NJC J2 H2 Chemistry, 2024 A-Level Paper 2, Questions 4–6
Difficulty: Intermediate to Advanced | Prerequisites: Organic functional groups, chirality basics, acid-base equilibria, Ksp.
Big picture: These three questions cover stereoisomerism in organic chemistry, analytical chemistry of sea water using EDTA titrations and solubility product calculations, and the carbonate buffer system that regulates ocean pH. Question 4 is the heaviest at 18 marks and requires you to identify chiral centres, distinguish enantiomers, draw SN2 mechanisms, and compare halide reactivity. Questions 5 and 6 are applied chemistry, testing your ability to handle multi-step calculations involving concentrations, equilibrium constants, and buffer chemistry in a real-world marine context.
Q4 tests everything about stereoisomers: chiral centres, enantiomers, optical activity, SN2 mechanisms, and halide hydrolysis rates. Q5 applies EDTA titration and Ksp calculations to sea water analysis. Q6 asks you to work through the CO₂/carbonate buffer system in the ocean, including pH calculations, partial pressures, and equilibrium expressions.
Chiral centre
A carbon atom bonded to four different groups. This is the structural requirement for optical isomerism. Think of it as a carbon with no two identical attachments.
Enantiomers
A pair of stereoisomers that are non-superimposable mirror images of each other. They have identical physical properties except for the direction in which they rotate plane-polarised light.
Plane-polarised light
Light that vibrates in a single plane. Chiral molecules rotate the plane of polarisation: one enantiomer rotates it clockwise (+), the other anticlockwise (-).
SN2 mechanism (nucleophilic substitution, bimolecular)
A one-step substitution where the nucleophile attacks from the opposite side to the leaving group (backside attack). The rate depends on the concentration of both the substrate and the nucleophile. Produces inversion of configuration at the chiral centre.
EDTA (ethylenediaminetetraacetate, EDTA⁴⁻)
A hexadentate ligand that forms stable 1:1 complexes with metal cations such as Ca²⁺ and Mg²⁺. Used in complexometric titrations to determine the concentration of metal ions in solution.
Solubility product (Ksp)
The equilibrium constant for the dissolution of a sparingly soluble salt. For Ca(OH)₂: Ksp = [Ca²⁺][OH⁻]². A larger Ksp means a more soluble salt.
Ionic product (Q)
The product of the ion concentrations at any given moment, not necessarily at equilibrium. If Q > Ksp, precipitation occurs. If Q < Ksp, the solution is unsaturated.
Buffer solution
A solution that resists changes in pH when small amounts of acid or base are added. The CO₃²⁻/HCO₃⁻ system in sea water acts as a natural buffer.
Partial pressure
The pressure exerted by one component of a gas mixture. Partial pressure = mole fraction × total pressure. Used here to calculate dissolved CO₂ concentration via Henry's law.
Acyl chloride (ethanoyl chloride)
A highly reactive organic compound with the general structure RCOCl. Reacts vigorously with water (hydrolysis) to form a carboxylic acid and HCl.
Identify the carbon atoms bonded to four different groups.
Vitamin E has three chiral centres. Look for carbons in the ring and in the side chain where branching creates four distinct substituents.
F and G are a pair of enantiomers (non-superimposable mirror images).
When H reacts with F, a reaction occurs (e.g. the reagent is chiral or stereospecific, so it distinguishes the two enantiomers).
When H reacts with G, no reaction occurs.
This tells you that H is itself chiral and reacts selectively with one enantiomer but not the other, or that the two enantiomers have different spatial arrangements that affect reactivity with a specific reagent.
C₄H₈O with one degree of unsaturation.
A secondary alcohol with an alkene group and a chiral carbon is required.
Example: but-3-en-2-ol, CH₂=CHCH(OH)CH₃, which has a chiral centre at C-2.
Distinguishing J and K (enantiomers): use a polarimeter. One enantiomer rotates plane-polarised light clockwise, the other anticlockwise, by the same magnitude.
Nucleophile: OH⁻(aq)
Substrate: 2-bromooctane (a secondary halogenoalkane, but SN2 is specified here)
Mechanism:
OH⁻ attacks the carbon bearing the Br from the opposite side (backside attack).
A transition state forms with five groups around the central carbon.
Br leaves as Br⁻.
The product has inverted configuration at the chiral centre (Walden inversion).
Key diagram elements: lone pair on OH⁻ shown attacking, dotted bonds to incoming OH and departing Br in the transition state, curly arrows from nucleophile to carbon and from C–Br bond to Br.
This test compares the rate of hydrolysis of C–X bonds (X = Cl, Br, I).
Warm the halogenoalkane with ethanolic silver nitrate.
The halide ion produced reacts with Ag⁺ to form a precipitate: AgCl (white), AgBr (cream), AgI (yellow).
Rate of hydrolysis: iodoalkane > bromoalkane > chloroalkane, because the C–I bond is weakest.
The faster the precipitate forms, the weaker the C–X bond.
Ethanoyl chloride + water → ethanoic acid + HCl. This is a vigorous, rapid hydrolysis producing white fumes of HCl.
Chlorobenzene does not react with cold water because the C–Cl bond in chlorobenzene is strengthened by delocalisation of the lone pair on Cl into the benzene ring (partial double-bond character). The carbon is also not electrophilic enough.
Ethanoyl chloride is far more reactive than chloroethane toward water because:
The C=O group in ethanoyl chloride is strongly electron-withdrawing, making the carbon much more electrophilic (δ⁺⁺).
The Cl in ethanoyl chloride is a better leaving group in this context because the resulting carboxylate/acid product is stabilised.
Chloroethane undergoes nucleophilic substitution much more slowly because the carbon is less electrophilic.
The indicator (HIn²⁻) reacts with Ca²⁺: HIn²⁻ + Ca²⁺ → H⁺ + [InCa]⁻
At the end-point, EDTA has complexed all the Ca²⁺ and Mg²⁺, so the indicator is released from the metal and changes colour.
The colour of the sea water sample at the end-point is the colour of the free indicator (not the metal-indicator complex).
EDTA forms 1:1 complexes with both Ca²⁺ and Mg²⁺.
Total moles of EDTA used = total moles of (Ca²⁺ + Mg²⁺).
A separate titration using a Ca²⁺-selective method (e.g. at higher pH where Mg(OH)₂ precipitates) gives moles of Ca²⁺ alone.
Moles of Mg²⁺ = total – moles of Ca²⁺.
Convert to concentrations using the volume of the sea water sample.
Q = [ion product] at current concentrations.
For SO₄²⁻ in ocean 2: calculate Q for CaSO₄ or BaSO₄ using the given ion concentrations.
If Q > Ksp, precipitation occurs. If Q < Ksp, the ions remain in solution.
Ca(OH)₂(s) ⇌ Ca²⁺(aq) + 2OH⁻(aq), so Ksp = [Ca²⁺][OH⁻]²
Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq), so Ksp = [Mg²⁺][OH⁻]²
Mg(OH)₂ has a much smaller Ksp than Ca(OH)₂, so Mg(OH)₂ is less soluble.
To calculate: use the given [OH⁻] or pH to find [OH⁻], then substitute into the Ksp expression.
Given two pH values (e.g. historical and current ocean surface pH), calculate [H⁺] for each using [H⁺] = 10^(-pH).
Percentage increase = ((new [H⁺] – old [H⁺]) / old [H⁺]) × 100%.
A small pH decrease corresponds to a large percentage increase in [H⁺] because the pH scale is logarithmic.
Steps:
Convert total pressure from bar to atm if needed (1 bar ≈ 0.987 atm, or use the conversion factor given).
Calculate partial pressure of CO₂: p(CO₂) = mole fraction of CO₂ × total pressure.
Use Henry's law or the given relationship to find the concentration of dissolved CO₂.
Convert mass concentration to molar concentration if needed: mol conc = mass conc / molar mass.
CO₂ dissolves in sea water and undergoes two equilibria:
K₁: CO₂(aq) + H₂O(l) ⇌ H⁺(aq) + HCO₃⁻(aq)
K₂: HCO₃⁻(aq) ⇌ H⁺(aq) + CO₃²⁻(aq)
The CO₃²⁻/HCO₃⁻ pair acts as a buffer:
When atmospheric CO₂ increases and dissolves, more H⁺ is produced.
CO₃²⁻ reacts with the excess H⁺: CO₃²⁻ + H⁺ → HCO₃⁻, which removes H⁺ and resists pH decrease.
The buffer has a limited capacity; large increases in CO₂ will eventually overwhelm it, leading to ocean acidification.
The overall equilibrium for CO₂(aq) + H₂O(l) ⇌ 2H⁺(aq) + CO₃²⁻(aq) has:
K_overall = K₁ × K₂
Calculate by multiplying the given K₁ and K₂ values.
Units: work out from the individual expressions. K₁ has units of mol dm⁻³; K₂ has units of mol dm⁻³; so K_overall has units of mol² dm⁻⁶.
At pH 8.1 (typical ocean surface), HCO₃⁻ is the most abundant inorganic carbon species.
This is because pH 8.1 sits between pK₁ and pK₂ of the carbonate system, where the bicarbonate ion dominates.
If pK₂ decreases as temperature increases, then K₂ increases with temperature.
By Le Chatelier's principle, an increase in temperature shifts equilibrium in the endothermic direction.
Since K₂ increases (equilibrium shifts to the right), the forward reaction is endothermic.
Students often confuse enantiomers with diastereomers. Enantiomers are mirror images; diastereomers are stereoisomers that are not mirror images. If a question says "pair of enantiomers," both molecules must have opposite configuration at every chiral centre.
A frequent error in SN2 mechanisms is forgetting to show inversion of configuration. The nucleophile must attack from the back side, and the product's stereochemistry flips at the carbon.
When comparing halide hydrolysis rates, students sometimes rank by electronegativity (Cl > Br > I) instead of bond strength (C–I < C–Br < C–Cl). The weakest bond breaks fastest, so iodoalkanes hydrolyse fastest.
In buffer questions, students often write that the buffer "prevents" pH change. It does not. It resists or minimises pH change. A buffer has a limited capacity and can be overwhelmed.
⚠️ Q4 is worth 18 marks, the highest in the paper. Stereoisomerism is clearly a priority topic. Practise drawing chiral centres and SN2 mechanisms until they are automatic.
⚠️ SN2 mechanism diagrams must include: curly arrows from the nucleophile lone pair to the electrophilic carbon, curly arrows from the C–X bond to the leaving group, a transition state with partial bonds, and clear indication of inversion.
⚠️ Halide hydrolysis with AgNO₃: you must state both the observation (precipitate colour and speed of formation) and the explanation (bond strength).
⚠️ EDTA titration calculations are multi-step. Write out moles clearly at each stage. A common mark-losing error is forgetting that EDTA reacts 1:1 with both Ca²⁺ and Mg²⁺.
⚠️ For the buffer question, examiners want you to write the specific equilibrium equations and show how the addition of H⁺ is countered by CO₃²⁻. A generic "buffer resists pH change" earns no marks without the chemistry.
⚠️ Deducing endothermic/exothermic from pK and temperature data is a 1-mark question that students often miss. Remember: K increases with temperature → forward reaction is endothermic.
True or false: Enantiomers have different melting points. (False – enantiomers have identical physical properties except for the direction of rotation of plane-polarised light.)
Fill in the blank: In an SN2 reaction, the nucleophile attacks from the ______ side relative to the leaving group. (opposite / back)
True or false: The C–I bond is stronger than the C–Cl bond. (False – C–I is weaker, which is why iodoalkanes hydrolyse fastest.)
Fill in the blank: At ocean pH 8.1, the most abundant inorganic carbon species is ______. (HCO₃⁻ / bicarbonate)
True or false: If Q < Ksp for a salt in solution, precipitation will occur. (False – precipitation occurs when Q > Ksp.)
Q: Draw the SN2 mechanism for the reaction of OH⁻(aq) with 2-bromooctane. State the stereochemical outcome.
A: OH⁻ attacks the carbon bearing the Br from the back side. A transition state forms with partial bonds to both OH and Br. Br departs as Br⁻. The configuration at the carbon is inverted (Walden inversion). The product is an alcohol with opposite spatial arrangement to the starting material.
Q: In a halide hydrolysis test with ethanolic AgNO₃, a cream precipitate forms more slowly than a yellow precipitate from another sample. Identify which halogenoalkane produces each precipitate.
A: The yellow precipitate is AgI, formed from the iodoalkane (fastest hydrolysis, weakest C–I bond). The cream precipitate is AgBr, formed from the bromoalkane. A white precipitate (AgCl) from the chloroalkane would form slowest.
Q: Explain why ethanoyl chloride reacts vigorously with cold water but chloroethane does not.
A: In ethanoyl chloride, the electron-withdrawing C=O group makes the carbon very electrophilic (strong δ⁺), so nucleophilic attack by water is rapid. In chloroethane, the carbon bonded to Cl is only weakly electrophilic, and there is no adjacent electron-withdrawing group to enhance its reactivity, so hydrolysis is very slow at room temperature.
Q: Write the equilibrium expression for the overall reaction CO₂(aq) + H₂O(l) ⇌ 2H⁺(aq) + CO₃²⁻(aq), in terms of K₁ and K₂.
A: K_overall = K₁ × K₂ = ([H⁺][HCO₃⁻] / [CO₂]) × ([H⁺][CO₃²⁻] / [HCO₃⁻]) = [H⁺]²[CO₃²⁻] / [CO₂]. The HCO₃⁻ terms cancel.
Q: The pK₂ for the bicarbonate equilibrium decreases as temperature rises. Is the forward reaction endothermic or exothermic?
A: Endothermic. A decrease in pK₂ means K₂ increases. By Le Chatelier's principle, increasing temperature favours the endothermic direction. Since K₂ increases, the forward reaction is endothermic.
Stereoisomerism (Q4) connects to Organic Mechanisms (SN1 vs SN2, especially in Paper 3) and to Transition Metals (ligand chirality in advanced contexts). The EDTA titration in Q5 connects to Transition Metals (complex formation, coordination number) and to Ionic Equilibrium (Ksp, precipitation). The CO₂ buffer system in Q6 ties directly to Ionic Equilibrium (buffer calculations, Ka expressions) and to Energetics (endothermic/exothermic deduction from K and temperature data).
Stereoisomers, chiral centre, enantiomers, optical activity, plane-polarised light, SN2 mechanism, Walden inversion, backside attack, 2-bromooctane, halogenoalkane hydrolysis, AgNO₃ test, AgCl white, AgBr cream, AgI yellow, ethanoyl chloride, acyl chloride hydrolysis, chlorobenzene, EDTA titration, complexometric titration, hexadentate ligand, Ksp, solubility product, ionic product Q, Ca(OH)₂, Mg(OH)₂, carbonate buffer, ocean acidification, CO₂ dissolving, Henry's law, partial pressure, pH calculation, K₁ K₂ equilibrium, HCO₃⁻ bicarbonate, pK and temperature, endothermic, H2 Chemistry, A-Level, NJC, 2024 Paper 2