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A-LevelChemistryHydrocarbons9701/43 · Oct/Nov 2023 · Q814 Marks

Capsaicin is found in chilli peppers. [Structure of capsaicin] You should assume the CH3O group is unreactive in the reactions involved in this question.

  1. (a)Name all the functional groups in capsaicin in addition to the CH3O group.[1]
  2. (b)Complete the equation for the reaction of capsaicin with an excess of Br2(aq) in the dark. Draw the structure of the organic product in the labelled box. [Equation with capsaicin + .........Br2 → organic product box + .........][3]
  3. (c)Capsaicin is heated with an excess of hydrogen gas in the presence of platinum metal. The six-membered ring reacts in the same way as benzene under these conditions. Draw the structure of the organic product formed.[2]
  4. (d)(i)When capsaicin is treated with reagent J under suitable conditions one of the products is methylpropanoic acid, CH3CH(CH3)COOH. Identify reagent J and any necessary conditions.[1]
  5. (d)(ii)There are three different peaks in the proton (¹H) NMR spectrum of CH3CH(CH3)COOH in CDCl3. [Table 8.1 with chemical shift ranges] Use Table 8.1 to complete Table 8.2 and state: • the typical proton (¹H) chemical shift values (δ) for the protons • the splitting pattern (singlet, doublet, triplet, quartet or multiplet) shown by each peak • the explanation for the splitting patterns of the CH3 protons and the CH proton. [Table 8.2 with environment, δ/ppm, splitting pattern, explanation for splitting pattern columns, and rows for CH3, CH, COOH][3]
  6. (e)(i)Capsaicin is heated with an excess of hot aqueous NaOH. Draw the structures of the two organic products H and K. [Boxes for H and K with molecular formulas given][2]
  7. (e)(ii)Name the two types of reaction occurring in (e)(i).[1]
  8. (f)Draw the structure of the organic product L formed when capsaicin is treated with LiAlH4 in dry ether. [Box for L][1]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

  1. [1] phenol, amide AND alkene / C=C all three needed
  2. [1] M1: HBr u/c
  3. [1] M2: structure of the organic product (benzene ring has Br added, alkene has Br added)
  4. [1] M3: correct balanced equation (capsaicin + 2 Br2 -> organic product + 2HBr) ecf
  5. [1] M1: ring reduced (benzene ring reduced to cyclohexane)
  6. [1] M2: alkene reduced (C=C becomes C-C)
  7. [1] hot AND concentrated acidified AND MnO4¯ / KMnO4 all
  8. [1] for δ values (CH3: 0.9-1.7, CH: 2.2-3.0, COOH: 9.0-13.0)
  9. [1] for splitting (CH3: doublet, CH: multiplet/heptet/septet, COOH: singlet)
  10. [1] for explanations (CH3: 1H on neighbouring C / next to CH / one vicinal proton; CH: 6H on neighbouring C / next to 2 x (CH3) / six vicinal protons)
  11. [1] H: Benzene ring with -CH2CH2NH2 and -OCH3, with -O-Na+ instead of -OH (sodium 4-(2-aminoethyl)phenoxide-3-methoxide)
  12. [1] K: CH3CH(CH3)COO-Na+ and the rest of the amide alkyl chain ending in C=C (sodium 8-methylnon-6-enoate)
  13. [1] hydrolysis AND neutralisation / acid-base
  14. [1] L: Benzene ring with -CH2CH2NHCH2-(CH2)6-CH(CH3)-CH2-OH (amide reduced to amine, alkene reduced to alkane, ketone reduced to alcohol, phenol group remains)

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About this question

Topic
Hydrocarbons, A-Level Chemistry (syllabus 9701). Structured question, 14 marks.
Source
A-Level Chemistry, Oct/Nov 2023 Paper 4 Variant 3, question 8.

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