Chemistry: Unit-wise Subjective Question Bank

1. (a)

(b) Primary: oxalic acid crystals, anhydrous Na₂CO₃ (also K₂Cr₂O₇). Secondary: NaOH, HCl (also KMnO₄, Na₂S₂O₃). (c) NaOH is hygroscopic and absorbs CO₂, so the weighed mass is not the true mass of NaOH; oxalic acid crystals have a fixed composition, are stable and need no correction.

2. (a) Equivalent weight of oxalic acid crystals = 126/2 = 63.

(b) Apply the normality equation at the end point, N₁V₁ = N₂V₂, with N₁ = 0.10 N, V₁ = 500 mL, V₂ = 400 mL:

Mass per litre = normality × equivalent weight = 0.125 × 31.5 = 3.94 g L⁻¹. (c) Dilute H₂SO₄ supplies the H⁺ that the half-reaction MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O needs; without it the brown MnO₂ forms. HCl would be oxidised and HNO₃ would itself oxidise, so neither can be used. The mixture is warmed to 60–70 °C because the reaction between MnO₄⁻ and oxalate is slow at room temperature; once Mn²⁺ forms it catalyses the reaction.

3. (a) In the titration of Na₂CO₃ against H₂SO₄ methyl orange is used: it turns from yellow to orange-pink at the end point, when all the carbonate has become H₂CO₃. (b) Yes. Anhydrous Na₂CO₃ is pure, stable, non-hygroscopic when dry and has a known formula, so a standard solution can be made by direct weighing. (c) Equivalent weight of Na₂CO₃ = 106/2 = 53.

4. (a) Molarity = (17.7/118) mol per litre = 0.15 M. (b) Mass of 1 L of solution = 1000 × 1.0077 = 1007.7 g. Mass of solvent = 1007.7 − 17.7 = 990.0 g = 0.990 kg.

(c) A dibasic acid has two replaceable H, so its n-factor is 2: N = 0.15 × 2 = 0.30 N.

5. (a) One litre of the solution contains 2 mol (73 g) of HCl. (b) Molarity is moles of solute per litre of solution, and it changes with temperature because the volume changes. Molality is moles of solute per kilogram of solvent, and it does not change with temperature because mass does not. (c) The normality factor is the ratio of the actual strength of the solution to the strength it was meant to have. It lets the student use a solution that is not exactly as planned: the volume is simply multiplied by the factor, so there is no need to remake the solution. (d) Take 1 L of liquor ammonia. Mass = 1000 × 0.91 = 910 g. Mass of NH₃ = 0.25 × 910 = 227.5 g. Moles = 227.5/17 = 13.4.

6. (a) A primary standard is a substance pure enough, and stable enough, that its standard solution can be made by dissolving an accurately weighed mass in a known volume, with no need to standardise it afterwards. (b) In a weak acid–weak base titration the pH changes gradually near the equivalence point instead of jumping, so no indicator gives a sharp colour change and the end point cannot be seen. (c) True normality of oxalic acid = 0.1 × 1.01 = 0.101 N. At the end point N₁V₁ = N₂V₂:

Strength of NaOH = 0.101 × 40 = 4.04 g L⁻¹ = 0.404 g per 100 mL, so x = 0.404 %.

1. (a) Law of equivalence: at the end point, equal numbers of gram equivalents of the two reactants have reacted. For a solution of normality N and volume V (in mL), the number of milli-equivalents is N × V.

(b) (i) Redox (permanganate) titration of oxalic acid against KMnO₄. (ii) MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, so the n-factor is 5 and E = 158/5 = 31.6. (iii) The first reading (11.5 mL) is a rough value; the concordant readings are 11.0 mL and 11.0 mL, so the mean is 11.0 mL.

(iv) Dilute H₂SO₄ gives the H⁺ the permanganate half-reaction needs and keeps the medium acidic, so that Mn²⁺ and not brown MnO₂ is the product. Neither HCl nor HNO₃ can be used, because the first is oxidised by MnO₄⁻ and the second is itself an oxidant. (v) Titrant (in the burette): KMnO₄ solution. Titrand (in the conical flask): oxalic acid solution. KMnO₄ acts as its own indicator, so no other indicator is added.

2. (a) No. NaOH is hygroscopic and absorbs CO₂ from air, so its mass cannot be relied on when weighed out; its solution must be standardised against a primary standard, which makes NaOH a secondary standard. (b) The equivalence point is the theoretical stage at which the reactants have combined in exactly equivalent amounts. The end point is the stage at which the indicator changes colour and the titration is stopped. A well-chosen indicator makes the two almost coincide, but they are not identical and the small difference is the titration error. (c) Equivalent weight of oxalic acid crystals = 63.

1. (a) The ionic product of water is the product of the molar concentrations of H⁺ and OH⁻ ions in pure water (or in any dilute aqueous solution) at a given temperature: Kw = [H⁺][OH⁻] = 1 × 10⁻¹⁴ at 25 °C. (b) Take logarithms of Kw = [H⁺][OH⁻] and change sign:

(c) HCl is a strong acid, so [H⁺] = 0.01 M.

(d) The ionisation of water is endothermic, so raising the temperature shifts the equilibrium to the right and Kw increases (about 5.5 × 10⁻¹⁴ at 50 °C); the pH of pure water falls below 7 although it stays neutral.

2. (a) The pH of a solution is the negative logarithm to the base 10 of its hydrogen-ion concentration, pH = −log[H⁺]. (b) pOH = 14 − 4.7 = 9.3.

(c) NaOH is a strong base, so 1 mol NaOH gives 1 mol OH⁻ (the water's own OH⁻ is ignored, as the question intends).

3. (a)

Every B–L base is a Lewis base, but the Lewis concept is wider: it includes reactions that involve no proton. (b) Lewis acid: BF₃ (or AlCl₃, H⁺). Lewis base: NH₃ (or OH⁻, H₂O). (c) NH₃ (base) / NH₄⁺ (its conjugate acid); H₂O (acid) / OH⁻ (its conjugate base).

4. After mixing, the total volume is 100 mL and each concentration is diluted.

Since Qsp (1.6 × 10⁻⁸) is greater than Ksp (1.8 × 10⁻¹⁰), the solution is supersaturated and AgCl precipitates.

5. A buffer solution resists a change in pH when small amounts of acid or alkali are added. An acidic buffer is a weak acid with its salt, for example CH₃COOH + CH₃COONa. For HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻]/[HA], so

Take −log of both sides. The salt is fully ionised, so [A⁻] ≈ [salt]; the acid is weakly ionised and suppressed by the common ion, so [HA] ≈ [acid]:

1. (a) A weak electrolyte is a substance that ionises only partly in solution, so that ions and unionised molecules are in equilibrium (ethanoic acid, NH₄OH). (b) Let 1 mol of weak acid HA be dissolved in V litres, and let α be the degree of ionisation.

Concentrations are these amounts divided by V, where C = 1/V:

This is Ostwald's dilution law. For a weak electrolyte α is very small, so 1 − α ≈ 1:

Since α ∝ 1/√C, the degree of ionisation increases on dilution. For a weak base replace Ka by Kb. (c) Limitations: (i) it fails for strong electrolytes, which are completely ionised, so the "constant" changes with concentration; (ii) it fails for weak electrolytes at high concentration (inter-ionic attractions); (iii) it does not apply without modification when a common ion is present. Strong electrolytes do not obey it. (d) α = 1.34 % = 0.0134 and C = 0.1 M.

2. (a) Solubility product. In group II the sulphides of Pb²⁺, Cu²⁺, Cd²⁺ and others have very small Ksp, so even the tiny [S²⁻] from H₂S in acid is enough to precipitate them, while the more soluble ZnS and NiS (larger Ksp) stay in solution. In group IV, NH₄OH/NH₄Cl is added so that [S²⁻] is large enough to exceed Ksp of ZnS, NiS, CoS and MnS. A precipitate forms only when the ionic product exceeds Ksp. Common ion effect. (i) Dilute HCl is added in group II: H⁺ is the common ion for H₂S ⇌ 2H⁺ + S²⁻, so [S²⁻] is kept low and ZnS and NiS do not precipitate. (ii) In group III, NH₄Cl is added with NH₄OH: the common ion NH₄⁺ suppresses the ionisation of NH₄OH, so [OH⁻] is small and only Fe(OH)₃, Al(OH)₃ and Cr(OH)₃ (very small Ksp) precipitate, not Mg(OH)₂. (iii) In group V, (NH₄)₂CO₃ supplies CO₃²⁻ for Ca²⁺, Sr²⁺, Ba²⁺. (b) See the table in short answer 3 above: a Brønsted–Lowry base accepts a proton; a Lewis base donates an electron pair, and is not limited to reactions involving H⁺. (c) α = √(Ka/C) = √(4.8 × 10⁻¹⁰/0.01) = √(4.8 × 10⁻⁸) = 2.2 × 10⁻⁴. [H⁺] = Cα = 0.01 × 2.19 × 10⁻⁴ = 2.2 × 10⁻⁶ M, and pH = −log(2.19 × 10⁻⁶) = 5.66.

3. (a)

(b) It fails for strong electrolytes (they are completely ionised). (c) NaCl is the salt of a strong acid (HCl) and a strong base (NaOH). Neither Na⁺ nor Cl⁻ reacts with water, so [H⁺] = [OH⁻] and the solution is neutral. (d) (i) After mixing equal volumes each concentration is halved: [Ba²⁺] = 1 × 10⁻³ M and [SO₄²⁻] = 1 × 10⁻⁴ M.

1 × 10⁻⁷ > 1 × 10⁻¹⁰, so BaSO₄ precipitates. (ii) [Ba²⁺] = 1 × 10⁻⁸ M and [SO₄²⁻] = 1 × 10⁻³ M.

1 × 10⁻¹¹ < 1 × 10⁻¹⁰, so no precipitate forms.

4. (a) Saturated Zn(OH)₂. Zn(OH)₂ ⇌ Zn²⁺ + 2OH⁻, so Ksp = (s)(2s)² = 4s³.

So in the saturated solution [Zn²⁺] = 1.65 × 10⁻⁵ M and [OH⁻] = 3.30 × 10⁻⁵ M. Added alkali. 0.22 g NaOH = 0.22/40 = 5.5 × 10⁻³ mol of OH⁻. After mixing, the volume is 1.1 L.

Since 3.8 × 10⁻¹⁰ exceeds Ksp = 1.8 × 10⁻¹⁴, Zn(OH)₂ is precipitated. The added OH⁻ is a common ion, which has raised the ionic product far beyond the solubility product. (b) pH 8 means pOH = 6, so [OH⁻] = 10⁻⁶ M. A 1000-fold dilution would give 10⁻⁹ M, but that is far less than the 10⁻⁷ M OH⁻ that pure water already supplies, so water dominates.

pOH = 6.998, so pH = 7.002, just above 7: a base, however much diluted, cannot give an acidic solution.

1. The instantaneous rate is the rate of the reaction at one particular moment: the slope of the tangent to the concentration–time curve at that time, rate = ±d[C]/dt as Δt → 0. For 2N₂O₅ → 4NO₂ + O₂ each rate is divided by the stoichiometric coefficient:

Moles of O₂ formed = 2.24/22.4 = 0.10 mol in 30 min. From the equation 1 mol O₂ is formed when 2 mol N₂O₅ react, so N₂O₅ used = 0.20 mol.

Dividing by the volume of the vessel gives the rate in mol L⁻¹ min⁻¹.

2. (a) The rate law is the expression that relates the rate of a reaction to the molar concentrations of the reactants, each raised to the power that is found by experiment: rate = k[P]ᵐ[Q]ⁿ. (b) The powers m and n are not the stoichiometric coefficients; they depend on the mechanism, in particular on the slowest step, and so can only be found by measuring how the rate changes when the concentrations are changed. (c) The rate constant k is the rate of the reaction when every reactant concentration is 1 mol L⁻¹. It depends on temperature and catalyst, not on concentration. (d)

3. (a) For A → products, let the initial concentration be a and let x have reacted in time t. The rate is proportional to the concentration left:

At t = 0, x = 0, so I = −ln a.

At t = t½, x = a/2, so a − x = a/2:

The initial concentration a has cancelled, so t½ is independent of a. (b) k = 0.693/10 = 0.0693 s⁻¹. For 90 % completion, a − x = 0.1a:

4. (a) The half-life is the time taken for the concentration of a reactant to fall to half of its initial value. (b) For zero order, rate = k[A]⁰ = k, so −d[A]/dt = k. Integrating: a − [A] = kt. At t = t½, [A] = a/2:

so t½ ∝ a, the initial concentration. (c)

5. (a) The order is the sum of the powers to which the concentrations are raised in the experimental rate law. (b) rate = k[A]¹[B]⁰ = k[A]. (c) Only A appears in the rate law. Doubling [A] doubles the rate; doubling [B] has no effect. The rate increases 2 times. (d) The rate roughly doubles for every 10 °C rise, so a 20 °C rise multiplies it by 2² = 4 times.

1. (a) Order is found by experiment and may be zero, fractional or whole; molecularity is a theoretical count of colliding molecules in one elementary step and is always a whole number (1, 2 or 3). Order refers to the overall reaction; molecularity refers to a single elementary step. (b) See the figure: a straight line sloping downwards from [A]₀, with slope −k. (c) Compare I and II: [X] is constant and [Y] is doubled; the rate doubles (1.6 × 10⁻⁴ → 3.2 × 10⁻⁴), so the order in Y is 1. Compare II and III: [Y] is constant and [X] is doubled, and the rate does not change, so the order in X is 0. Rate law: rate = k[Y].

The unit s⁻¹ confirms a first-order reaction overall. (d) Zero order with respect to X.

2. (a) The rate law is rate = k[P] (Q is zero order). From experiment I: k = 2 × 10⁻²/0.1 = 0.2 s⁻¹. Experiment II: 4 × 10⁻² = 0.2 × A, so A = 0.2 M. Experiment IV: 2 × 10⁻² = 0.2 × B, so B = 0.1 M. Experiment III: C = 0.2 × 0.4 = 8 × 10⁻² M s⁻¹. (b) k = 0.2 s⁻¹ (see (a)); the unit s⁻¹ shows overall first order. (c) The rate law expresses the rate as the product of the concentrations of the reactants raised to the powers m and n. m and n are found only by measuring the rate at different concentrations because the rate depends on the slow step of the mechanism, which cannot be seen in the balanced equation. (d) Order is the sum of experimental powers, and a reaction with a complex mechanism can have fractional powers, for example rate = k[A]³ᐟ² for the pyrolysis of ethanal. Molecularity counts molecules in one elementary collision, and one cannot have half a molecule colliding, so it is always a whole number.

3. (a) Compare 1 and 2: [X] constant, [Y] × 4, rate × 4, so order in Y = 1. Compare 2 and 3: [Y] constant, [X] × 2, rate × 2, so order in X = 1. Rate law: rate = k[X][Y].

Experiment 4: 1.8 × 10⁻² = 0.10 × 0.5 × A, so A = 0.36 mol L⁻¹. (b) A pseudo-first-order reaction is a reaction of higher molecularity that follows a first-order rate law because one reactant is present in such large excess that its concentration stays constant. Example: acid hydrolysis of ethyl ethanoate, CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH, where water is in large excess and rate = k′[ester]. (c) An effective collision is one in which the molecules carry at least the threshold (activation) energy and meet with the proper orientation: the reactive atom of one molecule strikes the reactive site of the other and bond-breaking and bond-making take place. A collision with the wrong orientation, or with energy below Ea, is ineffective and the molecules separate unchanged.

1. (a) Hess's law: the total enthalpy change of a reaction is the same whether it takes place in one step or in several steps, provided the initial and final conditions are the same; it depends only on the initial and final states. (b) The enthalpy of a reaction is the heat change at constant pressure when the number of moles of reactants shown in the balanced equation react completely. (c) ΔH = Σ ΔH_f (products) − Σ ΔH_f (reactants); ΔH_f of an element (O₂) is zero.

The reaction is exothermic: −196 kJ for 2 mol H₂O₂.

2. The standard heat of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states at 298 K and 1 atm. Formation reaction: 2C(s) + 2H₂(g) + O₂(g) → CH₃COOH(l). The three given equations are used as follows: (i) 2C + 2O₂ → 2CO₂, ΔH = 2(−99) = −198; (ii) 2H₂ + O₂ → 2H₂O, ΔH = 2(−68) = −136; (iii) 2CO₂ + 2H₂O → CH₃COOH + 2O₂, ΔH = +208 (the combustion equation reversed). Adding (i), (ii) and (iii) gives 2C + 2H₂ + O₂ → CH₃COOH.

3. (a) The Gibbs free energy change ΔG is the part of the enthalpy change that is available to do useful work at constant T and P: ΔG = ΔH − TΔS; a process is spontaneous when ΔG is negative. (b) ΔH positive makes ΔG more positive, and ΔS positive makes −TΔS negative. At low temperature the ΔH term dominates, so ΔG > 0 and the process is non-spontaneous; at high temperature the TΔS term dominates, so ΔG < 0 and the process becomes spontaneous. (Melting of ice and evaporation of water follow this pattern.) (c) T = 15 + 273 = 288 K and ΔS = 0.300 kJ mol⁻¹ K⁻¹.

ΔG is negative, so the reaction is spontaneous at 15 °C.

4. The heat of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states. Formation: 2C + 3H₂ + ½O₂ → C₂H₅OH. The heats of combustion of carbon and hydrogen are the heats of formation of CO₂ and H₂O. Combustion: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O, ΔH_c = −330.

1. (a) Hess's law states that the total enthalpy change of a reaction is the same whether it takes place in one step or in several steps. Illustration: the formation of CO₂ from carbon. Path 1 (direct): C + O₂ → CO₂, ΔH = −394 kJ. Path 2 (two steps): C + ½O₂ → CO, ΔH₁ = −111 kJ; CO + ½O₂ → CO₂, ΔH₂ = −283 kJ. Sum: −111 + (−283) = −394 kJ, the same as the direct path. (b) The formation equation is 2C(s) + H₂(g) → C₂H₂(g). Data: (i) C + O₂ → CO₂, −394; (ii) H₂ + ½O₂ → H₂O, −286; (iii) C₂H₂ + 2½O₂ → 2CO₂ + H₂O, −1300. 2 × (i): 2C + 2O₂ → 2CO₂, ΔH = −788. (ii): H₂ + ½O₂ → H₂O, ΔH = −286. Reverse of (iii): 2CO₂ + H₂O → C₂H₂ + 2½O₂, ΔH = +1300. Adding gives 2C + H₂ → C₂H₂ with

Acetylene is formed endothermically, which is why it is so reactive. (c) See the figure. Exothermic: products at lower enthalpy than reactants, ΔH negative; endothermic: products at higher enthalpy, ΔH positive; Ea is the height from the reactants to the top of the hump in both cases.

2. (a) First law: energy can be neither created nor destroyed, but only converted from one form to another; for a system ΔU = q + w. If the only work is expansion against a constant pressure P, w = −PΔV, so at constant pressure q_p = ΔU + PΔV. The heat absorbed at constant pressure is defined as the enthalpy change ΔH:

(b) ΔG° = −2.303RT log K. The more negative ΔG°, the larger K and the more completely the reaction goes to products. (c) T = 298 K; ΔS = −0.199 kJ K⁻¹.

The reaction is spontaneous at 298 K even though ΔS is negative, because the exothermic ΔH outweighs the −TΔS term.

1. (a) The standard electrode potential of an electrode is its potential when it is in contact with a 1 M solution of its ions at 25 °C (and 1 atm for gases), measured against the standard hydrogen electrode whose potential is taken as zero. (b) Mg has the more negative potential, so it is the stronger reducing agent: Mg is the anode (oxidation) and Zn the cathode (reduction).

(c) Anode: Mg → Mg²⁺ + 2e⁻. Cathode: Zn²⁺ + 2e⁻ → Zn. Cell reaction: Mg + Zn²⁺ → Mg²⁺ + Zn. (d) E°cell = E°cathode − E°anode = (−0.76) − (−2.38) = +1.62 V.

2. (a) The SHE has a potential arbitrarily assigned as 0.00 V at all temperatures. The potential of any other electrode is found by combining it with the SHE, so the SHE is the standard against which every electrode potential is measured. (b) The emf is positive, so the combination works as written. Zn is oxidised, so Zn is the anode and the SHE is the cathode:

(c) Zn + 2H⁺ → Zn²⁺ + H₂. (d) E°cell = E°(SHE) − E°(Zn²⁺/Zn) = 0 − E° = +0.76 V, so E°(Zn²⁺/Zn) = −0.76 V.

3. (a) Cu is the stronger reducing agent (lower E°), so Cu is the anode and Ag the cathode:

(b) E°cell = 0.80 − 0.34 = +0.46 V. (c) No. The cell reaction Cu + 2Ag⁺ → Cu²⁺ + 2Ag has a positive emf, so it is spontaneous: copper would displace silver from the solution, the vessel would be eaten away and silver would deposit.

4. (a) See the figure: platinum foil coated with platinum black, dipping in 1 M HCl; pure H₂ gas at 1 atm bubbled over it; a side tube to a salt bridge. (b) Its potential is fixed at zero by convention, so every other electrode potential is measured relative to it. (c) Zn (more negative) is the anode and Pb the cathode.

E°cell = −0.13 − (−0.76) = +0.63 V.

5. (a) C is displaced by none, so it is the most reactive. D and C displace hydrogen, so both are above H. B and A are below hydrogen (they do not displace it), and B displaces A but not the reverse. Series in decreasing activity: C > D > H > B > A. (b) Zn is oxidised, and Fe³⁺ is reduced to Fe²⁺ at an inert platinum electrode:

Cell reaction: Zn + 2Fe³⁺ → Zn²⁺ + 2Fe²⁺. E°cell = 0.77 − (−0.76) = +1.53 V.

1. (a) The standard reduction potential is the potential of a half-cell, when the reduction half-reaction takes place at it in contact with a 1 M solution of its ions at 25 °C, measured against the SHE. (b) The smaller (more negative) the E°, the stronger the reducing power: A > B > C (−0.76, +0.34, +0.80). (c) B (E° = +0.34 V) is higher than hydrogen (0.00 V), so B is reduced and the SHE is the anode:

(d) Anode: H₂ → 2H⁺ + 2e⁻. Cathode: B²⁺ + 2e⁻ → B. Cell reaction: H₂ + B²⁺ → 2H⁺ + B. (e) E°cell = 0.34 − 0.00 = +0.34 V.

2. (a) Lead storage battery. Anode: spongy lead (Pb); cathode: lead dioxide (PbO₂) on a lead grid; electrolyte: about 38 % H₂SO₄. On discharge: Pb + SO₄²⁻ → PbSO₄ + 2e⁻ (anode); PbO₂ + 4H⁺ + SO₄²⁻ + 2e⁻ → PbSO₄ + 2H₂O (cathode). Overall: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O. Sulphuric acid is consumed, so its density falls. On charging (the current is reversed) the reactions are reversed: 2PbSO₄ + 2H₂O → Pb + PbO₂ + 2H₂SO₄. One cell gives about 2 V, so a 12 V battery has six cells in series. (b) H₂–O₂ fuel cell. Porous carbon electrodes carrying a Pt or Ni catalyst dip in hot KOH. H₂ is fed to the anode and O₂ to the cathode. Anode: 2H₂ + 4OH⁻ → 4H₂O + 4e⁻. Cathode: O₂ + 2H₂O + 4e⁻ → 4OH⁻. Overall: 2H₂ + O₂ → 2H₂O. (c) It supplies electricity with water as the only product, and the water is drinkable by the crew; it is efficient and light for the power it gives, so it provides power for heat and light in spacecraft.

1. A d–d transition is the excitation of an electron from a lower-energy d-orbital to a higher-energy d-orbital of the same metal ion (for an octahedral complex, from t₂g to e_g) when the ion absorbs a photon of visible light of energy equal to the crystal-field splitting Δo. The complementary colour is transmitted, so the compound is coloured. Characteristics of transition metals: 1. Variable oxidation states, because the (n−1)d and ns electrons have similar energy (Mn shows +2 to +7). 2. Coloured compounds, from d–d transitions in ions with a partly filled d subshell. 3. Complex formation, because the ions are small, highly charged and have vacant d-orbitals ([Cu(NH₃)₄]²⁺, [Fe(CN)₆]⁴⁻). 4. Catalytic activity, from variable oxidation state and the ability to adsorb reactants on the surface (Fe in the Haber process, V₂O₅ in the contact process). 5. Paramagnetism, from unpaired d electrons; also high melting points, hardness, high density and the formation of alloys and interstitial compounds.

2. (a) Assumptions of crystal field theory: 1. The ligands are treated as point negative charges (or dipoles), and the bond between metal and ligand is purely electrostatic. 2. In the free ion the five d-orbitals have the same energy; the approaching ligands repel the d electrons, and the orbitals split into sets of different energy according to their orientation relative to the ligands. 3. The difference in energy, the crystal field splitting Δ, depends on the ligand and on the geometry, and electrons fill the split orbitals according to Δ and the pairing energy. (b) See the figure: in an octahedral complex the three t₂g orbitals are lowered by 0.4Δo and the two e_g orbitals are raised by 0.6Δo.

3. (a) Cu²⁺ is d⁹: the t₂g set is full and the e_g set holds three electrons, with unequal occupancy of d_z² and d_x²−y². Unequal repulsion on the ligands along z and in the xy plane makes the complex elongate along the z-axis (Jahn–Teller distortion), which lowers the total energy. (b) In [Cu(H₂O)₆]²⁺ the d–d transition absorbs light in the red-orange region, so the transmitted (complementary) colour is blue. (c) Fe³⁺ (d⁵) is more stable than Fe²⁺ (d⁶) because a half-filled d subshell gives extra exchange energy and symmetry. (d) They have variable oxidation states and vacant d-orbitals, so they form intermediate compounds with reactants and offer a surface for adsorption.

4. (a) Cr: [Ar] 3d⁵4s¹ (not 3d⁴4s²) and Cu: [Ar] 3d¹⁰4s¹ (not 3d⁹4s²). A half-filled (d⁵) or fully filled (d¹⁰) subshell has extra stability from symmetry and exchange energy, so one 4s electron moves into the 3d subshell. (b) Zn is [Ar] 3d¹⁰4s² and forms only Zn²⁺ (3d¹⁰): its d subshell is full in both the atom and its common ion, so it does not meet the definition of a transition element (an ion with a partly filled d subshell). (c) Sc³⁺ is 3d⁰ and Zn²⁺ is 3d¹⁰, so there is no partly filled d subshell and no d–d transition is possible. (d) Cr³⁺ is [Ar] 3d³: three d electrons.

1. (a) A transition element is one that has a partly filled d subshell in its atom or in any of its common ions. General configuration: (n−1)d¹⁻¹⁰ ns¹⁻² (in the 3d series, [Ar] 3d¹⁻¹⁰ 4s¹⁻²). (b) Manganese ([Ar] 3d⁵4s²) uses its 4s and 3d electrons and shows every state from +2 to +7: +2 in MnSO₄, +3 in Mn₂O₃, +4 in MnO₂, +6 in K₂MnO₄ and +7 in KMnO₄. The 4s and 3d energies are close, so successive electrons can be removed with small increases in ionisation energy; the highest state +7 equals the number of 3d + 4s electrons. (c) Ti³⁺ is 3d¹: in [Ti(H₂O)₆]³⁺ the single electron is in t₂g, and absorbing light of energy Δo (green–yellow, about 500 nm) promotes it to e_g, so the complex transmits violet-purple and is coloured. Sc³⁺ is 3d⁰ with no d electron to excite, so it is colourless. (d) (i) Iron in the Haber process: N₂ + 3H₂ ⇌ 2NH₃ (heterogeneous; reactants adsorbed on the Fe surface). (ii) V₂O₅ in the contact process: 2SO₂ + O₂ ⇌ 2SO₃ (V⁵⁺ ⇌ V⁴⁺ cycling). (Also Ni in hydrogenation of oils; MnO₂ in the decomposition of KClO₃.)

1. (a) A = copper (Cu), in the sulphate CuSO₄. With a little NH₄OH a blue precipitate B = Cu(OH)₂ forms; excess ammonia dissolves it to the deep blue C = tetraamminecopper(II) sulphate, [Cu(NH₃)₄]SO₄. CuSO₄ + 2NH₄OH → Cu(OH)₂↓ + (NH₄)₂SO₄ Cu(OH)₂ + (NH₄)₂SO₄ + 2NH₄OH → [Cu(NH₃)₄]SO₄ + 4H₂O (b) Cu: [Ar] 3d¹⁰4s¹. (c) Electrolytic refining (after poling): impure copper is the anode, pure copper the cathode, and acidified CuSO₄ is the electrolyte; 99.95 % pure copper deposits.

2. (a) See the figure: the chief ore is copper pyrite, CuFeS₂, and the roasted ore is smelted with coke and silica in a small water-jacketed blast furnace. (b) The reactions that occur in the furnace: FeO + SiO₂ → FeSiO₃ (slag; silica is the flux that removes iron as a fusible silicate) Cu₂O + FeS → Cu₂S + FeO The FeO formed in the second reaction is slagged by silica. The molten matte (Cu₂S + FeS, about 45 % Cu) sinks below the lighter slag and is tapped off for the Bessemer converter. Matte forms because copper has a greater affinity for sulphur than iron has, so the sulphur stays with copper and the iron passes into the slag as FeSiO₃.

3. Preparation: (i) directly, by heating mercury in excess chlorine, Hg + Cl₂ → HgCl₂; (ii) industrially, by heating mercury(II) sulphate with sodium chloride and subliming the product, HgSO₄ + 2NaCl → HgCl₂↑ + Na₂SO₄ (a little MnO₂ is added to oxidise any mercurous salt). Properties: white crystalline, poisonous, sublimes on heating, sparingly soluble in water and largely un-ionised (covalent). Reactions: (i) with KI, a scarlet precipitate that dissolves in excess KI: HgCl₂ + 2KI → HgI₂↓ + 2KCl; HgI₂ + 2KI → K₂[HgI₄]; (ii) with SnCl₂, a white precipitate that turns grey: 2HgCl₂ + SnCl₂ → Hg₂Cl₂↓ + SnCl₄, then Hg₂Cl₂ + SnCl₂ → 2Hg + SnCl₄; (iii) with NH₃, a white precipitate: HgCl₂ + 2NH₃ → Hg(NH₂)Cl↓ + NH₄Cl. Uses: as an antiseptic and disinfectant (1 : 1000), to preserve wood and anatomical specimens, to make Nessler's reagent and calomel, and as a catalyst (with HgSO₄) for the hydration of ethyne.

4. (a) Dissolve zinc (or ZnO or ZnCO₃) in dilute H₂SO₄ and crystallise below 30 °C: Zn + H₂SO₄ → ZnSO₄ + H₂; the solution gives ZnSO₄·7H₂O. (b) At red heat (about 800 °C) it loses its water and then decomposes: 2ZnSO₄ → 2ZnO + 2SO₂ + O₂. (c) A double salt is a compound formed by the crystallisation of a solution containing two simple salts in equimolar amounts; it exists in the solid state but dissociates into its ions in water. Example: potash alum, K₂SO₄·Al₂(SO₄)₃·24H₂O (or Mohr's salt). (d) By mixing solutions of zinc sulphate and barium sulphide: ZnSO₄ + BaS → ZnS↓ + BaSO₄↓. The mixed white precipitate of ZnS and BaSO₄ is lithopone, a white pigment. (e) The zinc atom ([Ar] 3d¹⁰4s²) and its only ion Zn²⁺ (3d¹⁰) both have a completely filled d subshell, so zinc does not show the properties of a typical transition element (variable oxidation states, coloured ions, paramagnetism).

5. (a) The open hearth process is a method of making steel from pig iron in a shallow, saucer-shaped furnace heated by producer gas and pre-heated air (the Siemens regenerative principle); the impurities are oxidised by haematite in the charge. (b) Fe₂O₃ + 3C → 2Fe + 3CO↑; 2Fe₂O₃ + 3Si → 4Fe + 3SiO₂; Fe₂O₃ + 3Mn → 2Fe + 3MnO; 5Fe₂O₃ + 6P → 10Fe + 3P₂O₅. Lime is added as a flux and the SiO₂ and P₂O₅ pass into the slag (CaSiO₃, Ca₃(PO₄)₂). (c) Spiegeleisen is an alloy of iron, manganese and carbon. It is added at the end to restore the required amount of carbon and, through manganese, to remove dissolved oxygen (as MnO), since excess oxygen makes steel brittle. (d) Stainless steel: about 74 % Fe, 18 % Cr, 8 % Ni (and about 0.1 % C).

6. (a) M is zinc, and its main ore is zinc blende, ZnS. (b) Zn²⁺ is 3d¹⁰ and the atom has a full d subshell, so there is no partly filled d subshell in either; it is a non-typical transition (d-block) element. (c) Vertical retort: a tall, narrow column of fireclay or silicon carbide heated from outside. Briquettes of roasted ore (ZnO) and coke are fed in at the top, and zinc vapour and CO leave from the top to a condenser. Roasting: 2ZnS + 3O₂ → 2ZnO + 2SO₂. Reduction: ZnO + C → Zn + CO at about 1673 K; the zinc distils as vapour (b.p. 907 °C). (d) In moist air a thin protective layer of basic zinc carbonate forms, 2Zn + O₂ + CO₂ + H₂O → ZnCO₃·Zn(OH)₂, which stops further corrosion (the basis of galvanising).

7. Rusting is the slow corrosion of iron in moist air to form hydrated iron(III) oxide, Fe₂O₃·xH₂O, a brown flaky solid. Electrochemical theory. The impure iron surface with water containing dissolved CO₂ and O₂ acts as a tiny galvanic cell. - Anode (oxygen-poor region): Fe → Fe²⁺ + 2e⁻. - Cathode (oxygen-rich region): O₂ + 4H⁺ + 4e⁻ → 2H₂O. The electrons flow through the metal and the ions through the moisture. - Fe²⁺ is further oxidised by air and water: 4Fe²⁺ + O₂ + 4H₂O → 2Fe₂O₃ + 8H⁺, and the oxide hydrates to rust, Fe₂O₃·xH₂O.

Prevention: (i) painting, oiling or plastic coating to keep out air and water; (ii) galvanising (coating with zinc, which also acts as a sacrificial anode); (iii) alloying to stainless steel; (iv) cathodic protection with magnesium blocks.

1. Ore: copper pyrite, CuFeS₂ (2–3 % Cu). 1. Concentration by froth flotation. The powdered ore is agitated with water, pine oil and sodium ethyl xanthate while air is blown through; the sulphide particles are wetted by oil and rise in the froth, the earthy gangue sinks. 2. Roasting in a reverberatory furnace in a limited supply of air at about 1070 K, which drives off moisture, As and Sb as oxides, and burns part of the sulphur: 2CuFeS₂ + O₂ → Cu₂S + 2FeS + SO₂; 2FeS + 3O₂ → 2FeO + 2SO₂; 2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂. 3. Smelting of the roasted ore with coke and silica in a water-jacketed blast furnace: FeO + SiO₂ → FeSiO₃ (slag), Cu₂O + FeS → Cu₂S + FeO. The molten product is copper matte, Cu₂S + FeS (about 45 % Cu). 4. Bessemerisation. Molten matte is poured into a Bessemer converter lined with silica; hot air is blown through. 2FeS + 3O₂ → 2FeO + 2SO₂, FeO + SiO₂ → FeSiO₃; 2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂; and then auto-reduction: 2Cu₂O + Cu₂S → 6Cu + SO₂↑. The product is blister copper (98 % pure; the escaping SO₂ makes blisters). 5. Refining. First poling (green wood poles reduce residual Cu₂O), then electrolytic refining: anode of impure copper, cathode of pure copper, electrolyte CuSO₄ + dilute H₂SO₄. Anode: Cu → Cu²⁺ + 2e⁻; cathode: Cu²⁺ + 2e⁻ → Cu. The pure copper (99.95 %) deposits and the impurities fall as anode mud (Ag, Au).

2. (a) Basic Oxygen Process (LD process). Vessel: a pear-shaped converter lined with basic dolomite (CaO·MgO). Charge: molten pig iron, up to 30 % scrap, and quicklime as flux. A water-cooled lance blows 99.5 % pure oxygen at about 10 atm for 20 minutes; no fuel is needed because the oxidation of the impurities supplies the heat. C: 2C + O₂ → 2CO↑ (burns at the mouth); Si: Si + O₂ → SiO₂, then SiO₂ + CaO → CaSiO₃ (slag); Mn: 2Mn + O₂ → 2MnO, then MnO + SiO₂ → MnSiO₃; P: 4P + 5O₂ → 2P₂O₅, then P₂O₅ + 3CaO → Ca₃(PO₄)₂ (slag); S: S + O₂ → SO₂↑. When the carbon is down to the required level, ferro-manganese or other alloying metals are added. (b) Open Hearth Process. Vessel: shallow, saucer-shaped hearth with a basic (dolomite) lining for phosphoric iron; heated by producer gas and pre-heated air through regenerators. The oxidising agent is haematite in the charge: Fe₂O₃ + 3C → 2Fe + 3CO; 2Fe₂O₃ + 3Si → 4Fe + 3SiO₂; 5Fe₂O₃ + 6P → 10Fe + 3P₂O₅; Fe₂O₃ + 3Mn → 2Fe + 3MnO. Lime takes SiO₂ and P₂O₅ into the slag. A charge takes 8–10 hours. (c) Advantage: because it is slow, samples can be analysed and the composition adjusted, so alloy and special steels of accurately controlled composition can be made.

3. Ore: argentite (silver glance), Ag₂S. 1. Concentration by froth flotation. 2. Leaching. The concentrate is agitated with a dilute (0.1–0.5 %) NaCN solution while air is blown through: Ag₂S + 4NaCN ⇌ 2Na[Ag(CN)₂] + Na₂S The reaction is reversible. The air oxidises the sodium sulphide as it forms, 2Na₂S + 2O₂ + H₂O → Na₂S₂O₃ + 2NaOH, which removes a product and drives the equilibrium to the right (Le Chatelier's principle). The silver dissolves as sodium dicyanoargentate(I). 3. Precipitation. Zinc dust displaces silver because zinc is more electropositive: 2Na[Ag(CN)₂] + Zn → Na₂[Zn(CN)₄] + 2Ag↓ 4. Refining by the Moebius electrolytic process: anode of impure silver, cathode of pure silver, electrolyte of AgNO₃ with about 1 % HNO₃. Pure silver (99.9 %) collects as crystals at the cathode and falls into a bag below the anode; gold and platinum remain in the anode mud.

1. (a) B = propan-1-ol, CH₃CH₂CH₂OH. (b) SN2. 1-Bromopropane is a primary halide with little steric hindrance, and OH⁻ is a strong nucleophile, so it attacks the carbon from the back in a single step with inversion and no carbocation forms (a primary carbocation is unstable). (c) Alcoholic KOH gives elimination (dehydrohalogenation): CH₃CH₂CH₂Br + KOH(alc.) → CH₃CH=CH₂ + KBr + H₂O, that is propene. (d) By heating with silver nitrite: CH₃CH₂CH₂Br + AgNO₂ → CH₃CH₂CH₂NO₂ + AgBr, giving 1-nitropropane.

2. (a) A = chloroform (trichloromethane), CHCl₃. (b) Ethanol is oxidised by bleaching powder and the ethanal formed is chlorinated to chloral, which is hydrolysed: CaOCl₂ + H₂O → Ca(OH)₂ + Cl₂ CH₃CH₂OH + Cl₂ → CH₃CHO + 2HCl; CH₃CHO + 3Cl₂ → CCl₃CHO + 3HCl 2CCl₃CHO + Ca(OH)₂ → 2CHCl₃ + (HCOO)₂Ca (c) Conc. HNO₃ gives chloropicrin (nitrochloroform), a component of tear gas: CHCl₃ + HNO₃ → CCl₃NO₂ + H₂O. (d) Heated with silver powder, chloroform gives acetylene: 2CHCl₃ + 6Ag → HC≡CH + 6AgCl.

3. P is 1-bromopropane, Q is hexane, R is methanal, S is ethanal and T is propan-1-ol. (a) Wurtz reaction: 2CH₃CH₂CH₂Br + 2Na → CH₃(CH₂)₄CH₃ + 2NaBr (dry ether). (b) T → P: CH₃CH₂CH₂OH + PBr₃ → CH₃CH₂CH₂Br (or HBr/H₂SO₄). (c) P → propene with alcoholic KOH: CH₃CH₂CH₂Br + KOH → CH₃CH=CH₂ + KBr + H₂O. Ozonolysis of propene: CH₃CH=CH₂ + O₃ → ozonide, which on hydrolysis with Zn/H₂O gives CH₃CHO (S) + HCHO (R).

4. A = chloroform, CHCl₃; with silver powder, 2CHCl₃ + 6Ag → B = ethyne (acetylene), HC≡CH + 6AgCl. Hydration: HC≡CH + H₂O → (dil. H₂SO₄, HgSO₄, 333 K) CH₃CHO, so C = ethanal. (C is also the product of the Rosenmund reduction: CH₃COCl + H₂ → CH₃CHO + HCl over Pd/BaSO₄.) Ethanal with dilute alkali undergoes aldol condensation: 2CH₃CHO → CH₃CH(OH)CH₂CHO, so D = 3-hydroxybutanal (aldol).

5. (a)

(b) Saytzeff's rule: in dehydrohalogenation, the alkene with the more highly substituted double bond is the major product. CH₃CH₂CHBrCH₃ + KOH(alc.) → CH₃CH=CHCH₃ (but-2-ene, major) + CH₃CH₂CH=CH₂ (but-1-ene, minor).

1. A is chloroform, CHCl₃ (anaesthetic vapour; conc. HNO₃ gives chloropicrin, the tear-gas component). (a) The isomer is ethanol (the other isomer, methoxymethane, cannot be used). CaOCl₂ + H₂O → Ca(OH)₂ + Cl₂; CH₃CH₂OH + Cl₂ → CH₃CHO + 2HCl; CH₃CHO + 3Cl₂ → CCl₃CHO + 3HCl; 2CCl₃CHO + Ca(OH)₂ → 2CHCl₃ + (HCOO)₂Ca. (b) (i) In air and light chloroform is oxidised to the poisonous phosgene: 2CHCl₃ + O₂ → 2COCl₂ + 2HCl. (ii) With silver powder it gives ethyne: 2CHCl₃ + 6Ag → HC≡CH + 6AgCl. (iii) With acetone in the presence of a base it gives chloretone, a hypnotic: (CH₃)₂CO + CHCl₃ → (CH₃)₂C(OH)CCl₃. (iv) With aniline and alcoholic KOH it gives phenyl isocyanide (carbylamine reaction), a foul-smelling liquid: C₆H₅NH₂ + CHCl₃ + 3KOH → C₆H₅NC + 3KCl + 3H₂O. (c) Ether is stored over iron wire because, on standing in air and light, ether forms explosive peroxides; iron reduces these peroxides. Ethanol dissolves in water because its –OH group forms hydrogen bonds with water molecules, so the small non-polar ethyl group cannot prevent miscibility.

2. (a) Markovnikov's rule: when an unsymmetrical reagent HX adds to an unsymmetrical alkene, the hydrogen adds to the carbon that already carries more hydrogens and the halogen to the carbon with fewer hydrogens. (b) CH₃CH=CH₂ + HBr → A = 2-bromopropane, CH₃CHBrCH₃. A + aq. KOH → B = propan-2-ol, CH₃CH(OH)CH₃. B + PBr₃ → C = 2-bromopropane, CH₃CHBrCH₃ (3CH₃CH(OH)CH₃ + PBr₃ → 3CH₃CHBrCH₃ + H₃PO₃). C + AgCN(alc.) → D = isopropyl isocyanide (2-isocyanopropane), (CH₃)₂CH–NC + AgBr: silver cyanide is covalent, so the attack is through nitrogen. D + LiAlH₄ (or H₂/Ni) → E = N-methylpropan-2-amine, (CH₃)₂CH–NH–CH₃. (c) E is a secondary amine; with HNO₂ (NaNO₂ + HCl) it gives a yellow oily N-nitrosoamine: (CH₃)₂CH–NH–CH₃ + HNO₂ → (CH₃)₂CH–N(NO)–CH₃ + H₂O. (d) B → C (with PBr₃) → propane by reduction: CH₃CHBrCH₃ + Zn + HCl → CH₃CH₂CH₃ + ZnBrCl (or heat B with HI and red phosphorus).

1. A is chlorobenzene, C₆H₅Cl. (a) C₆H₅N₂Cl + CuCl/HCl → C₆H₅Cl + N₂ (Sandmeyer reaction); Cu powder and HCl give the same product (Gattermann reaction). (b) B = DDT, 2C₆H₅Cl + CCl₃CHO → (conc. H₂SO₄) (ClC₆H₄)₂CH–CCl₃ + H₂O (1,1,1-trichloro-2,2-bis(4-chlorophenyl)ethane). C = 4-chloroacetophenone, ClC₆H₄COCH₃ (Friedel–Crafts acylation; the para isomer is the major product). (c) In chlorobenzene the carbon–chlorine bond is shorter and stronger than in chloroethane: the carbon is sp² hybridised, and the lone pair on chlorine is delocalised into the ring by resonance, giving the C–Cl bond partial double-bond character. The electron-rich ring also repels the approaching OH⁻ nucleophile, and the phenyl cation needed for SN1 is unstable. The haloalkane has none of these effects, so substitution is easy.

2. (a) Phenol → benzene → chlorobenzene → DDT. C₆H₅OH + Zn → C₆H₆ + ZnO (distil with zinc dust) C₆H₆ + Cl₂ → (anhyd. FeCl₃) C₆H₅Cl + HCl 2C₆H₅Cl + CCl₃CHO → (conc. H₂SO₄) DDT + H₂O (b) Aniline → diazonium salt → chlorobenzene (Sandmeyer): C₆H₅NH₂ + NaNO₂ + 2HCl → (273–278 K) C₆H₅N₂Cl + NaCl + 2H₂O; C₆H₅N₂Cl + CuCl → C₆H₅Cl + N₂.

3. (a) Fittig reaction: 2C₆H₅Br + 2Na → C₆H₅–C₆H₅ (biphenyl) + 2NaBr (dry ether). (b) Wurtz–Fittig reaction: C₆H₅Br + CH₃Br + 2Na → C₆H₅CH₃ (toluene) + 2NaBr. (c) C₆H₅Cl + HNO₃ → (conc. H₂SO₄) 4-nitrochlorobenzene (major) + 2-nitrochlorobenzene (minor) + H₂O. Chlorine is deactivating by its −I effect but ortho–para directing because its lone pairs donate by resonance.

1. (a) The lower reactivity of chlorobenzene has three causes. (i) Resonance: the lone pair of chlorine is delocalised into the benzene ring, so the C–Cl bond has partial double-bond character and is stronger and shorter (about 169 pm against 177 pm in chloroethane). (ii) Hybridisation: the carbon bearing chlorine is sp² (more electronegative, holding the bond electrons closer) and not sp³. (iii) The electron-rich π cloud repels an incoming nucleophile, and the phenyl cation needed for SN1 is unstable, while backside attack for SN2 is blocked by the ring. (b) From benzene: C₆H₆ + Cl₂ → (anhydrous FeCl₃, dark) C₆H₅Cl + HCl. From benzenediazonium chloride (Sandmeyer): C₆H₅N₂Cl + CuCl/HCl → C₆H₅Cl + N₂; (Gattermann): C₆H₅N₂Cl + Cu/HCl → C₆H₅Cl + N₂. (c) 2C₆H₅Cl + CCl₃CHO → (conc. H₂SO₄) (ClC₆H₄)₂CH–CCl₃ + H₂O. Use: an insecticide used against mosquitoes and in malaria control. Harm: it is not biodegradable, accumulates in the food chain, and thins the eggshells of birds; its use has been restricted.

1. Primary: ethanol, CH₃CH₂OH. Secondary: propan-2-ol, (CH₃)₂CHOH. Victor Meyer's test. The alcohol is converted into its iodide, then the nitroalkane, then treated with nitrous acid and made alkaline:

For a 1° alcohol, CH₃CH₂OH → CH₃CH₂I → CH₃CH₂NO₂ → CH₃C(NO₂)=NOH (ethyl nitrolic acid) → red sodium salt. For a 2° alcohol, (CH₃)₂CHOH → (CH₃)₂CHI → (CH₃)₂CHNO₂ → (CH₃)₂C(NO₂)NO (pseudonitrole) → blue. The red or blue colour therefore tells primary from secondary.

2. (a) Fermentation is the slow decomposition of complex organic compounds such as sugars into simpler ones, with the evolution of a gas, by the action of enzymes from micro-organisms. (b) Molasses is the dark, viscous mother liquor left after the crystallisation of sugar from sugar-cane juice; it still holds about 40–50 % fermentable sugar. (c) Yeast supplies the enzymes invertase and zymase that catalyse the conversion. (d) C₁₂H₂₂O₁₁ + H₂O → (invertase) C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (fructose); C₆H₁₂O₆ → (zymase) 2C₂H₅OH + 2CO₂. (e) Absolute alcohol is 100 % ethanol; denatured alcohol is ethanol made unfit for drinking by adding methanol, pyridine or a dye, so it is sold without the heavy excise duty.

3. The primary alcohol of M = 46 is ethanol. (a) C₂H₅OH + 4I₂ + 6NaOH → CHI₃↓ (yellow) + HCOONa + 5NaI + 5H₂O (iodoform test). (b) CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O (conc. H₂SO₄, warm): ethyl ethanoate, a fruity-smelling ester. (c) With conc. H₂SO₄ at 443 K, dehydration gives ethene: C₂H₅OH → CH₂=CH₂ + H₂O; at 413 K with excess alcohol, ethoxyethane: 2C₂H₅OH → C₂H₅OC₂H₅ + H₂O. (d) Ethanol gives the yellow iodoform precipitate with I₂/NaOH, methanol does not (it has no CH₃CH(OH)– group).

4. The tertiary alcohol of C₄H₁₀O is (CH₃)₃C–OH, 2-methylpropan-2-ol. The isomers of C₃H₈O that are alcohols are CH₃CH₂CH₂OH, propan-1-ol, and CH₃CH(OH)CH₃, propan-2-ol (the third isomer, methoxyethane, is an ether). They are distinguished by Victor Meyer's test: propan-1-ol gives a blood-red colour (nitrolic acid sodium salt) and propan-2-ol a blue colour (pseudonitrole).

5. (a) The oxo process converts an alkene to an aldehyde with carbon monoxide and hydrogen, and then reduces the aldehyde: CH₃CH=CH₂ + CO + H₂ → (Co₂(CO)₈, 420 K, 100 atm) CH₃CH₂CH₂CHO → (H₂, Ni) CH₃CH₂CH₂CH₂OH (butan-1-ol); with ethene the product is propan-1-ol. (b) Hydroboration–oxidation: 6CH₂=CH₂ + B₂H₆ → 2(CH₃CH₂)₃B; (CH₃CH₂)₃B + 3H₂O₂ + NaOH → 3CH₃CH₂OH + Na₃BO₃ (anti-Markovnikov addition of water). (c) Absolute alcohol: 100 % ethanol. Power alcohol: a mixture of ethanol (about 20 %) and petrol used as a motor fuel. Rectified spirit: ethanol of about 95 % purity (the azeotrope with water) obtained by fractional distillation.

1. (a) (i) C₂H₅Br + KOH(aq) → C₂H₅OH + KBr. (ii) CH₃CHO + H₂ → (Ni) C₂H₅OH, or with LiAlH₄. (iii) C₆H₁₂O₆ → (zymase, 303 K) 2C₂H₅OH + 2CO₂. (b) (i) Sodium: 2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂↑ (sodium ethoxide; shows the weak acidic nature). (ii) PCl₅: C₂H₅OH + PCl₅ → C₂H₅Cl + POCl₃ + HCl↑ (white fumes; a test for –OH). (iii) Conc. H₂SO₄: at 443 K, C₂H₅OH → CH₂=CH₂ + H₂O; at 413 K with excess alcohol, 2C₂H₅OH → C₂H₅OC₂H₅ + H₂O. (iv) Acidified K₂Cr₂O₇: a primary alcohol is oxidised to the aldehyde and then the acid: CH₃CH₂OH → (O) CH₃CHO → (O) CH₃COOH (the orange Cr₂O₇²⁻ turns green Cr³⁺). (v) Heated copper at 573 K: catalytic dehydrogenation, CH₃CH₂OH → CH₃CHO + H₂ (a secondary alcohol gives a ketone and a tertiary alcohol gives an alkene).

1. (a) Dow's process: C₆H₅Cl + 2NaOH → (623 K, 300 atm) C₆H₅ONa + NaCl + H₂O; C₆H₅ONa + HCl → C₆H₅OH + NaCl. (b) The –OH group is a strong activating group: its lone pair is delocalised into the ring (+M effect), raising the electron density at the ortho and para positions. The ring therefore attacks the NO₂⁺ electrophile more readily, and phenol is nitrated by dilute HNO₃ at room temperature, whereas benzene needs a conc. HNO₃–H₂SO₄ mixture. (c) Two molecules of phenol condense with phthalic anhydride in conc. H₂SO₄: 2C₆H₅OH + C₆H₄(CO)₂O → (conc. H₂SO₄) phenolphthalein + H₂O.

2. (a) Acidity order: ethanoic acid > phenol > ethanol. In ethanol the ethoxide ion has the charge on oxygen with an electron-releasing ethyl group, so it is unstable and ethanol is very weakly acidic. In phenol the phenoxide ion is stabilised by resonance, the negative charge being spread over the ring, so phenol (pKa ≈ 10) is more acidic than ethanol (pKa ≈ 16) and dissolves in NaOH. In ethanoate the charge is shared equally by two oxygen atoms (resonance), a more effective delocalisation, so ethanoic acid (pKa 4.76) is stronger still. (b) NaHCO₃ test: a carboxylic acid liberates CO₂ with effervescence; phenol is too weak an acid and gives none. (Phenol gives a violet colour with neutral FeCl₃, which carboxylic acids do not.)

3. (a) Reimer–Tiemann: C₆H₅OH + CHCl₃ + 3NaOH → o-HOC₆H₄CHO (salicylaldehyde) + 3NaCl + 2H₂O. (b) Kolbe: C₆H₅ONa + CO₂ → (400 K, 4–7 atm) o-HOC₆H₄COONa → (H⁺) salicylic acid. (c) Coupling: C₆H₅N₂Cl + C₆H₅OH → (alkaline, 273 K) p-HOC₆H₄–N=N–C₆H₅ + HCl (p-hydroxyazobenzene, an orange dye). (d) C₆H₅OH + Zn → C₆H₆ + ZnO. (e) C₆H₅OH + 3Br₂ → 2,4,6-tribromophenol (white precipitate) + 3HBr.

4. FeCl₃ test: a few drops of neutral FeCl₃ with phenol give a violet (purple) colour because of a complex, [Fe(OC₆H₅)₆]³⁻. Bromine water test: aqueous bromine is decolourised and a white precipitate of 2,4,6-tribromophenol forms at once. Liebermann test: phenol is warmed with conc. H₂SO₄ and a crystal of NaNO₂; a deep green or blue colour forms that turns red on dilution with water and green or blue again on adding alkali.

1. (a) (i) From chlorobenzene: C₆H₅Cl + 2NaOH → (623 K, 300 atm) C₆H₅ONa + NaCl + H₂O, then HCl gives phenol. (ii) From diazonium salt: C₆H₅N₂Cl + H₂O → (warm, 283 K) C₆H₅OH + N₂ + HCl. (iii) From benzene sulphonic acid: C₆H₅SO₃H + 2NaOH → (fusion, 573 K) C₆H₅ONa + Na₂SO₃ + H₂O, then C₆H₅ONa + HCl → C₆H₅OH. (b) Phenol reacts with NaOH and with sodium: C₆H₅OH + NaOH → C₆H₅ONa + H₂O, but not with NaHCO₃. Phenol is more acidic than water and alcohol because its conjugate base, the phenoxide ion, is stabilised by resonance; in an alkoxide the negative charge on oxygen is intensified by the electron-releasing alkyl group. Water has pKa 15.7, ethanol 16, phenol about 10. Phenol is a weaker acid than carbonic acid, so it does not liberate CO₂ from carbonates. (c) Nitration: C₆H₅OH + dil. HNO₃ → o-nitrophenol + p-nitrophenol + H₂O (conc. HNO₃/H₂SO₄ gives picric acid, 2,4,6-trinitrophenol). Bromination: C₆H₅OH + 3Br₂ → 2,4,6-tribromophenol↓ + 3HBr (in water); in CS₂ at low temperature, p-bromophenol is the main product.

1. Williamson's synthesis: C₂H₅Br + C₂H₅ONa → C₂H₅OC₂H₅ + NaBr, with sodium ethoxide from 2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂. (a) Heated with excess HI, ethoxyethane gives two molecules of iodoethane: C₂H₅OC₂H₅ + 2HI → 2C₂H₅I + H₂O. (b) On exposure to air and light ether forms explosive peroxides, which have higher boiling points and concentrate in the residue; if the ether is distilled to dryness the peroxide residue may explode. (c) Phenol with NaOH gives sodium phenoxide, which with CH₃Br (Williamson) gives anisole: C₆H₅OH + NaOH → C₆H₅ONa + H₂O; C₆H₅ONa + CH₃Br → C₆H₅OCH₃ + NaBr.

2. (a) Ether forms peroxides by auto-oxidation in air; iron wire reduces the peroxides as they form. Ethanol dissolves in water because it forms hydrogen bonds with water through its –OH group. (b) Cold HI: C₂H₅OC₂H₅ + HI → C₂H₅OH + C₂H₅I. Conc. H₂SO₄ (cold): ether dissolves as an oxonium salt, [(C₂H₅)₂OH]⁺HSO₄⁻, because the oxygen lone pair can accept a proton. Cl₂ in sunlight: substitution at the carbon next to oxygen, C₂H₅OC₂H₅ + 10Cl₂ → C₂Cl₅OC₂Cl₅ + 10HCl (perchloroethoxyethane with excess chlorine).

3. One correct sequence: ethoxyethane → iodoethane → ethyl cyanide → propanamine, with ethanol as the second product of the first step. C₂H₅OC₂H₅ + HI(cold) → C₂H₅I + C₂H₅OH C₂H₅I + KCN(alc.) → C₂H₅CN + KI C₂H₅CN + 4[H] → (LiAlH₄ or H₂/Ni) C₂H₅CH₂NH₂ (propanamine) Ethanol is also the starting point for iodoethane: C₂H₅OH + HI → C₂H₅I + H₂O, which closes the list.

1. (a) Williamson's synthesis is the reaction of a sodium alkoxide (or phenoxide) with a primary alkyl halide, which gives an ether by SN2 substitution: R–ONa + R′–X → R–O–R′ + NaX. Ethoxyethane: C₂H₅ONa + C₂H₅Br → C₂H₅OC₂H₅ + NaBr. Anisole: C₆H₅ONa + CH₃Br → C₆H₅OCH₃ + NaBr (the phenoxide from phenol and NaOH). A tertiary halide cannot be used, because the strongly basic alkoxide would eliminate HX to give an alkene. (b) (i) C₂H₅OC₂H₅ + 2HI → (hot, excess) 2C₂H₅I + H₂O. (ii) Cold conc. H₂SO₄ dissolves ether as an oxonium salt, (C₂H₅)₂O + H₂SO₄ → [(C₂H₅)₂OH]⁺HSO₄⁻, which is how ether is separated from alkanes. (iii) Air and light: ether slowly forms explosive peroxides, so it is stored over iron wire and away from light. (iv) Chlorine in sunlight substitutes the hydrogen on the carbons next to oxygen, finally giving C₂Cl₅OC₂Cl₅. (c) Ethers are used as solvents for fats, oils and Grignard reagents, and diethyl ether was the first general anaesthetic; it is also used as a fuel starter for engines.

1. (a) M = propanone (acetone), CH₃COCH₃ (it has the CH₃CO– group). (b) Oxidation of propan-2-ol: CH₃CH(OH)CH₃ + [O] → (K₂Cr₂O₇/H₂SO₄) CH₃COCH₃ + H₂O; or dehydrogenation over hot copper at 573 K. (c) Clemmensen reduction: CH₃COCH₃ + 4[H] → (Zn–Hg, conc. HCl) CH₃CH₂CH₃ + H₂O (or Wolff–Kishner). (d) CH₃COCH₃ + CH₃MgBr → (CH₃)₂C(OMgBr)CH₃; H₂O/H⁺ → (CH₃)₃C–OH, 2-methylpropan-2-ol, a tertiary alcohol. (e) 2,4-DNP test: propanone gives an orange-yellow precipitate of 2,4-dinitrophenylhydrazone (all carbonyl compounds give it).

2. (i) Carbonylation: CH₃ONa + CO → (425 K, 3 atm) CH₃COONa. (ii) Fehling's test: CH₃CHO + 2Cu²⁺ + 5OH⁻ → CH₃COO⁻ + Cu₂O↓ (red) + 3H₂O. (iii) Aldol: 2CH₃CHO → (dil. NaOH) CH₃CH(OH)CH₂CHO. (iv) Williamson: C₂H₅ONa + C₂H₅Br → C₂H₅OC₂H₅ + NaBr. (v) 2,4-DNP: CH₃CHO + H₂N–NH–C₆H₃(NO₂)₂ → CH₃CH=N–NH–C₆H₃(NO₂)₂ + H₂O (yellow-orange precipitate).

3. (a) Tollens': CH₃CHO + 2[Ag(NH₃)₂]OH → CH₃COONH₄ + 2Ag↓ (silver mirror) + 3NH₃ + H₂O. (b) Cannizzaro: 2HCHO + NaOH(conc.) → CH₃OH + HCOONa. (c) Decarbonylation (loss of CO): CH₃CH₂CHO → (Pd or Rh catalyst, heat) CH₃CH₃ + CO. (d) Reimer–Tiemann: C₆H₅OH + CHCl₃ + 3NaOH → o-HOC₆H₄CHO + 3NaCl + 2H₂O. (e) Oxo process: CH₂=CH₂ + CO + H₂ → (Co₂(CO)₈, 420 K, 100 atm) CH₃CH₂CHO, which is reduced to propan-1-ol.

4. (a) Clemmensen: propane, CH₃CH₂CH₃. (b) Wolff–Kishner: propane (via the hydrazone, with loss of N₂). (c) LiAlH₄: propan-2-ol, CH₃CH(OH)CH₃. (d) PCl₅: 2,2-dichloropropane, CH₃CCl₂CH₃ + POCl₃. (e) NaHSO₃: the white crystalline bisulphite addition compound, (CH₃)₂C(OH)SO₃Na.

5. (a) X = ethanal, CH₃CHO; Y = propan-2-ol, CH₃CH(OH)CH₃; Z = propanone, CH₃COCH₃. CH₃CHO + CH₃MgBr → CH₃CH(OMgBr)CH₃ (dry ether); CH₃CH(OMgBr)CH₃ + H₂O → CH₃CH(OH)CH₃ + Mg(OH)Br. CH₃CH(OH)CH₃ + [O] → (acidified KMnO₄) CH₃COCH₃ + H₂O. Each compound contains CH₃CO– (ethanal, propanone) or CH₃CH(OH)– (propan-2-ol), so all give yellow iodoform. (b) Propanone with dilute NaOH undergoes aldol condensation: 2CH₃COCH₃ → (CH₃)₂C(OH)CH₂COCH₃, diacetone alcohol (4-hydroxy-4-methylpentan-2-one).

6. Benzene → benzenediazonium chloride → phenol → benzene → toluene → benzaldehyde → benzoin: 1. C₆H₆ → (conc. HNO₃ + conc. H₂SO₄, 330 K) C₆H₅NO₂ → (Sn/HCl) C₆H₅NH₂ → (NaNO₂/HCl, 273 K) C₆H₅N₂Cl. 2. C₆H₅N₂Cl + H₂O → (warm) C₆H₅OH + N₂ + HCl. 3. C₆H₅OH + Zn (dust, heat) → C₆H₆ + ZnO. 4. C₆H₆ + CH₃Cl → (anhydrous AlCl₃) C₆H₅CH₃ + HCl (Friedel–Crafts alkylation). 5. C₆H₅CH₃ + CrO₂Cl₂ → (CS₂) chromium complex → (H₃O⁺) C₆H₅CHO (Etard reaction). 6. 2C₆H₅CHO → (alcoholic KCN, warm) C₆H₅CH(OH)COC₆H₅, benzoin (benzoin condensation).

1. (a) Hydroboration–oxidation: 6CH₃CH=CH₂ + B₂H₆ → 2(CH₃CH₂CH₂)₃B; (CH₃CH₂CH₂)₃B + 3H₂O₂ + NaOH → 3CH₃CH₂CH₂OH + Na₃BO₃ (anti-Markovnikov hydration). (b) Decarbonylation: CH₃CH₂CHO → (Pd or Rh, heat) CH₃CH₃ + CO. (c) Sandmeyer: C₆H₅N₂Cl + CuCl/HCl → C₆H₅Cl + N₂. (d) Iodoform: CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃↓ (yellow) + CH₃COONa + 3NaI + 3H₂O. (e) Elimination: CH₃CH₂Br + KOH(alc.) → CH₂=CH₂ + KBr + H₂O (dehydrohalogenation); or ethanol with conc. H₂SO₄ at 443 K. (f) Cannizzaro: 2C₆H₅CHO + NaOH(conc.) → C₆H₅CH₂OH + C₆H₅COONa. (g) Reimer–Tiemann: C₆H₅OH + CHCl₃ + NaOH → salicylaldehyde. (h) Friedel–Crafts alkylation: C₆H₆ + CH₃Cl → (anhydrous AlCl₃) C₆H₅CH₃ + HCl.

2. (a) X = ethanal, Y = propan-2-ol, Z = propanone. CH₃CHO + CH₃MgBr → (dry ether) CH₃CH(OMgBr)CH₃; + H₂O/H⁺ → CH₃CH(OH)CH₃ + Mg(OH)Br. Then CH₃CH(OH)CH₃ + [O] → (acidified K₂Cr₂O₇ or KMnO₄) CH₃COCH₃ + H₂O. (b) The iodoform test is given by compounds with the CH₃CO– group, or the CH₃CH(OH)– group, which the hypoiodite (NaOH + I₂) first oxidises to CH₃CO–. The three α-hydrogens are replaced by iodine to give –COCI₃, which is cleaved by alkali into CHI₃ (yellow) and a carboxylate. Ethanal (CH₃CO–H), propan-2-ol (CH₃CH(OH)–) and propanone (CH₃CO–) all meet this requirement. (c) Propan-2-ol → 2-iodopropane → 2-nitropropane → pseudonitrole: CH₃CH(OH)CH₃ + I₂/red P → CH₃CHICH₃; CH₃CHICH₃ + AgNO₂ → CH₃CH(NO₂)CH₃ + AgI; CH₃CH(NO₂)CH₃ + HNO₂ → (CH₃)₂C(NO₂)(NO) + H₂O. The pseudonitrole dissolves in NaOH with a blue colour. (d) Propanone can be prepared by (i) oxidation of propan-2-ol with acidified K₂Cr₂O₇; (ii) catalytic dehydrogenation of propan-2-ol over copper at 573 K; (iii) dry distillation of calcium ethanoate, (CH₃COO)₂Ca → CH₃COCH₃ + CaCO₃; or (iv) hydration of propyne with HgSO₄/dilute H₂SO₄.

1. Work backwards: E is a primary alcohol giving iodoform, so E = ethanol, CH₃CH₂OH, which comes from ethylamine D by nitrous acid. Ethylamine comes by Hofmann bromamide degradation (one carbon lost) from propanamide C, so B = propanoyl chloride and A = propanoic acid. A: CH₃CH₂COOH + SOCl₂ → B: CH₃CH₂COCl + SO₂↑ + HCl↑ B + 2NH₃ → C: CH₃CH₂CONH₂ (propanamide) + NH₄Cl C + Br₂ + 4KOH → D: CH₃CH₂NH₂ (ethanamine) + K₂CO₃ + 2KBr + 2H₂O D + HNO₂ (NaNO₂ + HCl, 273 K) → E: CH₃CH₂OH + N₂↑ + H₂O

2. The Grignard reagent adds to solid CO₂ (dry ice) and the product is hydrolysed: CH₃MgBr + CO₂ → (dry ether) CH₃COOMgBr; CH₃COOMgBr + H₂O/H⁺ → CH₃COOH + Mg(OH)Br. (i) CH₃COOH + NH₃ → CH₃COONH₄ → (heat) CH₃CONH₂ + H₂O (ethanamide). (ii) CH₃COOH + SOCl₂ → CH₃COCl + SO₂ + HCl (ethanoyl chloride). (iii) 2CH₃COOH → (P₂O₅, heat) (CH₃CO)₂O + H₂O (ethanoic anhydride). (iv) CH₃COOH + 4[H] → (LiAlH₄, dry ether) CH₃CH₂OH + H₂O.

3. (a) (i) CH₃MgBr + CO₂ → CH₃COOMgBr → (H₃O⁺) CH₃COOH. (ii) CH₃CN + 2H₂O → (H⁺ or OH⁻, heat) CH₃COOH + NH₃. (iii) CH₃ONa + CO → (425 K, 3 atm) CH₃COONa; CH₃COONa + HCl → CH₃COOH + NaCl. (b) CH₃COOH → (NH₃, heat) CH₃CONH₂ → (Br₂/KOH) CH₃NH₂ → (HNO₂) CH₃OH → (K₂Cr₂O₇/H⁺) HCOOH. (Each step loses or oxidises one carbon unit: Hofmann removes the carboxyl carbon, and oxidation of methanol gives methanoic acid.)

4. (a) FCH₂COOH > ClCH₂COOH > C₆H₅COOH > CH₃COOH. Electron-withdrawing groups stabilise the carboxylate ion by the −I effect and increase acidity: F (pKa of FCH₂COOH 2.66) is more electronegative than Cl (2.86). The phenyl group is weakly electron-withdrawing (sp² carbon), so benzoic acid (4.20) is stronger than ethanoic acid (4.76), where the methyl group releases electrons (+I effect) and destabilises the ion. (b) Tollens' test (or Fehling's, or HgCl₂): methanoic acid has an aldehyde group and gives a silver mirror; ethanoic acid does not. (c) HVZ reaction: α-halogenation of a carboxylic acid with Cl₂ or Br₂ in the presence of red phosphorus: CH₃COOH + Cl₂ → (red P) ClCH₂COOH + HCl.

5. (a) CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O (a primary amine with one carbon fewer). (b) An amide can accept a proton on its nitrogen lone pair (weakly basic) and can lose an N–H proton to a strong base such as sodium, forming a salt (weakly acidic); it is amphoteric. (c) Acid chloride > anhydride > ester > amide. (d) Claisen condensation: two molecules of an ester with α-hydrogens condense in the presence of sodium ethoxide to give a β-keto ester: 2CH₃COOC₂H₅ → (C₂H₅ONa) CH₃COCH₂COOC₂H₅ + C₂H₅OH (ethyl acetoacetate).

1. (e) A = ethanoic acid, CH₃COOH (acidic, turns blue litmus red) and B = methyl methanoate, HCOOCH₃ (an ester with a fruity smell). (a) CH₃MgBr + CO₂ → CH₃COOMgBr → (H₂O/H⁺) CH₃COOH + Mg(OH)Br. (b) Ethanoic acid is first converted to methanol, and methanoic acid is made from it: CH₃COOH → (NH₃, heat) CH₃CONH₂ → (Br₂/KOH) CH₃NH₂ → (HNO₂) CH₃OH. Part of the methanol is oxidised, CH₃OH → (K₂Cr₂O₇/H⁺) HCOOH, and the rest is esterified with it: HCOOH + CH₃OH ⇌ (conc. H₂SO₄) HCOOCH₃ + H₂O. (Any valid route scores.) (c) Tollens' test: methanoic acid gives a silver mirror; ethanoic acid does not. (Methanoic acid also decolourises warm KMnO₄ and gives a white precipitate then Hg with HgCl₂.) (d) FCH₂COOH > ClCH₂COOH > C₆H₅COOH > CH₃COOH (the stronger the electron-withdrawing group, the stronger the acid; the +I methyl group weakens ethanoic acid).

2. (a) Acid chloride: CH₃COOH + SOCl₂ → CH₃COCl + SO₂ + HCl (or PCl₅, PCl₃). Amide: CH₃COOH + NH₃ → CH₃COONH₄ → (heat) CH₃CONH₂ + H₂O. Ester: CH₃COOH + C₂H₅OH ⇌ (conc. H₂SO₄) CH₃COOC₂H₅ + H₂O. Anhydride: 2CH₃COOH → (P₂O₅, heat) (CH₃CO)₂O + H₂O; or CH₃COCl + CH₃COONa → (CH₃CO)₂O + NaCl. (b) Hydrolysis: CH₃COCl + H₂O → CH₃COOH + HCl. Ammonolysis: CH₃COCl + 2NH₃ → CH₃CONH₂ + NH₄Cl. Alcoholysis: CH₃COCl + C₂H₅OH → CH₃COOC₂H₅ + HCl. (c) Acid chloride > anhydride > ester > amide — the better the leaving group (Cl⁻, RCOO⁻, RO⁻, NH₂⁻) and the less the lone pair donates into the carbonyl, the more reactive the derivative.

1. (a) The NO₂ group withdraws electrons by both the −I and −M effects. In the resonance structures of nitrobenzene the ortho and para carbons carry a positive charge, so they are less electron-rich than the meta carbons. The electrophile therefore attacks the meta position, and the ring is deactivated towards substitution. (b) (i) Zn/NH₄Cl (neutral): C₆H₅NO₂ + 4[H] → C₆H₅NHOH + H₂O (N-phenylhydroxylamine). (ii) LiAlH₄: 2C₆H₅NO₂ + 8[H] → C₆H₅N=NC₆H₅ + 4H₂O (azobenzene). (c) C₂H₇N has two isomers: CH₃CH₂NH₂ (ethanamine, primary) and CH₃NHCH₃ (N-methylmethanamine, secondary).

2. Nitration: C₆H₆ + HNO₃ → (conc. H₂SO₄, 330 K) C₆H₅NO₂ + H₂O. (i) Azoxybenzene: 2C₆H₅NO₂ + 6[H] → (Na₃AsO₃/NaOH, weakly alkaline) C₆H₅N(O)=NC₆H₅ + 3H₂O. (ii) p-Aminophenol: C₆H₅NO₂ + 4[H] → (electrolytic reduction in conc. H₂SO₄; the intermediate phenylhydroxylamine rearranges) p-H₂NC₆H₄OH + H₂O. (iii) Aniline: C₆H₅NO₂ + 6[H] → (Sn/conc. HCl, acidic medium) C₆H₅NH₂ + 2H₂O. (iv) Azobenzene: 2C₆H₅NO₂ + 8[H] → (Zn/NaOH, limited, or LiAlH₄) C₆H₅N=NC₆H₅ + 4H₂O.

3. (a) Nitroethane: C₂H₅Br + AgNO₂ → C₂H₅NO₂ + AgBr (the nitro compound is the main product with silver nitrite). Nitrobenzene: nitration of benzene with conc. HNO₃ + conc. H₂SO₄ at 330 K. (b) C₂H₅NO₂ + 6[H] → (Sn/HCl) C₂H₅NH₂ + 2H₂O (ethanamine). (c) Electrolytic reduction in conc. H₂SO₄ gives p-aminophenol, the raw material for paracetamol. (d) Nitrobenzene (oil of mirbane) is used as a solvent, in shoe polish and soap perfume, and above all to make aniline for dyes.

1. The product depends on the medium:

In acid, reduction goes all the way to the amine. In neutral solution it stops half-way at the hydroxylamine. In alkaline media the intermediates couple so that two rings are joined (azoxy, then azo, then hydrazo). In strongly acidic electrolytic reduction the phenylhydroxylamine rearranges to put the –OH para to the nitrogen.

1. Hoffmann's method uses diethyl oxalate, (COOC₂H₅)₂, which reacts differently with the three classes. - Primary amine: (COOC₂H₅)₂ + 2RNH₂ → (CONHR)₂ + 2C₂H₅OH, a solid dialkyl oxamide. - Secondary amine: (COOC₂H₅)₂ + R₂NH → C₂H₅OOC–CO–NR₂ + C₂H₅OH, a liquid oxamic ester. - Tertiary amine: no reaction (no N–H), so it is recovered as such. Separation: the mixture is heated with diethyl oxalate and the products are distilled. The tertiary amine distils first (it is volatile). The liquid oxamic ester (b.p. higher) distils next, leaving the solid oxamide in the flask, or the solid is filtered off. The oxamide and the ester are then each hydrolysed with aqueous KOH to give back the primary amine, (CONHR)₂ + 2KOH → 2RNH₂ + (COOK)₂, and the secondary amine, C₂H₅OOC–CO–NR₂ + 2KOH → R₂NH + (COOK)₂ + C₂H₅OH.

2. (a) Heat the mixture with diethyl oxalate. Propylamine (A) forms the solid N,N′-dipropyloxamide, (CONHC₃H₇)₂, and N-methylethanamine (B) forms the liquid oxamic ester, C₂H₅OOC–CO–N(CH₃)C₂H₅. The liquid is distilled off (or the solid is filtered), and each is hydrolysed with KOH to regenerate the amine. (b) Nitrous acid test (NaNO₂ + HCl): A, a primary amine, gives colourless gas bubbles of N₂ and the alcohol: CH₃CH₂CH₂NH₂ + HNO₂ → CH₃CH₂CH₂OH + N₂↑ + H₂O. B, a secondary amine, gives a yellow oily liquid, the N-nitrosoamine, with no gas: CH₃CH₂NHCH₃ + HNO₂ → CH₃CH₂N(NO)CH₃ + H₂O.

3. Benzene → nitrobenzene → aniline → diazonium chloride → phenol, and diazonium chloride + aniline → p-aminoazobenzene: C₆H₆ + HNO₃ → (conc. H₂SO₄, 330 K) C₆H₅NO₂ + H₂O C₆H₅NO₂ + 6[H] → (Sn/HCl) C₆H₅NH₂ + 2H₂O C₆H₅NH₂ + NaNO₂ + 2HCl → (273–278 K) C₆H₅N₂Cl + NaCl + 2H₂O C₆H₅N₂Cl + H₂O → (warm) C₆H₅OH + N₂ + HCl C₆H₅N₂Cl + C₆H₅NH₂ → (pH 4–5, 273 K) p-H₂NC₆H₄–N=N–C₆H₅ (p-aminoazobenzene) + HCl (coupling).

4. (a) Basic strength: ethylamine > ammonia > aniline. In ethylamine the ethyl group is electron-releasing (+I), so the lone pair on nitrogen is more available for protonation than in ammonia. In aniline the lone pair on nitrogen is delocalised into the benzene ring by resonance, so it is much less available and aniline (Kb ≈ 4 × 10⁻¹⁰) is a weaker base than ammonia (Kb ≈ 1.8 × 10⁻⁵). (b) Azo-dye test: treat the amine with NaNO₂/HCl at 273 K and then with alkaline β-naphthol. An aryl amine gives a stable diazonium salt that couples to give an orange-red dye; an alkyl primary amine gives an unstable salt that decomposes to an alcohol and nitrogen, and no dye forms.

5. (a) Carbylamine: C₂H₅NH₂ + CHCl₃ + 3KOH → C₂H₅NC (foul-smelling ethyl isocyanide) + 3KCl + 3H₂O. (b) Conc. HCl: C₂H₅NH₂ + HCl → C₂H₅NH₃⁺Cl⁻. (c) Acylation: C₂H₅NH₂ + CH₃COCl → CH₃CONHC₂H₅ + HCl. (d) Nitrous acid: C₂H₅NH₂ + HNO₂ → C₂H₅OH + N₂ + H₂O. (e) Alkylation: C₂H₅NH₂ + 3CH₃Br → C₂H₅N⁺(CH₃)₃Br⁻ + 2HBr (quaternary salt with excess halide).

1. (a) Azo-dye test with NaNO₂/HCl at 273 K followed by alkaline β-naphthol: an aryl amine gives an orange-red azo dye, an alkyl amine does not (the alkyl diazonium salt decomposes at once to an alcohol and N₂). (b) Ethylamine (primary) and dimethylamine (secondary) are heated with diethyl oxalate. Ethylamine gives the solid N,N′-diethyloxamide, (CONHC₂H₅)₂; dimethylamine gives the liquid ester C₂H₅OOC–CO–N(CH₃)₂. The liquid is distilled from the solid. Each is boiled with aqueous KOH to regenerate the amine: (CONHC₂H₅)₂ + 2KOH → 2C₂H₅NH₂ + (COOK)₂. (c) In water the order is trimethylamine < propylamine < ethylmethylamine (3° < 1° < 2°): the +I effect raises the basicity, but in the tertiary amine the three methyl groups hinder solvation of the ion, so it is the weakest of the three. (d) CH₃CH₂NH₂ → (HNO₂) CH₃CH₂OH → (K₂Cr₂O₇/H⁺) CH₃COOH → (NH₃, heat) CH₃CONH₂ → (Br₂/KOH) CH₃NH₂ (Hofmann bromamide, one carbon fewer).

2. (a) (i) Reduction of nitrobenzene: C₆H₅NO₂ + 6[H] → (Sn/HCl, then NaOH) C₆H₅NH₂ + 2H₂O. (ii) From phenol: C₆H₅OH + NH₃ → (anhydrous ZnCl₂, 573 K, pressure) C₆H₅NH₂ + H₂O. (b) (i) C₆H₅NH₂ + CH₃COCl → C₆H₅NHCOCH₃ (acetanilide) + HCl. (ii) C₆H₅NH₂ + 3Br₂ → 2,4,6-tribromoaniline (white precipitate) + 3HBr. (iii) C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂Cl + NaCl + 2H₂O. (iv) C₆H₅NH₂ + CHCl₃ + 3KOH → C₆H₅NC (phenyl isocyanide) + 3KCl + 3H₂O. (c) Coupling is the reaction of a diazonium salt with a phenol or an aromatic amine to form an azo compound, the parent of the azo dyes: C₆H₅N₂Cl + C₆H₅OH → (NaOH, 273 K) p-HOC₆H₄–N=N–C₆H₅ + HCl (p-hydroxyazobenzene, orange).

1. An organometallic compound contains at least one direct bond between a metal and a carbon atom. CH₃MgI (Mg–C), Ni(CO)₄ (Ni–C of the carbonyl), C₂H₅Li (Li–C) and (C₄H₉)₂SnCl₂ (Sn–C) are organometallic. C₂H₅ONa is not: sodium is bonded to oxygen (a metal alkoxide), so there is no metal–carbon bond.

2. (a) From a haloalkane: CH₃CH₂Br + Mg → (dry ether) CH₃CH₂MgBr. From a haloarene: C₆H₅Br + Mg → (dry ether, or THF) C₆H₅MgBr. (b) Grignard reagents are strongly basic and are destroyed by any water or alcohol; ether also dissolves the reagent by co-ordinating to magnesium. (c) CH₃MgBr + H₂O → CH₄ + Mg(OH)Br. (d) With HCN the Grignard reagent takes up the acidic hydrogen: CH₃MgBr + HCN → CH₄ + Mg(CN)Br (so HCN is a proton source; addition to a nitrile carbon needs RCN, not HCN).

3. Grignard reagent + carbonyl compound, then hydrolysis: - Primary alcohol: with methanal, RMgX + HCHO → RCH₂OMgX → (H₂O) RCH₂OH. Example: CH₃MgBr + HCHO → CH₃CH₂OH (ethanol) after hydrolysis. - Secondary alcohol: with any other aldehyde, RMgX + R′CHO → R(R′)CHOH. Example: CH₃MgBr + CH₃CHO → CH₃CH(OH)CH₃ (propan-2-ol). - Tertiary alcohol: with a ketone, RMgX + R′COR″ → R(R′)(R″)COH. Example: CH₃MgBr + CH₃COCH₃ → (CH₃)₃COH (2-methylpropan-2-ol).

4. (a) Organolithium: RLi, for example CH₃Li or C₄H₉Li. Organocopper: R₂CuLi (a Gilman reagent), for example (CH₃)₂CuLi. Organocadmium: R₂Cd, for example (CH₃)₂Cd, which converts an acid chloride into a ketone: R′COCl + R₂Cd → 2R′COR + CdCl₂. (b) The bond is polar covalent, with carbon the negative end. It is more ionic for Li, Na, K, Mg (electropositive metals) and more covalent for Cu, Cd, Sn, Pb. (c) Zeise's salt, K[PtCl₃(C₂H₄)], in which ethene binds to platinum through its π bond (the first organometallic compound made); ferrocene, Fe(C₅H₅)₂, is another.

1. A Grignard reagent is an alkyl or aryl magnesium halide, RMgX, made by reacting a haloalkane or haloarene with magnesium turnings in dry ether: C₂H₅Br + Mg → C₂H₅MgBr. The C–Mg bond is polar, C^δ−–Mg^δ+, so the carbon acts as a strong nucleophile and a strong base. (a) Water: C₂H₅MgBr + H₂O → C₂H₆ + Mg(OH)Br. (b) Carbon dioxide: C₂H₅MgBr + CO₂ → C₂H₅COOMgBr → (H₃O⁺) C₂H₅COOH (propanoic acid). (c) Ethanal: C₂H₅MgBr + CH₃CHO → CH₃CH(OMgBr)C₂H₅ → (H₂O) CH₃CH(OH)C₂H₅ (butan-2-ol, secondary). (d) Propanone: C₂H₅MgBr + CH₃COCH₃ → (CH₃)₂C(OMgBr)C₂H₅ → (H₂O) (CH₃)₂C(OH)C₂H₅ (2-methylbutan-2-ol, tertiary). (e) Ethanenitrile: C₂H₅MgBr + CH₃CN → CH₃C(=NMgBr)C₂H₅ → (H₃O⁺) CH₃COC₂H₅ (butanone) + NH₃. (f) Ethanoyl chloride: C₂H₅MgBr + CH₃COCl → CH₃COC₂H₅ + MgBrCl (butanone); with excess Grignard reagent this ketone goes on to the tertiary alcohol, 3-methylpentan-3-ol.

1. (a) Condensation polymerisation is the joining of many molecules of one or more monomers, each with two or more functional groups, with the loss of a small molecule such as water or HCl at every step. Bakelite is formed from phenol (C₆H₅OH) and methanal (HCHO). (b) Addition polymers are formed by the repeated addition of unsaturated monomers (alkenes) with no loss of atoms: n CH₂=CH₂ → (–CH₂–CH₂–)n, polythene. Condensation polymers are formed from bifunctional monomers with elimination of water: n H₂N(CH₂)₆NH₂ + n HOOC(CH₂)₄COOH → nylon 6,6 + 2n H₂O.

2. (a) A homopolymer is made of one kind of monomer: polythene from ethene, (–CH₂–CH₂–)n. A copolymer is made from two or more different monomers: Buna-S from buta-1,3-diene and styrene. (b) Polystyrene: styrene (phenylethene), C₆H₅CH=CH₂, molecular formula C₈H₈. Bakelite: phenol, C₆H₅OH (C₆H₆O), and methanal, HCHO (CH₂O). (c) Polystyrene is used for disposable cups, packaging foam and thermal insulation.

3. (a) Nylon 6,6: hexane-1,6-diamine, H₂N–(CH₂)₆–NH₂, and hexanedioic acid (adipic acid), HOOC–(CH₂)₄–COOH. Use: textile fibres, ropes, fishing nets, parachutes. (b) Polystyrene: styrene, C₆H₅–CH=CH₂. Use: packaging, toys, insulation. (c) A fibre is a polymer with strong intermolecular forces (hydrogen bonds in nylon) that can be drawn into long threads; nylon and terylene are synthetic fibres, and cotton and silk are natural.

4. (a) Aspirin (2-acetoxybenzoic acid): a benzene ring with –COOH and –OCOCH₃ on adjacent carbons, C₆H₄(OCOCH₃)COOH. DDT: two p-chlorophenyl groups on one carbon that also carries a CCl₃ group, (ClC₆H₄)₂CH–CCl₃. (b) An antipyretic lowers raised body temperature (paracetamol); an analgesic relieves pain without loss of consciousness (aspirin, ibuprofen). Paracetamol acts as both. (c) A habit-forming drug causes physical or psychological dependence, so the user needs more and more of it (morphine, heroin, cocaine, alcohol, tobacco).

1.

Elastomers are polymers with weak intermolecular forces and a coiled structure, so they stretch and return to their shape: natural rubber (polyisoprene) and Buna-S.

2. (a) An ideal drug (i) acts on the target site only, (ii) has few or no side effects, (iii) is effective in a small dose, and (iv) is non-addictive and stable. (b) Classification: analgesics (aspirin, paracetamol); antipyretics (paracetamol); antibiotics (penicillin); antiseptics and disinfectants (dettol, boric acid, chlorine); tranquillisers (diazepam); antacids (Mg(OH)₂, NaHCO₃); anaesthetics (chloroform, ether, N₂O). (c) Drug addiction is physical and mental dependence on a drug such that stopping it causes withdrawal symptoms. (d) Insecticides kill insects (DDT, malathion); herbicides kill weeds (2,4-D); fungicides kill fungi (Bordeaux mixture, CuSO₄ + Ca(OH)₂).

1. (a) Portland cement is a fine grey powder made by burning a mixture of limestone (CaCO₃) and clay to clinker and grinding it with a little gypsum; it sets and hardens by hydration, even under water (a hydraulic cement). (b) The major components are calcium oxide (lime, 60–67 %), silica (17–25 %), alumina (3–8 %), iron oxide (0.5–6 %), magnesia and SO₃, present in the clinker as tricalcium silicate (C₃S), dicalcium silicate (C₂S), tricalcium aluminate (C₃A) and tetracalcium aluminoferrite (C₄AF). (c) Gypsum (2–5 %) acts as a retarder: without it, tricalcium aluminate reacts with water at once and the cement undergoes flash set; gypsum forms calcium sulphoaluminate that slows the reaction, so the cement has time to be placed. (d) Vicat apparatus (initial and final setting time), Le Chatelier apparatus (soundness), and the Blaine air-permeability apparatus (fineness) or a compression-testing machine (strength).

2. See the flow sheet. Steps: (i) Crushing and grinding: limestone and clay are crushed and ground to a fine raw mix in a ball mill, and the proportions are checked (lime : silica : alumina). (ii) Strong heating: the mix passes through a rotary kiln inclined at 1 in 30 and is heated by a flame at the lower end. It is dried, then calcined (CaCO₃ → CaO + CO₂ at about 900 °C) and finally clinkered at 1400–1500 °C, where lime combines with silica, alumina and iron oxide to give greenish-black nodules of clinker. (iii) Cooling and final grinding: the clinker is cooled in air, 2–5 % gypsum is added, and the mixture is ground to a very fine powder in a ball mill, then stored in silos and bagged.

3. (a)

(b) On the basis of setting and hardening: hydraulic cements set and harden by reaction with water even under water (Portland cement, pozzolana cement); non-hydraulic cements harden only in air, by drying and absorbing CO₂ (lime mortar, gypsum plaster).

4. Quality control is exercised at every stage. (i) Raw materials: samples of limestone and clay are analysed chemically and the lime saturation and the hydraulic and silica moduli are checked so that the raw mix has the correct composition. (ii) Kiln: temperature, flame and the amount of free lime in the clinker are monitored, since excess free lime causes unsoundness. (iii) Final grinding: the gypsum dose (2–5 %) and the fineness are controlled (Blaine apparatus, sieve test). (iv) Finished cement: the Vicat apparatus tests the setting time (initial not less than 30 minutes, final not more than 600 minutes), the Le Chatelier apparatus tests soundness, and cubes are cast and crushed at 3, 7 and 28 days for compressive strength.

1. Raw materials: limestone (about 75 %, the source of CaO), clay or shale (SiO₂, Al₂O₃, Fe₂O₃), with sand, bauxite or iron ore to correct the composition, gypsum as a retarder, and coal or fuel oil as fuel. Steps (see the flow sheet in short answer 2): 1. Crushing and grinding. Limestone and clay are crushed and ground, dry or with water, to a fine mix (raw meal or slurry), blended so that the lime content is correct. 2. Strong heating in the rotary kiln (60–150 m long, inclined 1 in 30, turning 1–2 rpm): the charge is dried; at about 900 °C the limestone decomposes, CaCO₃ → CaO + CO₂; and at 1400–1500 °C the lime combines with SiO₂, Al₂O₃ and Fe₂O₃ and about a quarter of the charge melts to form clinker nodules of 5–25 mm. 3. Final grinding. Clinker is cooled, mixed with 2–5 % gypsum and ground in a ball mill to a fine powder, which is bagged. Composition of the clinker: CaO 60–67 %, SiO₂ 17–25 %, Al₂O₃ 3–8 %, Fe₂O₃ 0.5–6 %, MgO 0.1–4 %, SO₃ 1–3 %, combined as C₃S (alite, the main compound), C₂S (belite), C₃A and C₄AF. Role of gypsum: it retards the very rapid hydration of C₃A and so prevents flash set; the cement then sets in about 30–600 minutes.

1. (a) Raw materials are (i) fibrous: wood (spruce, pine), bamboo, straw, bagasse, cotton rags and waste paper, which supply cellulose fibres; and (ii) non-fibrous additives: fillers (china clay, talc, CaCO₃, TiO₂), sizing agents (rosin with alum, starch), dyes and water. (b)

(c) Lokta (the inner bark of Daphne shrubs), a bast fibre, is the traditional Nepali raw material; bamboo, sabai grass and rice straw are also used.

2. (a)

(b) Lignin is the brown cementing material between fibres; it makes paper weak and yellow. Removing it frees the cellulose fibres, and since the fibres can then lie closer it increases fibre-to-fibre bonding and the strength of the paper. (c) The sulphate (Kraft) process, with white liquor NaOH + Na₂S at about 170 °C, and the sulphite process, with Ca(HSO₃)₂ or Mg(HSO₃)₂ containing SO₂.

3. Raw material (wood, bamboo) → debarking and chipping → pulping (cooking the chips with chemicals, for example white liquor in the Kraft process at about 170 °C, to dissolve the lignin) → washing → bleaching (Cl₂, ClO₂ or H₂O₂ removes the colour) → beating and refining (the fibres are fibrillated in water) → addition of filler, size and dye → sheet formation on a moving wire screen (Fourdrinier machine) → pressing → drying on heated rollers → calendering (smoothing between rollers) → reeling and cutting.

1. Raw materials: wood (spruce, pine, eucalyptus) or bamboo as the fibrous material; white liquor (NaOH + Na₂S), bleach (ClO₂ or H₂O₂), filler (china clay), size (rosin + alum) and water. Stages: (1) Debarking and chipping: logs are debarked and cut into chips of 2–3 cm. (2) Pulping: chips are cooked in a digester with white liquor at about 170 °C and 7–8 atm for 2–4 hours; the alkali and HS⁻ break the lignin into soluble sodium salts and leave the cellulose fibres; the spent dark liquor (black liquor) is burnt to recover the chemicals. (3) Washing and screening remove the liquor and uncooked knots. (4) Bleaching with chlorine dioxide or hydrogen peroxide gives a white pulp. (5) Beating and refining roughen the fibres so they bond; filler, size and dye are added. (6) Sheet formation: the dilute stock is spread on a moving wire belt (Fourdrinier), water drains away and the fibres mat together. (7) Pressing, drying and calendering give the smooth finished paper, which is reeled. Quality of paper is judged by (i) strength (tensile and tear strength, which depend on fibre length and bonding) and (ii) brightness and opacity, together with smoothness, ink absorbency and pH (acid-free paper lasts longer).

1. (a)

(b) Artificial (induced) radioactivity is radioactivity produced in a stable nucleus by bombarding it with particles (discovered by Irène Curie and Frédéric Joliot). Example: ²⁷₁₃Al + ⁴₂He → ³⁰₁₅P + ¹₀n, and ³⁰₁₅P → ³⁰₁₄Si + ⁰₊₁e (positron). (c) The becquerel (Bq) = 1 disintegration per second (SI unit); the curie (Ci) = 3.7 × 10¹⁰ Bq.

2. (a) Fusion needs a temperature of about 10⁷–10⁸ K to give the nuclei enough kinetic energy to overcome the electrostatic repulsion between them, so it is called a thermonuclear reaction (as in the Sun). (b) Fusion: ²₁H + ³₁H → ⁴₂He + ¹₀n + 17.6 MeV. Fission: ²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3¹₀n + energy (about 200 MeV per nucleus). (c) The fusion fuel is made of very light nuclei, so one kilogram contains many more nuclei than a kilogram of uranium; although fission releases more energy per nucleus, fusion releases about three to four times as much energy per unit mass.

3. (a) Natural radioactivity is the spontaneous emission of α, β or γ rays by heavy unstable nuclei found in nature (U, Th, Ra); artificial radioactivity is produced by bombarding a stable nucleus with particles, giving unstable isotopes (³⁰P, ²⁴Na) that then decay. (b) 25 % = ¼ = (½)², so two half-lives have elapsed.

Check with the decay law: N/N₀ = (½)^(t/t½) = (½)² = 0.25.

1. (a) Nuclear fission is the splitting of a heavy nucleus into two lighter nuclei, with a large release of energy, when it absorbs a neutron: ²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3¹₀n + 200 MeV. The three neutrons released each split another uranium nucleus, so a chain reaction is set up. If the chain is uncontrolled the energy is released in a flash (an atomic bomb); if the neutrons are absorbed by control rods (cadmium or boron) it is controlled, as in a reactor. (b) In a nuclear power plant a controlled fission chain heats water, the steam drives turbines and the turbines drive generators; a moderator (graphite or heavy water) slows the neutrons. A nuclear weapon uses an uncontrolled chain reaction. (c) Medical: cobalt-60 for cancer radiotherapy; iodine-131 for the diagnosis and treatment of thyroid disease; sterilisation of instruments. Industrial: radioactive tracers for detecting leaks in pipes; thickness gauges; irradiation to preserve food; radiocarbon dating of fossils. (d) Harmful effects: radiation damages cells and DNA, causing burns, cancer, genetic mutations and, at a high dose, death; radioactive waste stays hazardous for many years.