Benzene worksheet answers for A Level chemistry (free PDF download)

Problem 1 Describing benzene

1. C₆H₆.
2. 120°.
3. Six π electrons.
4. Planar (flat) hexagon.

Problem 2 Bond lengths

1. Benzene’s C–C bonds are 0.140 nm, halfway between the single and double bond lengths.
2. All six bonds are equal and intermediate in length, so the ring has delocalised electrons, not separate single and double bonds.
3. The Kekulé model has three short C=C bonds and three longer C–C bonds, so the lengths would alternate.

Problem 3 Naming arenes

1. Nitrobenzene.
2. Bromobenzene.
3. Methylbenzene.
4. Phenylamine.

Problem 4 Reagents for substitution

1. Concentrated HNO₃ and concentrated H₂SO₄ (catalyst), at about 50 °C.
2. Br₂ with FeBr₃ (or Fe) as a halogen carrier at room temperature.
3. CH₃COCl with anhydrous AlCl₃ as catalyst (Friedel–Crafts acylation).

Problem 5 Why electrophiles?

1. An electrophile is an electron-pair acceptor.
2. The delocalised π electrons give a region of high electron density above and below the ring, which attracts positive or electron-deficient species.

Problem 6 Stabilisation energy

1. 3 × (−120) = −360 kJ/mol.
2. −208 − (−360) = 152 kJ/mol.
3. It shows that the delocalised ring is more stable than three separate C=C bonds.

Problem 7 Bromine water test

1. Electrophilic addition to the C=C bond.
2. Cyclohexene has a localised π bond with high electron density that polarises bromine. Benzene’s electrons are spread over the ring and held more strongly, so it cannot polarise bromine enough to react.
3. Add a halogen carrier such as FeBr₃, or iron, with liquid bromine.

Problem 8 Making the nitronium ion

1. HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻.
2. Sulfuric acid is a stronger acid and protonates nitric acid, which then loses water to form NO₂⁺; it is regenerated at the end.
3. C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O.

Problem 9 Describing the mechanism

1. The positive charge is spread (delocalised) over the five carbon atoms that are not bonded to both H and Br.
2. A pair of π electrons from the ring forms a bond to the electrophile Br⁺, so the ring loses its full delocalisation.
3. The C–H bond electrons return to the ring while H⁺ leaves, restoring the delocalised π system.
4. Bromine alone is not polarised enough; FeBr₃ accepts a Br⁻ to create a stronger electrophile, Br⁺ (or polarised Br₂).

Problem 10 Halogen carrier

1. Br₂ + FeBr₃ → Br⁺ + FeBr₄⁻.
2. C₆H₆ + Br₂ → C₆H₅Br + HBr.
3. FeBr₃ is regenerated when FeBr₄⁻ gives up Br⁻ to H⁺, forming HBr and FeBr₃, so it is not used up.

Problem 11 Yield of nitrobenzene

1. n = 15.6 ÷ 78.0 = 0.200 mol.
2. 0.200 × 123.0 = 24.6 g.
3. 18.5 ÷ 24.6 × 100 = 75.2 %.

Problem 12 Friedel–Crafts acylation

1. C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl.
2. Anhydrous AlCl₃; water would react with AlCl₃ and with CH₃COCl, destroying the catalyst and the reagent.
3. The acylium ion, CH₃CO⁺.

Problem 13 Kekulé and isomers

1. In the Kekulé structure the two carbons carrying Br would be joined by C=C in one form and C–C in the other, so two different isomers should exist. Only one has ever been found.
2. In the delocalised model all C–C bonds are the same, so there is just one 1,2-dibromobenzene.

Problem 14 Reading a table of reactivity

1. Benzene does not react with bromine water or KMnO₄, so it has no localised C=C bonds.
2. The reaction with Br₂ and FeBr₃ gives HBr, which means a hydrogen atom is replaced: electrophilic substitution.

Problem 15 Two-step synthesis

1. Concentrated HNO₃ with concentrated H₂SO₄ at not above 55 °C (about 50 °C).
2. C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O.
3. At higher temperature further substitution gives dinitrobenzene, lowering the yield of mononitrobenzene.

Problem 16 Atom economy

1. Atom economy = 123.0 ÷ (78.0 + 63.0) × 100 = 123.0 ÷ 141.0 × 100 = 87.2 %.
2. Some benzene forms dinitrobenzene, some product is lost in washing and separation, and the reaction may not go to completion.

Problem 17 Dinitration

1. C₆H₅NO₂ + HNO₃ → C₆H₄(NO₂)₂ + H₂O.
2. 0.100 × 168.0 = 16.8 g.
3. At higher temperature substitution happens more than once, so the product is not pure nitrobenzene.

Problem 18 Why substitution wins

1. Loss of H⁺ regenerates the delocalised ring and gives a substituted benzene; addition would leave a non-aromatic product.
2. Addition would keep the energy cost of losing 152 kJ/mol of stabilisation, while substitution recovers it.
3. Aromatic compounds contain a planar ring with delocalised π electrons (6 in benzene).

Problem 19 Hydrogenation data

1. 2 × (−120) = −240 kJ/mol.
2. The diene is only 8 kJ/mol more stable than predicted (−232 compared with −240), whereas benzene is 152 kJ/mol more stable.
3. The unusual gain shows that benzene has exceptional delocalisation, which is the basis of its aromatic stability.

Problem 20 Is benzene unreactive?

Benzene is not truly unreactive, but it is less reactive than an alkene and prefers a different type of reaction. Its delocalised ring is 152 kJ/mol more stable than the Kekulé structure, so it does not decolourise bromine water and needs a strong electrophile such as NO₂⁺ or Br⁺ produced with a catalyst. The mechanism begins by sacrificing the delocalisation, which gives an activation barrier, but it ends by losing H⁺ to restore the ring, giving substitution. Therefore: benzene is resistant to addition but reacts readily by electrophilic substitution under suitable conditions.