Chemistry (Theory)

Polymers and proteins have very large molar masses, so a normal solution gives tiny ΔTb, ΔTf or vapour-pressure changes that are hard to measure.
Osmotic pressure is measurable even for very dilute solutions, is measured at room temperature (proteins are not denatured by heating) and uses molarity instead of molality.
Hence osmotic pressure is the method of choice.

By Henry's law, the solubility of a gas in a liquid is proportional to its partial pressure above the liquid (p = K_H·x).
At high altitude the atmospheric pressure, and so the partial pressure of O₂, is low.
Therefore less O₂ dissolves in blood → anoxia (weakness, inability to think clearly).

Zn is oxidised (Zn → Zn²⁺), so the zinc electrode is the anode and is written on the left as Zn | Zn²⁺.
Ag⁺ is reduced (Ag⁺ → Ag), so the silver electrode is the cathode, written on the right as Ag⁺ | Ag.
The salt bridge is shown by ||: Zn | Zn²⁺ || Ag⁺ | Ag.
Stoichiometric coefficients are never written in cell notation.

ΔG = −nFE_cell.
A spontaneous reaction has ΔG < 0; since n and F are positive, E_cell must be > 0.
So ΔG is negative and E_cell is positive.

A catalyst provides an alternative pathway with a lower activation energy (Ea).
ΔH, ΔG and ΔS are state functions that depend only on the initial and final states, so a catalyst does not change them (nor the equilibrium constant).

The photochemical combination of H₂ and Cl₂ (over water) is the standard textbook example of a zero-order reaction.
Its rate depends on the intensity of light absorbed, not on the concentrations of H₂ and Cl₂: rate = k.
(Note: the official key accepted any option for this question, but the textbook answer is zero order.)

Lanthanoids lose the two 6s electrons and one 4f/5d electron easily, so +3 is their characteristic oxidation state.
+2 (e.g. Eu²⁺, Yb²⁺) and +4 (e.g. Ce⁴⁺) appear only where they give empty, half-filled or full f subshells.

Both compounds have the same formula; the ligand inside the sphere and the counter-ion outside it are exchanged (SO₄²⁻ ↔ Br⁻).
So in water one gives Br⁻ ions and the other gives SO₄²⁻ ions → ionisation isomers.

In the Swarts reaction an alkyl chloride/bromide is heated with a metallic fluoride such as AgF, Hg₂F₂, CoF₂ or SbF₃: CH₃Br + AgF → CH₃F + AgBr.
Finkelstein gives alkyl iodides (NaI/acetone); Sandmeyer gives aryl halides from diazonium salts.

HCl/ZnCl₂ (Lucas reagent) reacts through a carbocation (SN1).
Carbocation stability: 3° > 2° > 1°, so tertiary alcohols react fastest (turbidity at once), secondary in about 5 minutes, primary only on heating.
Reactivity: 1° < 2° < 3°.

This is the Hoffmann bromamide degradation: RCONH₂ + Br₂ + 4NaOH → RNH₂ + Na₂CO₃ + 2NaBr + 2H₂O.
The amine formed has one carbon less than the amide, so acetamide (2 C) gives methanamine CH₃NH₂ (1 C).

In aniline the lone pair on N is delocalised into the benzene ring by resonance, so it is less available to accept a proton.
Alkylamines are stronger than NH₃ because the +I effect of alkyl groups increases electron density on N.
So aniline (pKb ≈ 9.4) is the least basic.

Amylose is an unbranched chain of α-D-glucose units joined by C1–C4 α-glycosidic bonds.
Amylopectin has the same C1–C4 α chain plus C1–C6 α links at the branch points.
Cellulose uses C1–C4 β links.
(Because of the printing error the official key gave full marks for any attempt.)

The α-helix is a secondary structure of proteins: the polypeptide chain coils into a right-handed helix held by hydrogen bonds between –NH and C=O groups of the peptide links (every 4th residue).

A is true: the lone pair of –NH₂ is donated into the ring (+R effect), raising electron density at o- and p-positions.
R is true: aniline (a Lewis base) forms a salt with the Lewis acid AlCl₃, and the N then carries a positive charge, deactivating the ring.
But R does not explain why –NH₂ is o/p-directing.

In acetanilide the N lone pair is also shared with the C=O group (resonance), so it is less available to the ring.
Hence –NHCOCH₃ is a weaker activator than –NH₂, and substitution can be controlled to give a single (para) product.
So A is true and R is false (the activating effect is LESS, not more).

Molecularity = number of reacting species that collide simultaneously in an elementary step.
As written, one H₂ and one Br₂ molecule collide, so the molecularity appears to be 2, and this is exactly what R states.
So the official answer is (a). (Remember: the experimental rate law of this reaction is complex, which is why A only says 'appears'.)

Δt ≈ (4/9)Δo, so the tetrahedral splitting is small and nearly always less than the pairing energy P.
Electrons therefore prefer to occupy the higher orbitals singly rather than pair up → high-spin complexes.
So low-spin tetrahedral complexes are rare, and R is the correct reason.

Henry's law: at constant temperature, the partial pressure of a gas in vapour phase is directly proportional to its mole fraction in the solution: p = K_H·x (K_H = Henry's law constant). Equivalently, solubility of a gas ∝ its pressure.
Applications (any one):
• Soft drinks/soda water are bottled under high pressure to increase CO₂ solubility.
• Scuba divers use air diluted with helium to reduce the 'bends' caused by N₂ dissolving in blood at high pressure.
• At high altitude, low O₂ partial pressure → less O₂ in blood (anoxia).

For a strong electrolyte, it is already almost fully ionised, so Λm rises only slightly on dilution (Λm = Λ°m − A√C, interionic attraction decreases). → B (1.5 times).
For a weak electrolyte, the degree of dissociation increases sharply on dilution, so Λm rises very steeply → A (25 times).
Graph: plot Λm (y-axis) against √C (x-axis). B gives a nearly straight line with small negative slope, which can be extrapolated to Λ°m at √C = 0. A gives a curve that stays low at higher √C and shoots up steeply as √C → 0 (cannot be extrapolated).

Area A = πr² = 3.14 × (0.5 cm)² = 0.785 cm²; l = 50 cm.
κ = l / (R × A) = 50 / (5.55 × 10³ × 0.785) S cm⁻¹
= 50 / 4356.75 = 1.147 × 10⁻² S cm⁻¹ (≈ 11.47 × 10⁻³ S cm⁻¹).
Common mistake: using diameter instead of radius in πr².

(a) In acid, MnO₄⁻ (Mn +7) is reduced to Mn²⁺, gaining 5e⁻ each; NO₂⁻ (N +3) is oxidised to NO₃⁻ (N +5), losing 2e⁻ each. 2 × 5 = 5 × 2 electrons balance.
2MnO₄⁻ + 5NO₂⁻ + 6H⁺ → 2Mn²⁺ + 5NO₃⁻ + 3H₂O
(b) Dichromate (Cr +6) gains 3e⁻ per Cr in acid:
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Check: O 7 = 7, H 14 = 14, charge (−2 + 14 − 6 = +6) = 2 × (+3).

(a) PCl₅ replaces –OH by –Cl: A = CH₃CHClCH₃ (2-chloropropane).
AgCN is mainly covalent, so attack occurs through N: B = (CH₃)₂CH–N≡C (isopropyl isocyanide).
(b) Alcoholic KOH causes dehydrohalogenation (β-elimination): A = CH₃–CH=CH₂ (propene).
HBr adds by Markovnikov's rule (Br to the carbon with fewer H): B = CH₃–CHBr–CH₃ (2-bromopropane).

(i) On losing H⁺, phenol forms phenoxide ion, whose negative charge is delocalised over the benzene ring (resonance) → stable ion → phenol ionises more readily. The alkoxide ion has no such delocalisation; the +I effect of the alkyl group even intensifies its charge.
(ii) Branching makes the molecule more compact (nearly spherical), so its surface area decreases. Van der Waals attractions decrease, so less energy is needed to boil it → lower boiling point.

(i) Step 1 – protonation of alcohol: CH₃CH₂–OH + H⁺ ⇌ CH₃CH₂–OH₂⁺ (fast).
Step 2 – loss of water forms carbocation (slow, rate-determining): CH₃CH₂–OH₂⁺ → CH₃CH₂⁺ + H₂O.
Step 3 – loss of a β-proton: CH₃CH₂⁺ → CH₂=CH₂ + H⁺.
(ii) C₆H₅OH + CHCl₃ + aq NaOH → (intermediate o-substituted benzal chloride) → sodium salicylaldehyde; acidification (H⁺) gives salicylaldehyde, o-HO–C₆H₄–CHO.

(a) A primary amine (aliphatic or aromatic) heated with chloroform and alcoholic KOH gives a foul-smelling isocyanide (carbylamine). It is a test for 1° amines; 2° and 3° amines do not respond.
R–NH₂ + CHCl₃ + 3KOH (alc.) → R–N≡C + 3KCl + 3H₂O
(b) Phthalimide + ethanolic KOH → potassium phthalimide; heated with an alkyl halide R–X → N-alkylphthalimide; alkaline hydrolysis (aq NaOH) → primary amine R–NH₂ + sodium phthalate.
It gives pure 1° aliphatic amines; aromatic amines cannot be made this way because aryl halides do not undergo nucleophilic substitution with the phthalimide anion.

(a) On prolonged heating with HI, glucose is reduced to n-hexane: C₆H₁₂O₆ —HI, Δ→ CH₃–CH₂–CH₂–CH₂–CH₂–CH₃. Since the product is a straight-chain alkane, all six carbons of glucose are in a straight chain.
(b) Peptide linkage – the amide bond –CO–NH– formed between –COOH of one amino acid and –NH₂ of the next, with loss of water.

(a) Ideal: obeys Raoult's law at all concentrations; ΔmixH = 0 and ΔmixV = 0; A–B interactions equal to A–A and B–B.
Non-ideal: does not obey Raoult's law (positive or negative deviation); ΔmixH ≠ 0, ΔmixV ≠ 0.
(b) n(urea) = 30/60 = 0.5 mol; n(water) = 846/18 = 47 mol.
x(water) = 47 / (47 + 0.5) = 47/47.5.
p = x(water) × p° = (47/47.5) × 23.8 = 23.55 ≈ 23.5 mm Hg.
(Relative lowering = 0.5/47.5 = 0.0105.)

(a) Finkelstein reaction: CH₃Cl + NaI → CH₃I + NaCl; NaCl is insoluble in acetone, which drives the reaction forward.
(b) Three –NO₂ groups at o/p positions strongly withdraw electrons, so the Cl is replaced by –OH just with warm water: product 2,4,6-trinitrophenol (picric acid).
(c) Dehydrohalogenation: CH₃CH₂CH₂CH₂Cl + KOH (alc.) → CH₃CH₂CH=CH₂ + KCl + H₂O (only one kind of β-H, so but-1-ene).

(a) Phenol + conc. HNO₃ (with conc. H₂SO₄) → 2,4,6-trinitrophenol (picric acid). (Industrially phenol is first sulphonated, then nitrated.)
(b) CH₃COCH₃ + CH₃MgBr (dry ether) → (CH₃)₃C–OMgBr —H₃O⁺→ (CH₃)₃C–OH (2-methylpropan-2-ol).
(c) C₆H₅OH + NaOH → C₆H₅ONa; C₆H₅ONa + CH₃I → C₆H₅OCH₃ (anisole) + NaI (Williamson synthesis).
(d) Hydroboration–oxidation: CH₃CH=CH₂ + (BH₃)₂ → (CH₃CH₂CH₂)₃B —H₂O₂/OH⁻→ CH₃CH₂CH₂OH (anti-Markovnikov product).
Mistake to avoid: acid-catalysed hydration of propene gives propan-2-ol, not propan-1-ol.

(i) –COOH withdraws electrons from the ring (−I and −R). The o- and p-positions become more electron-poor than the meta position, so electrophilic substitution occurs at meta.
(ii) Aldehydes/methyl ketones add NaHSO₃ to form crystalline addition products that can be separated from impurities; the adduct is then decomposed with dilute acid or alkali to regenerate the pure carbonyl compound.
(iii) In –COOH, the lone pair of the –OH oxygen is in resonance with C=O, so the carbonyl carbon is much less electrophilic; nucleophilic additions (e.g. with HCN, NH₂OH) typical of aldehydes/ketones do not occur.

Cu at 573 K dehydrogenates alcohols: 1° → aldehyde, 2° → ketone.
B does not reduce Fehling's → B is a ketone, so A is a 2° alcohol.
The only 2° alcohol C₃H₈O is CH₃CH(OH)CH₃ (A, propan-2-ol).
CH₃CH(OH)CH₃ —Cu, 573 K→ CH₃COCH₃ (B, acetone) + H₂.
CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃ (C, yellow ppt of iodoform) + CH₃COONa + 3NaI + 3H₂O.

(a) (i) Lactose (milk sugar) → β-D-galactose + β-D-glucose.
(ii) Maltose → 2 molecules of α-D-glucose.
(b) Starch is a polymer of α-D-glucose (amylose: C1–C4 α links; amylopectin: also C1–C6 α branches), whereas cellulose is a straight-chain polymer of β-D-glucose joined by C1–C4 β-glycosidic links.

(i) Average rate = change in concentration of a reactant or product divided by the time interval: r_av = −Δ[R]/Δt = +Δ[P]/Δt.
(ii) Any two: concentration of reactants, temperature, presence of a catalyst, surface area (and light for photochemical reactions).
(iii) (1) For a zero-order reaction rate = k[R]⁰ = k, so the rate is independent of concentration and remains constant until the reactant is used up.
(2) Units of k = units of rate = mol L⁻¹ s⁻¹.

(i) No AgCl precipitate means no ionisable Cl⁻; both Cl are inside the coordination sphere: [Pt(NH₃)₂Cl₂] (a neutral square-planar complex).
(ii) en (ethane-1,2-diamine) is bidentate; 3 en × 2 donor atoms = 6. Secondary valency (coordination number) = 6.
(iii) (1) Cation Fe³⁺, anion [Fe(CN)₆]⁴⁻; balance charges 4 × (+3) = 3 × (−4) → Fe₄[Fe(CN)₆]₃ (Prussian blue).
(2) [Co(NH₃)₅Cl]Cl₂: ligands alphabetically (ammine before chlorido), Co oxidation state x + 0 − 1 = +2 → x = +3. Name: pentaamminechloridocobalt(III) chloride.

(i) Kohlrausch's law: the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its cation and anion, each ion contributing independently.
Λ°m(CH₃COOH) = λ°(H⁺) + λ°(CH₃COO⁻).
(In practice: Λ°m(CH₃COOH) = Λ°m(CH₃COONa) + Λ°m(HCl) − Λ°m(NaCl).)
(ii) E°cell = E°cathode − E°anode = 0.80 − (−0.25) = 1.05 V; n = 2.
ΔG° = −nFE° = −2 × 96500 × 1.05 = −202650 J mol⁻¹.
Maximum work = −ΔG° = 202650 J mol⁻¹ = 202.65 kJ mol⁻¹.
log Kc = nE°/0.059 = (2 × 1.05)/0.059 = 2.10/0.059 ≈ 35.6.

(i) Faraday's first law: the mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed: m = Z·Q = Z·I·t.
Cu²⁺ + 2e⁻ → Cu, so 1 mol Cu²⁺ needs 2 mol electrons = 2 F (= 193000 C).
(ii) Cell reaction: Mg + Cu²⁺ → Mg²⁺ + Cu; n = 2.
E = E° − (0.059/2) log([Mg²⁺]/[Cu²⁺])
= 2.71 − 0.0295 × log(0.1/0.01)
= 2.71 − 0.0295 × log 10 = 2.71 − 0.0295
= 2.68 V.
Mistake to avoid: putting [Cu²⁺] in the numerator – products go on top.

(i) Mn is 3d⁵4s²; it can use all seven (3d + 4s) electrons in bonding (e.g. MnO₄⁻, Mn₂O₇), the largest number available in the series.
(ii) They show variable oxidation states (can form unstable intermediates), provide a large surface for adsorption, and form complexes; e.g. V₂O₅ in contact process, Fe in Haber process.
(iii) Cr²⁺ (d⁴) is oxidised to Cr³⁺ (d³ = half-filled t₂g³), which is stable → reducing agent. Mn³⁺ (d⁴) is reduced to Mn²⁺ (d⁵, half-filled), which is extra stable → oxidising agent.
(iv) Zn (3d¹⁰4s²) has no unpaired d electrons, so metallic bonding (interatomic attraction) is weakest → lowest enthalpy of atomisation.
(v) In water Cu⁺ disproportionates: 2Cu⁺(aq) → Cu²⁺(aq) + Cu(s), because the very high hydration enthalpy of Cu²⁺ more than compensates for the second ionisation enthalpy.

(i)(1) 2CH₃CHO —dil. NaOH→ CH₃–CH(OH)–CH₂–CHO (aldol) —heat, −H₂O→ CH₃–CH=CH–CHO (but-2-enal).
(2) Hell–Volhard–Zelinsky: CH₃CH₂COOH —(i) Cl₂/red P, (ii) H₂O→ CH₃CH(Cl)COOH (Cl enters at the α-carbon).
(ii) B gives Fehling's (an aldehyde) and iodoform (has CH₃CO– group) → B = CH₃CHO (ethanal, 2 C).
C is a ketone that gives iodoform; it must have 5 − 2 = 3 C → C = CH₃COCH₃ (propanone).
Joining the two carbonyl carbons with a C=C: A = (CH₃)₂C=CH–CH₃ (2-methylbut-2-ene).
Check: (CH₃)₂C=CHCH₃ —O₃, Zn/H₂O→ (CH₃)₂C=O + CH₃CHO.

(i)(1) Iodoform test: warm with I₂/NaOH. Butanone (has CH₃CO–) gives a yellow precipitate of CHI₃; butanal does not. (Tollens' test also works: butanal gives a silver mirror.)
(2) Add NaHCO₃ solution: ethanoic acid gives brisk effervescence of CO₂; ethanal does not. (Or Tollens'/Fehling's: only ethanal responds.)
(ii) Acetone oxime: (CH₃)₂C=N–OH, formed from (CH₃)₂C=O + NH₂OH with loss of H₂O.
(iii) A: CH₃COOH + PCl₅ → CH₃COCl (acetyl chloride).
B: Rosenmund reduction, CH₃COCl + H₂ (Pd/BaSO₄) → CH₃CHO (ethanal).
C: CH₃CHO + CH₃MgBr then H₃O⁺ → CH₃CH(OH)CH₃ (propan-2-ol).
D: CH₃CHO —LiAlH₄→ CH₃CH₂OH (ethanol).