Hydrocarbons • Topic 3 of 3

Aromatic Hydrocarbons

Aromatic hydrocarbons (arenes) contain a benzene ring or related conjugated cyclic system. Benzene, C6H6, is the parent. Despite a high degree of unsaturation it is unusually stable and prefers substitution over addition — a hallmark of aromatic character.

Structure of benzene

Kekulé structure: a six-membered ring with alternating single and double bonds. But all six C–C bonds in benzene are equal (139 pm, between a single 154 pm and double 134 pm bond), which a fixed Kekulé picture cannot explain. Resonance: benzene is a resonance hybrid of two Kekulé forms; the true structure has delocalised electrons, and its measured stability exceeds the calculated value by the resonance energy (~150 kJ mol-1). Molecular-orbital / orbital picture: each carbon is sp2 hybridised and planar; the six unhybridised p-orbitals overlap sideways to form two continuous π electron clouds above and below the ring, holding six delocalised π electrons.

Hückel’s rule and aromaticity

A ring is aromatic if it is (i) cyclic, (ii) planar, (iii) fully conjugated and (iv) contains (4n + 2) π electrons (n = 0, 1, 2, …) — Hückel’s rule. Benzene has 6 π electrons (n = 1), so it is aromatic. Rings with 4n π electrons (e.g. cyclobutadiene) are antiaromatic and unstable.

Preparation of benzene

  • Cyclic polymerisation of ethyne: 3 HC≡CH ⟶[red-hot Cu tube, 873 K] C6H6.
  • Decarboxylation: C6H5COONa + NaOH ⟶[CaO] C6H6 + Na2CO3.
  • Reduction of phenol: phenol vapour + Zn dust → benzene + ZnO.

Electrophilic aromatic substitution (EAS)

Because the ring is electron-rich, it reacts with electrophiles E+ by substitution, preserving aromaticity. The general mechanism has two steps: the electrophile adds to give a resonance-stabilised carbocation called the arenium ion (sigma complex), then a proton is lost to restore the aromatic ring. Key reactions:

  • Nitration: conc. HNO3 + conc. H2SO4 generates NO2+; benzene → nitrobenzene.
  • Halogenation: Cl2/FeCl3 (Lewis acid) gives chlorobenzene.
  • Sulphonation: fuming H2SO4 (SO3) gives benzenesulphonic acid (reversible).
  • Friedel–Crafts alkylation: R–Cl/AlCl3 gives an alkylbenzene.
  • Friedel–Crafts acylation: RCOCl/AlCl3 gives an aryl ketone.

Directive influence of substituents

A group already on the ring decides where the next one goes. Ortho/para directors (activating, electron-donating: –CH3, –OH, –NH2, –OR; the halogens are o/p but deactivating) push the new group to the 2- and 4-positions. Meta directors (deactivating, electron-withdrawing: –NO2, –COOH, –CHO, –SO3H, –C≡N) send it to the 3-position. The effect is explained by which positions give the most stable arenium-ion intermediate.

Carcinogenicity and toxicity

Benzene and many polynuclear aromatic hydrocarbons (e.g. benzo[a]pyrene in tobacco smoke and coal tar) are carcinogenic — chronic exposure damages bone marrow and can cause leukaemia. Benzene must therefore be handled with strict ventilation and care.

Electrophilic aromatic substitution: benzene plus E+ forms the arenium ion, then loses H++E+→EH+arenium ion (sigma complex)→Eproduct + H+ (ring restored)
Solved Examples
1
Worked Example
Using Huckel's rule, explain why benzene is aromatic but cyclobutadiene is not.
Solution
  1. An aromatic ring must be cyclic, planar, fully conjugated and have (4n+2) π electrons.
  2. Benzene has 6 π electrons; setting 4n+2 = 6 gives n = 1 (a whole number), so it is aromatic.
  3. Cyclobutadiene has 4 π electrons; 4n+2 = 4 gives n = 1/2 (not a whole number). It instead fits 4n, making it antiaromatic.

Answer: Benzene satisfies (4n+2) with n=1, so it is aromatic; cyclobutadiene has 4n π electrons and is antiaromatic.

2
Worked Example
Write the electrophile and the product for the nitration of benzene.
Solution
  1. Conc. HNO3 + conc. H2SO4 generate the nitronium ion: HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4-.
  2. NO2+ attacks the ring to give the arenium ion, which then loses H+.
  3. The product is nitrobenzene, C6H5NO2.

Answer: Electrophile = NO2+ (nitronium ion); product = nitrobenzene.

3
Worked Example
Predict the major products when toluene (methylbenzene) undergoes nitration.
Solution
  1. The methyl group is electron-donating, so it is an activating ortho/para director.
  2. The incoming NO2+ therefore enters at the 2- (ortho) and 4- (para) positions.
  3. Steric factors favour para, so the major product is the para isomer.

Answer: o-nitrotoluene and p-nitrotoluene (para predominates); the m-isomer is minor.

4
Worked Example
Where does the second substituent enter when nitrobenzene is nitrated again, and why?
Solution
  1. –NO2 is strongly electron-withdrawing, so it is a deactivating meta director.
  2. It destabilises the arenium ion most at the ortho and para positions, so attack there is disfavoured.
  3. The least-destabilised position is meta (3-position).

Answer: The new NO2 enters at the meta position, giving m-dinitrobenzene.

5
Worked Example
Outline the two-step mechanism of electrophilic aromatic substitution.
Solution
  1. Step 1: the π cloud of benzene attacks the electrophile E+, forming a resonance-stabilised carbocation, the arenium ion (sigma complex). Aromaticity is temporarily lost.
  2. Step 2: a base removes the proton from the carbon bearing E, restoring the aromatic sextet.
  3. The net result is replacement of one H by E with the ring intact.

Answer: Electrophile addition to form the arenium ion, then loss of H+ to regenerate the aromatic ring.

6
Worked Example
Name the reaction and reagents that convert benzene to acetophenone (C6H5COCH3).
Solution
  1. Acetophenone is an aryl methyl ketone, so an acyl group must be added to the ring.
  2. This is Friedel–Crafts acylation, using an acyl chloride and a Lewis-acid catalyst.
  3. C6H6 + CH3COCl ⟶[anhyd. AlCl3] C6H5COCH3 + HCl.

Answer: Friedel–Crafts acylation with CH3COCl and anhydrous AlCl3.

Key Points

  • Benzene (C6H6) is a resonance hybrid of two Kekulé forms; all six C–C bonds are equal (139 pm) and six delocalised π electrons give a large resonance energy.
  • Huckel's rule: a cyclic, planar, fully conjugated ring with (4n+2) π electrons is aromatic; benzene (6 π, n=1) is aromatic, while 4n systems are antiaromatic.
  • Arenes undergo electrophilic aromatic substitution (not addition) to keep aromaticity: E+ adds to form the arenium ion, then H+ is lost.
  • Key EAS reactions: nitration (NO2+), halogenation (X2/FeX3), sulphonation (SO3), Friedel–Crafts alkylation and acylation (AlCl3).
  • Directive influence: electron-donating groups (–CH3, –OH, –NH2) are activating o/p directors; electron-withdrawing groups (–NO2, –COOH) are deactivating m directors; benzene and PAHs are carcinogenic.
Quick Check — 4 MCQs
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Q1.According to Huckel's rule, a planar conjugated ring is aromatic if it has:
Explanation: Aromatic systems contain (4n+2) π electrons; benzene has 6 (n=1).
Q2.The reactive electrophile in the nitration of benzene is:
Explanation: Conc. HNO3/H2SO4 generate the nitronium ion NO2+.
Q3.Which group directs an incoming electrophile mainly to the meta position?
Explanation: –NO2 is electron-withdrawing and deactivating, so it is a meta director.
Q4.The intermediate formed when an electrophile adds to benzene is the:
Explanation: The electrophile adds to give a resonance-stabilised carbocation, the arenium ion, before H+ is lost.
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