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A-LevelChemistryEquilibria9701/42 · Feb/Mar 2020 · Q521 Marks

Gallic acid, C₇H₆O₅, is a naturally occurring aromatic molecule. [Figure 5.1]

  1. (a)Gallic acid contains the carboxylic acid and phenol functional groups. State and explain the relative acid strength of these two functional groups.[2]
  2. (b)(i)A buffer solution was prepared by dissolving 2.04g of gallic acid in 250cm³ of a solution containing 0.0600 moldm⁻³ of gallate ions, C₇H₅O₅⁻. C₇H₆O₅ ⇌ C₇H₅O₅⁻ + H⁺ Ka = 3.89 × 10⁻⁵ mol dm⁻³ at 298 K Define the term buffer solution.[2]
  3. (b)(ii)Calculate the pH of this buffer solution. pH =[3]
  4. (b)(iii)Write two equations to show how a solution containing gallic acid, C₇H₆O₅, and gallate ions, C₇H₅O₅⁻, acts as a buffer.[2]
  5. (c)(i)Construct an equation for the formation of NO₂⁺ by this method.[1]
  6. (c)(ii)Complete the mechanism and draw the intermediate of step 1. Include all relevant charges and curly arrows to show the movement of electron pairs. [Figure 5.2][2]
  7. (c)(iii)State the name of the mechanism in (c)(ii).[1]
  8. (c)(iv)State the reagents and conditions for step 4.[2]
  9. (c)(v)Suggest the structure of compound Q in the box provided. [Figure 5.3][2]
  10. (c)(vi)State the reagents and conditions for step 6.[1]
  11. (d)(i)State the number of peaks that would be observed in the ¹³C NMR spectrum of gallic acid.[1]
  12. (d)(ii)The proton NMR spectrum of gallic acid dissolved in D₂O is recorded. • Predict the number of peaks observed and any expected splitting pattern. • State the expected chemical shift range (δ) of each peak predicted.[2]

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Mark scheme

Marking points from the official mark scheme (Feb/Mar 2020). The mark or mark type is shown in brackets.

  1. [2] M1 COOH is more acidic than phenol AND because the O-H bond in acid is weaker OR carboxylate ion is more stable
  2. M2 O-H bond weakened / loses proton more easily AND by negative inductive effect of C=O / due to electronegative C=O OR carboxylate ion / anion is more stable AND due to delocalisation of minus charge by C=O/2O
  3. [2] M1 a solution which resists changes in pH / controls pH / keeps pH within a small range
  4. M2 when small amounts of H+ or OH- are added
  5. [3] M1 no. of mol of gallic acid = 2.04 ÷ 170.0 OR 0.012(0) mol
  6. M2 [H+] = Ka[HA] / [A-] = 3.89 × 10–5 × 0.012/0.250 + 0.0600 = 3.89 × 10-5 × (0.048 ÷ 0.0600) = 3.112 x 10-5
  7. M3 pH = -log10 (3.112 × 10–5) = 4.5(1) min 2sf
  8. [2] M1 C7H6O5 + OH- → C7H5O5¯ + H2O OR H+ + OH- → H2O AND C7H6O5 → C7H5O5¯ + H+
  9. M2 C7H5O5 + H+ → C7H6O5
  10. [1] HNO3 + 2H2SO4 → H3O+ + NO2+ + 2HSO4 OR HNO3 + H2SO4 → H2O + NO2+ + HSO4-
  11. [2] M1 curly arrow from ring to N of NO2+
  12. M2 correct intermediate AND curly arrow from C—H back to ring
  13. [1] electrophilic substitution
  14. [2] M1 Sn and HCl
  15. M2 heat and concentrated (dependent on metal (Fe / Sn) and acid seen for M1)
  16. [2] M1 1 × diazonium salt with ester group unchanged
  17. M2 3 × diazonium salt (to match formula)
  18. [1] warm / T ≥ 30 °C AND H2O / named aqueous acid
  19. [1] 5
  20. [2] M1 only one peak
  21. M2 singlet at δ 6.0–9.0 ppm OR M1 singlet(s) only M2 only one peak at δ 6.0–9.0 ppm

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About this question

Topic
Equilibria, A-Level Chemistry (syllabus 9701). Structured question, 21 marks.
Source
A-Level Chemistry, Feb/Mar 2020 Paper 4 Variant 2, question 5.

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