A-LevelChemistryHydrocarbons9701/42 · May/June 2020 · Q419 Marks

When benzene undergoes nitration a nitro group substitutes at a carbon atom. State the shape (geometry) around the substituted carbon atom • in the benzene molecule, • in the intermediate complex, • in the nitrobenzene product.

(a)(i)When benzene undergoes nitration a nitro group substitutes at a carbon atom. State the shape (geometry) around the substituted carbon atom • in the benzene molecule, • in the intermediate complex, • in the nitrobenzene product.[2]
(a)(ii)Naphthalene, C₁₀H₈, is an arene hydrocarbon. naphthalene [Figure for naphthalene] When naphthalene undergoes nitration, a mixture of two organic compounds is formed. Each compound contains one nitro group. Suggest the structures of these compounds. structure 1 structure 2[1]
(b)Naphthalene can be oxidised under certain conditions to phthalic anhydride, C₈H₄O₃, carbon dioxide and water. Construct an equation for this reaction. Use [O] to represent an atom of oxygen from the oxidising agent.[1]
(c)The indicator, phenolphthalein, can be synthesised from phthalic anhydride and phenol under certain conditions. phenol phthalic anhydride phenolphthalein [Reaction diagram] Deduce the type of reaction shown by this equation.[1]
(d)(i)Name the functional groups, in addition to the benzene ring, present in a phenolphthalein molecule.[1]
(d)(ii)Phenolphthalein reacts separately with the two reagents shown in the table. Complete the table by: • drawing the structures of the organic products formed (part of the structure has been given for you) • stating the types of reaction. reagent an excess of hot NaOH(aq) organic product structure [Partial structure for NaOH product] an excess of Br₂(aq) [Partial structure for Br₂ product] type of reaction[4]
(e)Phenolphthalein is an indicator and is represented by the formula HIn. Phenolphthalein, HIn, is a weak acid. HIn ⇌ H⁺ + In⁻ The Ka value for phenolphthalein is 5.0 × 10⁻¹⁰ mol dm⁻³ at 298 K. This indicator changes colour at a pH of approximately 8.8. Calculate the ratio [In⁻]/[HIn] at pH 8.8. ratio [In⁻]/[HIn] = [2][2]
(f)(i)On the structure of methyl orange, circle the bond or bonds that make this compound a dye. The colour of this indicator changes between pH 3.2 and pH 4.4.[1]
(f)(ii)Suggest the structure of methyl orange at pH 3.0. Assume the -SO₃⁻Na⁺ group is unreactive.[1]
(g)(i)Deduce the identities of compounds R and S and draw their structures in the boxes.[2]
(g)(ii)Suggest reagents and conditions for step 1 and step 2. step 1 step 2[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 (May/June 2020). The mark or mark type is shown in brackets.

[2] • trigonal planar • tetrahedral. • trigonal planar Award one mark for two correct statements, award two marks for three correct statements
[1] Both structures required (1-nitronaphthalene and 2-nitronaphthalene)
[1] C₁₀H₈ + 9[O] → C₈H₄O₃ + 2CO₂ + 2H₂O
[1] condensation/ addition-elimination
[1] phenol AND ester
[4] M1 correct hydrolysis product of ester M2 (di)phenoxide salt M3 bromination of both phenol rings at position 2 or/and 6 M4 hydrolysis AND (electrophilic) substitution
[2] M1 [H⁺] = 10⁻⁸˙⁸ = 1.585 x10⁻⁹ M2 [In⁻]/[HIn] = 5.0 x 10⁻¹⁰/1.585 x10⁻⁹ = 0.315
[1] bond circled between the two Ns, or N=N or –N=N-
[1] (Structure for protonated methyl orange: (H₃C)₂N - N⁺H=N - C₆H₄SO₃⁻Na⁺)
[2] R: H₂N-C₆H₄-SO₃⁻Na⁺ (4-aminobenzenesulfonate) S: ⁺N≡N-C₆H₄-SO₃⁻Na⁺ (4-diazoniumbenzenesulfonate) Award one mark for each correct structure
[3] M1 step 1 Sn and HCI conc. and heat M2 step 2 NaNO₂ and HCl and 0-10 °C