A-LevelChemistryHydrocarbons9701/41 · Oct/Nov 2024 · Q614 Marks

Benzene, C6H6, reacts with chloroethane, C2H5Cl, in the presence of a suitable catalyst to form ethylbenzene, C6H5C2H5. In the presence of the catalyst, the ion C2H5⁺ is formed. This ion reacts with benzene.

(a)Complete the equation for the reaction of C2H5Cl with this catalyst to form C2H5+ as one product. C2H5Cl + ______ → C2H5+ + ______[1]
(b)Ethylbenzene reacts with more C2H5Cl, forming a mixture containing 1,2-diethylbenzene and 1,4-diethylbenzene.
(b)(i)Draw the structures of 1,2-diethylbenzene and 1,4-diethylbenzene. 1,2-diethylbenzene 1,4-diethylbenzene[1]
(b)(ii)Explain why there is very little 1,3-diethylbenzene in the product mixture.[1]
(c)1,2-diethylbenzene can be oxidised to benzene-1,2-dioic acid, C6H4(COOH)2. benzene-1,2-dioic acid COOH HOOO.
(c)(i)State the reagent and conditions used for this reaction.[1]
(c)(ii)Complete the overall equation for this reaction. An atom of oxygen from the oxidising agent is represented as [O]. All of the atoms in the two ethyl groups are fully oxidised in this reaction. + [O] → C6H4(COOH)2 + (1,2-diethylbenzene)[2]
(c)(iii)Predict the number of peaks in the carbon-13 NMR spectrum of benzene-1,2-dioic acid.[1]
(d)The proton (1H) NMR spectra of ethylbenzene, C6H5C2H5, in CDCl3 and of benzene-1,2-dioic acid, C6H4(COOH)2, in CDCl3 are shown. They have not been identified. [Figure 6.1] [Figure 6.2]
(d)(i)Explain the use of CDCl3, instead of CHCl3, as the solvent when obtaining these spectra.[1]
(d)(ii)Identify the substance shown by the spectrum in Fig. 6.1, and complete Table 6.1. substance Table 6.1 peak at δ = 1.2 peak at δ = 2.6 name of splitting pattern group responsible for peak explanation of splitting pattern[3]
(d)(iii)Identify the substance shown by the spectrum in Fig. 6.2, and complete Table 6.2. substance Table 6.2 peak at δ = 7.8 peak at δ = 13.1 group responsible for peak[1]
(d)(iv)When D₂O is used as a solvent, the spectrum obtained is different from the spectrum in Fig. 6.2. Describe this difference and explain your answer.[1]
(e)Benzene-1,2-dioic acid can be used to produce K. [benzene-1,2-dioic acid structure] heat [structure K] Suggest the name of this type of reaction.[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 2024). The mark or mark type is shown in brackets.

[1] C2H5Cl + AlCl3 → C2H5+ + A1C14- OR C2H5Cl + FeCl3 → C2H5+ + FeCl4-
[1] Structure of 1,2-diethylbenzene AND structure of 1,4-diethylbenzene
[1] alkyl / ethyl group is 2,4 directing OR ethyl group is electron donating / positive inductive effect
[1] hot alkaline KMnO4 / MnO4- (followed by acid)
[1] Correct reactants and products: C10H14 (1,2-diethylbenzene) + [O] → C6H4(COOH)2 + CO2 + H2O
[1] Correct stoichiometry including 12 [O] AND correct balancing of CO2 and H2O
[1] 4
[1] CDCl3 does not cause a peak OR does not interfere with spectrum / peaks
[1] Substance for Fig. 6.1: ethylbenzene / C6H5C2H5
[1] For peak at δ=1.2: name of splitting pattern is triplet
[1] For peak at δ=1.2: group responsible is CH3
[1] For peak at δ=1.2: explanation of splitting pattern is 2H on neighbouring C / next to CH2
[1] For peak at δ=2.6: name of splitting pattern is quartet / quadruplet
[1] For peak at δ=2.6: group responsible is CH2
[1] For peak at δ=2.6: explanation of splitting pattern is 3H on neighbouring C / next to CH3 / coupling by CH3
[1] Substance for Fig. 6.2: benzene-1,2-dioic acid / C6H4(COOH)2; Group responsible for peak at δ=7.8: H on the benzene / C6H4; Group responsible for peak at δ=13.1: COOH/carboxylic acid (all three needed for this mark)
[1] COOH peak disappears OR remove peak at 13.1 (ecf (d)(iii) if δ value used) AND as proton / hydrogen exchanges with deuterium / D
[1] dehydration / elimination / (auto)condensation