Edexcel A-Level Chemistry Paper 2, November 2021: Question 10

✅ Correct Answer

C₄H₉NH₂ + H₂O ⇌ C₄H₉NH₃⁺ + OH⁻

The equation shows butylamine accepting a proton from water to produce hydroxide ions, which creates an alkaline solution (pH > 7).

❌ Common Errors

• Combining product ions into a single uncharged molecule or failing to show the positive charge on the protonated amine.
• Omitting equilibrium arrows or writing irreversible reaction arrows.

💡 Key Knowledge

• Phenylamine: The lone pair of electrons on the nitrogen atom interacts with (or delocalises into) the pi-electron system of the benzene ring.
• Butylamine: The alkyl group has a positive inductive effect, pushing electron density towards the nitrogen atom.
• Conclusion: The nitrogen lone pair in phenylamine is less available to accept a proton compared to butylamine.

🧠 Exam Technique

To secure all 3 marks, you must explicitly mention: (1) the nitrogen lone pair, (2) its delocalisation into the benzene ring (for phenylamine) or inductive push (for butylamine), and (3) the resulting difference in proton-accepting ability.

✅ Correct Answers

Name: Propanoyl chloride (or propanoic anhydride)

Structural Formula: CH₃CH₂COCl (or (CH₃CH₂CO)₂O )

❌ Common Errors

Students frequently confuse acyl chlorides with carboxylic acids or standard haloalkanes, resulting in incorrect carbon chain lengths or missing functional groups. Ensure the acyl chloride has the exact carbon count required to match the product CH₃CH₂CONHC₄H₉ .

✅ Correct Answer

A - blue solution

Excess butylamine undergoes a ligand exchange reaction with aqueous copper(II) sulfate, forming the complex ion [Cu(C₄H₉NH₂)₄(H₂O)₂]²⁺ or [Cu(C₄H₉NH₂)₄]²⁺ , which remains in solution as a deep blue solution.

💡 Examiner Insight

Candidates often mix up ligand exchange reactions involving uncharged amines (which yield deep blue solutions via replacement of water ligands) with dropwise addition of alkali/hydroxide (which yields precipitates).

✅ Correct Answer

D - nucleophilic substitution

The lone pair on the nitrogen of the primary amine acts as a nucleophile, attacking the electron-deficient carbon bonded to the halogen in the bromoethane molecule.

❌ Common Errors

Confusing nucleophilic substitution with nucleophilic addition or electrophilic substitution. Always identify the attacking species (amine = nucleophile) and the reaction outcome (substitution of the halogen atom).

🧠 Mechanism Requirements

• Arrow 1: Curly arrow starting from the lone pair on the nitrogen atom of butylamine pointing directly to the δ+ carbon atom of bromoethane.
• Arrow 2: Curly arrow starting from the C–Br bond and pointing to the bromine atom, showing heterolytic fission to form Br⁻ .
• Intermediate: Show the positively charged nitrogen intermediate along with the displaced Br⁻ ion.
• Arrow 3: Curly arrow from the N–H bond of the intermediate back onto the positively charged nitrogen atom to remove a proton and complete the formation of the secondary amine.

❌ Common Pitfalls

Make sure curly arrows start precisely from lone pairs or bond lines—not from floating spaces or atom symbols. Ensure dipoles ( δ+ and δ- ) are clearly indicated on the halogenoalkane.