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A-LevelChemistryAnalytical techniques9701/42 · Feb/Mar 2018 · Q914 Marks

Two molecules of compound F react together under suitable conditions to form compound G. Some information about compounds F and G is given. • The mass spectrum of F has a peak due to the molecular ion at m/e = 106, and a peak at m/e = 107 with an abundance 8% of the 106 peak. • The mass spectrum of G has a peak due to the molecular ion at m/e = 212, and major peaks at m/e = 91 and m/e = 105. • Both F and G contain oxygen and are neutral compounds which are insoluble in water. • The ¹H NMR spectrum of F includes a singlet peak at δ = 10.0 due to one proton. • The ¹H NMR spectrum of G includes a singlet peak at δ = 5.1 due to two protons. • When G is heated with dilute sulfuric acid, benzoic acid, C₆H₅CO₂H, and phenylmethanol, C₆H₅CH₂OH, are produced. Use this information to answer (i)–(vi).

  1. (a)(i)Explain how the mass spectrum of F shows that a molecule of F contains seven carbon atoms. Show your working.[1]
  2. (a)(ii)Suggest the molecular formula of the fragment of G at m/e = 91.[1]
  3. (a)(iii)Suggest the molecular formulae of F and G. F G[2]
  4. (a)(iv)Suggest structures for compounds F and G. [Boxes for structures F and G][2]
  5. (a)(v)On the structures you have drawn in (iv), circle the protons responsible for the ¹H NMR peaks at δ = 10.0 in F and δ = 5.1 in G.[1]
  6. (a)(vi)State the type of reaction that G undergoes when heated with dilute sulfuric acid.[1]
  7. (b)Describe and explain the relative acidities of benzoic acid, phenylmethanol and 4-methylphenol. [Figures of benzoic acid, phenylmethanol, and 4-methylphenol][3]
  8. (c)The ester 4-methylphenyl benzoate is used in the manufacture of perfumes. [Figure of 4-methylphenyl benzoate] Suggest a two-step route for the synthesis of 4-methylphenyl benzoate from 4-methylphenol and benzoic acid. Include reagents and conditions for each step, and the structure of the intermediate compound.[3]

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Mark scheme

Marking points from the official mark scheme (Feb/Mar 2018). The mark or mark type is shown in brackets.

  1. [1] n = (100 / 1.1)(8 / 100) = 7.3 (⇒ 7 C atoms)
  2. [1] C7H7+
  3. [2] F is C7H6O G is C14H12O2
  4. [2] [Structures: F is C6H5CHO (benzaldehyde); G is C6H5COOCH2C6H5 (benzyl benzoate)]
  5. [1] [Circled protons: F - aldehyde proton (-CHO); G - methylene protons (-CH2-) between O and benzene ring.]
  6. [1] hydrolysis
  7. [3] benzoic acid > methylphenol > phenylmethanol methylphenoxide anion has delocalisation of the lone pair on oxygen over the ring benzoic acid has an (extra) electronegative oxygen or electron withdrawing C=O
  8. [3] step 1 treat benzoic acid with SOCl2 or PCl5 to make the acyl chloride formula is C6H5COCI step 2 dissolve the methylphenol in NaOH(aq) (and shake with the benzoyl chloride)

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About this question

Topic
Analytical techniques, A-Level Chemistry (syllabus 9701). Structured question, 14 marks.
Source
A-Level Chemistry, Feb/Mar 2018 Paper 4 Variant 2, question 9.

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