Edexcel A-Level Chemistry Paper 3, June 2025: Question 6

✅ Correct Structures & Reagents

• Compound A: Benzyl bromide (bromomethylbenzene)
  C₆H₅–CH₂–Br
  Structure: A benzene ring with a –CH₂Br side chain.
• Compound B: Benzylmagnesium bromide
  C₆H₅–CH₂–MgBr
  Structure: A benzene ring with a –CH₂MgBr side chain.
• Compound C: Magnesium alkoxide intermediate
  C₆H₅–CH₂–C(CH₃)₂–OMgBr
  Structure: Benzyl group bonded to the central tertiary carbon of propanone, bearing two –CH₃ groups and an –OMgBr (or –O⁻) group.
• Reagent D: Dilute aqueous strong acid, e.g. HCl(aq) or H₂SO₄(aq) .

💡 Reaction Steps & Chemistry

• Alcohol → Bromoalkane: Phenylmethanol reacts with KBr and 50% H₂SO₄ to generate HBr in situ, substituting –OH with –Br.
• Grignard Formation: Refluxing benzyl bromide with Mg in dry ether forms the organomagnesium compound (Grignard reagent, R–MgBr).
• Nucleophilic Addition: The carbanion-like benzylic carbon (δ⁻) attacks the electrophilic carbonyl carbon (δ⁺) of propanone, giving a halomagnesium alkoxide salt.
• Hydrolysis: Reagent D protonates the alkoxide to form the final tertiary alcohol and basic magnesium halide salts.

🧠 Exam Technique & Guidance

• Kekulé or Delocalised: You may draw benzene rings with the circle or alternating single/double Kekulé bonds, but maintain aromaticity properly. Missing delocalised circles will lose marks.
• Charges: You may write the alkoxide as –O⁻ or –OMgBr . However, do not assign full ionic charges like Mg²⁺ or Br⁻ unless specifically showing complete ionic dissociation. Partial charges (δ⁺/δ⁻) are ignored even if incorrect.
• Reagent D: Must state an actual dilute acid such as HCl(aq) or dilute hydrochloric acid . Examiners do not accept bare shorthand like H⁺ or H₃O⁺.

❌ Common Errors to Avoid

• Brominating the ring: Mistaking benzyl alcohol for phenol and substituting onto the benzene ring (e.g., 2-bromophenylmethanol) instead of replacing the benzylic –OH.
• Losing the –CH₂– carbon: Drawing compound B as phenylmagnesium bromide (C₆H₅MgBr) rather than benzylmagnesium bromide (C₆H₅CH₂MgBr).
• Leaving the alkoxide unhydrolysed: Forgetting reagent D or failing to specify dilute/aqueous conditions.

📐 Step-by-Step Peak Deductions

1. Environment K (benzylic –CH₂–):
   • Area: 2 protons → area = 2.
   • Splitting: Flanked by an aromatic carbon (no H) and a quaternary carbon (no H). Neighbouring H = 0 → n + 1 = singlet.
   • Shift: R–CH₂–Ar group is typically around 2.3 ppm (Data booklet range: 1.6 – 3.0 ppm).
2. Environment L (two equivalent –CH₃ groups):
   • Area: Two identical methyl groups = 6 protons → area = 6.
   • Splitting: Attached to quaternary carbon with no adjacent hydrogens → n + 1 = singlet.
   • Shift: Alkyl R–CH₃ range from booklet: –0.2 – 2.0 ppm.
3. Environment M (–O–CH₂–CH₃):
   • Area: 2 protons → area = 2.
   • Splitting: Adjacent to the methyl group (3 protons) → 3 + 1 = quartet.
   • Shift: Deshielded by direct bond to oxygen: R–O–CH₂– falls in 2.8 – 4.4 ppm.
4. Environment N (–O–CH₂–CH₃):
   • Area: 3 protons → area = 3.
   • Splitting: Adjacent to the –CH₂– group (2 protons) → 2 + 1 = triplet.
   • Shift: Alkyl terminal –CH₃ falls in –0.2 – 2.0 ppm.

❌ Common Mistakes & Mark Scheme Penalties

• Numbers instead of words: Do not write numbers for splitting patterns (e.g. "1" instead of "singlet", "4" instead of "quartet"). The mark scheme penalises numerical entries up to a maximum deduction of 2 marks.
• Quaternary vs Quartet: "Quaternary" refers to carbon substitution, not NMR multiplicity. The mark scheme explicitly states: "Do not award quaternary". Always write quartet.
• Confusing environments K and M: Both are –CH₂– groups (area 2), but M is bonded to oxygen (much higher chemical shift: 2.8–4.4 ppm) and split into a quartet by the ethyl –CH₃. K is benzylic and a singlet.
• Environment L integration: Forgetting that both methyl groups attached to the tertiary carbon are chemically equivalent by symmetry, giving an area of 6, not 3.