AQA A-Level Chemistry Paper 3, 2017: Question 2

✅ Correct Answer [1 Mark]

Electrophilic addition

🧠 Exam Technique

Always state both the attack type and the reaction type: "electrophilic" (the C=C double bond attacks the electrophilic δ+ H of H₂SO₄) and "addition" (one product is formed across the double bond).

💡 Mechanism Architecture & Arrow Details

• Representation of sulfuric acid: Display as H–OSO₂OH or H–OSO₃H so the reactive O–H bond is visible.
• Arrow 1 (M1): From the centre of the C=C double bond of propene ( CH₃–CH=CH₂ ) pointing directly to the H atom of the H–O bond of H₂SO₄.
• Arrow 2 (M2): From the H–O single bond onto the oxygen atom of the hydrogensulfate group, showing heterolytic fission to generate ⁻:OSO₂OH .
• Intermediate (M3): Draw the secondary carbocation correctly: CH₃–CH⁺–CH₃ .
• Arrow 3 (M4): From a lone pair on the negatively charged oxygen atom of ⁻:OSO₂OH (or ⁻:OSO₃H ) directly to the positively charged C⁺ atom.

❌ Common Errors

• Drawing an arrow to a lone proton H⁺ without showing the H–O bond breaking in H₂SO₄.
• Forming the primary carbocation ( CH₃CH₂CH₂⁺ ) instead of the secondary carbocation. The question explicitly demands the major product!
• Drawing the arrow in M4 from the sulfur atom or an uncharged oxygen instead of the lone pair on the negatively charged oxygen.
• Writing HSO₄ without the negative charge and lone pair.

✅ Correct Answer [1 Mark]

CH₃CH₂CH₂OSO₃H  (or CH₃CH₂CH₂OSO₂OH / propyl hydrogensulfate)

Structural, displayed, or skeletal formula showing the hydrogensulfate group bonded to the terminal carbon atom (C1).

🧠 Exam Technique

The minor product is derived from the primary carbocation intermediate ( CH₃CH₂CH₂⁺ ). Make sure the oxygen is explicitly linked to C1 (e.g. –O–SO₃H ), not the sulfur directly bonded to carbon.

✅ Model Answer [2 Marks]

• Mark 1: The major product is formed via the more stable secondary carbocation (whereas the minor product forms via a less stable primary carbocation).
• Mark 2: Due to the positive inductive effect (or electron-releasing character) of two alkyl / methyl groups (compared to only one alkyl group in the primary carbocation), which reduces the charge density on the positive carbon.

❌ Common Errors

• Saying "the secondary product is more stable" rather than the secondary carbocation intermediate. Marks are only awarded for carbocation stability!
• Failing to mention the number of alkyl groups (two vs one) when explaining the inductive effect.

💡 Why Three Products?

Elimination of H₂O from butan-2-ol ( CH₃–CH(OH)–CH₂–CH₃ ) can remove an H from C1 to give but-1-ene, or from C3 to give but-2-ene. Because rotation is restricted around the C=C bond, but-2-ene exists as stereoisomers: (E)-but-2-ene and (Z)-but-2-ene.

🧠 Exam Technique: Skeletal Drawings

Make sure you draw strictly skeletal formulas (no C or H letters along the backbone). Ensure the cis/trans geometry of the E and Z isomers is clearly distinguishable.

✅ Correct Answers [3 Marks]

• By-product: butanone (allow butan-2-one) [1 mark]
• Role of sulfuric acid: oxidising agent (allow electron acceptor) [1 mark]
• Separation method: (fractional) distillation (allow gas chromatography) [1 mark]

❌ Common Errors & Pitfalls

• Wrong role: Stating "catalyst" or "dehydrating agent". In the formation of butanone, hot concentrated H₂SO₄ actually oxidises the secondary alcohol; hence its role is an oxidising agent.
• Saying "electron pair acceptor" (that describes a Lewis acid, not an oxidising agent).
• Suggesting "filtration" or "separating funnel" instead of distillation to separate miscible organic liquids with distinct boiling points.

✅ Correct Answers [2 Marks]

• Observation: White fumes / misty fumes / steamy fumes [1 mark]
• Role of sulfuric acid: Acid / proton donor [1 mark]

💡 Chemistry Behind the Reaction

NaCl(s) + H₂SO₄(l) → NaHSO₄(s) + HCl(g)

Chloride ( Cl⁻ ) is a weak reducing agent and cannot reduce concentrated sulfuric acid. Therefore, this is purely an acid-base reaction where H₂SO₄ acts as a Brønsted-Lowry acid, releasing HCl gas which fumes in moist air.

✅ Correct Deductions [3 Marks]

🧠 Redox Summary of Iodide with H₂SO₄

Iodide ( I⁻ ) is a very strong reducing agent. It reduces sulfuric acid through multiple stages:

• H₂SO₄ (+6) → SO₂ (+4) (colourless, choking gas)
• H₂SO₄ (+6) → S (0) (yellow solid)
• H₂SO₄ (+6) → H₂S (-2) (gas with bad egg smell)
• Meanwhile, 2I⁻ → I₂ + 2e⁻ produces iodine (purple vapour / black solid).