AQA A-Level Chemistry Paper 3, June 2025: Question 7

✅ Correct Answer

B: (C₂H₅)₃N⁺CH₃

The original methylamine starts with one methyl group (–CH₃) attached to the nitrogen. Successive substitution by excess bromoethane introduces three ethyl groups (–C₂H₅), yielding a total of four alkyl groups attached to the central N⁺.

🧠 Exam Technique: Tracking the Groups

Always identify:

1. The starting amine: Methylamine ( CH₃NH₂ ) contains 1 methyl group and 2 replaceable H atoms.
2. The reagent in excess: Bromoethane ( C₂H₅Br ) provides ethyl groups.
3. The end-point: A quaternary cation must have a total of 4 alkyl groups bonded to N. Therefore, it must retain the original 1 methyl group and gain 3 ethyl groups: (C₂H₅)₃N⁺CH₃ .

Step 1: Primary Amine → Secondary Amine
CH₃NH₂ + C₂H₅Br → (C₂H₅)(CH₃)NH + HBr
One H on the nitrogen is replaced by one ethyl group.

Step 2: Secondary Amine → Tertiary Amine
(C₂H₅)(CH₃)NH + C₂H₅Br → (C₂H₅)₂(CH₃)N + HBr
The second H on the nitrogen is replaced by another ethyl group.

Step 3: Tertiary Amine → Quaternary Ammonium Salt
(C₂H₅)₂(CH₃)N + C₂H₅Br → [(C₂H₅)₃N⁺CH₃]Br⁻
The nitrogen's lone pair donates to a third ethyl group, creating a permanent positive charge on nitrogen.

💡 Key Knowledge

• Nucleophilicity of Amines: Ammonia and amines act as nucleophiles because of the lone pair of electrons on the nitrogen atom.
• Further Substitution: Primary, secondary, and tertiary amines retain lone pairs, allowing successive nucleophilic substitutions until the quaternary ammonium salt is formed.
• Excess Haloalkane: Ensures the reaction goes all the way to completion to produce the quaternary ammonium salt as the major product.
• Excess Amine (Contrast): If the amine were in excess, the reaction would favour the primary or secondary amine instead of further alkylation.

❌ Common Distractor Traps

• Selecting A: (C₂H₅)₄N⁺
  Students confuse this with the reaction between ammonia (NH₃) and excess bromoethane. Ammonia has no pre-existing alkyl groups, yielding tetraethylammonium. Methylamine already has a –CH₃ group that cannot be displaced.
• Selecting C or D:
  These correspond to having 2 or 3 methyl groups on the nitrogen. Because the bromoalkane in excess is bromoethane, additional groups added must be ethyl, not methyl.