Edexcel A-Level Chemistry Paper 2, June 2025: Question 7

17 marks · Hard difficulty · Synoptic Questions

Assess methods for preparing butylamine, write an equation for reducing 1,3-dinitrobenzene, explain the relative basicities of aromatic and aliphatic diamines using pKb data, and add curly arrows to an addition-elimination mechanism.

Practise this question

Question

Question 7 comprises parts (a) to (c) totalling 17 marks. Part (a) asks students to assess two methods for the preparation of butylamine from halogenoalkanes and nitriles, considering reagents, conditions, atom economy, reaction type, and equations (7 marks). Part (b)(i) asks for a balanced equation and conditions for preparing benzene-1,3-diamine from 1,3-dinitrobenzene (3 marks). Part (b)(ii) gives a table with pKb1 values for benzene-1,3-diamine (8.89) and propane-1,3-diamine (3.83) alongside an equilibrium equation for protonation, asking students to explain factors affecting basicity (4 marks). Part (c) displays an incomplete nucleophilic addition-elimination mechanism between ethanoyl chloride and methylamine, asking students to add curly arrows across the three sequential stages (3 marks).

Mark scheme

Show the mark scheme Mark scheme for Question 7. Part (a) awards marks for equations, excess ammonia in sealed tube, nucleophilic substitution classification, LiAlH4 in dry ether or catalytic hydrogenation for the nitrile reduction, and evaluating yield/atom economy. Part (b)(i) provides the balanced reduction equation with 12[H] producing 4H2O, requiring tin and concentrated HCl with heating under reflux. Part (b)(ii) defines base as a proton acceptor/lone pair donor, links lower pKb to stronger base, notes lone pair delocalisation into the benzene pi ring reducing availability, compared to electron-releasing alkyl groups. Part (c) details the six required curly arrows: amine nitrogen lone pair attacking carbonyl carbon, C=O pi bond breaking to oxygen, oxygen lone pair reforming C=O while C-Cl bond breaks, and chloride ion removing the proton from nitrogen.

How to answer it

Synthesis, Basicity & Acylation of Amines

OVERVIEW

What This Question Tests

A comprehensive assessment of organic nitrogen chemistry, combining synthetic route evaluation, redox equations, quantitative pKb basicity analysis, and arrow-pushing mechanisms.

  • Synthetic Pathways to Amines: Nucleophilic substitution of halogenoalkanes vs reduction of nitriles (reagents, conditions, equations, and evaluation of atom economy/yield).
  • Reduction of Aromatic Nitro Compounds: Converting dinitrobenzene to diamines using Sn / conc. HCl.
  • Relative Basicity: Interpreting pKb values and explaining basicity differences in aliphatic vs aromatic amines using lone-pair delocalisation and inductive effects.
  • Nucleophilic Addition-Elimination Mechanism: Detailed electron movements (curly arrows) for the reaction of acyl chlorides with primary amines.
PART (a) • 7 MARKS

Preparation of Butylamine: Halogenoalkane vs Nitrile Routes

Comprehensive assessment of reagents, conditions, atom economy, and reaction types

✅ Route 1: From 1-Halogenobutane (3 Marks)

  • Equation:
    C₄H₉Br + 2NH₃ → C₄H₉NH₂ + NH₄Br
    (Also accepted: C₄H₉Br + NH₃ → C₄H₉NH₂ + HBr)
  • Conditions: Excess ammonia, alcoholic/ethanolic solvent, heated in a sealed tube (or under pressure).
  • Reaction Type: Nucleophilic substitution.

✅ Route 2: From Butanenitrile (3 Marks)

  • Equation:
    C₃H₇CN + 4[H] → C₄H₉NH₂
    (Or with hydrogen gas: C₃H₇CN + 2H₂ → C₄H₉NH₂)
  • Reagent & Conditions: LiAlH₄ in dry ether (or H₂ with Ni / Pt / Pd catalyst).
  • Reaction Type: Reduction (or hydrogenation / redox).

✅ Comparative Assessment & Evaluation (1 Mark — Any 1 of the following)

  • Yield / Purity: Halogenoalkane route yields a mixture of primary, secondary, and tertiary amines (plus quaternary ammonium salts) due to further substitution, resulting in a low yield of butylamine; the nitrile route gives only the primary amine.
  • Atom Economy: The nitrile route (especially with H₂) has a higher atom economy (up to 100%) because only one product is formed, whereas halogenoalkane substitution produces an inorganic salt by-product (e.g. NH₄Br).
  • Practicality: Heating in a sealed tube under pressure is hazardous/difficult on an industrial scale; handling LiAlH₄ requires strictly anhydrous conditions because it reacts violently with water.

❌ Common Errors in Part (a)

  • Incorrect Carbon Count: Using butanenitrile for 4 carbons requires C₃H₇CN, not C₄H₉CN (which would yield pentylamine). Penalised once.
  • Writing "Reflux" for Ammonia: Ammonia is a gas at room temperature and would boil straight out of an open condenser. You must state sealed tube or under pressure.
  • Omitting "Excess" Ammonia: Failing to specify excess NH₃ leads to further alkylation and lower marks.
  • Missing "Dry Ether": Stating LiAlH₄ without specifying dry ether as the solvent.

🧠 Examiner Tip: Structured Comparison

Organise your 6-7 mark extended response methodically with clear subheadings:

  1. Route 1 (Reagents, Conditions, Equation, Type)
  2. Route 2 (Reagents, Conditions, Equation, Type)
  3. Evaluation (Atom economy, purity, ease of procedure)

This guarantees you hit every bullet point outlined in the question prompt.

Mark Scheme Breakdown: [1] Halogenoalkane equation • [1] Excess NH₃, alcoholic, sealed tube/pressure • [1] Nucleophilic substitution • [1] Nitrile equation • [1] LiAlH₄ in dry ether (or H₂/Ni) • [1] Reduction • [1] Valid comparative evaluation point.
PART (b)(i) • 3 MARKS

Reduction of Aromatic Dinitro Compounds

Synthesis of benzene-1,3-diamine

✅ Balanced Equation & Reaction Conditions

C₆H₄(NO₂)₂ + 12[H] → C₆H₄(NH₂)₂ + 4H₂O

Note: Structural/skeletal representations with 1,3-disubstituted benzene rings are fully acceptable. You may also balance with 6H₂ instead of 12[H] .

Conditions:

  • Reagents: Tin ( Sn ) and concentrated hydrochloric acid ( HCl ). (Zinc and HCl is also accepted)
  • Temperature/Setup: Heat under reflux.

🧠 Balancing Reduction Equations

Each nitro group ( -NO₂ ) has 2 oxygens. Removing 2 oxygens needs 4[H] to make 2H₂O, and adding 2 hydrogens to make -NH₂ needs 2[H].

Total per nitro group: 6[H] forming 2H₂O.

Since this is a dinitro compound, you must multiply by 2: 12[H] producing 4H₂O .

❌ Common Errors

  • Writing "dilute HCl" — concentrated acid is required.
  • Balancing with only 6[H] because candidates forget there are two nitro groups.
  • Omitting the water product ( 4H₂O ).
  • Using molecular formulae (e.g. C₆H₈N₂) for organic structures when skeletal/structural is standard.
Mark Scheme Breakdown: [1] Correct organic reactant & product structures • [1] Correct balancing (+ 12[H] → + 4H₂O) • [1] Sn + conc. HCl, heat under reflux.
PART (b)(ii) • 4 MARKS

Explaining Amine Basicity from pKb Data

Benzene-1,3-diamine (pKb1 = 8.89) vs Propane-1,3-diamine (pKb1 = 3.83)

💡 Key Knowledge: Definition & Logarithmic Scales

  • Base Definition: A base is a proton (H⁺) acceptor or a lone pair donor.
  • Understanding pKb: Like pH and pKa, a lower pKb means a higher Kb, which indicates a stronger base.

✅ Model Answer (4 Marks)

  1. Base Definition [1 Mark]: A base acts as a proton (H⁺) acceptor using a lone pair of electrons on the nitrogen atom.
  2. Data Deduction [1 Mark]: Propane-1,3-diamine has a significantly lower pKb1 (3.83) than benzene-1,3-diamine (8.89), meaning propane-1,3-diamine is the stronger base (or benzene-1,3-diamine is the weaker base).
  3. Aromatic Ring Effect [1 Mark]: In benzene-1,3-diamine, the lone pair of electrons on the nitrogen atom is delocalised into the benzene π-system. (Alternatively: alkyl groups in propane-1,3-diamine are electron-releasing via a positive inductive effect).
  4. Availability of Lone Pair [1 Mark]: Therefore, the lone pair on the nitrogen of benzene-1,3-diamine is less readily available to accept a proton (or in propane-1,3-diamine, it is more available).

❌ Examiner Red Flags

  • Confusing pKb with base strength: Incorrectly stating "higher pKb = stronger base". Remember: lower pK = stronger!
  • Phrasing: Saying the lone pair is "donated into the benzene ring" was rejected by the mark scheme. You must use the word delocalised into the π-system / ring.
  • Forgetting "Lone Pair": Writing that "the nitrogen is less available" rather than explicitly stating the lone pair on nitrogen is less available loses the mark.
Mark Scheme Breakdown: [1] Base = proton acceptor / lone pair donor • [1] Identification of relative base strength using pKb • [1] Lone pair delocalisation into π-ring (or inductive release in aliphatic) • [1] Availability of lone pair for protonation linked to strength.
PART (c) • 3 MARKS

Nucleophilic Addition-Elimination Mechanism

Reaction of Ethanoyl Chloride with Methylamine to form an N-Substituted Amide

💡 The 6 Essential Curly Arrows

The mechanism proceeds via a tetrahedral intermediate in three key stages:

1 Nucleophilic Attack:

  • Arrow 1: From the lone pair on the nitrogen atom of CH₃NH₂ to the carbonyl carbon atom ( C=O ).
  • Arrow 2: From the C=O double bond onto the oxygen atom (forming C–O⁻ ).

2 Elimination of Chloride:

  • Arrow 3: From a lone pair on the negative oxygen ( :O⁻ ) back into the C–O bond to reform the C=O double bond.
  • Arrow 4: From the C–Cl single bond to the chlorine atom (releasing :Cl⁻ ).

3 Deprotonation:

  • Arrow 5: From the lone pair of the chloride ion ( :Cl⁻ ) to a hydrogen atom bonded to the positively charged nitrogen ( N⁺–H ).
  • Arrow 6: From the N–H bond back onto the nitrogen atom (neutralising the N⁺ ).

🧠 Precision Arrow-Pushing Rules

  • Every arrow represents the movement of an electron pair and must be double-headed. Half-headed fishhook arrows lose marks.
  • Arrows must start precisely on a lone pair or the centre of a bond.
  • Arrows must point cleanly to an atom (forming a lone pair/coordination) or to a bond (forming a multiple bond).

❌ Common Mistakes in this Mechanism

  • Simultaneous One-step Substitution: Drawing the amine attacking while the chloride leaves directly (SN2 style). Acyl chlorides always react via addition-elimination through a tetrahedral intermediate!
  • Incorrect arrow origin on deprotonation: Drawing Arrow 5 starting from the negative charge sign rather than the lone pair on :Cl⁻ .
  • Extra Arrows: The mark scheme subtracts 1 mark for each superfluous incorrect arrow drawn beyond the required 6.
Mark Scheme Breakdown: 6 correct curly arrows = [3] marks • 4 or 5 correct arrows = [2] marks • 2 or 3 correct arrows = [1] mark.

Topics

Organic Chemistry · Topic 6: Organic Chemistry I · Topic 18: Organic Chemistry III

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.