Chemistry (Theory)

Nernst equation: \( E = E^\circ - \frac{0.059}{2}\log\frac{[Zn^{2+}]}{[Cu^{+}]^2} \).
As the cell runs, Zn²⁺ is produced (numerator grows) and Cu⁺ is used up (denominator shrinks).
So the log term grows and E falls; at equilibrium E becomes zero.
Mistake to avoid: [Cu⁺] decreases during the reaction, it does not increase.

Colour in transition-metal ions comes from d–d transitions, which need partly filled d orbitals.
Sc³⁺ = [Ar]3d⁰ – no d electron, so no d–d transition → colourless.
Fe³⁺ (3d⁵), Cr³⁺ (3d³) and Co³⁺ (3d⁶) all have unpaired d electrons and are coloured.

Hoffmann bromamide degradation needs a primary amide: RCONH₂ + Br₂ + 4NaOH → RNH₂ + Na₂CO₃ + 2NaBr + 2H₂O.
So ArCONH₂ (an aromatic amide) gives ArNH₂, an amine with one carbon less.

The anomeric carbon is the former carbonyl carbon that becomes a new chiral centre on ring closure (the hemiacetal carbon).
In an aldohexose like glucose this is C-1: it is bonded to both the ring oxygen and an OH group.
The α and β forms (anomers) differ only in the OH position at this carbon.

Henry's law: p = K_H·x, so x = p/K_H.
At the same partial pressure, a larger K_H means a smaller mole fraction x in solution, i.e. lower solubility.

(A) La (5d¹6s²) is the first member of the 5d transition series – true.
(B) Lanthanoid contraction makes Hf almost the same size as Zr – true.
(C) Ionic radius falls steadily from La³⁺ to Lu³⁺ – true.
(D) Only promethium (Pm) is radioactive; lanthanoids in general are not – false.

A chiral molecule needs a carbon with four different groups.
2-Bromobutane: C-2 carries H, Br, CH₃ and C₂H₅ – four different groups → chiral.
2-Bromopropane: C-2 has two CH₃ groups → not chiral. The 1-bromo compounds have no such carbon.

Step 1 (diazotisation): C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ (X) + NaCl + 2H₂O.
Step 2: ethanol (a mild reducing agent) replaces –N₂⁺ by –H: C₆H₅N₂⁺Cl⁻ + CH₃CH₂OH → C₆H₆ + N₂ + CH₃CHO + HCl.
So Y = benzene. (Warming X with water would give phenol instead.)

In neutral or faintly alkaline solution permanganate oxidises iodide to iodate:
2MnO₄⁻ + H₂O + I⁻ → 2MnO₂ + 2OH⁻ + IO₃⁻.
In acidic medium the product would be I₂ instead.

DDT (p,p'-dichlorodiphenyltrichloroethane) is an insecticide.
It is chemically very stable and fat-soluble, so it is not broken down and builds up in food chains – hence non-biodegradable.

Oxidation state of Mn changes from +7 (MnO₄⁻) to +2 (Mn²⁺): 5 electrons per ion.
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
1 mol MnO₄⁻ needs 5 mol electrons = 5 F.

Mechanism of acid-catalysed dehydration:
Step 1: the –OH oxygen takes a proton → protonated alcohol (R–OH₂⁺).
Step 2: water leaves → carbocation (slow step).
Step 3: a β-hydrogen is lost as H⁺ → alkene.

A is true: H₂ forms at the cathode, Cl₂ at the anode.
R is false: thermodynamically water is easier to oxidise (E° for O₂ formation 1.23 V vs 1.36 V for Cl₂). Cl₂ is liberated only because oxygen evolution has a high overpotential.

Cu⁺ = [Ar]3d¹⁰ – all d electrons paired, no unpaired electron.
No unpaired electrons → diamagnetic. So R correctly explains A.

For the same alkyl group, boiling point rises with the size and mass of the halogen: RI > RBr > RCl > RF, because van der Waals forces grow. So n-butyl bromide boils higher – A is false.
R is true: Cl is more electronegative than Br, so C–Cl is more polar.

In acetanilide (C₆H₅NHCOCH₃) the nitrogen lone pair is delocalised onto the C=O group (resonance with the acetyl group).
So less electron density is available on N to accept a proton → weaker base than aniline. R is the reason for A.

[A] halves every hour: 0.40 → 0.20 → 0.10 → 0.05 M.
A half-life that stays constant (1 h) and does not depend on the starting concentration is the mark of a first-order reaction.
Rate law: Rate = k[A].
k = 0.693/t½ = 0.693/1 h = 0.693 h⁻¹.

(i) Adding H₂ raises [H₂] in the fixed volume; since rate ∝ [H₂], the rate increases.
(ii) The rate constant depends only on temperature (and catalyst), not on concentration. Temperature is constant, so k does not change.

No AgCl precipitate → no free Cl⁻ ions; all six Cl are bound to Pt inside the coordination sphere.
Formula: K₂[PtCl₆] (ionises into 2K⁺ and [PtCl₆]²⁻).
Oxidation state of Pt: x − 6 = −2 → x = +4.
IUPAC name: potassium hexachloridoplatinate(IV).

Definition: a fuel cell is a galvanic cell in which the energy of combustion of fuels such as hydrogen or methane is converted directly into electrical energy, the reactants being supplied continuously.
Example: H₂–O₂ cell, 2H₂ + O₂ → 2H₂O.
Advantages: (1) high efficiency (about 70%, vs about 40% for thermal power plants); (2) pollution-free – product is water; also it runs continuously as long as fuel is fed.

(a) NaBH₄ is a mild reducing agent: it reduces the ketone C=O to CH–OH but does not reduce the ester group. Product: 2-hydroxycyclohexyl ring with the –CH₂CH₂COOCH₃ side chain unchanged.
(b) Alcohols are acidic enough to react with active metals, releasing hydrogen: 6(CH₃)₃COH + 2Al → 2Al[OC(CH₃)₃]₃ (aluminium tert-butoxide) + 3H₂.

Essential amino acids: those that cannot be synthesised in the body and must be taken in through food (e.g. valine, leucine, lysine).
Amphoteric nature: an amino acid contains an acidic –COOH group and a basic –NH₂ group; in solution it exists as a zwitter ion H₃N⁺–CHR–COO⁻, so it reacts with both acids (COO⁻ takes H⁺) and bases (NH₃⁺ gives H⁺).

(i) Allyl chloride ionises to the allyl carbocation CH₂=CH–CH₂⁺, which is stabilised by resonance; n-propyl carbocation has no such stabilisation. So allyl chloride hydrolyses more readily.
(ii) AgCN is mainly covalent: the carbon is tied to Ag, so the lone pair on nitrogen attacks the alkyl carbon → R–N≡C (isocyanide). (KCN is ionic and gives nitriles.)
(iii) In SN2 the nucleophile attacks from the back side. In CH₃Cl the carbon is unhindered; in (CH₃)₃CCl three bulky methyl groups block the approach, so CH₃Cl reacts faster.

(i) With peroxide, HBr adds anti-Markovnikov (Kharasch effect): A = 1-bromopropane CH₃CH₂CH₂Br. Aqueous KOH substitutes Br by OH: B = propan-1-ol.
(ii) Alcoholic KOH causes dehydrohalogenation; by Saytzeff's rule the more substituted alkene forms: A = but-2-ene CH₃CH=CHCH₃. HBr adds to give B = 2-bromobutane CH₃CH₂CHBrCH₃.
(iii) Grignard reagent CH₃CH₂MgCl comes from A = chloroethane CH₃CH₂Cl. Water protonates the carbanion part: B = ethane CH₃CH₃ (+ Mg(OH)Cl).

Al has the lower E°, so Al is the anode: 2Al + 3Ni²⁺ → 2Al³⁺ + 3Ni; n = 6.
E°cell = E°cathode − E°anode = −0.25 − (−1.66) = 1.41 V.
Nernst: \( E = E^\circ - \frac{0.059}{6}\log\frac{[Al^{3+}]^2}{[Ni^{2+}]^3} \).
Q = (10⁻³)²/(10⁻¹)³ = 10⁻⁶/10⁻³ = 10⁻³, log Q = −3.
E = 1.41 − (0.059/6)(−3) = 1.41 + 0.0295 = 1.4395 ≈ 1.44 V.
Mistake to avoid: use n = 6 (LCM of 2 and 3) and raise concentrations to their coefficients.

(a) Mn³⁺ (3d⁴) gains an electron to become Mn²⁺ (3d⁵), an extra-stable half-filled configuration → it is easily reduced, so it oxidises others. Cr²⁺ (3d⁴) loses an electron to become Cr³⁺ (3d³ = half-filled t₂g³), which is very stable → it reduces others.
(b) 5f orbitals are more diffuse and shield the outer electrons more poorly than 4f orbitals, so the effective nuclear charge rises more per element → larger contraction in actinoids.
(c) H, B, C and N atoms are very small and get trapped in the empty spaces (interstitial sites) of the metal crystal lattice, forming interstitial compounds like TiC, Fe₃H.

Normal molar mass of NaOH = 23 + 16 + 1 = 40 g mol⁻¹.
Van't Hoff factor i = normal molar mass / observed molar mass = 40/25 = 1.6.
NaOH → Na⁺ + OH⁻, so n = 2 ions: α = (i − 1)/(n − 1) = (1.6 − 1)/(2 − 1) = 0.6.
Percentage dissociation = 60%.

(a) Higher pKb = weaker base. Aryl amines are weak (lone pair delocalised in ring); N-methyl group makes C₆H₅NHCH₃ slightly stronger than aniline. Aliphatic amines are stronger; in water (C₂H₅)₂NH is the strongest. So pKb: C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂NH.
(b) Secondary amine has one N–H (weaker H-bonding) < primary amine with two N–H < alcohol (O–H bond more polar, stronger H-bonds).
(c) Solubility depends on H-bonding with water versus the size of the hydrocarbon part: aniline (large phenyl) least; diethylamine (bigger alkyl, one N–H) next; ethylamine most soluble.

Positive 2,4-DNP → carbonyl compound. No Tollens'/Fehling's → not an aldehyde, so a ketone.
Iodoform (yellow ppt) with NaOI → contains a CH₃CO– group.
C₈H₈O aromatic ketone with CH₃CO– = acetophenone C₆H₅COCH₃ (A), IUPAC: 1-phenylethan-1-one.
C₆H₅COCH₃ + 3NaOI → C₆H₅COONa + CHI₃↓ (B = triiodomethane, iodoform) + 2NaOH.
Vigorous oxidation removes the side chain to –COOH: C = benzoic acid, C₆H₅COOH (IUPAC: benzenecarboxylic acid).

(a) No. In Williamson synthesis the halide must be primary. Ethoxide is a strong base as well as a nucleophile; with tertiary t-butyl chloride it removes a β-H, so elimination wins and 2-methylpropene forms instead of the ether.
Correct route: (CH₃)₃CONa + CH₃CH₂Cl → (CH₃)₃C–O–CH₂CH₃ + NaCl (SN2 on the primary halide).
(b) (CH₃)₃C–O–C₂H₅ = 2-ethoxy-2-methylpropane.

(a) Example: chloroform + acetone (also HNO₃ + water). Reason: new A–B attractions (here H-bond between CHCl₃ and the C=O of acetone) are stronger than A–A and B–B, so molecules escape less and vapour pressure is lower than predicted.
(b) Raoult's law: in a solution of volatile liquids, the partial vapour pressure of each component equals its mole fraction times the vapour pressure of the pure component, p₁ = p₁°x₁.
(c) Ideal solution: ΔmixH = 0, ΔmixV = 0 (and it obeys Raoult's law at all concentrations).

(a) Complete hydrolysis of DNA gives 2-deoxy-D-ribose, phosphoric acid and the bases adenine, guanine, cytosine and thymine.
Structural difference: DNA is a double-stranded helix; RNA is single-stranded (also DNA sugar is deoxyribose, RNA has ribose; DNA has T, RNA has U).
(b) Nucleoside = pentose sugar + nitrogenous base. Nucleotide = nucleoside + phosphate group (sugar + base + phosphate).
(c) Functions: DNA stores and passes on hereditary (genetic) information; nucleic acids (RNA) carry out protein synthesis.

(i)(I) Carbonyl + semicarbazide → semicarbazone (loss of H₂O): C₆H₁₀=N–NH–CO–NH₂ (cyclohexanone semicarbazone).
(II) Dialkylcadmium converts acid chlorides to ketones: (CH₃)₂Cd + 2CH₃COCl → 2CH₃COCH₃ (propanone) + CdCl₂.
(III) Rosenmund reduction: C₆H₅COCl + H₂ (Pd/BaSO₄) → C₆H₅CHO (benzaldehyde) + HCl. BaSO₄ poisons Pd so the aldehyde is not reduced further.
(ii)(I) Add NaHCO₃ solution: benzoic acid gives brisk effervescence of CO₂; ethyl benzoate (ester) does not.
(II) Warm with Tollens' reagent: propanal gives a silver mirror; propanone does not. (Iodoform test also works: propanone gives yellow CHI₃, propanal does not.)

(i)(I) Alkaline KMnO₄ oxidises the whole side chain of an alkylbenzene to –COO⁻; acidifying gives benzoic acid C₆H₅COOH.
(II) Hydroboration–oxidation adds water anti-Markovnikov: (BH₃)₂ then H₂O₂/OH⁻ gives cyclohexylmethanol C₆H₁₁CH₂OH; PCC then oxidises the primary alcohol to the aldehyde C₆H₁₁CHO.
(III) Tollens' reagent oxidises only the aldehyde group; the ketone is untouched → 4-oxocyclohexane-1-carboxylic acid (as its salt, with a silver mirror).
(ii)(I) C₆H₅CHO + C₆H₅MgBr, then H₃O⁺ → (C₆H₅)₂CHOH (diphenylmethanol); oxidise with PCC or CrO₃ → (C₆H₅)₂CO (benzophenone).
(II) Cross aldol (Claisen–Schmidt): C₆H₅CHO + CH₃CHO, dil. NaOH, heat → C₆H₅CH=CHCHO (cinnamaldehyde); H₂/Ni reduces both C=C and C=O → C₆H₅CH₂CH₂CH₂OH (3-phenylpropan-1-ol).

(i)(I) In [Ni(CO)₄], Ni is 0 (3d⁸4s²). CO is a strong field ligand: the 4s electrons are pushed into 3d → 3d¹⁰, sp³ hybrid, no unpaired electron → diamagnetic. In [NiCl₄]²⁻, Ni²⁺ is 3d⁸; Cl⁻ is weak field and causes no pairing → 2 unpaired electrons, sp³ → paramagnetic.
(II) CO donates a lone pair (σ bond) and also accepts metal d electrons into its empty π* orbitals (π back-bonding). This synergic bonding makes the M–CO bond strong; NH₃ forms only a σ bond.
(III) In trans-[Co(en)₂Cl₂]⁺ the two Cl are opposite; the ion has a plane of symmetry, so its mirror image is superimposable → optically inactive.
(ii) Fe³⁺ = 3d⁵.
(I) Strong field (Δo > P): t₂g⁵ eg⁰ → 1 unpaired electron (low spin).
(II) Weak field (Δo < P): t₂g³ eg² → 5 unpaired electrons (high spin).

(i)(I) Coordination isomerism – ligands swap between the cationic and anionic metal centres: isomer [Cr(NH₃)₆][Co(CN)₆].
(II) Optical isomerism – [Co(en)₃]³⁺ has no plane of symmetry; the three en rings can twist either way, giving two non-superimposable mirror images, d (dextro) and l (laevo).
(III) Geometrical (fac–mer) isomerism – fac: the three NH₃ (and three NO₂) on one triangular face of the octahedron; mer: the three like ligands lie in a plane around the meridian.
(ii) Weak field ligands (I⁻, Cl⁻, F⁻, H₂O) cause small splitting of d orbitals (small Δo); strong field ligands (CN⁻, CO, en, NH₃) cause large splitting (large Δo).
Effect on spin: with strong field ligands Δo > P, electrons pair in t₂g → fewer unpaired electrons → low-spin complex. With weak field ligands Δo < P, electrons enter eg before pairing → more unpaired electrons → high-spin complex.

(i) \( k = \frac{2.303}{t}\log\frac{[R]_0}{[R]} \).
[R]₀/[R] = (1.2 × 10⁻²)/(0.2 × 10⁻²) = 6.
k = (2.303/60 min) × log 6 = (2.303 × 0.778)/60 = 1.7917/60 = 0.0299 min⁻¹ ≈ 2.99 × 10⁻² min⁻¹.
(ii)(I) Order depends on the actual mechanism (the slow step), not on the stoichiometric coefficients; it must be found experimentally.
(II) If one reactant is taken in large excess (e.g. water in hydrolysis of an ester), its concentration stays nearly constant, so rate depends on the other reactant only → pseudo first-order reaction.

(i) \( \log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{T_2 - T_1}{T_1T_2}\right) \), with k₂/k₁ = 2, T₁ = 298 K, T₂ = 308 K.
0.3 = (Ea/19.15) × (10/(298 × 308)).
Ea = 0.3 × 19.15 × 298 × 308 / 10 = 0.3 × 19.15 × 9178.4 = 52 730 J mol⁻¹ ≈ 52.7 kJ mol⁻¹.
(ii)(1) The slow step decides the rate: Rate = k[H₂O₂][I⁻].
(2) Overall order = 1 + 1 = 2; the slow step involves two species, so molecularity = 2 (bimolecular).
Note: I⁻ is a catalyst – used in step I and regenerated in step II.