WJEC A-Level Chemistry AS Unit 2, June 2025: Question 10

✅ Correct Answer

• Ethanolic / dissolved in ethanol [1]
• Reflux / heat [1]

🧠 Exam Technique

Always name both the solvent/medium and the temperature condition when asked for "conditions" for haloalkane reactions:

• Ethanolic NaOH + Heat/Reflux = Elimination (forms alkene)
• Aqueous NaOH + Warm = Nucleophilic Substitution (forms alcohol)

❌ Common Errors

• Stating "aqueous" or "dilute" — this triggers substitution to produce butan-1-ol, giving 0 marks.
• Omitting "heat" or "reflux" — room temperature is insufficient for a good elimination rate.

✅ Correct Answer

Hydrogen bromide / HBr [1]

💡 Key Knowledge

An elimination reaction removes atoms from adjacent carbons to form a C=C double bond. A hydrogen atom is lost from C2 and a bromine atom from C1, together constituting HBr (which is neutralised by NaOH to give NaBr + H₂O).

❌ Common Errors

• Writing Br₂ (bromine) or H₂ (hydrogen).
• Writing ions such as Br⁻ or H⁺ (the question explicitly asks for the molecule).
• Writing H₂O or NaBr (these are overall mixture products, but the fragment eliminated directly from 1-bromobutane is HBr).

Step 1: Electrophilic attack on H–Br H H H H | | | | H - C - C - C = C - H | | | | H H H \__ \ H^δ+ — Br^δ− | ^ \_____/ (arrow from H-Br bond to Br) Step 2: Secondary carbocation intermediate & bromide attack H H H H | | | | H - C - C - C⁺ - C - H :Br⁻ | | | | / H H H H / (arrow from lone pair to C+) v Step 3: Final product (2-bromobutane) H H H H | | | | H - C - C - C - C - H | | | | H H :Br: H

✅ Mark Allocation

• Mark 1: Correct but-1-ene and H–Br structures with dipoles correctly labelled: H^δ+–Br^δ−.
• Mark 2: Correct secondary carbocation intermediate ( CH₃CH₂CH⁺CH₃ ) with positive charge on C2, plus :Br⁻ with its negative charge and lone pair.
• Mark 3: All three curly arrows drawn accurately:
  1. From C=C double bond to H^δ+.
  2. From H–Br bond onto Br^δ−.
  3. From lone pair on :Br⁻ to the carbocation C⁺.

❌ Common Errors & Penalties

• Forming the primary carbocation: If your arrows lead to 1-bromobutane instead of 2-bromobutane, you are penalised 1 mark.
• Vague arrow origins: Starting the curly arrow near an atom rather than squarely from the middle of the C=C bond or from the lone pair on :Br⁻.
• Missing charges/lone pairs: Omitting the negative charge or lone pair on the bromide ion.
• Inverted dipoles: Drawing H^δ−–Br^δ+ loses the first mark immediately.

✅ Correct Intermediate Structures

Alkene Intermediate Y: 2-methylbut-2-ene

Alkene Intermediate Z: 2-methylbut-1-ene

🧠 Deductive Exam Technique

Look at the downstream addition products given at the bottom:

• Intermediate Y leads to A (2-bromo-3-methylbutane) and B. In A, the Br is on C3 and H on C2, which means the C=C bond must have been between C2 and C3!
• Intermediate Z leads to B and C (1-bromo-2-methylbutane). In C, the Br is on C1, which means the C=C bond must have been between C1 and C2 (or terminal carbons)!

✅ Model Answer

• Product with greatest yield: B [1]
• Reason: Product B is formed via the most stable carbocation [1]
• Explanation: The intermediate leading to B is a tertiary carbocation (the positively charged carbon is bonded to three alkyl / electron-releasing groups) [1]

💡 Key Knowledge: Carbocation Stability

Carbocation stability order:

Tertiary (3°) > Secondary (2°) > Primary (1°)

Alkyl groups are electron-donating (positive inductive effect, +I). They push electron density towards the C⁺ atom, delocalising and spreading the positive charge and thereby stabilising the ion.

Alternative valid marking route: Award [2] for explaining that B is formed by electrophilic addition to both intermediates Y and Z.

❌ Common Errors

• Failing to explicitly name the product as B.
• Simply stating "Markovnikov's rule" without explaining carbocation stability—the question explicitly directs you to "explain your reasoning in terms of the stability of any carbocation intermediates".
• Confusing the classification: stating B is formed from a secondary carbocation rather than a tertiary carbocation.