Topic 18 · page 5 of 10

Reactions of benzene at a glance

Benzene burns with a smoky flame, and reacts by substitution with five reagent combinations. Know the reagents, conditions and organic product of each, then learn how they work.

13 min read18.4

Read first: Why benzene resists bromination

The reactions in this topic 18.4

The specification limits the reactions of benzene to one combustion reaction and four types of substitution. Page 5 is the map. Pages 6 to 8 explain each reaction.

A hub diagram. Benzene in the centre. Burning in oxygen gives carbon dioxide, water and soot. Bromine with FeBr3 gives bromobenzene. Concentrated nitric and sulfuric acids at 50 degrees give nitrobenzene. Fuming sulfuric acid gives benzenesulfonic acid. Chloroethane with AlCl3 gives ethylbenzene. Ethanoyl chloride with AlCl3 gives phenylethanone.
Every substitution replaces one H atom on the ring with a new group.
Reaction Reagents Conditions Organic product
combustion oxygen in air burn CO₂ and H₂O, with soot
bromination Br₂ and an FeBr₃ catalyst room temperature, dry bromobenzene, C₆H₅Br
nitration concentrated HNO₃ and concentrated H₂SO₄ about 50 °C nitrobenzene, C₆H₅NO₂
sulfonation fuming sulfuric acid warm benzenesulfonic acid, C₆H₅SO₃H
Friedel-Crafts alkylation halogenoalkane, AlCl₃ catalyst heat under reflux, dry alkylbenzene, such as ethylbenzene
Friedel-Crafts acylation acyl chloride, AlCl₃ catalyst heat under reflux, dry aromatic ketone, such as phenylethanone

In every substitution one hydrogen atom on the ring is replaced and the ring survives. The products of halogenation, nitration and the two Friedel-Crafts reactions all form by one mechanism, covered next.

Combustion and the smoky flame 18.4

Benzene burns in air. With plenty of oxygen, combustion is complete:

2 CX6HX6(l)+15 OX2(g)→12 COX2(g)+6 HX2O(l) \ce{2C6H6(l) + 15O2(g) -> 12CO2(g) + 6H2O(l)}

Benzene is a hydrocarbon with a high carbon to hydrogen ratio, one carbon for each hydrogen. In an ordinary flame there is not enough oxygen for all that carbon to burn completely, so combustion is incomplete. Unburnt carbon is released as soot, and the glowing particles make the flame yellow and smoky.

2 CX6HX6(l)+3 OX2(g)→12 C(s)+6 HX2O(l) \ce{2C6H6(l) + 3O2(g) -> 12C(s) + 6H2O(l)}

Carbon monoxide can also form in incomplete combustion. It is toxic.

The proportion of carbon by mass shows why benzene is smoky:

Hydrocarbon Formula Carbon by mass
hexane C₆H₁₄ 72.0 ÷ 86.0 = 83.7%
cyclohexene C₆H₁₀ 72.0 ÷ 82.0 = 87.8%
benzene C₆H₆ 72.0 ÷ 78.0 = 92.3%

Worked exampleComplete combustion of benzene

What volume of oxygen at RTP, and what mass of carbon dioxide, results from the complete combustion of 1.00 g of benzene?

  1. n(C₆H₆) = 1.00 ÷ 78.0 = 0.01282 mol
  2. From the equation C₆H₆ + 7½O₂ → 6CO₂ + 3H₂O, n(O₂) = 7.5 × 0.01282 = 0.09615 mol
  3. V(O₂) = 0.09615 × 24.0 = 2.31 dm³
  4. n(CO₂) = 6 × 0.01282 = 0.07692 mol, so m(CO₂) = 0.07692 × 44.0 = 3.38 g

Benzene is carcinogenic. In school, a demonstration of the smoky flame uses a small sample of an arene such as methylbenzene, in a fume cupboard.

Check yourself

  1. Write the equation for the complete combustion of one mole of benzene.
  2. Explain why benzene burns with a smoky flame, but hexane burns with a cleaner one.
  3. 0.390 g of benzene burns completely in excess oxygen. Calculate the volume of oxygen used, at RTP, and the mass of carbon dioxide formed.
  4. Give the reagents and the organic product for (a) nitration, (b) sulfonation, (c) bromination of benzene.
  5. What type of reaction do all of the reactions in the table (apart from combustion) have in common?

Answers cover these until you have tried the questions

  1. One mole needs 7½ mol of oxygen, so it is neater to write two moles: 2 CX6HX6(l)+15 OX2(g)→12 COX2(g)+6 HX2O(l)\ce{2C6H6(l) + 15O2(g) -> 12CO2(g) + 6H2O(l)}
  2. Benzene's carbon to hydrogen ratio is 1 : 1 (92.3% carbon by mass), much higher than hexane's (83.7%). There is not enough oxygen for all of the carbon, so combustion is incomplete and unburnt carbon (soot) is released, giving a smoky flame.
  3. n(C₆H₆) = 0.390 ÷ 78.0 = 0.00500 mol. n(O₂) = 7.5 × 0.00500 = 0.0375 mol, so V = 0.0375 × 24.0 = 0.900 dm³. n(CO₂) = 6 × 0.00500 = 0.0300 mol, so m = 0.0300 × 44.0 = 1.32 g.
  4. (a) Concentrated HNO₃ and concentrated H₂SO₄, giving nitrobenzene. (b) Fuming sulfuric acid, giving benzenesulfonic acid. (c) Br₂ with an FeBr₃ catalyst, giving bromobenzene.
  5. Electrophilic substitution (one H on the ring is replaced). The mechanism of sulfonation is not required.