Examiners' Report - Paper 3 Chemistry

• • Students did not recognise that enthalpy change could not be measured directly.
  • In this case, the enthalpy change was for the formation of a hydrated salt from an anhydrous one –
  • with the expected answer:
  • it would be impossible to prevent some salt dissolving during the addition of water to the anhydrous salt.
  • An alternative answer, related to the difficulty in measuring the temperature of a solid, was also allowed.

  
• • Some students also seemed to believe that calorimetry is only possible if something is being burned. (NOT true - misconception).
• • This question was marked using a ‘levels of response’ mark scheme.
  • The key to success was for students to concentrate first on the inclusion of as much correct chemistry as possible to ensure access to Level 3 (worth 5 or 6 marks).
  • Within a level, the mark awarded depended on the clarity and coherence of an answer, together with a clear, logical progression through the description.

  
• • Appropriate apparatus and quantities should have been mentioned as necessary.
  • For example, rather than writing ‘add water to a container’, a good start to the answer would be to write
  • ‘A measuring cylinder was used to measure 50 cm3 of water into a polystyrene cup’
• ELECTROPHILIC ADDITION WITH CONCENTRATED SULFURIC ACID.

  • The most common omissions were the
  • curly arrow to show breaking of the H-O bond in H2SO4
  • the negative charge on the HSO4 − ion.

  
• Explain why the product shown in your answer to Question 2.2 is the major product. [2 marks]

  • A simple statement that a secondary carbocation is more stable than a primary was sufficient for the first mark (although many students mistakenly referred to the greater stability of the final product instead).
  • For the second mark, the explanation needed to make clear that the positive inductive effect was GREATER in the secondary carbocation due to it having MORE alkyl groups attached to the C+.
• • Butanal was a common wrong answer - because butan-2-ol is a secondary alcohol, butanone (when it is oxidised) is the correct answer.
  • for the by-product and the fact that sulfuric acid acts as an oxidising agent was not well recognised.

  
• • This question, and others in question 3, illustrate the importance of students paying attention to WHY! certain techniques and processes are used during practicals.
  • The role of anti-bumping granules was not as well understood as expected = it allows smaller BUBBLES to form / prevents the formation of (very) large BUBBLES

  
• • Nearly 60% of students gained this mark but a lack of specific clarity cost many dearly
  • They did not appear to recall, from the stem of the question, that the reaction was between NaOH and ethyl benzoate.
  • This recall should have enabled students to make the SPECIFIC STATEMENT that the excess of sodium hydroxide was to
  • ensure that all the ethyl benzoate would react.

  
• • Answers here often incorrectly referred to the ‘control’ of the temperature rather than recognising the risk of flammability of organic reagents.
  • Clarity is again key, because any suggestion that NaOH is flammable is incorrect

  
• State why reflux is used in this hydrolysis - 2017

  • As was also evident in question 02.6, some students thought that reflux is a separation method.
  • A specific answer was again the key here, with a need to mention the fact that reactant/organic VAPOURS are returned to the reaction mixture.
• Suggest why sodium benzoate is soluble in cold water but benzoic acid is insoluble in cold water. [2 marks]

  • This proved to be the trickiest question on the paper, with only 1% of students earning both marks.
  • Most could state that sodium benzoate is ionic – although there were also many incorrect references to it being ‘polar’ or ‘a molecule’.
• Suggest why sodium benzoate is soluble in cold water but benzoic acid is insoluble in cold water. [2 marks]

  • However, a proper description of why benzoic acid is insoluble was beyond most students.
  • Many stated that it is non-polar, while others suggested that its ability to hydrogen-bond with itself is what prevents it dissolving.
  • Very few were able to clearly explain that, despite the polarity (and hence the ability to form hydrogen bonds with water),
  • the large non-polar benzene ring prevented dissolving.
• There was often confusion in the order of the steps, which prevented answers gaining full marks.

  • The steps most often missing were the need for hot filtration after dissolving the impure solid in hot solvent
  • and the final wash and dry after Buchner filtration

  
• • It seemed that many were thrown by the realisation that the answer was over 100%, with many students responding by doing the percentage calculation the wrong way round, because they seemed to feel that the answer must be less than 100%.
  • Relative molecular masses were often calculated incorrectly despite that for ethyl benzoate being given in the stem.

  
• Calculate concentration of HX in original solution

  • A significant number of students attempted to answer this question using the expression for Ka, instead of recognising that this is a titration and that
  • the end-point allows the amount of NaOH to be calculated, and hence the amount and concentration of the acid.
  • Perhaps students did not recognise that that the end point of the titration could be obtained easily from Figure 1.

  
• • A mark was often lost as a result of students assuming that half-equivalence was at 10 or 15 cm3 and so plotting incorrectly. (The equivalence at Figure 1 is 24cm3!)
  • Many also failed to remember the need to show an initial steep section before buffering ‘levels off’ the curve.