Aromatic Compounds
Kekulé structure of benzene, aromaticity and Hückel's rule (4n+2), electrophilic aromatic substitution (nitration, halogenation, Friedel-Crafts), ortho/meta/para orientation and activating/deactivating groups.
Complete Theory
4Benzene (C₆H₆) is the parent compound of aromatic compounds. Its structure was proposed by August Kekulé (1865): a hexagonal ring of 6 carbon atoms with alternating single and double bonds (three conjugated double bonds).
However, benzene does not behave like a normal conjugated triene:
Aromaticity confers thermodynamic and chemical stability: aromatic compounds tend to keep the system intact, preferring substitution (which preserves aromaticity) over addition (which destroys it).
Real-world applications: benzene is a crucial industrial intermediate (production of styrene, phenol, cyclohexane for nylon, aniline). However, it is carcinogenic (leukemia), so its use is strictly regulated.
However, benzene does not behave like a normal conjugated triene:
- It does not undergo electrophilic addition (like alkenes), but electrophilic aromatic substitution
- All C—C bonds in benzene are identical (length 139.7 pm, intermediate between C—C single 154 pm and C=C double 134 pm)
- The resonance energy of benzene is about 152 kJ/mol: it is 152 kJ/mol more stable than the hypothetical cyclohexatriene with localized double bonds
Aromaticity confers thermodynamic and chemical stability: aromatic compounds tend to keep the system intact, preferring substitution (which preserves aromaticity) over addition (which destroys it).
Real-world applications: benzene is a crucial industrial intermediate (production of styrene, phenol, cyclohexane for nylon, aniline). However, it is carcinogenic (leukemia), so its use is strictly regulated.
Hückel's rule (Erich Hückel, 1931) establishes the criteria for aromaticity: a planar cyclic compound is aromatic if it possesses delocalized electrons, where is an integer (0, 1, 2, 3...).
Necessary conditions for aromaticity:
Aromaticity is not limited to hydrocarbons: heterocyclic compounds such as pyridine (C₅H₅N, 6 electrons), pyrrole (C₄H₅N, 6 electrons), thiophene (C₄H₄S) are aromatic (the heteroatom contributes a lone pair to the system).
Real-world applications: heterocyclic aromatic compounds are ubiquitous in biochemistry: the nitrogenous bases of DNA/RNA (purines, pyrimidines), chlorophyll (porphyrin ring with 22 electrons, n=5), the heme of hemoglobin.
Necessary conditions for aromaticity:
- Cyclic structure (closed ring of atoms)
- Planarity (all ring atoms must allow continuous overlap)
- Every atom of the ring must have a p orbital perpendicular to the ring plane (sp² or sp hybridization)
- delocalized electrons (Hückel's rule)
- Benzene (C₆H₆): 6 electrons (n=1: 4·1+2=6) → aromatic
- Naphthalene (C₁₀H₈): 10 electrons (n=2: 4·2+2=10) → aromatic (bicyclic)
- Anthracene (C₁₄H₁₀): 14 electrons (n=3) → aromatic (tricyclic)
- Cyclopentadienyl (C₅H₅⁻): 6 electrons (anion) → aromatic
- Cycloheptatrienyl (C₇H₇⁺): 6 electrons (tropylium cation) → aromatic
- Cyclobutadiene (C₄H₄): 4 electrons (n=0.5, not integer) → antiaromatic (unstable, non-planar)
Aromaticity is not limited to hydrocarbons: heterocyclic compounds such as pyridine (C₅H₅N, 6 electrons), pyrrole (C₄H₅N, 6 electrons), thiophene (C₄H₄S) are aromatic (the heteroatom contributes a lone pair to the system).
Real-world applications: heterocyclic aromatic compounds are ubiquitous in biochemistry: the nitrogenous bases of DNA/RNA (purines, pyrimidines), chlorophyll (porphyrin ring with 22 electrons, n=5), the heme of hemoglobin.
Electrophilic aromatic substitution (SEAr) is the characteristic reaction of aromatic compounds. An electrophile replaces a hydrogen atom on the ring, preserving aromaticity. The general two-step mechanism:
- Electrophilic attack (slow, rate-determining step): the electrophile E⁺ attacks the cloud of the ring, forming a Wheland intermediate ( complex, or arenium ion). This intermediate is a delocalized carbocation (non-aromatic, less stable) stabilized by resonance over three ring positions.
- Proton loss (fast step): a base (typically the counterion of the electrophile, e.g., AlCl₄⁻) removes the hydrogen from the attacked carbon, restoring the aromatic system.
- Nitration: . The electrophile is the nitronium ion NO₂⁺, generated from the sulfuric-nitric acid mixture. Product: nitrobenzene (intermediate for aniline and azo dyes).
- Halogenation: . The electrophile is Cl⁺ (polarized by FeCl₃ or AlCl₃). Bromination uses Br₂/FeBr₃, iodination is more difficult (I₂/HNO₃).
- Sulfonation: . The electrophile is SO₃ (or HSO₃⁺). It is reversible: desulfonation is possible with hot dilute acid.
- Friedel-Crafts alkylation: . The electrophile is the carbocation R⁺ (generated from RCl + AlCl₃). Limitations: possible carbocation rearrangements and polyalkylation.
- Friedel-Crafts acylation: . The electrophile is the acylium ion RCO⁺. No rearrangements occur and it produces aromatic ketones (e.g., acetophenone).
When a monosubstituted benzene (C₆H₅—G) undergoes a second electrophilic substitution, the attack position (ortho, meta, or para) is determined by the nature of the existing substituent G. Substituents fall into three classes:
1. Ortho-para directing activators: accelerate the reaction (compared to benzene) and direct the electrophile to ortho and para positions.
Real-world applications: knowledge of orientation is fundamental in the synthesis of polysubstituted aromatic compounds. For example, to obtain p-nitrotoluene, toluene is nitrated (methyl activating ortho/para) and the ortho isomer is separated. To introduce a group at the meta position, a nitro group (meta-directing deactivator) is used first, then reduced to an amine.
1. Ortho-para directing activators: accelerate the reaction (compared to benzene) and direct the electrophile to ortho and para positions.
- Groups with lone pairs donating by resonance: —NH₂, —OH, —OR (strongly activating). The lone pair delocalizes into the ring, increasing electron density especially at ortho and para positions.
- Alkyl groups (—CH₃, —C₂H₅): weakly activating by inductive effect (+I) and hyperconjugation. They are ortho/para directors. Toluene nitrates faster than benzene, forming predominantly o- and p-nitrotoluene.
- Electron-withdrawing groups by resonance and inductive effect (−M, −I): —NO₂, —CN, —COR, —COOH, —SO₃H. They withdraw electron density from the ring, deactivating it.
- Halogens (—F, —Cl, —Br, —I): they are deactivating (strong −I inductive effect) but ortho/para directing (by resonance +M effect: the lone pairs delocalize into the ring). Unique case: they deactivate but direct ortho/para.
Real-world applications: knowledge of orientation is fundamental in the synthesis of polysubstituted aromatic compounds. For example, to obtain p-nitrotoluene, toluene is nitrated (methyl activating ortho/para) and the ortho isomer is separated. To introduce a group at the meta position, a nitro group (meta-directing deactivator) is used first, then reduced to an amine.
Worked Examples
2Example 1Nitration of Toluene
Given
Toluene: C₆H₅—CH₃
Nitrating mixture: HNO₃/H₂SO₄
Find
Nitration products
Ortho/meta/para distribution
Step-by-step solution
1The methyl group (—CH₃) is a weak activator by inductive effect (+I) and hyperconjugation. It directs the attack of the electrophile NO₂⁺ to ortho and para positions (relative to CH₃).
2The typical product distribution: about 58–63% ortho-nitrotoluene (2-nitrotoluene), 4% meta-nitrotoluene (3-nitrotoluene), 35–37% para-nitrotoluene (4-nitrotoluene). The ortho preference is due to the proximity of the methyl group, but steric hindrance partially limits ortho.
3The ortho and para products have different properties: o-nitrotoluene is a yellow oil (mp −4°C), p-nitrotoluene is a crystalline solid (mp 52°C). They are separated by fractional crystallization. Both are intermediates for the production of toluidines (reduction) and azo dyes.
✓ Final result: Mixture: 60% ortho, 4% meta, 36% para-nitrotoluene (CH₃ activates and directs ortho/para)
Example 2Aromaticity Check with Hückel's Rule
Given
Pyridine: C₅H₅N (6-membered ring: 5 C + 1 N)
Cyclooctatetraene: C₈H₈ (8-membered ring)
Find
Is pyridine aromatic?
Is cyclooctatetraene aromatic?
Step-by-step solution
1Pyridine: planar 6-membered ring, each atom contributes 1 electron (sp² carbons) → 5 electrons from C. Nitrogen has a p orbital perpendicular to the plane with 1 electron (the other lone pair is in an sp² orbital in the plane, not part of the system). Total: 6 electrons = 4·1+2 → aromatic.
2Cyclooctatetraene: 8-membered ring with 4 alternating double bonds. That would be 8 electrons = 4·2 (n=2, but 8 is not of the form 4n+2). To avoid antiaromaticity, the molecule adopts a tub (non-planar) conformation, losing continuous delocalization. It is not aromatic; it is a non-planar conjugated polyene.
✓ Final result: Pyridine: yes, aromatic (6 e⁻, planar). Cyclooctatetraene: no, non-planar (8 e⁻, adopts tub conformation)
Exercises with Solutions
4Exercise 1Structure of benzeneMedium
Problem to solve
Explain why benzene has all C—C bonds of equal length (139.7 pm) despite the Kekulé formula showing alternating single and double bonds.
Given data
Benzene: C₆H₆C—C bond length: 154 pm (single), 134 pm (double)
Step-by-step solution
1The Kekulé formula shows two equivalent resonance structures with alternating double bonds. Neither alone represents the real structure.
2The six electrons are delocalized over the entire ring (resonance hybrid), uniformly distributing electron density among all bonds.
3Each C—C bond has bond order 1.5 (average between single and double), corresponding to a length of 139.7 pm, intermediate between 154 and 134 pm. The resonance energy (152 kJ/mol) further stabilizes the molecule.
✓ Final answer: Delocalization of π electrons → bond order 1.5 → intermediate length 139.7 pm
Exercise 2SEAr orientationHard
Problem to solve
Predict the main product of the nitration of benzoic acid (C₆H₅—COOH).
Given data
Benzoic acid: benzene with —COOH group (carboxyl)
Step-by-step solution
1The —COOH group is a strong meta-directing deactivator, by resonance effect (−M, the carbonyl attracts electrons) and inductive effect (−I).
2The electron density of the ring is reduced, especially at ortho and para positions (the carboxyl group attracts electrons). The meta position is less deactivated.
3The main product is m-nitrobenzoic acid (3-nitrobenzoic acid). The yield is lower compared to nitration of toluene (COOH deactivates the ring).
✓ Final answer: m-nitrobenzoic acid (—COOH group meta-directing deactivator)
Exercise 3Friedel-CraftsHard
Problem to solve
What product is obtained from the Friedel-Crafts alkylation of benzene with 2-chloropropane in the presence of AlCl₃? Why?
Given data
Benzene + CH₃CHClCH₃ (2-chloropropane) + AlCl₃
Step-by-step solution
12-chloropropane (CH₃—CHCl—CH₃) reacts with AlCl₃ (Lewis acid catalyst) forming an isopropyl carbocation: CH₃—C⁺H—CH₃ (secondary, stable).
2An n-propyl carbocation (primary, less stable) could also form, but it would immediately rearrange by 1,2-hydride migration to the more stable isopropyl carbocation.
3The isopropyl carbocation attacks the benzene ring → cumene (isopropylbenzene). Cumene is an important intermediate (cumene process: oxidation to hydroperoxide, then cleavage into phenol and acetone).
✓ Final answer: Cumene (isopropylbenzene) — the primary carbocation rearranges to secondary (isopropyl)
Exercise 4Orientation-guided synthesisVery Hard
Problem to solve
Design the synthesis of p-bromoacetanilide starting from benzene.
Given data
Target: p-bromoacetanilide (Br—C₆H₄—NHCOCH₃) starting from benzene
Step-by-step solution
1Step 1: Nitration of benzene (HNO₃/H₂SO₄) → nitrobenzene.
2Step 2: Reduction of nitrobenzene to aniline (C₆H₅NH₂): use Sn/HCl or Fe/HCl. Aniline has —NH₂ (strongly activating ortho/para-directing).
3Step 3: Acetylation of aniline: reaction with acetic anhydride (CH₃CO)₂O → acetanilide (C₆H₅NHCOCH₃). Acetylation reduces the reactivity of aniline (avoids polybromination) and still directs ortho/para.
4Step 4: Bromination of acetanilide (Br₂/FeBr₃ or Br₂/acetic acid). The —NHCOCH₃ group is activating ortho/para-directing. For steric reasons (the acetamido group is bulky), bromination occurs preferentially at para → p-bromoacetanilide.
✓ Final answer: Benzene → nitrobenzene → aniline → acetanilide → p-bromoacetanilide (following group orientation)
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