A-LevelChemistryGroup 179701/22 · Feb/Mar 2022 · Q320 Marks

The hydrogen halides HCl, HBr and HI are all colourless gases at room temperature.

(a)The hydrogen halides can be formed by reacting the halogens with hydrogen. Describe and explain the relative reactivity of the halogens down the group when they react with hydrogen to form HCl, HBr and HI.[2]
(b)(i)HCl is a product of several different reactions. Some of these are shown in [Figure 3.1]. Write an equation for reaction 1.[1]
(b)(ii)In reaction 2, NaCl reacts with concentrated H₂SO₄ to form HCl and NaHSO₄. When NaBr reacts with concentrated H₂SO₄, the products include Br₂ and SO₂. Identify the type(s) of reaction that occur in each case by completing Table 3.1. Explain the difference in these reactions. Table 3.1 reactants type(s) of reaction NaCl and concentrated H₂SO₄ NaBr and concentrated H₂SO₄ explanation[3]
(c)When heated with a Bunsen burner, HCl does not decompose, whereas HI forms H₂ and I₂. Explain the difference in the effect of heating on HCl and HI.[1]
(d)(i)The hydrogen halides dissolve in water to form strong Brønsted-Lowry acids. The concentration of a strong acid can be determined by titration. State what is meant by strong Brønsted-Lowry acid.[2]
(d)(ii)On [Figure 3.2], sketch the pH titration curves produced when: • 0.1 mol dm⁻³ NaOH(aq) is added to 25 cm³ of 0.1 mol dm⁻³ HBr(aq), to excess • 0.1 mol dm⁻³ NH₃(aq) is added to 25 cm³ of 0.1 mol dm⁻³ HBr(aq), to excess.[3]
(e)(i)HBr reacts with propene to form two bromoalkanes, CH₃CH₂CH₂Br and (CH₃)₂CHBr. Complete the diagram to show the mechanism of the reaction of HBr and propene to form the major organic product. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate. Draw the structures of the intermediate and the major organic product. [Diagram fragment with propene and H-Br][4]
(e)(ii)Explain why the two bromoalkanes are not produced in equal amounts by this reaction.[2]
(e)(iii)The reaction of CH₃CH₂CH₂Br and NaOH is different depending on whether water or ethanol is used as a solvent. Complete Table 3.2 to identify the organic and inorganic products of the reaction of CH₃CH₂CH₂Br and NaOH in each solvent. Table 3.2 solvent organic product(s) inorganic product(s) water ethanol[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 2022). The mark or mark type is shown in brackets.

[1] M1: reaction less vigorous (down the group)
[1] M2: Any two of the following for one mark: electronegativity decreases / less attractive to e- addition / weaker oxidising agent / greater nuclear charge outweighing increased shielding (ENC argument)
[1] SiCl₄ + 2H₂O → SiO₂ + 4HCl
[2] M1: All three correct: row 1: acid-base; row 2: acid-base, redox
[1] M2: explanation: H₂SO₄ is strong enough to oxidise / is an oxidising agent with NaBr / HBr / bromide
[1] H-Cl bond is stronger than H—I / BDE decreases down the group
[1] proton / H⁺ donor
[1] fully dissociates (in aqueous solution / water / solvent)
[1] M1: correct sigmoid shape with vertical section at 25 cm³ for both
[1] M2: both curves show initial pH < 2
[1] M3: (with NaOH) heading to pH > 12 (with NH₃) heading to pH 8–12
[1] M1: curly arrow from C=C to H of HBr
[1] M2: correct dipole (δ⁺H—Brδ⁻) AND curly arrow from H—Br to Br
[1] M3: curly arrow from lone pair on :Br⁻ to carbocation
[1] M4: correct intermediate AND product (2-bromopropane)
[1] 2° carbocation more stable (than 1°) (or reverse argument)
[1] greater positive inductive effect of two methyl groups (cf. one ethyl) (or reverse argument)
[1] M1: (water solvent) CH₃CH₂CH₂OH / propan-1-ol AND NaBr/sodium bromide
[1] M2: (ethanol solvent) CH₃CH=CH₂ / propene AND H₂O/water AND NaBr/sodium bromide