Chemistry (Theory)

(i) Propanal to propane: reduce the C=O group to CH₂ by Clemmensen reduction.
CH₃CH₂CHO + 4[H] —Zn-Hg / conc. HCl→ CH₃CH₂CH₃ + H₂O
(Wolff–Kishner reduction with NH₂NH₂ then KOH/ethylene glycol, heat, is also accepted.)
(ii) Ethanal to but-2-enal: ethanal has α-hydrogens, so it undergoes aldol condensation.
2 CH₃CHO —dil. NaOH→ CH₃CH(OH)CH₂CHO (3-hydroxybutanal, the aldol)
CH₃CH(OH)CH₂CHO —heat, −H₂O→ CH₃CH=CHCHO (but-2-enal)
(iii) Ethanoic acid to ethanamide:
CH₃COOH + NH₃ → CH₃COO⁻NH₄⁺ (ammonium ethanoate)
CH₃COO⁻NH₄⁺ —heat→ CH₃CONH₂ + H₂O
(Alternative: CH₃COOH —SOCl₂→ CH₃COCl —NH₃→ CH₃CONH₂.)
Common mistake: in (ii) stopping at the aldol – the question asks for the unsaturated aldehyde, so dehydration on heating must be shown.

Rate of reaction = −Δ[N₂]/Δt = −(1/3) Δ[H₂]/Δt = +(1/2) Δ[NH₃]/Δt
(i) Rate = (1/2) × 3.6 × 10⁻⁴ = 1.8 × 10⁻⁴ mol L⁻¹ s⁻¹
(ii) −Δ[H₂]/Δt = 3 × rate = 3 × 1.8 × 10⁻⁴ = 5.4 × 10⁻⁴ mol L⁻¹ s⁻¹
Common mistake: forgetting to divide by the stoichiometric coefficient (2 for NH₃) when writing the rate of reaction.

(i) Kohlrausch's law: at infinite dilution, every ion migrates independently, so the limiting molar conductivity of an electrolyte equals the sum of the individual contributions of its cation and anion: Λ°m = ν₊λ°₊ + ν₋λ°₋.
(ii) Degree of dissociation α = Λm / Λ°m
α = 48 / 400 = 0.12 (i.e. 12% dissociated)

(i) [NiCl₄]²⁻: Ni²⁺ = 3d⁸. Cl⁻ is a weak-field ligand, so the 3d electrons are not paired; two unpaired electrons remain. The four Cl⁻ use 4s + three 4p orbitals → sp³ (tetrahedral), paramagnetic.
(ii) [Co(NH₃)₆]³⁺: Co³⁺ = 3d⁶. NH₃ is strong-field here, so all six d electrons pair up in three 3d orbitals, leaving two 3d orbitals empty → d²sp³ (inner-orbital, octahedral); no unpaired electrons → diamagnetic.
(iii) [FeF₆]³⁻: Fe³⁺ = 3d⁵. F⁻ is weak-field, the five electrons stay unpaired; 4s, 4p and two 4d orbitals are used → sp³d² (outer-orbital, octahedral), paramagnetic (5 unpaired electrons).

(i) Ligands in alphabetical order: ammine (×4 → tetraammine), aqua, chlorido. Charge on Co: x + 0 + 0 − 1 = +2 → x = +3.
Name: tetraammineaquachloridocobalt(III) chloride.
(ii) Ambidentate ligand: has two different donor atoms but attaches to the metal through only one of them at a time, e.g. NO₂⁻ (through N or O), SCN⁻ (through S or N).
Bidentate ligand: attaches to the metal through two donor atoms at the same time, e.g. ethane-1,2-diamine, C₂O₄²⁻.
(iii) [Fe(C₂O₄)₃]³⁻ is more stable. Oxalate is a bidentate (chelating) ligand that forms five-membered rings with Fe³⁺; chelate complexes are much more stable than complexes of comparable monodentate ligands (chelate effect). NH₃ is monodentate.

Cell reaction: Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s); n = 2
E°cell = E°cathode − E°anode = 0.80 − (−0.76) = 1.56 V
Nernst equation: Ecell = E°cell − (0.059/n) log ([Zn²⁺]/[Ag⁺]²)
[Zn²⁺]/[Ag⁺]² = 0.01 / (0.001)² = 10⁻² / 10⁻⁶ = 10⁴
Ecell = 1.56 − (0.059/2) × log 10⁴ = 1.56 − 0.0295 × 4 = 1.56 − 0.118
Ecell = 1.442 V
Common mistake: not squaring [Ag⁺] (coefficient 2 in the balanced equation) or using n = 1.

(i) –CHO is an electron-withdrawing, meta-directing group, so nitration gives m-nitrobenzaldehyde (3-nitrobenzaldehyde).
(ii) Heating a sodium carboxylate with soda lime removes CO₂ (decarboxylation): C₆H₁₁COONa + NaOH —CaO, Δ→ C₆H₁₂ (cyclohexane) + Na₂CO₃.
(iii) Aldehydes react with NH₂OH (addition–elimination, loss of H₂O): CH₃CHO + NH₂OH → CH₃CH=N–OH + H₂O; product is ethanal oxime (acetaldoxime).

(i) Propanal has an H atom on the carbonyl carbon; on oxidation only this C–H bond breaks to give propanoic acid. Propanone has no such H; oxidation requires breaking a stronger C–C bond, so it needs vigorous conditions.
(ii) Iodoform test: warm each with NaOH + I₂. Acetophenone (C₆H₅COCH₃, has a CH₃CO– group) gives a yellow precipitate of iodoform (CHI₃). Benzophenone (C₆H₅COC₆H₅) gives no precipitate.
(iii) Propanone + 2,4-dinitrophenylhydrazine → (CH₃)₂C=N–NH–C₆H₃(NO₂)₂ + H₂O, i.e. the C=O oxygen is replaced by =N–NH– attached to a benzene ring carrying NO₂ groups at positions 2 (ortho) and 4 (para) to the NH.

For first order: k = (2.303/t) log ([R]₀/[R])
75% complete → [R] = 25% of [R]₀, so [R]₀/[R] = 100/25 = 4
k = (2.303/30) × log 4 = (2.303/30) × 0.6 = 0.046 min⁻¹
t₁/₂ = 0.693/k = 0.693/0.046 ≈ 15 min
Shortcut check: 75% completion = two half-lives (100 → 50 → 25), so t₁/₂ = 30/2 = 15 min.
Common mistake: taking [R] = 75 instead of the amount left (25).

Any three of these differences earn the marks (1 mark each):
1. Affinity: in a lyophobic sol the dispersed particles have little attraction for the medium (solvent-hating); in a lyophilic sol they attract the medium strongly (solvent-loving).
2. Reversibility: a lyophobic sol, once precipitated, cannot be re-formed simply by adding the medium (irreversible); a lyophilic sol re-forms on adding the medium (reversible).
3. Stability: lyophobic sols are unstable and need stabilising ions; lyophilic sols are stable because the particles are solvated.
4. Coagulation: lyophobic sols coagulate easily with a small amount of electrolyte; lyophilic sols are difficult to coagulate.

(i) Milk is an emulsion: liquid fat droplets (dispersed phase = liquid) spread through water (dispersion medium = liquid).
(ii) Molecules of the dispersion medium keep striking a colloidal particle from all sides; the hits are not balanced at every instant, so the particle is pushed along a random zig-zag path – this is Brownian movement.
(iii) Physical adsorption is an exothermic process. By Le Chatelier's principle, raising the temperature shifts the equilibrium towards desorption; also the weak van der Waals attraction between adsorbate and adsorbent is easily broken by the extra thermal energy.

(i) Mn²⁺ is 3d⁵ – a stable half-filled configuration – so Mn readily forms Mn²⁺; in addition Mn has a low enthalpy of atomisation. Hence E° is highly negative.
(ii) E°(M²⁺/M) depends on enthalpy of atomisation + (IE₁ + IE₂) − hydration enthalpy. The sum of the first and second ionisation enthalpies and the sublimation enthalpies do not change regularly along the series (e.g. Mn and V have relatively low sublimation enthalpies), so E° values are irregular.
(iii) Converting Cu(s) to Cu²⁺(aq) needs a high enthalpy of atomisation plus a high IE₂; the hydration enthalpy of Cu²⁺ is not large enough to compensate. So Cu has a positive E° and does not liberate H₂ from acids.

(i) Rosenmund reduction: an acyl chloride is hydrogenated over palladium poisoned with BaSO₄ to give an aldehyde.
C₆H₅COCl + H₂ —Pd/BaSO₄→ C₆H₅CHO + HCl
(ii) Etard reaction: toluene is oxidised by chromyl chloride (in CS₂) to a chromium complex, which on hydrolysis gives benzaldehyde.
C₆H₅CH₃ + CrO₂Cl₂ (CS₂) → C₆H₅CH(OCrOHCl₂)₂ —H₃O⁺→ C₆H₅CHO
(iii) Stephen reaction: a nitrile is reduced by SnCl₂ + HCl to an imine (as its salt), which is hydrolysed to an aldehyde.
CH₃CN + SnCl₂ + HCl → CH₃CH=NH —H₃O⁺→ CH₃CHO
Tip: full marks need the equation with reagents/conditions, not just a description.

(i)(I) FCH₂COOH is stronger. Fluorine pulls electron density (–I effect), dispersing the negative charge of the carboxylate ion FCH₂COO⁻ and stabilising it.
(II) CH₃COOH is stronger. In the acetate ion the negative charge is shared equally by two electronegative O atoms (equivalent resonance structures); in phenoxide the charge is spread onto less electronegative ring carbons, so it is less stabilised.
(ii) Warm each with NaOH + I₂ (iodoform test). Pentan-2-one (CH₃COCH₂CH₂CH₃, has a CH₃CO– group) gives a yellow precipitate of iodoform; pentan-3-one (CH₃CH₂COCH₂CH₃) does not.

(i) Transition metals can show several oxidation states and form unstable intermediate compounds/complexes with reactants; they also provide a large surface area for adsorption, using (n−1)d and ns electrons for bonding. This lowers the activation energy (e.g. Fe in Haber process, V₂O₅ in contact process).
(ii) Zn, Cd, Hg have a completely filled (n−1)d¹⁰ configuration both in the ground state and in their common +2 ions, so they do not have partly filled d orbitals – they do not fit the definition of a transition element.
(iii) Hf (5d series) comes after the 14 lanthanoids. The 4f electrons shield the outer electrons poorly, so the size shrinks steadily (lanthanoid contraction); this cancels the expected increase in size from Zr to Hf, making their radii almost identical (≈160 and 159 pm).

(i) The catalyst AlCl₃ is a Lewis acid; aniline, a Lewis base, gives a salt with it. Nitrogen then carries a positive charge, which strongly deactivates the ring, so no Friedel–Crafts reaction occurs.
(ii) Lower pKb = stronger base. In water: C₂H₅NH₂ (pKb ≈ 3.29) > (CH₃)₃N (≈ 4.22) > NH₃ (4.75) > C₆H₅NH₂ (9.38) in basic strength. Increasing pKb: C₂H₅NH₂ < (CH₃)₃N < NH₃ < C₆H₅NH₂. (Gas-phase order (CH₃)₃N < C₂H₅NH₂ < NH₃ < C₆H₅NH₂ was also accepted since no medium was given.)
(iii) Treat both with benzenesulphonyl chloride (Hinsberg reagent). Ethylamine (1°) forms N-ethylbenzenesulphonamide, which still has an acidic H on N and dissolves in alkali. Diethylamine (2°) forms N,N-diethylbenzenesulphonamide with no H on N – it is insoluble in alkali.
(iv)(I) Sn/HCl reduces –NO₂ to –NH₂: A = aniline (C₆H₅NH₂). Bromine water brominates all o- and p-positions of the strongly activated ring: B = 2,4,6-tribromoaniline (white ppt).
(II) Hoffmann bromamide degradation removes one carbon: CH₃CH₂CONH₂ → A = ethanamine CH₃CH₂NH₂. Acetylation with CH₃COCl/pyridine gives B = N-ethylethanamide, CH₃CH₂NHCOCH₃.