A-LevelChemistryChemical bonding9701/42 · Feb/Mar 2021 · Q512 Marks

Carboplatin and satraplatin are used as anticancer drugs instead of cisplatin. [Figure 5.1]

(a)(i)Describe the action of cisplatin as an anticancer drug.[2]
(a)(ii)Suggest the geometry of the platinum centre in the carboplatin complex.[1]
(a)(iii)Suggest why carboplatin does not show cis-trans isomerism.[1]
(a)(iv)Satraplatin is a neutral complex, containing the ligands CH₃CO₂⁻, C₆H₁₁NH₂, Cl⁻ and NH₃. Deduce the oxidation state of platinum in satraplatin.[1]
(b)(i)Compound M is made from 1,3-dimethylbenzene in a two-step synthesis. [Figure 5.2] Draw the structure of L.[1]
(b)(ii)Suggest reactants and conditions for each step of this synthesis. step 1 step 2[2]
(b)(iii)Write an equation for step 2.[1]
(b)(iv)A student investigates a possible synthesis of M directly from benzene using COCl₂ in the presence of an AlCl₃ catalyst. Benzene initially reacts with COCl₂ as shown. reaction 1 COCl₂ + AlCl₃ → AlCl₄⁻ + Cl−C=O⁺ reaction 2 C₆H₆ + Cl−C=O⁺ → C₆H₅COCl + H⁺ Reaction 2 is the electrophilic substitution of Cl−C=O⁺ for H⁺ in benzene. Suggest a mechanism for reaction 2. [Figure 5.3][3]

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Mark scheme

Marking points from the official mark scheme (Feb/Mar 2021). The mark or mark type is shown in brackets.

[2] M1: Cisplatin binds/bonds to DNA (specifically to guanine bases).
[2] M2: This binding causes distortion of the DNA helix, preventing DNA replication / cell division / mitosis / transcription (and formation of mRNA) in cancer cells.
[1] square planar
[1] The bidentate carboxylate ligand bridges two coordination sites, forcing a cis arrangement OR the two coordinating oxygen atoms are part of the same ligand, preventing them from being trans to each other.
[1] +4
[1] Structure of isophthalic acid (benzene with two meta -COOH groups).
[2] M1 (step 1): Reactants: acidified / alkaline KMnO₄(aq). Conditions: heat/reflux. (Followed by acidification if alkaline KMnO₄ is used).
[2] M2 (step 2): Reactants: PCl₅ OR SOCl₂. Conditions: (heat with) PCl₃ (if PCl₃ is used).
[1] C₈H₆O₄ + 2PCl₅ → C₈H₄O₂Cl₂ + 2POCl₃ + 2HCl (OR corresponding equation with SOCl₂ or PCl₃)
[3] M1: Curly arrow from the delocalised π-electron ring of benzene attacking the carbon atom of the electrophile (Cl−C=O⁺).
[3] M2: Formation of the correct carbocation intermediate (sigma complex) with positive charge delocalised around the ring, and the new bond to the electrophile formed.
[3] M3: Curly arrow from a C-H bond within the ring attacking the positive charge on the ring, leading to the loss of H⁺ and reformation of the aromatic ring.

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