A-LevelChemistryOrganic synthesis9701/42 · Feb/Mar 2022 · Q418 Marks

Compounds F and J are shown in Fig. 4.1. [Figure 4.1 showing structures of F and J]

(a)(i)Identify the other functional groups in F and J.[2]
(a)(ii)State the number of chiral centres in a molecule of F and in a molecule of J.[1]
(b)(i)A student proposes a multi-step synthesis of F from benzene, as shown in Table 4.1. Complete Table 4.1 by providing relevant details of the reagents and conditions for steps 1 and 4, and the structure of product D. Table 4.1 step organic reactant reagent(s) and conditions organic product 1 [Benzene ring] ... [Benzene ring with CH3] 2 [Benzene ring with CH3] concentrated HNO3 and concentrated H2SO4 [Benzene ring with CH3 and NO2] 3 D hot alkaline KMnO4 then dilute H2SO4 E [Benzene ring with COOH and NO2] 4 [Benzene ring with COOH and NO2] ... F [Benzene ring with COOH and NH2][3]
(b)(ii)In a second multi-step synthesis, the student changes the order in which the reagents and conditions are used. The reaction scheme is shown in Fig. 4.2. G is the major product of this synthesis. [Figure 4.2] Draw the structure of G. Explain why G is the major product of the synthesis rather than E.[2]
(c)(i)J reacts under suitable conditions with NaOH(aq). After acidification of the reaction mixture, compounds K and L form. [Figure 4.3 showing J reacting to form K and L] Give the molecular formula of L.[1]
(c)(ii)State the two types of reaction that occur when J reacts with NaOH(aq). 1 2[2]
(d)(i)K can also be synthesised from phenol, C6H5OH. Fig. 4.4 shows several reactions of phenol. [Figure 4.4 showing reactions of phenol to M, N, K] Write an equation for the formation of M in reaction 1.[1]
(d)(ii)Draw N, the product of reaction 2.[1]
(d)(iii)Explain why phenol is a weaker acid than K.[2]
(e)Phenol and benzene both react with nitric acid, as shown in Fig. 4.5. [Figure 4.5 showing nitration of phenol and benzene] Explain why the reagents and conditions for these two reactions are different.[3]

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 2022). The mark or mark type is shown in brackets.

[2] In F: (phenyl)amine AND carboxylic acid ; In J: phenol AND ester (Any two for one mark, All four for two marks)
[1] 0 (zero) in F AND 2 (two) in J
[3] step 1 CH3C1 AND AlCl3 [1] ; step 2 D = [structure of CH3-C6H4-COOH with CH3 para to COOH] [1] ; step 4 (hot) Sn AND concentrated AND HCI [1]
[2] [structure of G: HOOC-C6H3(NO2)-CH3 with NO2 meta to COOH] [1] ; COOH group is electron-withdrawing group and 3,5-/meta- directing [1]
[1] C9H18O
[2] hydrolysis [1] ; acid-base / neutralisation [1]
[1] C6H5OH + Na → C6H5O(-)Na(+) + ½H2
[1] [structure of N: 2,4,6-tribromophenol] ; Br-C6H2(OH)-Br2
[2] • (CO)O-H bond weaker / more easy to donate H+ in K ; • owing to negative inductive / electron withdrawing effect of C=O / COOH group ; • carboxylate anion stabilised / phenoxide anion is less stabilised (All three for two marks)
[3] p-orbital on oxygen overlaps with ring / π system OR lone pair of e¯ on oxygen is delocalised into the ring [1] ; electron density in ring increases [1] ; attracts/polarises electrophile better [1]