A-Level Chemistry October/November 2024 Q6(c)(ii): Complete the overall equation for this reaction. An atom of oxygen from the oxidising a…

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  Step 1: Identify Reactant and Products

  The starting material is a benzene ring with two ethyl groups, which can be written as C6H4(C2H5)2 or C10H14.

  The question states that "All of the atoms in the two ethyl groups are fully oxidised".

  • The carbon atom attached to the ring in each ethyl group is oxidised to a carboxylic acid group, -COOH. This gives the product C6H4(COOH)2.
  • The terminal CH3 group in each ethyl group is also fully oxidised. The most oxidised form of carbon is carbon dioxide, CO2. Since there are two ethyl groups, this produces 2CO2.
  • All hydrogen atoms from the original molecule must be accounted for. The product of oxidising hydrogen is water, H2O.

  This gives the unbalanced equation with all species identified: C6H4(C2H5)2 + [O] → C6H4(COOH)2 + CO2 + H2O

  Step 2: Balance the Equation

  We balance the equation by conserving atoms, element by element.

  1. Carbon (C):

     • Reactant C6H4(C2H5)2: 6 (in ring) + 2 × 2 (in ethyls) = 10 C atoms.
     • Products: C6H4(COOH)2 has 6 (in ring) + 2 × 1 (in COOH) = 8 C atoms. To balance the 10 C atoms from the reactant, we need 2 more C atoms on the product side. Therefore, we need 2CO2. C6H4(C2H5)2 + [O] → C6H4(COOH)2 + 2CO2 + H2O

  2. Hydrogen (H):

     • Reactant C6H4(C2H5)2: 4 (on ring) + 2 × 5 (in ethyls) = 14 H atoms.
     • Products: C6H4(COOH)2 has 4 (on ring) + 2 × 1 (in COOH) = 6 H atoms. The remaining 14 - 6 = 8 H atoms must form water. This requires 4H2O. C6H4(C2H5)2 + [O] → C6H4(COOH)2 + 2CO2 + 4H2O

  3. Oxygen (O):

     • Now we balance the oxygen atoms using [O].
     • Products: C6H4(COOH)2 has 2 × 2 = 4 O atoms. 2CO2 has 2 × 2 = 4 O atoms. 4H2O has 4 × 1 = 4 O atoms.
     • Total O atoms in products = 4 + 4 + 4 = 12 O atoms.
     • Therefore, we need 12[O] on the reactant side.

  Final Answer

  C6H4(C2H5)2 + 12[O] → C6H4(COOH)2 + 2CO2 + 4H2O

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  How the marks are awarded

  • B1 — Correctly identifying all species in the reaction. The reactant is C6H4(C2H5)2 (or C10H14), and the products from the complete oxidation of the side-chains are C6H4(COOH)2, CO2, and H2O.
  • B1 — Correctly balancing the entire equation, specifically by determining the stoichiometric coefficients of 12 for [O], 2 for CO2, and 4 for H2O.

  Common mistakes

  • Omitting CO2 as a product, by incorrectly assuming only the carbon attached to the ring is oxidised.
  • Incorrectly calculating the number of H atoms in the reactant or products, leading to an incorrect coefficient for H2O.
  • Miscalculating the total number of oxygen atoms required on the product side, leading to an incorrect coefficient for [O].
  • Only partially balancing the equation, for example, balancing C and H but not O.

  Examiner tip: When balancing oxidation reactions, systematically account for the fate of every atom (C, H, etc.) from the reactant to determine all products before attempting to balance the oxygen atoms.

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