AQA A-Level Chemistry Paper 2, 2019: Question 1

✅ Correct Answer & Equation

Br-(CH₂)₆-Br + 4NH₃ → H₂N-(CH₂)₆-NH₂ + 2NH₄Br

(Skeletal or displayed formula for both organic compounds is fully accepted.)

❌ Common Errors

• Forming HBr instead of NH₄Br: Because ammonia is in excess and basic, it reacts with the acidic by-product HBr to form ammonium bromide ( NH₄Br ). Writing + 2 HBr forfeits the balancing mark.
• Incorrect stoichiometry: Forgetting that each substituted bromine requires two ammonia molecules (one to form the C–N bond, one to remove the proton). Thus, a dihaloalkane requires a 1:4 molar ratio.
• Using molecular formula: Writing molecular formulae such as C₆H₁₂Br₂ or C₆H₁₆N₂ loses M1. You must show the functional groups clearly.

✅ Mechanism Requirements & Product

Step-by-step mechanism description:

• M1: Arrow from lone pair on the N of :NH₃ to the terminal carbon atom bonded to Br. A second arrow from the C–Br bond onto the Br atom (showing loss of Br⁻ ).
• M2: Correct intermediate structure: H₃N⁺–(CH₂)₆–NH₂ showing the formal positive charge on the quaternary/ammonium nitrogen.
• M3: Arrow from the lone pair of a second :NH₃ molecule to one of the H atoms on the –N⁺H₃ group, with a concomitant arrow from the N–H bond to the positive N atom.
• Cyclic secondary amine: A 7-membered ring containing 6 carbons and 1 NH group (azepane / hexamethyleneimine). Also accepted: 14-membered cyclic diamine dimer.

🧠 Exam Technique & Pitfalls

• Arrow precision: Always ensure the arrow starts clearly on the lone pair or the center of the covalent bond, and terminates directly at the receiving atom.
• Don't deprotonate with bromide: The second deprotonation must be carried out by another :NH₃ molecule. Using Br⁻ to remove the proton is penalised.
• Why does cyclisation happen? 6-bromohexylamine has both an electrophilic C–Br end and a nucleophilic –NH₂ end. Intramolecular nucleophilic substitution loops the chain around to form a stable ring.

✅ Correct Reagents & Conditions

Stage 1 (Nucleophilic substitution to extend chain from 4C to 6C):

• Reagent [M1]: KCN or NaCN
• Condition [M2]: Aqueous alcohol (aqueous ethanol)

Stage 2 (Reduction of dinitrile to diamine):

• Reagent & Condition [M3]: H₂ with a Ni (or Pt / Pd ) catalyst.
  Alternative accepted: LiAlH₄ in dry ether followed by dilute acid.

❌ Common Errors & Examiner Warnings

• Using HCN: Writing HCN or mentioning any acid in Stage 1 causes an immediate Contradiction Error (CE) and loses both M1 and M2. Hydrogen cyanide is too weak a nucleophile and generates toxic fumes.
• Inappropriate reducing agents: NaBH₄ cannot reduce nitriles! Neither can Sn / HCl or Fe / HCl (which are specifically for reducing nitroarenes to aromatic amines).
• Missing conditions for LiAlH₄: If using LiAlH₄ , specifying aqueous or acidic conditions in the first step is an instant CE.

✅ Model Answer

In 3-aminopentane:

• The lone pair of electrons on the nitrogen atom is more available / accepts a proton (H⁺) more readily. [1 mark]
• Due to the positive inductive effect (electron-releasing nature) of the attached alkyl groups. [1 mark]

💡 Key Knowledge

• Definition of a base: A Brønsted-Lowry base is a proton acceptor. Base strength depends entirely on how available the lone pair on nitrogen is to bind a H⁺.
• Alkyl groups push electrons: The alkyl chains on 3-aminopentane push electron density toward the nitrogen atom (+I inductive effect), increasing its electron density compared to ammonia.
• Examiner tip: Always specify the exact location: say "lone pair on the nitrogen atom", not merely "nitrogen has more electrons".

✅ Model Answer

No carbon atom is bonded to four different groups.

(Also accepted: The central carbon atom / C3 is bonded to two identical ethyl groups [ –CH₂CH₃ ], making the molecule symmetrical.)

🧠 Exam Technique

Structure of 3-aminopentane:

CH₃–CH₂–CH(NH₂)–CH₂–CH₃

Look at carbon-3: it is bonded to –H , –NH₂ , and two identical ethyl groups ( –CH₂CH₃ ). Because it does not have four different groups attached, it is achiral and optically inactive.