrated sulfuric acid ($H_2{SO}_4$).
• Conditions: $50{}^{\circ }C$.

Generating the electrophile

The actual electrophile in this reaction is the nitronium ion, ${NO}_2^{+}$. Nitric acid alone is not a strong enough acid to generate this effectively, which is why we add concentrated sulfuric acid.

Here, sulfuric acid is a stronger acid than nitric acid, so it forces nitric acid to act as a base (a proton acceptor):

$${HNO}_3+H_2{SO}_4\rightleftharpoons H_2{NO}_3^{+}+{HSO}_4^{-}$$

The protonated nitric acid is unstable and immediately decomposes to form water and the powerful nitronium ion electrophile:

$$H_2{NO}_3^{+}\to {NO}_2^{+}+H_{2}O$$

Combining these gives the overall generation equation you should memorise:

$${HNO}_3+H_2{SO}_4\to {NO}_2^{+}+{HSO}_4^{-}+H_{2}O$$

The Nitration Mechanism

Here is exactly what is happening in the mechanism steps:

1. Attack: A pair of electrons from the delocalised $\pi$ ring moves to the positively charged nitrogen atom on the ${NO}_2^{+}$ ion.
2. Intermediate: The delocalised ring is temporarily broken. The intermediate is drawn with a "horseshoe" shaped partial ring and a positive charge in the centre. Both the new ${-NO}_2$ group and the old hydrogen atom are currently attached to the same top carbon.
3. Restoration: The electrons from the $C-H$ bond move back into the ring. The delocalised system is restored, nitrobenzene is formed, and an $H^{+}$ ion is released.

---

Reaction 2: Friedel–Crafts Acylation

Friedel–Crafts acylation replaces a hydrogen atom on the benzene ring with an acyl group ($R-C=O$).

Reagents and conditions

• Reagents: An acyl chloride (e.g., ethanoyl chloride, ${CH}_{3}COCl$) and an aluminium chloride catalyst (${AlCl}_3$).
• Conditions: Anhydrous (dry) conditions.

Generating the electrophile

The electrophile is an acylium ion (e.g., ${CH}_3{CO}^{+}$). The polar $C-Cl$ bond in the acyl chloride is not reactive enough on its own to break the stable benzene ring.

We use ${AlCl}_3$ as a halogen carrier. The aluminium atom in ${AlCl}_3$ is electron-deficient, so it aggressively accepts a lone pair from the chlorine atom of the acyl chloride, breaking the $C-Cl$ bond:

$${CH}_{3}COCl+{AlCl}_3\to {CH}_3{CO}^{+}+{AlCl}_4^{-}$$

The Acylation Mechanism

The mechanism follows the exact same logic as nitration:

1. Electrons from the delocalised ring attack the positively charged carbon on the acylium ion.
2. The horseshoe intermediate forms.
3. The $C-H$ bond breaks, throwing its electrons back into the ring to restore stability, ejecting an $H^{+}$ ion.

Regenerating the catalyst

Because ${AlCl}_3$ is a catalyst, it must be regenerated at the end of the reaction. The $H^{+}$ ion kicked out of the benzene ring reacts with the ${AlCl}_4^{-}$ ion formed earlier:

$$H^{+}+{AlCl}_4^{-}\to {AlCl}_3+HCl$$

Notice that misty white fumes of hydrogen chloride ($HCl$) gas are produced as a byproduct of this reaction.

---

A Note on Practical Skills

You may carry out a nitration in the lab — a common experiment is the nitration of methyl benzoate to produce methyl 3-nitrobenzoate. Because the product is a solid at room temperature, it provides an excellent opportunity to practice purification.

You would purify the crude product using recrystallisation (dissolving the product in the minimum volume of hot solvent, letting it cool to form pure crystals, and filtering under reduced pressure). Finally, you would determine its melting point to check its purity; a narrow melting point range that matches the data book value indicates a pure sample.

---