AQA A-Level Chemistry Paper 2, 2017: Question 11

✅ Mark Scheme Breakdown (6 Marks)

• [M1] Fundamental Principle: Base strength depends on the availability of the lone pair of electrons on the nitrogen atom.
• [M2] Delocalisation in E: In E, the nitrogen lone pair is delocalised into the benzene π-system (or ring).
• [M3] Effect on E: The lone pair is therefore less available to donate to / accept a proton (H⁺).
• [M4] Inductive Effect in F/G: In F (and alkyl group in G), the alkyl group exerts a positive inductive effect (pushes electrons towards nitrogen).
• [M5] Effect on F: The lone pair on nitrogen is more available to donate to / accept a proton (H⁺).
• [M6] Correct Order: E < G < F (or stated as: F is the most basic and E is the least basic).

💡 Key Knowledge

• Base Definition: Amines act as Brønsted-Lowry bases because the lone pair on the nitrogen can accept a proton: R-NH₂ + H⁺ → R-NH₃⁺ .
• Aromatic Amines (E - Phenylamine): The lone pair on N is in a p-orbital that overlaps with the delocalised π-system of the benzene ring. This lowers electron density on N, significantly weakening basicity.
• Aliphatic Amines (F - 2-phenylethylamine): The benzene ring is insulated from N by two -CH₂- groups. The alkyl chain donates electron density via the positive inductive effect, making the lone pair readily available.
• Secondary Mixed Amine (G - N-ethylphenylamine): Nitrogen is directly attached to the ring (lone pair delocalised, reducing basicity), but also has an ethyl group offering a weak inductive boost. Hence, basicity lies between E and F.

🧠 Exam Technique: Structuring 6-Mark Explanations

Always structure amine basicity questions in three distinct logical tiers:

1. Define the criterion: State clearly that basicity depends on nitrogen lone pair availability to accept an H⁺ ion.
2. Explain both extremes:
   • Ring attachment: delocalisation of lone pair into ring → less available.
   • Alkyl attachment: positive inductive effect of alkyl group → more available.
3. State the final relative order clearly: Always verify if the question asks for increasing (weakest to strongest: E < G < F) or decreasing order.

❌ Common Examiner Traps

• Missing the lone pair: Stating simply "nitrogen accepts a proton" without mentioning the lone pair on nitrogen loses M1.
• Confusing F with an aromatic amine: Assuming F has delocalisation because a ring is present. Note the -CH₂CH₂- spacer! The lone pair cannot delocalise into the ring.
• Reversing the inequality: Writing E > G > F when asked for increasing base strength loses M6.
• Vague inductive wording: Saying "alkyl groups are electronegative" instead of stating they have an electron-releasing / positive inductive effect.

📐 Step-by-Step Synthetic Pathway

✅ Mark Allocation (5 Marks)

• Intermediate 1 (1 Mark): C₆H₅CH₂Cl or C₆H₅CH₂Br .
• Intermediate 2 (1 Mark): C₆H₅CH₂CN .
• Step 1 Reagents (1 Mark): Cl₂ and UV (or Br₂ & UV).
• Step 2 Reagents (1 Mark): KCN (or NaCN) with alcoholic and aqueous solvents both explicitly specified.
• Step 3 Reagents (1 Mark): H₂ with Ni / Pt / Pd catalyst (or LiAlH₄ in dry ether).

❌ Critical Synthetic Traps in 11.2

• Using AlCl₃ / Fe / FeCl₃ in Step 1: Using a halogen carrier causes electrophilic aromatic substitution on the benzene ring (giving 2- or 4-chloromethylbenzene), rather than substitution on the methyl side chain! Side chain chlorination requires free-radical conditions: UV light.
• Forgetting solvent conditions for cyanide: Simply writing "KCN" loses the mark. The AQA mark scheme strictly demands alcoholic & aqueous (or aqueous ethanol).
• Wrong reducing agents in Step 3:
  • NaBH₄ cannot reduce nitriles (not powerful enough).
  • Sn / HCl or Fe / HCl reduces aromatic nitro groups ( -NO₂ → -NH₂ ), not nitriles!
• Direct amination trap: Attempting to react C₆H₅CH₂Cl directly with NH₃ gives 1-phenylmethanamine ( C₆H₅CH₂NH₂ ), which is missing one carbon atom compared to Amine F ( C₆H₅CH₂CH₂NH₂ ). The cyanide step is essential to extend the chain by one carbon.

🧠 Top-Grade Synthesis Checklist

• Carbon Count First: Always count the carbons in starting material vs product. Methylbenzene has 7 carbons; Amine F has 8 carbons. An increase of exactly 1 carbon is your direct clue that a nitrile intermediate ( -C≡N ) must be used.
• Structural Formula Accuracy: When drawing or writing intermediates, make sure formulas are unambiguous: write C₆H₅CH₂Cl or display the full structure clearly so bonds attach to carbon, not nitrogen or hydrogen.