Aromatic Nitration

Introduction

Aromatic nitration replaces a ring hydrogen with a nitro group, –NO₂. The standard reagent mixture is concentrated nitric acid and sulfuric acid. Sulfuric acid helps generate the strongly electrophilic nitronium ion, NO₂⁺, from nitric acid. Benzene attacks this species, forms a nonaromatic sigma complex, and then loses a proton to regain aromaticity and yield nitrobenzene.

Core explanation

Nitric acid alone contains nitrogen in an electron-poor environment, but the mixed acid generates a more reactive electrophile. Sulfuric acid protonates nitric acid; the protonated species can lose water to give NO₂⁺. Nitronium is commonly drawn as O=N⁺=O, with positive charge associated with nitrogen in the electrophilic representation. The exact acid equilibria are more detailed than a single arrow, but the key particle that attacks the aromatic ring is NO₂⁺, not neutral nitrate or a free oxygen atom.

In the ring step, a pair of benzene pi electrons forms a bond to the nitrogen of NO₂⁺. The attacked ring carbon now bears both H and NO₂, and the ring temporarily loses aromatic conjugation. Positive charge in the resulting sigma complex is delocalised over several ring carbons by resonance. Draw at least two contributors by shifting the remaining pi bonds; do not move atoms or the attached nitro group between contributors. The sigma complex is high in energy because aromatic stabilisation has been interrupted.

A base in the mixture, often represented as HSO₄⁻, removes the hydrogen from the same carbon that received NO₂. The C–H bonding pair returns to the ring pi system, restoring aromaticity. The product is nitrobenzene. The overall change is benzene C–H → benzene C–NO₂, not addition of HNO₃ across one benzene double bond. This final deprotonation is the characteristic rearomatisation step shared by many electrophilic aromatic substitutions.

On a substituted aromatic ring, the existing group affects rate and position. An electron-donating group may speed nitration and direct nitro placement to particular ring positions, while strongly electron-withdrawing groups usually deactivate the ring. The nitro group itself is strongly electron-withdrawing and deactivates further electrophilic substitution; it is generally meta-directing for subsequent EAS. Multiple nitrations therefore require more forcing conditions and position analysis. These are mechanistic consequences of sigma-complex stability, not arbitrary naming rules.

Nitration is useful because a nitro group can be reduced to an amino group, –NH₂, by suitable reducing systems. Benzene can thus be converted through nitrobenzene to aniline, an important aromatic amine building block. In a synthesis plan, nitration determines where nitrogen-based functionality enters the ring. The subsequent reduction changes N–O bonds but ordinarily preserves the ring C–N attachment.

The reaction involves concentrated corrosive acids and is strongly exothermic; actual laboratory handling requires controlled procedures. For mechanism exercises, focus on correct electrophile generation, electron flow and aromaticity restoration rather than treating the two acid names as a product recipe. If the substrate bears multiple activating groups, a product mixture or overreaction can occur, so prediction should include the ring's substitution pattern and conditions.

Step-by-step reasoning

Write HNO₃/H₂SO₄ and identify NO₂⁺ as the electrophile. Choose the ring carbon that will receive the nitro group, considering any existing substituent's directing effect. Draw a pi-bond arrow to nitronium nitrogen, yielding the sigma complex with H still attached to that carbon. Show resonance-delocalised positive charge. Then let base remove that H and return the C–H electron pair into the ring to restore aromaticity.

Visual explanation

Draw a small box where HNO₃ plus acid gives NO₂⁺ and water. An arrow leads to benzene with one pi bond reaching toward the nitrogen of NO₂⁺. In the next picture, the chosen carbon bears both H and NO₂ and the ring is marked nonaromatic and positively charged. A final arrow removes H and draws the circle or three alternating pi bonds back inside the ring.

Real-world analogy

A six-person ring passes a baton in a coordinated pattern. A new participant holding a nitro marker enters and temporarily interrupts the pattern at one place. The original participant at that place exits, allowing the coordinated pattern to resume. The temporary interruption is the sigma complex; the restored pattern is nitrobenzene's aromatic pi system.

Real-world example

Nitrobenzene can be reduced to aniline, a precursor for dyes and other aromatic chemicals. Starting from a substituted benzene, choosing nitration conditions and understanding directing effects can place the future amino group at a useful position. A synthetic chemist plans the sequence with the final amine in mind, not just the immediate nitro product.

Why?

Why is sulfuric acid included with nitric acid? It promotes formation of the highly electrophilic nitronium ion by protonating nitric acid and enabling water loss. Benzene is sufficiently stabilised by aromaticity that a strong electrophile is needed for attack at a practical rate. The final proton loss restores aromatic stabilisation after the difficult C–N bond-forming step.

Common misconception

"Nitrate ion, NO₃⁻, is the electrophile in benzene nitration." Nitrate is negatively charged and not the strongly electron-poor species that accepts the ring pi pair. The key electrophile is NO₂⁺, nitronium, generated in the mixed acid. The nitro group bonds to the ring through nitrogen.

Worked example

Question: Identify the electrophile and outline the ring steps in converting benzene to nitrobenzene with HNO₃/H₂SO₄.

Reasoning: Mixed acid generates NO₂⁺. Benzene pi electrons bond to nitronium nitrogen, forming a sigma complex with one ring carbon bearing H and NO₂. HSO₄⁻ or another base removes that H while the C–H electrons restore aromaticity.

Answer: The electrophile is NO₂⁺; electrophilic attack followed by rearomatising deprotonation gives nitrobenzene.

Quick check

1. Which atom of NO₂⁺ becomes bonded to the aromatic ring during nitration? Answer: Nitrogen, producing a ring C–N bond to the nitro group.

Exam focus

Show nitronium generation from nitric acid in sulfuric acid before ring attack. Draw the C–N bond, not a C–O bond, and keep H on the attacked carbon until the deprotonation step. Include sigma-complex resonance and restoration of aromaticity; on substituted rings, justify position by the existing group's directing effect.

Advanced insight

The nitro group's strong electron withdrawal through both inductive and resonance effects destabilises many later sigma complexes, slowing further EAS and favouring meta substitution. This creates an important planning constraint: nitrating a ring early can make subsequent electrophilic substitutions harder or change their positions. Reduction to an amino group later reverses the ring's electronic character substantially.

Summary

Mixed nitric and sulfuric acids generate nitronium ion, NO₂⁺. An aromatic ring bonds to its nitrogen, forming a positively charged nonaromatic sigma complex, then loses the H from that carbon to restore aromaticity. Benzene gives nitrobenzene, whose nitro group can later be reduced to an amine. Existing ring substituents govern nitration rate and position through their effects on sigma-complex stability.

Practice questions

1. What is the electrophile in ordinary benzene nitration? Answer: Nitronium ion, NO₂⁺, formed from nitric acid in sulfuric acid. 2. Does the nitro group attach through N or O? Answer: Through nitrogen, creating a ring C–N bond. 3. Why does the sigma complex lose H after nitronium attack? Answer: Loss of H lets the C–H electron pair restore the aromatic pi system. 4. What useful aromatic amine can nitrobenzene form after reduction? Answer: Aniline, C₆H₅NH₂, with the ring C–N bond retained.