AQA A-Level Chemistry Paper 2, 2024: Question 9

✅ Correct answers

• Missing reagent in Step 1: CH₃NH₂ (methylamine). Equation forms the salt CH₃CH₂NH₂CH₃⁺ Br⁻ (ethylmethylammonium bromide).
• Secondary amine from Step 2 (after deprotonation): N-methyl ethylamine (a.k.a. N-methyl ethanamine).

💡 Key knowledge

Haloalkane + ammonia gives an alkylammonium salt first; base removes H⁺ to give the amine. Using a primary amine instead of NH₃ gives an N-alkylated product.

❌ Common errors

• Naming the secondary amine as “ethylmethylamine bromide” (that’s the salt from Step 1, not the amine).
• Forgetting the positive charge and Br⁻ in the Step-1 salt.

✅ Correct answers (skeletal acceptable)

• Excess NH₃: butan-2-amine (2-aminobutane) as the major product.
• Excess CH₃CHBrCH₂CH₃: predominantly a trialkylammonium salt, i.e. a tetra-alkylammonium bromide after over-alkylation [(sec-C₄H₉)₃NH]⁺ Br⁻ (structure shown as a quaternary ammonium salt is credited).

🧠 Exam technique

With excess NH₃, nucleophile is NH₃ → favours primary amine. With excess haloalkane, the amine formed acts as a nucleophile and gets further alkylated to secondary/tertiary and finally a quaternary ammonium salt.

✅ Mechanism & product

• Step 1 (S_N2): 2-bromopropane + KCN/NaCN in ethanol, reflux → (CH₃)₂CHCN + Br⁻. Curly arrows: CN⁻ → C–Br carbon; C–Br bond → Br.
• Step 2 (reduction): (CH₃)₂CHCN + 2 H₂ → (CH₃)₂CHCH₂NH₂ (amine G, 2-methylpropan-1-amine).

💡 Key knowledge

CN⁻ substitution increases the carbon chain by one. Hydrogenation of a nitrile adds across C≡N to give a primary amine. In full display, remember the extra CH₂ between the former nitrile carbon and –NH₂.

❌ Common errors

• Missing hydrogens in the fully displayed amine G or drawing too few carbons.
• Using aqueous CN⁻ (promotes hydrolysis) instead of ethanol solvent for S_N2.

✅ Explanation

In P the nitrogen has two alkyl groups. Alkyl groups are electron-donating (+I inductive effect), which pushes electron density onto N, making its lone pair more available to accept a proton (H⁺). Therefore P is the stronger Brønsted–Lowry base than Q.

❌ Common errors

• Saying “more electrons” without linking to availability of the lone pair.
• Confusing nucleophilicity with basicity (both increase with electron donation, but here the mark is for accepting H⁺).

💡 Key knowledge

• Ammonia/amine + haloalkane → substitution via S_N2 (primary) or S_N1 (tertiary).
• Excess NH₃ limits over-alkylation; excess haloalkane promotes it → quaternary ammonium salt.
• CN⁻ substitution followed by reduction is a reliable route to longer-chain primary amines.

🧠 Exam technique

• Draw charges and counter-ions for ammonium salts (e.g. Br⁻) in Step 1.
• Count carbons carefully when reducing nitriles—add a CH₂ before –NH₂.