Skip to main content
A-LevelChemistryHydrocarbons9701/42 · Feb/Mar 2025 · Q519 Marks

Cumene is an aromatic hydrocarbon used in the synthesis of other useful chemicals.

  1. (a)Complete Table 5.1 to show the number of sp² and sp³ hybridised carbon atoms that are present in a molecule of cumene. [Figure 5.1] Table 5.1 type of hybridisation sp2 sp3 number of carbon atoms[1]
  2. (b)(i)Cumene can be synthesised via a Friedel-Crafts alkylation reaction, as shown in Fig. 5.2. [Figure 5.2] Name the mechanism involved in the Friedel-Crafts alkylation shown in Fig. 5.2.[1]
  3. (b)(ii)The first step of the reaction forms the (CH3)2CH+ cation. Identify a suitable reagent for the formation of this cation from 2-bromopropane, (CH3)2CHBr.[1]
  4. (b)(iii)Complete Fig. 5.3 to show the mechanism for the reaction of benzene with the (CH3)2CH+ cation. Include all relevant curly arrows and charges. [Figure 5.3] (CH3)2CH+ intermediate +[3]
  5. (b)(iv)The Friedel–Crafts alkylation of benzene by 1-bromopropane, CH3CH2CH2Br, also produces cumene as the major product. The CH3CH2CH2+ cation formed in the first step quickly rearranges to form the (CH3)2CH+ cation. Suggest why this is the case. Explain your answer.[1]
  6. (c)(i)Cumene oxidises in air to form phenol, C6H5OH, and propanone, CH3COCH3. reaction 1 C6H5CH(CH3)2 + O2 → C6H5OH + CH3COCH3 ΔH° = −371 kJ mol-1 Table 5.2 gives some relevant standard entropies for reaction 1. Table 5.2 compound C6H5CH(CH3)2 O2 C6H5OH CH3COCH3 standard entropy, S°/JK-1 mol-1 278 205 146 200 Calculate the standard entropy change, ΔS°, of reaction 1. ΔS° = ...................................... JK-1 mol-1[1]
  7. (c)(ii)Show that reaction 1 is feasible at 25°C.[2]
  8. (d)(i)Fig. 5.4 shows two reactions of phenol. [Figure 5.4] State the conditions for the bromination of phenol in reaction 2. Explain why these conditions are different from those for the bromination of benzene. conditions ........................................................................................................................................................... explanation ...........................................................................................................................................................[4]
  9. (d)(ii)Draw the structure of organic compound Q in Fig. 5.4.[1]
  10. (d)(iii)Identify organic reagent R.[1]
  11. (d)(iv)Name the functional group that is formed in reaction 3.[1]
  12. (d)(v)The reaction of phenol with HNO3 produces a mixture of isomers with molecular formula C6H5NO3. Identify the two isomers that are produced in the largest quantities. Explain your answer. isomer 1 ........................................................................................................................................................... isomer 2 ........................................................................................................................................................... explanation .......................................................................................................................................................................[2]

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 (Feb/Mar 2025). The mark or mark type is shown in brackets.

  1. [1] sp2 = 6 / six sp³ = 3 / three
  2. [1] electrophilic substitution
  3. [1] AlBr3
  4. [3] M1: curly arrow from inside the hexagon ring AND towards positively charged carbon atom (in the electrophile (CH3)2CH+)
  5. M2: structure of the intermediate
  6. M3: curly arrow from C–H bond into the ring AND formation / loss of H+
  7. [1] 2° carbocation / (CH3)2CH+ AND is more stable due to greater positive inductive effect (of alkyl groups)
  8. [1] ΔS = 146 + 200 – 205 – 278 = –137 (J K-1 mol-1)
  9. [2] ΔG = ΔH – TΔS AND T = 298 / 298.15 used
  10. ΔG = −371 – (25+273)×(-0.137) = −330 (kJ mol-1) [so feasible] ecf M1
  11. [4] M1: conditions for phenol = aqueous / aq, bromine water, no AlBr3 needed
  12. M2: lone pair (of e¯)/p-orbital on oxygen AND overlap / delocalised with ring / π system
  13. M3: electron density in ring increases / ring becomes more electron rich (and attracts electrophiles)
  14. M4: (phenol is more reactive so) polarises Br2 / bromine / electrophiles more
  15. [1] [Structure of 2,4,6-tribromophenol]
  16. [1] (C6H5N2+)Cl- / benzene diazonium chloride
  17. [1] azo
  18. [2] structure / name of 2-nitrophenol and 4-nitrophenol
  19. -OH / hydroxyl group is electron donating group AND 2,4-directing / ortho / para directing

Write your answer and get it marked

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.

Sign up free →

About this question

Topic
Hydrocarbons, A-Level Chemistry (syllabus 9701). Structured question, 19 marks.
Source
A-Level Chemistry, Feb/Mar 2025 Paper 4 Variant 2, question 5.

Questions and mark schemes are from past examination papers and remain the copyright of the exam board. How Oracle Prep builds its content

Related Chemistry questions

© 2026 Oracle Prep