CAPE Chemistry Unit 2 — Paper 2 Solutions

Structural isomerism is the phenomenon where compounds possess the same molecular formula but the atoms are bonded together (linked) differently — i.e. the compounds have different structural formulas.

1. Chain isomerism — isomers differ in the length / branching of the carbon chain.
2. Positional isomerism — isomers differ in the position of the functional group (or substituent / multiple bond) on the carbon skeleton.
3. Functional-group isomerism — isomers differ in the functional-group class (e.g. an alcohol vs an ether of the same molecular formula).

Chain isomerism — the components of the same molecular formula display different branched structures and different chain lengths. Commonly, chain isomers differ in the branching of the carbon backbone.

1. 2-methylheptane: CH₃—CH(CH₃)—CH₂—CH₂—CH₂—CH₂—CH₃
2. 3-methylheptane: CH₃—CH₂—CH(CH₃)—CH₂—CH₂—CH₂—CH₃
3. Other accepted answer: 4-methylheptane.

1. 2,3-dimethylhexane: CH₃—CH(CH₃)—CH(CH₃)—CH₂—CH₂—CH₃
2. 3,4-dimethylhexane: CH₃—CH₂—CH(CH₃)—CH(CH₃)—CH₂—CH₃
3. 2,3,4-trimethylpentane: CH₃—CH(CH₃)—CH(CH₃)—CH(CH₃)—CH₃

In each, every non-methyl carbon is bonded to two or three other carbons (no primary main-chain carbons aside from the methyls).

The final product (drawn in Figure 1) is a vicinal diol — two OH groups on adjacent carbons, each carbon also bearing methyl substituents. Working backwards through the scheme:

• Compound A — 2,3-dimethylbutane.  CH₃—CH(CH₃)—CH(CH₃)—CH₃. The Cl₂/hv step is free-radical halogenation, which substitutes H atoms at the two tertiary carbons in preference. (2 marks)
• Compound B — 2,3-dichloro-2,3-dimethylbutane.  CH₃—CCl(CH₃)—CCl(CH₃)—CH₃. (2 marks)
• Reagent X — NaOH(aq) (or NaOH dissolved in ethanol). The hydroxide-ion nucleophile substitutes both Cl atoms via S_N chemistry, giving the vicinal diol Compound C. (1 mark)

The mechanism is free-radical substitution initiated by UV light. The three named steps are:

1. Initiation — homolytic fission of the Cl–Cl bond by UV light to give two chlorine radicals.
2. Propagation — Cl• abstracts an H from the alkane (giving HCl + alkyl radical), then the alkyl radical reacts with another Cl₂ to give an alkyl chloride and a fresh Cl•.
3. Termination — two radicals combine to end the chain (e.g. Cl• + Cl• → Cl₂; R• + Cl• → R–Cl; R• + R• → R–R).

Initiation is homolytic fission: each electron of the Cl–Cl σ bond moves independently to one chlorine atom. This is shown with two single-headed (fish-hook) curved arrows.

    Cl ⤴⤵ Cl    ⟶   Cl• + Cl•   (UV light)

The two single-headed arrows start at the centre of the bond and curl outwards, one to each chlorine atom; the products are two neutral chlorine radicals each carrying one of the bonding electrons.

If the chlorine is in excess, more than two H atoms can be replaced — for example a tetrachlorinated derivative such as 2,2,3,3-tetrachloro-2,3-dimethylbutane or further-chlorinated by-products at the methyl carbons, e.g.:

Cl—CH₂—CCl(CH₃)—CCl(CH₃)—CH₂—Cl   (an isomeric tetrachloride)

Compound C is a tertiary diol — its two OH groups sit on tertiary carbons. Tertiary alcohols are weaker acids than primary alcohols (such as ethanol) because:

• The alkyl groups attached to the tertiary carbon are electron-donating (positive inductive effect).
• This destabilises the conjugate base (the alkoxide R–O⁻) by intensifying the negative charge on oxygen.
• A less stable conjugate base means a weaker acid.

Compound C therefore has a HIGHER pK_a than ethanol (pK_a > 16). The weaker the acid, the higher the pK_a.

The chlorination is a photochemical / free-radical reaction and only initiates when UV light homolytically cleaves Cl₂. In a closed vessel with no light there is no source of energy to start the reaction.

No colour change is expected — the yellow-green colour of Cl₂ remains because Cl₂ is not consumed (the reaction does not occur without light).

Both ethanol and Compound C are alcohols; both are extremely weak acids in water (slight ionisation only). Universal indicator therefore sits very close to neutral but slightly on the acidic side.

• Test tube 1 (ethanol): light yellow-green — pH is essentially neutral but very slightly acidic.
• Test tube 2 (Compound C): a slightly darker yellow-green — Compound C is a marginally weaker acid than ethanol (higher pK_a), so the ionisation is even less and the colour sits even closer to neutral / slightly green-tinged.

From Figure 2, where wavelength increases left to right (and frequency decreases):

• A: Radio waves (longest wavelength, lowest frequency)
• B: Infrared
• C: Visible
• D: X-rays (shortest wavelength, highest frequency)

Use c = λν  ⇒  λ = c ⁄ ν, with c = 3.0 × 10⁸ m s⁻¹:

λ = (3.0 × 10⁸) ⁄ (9.4 × 10¹⁴)

λ ≈ 3.19 × 10⁻⁷ m  (≈ 319 nm)

3.19 × 10⁻⁷ m = 319 nm sits just above the visible (~400 nm) region.

UV / Visible region (specifically near-UV, just outside the visible range).

Use Planck's relation, E = hν, with h = 6.63 × 10⁻³⁴ J s and ν = 9.4 × 10¹⁴ Hz:

E = (6.63 × 10⁻³⁴)(9.4 × 10¹⁴)

E ≈ 6.23 × 10⁻¹⁹ J

Option 1 is a back-titration: a known excess of NaOH is added (which reacts with the acetylsalicylic acid), and the unreacted NaOH is then titrated with a standard strong acid.

HCl(aq)  (or any strong acid such as dilute H₂SO₄).

This is a strong-base / strong-acid titration in reverse (acid added to base). Expected shape: pH starts very high (~13), drops slowly with each addition of HCl, falls sharply through pH 7 at the equivalence point, then levels off near pH 1.

This is a weak-acid / strong-base direct titration. Expected shape: pH starts low (~3), rises gently through a buffer region (with half-neutralisation = pK_a of aspirin ~3.5), then climbs sharply through pH ~8.7 at the equivalence point, before levelling off near pH 13.

The chosen indicator must change colour within the steep / vertical part of the titration curve (i.e. across the equivalence-point pH).

Option 1 — strong base + strong acid (steep section ≈ pH 3–11):

Cannot use Methyl violet (range 0.0–1.0). The pH never enters that range during the steep drop, so methyl violet would not register the end-point colour change.

Reason: the curve changes rapidly between pH ~11 and ~3, so any indicator that changes colour inside that window (methyl orange, bromothymol blue or phenolphthalein) is fine, but methyl violet's range is too acidic and will not distinguish the end-point.

Option 2 — weak acid + strong base (steep section ≈ pH 7–11):

Cannot use Methyl violet (0.0–1.0), Methyl orange (3.1–4.4) and Bromothymol blue (6.0–7.6). Only phenolphthalein (8.2–10.0) changes colour fully within the steep portion of the weak-acid / strong-base curve.

moles NaOH = c × V = 0.10 × (19.25 ÷ 1000) = 1.925 × 10⁻³ mol

Aspirin reacts with NaOH 1 : 1 (one −COOH per molecule), so:

moles aspirin = 1.925 × 10⁻³ mol

mass aspirin = n × M_r = 1.925 × 10⁻³ × 180.06 = 0.347 g

≈ 347 mg of aspirin in the 25.0 cm³ aliquot

The 250 cm³ stock contained 10 tablets, so a 25.0 cm³ aliquot contains 1 tablet's worth of aspirin. The titration calculation gave 347 mg per tablet, which is above the acceptable upper limit of 315 mg.

The information in Table 3 is NOT supported by the titration result.

Reason: the 347 mg figure exceeds the upper bound of 315 mg per tablet. The discrepancy is most likely due to experimental error introduced when grinding the tablets and dissolving them — aspirin is poorly soluble in cold water and may not have completely dissolved, so the back-titration / direct-titration data over-reads. (Other accepted reasons: indicator end-point error, inaccurate volumetric work, inactive binders treated as active ingredient.)

• A — N₂(g)  (nitrogen, fed in from the air separation unit)
• B — H₂(g)  (hydrogen, fed in from steam-reforming of natural gas)
• C — unreacted N₂(g) and H₂(g)  (recycled back into the converter)
• D — liquid ammonia, NH₃(l)  (the product, stored / shipped under refrigeration)

N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g)    ΔH = −92 kJ mol⁻¹

Vessel E is the catalytic converter where the equilibrium reaction takes place over an iron catalyst at high pressure and temperature.

Vessel F cools / compresses the gas mixture so that ammonia condenses out and is separated from the unreacted N₂ and H₂:

NH₃(g) → NH₃(l)

Le Chatelier's principle: if a system at equilibrium is subjected to a change in conditions (concentration, pressure or temperature), the equilibrium will shift in the direction that opposes the change and re-establishes equilibrium.

Explanation for high pressure favouring NH₃ formation:

• The forward reaction goes from 4 moles of gas (1 mol N₂ + 3 mol H₂) to 2 moles of gas (2 NH₃).
• Increasing the total pressure forces the system to relieve the extra pressure by shifting toward the side with fewer gas molecules — the product side.
• The yield of ammonia therefore increases at higher pressures (typically operated at ~200 atm).

The very strong N≡N triple bond in N₂ and the H–H bond in H₂ are broken to form 6 strong N–H bonds in 2 NH₃. The total energy released by formation of the six N–H bonds is greater than the total energy absorbed in breaking N≡N and 3 H–H bonds — so the reaction is exothermic (ΔH = −92 kJ mol⁻¹).

• Pressure: ~200 atm
• Temperature: ~450 °C  (a compromise — high enough for a useable rate, low enough that yield is not too suppressed)
• Catalyst: finely-divided iron (often promoted with iron(III) oxide / aluminium oxide / potassium carbonate)

• Pressure 200 atm: compresses the gas mixture so reactant molecules are much closer together — there are more frequent collisions per unit volume.
• Temperature 450 °C: molecules have greater kinetic energy and move faster, colliding more often and with a higher fraction of collisions exceeding the activation energy.
• Iron catalyst: provides an alternative reaction pathway with a lower activation energy, so a much larger fraction of collisions are successful at a given temperature.

1. Steam reforming of natural gas to produce H₂:  CH₄(g) + H₂O(g) → CO(g) + 3 H₂(g)  and  CO(g) + H₂O(g) → CO₂(g) + H₂(g)  (water-gas shift). The CO₂ released here is the largest single source.
2. Combustion of fossil fuels to provide the high-temperature, high-pressure energy needed to compress and heat the gases through the converter — releases additional CO₂.

1. Effect 1 — too low a temperature: the reaction would be far too slow to be commercially viable. Low T also disfavours collisions exceeding the activation energy.
2. Effect 2 — too low a pressure: by Le Chatelier's principle the equilibrium shifts to the reactant side (more gas moles), significantly reducing the yield of ammonia.

1. Eutrophication. Excess ammonia / ammonium fertiliser runs off agricultural land into water bodies; it is converted by bacteria to nitrate, which fuels algal blooms, depletes dissolved oxygen and kills aquatic life.
2. Smog formation / particulate matter. Ammonia in the atmosphere reacts with NO_x and SO₂ to form ammonium-nitrate and ammonium-sulfate fine particulates (PM_2.5) — a major contributor to urban smog and respiratory disease.

(i) State the reason for this. (1 mark)

If used in its aqueous (or gaseous) form, ammonia would readily evaporate from the soil and would not be very effective as a fertiliser. A solid ammonium salt fixes the N in a form that the plant can take up over time without volatile losses.

(ii) Give the formula of any ONE of the most commonly used ammonium salts in fertilisers. (2 marks)

Ammonium nitrate, NH₄NO₃  (34 % N by mass — one of the highest-N solid fertilisers).

Other accepted answers: ammonium sulfate (NH₄)₂SO₄; ammonium phosphate (NH₄)₃PO₄; urea CO(NH₂)₂.