CAPE Chemistry Unit 2 — Paper 2 Solutions

• Compound A: 2,2,3-trimethylbutane — branched C₇H₁₆.
• Compound B: heptane (n-heptane) — straight-chain C₇H₁₆.

Structural isomerism is the phenomenon where compounds have the same molecular formula (i.e. the same number and type of atoms) but differ in how the atoms are linked or connected.

Both A and B share the molecular formula C₇H₁₆, but the seven carbons are joined in different ways.

Chain isomerism — A and B have the same molecular formula but different lengths/branching of the carbon chain.

Type 1 — Positional isomerism. Same molecular formula and same functional group, but the functional group, substituent or multiple bond is in a different position on the carbon chain. Example: propan-1-ol vs propan-2-ol — both C₃H₈O alcohols, OH on different carbons.

CH₃—CH₂—CH₂—OH   and   CH₃—CH(OH)—CH₃

Type 2 — Functional-group isomerism. Same molecular formula, but the compounds belong to different functional-group classes. Example: ethanol (an alcohol) vs methoxymethane (an ether) — both C₂H₆O.

CH₃—CH₂—OH   and   CH₃—O—CH₃

(i) Complete Table 1 — observations and inferences. (3 marks)

(ii) Predict the functional groups for R and S. (2 marks)

Product R: carboxylic acid (—COOH).    Product S: ketone (C=O within a chain).

(iii) Draw the displayed structure of either R or S. (1 mark)

Product S — propanone:

         O
         ‖
  H3C — C — CH3           (propanone, S)

(Acceptable alternative — Product R, ethanoic acid: CH₃—COOH.)

(iv) Name and draw the displayed structure of the C₅H₁₀ hydrocarbon that was oxidised. (2 marks)

Hot acidified KMnO₄ cleaves the C=C bond of an alkene. The carbon of the alkene that bears no H atoms becomes a ketone; the carbon that bears one H atom becomes a carboxylic acid. To make propanone (no H on its α-carbon) and ethanoic acid (one H on its α-carbon), the alkene must be:

2-methylbut-2-ene:    (CH₃)₂C=CH—CH₃

     CH3          H
       \\         /
        C   =   C
       /         \\
     CH3          CH3     (2-methylbut-2-ene, C5H10)

(i) Deduce the monomer used to form the polymer in Figure 2. (1 mark)

The monomer is phenylethene (styrene), C₆H₅—CH=CH₂.

(ii) Using Figure 2 and your response to (c)(i), explain how addition polymers are formed. (2 marks)

Addition polymerisation occurs when many unsaturated monomer molecules (in this case phenylethene) join by the cleavage of the C=C double bond. The double bond breaks and forms new single covalent bonds, linking many monomer units into a long chain to give poly(phenylethene). No small molecules are eliminated.

(iii) Identify the reagent and describe the colour changes used to distinguish the polymer from the monomer. (3 marks)

Use bromine water (Br₂(aq)), the standard test for the C=C of an alkene.

• Monomer (phenylethene): the bromine water is decolourised — brown / orange-yellow → colourless. (Br₂ adds across the C=C.)
• Polymer (poly(phenylethene)): no observable change — the bromine water remains brown / orange-yellow because the polymer has no C=C bonds left.

(iv) State TWO characteristics of condensation polymers. (2 marks)

1. They are formed from monomers that contain two functional groups (bifunctional monomers).
2. Their formation involves the elimination of a small molecule (usually H₂O or HCl) at each linkage.
3. Other accepted: they can be hydrolysed back to the monomers; they typically have ester or amide linkages.

(v) List THREE examples of condensation polymers. (3 marks)

1. Polyesters — for example terylene (PET).
2. Polyamides — for example nylon-6,6.
3. Proteins / polypeptides — for example keratin.

The spectrum is classified by wavelength (or frequency). Shorter wavelength ⇒ higher frequency ⇒ higher energy.

Apply Planck's relation, E = hν, with h = 6.63 × 10⁻³⁴ J s.

E = (6.63 × 10⁻³⁴)(1.01 × 10¹⁶)

E ≈ 6.70 × 10⁻¹⁸ J

The three transitions accessible from low-lying filled levels to anti-bonding orbitals at UV/VIS energies are:

1. π → π*  (from the bonding π level to the anti-bonding π*)
2. n → π*  (from a non-bonding lone pair to the anti-bonding π*)
3. n → σ*  (from a non-bonding lone pair to the anti-bonding σ*)

The σ → σ* transition needs much more energy and lies in the far-UV — outside the normal UV/VIS region.

A pale-blue precipitate of Cu(OH)₂ may form first; it then dissolves in excess ammonia.

A deep-blue solution is formed — due to the tetraamminediaquacopper(II) complex [Cu(NH₃)₄(H₂O)₂]²⁺.

615 nm is the wavelength of maximum absorbance, λ_max, for the deep-blue copper-ammonia complex. Working at λ_max gives the largest absorbance signal and therefore the most sensitive and accurate measurement. (The deep-blue complex absorbs strongly in the orange-red region near 615 nm.)

1. Prepare several standard solutions of Cu²⁺ of known concentrations.
2. Add the same excess of ammonia to each standard (in the same way as for the unknown).
3. Measure the absorbance of each standard at 615 nm using the same cuvette.
4. Plot a graph of absorbance (y-axis) against concentration (x-axis) — this is the calibration curve.
5. Measure the absorbance of the unknown sample under the same conditions.
6. Read the corresponding concentration off the calibration curve (or use the line of best fit).

Beer–Lambert's law guarantees absorbance is proportional to concentration over the linear range, so the curve allows the unknown concentration to be read off directly.

Given concentration: 2.5 × 10⁻³ g dm⁻³. Convert g/dm³ → mol/dm³ by dividing by the molar mass of copper (M_r(Cu) = 63.5 g mol⁻¹).

c = 2.5 × 10⁻³ ÷ 63.5

c ≈ 3.94 × 10⁻⁵ mol dm⁻³

Beer–Lambert: A = εcℓ   ⇒   ε = A ⁄ (cℓ)

With ℓ = 1 cm and c = 3.94 × 10⁻⁵ mol dm⁻³:

ε = A ⁄ (3.94 × 10⁻⁵ × 1) = (A ÷ 3.94 × 10⁻⁵) dm³ mol⁻¹ cm⁻¹

ε = A / (3.94 × 10⁻⁵) dm³ mol⁻¹ cm⁻¹  (insert the recorded absorbance reading from the spectrophotometer to obtain the numerical value).

Both samples were measured under identical conditions (same ε, same ℓ = 1 cm), so concentration is directly proportional to absorbance:

c_unknown ⁄ c_standard = A_unknown ⁄ A_standard

c_unknown = (0.350 / A_standard) × 3.94 × 10⁻⁵ mol dm⁻³.

(Plug in the recorded standard absorbance to obtain the numerical answer.)

1. Prepare the sample in a suitable solvent. If necessary, add a reagent to form a coloured species (so that the analyte absorbs strongly in the UV/VIS).
2. Select an appropriate wavelength for measurement — usually λ_max, the wavelength of maximum absorbance.
3. Fill a clean cuvette with a blank (solvent only) and zero the spectrophotometer.
4. Place the sample in the cuvette and measure its absorbance at λ_max.
5. Compare the absorbance with standards (calibration curve) or use the Beer–Lambert relation A = εcℓ to calculate the concentration.

The blank corrects for any absorption by the solvent / cuvette, so the recorded absorbance is due only to the analyte.

• Molecules absorb infrared radiation when the radiation frequency matches the natural vibrational frequency of one of their bonds.
• The absorbed energy makes the bonds undergo stretching and bending vibrations. Different bonds absorb at characteristic wavenumbers, so IR spectra can identify functional groups.
• For a vibration to be IR-active, it must produce a change in dipole moment.

Reading from a typical IR correlation table:

• 1695 cm⁻¹  ⇒  C=O stretch (carbonyl group, conjugated; pure C=O is ~1715, the lower 1695 value indicates conjugation with C=C).
• 1619 cm⁻¹  ⇒  C=C stretch (alkene).

This is consistent with the structure of Compound D (an α,β-unsaturated ketone, ethyl vinyl ketone — pent-1-en-3-one).

1. Stretching vibrations — the bond length oscillates (symmetric or asymmetric stretching of, e.g., C=O and C=C).
2. Bending vibrations — the bond angle oscillates (scissoring, rocking, wagging, twisting of C–H bonds).

1. Availability of raw materials nearby.
2. Availability of energy / power supply.
3. Access to transport (roads, rail, ports for export of product).
4. Availability of a reliable water supply.
5. Other accepted answers: availability of skilled labour, distance from markets, environmental and safety regulations, waste-disposal facilities.

Step 1 — Formation of sulfur dioxide. Sulfur (or sulfide ores) is burned in air:

S(s) + O₂(g) → SO₂(g)

or, by roasting zinc-sulfide ore:

2 ZnS(s) + 3 O₂(g) → 2 ZnO(s) + 2 SO₂(g)

Step 2 — Catalytic oxidation of SO₂ to SO₃:

2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g)

Conditions: V₂O₅ catalyst, ~450 °C, ~1–2 atm pressure.

Sulfur dioxide (SO₂) is used as a preservative in foods such as dried fruits, fruit juices and wines (it inhibits microbial growth and prevents browning).

1. Acid rain. SO₂ reacts with moisture in the atmosphere to form sulfurous / sulfuric acid (SO₂ + H₂O → H₂SO₃; further oxidation gives H₂SO₄) which damages buildings, vegetation and aquatic ecosystems.
2. Respiratory irritation. SO₂ causes respiratory problems and irritates the eyes, throat and lungs in humans.

This is the Bayer process. Bauxite ore contains aluminium oxide together with impurities such as iron(III) oxide and silicon dioxide.

Step 1 — Dissolve the aluminium oxide in hot, concentrated NaOH:

Al₂O₃(s) + 2 NaOH(aq) + 3 H₂O(l) → 2 NaAl(OH)₄(aq)

The aluminium oxide dissolves as sodium aluminate; impurities such as Fe₂O₃ remain as solid "red mud" and are filtered off.

Step 2 — Cool / seed the solution to precipitate Al(OH)₃:

NaAl(OH)₄(aq) → Al(OH)₃(s) + NaOH(aq)

Step 3 — Heat (calcine) the aluminium hydroxide to give pure alumina:

2 Al(OH)₃(s) → Al₂O₃(s) + 3 H₂O(g)

The pure Al₂O₃ is dissolved in molten cryolite (Na₃AlF₆) and electrolysed using carbon electrodes.

At the cathode — Al³⁺ ions are reduced to molten aluminium metal:

Al³⁺ + 3 e⁻ → Al(l)

At the anode — O²⁻ ions are oxidised:

2 O²⁻ → O₂(g) + 4 e⁻

The liberated oxygen reacts with the hot carbon anode (which is consumed and must be replaced):

C(s) + O₂(g) → CO₂(g)

Overall: 2 Al₂O₃ + 3 C → 4 Al + 3 CO₂

The Bayer process produces large volumes of "red mud" — a highly alkaline solid waste rich in iron oxide, residual NaOH and trace heavy metals. If not properly contained, red mud can leach into surrounding soil and waterways, causing severe alkaline contamination of land and freshwater systems.

Other accepted answers: dust and red-mud spills smother vegetation; CO₂ emissions from the carbon anodes contribute to climate change; high energy demand of smelting drives fossil-fuel use.

At the high temperatures inside a combustion engine, atmospheric nitrogen and oxygen react:

N₂(g) + O₂(g) → 2 NO(g)

Once in the exhaust, NO is further oxidised in air:

2 NO(g) + O₂(g) → 2 NO₂(g)

Add a small piece of aluminium foil (or Devarda's alloy) and aqueous NaOH to the unknown solution; warm gently. If nitrate ions are present, the aluminium reduces them and ammonia gas (NH₃) is produced.

Test the gas above the mixture with damp red litmus paper — it will turn blue, confirming NH₃ and therefore the presence of nitrate ions.

The school-lab test is NOT suitable for determining nitrate as a pollutant.

Reason: the school test is purely qualitative — it shows only whether nitrate is present or absent. It does not measure the concentration of nitrate. To monitor pollution levels you need a quantitative method such as the cadmium-reduction method (or colorimetric / ion-chromatography techniques) which gives an actual numerical concentration.

Ammonium molybdate, (NH₄)₆Mo₇O₂₄, in acidic conditions. A yellow precipitate (ammonium phosphomolybdate) confirms the presence of phosphate ions.