Class 12 – Chemistry
Master 30 advanced topics covering solutions, electrochemistry, kinetics, transition elements, coordination chemistry, advanced organic reactions, biomolecules and applied chemistry. Every topic now includes detailed theory, key formulas or reactions, a fully worked example, common mistakes and practice questions.
📘 How to use this page
Read the theory, learn the formula or reaction box, then attempt the worked example yourself before reading the solution. Use the “Common mistake” boxes to avoid frequent board-exam errors, and finish each topic with the practice questions. Chemistry rewards regular revision: keep a separate notebook for named reactions, exceptions and colour or precipitate tests.
🧪 Solutions and Electrochemistry
Concentration of Solutions and Composition
A solution is a homogeneous mixture of two or more components. The component present in larger amount is the solvent; the other is the solute. Composition can be expressed in several ways, and choosing the right unit matters because some depend on volume (and therefore on temperature) while others depend only on mass or moles.
Molarity (moles of solute per litre of solution) changes with temperature because the volume of a solution expands on heating. Molality (moles of solute per kilogram of solvent), mole fraction and mass percentage use only masses or moles, so they are temperature independent. For very dilute solutions such as pollutants, parts per million (ppm) is used. The sum of all mole fractions in a mixture is always 1.
Molarity M = nsolute / Vsolution(L) | Molality m = nsolute / masssolvent(kg)
Mole fraction xA = nA/(nA + nB) | Mass % = (mass of component / total mass) × 100
ppm = (mass of component / total mass) × 10⁶ | M = (10 × mass % × density) / molar mass
n(glucose) = 18/180 = 0.1 mol. Molality = 0.1/0.5 kg = 0.2 m. n(water) = 500/18 = 27.78 mol, so x(glucose) = 0.1/(0.1 + 27.78) ≈ 0.0036.
Second example: concentrated H₂SO₄ is 98% by mass with density 1.84 g mL⁻¹. M = 10 × 98 × 1.84 / 98 = 18.4 M.
- Compare molarity and molality and identify which is preferred when temperature changes.
- Find the molarity of a solution containing 5.3 g of Na₂CO₃ in 250 mL (answer: 0.2 M).
- Calculate the mole fraction of ethanol in a solution of 46 g ethanol and 54 g water.
Vapour Pressure and Raoult’s Law
Every liquid has a vapour pressure at a given temperature, the pressure of vapour in equilibrium with the liquid. In a solution of volatile liquids, Raoult’s law states that the partial vapour pressure of each component is proportional to its mole fraction in the solution, pA = xAp°A. The total pressure is the sum of partial pressures (Dalton’s law). For a non-volatile solute, the vapour pressure lowering equals the mole fraction of the solute.
An ideal solution obeys Raoult’s law at all concentrations; its mixing is accompanied by ΔmixH = 0 and ΔmixV = 0, because A–B interactions equal A–A and B–B interactions (for example n-hexane and n-heptane). Positive deviation occurs when A–B interactions are weaker (ethanol + acetone), giving a higher vapour pressure, ΔmixH > 0 and ΔmixV > 0, and minimum-boiling azeotropes. Negative deviation occurs when A–B interactions are stronger (chloroform + acetone, HNO₃ + water), giving lower vapour pressure, negative ΔmixH and ΔmixV, and maximum-boiling azeotropes. Azeotropes cannot be separated by fractional distillation.
For gases dissolved in liquids, Henry’s law states that solubility is proportional to the partial pressure: p = KHx. Higher temperature lowers gas solubility, explaining why aquatic life is stressed in warm water and why scuba divers can suffer “bends”.
pA = xAp°A | Ptotal = xAp°A + xBp°B | Vapour-phase mole fraction yA = pA/Ptotal
Relative lowering: (p° − p)/p° = xsolute | Henry: p = KHx
xA = 0.25, xB = 0.75. pA = 50, pB = 75, so P = 125 mmHg. yA = 50/125 = 0.40: the vapour is richer in the more volatile component A.
- Explain positive and negative deviations from ideal behaviour in terms of intermolecular forces.
- Why do azeotropic mixtures boil at constant composition?
- The vapour pressure of pure water at 300 K is 23.8 mmHg. Estimate the lowering when 0.1 mol of urea is dissolved in 9 mol of water.
Colligative Properties and Molar Mass
Colligative properties depend on the number of solute particles in a given amount of solvent, not on their nature. There are four: relative lowering of vapour pressure, elevation of boiling point (ΔTb), depression of freezing point (ΔTf) and osmotic pressure (π). Each can be used to determine the molar mass of a solute, but osmotic pressure is especially valuable for macromolecules and for solutions measured at room temperature because it gives a large, measurable value even for dilute solutions.
Osmosis is the flow of solvent through a semi-permeable membrane from lower to higher solute concentration. Solutions with equal osmotic pressure are isotonic; a more concentrated solution is hypertonic. When pressure larger than π is applied to the solution side, solvent flows backward: this is reverse osmosis, used in desalination.
Electrolytes dissociate (NaCl → 2 particles) and some solutes associate (ethanoic acid dimerises in benzene), changing particle number. The van ’t Hoff factor i corrects for this: i = observed colligative property / calculated value. For dissociation, i = 1 + (n − 1)α; for association, i = 1 − α(1 − 1/n).
ΔTb = Kbm | ΔTf = Kfm | π = CRT (R = 0.0821 L atm K⁻¹ mol⁻¹)
M₂ = (Kb × 1000 × w₂)/(ΔTb × w₁) | M₂ = (Kf × 1000 × w₂)/(ΔTf × w₁) (w₁ = solvent in g)
With electrolytes multiply each colligative property by i. For water: Kb = 0.52 K kg mol⁻¹, Kf = 1.86 K kg mol⁻¹
m = (6/60)/0.2 = 0.5 mol kg⁻¹. ΔTf = 1.86 × 0.5 = 0.93 K, so the solution freezes at −0.93 °C. ΔTb = 0.52 × 0.5 = 0.26 K. A 0.1 M glucose solution at 300 K has π = 0.1 × 0.0821 × 300 ≈ 2.46 atm.
- Predict how dissociation of an electrolyte affects particle count and colligative behaviour.
- 0.1 m NaCl and 0.1 m glucose: which has the lower freezing point, and why?
- Why is ethylene glycol added to car radiators in cold climates?
Electrochemical and Galvanic Cells
A galvanic (voltaic) cell converts the chemical energy of a spontaneous redox reaction into electrical energy by keeping the oxidation and reduction half-reactions in separate compartments. Oxidation occurs at the anode (negative terminal) and reduction at the cathode (positive terminal). Electrons travel through the external wire from anode to cathode, so conventional current flows the opposite way. A salt bridge (for example KNO₃ in agar gel) completes the circuit by letting ions migrate, keeping each half-cell electrically neutral and preventing liquid junction potential.
In the Daniell cell, Zn dips in ZnSO₄ and Cu in CuSO₄. The cell is written as Zn | Zn²⁺ (1 M) || Cu²⁺ (1 M) | Cu, with the anode on the left. Anode: Zn → Zn²⁺ + 2e⁻. Cathode: Cu²⁺ + 2e⁻ → Cu. Its standard EMF is 1.10 V.
An electrolytic cell does the reverse: an external source forces a non-spontaneous reaction. Here the anode is positive and the cathode negative, but oxidation is still at the anode and reduction at the cathode. Faraday’s laws link the amount of substance deposited with charge passed: m = ZIt, i.e. the mass deposited is proportional to the charge, and equal charge deposits masses proportional to equivalent masses.
1 F = 96500 C mol⁻¹ | Q = It | m = (M/nF) × It
Daniell cell reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Q = 2 × 1800 = 3600 C. Moles of electrons = 3600/96500 = 0.0373. Cu²⁺ + 2e⁻ → Cu needs 2 mol e⁻ per mol Cu, so n(Cu) = 0.01865 mol and mass = 0.01865 × 63.5 ≈ 1.18 g.
- Identify anode, cathode and electron-flow direction in a galvanic cell.
- Why can a KCl salt bridge not be used with silver nitrate solutions?
- How many faradays are needed to deposit 1 mol of Al from Al³⁺?
Electrode Potential and Cell EMF
When a metal is placed in a solution of its own ions, a potential difference develops at the interface. Since only differences can be measured, all potentials are quoted against the standard hydrogen electrode (SHE), assigned exactly 0.00 V (Pt, H₂ at 1 bar, 1 M H⁺, 298 K). Standard electrode potentials (E°) are written as reduction potentials: the more positive the value, the stronger the oxidising agent and the weaker the reducing agent of the reduced form.
The electrochemical series ranks half-reactions by E°. Any species can oxidise those below it in the reduction-potential table; a metal displaces ions of metals with higher (more positive) E°. The cell EMF is the difference between cathode and anode potentials, and it connects to thermodynamics through free energy: ΔG° = −nFE°cell = −RT ln K. A positive E°cell means a spontaneous reaction.
| Half-reaction (reduction) | E° (V) |
|---|---|
| Li⁺ + e⁻ → Li | −3.05 |
| Mg²⁺ + 2e⁻ → Mg | −2.37 |
| Al³⁺ + 3e⁻ → Al | −1.66 |
| Zn²⁺ + 2e⁻ → Zn | −0.76 |
| Fe²⁺ + 2e⁻ → Fe | −0.44 |
| 2H⁺ + 2e⁻ → H₂ | 0.00 |
| Cu²⁺ + 2e⁻ → Cu | +0.34 |
| Ag⁺ + e⁻ → Ag | +0.80 |
| Cl₂ + 2e⁻ → 2Cl⁻ | +1.36 |
| F₂ + 2e⁻ → 2F⁻ | +2.87 |
E°cell = E°cathode − E°anode (both as reduction potentials)
ΔG° = −nFE°cell | ΔG° = −2.303RT log K | log K = nE°/0.059 (298 K)
E°cell = 0.34 − (−0.76) = 1.10 V. ΔG° = −2 × 96500 × 1.10 = −212.3 kJ mol⁻¹. log K = 2 × 1.10/0.059 ≈ 37.3, so K ≈ 2 × 10³⁷, an extremely favourable reaction.
- Use two standard reduction potentials to identify the stronger oxidising agent.
- Will copper liberate hydrogen from dilute HCl? Justify using E°.
- Which of Zn and Fe is a better reducing agent, and what does this imply for galvanisation?
Nernst Equation and Non-Standard Cells
Real cells rarely operate at 1 M concentrations. The Nernst equation gives the potential at any concentration by correcting E° with the reaction quotient Q. As reactants are consumed and products accumulate, Q rises, E falls, and the cell finally reaches equilibrium where E = 0 and Q = K (a “dead” battery). This also explains why a more concentrated cathode solution or a more dilute anode solution raises the EMF.
A concentration cell uses the same electrode material in both half-cells but different ion concentrations; its EMF arises purely from the concentration difference. The Nernst relation also gives the electrode potential of the hydrogen electrode as a function of pH, the principle of the glass electrode in a pH meter: E(H⁺/H₂) = −0.059 pH volts at 298 K.
E = E° − (RT/nF) ln Q = E° − (0.059/n) log Q
For Zn + Cu²⁺ → Zn²⁺ + Cu: Q = [Zn²⁺]/[Cu²⁺]
Concentration cell: E = (0.059/n) log(c₂/c₁) | At equilibrium: E = 0, E° = (0.059/n) log K
Q = 0.1/0.01 = 10. E = 1.10 − (0.059/2) log 10 = 1.10 − 0.0295 ≈ 1.07 V. Product concentration is higher, so the voltage is slightly lower than standard.
- Explain qualitatively how increasing product concentration affects cell potential.
- Find the EMF of Ag | Ag⁺(0.01 M) || Ag⁺(1 M) | Ag at 298 K.
- What does E = 0 for a cell indicate about the reaction?
Conductance of Electrolyte Solutions
Electrolytic solutions conduct because ions move under an electric field. Resistance R and its reciprocal conductance G = 1/R depend on the cell geometry; to compare solutions we use conductivity κ = G × (l/A), where l/A is the cell constant. Its unit is S m⁻¹ (or S cm⁻¹). Molar conductivity Λm = κ/c is the conductivity of all ions produced by one mole of electrolyte between electrodes one metre (or centimetre) apart.
On dilution, κ falls because there are fewer ions per unit volume, but Λm rises because dissociation increases (weak electrolytes) and inter-ionic attraction decreases (strong electrolytes). For strong electrolytes Λm varies linearly with √c (Debye–Hückel–Onsager/Kohlrausch equation), and the intercept is the limiting molar conductivity Λ°m. Weak electrolytes show a steep rise at low concentration and cannot be extrapolated, so Λ° is obtained using Kohlrausch’s law of independent migration of ions: Λ°m is the sum of the individual contributions of cation and anion.
Temperature raises electrolytic conductance (ions move faster, viscosity falls), unlike metallic conductance, which decreases. Hydrogen and hydroxide ions have unusually high mobility due to proton hopping through the hydrogen-bonded network.
κ = G × (l/A) | Λm = κ × 1000/c (κ in S cm⁻¹, c in mol L⁻¹, Λm in S cm² mol⁻¹)
Λm = Λ°m − A√c (strong electrolytes) | α = Λm/Λ°m | Ka = cα²/(1 − α)
Λ°(CH₃COOH) = Λ°(CH₃COONa) + Λ°(HCl) − Λ°(NaCl)
α = 39.05/390.5 = 0.10. Ka = cα²/(1 − α) = 10⁻³ × 0.01 / 0.9 ≈ 1.1 × 10⁻⁵, close to the accepted value.
- Compare the dilution behaviour of strong and weak electrolytes.
- Why does Λm of a weak electrolyte rise sharply near infinite dilution?
- Why is alternating current used when measuring the conductance of an electrolyte?
Batteries, Fuel Cells and Corrosion
Primary batteries cannot be recharged because their reactions are irreversible. The dry (Leclanché) cell has a zinc anode, a graphite rod cathode surrounded by MnO₂ and carbon, and an NH₄Cl–ZnCl₂ paste; it gives about 1.5 V. The mercury cell gives a constant 1.35 V and suits watches and pacemakers. Secondary batteries can be recharged by reversing the reaction. The lead–acid storage battery has a lead anode, a PbO₂ cathode and about 38% H₂SO₄; each cell gives about 2 V and six give 12 V. Discharge: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O. Other examples are Ni–Cd cells (about 1.4 V) and lithium-ion cells, valued for high energy density.
Fuel cells convert the energy of combustion directly to electricity with efficiency around 70%, much higher than thermal plants. In the H₂–O₂ alkaline fuel cell: anode, 2H₂ + 4OH⁻ → 4H₂O + 4e⁻; cathode, O₂ + 2H₂O + 4e⁻ → 4OH⁻; overall 2H₂ + O₂ → 2H₂O. Unlike batteries, they run as long as fuel and oxidant are supplied and produce water as the only product.
Corrosion is an electrochemical process. In rusting, one spot of the iron surface acts as anode (Fe → Fe²⁺ + 2e⁻) and another as cathode (O₂ + 4H⁺ + 4e⁻ → 2H₂O); Fe²⁺ is further oxidised to hydrated Fe₂O₃·xH₂O (rust). Water, oxygen and electrolytes (acid rain, salt) accelerate rusting. Protection includes paint or oil coatings, galvanisation (zinc coating, which also protects when scratched), sacrificial anodes (Mg or Zn blocks attached to pipelines and ships) and alloying (stainless steel).
Zinc (E° = −0.76 V) is more reactive than iron (−0.44 V), so zinc oxidises first and protects iron cathodically. Tin (E° = −0.14 V) is less reactive than iron, so once scratched, iron becomes the anode and corrodes faster.
- Compare a rechargeable battery with a fuel cell.
- Write the anode and cathode reactions during discharge of a lead storage battery.
- Why is the density of the electrolyte used to check the charge of a lead-acid battery?
⏱️ Chemical Kinetics
Reaction Rate and Factors Affecting Rate
Chemical kinetics measures how fast reactants become products and what controls that speed. The average rate is the change in concentration over a time interval; the instantaneous rate is the slope of the concentration–time curve at one moment. Rates are expressed in mol L⁻¹ s⁻¹ and, for a general reaction aA + bB → cC + dD, are written so that all species give the same positive number: Rate = −(1/a)Δ[A]/Δt = −(1/b)Δ[B]/Δt = (1/c)Δ[C]/Δt = (1/d)Δ[D]/Δt.
Rate depends on concentration (or pressure for gases), temperature, nature of reactants, surface area of solids, light and catalysts. Collision theory explains this: reaction needs collisions with sufficient energy (greater than the threshold energy) and the correct orientation (steric factor). Raising concentration increases collision frequency; raising temperature increases the fraction of molecules with enough energy; a catalyst gives a lower-energy path. As a rough guide, a 10 K rise often doubles or triples the rate.
Average rate = −Δ[R]/Δt = +Δ[P]/Δt | Instantaneous rate = −d[R]/dt
Temperature coefficient = k(T + 10)/k(T) ≈ 2 to 3
Rate = ¼ d[NO₂]/dt = 4.0 × 10⁻³ mol L⁻¹ s⁻¹. −d[N₂O₅]/dt = 2 × rate = 8.0 × 10⁻³ mol L⁻¹ s⁻¹, and d[O₂]/dt = 4.0 × 10⁻³.
- Explain why raising temperature commonly increases reaction rate.
- Why does powdered marble react faster with HCl than a marble chip?
- Distinguish between average and instantaneous rate using a concentration–time graph.
Rate Laws, Order and Half-Life
The rate law expresses rate as a function of reactant concentrations: Rate = k[A]x[B]y. The exponents x and y are experimentally determined orders with respect to each reactant, and x + y is the overall order. The rate constant k is characteristic of the reaction and temperature, and its units depend on order: (mol L⁻¹)1−n s⁻¹. Molecularity is the number of species colliding in an elementary step; it is always a whole number (1, 2, rarely 3), unlike order, which may be zero or fractional. For a multistep reaction, the slowest step (rate-determining step) governs the rate law.
Integrated rate equations let us predict concentrations over time. A zero-order reaction has a constant rate independent of concentration (common for surface or enzyme-saturated reactions). A first-order reaction has an exponential decay and a half-life independent of initial concentration, as seen in radioactive decay. A pseudo-first-order reaction (such as ester hydrolysis in excess water) behaves as first order because one reactant is in large excess.
Zero order: [R] = [R]₀ − kt; t½ = [R]₀/2k; k units mol L⁻¹ s⁻¹
First order: k = (2.303/t) log([R]₀/[R]); t½ = 0.693/k; k units s⁻¹
Second order: units L mol⁻¹ s⁻¹ | Time for 99.9% completion (first order) = 10 t½ approx.
t½ = 0.693/0.005 = 138.6 s. For 90% completion [R] = 0.1[R]₀, so t = (2.303/k) log 10 = 2.303/0.005 ≈ 461 s.
Order from data: if doubling [A] quadruples the rate, the order in A is 2; if rate = k[A]²[B], overall order is 3 and k has units L² mol⁻² s⁻¹.
- Determine reaction order from a supplied rate expression, such as rate = k[A]1/2[B]3/2.
- Show that the half-life of a first-order reaction is independent of the initial concentration.
- A zero-order reaction has [R]₀ = 0.8 M and k = 0.02 M s⁻¹. Find its half-life.
Activation Energy and Arrhenius Equation
For reactants to become products they must cross an energy barrier, the activation energy (Ea), the minimum extra energy above the average that colliding molecules need. The Arrhenius equation shows how sensitively k depends on this barrier and on temperature. The exponential term e−Ea/RT is the fraction of molecules with energy at least Ea; because it is exponential, a small rise in temperature can increase the rate greatly. The pre-exponential factor A (frequency factor) represents the collision frequency and orientation requirements.
A plot of ln k (or log k) against 1/T is a straight line with slope −Ea/R (or −Ea/2.303R), which gives Ea experimentally. A catalyst provides an alternative mechanism with a lower Ea, increasing the rate of both forward and reverse reactions equally; it does not change ΔH, ΔG° or the equilibrium constant, and it is regenerated at the end of the reaction.
k = Ae−Ea/RT | log k = log A − Ea/(2.303RT)
log(k₂/k₁) = [Ea/(2.303R)] × [(T₂ − T₁)/(T₁T₂)] | R = 8.314 J K⁻¹ mol⁻¹
log(k₂/k₁) = [50000/(2.303 × 8.314)] × [10/(300 × 310)] = 2611 × 1.075 × 10⁻⁴ ≈ 0.281, so k₂/k₁ ≈ 1.9, roughly the “doubling per 10 K” rule.
- Sketch an energy profile showing catalysed and uncatalysed pathways.
- The rate constant of a reaction doubles when temperature rises from 298 K to 308 K. Estimate Ea (answer: about 52.9 kJ mol⁻¹).
- Why does a catalyst not alter the equilibrium constant?
🔬 Transition and Coordination Chemistry
d-Block Elements and Transition Behaviour
The d-block occupies groups 3–12 and has the general configuration (n−1)d1–10ns1–2. A transition element is strictly one with an incomplete d subshell in the atom or in a common ion, so Zn, Cd and Hg (d¹⁰ in every state) are usually not counted as typical transition elements. Notable exceptions in configuration are Cr (3d⁵4s¹) and Cu (3d¹⁰4s¹), arising from extra stability of half-filled and fully filled d subshells.
Characteristic properties include high melting points and enthalpies of atomisation (many unpaired d electrons give strong metallic bonding), variable oxidation states differing by one unit (because ns and (n−1)d energies are close), coloured ions (d–d transitions absorb visible light; d⁰ and d¹⁰ ions such as Sc³⁺, Zn²⁺ are colourless), paramagnetism (spin-only μ = √[n(n+2)] BM), catalytic activity (variable oxidation state and surface adsorption: V₂O₅ in contact process, Fe in Haber, Ni in hydrogenation), complex formation, and formation of interstitial compounds and alloys.
Two important oxidants are potassium dichromate and potassium permanganate. Dichromate: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (orange to green); in acid it oxidises Fe²⁺ to Fe³⁺, and chromate (yellow) and dichromate (orange) interconvert with pH: 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O. Permanganate: in acidic medium MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O; in neutral or weakly alkaline medium it is reduced to MnO₂ and in strongly alkaline medium to green MnO₄²⁻.
μ (spin only) = √[n(n+2)] BM, n = number of unpaired electrons
Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O | MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
Mn²⁺ is 3d⁵, n = 5, μ = √35 ≈ 5.92 BM. Cr³⁺ is 3d³, n = 3, μ = √15 ≈ 3.87 BM. Ti⁴⁺ (d⁰) and Zn²⁺ (d¹⁰) are colourless and diamagnetic, while Cu²⁺ (d⁹) is blue and paramagnetic.
- Connect variable oxidation states with transition-metal electronic structure.
- Why does Mn show the highest oxidation state (+7) in the 3d series?
- Why is Sc³⁺ colourless but Ti³⁺ is purple?
f-Block Elements and Lanthanoid Contraction
The f-block comprises the 14 lanthanoids (Ce to Lu, filling 4f) and the 14 actinoids (Th to Lr, filling 5f). Lanthanoids have the general configuration [Xe]4f1–145d0–16s². The most common oxidation state is +3; +2 (Eu²⁺, Yb²⁺, which are reducing) and +4 (Ce⁴⁺, a strong oxidising agent that is nonetheless stable in water) arise from stable empty, half-filled or filled 4f subshells. Many Ln³⁺ ions are coloured due to f–f transitions and most are paramagnetic.
The steady decrease in atomic and ionic radii from La to Lu is the lanthanoid contraction. Each added electron enters the 4f subshell, whose diffuse shape shields the nuclear charge poorly; effective nuclear charge increases and pulls the outer electrons inward. Consequences: (1) 4d and 5d elements of a group have almost equal radii (Zr 160 pm, Hf 159 pm), so Zr and Hf have extremely similar chemistry and are hard to separate; (2) decreasing basicity of Ln(OH)₃ from La to Lu; (3) a regular change in properties such as complexing ability.
Actinoids are all radioactive. They show a wider range of oxidation states (+3 to +7) because the 5f, 6d and 7s energies are close, and their 5f orbitals extend further than 4f, giving greater involvement in bonding and even poorer shielding (a larger actinoid contraction). An important alloy is Mischmetall (about 95% lanthanoids, 5% Fe), used in lighter flints and in making Mg alloys.
Between them lie fourteen 4f electrons, which shield poorly, so the lanthanoid contraction cancels the size increase expected from an extra shell. The result is almost identical radii and chemical properties.
- State two consequences of lanthanoid contraction.
- Compare the oxidation states of lanthanoids and actinoids and explain the difference.
- Why are most lanthanoid ions coloured?
Coordination Compounds and Nomenclature
A coordination compound has a central metal atom or ion bonded to a fixed number of molecules or ions called ligands through coordinate (dative) bonds. Werner’s theory proposed two kinds of valence: primary valence (ionisable, equal to the oxidation state) and secondary valence (non-ionisable, equal to the coordination number, e.g. six for most Co³⁺ and Cr³⁺ complexes). The metal and ligands together form the coordination sphere, written in square brackets; ions outside are counter ions.
Ligands are classified by the number of donor atoms: monodentate (NH₃, Cl⁻, H₂O), bidentate (ethane-1,2-diamine (en), oxalate), and polydentate (EDTA, hexadentate). Rings formed by bidentate or polydentate ligands are called chelates and are especially stable. Ambidentate ligands (NO₂⁻, SCN⁻) can bind through either of two different atoms.
IUPAC naming rules
- Cation first, then anion; within the complex, ligands are named first in alphabetical order, then the metal.
- Anionic ligands end in –o (chlorido, cyanido, oxalato); neutral ligands keep their names, except aqua (H₂O), ammine (NH₃), carbonyl (CO), nitrosyl (NO).
- Prefixes di, tri, tetra count simple ligands; bis, tris, tetrakis are used for complex ligands such as ethane-1,2-diamine.
- Oxidation state of the metal is given in Roman numerals; if the complex is an anion, the metal name ends in –ate (ferrate, cuprate, argentate, aurate).
[Co(NH₃)₆]Cl₃ — hexaamminecobalt(III) chloride
K₄[Fe(CN)₆] — potassium hexacyanidoferrate(II)
[Cu(NH₃)₄]SO₄ — tetraamminecopper(II) sulfate
[Pt(NH₃)₂Cl₂] — diamminedichloridoplatinum(II)
x + 6(−1) = −3, so x = +3. In [Co(NH₃)₅Cl]Cl₂ the complex ion [Co(NH₃)₅Cl]²⁺ plus two Cl⁻ gives three ions, and AgNO₃ precipitates only the two outer chlorides immediately.
- Identify the central metal, ligand and coordination number in [Ni(CN)₄]²⁻.
- Write the IUPAC name of [Co(en)₃]Cl₃ and K₃[Fe(C₂O₄)₃].
- Name two biological or industrial uses of coordination compounds.
Bonding, Geometry and Isomerism in Complexes
Valence bond theory explains geometry through hybridisation of metal orbitals that accept ligand lone pairs. Six-coordinate complexes are octahedral, using d²sp³ (inner-orbital, low-spin, e.g. [Co(NH₃)₆]³⁺) or sp³d² (outer-orbital, high-spin, e.g. [CoF₆]³⁻). Four-coordinate complexes are tetrahedral (sp³, e.g. [NiCl₄]²⁻) or square planar (dsp², e.g. [Ni(CN)₄]²⁻). Magnetic data decide between them.
Crystal field theory treats ligands as point charges that split the d orbitals. In an octahedral field, the d orbitals divide into lower t₂g (three) and upper eg (two) with splitting Δo. If Δo is smaller than the pairing energy P (weak-field ligands such as I⁻, Br⁻, Cl⁻, F⁻), electrons occupy eg before pairing: high spin. If Δo > P (strong-field ligands such as en, CN⁻, CO), electrons pair in t₂g first: low spin. Tetrahedral splitting is smaller, Δt ≈ (4/9)Δo, so tetrahedral complexes are nearly always high spin. Colour comes from d–d transitions absorbing part of the visible spectrum and showing the complementary colour.
The spectrochemical series (weak to strong field): I⁻ < Br⁻ < SCN⁻ < Cl⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NH₃ < en < CN⁻ < CO. Isomerism may be structural (ionisation, hydrate, linkage, coordination) or stereoisomerism (geometrical cis/trans, and optical isomers that are non-superimposable mirror images, as in [Co(en)₃]³⁺).
μ = √[n(n+2)] BM | Octahedral d⁴–d⁷: high spin if Δo < P, low spin if Δo > P | Δt ≈ 4/9 Δo
Metal carbonyls: Ni(CO)₄ tetrahedral, Fe(CO)₅ trigonal bipyramidal
F⁻ is a weak-field ligand, so the configuration is t₂g⁴eg² with four unpaired electrons: paramagnetic, μ = √24 ≈ 4.90 BM, sp³d² outer-orbital. NH₃ is strong field, so t₂g⁶eg⁰ with no unpaired electrons: diamagnetic, d²sp³ inner-orbital.
- Distinguish geometrical isomerism from linkage isomerism with one example of each.
- Why is [Ni(CN)₄]²⁻ diamagnetic while [NiCl₄]²⁻ is paramagnetic?
- Draw the cis and trans isomers of [Pt(NH₃)₂Cl₂] and say which one is the anticancer drug cisplatin.
🧬 Advanced Organic Chemistry
Haloalkanes and Haloarenes
Haloalkanes (R–X) and haloarenes (Ar–X) contain a polar C–X bond with the carbon carrying partial positive charge. Boiling points increase with molecular mass and with the size of the halogen (R–I > R–Br > R–Cl > R–F), while branching lowers boiling points. They are slightly soluble in water because they cannot hydrogen bond with it. Isomers can be primary, secondary or tertiary, and a carbon bearing four different groups is a chiral centre, giving enantiomers that rotate plane-polarised light equally in opposite directions; a 1:1 mixture is a racemic mixture with no net rotation.
Preparation of haloalkanes: from alcohols with SOCl₂ (best, gaseous by-products), with HX/ZnCl₂ or PCl₅; from alkenes by addition of HX (Markovnikov) and HBr/peroxide (anti-Markovnikov); free radical halogenation (poor selectivity); halogen exchange by the Finkelstein reaction (R–Cl + NaI in dry acetone → R–I + NaCl) and Swarts reaction (R–Br + AgF → R–F). Haloarenes are made by electrophilic halogenation (X₂/Fe) or by the Sandmeyer reaction from diazonium salts.
Haloarenes are much less reactive to nucleophilic substitution than haloalkanes because (i) the lone pair on halogen delocalises into the ring, giving the C–X bond partial double-bond character; (ii) the carbon is sp² hybridised and more electronegative, so the bond is shorter and stronger; and (iii) the phenyl cation is unstable and the electron-rich ring repels nucleophiles. Electron-withdrawing groups (–NO₂) at ortho/para positions activate the ring, so 2,4-dinitrochlorobenzene reacts readily. Reactions with metals: Wurtz (2R–X + 2Na → R–R), Fittig and Wurtz–Fittig, and Grignard reagent formation (R–X + Mg in dry ether → RMgX). Polyhalogens: chloroform (oxidises in air to poisonous phosgene, so stored in dark, full bottles), CCl₄ (fire extinguisher, now restricted), DDT (persistent pesticide) and freons (ozone depletion).
The dipole moments are 1.69 D (C₆H₅Cl) and 1.86 D (CH₃Cl). In chlorobenzene the carbon is sp², more electronegative than sp³, so the C–Cl bond is less polar, and resonance donates electron density toward the ring, opposing the dipole.
- Compare the carbon–halogen bond environment in chloromethane and chlorobenzene.
- Identify the chiral centre in 2-bromobutane and draw both enantiomers.
- Why is SOCl₂ the preferred reagent for converting alcohols to chloroalkanes?
Nucleophilic Substitution and Elimination
In nucleophilic substitution a nucleophile (Nu⁻) replaces the leaving group. There are two main mechanisms. SN2 is a single concerted step with backside attack and a five-coordinate transition state; its rate = k[RX][Nu⁻] (second order), configuration at a chiral carbon is inverted (Walden inversion), and the order of reactivity is CH₃X > 1° > 2° > 3°. It is favoured by strong nucleophiles and polar aprotic solvents (DMSO, acetone). SN1 is a two-step process with formation of a planar carbocation, then attack from either face; rate = k[RX] (first order), giving racemisation, with reactivity 3° > 2° > 1°. It is favoured by polar protic solvents (water, alcohol) and stable carbocations (benzylic, allylic, tertiary). For both, the leaving-group ability follows I⁻ > Br⁻ > Cl⁻ > F⁻.
Substitution competes with elimination (dehydrohalogenation). Alcoholic KOH, heat and bulky bases favour elimination to give alkenes; aqueous KOH favours substitution to give alcohols. When more than one alkene is possible, Saytzeff’s rule says the more substituted (more stable) alkene is the major product. Ambident nucleophiles show different products: KCN gives alkyl cyanide (R–CN) while AgCN gives isocyanide (R–NC).
| SN1 | SN2 | |
|---|---|---|
| Steps | Two (carbocation) | One (concerted) |
| Rate law | k[RX] | k[RX][Nu⁻] |
| Stereochemistry | Racemisation | Inversion |
| Best substrate | 3°, benzylic, allylic | CH₃, 1° |
| Solvent | Polar protic | Polar aprotic |
(a) Tertiary substrate with aqueous KOH: SN1 gives (CH₃)₃COH. (b) Alcoholic KOH: elimination gives isobutene, (CH₃)₂C=CH₂. 2-Bromobutane gives mainly but-2-ene (Saytzeff) with but-1-ene as the minor product. (R)-2-Bromobutane reacting with OH⁻ by SN2 gives (S)-butan-2-ol.
- Compare the products expected from substitution and elimination pathways.
- Why does SN1 give a racemic mixture?
- Arrange in order of SN2 reactivity: CH₃Br, (CH₃)₂CHBr, CH₃CH₂Br, (CH₃)₃CBr.
Alcohols: Preparation and Reactions
Alcohols contain an –OH group on an sp³ carbon and are classified as primary (1°), secondary (2°) or tertiary (3°) according to the carbon bearing –OH. Because of intermolecular hydrogen bonding, their boiling points are much higher than those of hydrocarbons, ethers or haloalkanes of similar mass (ethanol boils at 351 K but methoxymethane at 249 K), and lower members are miscible with water.
Preparation: acid-catalysed hydration of alkenes (Markovnikov); hydroboration–oxidation (anti-Markovnikov, gives 1° alcohol from terminal alkene); reduction of aldehydes and ketones (NaBH₄, LiAlH₄, H₂/Ni), of carboxylic acids and esters (LiAlH₄); hydrolysis of haloalkanes; and Grignard reagents: with HCHO gives 1° alcohols, with other aldehydes gives 2° alcohols and with ketones gives 3° alcohols. Ethanol is also made by fermentation of sugars.
Reactions: (1) O–H cleavage: with Na gives alkoxide and H₂; acidity order is water > 1° > 2° > 3° because alkyl groups push electron density and destabilise the alkoxide. (2) Esterification with acids or acid chlorides. (3) C–O cleavage: with HX (Lucas test: ZnCl₂/conc. HCl gives turbidity immediately for 3°, in about five minutes for 2°, and none at room temperature for 1°), with PCl₃, PCl₅, SOCl₂. (4) Dehydration: conc. H₂SO₄ at 443 K gives alkenes (ease 3° > 2° > 1°), while at 413 K gives ethers. (5) Oxidation: PCC gives aldehydes from 1° alcohols; KMnO₄ or K₂Cr₂O₇ gives carboxylic acids; 2° gives ketones; 3° resist oxidation under mild conditions. Cu at 573 K dehydrogenates 1° and 2° alcohols.
CH₃CH₂OH + Na → CH₃CH₂ONa + ½H₂ | CH₃CH₂OH → (conc. H₂SO₄, 443 K) CH₂=CH₂ + H₂O
RCH₂OH → (PCC) RCHO | RCH₂OH → (K₂Cr₂O₇/H⁺) RCOOH | R₂CHOH → R₂C=O
(a) Acid-catalysed hydration follows Markovnikov’s rule: CH₃CH=CH₂ + H₂O/H⁺ → CH₃CH(OH)CH₃. (b) Hydroboration–oxidation (B₂H₆, then H₂O₂/OH⁻) gives the anti-Markovnikov product, CH₃CH₂CH₂OH.
- Classify several alcohol structures as primary, secondary or tertiary.
- Explain the Lucas test and what you would observe for butan-1-ol, butan-2-ol and 2-methylpropan-2-ol.
- Why are alcohols less acidic than water but phenols more acidic than alcohols?
Phenols and Aromatic Reactivity
Phenols have –OH attached directly to a benzene ring. They can be made from chlorobenzene (NaOH, 623 K, 300 atm, then HCl), from benzenesulfonic acid (fusion with NaOH), from diazonium salts (warm water) and industrially by the cumene process, in which cumene is oxidised to cumene hydroperoxide and then converted to phenol and acetone.
Phenol is much more acidic than ethanol (pKa about 10 vs 16) because the phenoxide ion is resonance stabilised, spreading negative charge into the ring. Electron-withdrawing groups at ortho and para positions (–NO₂) increase acidity further; electron-donating groups (–CH₃) decrease it. Phenol dissolves in NaOH but not in NaHCO₃, since it is a weaker acid than carbonic acid. The –OH group is a strong activating, ortho/para-directing group, so phenols undergo electrophilic substitution under mild conditions.
Dilute HNO₃ → o- and p-nitrophenol (separable by steam distillation; the ortho isomer is volatile due to intramolecular H-bonding)
Conc. HNO₃ → 2,4,6-trinitrophenol (picric acid) | Bromine water → 2,4,6-tribromophenol (white ppt)
Br₂ in CS₂ at low temperature → mainly p-bromophenol
Kolbe’s reaction: PhONa + CO₂ (400 K, 4–7 atm), then H⁺ → salicylic acid
Reimer–Tiemann: PhOH + CHCl₃ + aq. NaOH (340 K) → salicylaldehyde
Zn dust distillation → benzene | Na₂Cr₂O₇/H₂SO₄ → benzoquinone | FeCl₃ test → violet colour
Ethanol < p-cresol < phenol < p-nitrophenol. The methyl group donates electrons and destabilises the phenoxide; the nitro group withdraws electrons by –I and –R effects and stabilises it. Salicylic acid on acetylation with acetic anhydride gives aspirin.
- Use resonance ideas to explain enhanced phenol acidity.
- Why does phenol give 2,4,6-tribromophenol with bromine water but not with Br₂/FeBr₃ needed for benzene?
- Write the equations of the Reimer–Tiemann reaction.
Ethers and Carbon–Oxygen Bond Chemistry
Ethers (R–O–R′) are classed as simple (symmetrical) or mixed (unsymmetrical). The C–O–C angle is about 111.7°, slightly larger than tetrahedral because of repulsion between alkyl groups. They have low boiling points (no O–H for hydrogen bonding), but the oxygen can accept hydrogen bonds, so the solubility of ethers in water is similar to that of alcohols of comparable mass.
Preparation: (1) dehydration of alcohols with excess alcohol at 413 K (works for primary alcohols; secondary and tertiary alcohols give alkenes); (2) the Williamson synthesis, an SN2 reaction between an alkoxide and a primary alkyl halide: R–ONa + R′–X → R–O–R′ + NaX. For an ether with a tertiary group, the tertiary part must come from the alkoxide, otherwise elimination gives an alkene.
Reactions: Ethers are fairly inert, which makes them good solvents. Excess HI cleaves them to alkyl iodides. With two primary or secondary alkyl groups, I⁻ attacks the smaller (less hindered) group by SN2; with a tertiary alkyl group the mechanism is SN1 and the tertiary iodide forms. Aryl alkyl ethers cleave to a phenol and an alkyl halide because the aryl C–O bond is stronger. Anisole undergoes electrophilic substitution (halogenation, nitration, Friedel–Crafts) at ortho and para positions. Ethers slowly form explosive peroxides in air and light, so they are stored carefully and tested before distillation.
2C₂H₅OH → (conc. H₂SO₄, 413 K) C₂H₅OC₂H₅ + H₂O
CH₃ONa + C₂H₅Cl → CH₃OC₂H₅ + NaCl | C₆H₅OCH₃ + HI → C₆H₅OH + CH₃I
Use sodium tert-butoxide with chloromethane: (CH₃)₃CONa + CH₃Cl → (CH₃)₃COCH₃. The reverse choice, CH₃ONa with (CH₃)₃CCl, mainly produces isobutene by elimination.
- Identify symmetrical and unsymmetrical ethers from structural formulas.
- Write the products of CH₃OC₂H₅ + HI (excess).
- Why is the boiling point of ethanol far higher than that of methoxymethane?
Aldehydes and Ketones
The carbonyl group (C=O) is planar and polar, making the carbon electrophilic. Aldehydes (R–CHO) and ketones (R–CO–R′) have boiling points higher than hydrocarbons but lower than alcohols (no intermolecular hydrogen bonds); lower members dissolve in water through hydrogen bonding with water.
Preparation of aldehydes: oxidation or dehydrogenation of 1° alcohols; Rosenmund reduction (RCOCl + H₂, Pd/BaSO₄); Stephen reaction (RCN + SnCl₂/HCl, then H₃O⁺); DIBAL-H reduction of nitriles/esters; Etard reaction (toluene + CrO₂Cl₂ → benzaldehyde); Gattermann–Koch (benzene + CO/HCl, AlCl₃) and hydration of ethyne (HgSO₄/H₂SO₄ gives ethanal). Ketones: oxidation of 2° alcohols, hydration of other alkynes, Friedel–Crafts acylation, and acid chloride with dialkylcadmium.
Nucleophilic addition is the characteristic reaction: HCN gives cyanohydrins; NaHSO₃ gives bisulphite adducts; alcohols/dry HCl give acetals; Grignard reagents give alcohols. Aldehydes are more reactive than ketones because of less steric hindrance and greater positive charge on carbon (only one electron-donating alkyl group). With ammonia derivatives (NH₂–Z), they give C=N compounds: oxime (NH₂OH), hydrazone, phenylhydrazone, semicarbazone and the orange-red 2,4-DNP derivative. Reductions: NaBH₄/LiAlH₄ give alcohols; Clemmensen (Zn–Hg/conc. HCl) and Wolff–Kishner (NH₂NH₂, KOH, ethylene glycol, heat) reduce C=O to CH₂.
Distinguishing tests: aldehydes reduce Tollens’ reagent (silver mirror) and Fehling’s solution (red-brown Cu₂O); ketones do not (aromatic aldehydes do not reduce Fehling’s). The iodoform test (I₂/NaOH → yellow CHI₃) is given by CH₃CO– groups and CH₃CH(OH)– groups. Compounds with α-hydrogens undergo aldol condensation (dilute NaOH, giving a β-hydroxy carbonyl that dehydrates on heating to an α,β-unsaturated compound). Compounds with no α-hydrogen (HCHO, benzaldehyde) undergo the Cannizzaro reaction (conc. NaOH: one molecule oxidised to acid salt, one reduced to alcohol).
2CH₃CHO → (dil. NaOH) CH₃CH(OH)CH₂CHO → (heat) CH₃CH=CHCHO (but-2-enal)
2HCHO + NaOH → CH₃OH + HCOONa | CH₃COCH₃ + HCN → (CH₃)₂C(OH)CN
Pentan-2-one contains the CH₃CO– group and gives a yellow precipitate of iodoform with I₂/NaOH; pentan-3-one (CH₃CH₂COCH₂CH₃) does not.
- Distinguish aldehyde and ketone functional groups in structural formulas.
- Give a chemical test to distinguish benzaldehyde from ethanal.
- Convert propanal into butan-2-ol using a Grignard reagent.
Carboxylic Acids and Their Reactions
Carboxylic acids (R–COOH) are prepared by oxidation of 1° alcohols and aldehydes, hydrolysis of nitriles, amides or esters, carbonation of Grignard reagents (RMgX + CO₂, then H₃O⁺), and vigorous oxidation of alkylbenzenes with KMnO₄ (any alkyl chain gives benzoic acid). They have strong intermolecular hydrogen bonding and form dimers, so their boiling points exceed those of alcohols of similar mass.
Carboxylic acids are much more acidic than alcohols and phenols because the carboxylate ion has two equivalent resonance structures with charge shared equally between two oxygens. Substituent effects: electron-withdrawing groups (–Cl, –F, –NO₂) stabilise the anion and raise acidity, and the effect increases with the number of groups and decreases with distance; electron-donating groups (alkyl) reduce acidity. Benzoic acid (pKa 4.2) is stronger than acetic acid (4.76).
RCOOH + NaHCO₃ → RCOONa + H₂O + CO₂↑ (test for –COOH)
Esterification: RCOOH + R′OH ⇌ RCOOR′ + H₂O (conc. H₂SO₄)
RCOOH + SOCl₂ → RCOCl | RCOOH → (LiAlH₄ or B₂H₆) RCH₂OH (NaBH₄ does not reduce acids)
HVZ reaction: RCH₂COOH + X₂/red P → RCHXCOOH
Decarboxylation: RCOONa + NaOH → (CaO, heat) RH + Na₂CO₃
Kolbe electrolysis: 2RCOO⁻ → R–R + 2CO₂
Cl₃CCOOH (pKa ≈ 0.7) > ClCH₂COOH (2.86) > HCOOH (3.75) > C₆H₅COOH (4.2) > CH₃COOH (4.76). The more electron-withdrawing groups, the more stable the anion.
- Compare the acidity of a carboxylic acid with a typical alcohol.
- Convert ethanoic acid into methane and into chloroethanoic acid.
- Why are carboxylic acids soluble in sodium bicarbonate while phenols are not?
Amines: Basicity and Reactions
Amines are derivatives of ammonia in which one or more hydrogens are replaced by alkyl or aryl groups, classified as primary (1°), secondary (2°) or tertiary (3°). Intermolecular hydrogen bonding makes boiling points follow 1° > 2° > 3° for isomers (tertiary amines have no N–H). Lower amines are water soluble and smell fishy.
Preparation: reduction of nitro compounds (Sn/HCl, Fe/HCl or H₂/Pd), nitriles (LiAlH₄) and amides; ammonolysis of alkyl halides (gives a mixture of 1°, 2°, 3° amines and quaternary salt); Gabriel phthalimide synthesis (pure 1° aliphatic amines; it fails for aryl amines because aryl halides do not undergo nucleophilic substitution); and Hofmann bromamide degradation (RCONH₂ + Br₂ + 4NaOH → RNH₂ + Na₂CO₃ + 2NaBr + 2H₂O), which gives an amine with one carbon fewer than the amide.
Basicity depends on the availability of the nitrogen lone pair. In the gas phase, basicity increases 3° > 2° > 1° > NH₃ because of +I effects, but in water solvation of the ammonium ion also matters, giving (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃ and (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃. Aromatic amines are weaker bases than ammonia because the lone pair is delocalised into the ring (aniline pKb ≈ 9.4); electron-donating groups on the ring increase basicity, and electron-withdrawing groups decrease it.
Carbylamine test: RNH₂ + CHCl₃ + 3KOH → RNC (foul smell) + 3KCl + 3H₂O (1° amines only)
Hinsberg test (C₆H₅SO₂Cl): 1° amine → sulphonamide soluble in alkali; 2° → insoluble; 3° → no reaction
Acylation: RNH₂ + (CH₃CO)₂O → RNHCOCH₃ | Aniline + Br₂(aq) → 2,4,6-tribromoaniline
1° aliphatic amine + HNO₂ → alcohol + N₂ | Aniline + NaNO₂/HCl (273–278 K) → diazonium salt
CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O (Hofmann). Gabriel needs an alkyl halide for SN2 with phthalimide anion; aryl halides resist nucleophilic substitution, so aniline cannot be prepared that way. To control bromination of aniline, first acetylate to acetanilide, brominate, then hydrolyse to get p-bromoaniline.
- Classify primary, secondary and tertiary amines from given structures.
- Arrange aniline, p-toluidine and p-nitroaniline in increasing basicity.
- How would you distinguish ethylamine from diethylamine?
Diazonium Salts and Synthetic Transformations
Arenediazonium salts (ArN₂⁺X⁻) form when a primary aromatic amine reacts with nitrous acid (NaNO₂ + HCl) at 273–278 K: diazotisation. Benzenediazonium chloride is stable only in cold solution because it decomposes above 5 °C, but the fluoroborate (ArN₂⁺BF₄⁻) is stable at room temperature. Aromatic diazonium ions are stabilised by resonance with the ring, unlike aliphatic diazonium ions, which decompose instantly.
Reactions are of two kinds. Replacement of N₂ (nitrogen is lost as gas): Sandmeyer (CuCl/HCl → Ar–Cl, CuBr/HBr → Ar–Br, CuCN/KCN → Ar–CN), Gattermann (Cu powder + HX), KI → Ar–I, Balz–Schiemann (HBF₄, heat → Ar–F), H₃PO₂ or ethanol → Ar–H (deamination), and warm water → phenol. Coupling reactions (nitrogen retained) form brightly coloured azo dyes: with phenol in mildly alkaline medium (pH 9–10) at the para position to give an orange p-hydroxyazobenzene, and with aniline in slightly acidic medium (pH 4–5) to give yellow p-aminoazobenzene. Methyl orange is a well-known azo dye used as an indicator.
Because the –NH₂ group can be removed or replaced by almost any group, diazonium salts allow preparations that direct substitution cannot do, such as aryl fluorides, iodides, nitriles and compounds with unusual substitution patterns.
Step 1: aniline + Br₂ water → 2,4,6-tribromoaniline. Step 2: NaNO₂/HCl at 273–278 K → 2,4,6-tribromobenzenediazonium chloride. Step 3: H₃PO₂ (or CH₃CH₂OH) replaces N₂⁺ by H → 1,3,5-tribromobenzene.
Another route: aniline → diazonium → CuCN → benzonitrile → H₃O⁺ → benzoic acid.
- Explain why diazonium salts are useful synthetic intermediates.
- Convert aniline into fluorobenzene and into iodobenzene.
- Write the coupling reaction of benzenediazonium chloride with β-naphthol (a red dye).
🌱 Biomolecules and Applied Chemistry
Carbohydrates and Molecular Structure
Carbohydrates are polyhydroxy aldehydes or ketones (or compounds that give them on hydrolysis). They are classified as monosaccharides (glucose, fructose, galactose; cannot be hydrolysed further), oligosaccharides (2–10 units, such as sucrose, maltose, lactose) and polysaccharides (starch, cellulose, glycogen). Monosaccharides are aldoses (glucose) or ketoses (fructose). Sugars that reduce Tollens’ or Fehling’s reagent are reducing sugars; fructose reduces them despite being a ketose, because it tautomerises in basic solution.
Glucose (C₆H₁₂O₆): its open-chain structure is established by the formation of n-hexane with HI (six-carbon straight chain), a cyanohydrin with HCN and gluconic acid with bromine water (an aldehyde), a pentaacetate with acetic anhydride (five –OH groups), and saccharic acid with HNO₃. It is D-(+)-glucose, where D refers to the configuration at C-5 relative to D-glyceraldehyde and (+) to dextrorotation. Glucose also exists as cyclic hemiacetals (pyranose rings); the new –OH at C-1 gives two anomers (α and β). In solution they interconvert through the open chain, so the specific rotation changes gradually to an equilibrium value: mutarotation.
Disaccharides: sucrose (α-D-glucose + β-D-fructose joined C1–C2; non-reducing because both anomeric carbons are involved), maltose (two α-D-glucose, C1–C4; reducing), lactose (β-D-galactose + β-D-glucose; reducing). Polysaccharides: starch is amylose (15–20%, unbranched α-1,4 chain) plus amylopectin (80–85%, branched with α-1,6 links); glycogen is a highly branched animal starch; cellulose has β-1,4 linkages which humans cannot digest.
Sucrose is dextrorotatory (+66.5°). On hydrolysis it forms glucose (+52.5°) and fructose (−92.4°). The mixture has a net rotation of about (52.5 − 92.4)/2 ≈ −20°, so the sign of rotation inverts from dextro to laevo.
- Distinguish monosaccharides, disaccharides and polysaccharides.
- Why is sucrose non-reducing while maltose is reducing?
- What is the structural difference between α-glucose and β-glucose, and between starch and cellulose?
Amino Acids, Peptides and Proteins
Amino acids contain both –NH₂ and –COOH; in biology the α-amino acids, R–CH(NH₂)–COOH, are the building blocks of proteins. In solution and in the solid state they exist as zwitterions (H₃N⁺–CHR–COO⁻), which explains their high melting points and water solubility. At the isoelectric point the net charge is zero and solubility is minimum. Ten are essential amino acids, which the body cannot synthesise and must obtain from food (valine, leucine, isoleucine, phenylalanine, tryptophan, methionine, lysine, threonine, histidine, arginine). Ninhydrin gives a blue-violet colour with free α-amino acids.
Amino acids join through peptide bonds (–CO–NH–), formed by condensation between the –COOH of one and the –NH₂ of another. Protein structure has four levels: primary (sequence of amino acids), secondary (α-helix and β-pleated sheet, stabilised by hydrogen bonds between backbone groups), tertiary (overall three-dimensional folding from hydrogen bonds, disulphide bridges, ionic and van der Waals forces), and quaternary (assembly of subunits as in haemoglobin). Fibrous proteins (keratin, myosin) are insoluble and structural; globular proteins (insulin, albumin) are water soluble.
Denaturation by heat, acid, base or heavy metals destroys the secondary and tertiary structure but not the primary structure (cooking an egg). Enzymes are biocatalysts, highly specific and efficient under mild conditions. Vitamins are essential organic nutrients: A, D, E and K are fat-soluble; B-complex and C are water-soluble (not stored, needed regularly). Deficiency diseases: A — night blindness; B₁ — beri-beri; C — scurvy; D — rickets; K — poor blood clotting; B₁₂ — pernicious anaemia.
Gly–Ala and Ala–Gly (the order defines which end has the free –NH₂), each formed with loss of one H₂O. Three different amino acids can form 3! = 6 tripeptides.
- Identify the amino and carboxyl groups in a general amino-acid structure.
- Differentiate between fibrous and globular proteins with examples.
- Why does an amino acid behave as both an acid and a base?
Nucleic Acids and Chemical Information
Nucleic acids are polymers of nucleotides. A nucleoside is a nitrogenous base joined to a pentose sugar (at C-1′); a nucleotide is a nucleoside with a phosphate group at C-5′. The sugar is β-D-2-deoxyribose in DNA and β-D-ribose in RNA. The bases are the purines adenine (A) and guanine (G), and the pyrimidines cytosine (C), thymine (T, only in DNA) and uracil (U, only in RNA). Nucleotides are connected by phosphodiester linkages between the 3′ carbon of one sugar and the 5′ carbon of the next.
DNA forms a right-handed double helix of two antiparallel strands held by hydrogen bonds between complementary bases: A pairs with T (two hydrogen bonds) and G with C (three). Chargaff’s rules follow: [A] = [T] and [G] = [C]. DNA stores genetic information and replicates semi-conservatively. RNA is usually single stranded and of three main kinds: messenger RNA (mRNA, carries the code), ribosomal RNA (rRNA, ribosome structure) and transfer RNA (tRNA, brings amino acids). In protein synthesis DNA is transcribed to mRNA, which is translated at ribosomes; each three-base codon specifies one amino acid. Changes in base sequence are mutations. DNA fingerprinting uses unique sequence patterns for identification.
T = 30%, so G + C = 40% with G = C = 20%. The complementary antiparallel strand is 3′-TACG-5′ (read 5′-CGTA-3′).
- List the three structural components of a nucleotide.
- Compare DNA and RNA in terms of sugar, bases and structure.
- Why does the double helix need A–T pairs to have fewer hydrogen bonds than G–C?
Polymers and Material Properties
Polymers are macromolecules made from many repeating monomer units. They can be natural (proteins, cellulose, rubber) or synthetic (nylon, polythene); linear, branched or cross-linked; and classified by the polymerisation process. Addition polymers form from unsaturated monomers without loss of atoms (chain-growth, often free-radical, cationic or anionic); condensation polymers form from bifunctional monomers with loss of small molecules such as H₂O (step-growth). By physical properties: elastomers (weak forces, stretch and recover: rubber), fibres (strong hydrogen bonding, high tensile strength: nylon-6,6, terylene), thermoplastics (soften on heating: polythene, PVC, polystyrene) and thermosetting polymers (cross-linked, harden irreversibly: Bakelite, melamine resin).
| Polymer | Monomer(s) | Type / note |
|---|---|---|
| Polythene (LDPE) | Ethene | Addition; high pressure (1000–2000 atm), 350–570 K, peroxide initiator; branched |
| HDPE | Ethene | Ziegler–Natta catalyst, 333–343 K, 6–7 atm; linear, high density |
| PVC | Vinyl chloride | Addition; pipes, insulation |
| Teflon | Tetrafluoroethene | Addition; non-stick coatings |
| Polyacrylonitrile (orlon) | Acrylonitrile | Addition; artificial wool |
| Nylon-6,6 | Hexamethylenediamine + adipic acid | Condensation; polyamide fibre |
| Nylon-6 | Caprolactam | Ring-opening, 533–543 K |
| Terylene (Dacron) | Ethylene glycol + terephthalic acid | Condensation; polyester |
| Bakelite | Phenol + formaldehyde | Condensation; thermosetting |
| Buna-S | 1,3-Butadiene + styrene | Copolymer; tyres |
| Neoprene | Chloroprene | Addition; oil-resistant rubber |
| PHBV | 3-Hydroxybutanoic + 3-hydroxypentanoic acids | Biodegradable polyester |
Natural rubber is cis-1,4-polyisoprene; it is soft and sticky, so it is vulcanised by heating with sulphur (373–415 K), which forms sulphur cross-links between chains, increasing strength and elasticity. A copolymer contains two or more different monomers.
n CH₂=CH₂ → –(CH₂–CH₂)ₙ– | n H₂N(CH₂)₆NH₂ + n HOOC(CH₂)₄COOH → –[NH(CH₂)₆NHCO(CH₂)₄CO]ₙ– + 2n H₂O
Average molecular mass M = (degree of polymerisation) × (monomer mass)
Monomer ethene has M = 28. n = 56,000/28 = 2000 repeat units.
- Compare addition and condensation polymerisation with examples.
- Why is Bakelite hard and infusible but polythene soft and meltable?
- What is the role of sulphur in vulcanisation?
Chemistry in Medicines, Food and Daily Life
Drugs act on targets such as enzymes (inhibitors) or receptors (agonists and antagonists). Main classes: analgesics (non-narcotic like aspirin and paracetamol, which also reduce fever; narcotic like morphine and codeine, used for severe pain); antipyretics; antibiotics (penicillin, ofloxacin, chloramphenicol; narrow- or broad-spectrum, bactericidal or bacteriostatic); antiseptics (applied to living tissue: dettol, tincture of iodine, 0.2% phenol) versus disinfectants (used on inanimate objects: 1% phenol, chlorine in water); antacids (NaHCO₃, Mg(OH)₂, and H₂-receptor antagonists cimetidine and ranitidine that reduce acid release); antihistamines (brompheniramine, terfenadine, which block histamine receptors); tranquillisers (equanil, barbiturates) and antifertility drugs (norethindrone).
Food chemicals: preservatives (sodium benzoate, salts of sorbic acid); antioxidants (BHA, BHT); and artificial sweeteners (saccharin is about 550 times sweeter than sucrose; aspartame is about 100 times sweeter but breaks down on cooking and is unsuitable for people with phenylketonuria; sucralose and alitame are stable at cooking temperature).
Cleansing agents: soaps are sodium or potassium salts of long-chain fatty acids, made by saponification of fats with NaOH. Soap has a hydrophobic tail dissolving in grease and a hydrophilic head dissolving in water, forming micelles around oil droplets. In hard water, Ca²⁺ and Mg²⁺ form insoluble scum with soap. Synthetic detergents work in hard water: anionic (sodium lauryl sulphate, alkylbenzenesulfonates), cationic (cetyltrimethylammonium bromide, used in hair conditioners) and non-ionic types. Detergents with unbranched hydrocarbon chains are biodegradable; branched chains persist and cause foaming in rivers.
Fat (triglyceride) + 3NaOH → 3 RCOONa (soap) + glycerol (saponification)
2RCOONa + Ca²⁺ → (RCOO)₂Ca↓ + 2Na⁺ (scum in hard water)
The calcium and magnesium salts of alkyl sulphates and alkylbenzenesulfonates are soluble, so no scum forms and the detergent anions remain free to form micelles, whereas soap anions precipitate as insoluble calcium and magnesium salts.
- Choose one household chemical product and identify the chemical principle behind its function.
- Why should antibiotics be taken for the full prescribed course?
- Differentiate between narcotic and non-narcotic analgesics.
Analytical Chemistry and Quantitative Laboratory Work
Volumetric analysis determines the concentration of an unknown solution by reacting it with a standard solution of known concentration. The equivalence point is where reactants are present in stoichiometric amounts; the end point is the observable change (usually an indicator colour change) and should coincide with it. Primary standards (oxalic acid, anhydrous Na₂CO₃, potassium hydrogen phthalate, K₂Cr₂O₇) are pure, stable and non-hygroscopic, so their solutions can be prepared by direct weighing; NaOH and KMnO₄ must be standardised.
Acid–base titration: for a reaction a A + b B, nB MAVA = nA MBVB. Phenolphthalein suits strong base–weak acid titrations, methyl orange suits strong acid–weak base. Redox titration with KMnO₄ is self-indicating (a faint permanent pink colour appears at the end point). Oxalic acid with permanganate: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O (about 333–343 K, acidified with dilute H₂SO₄, never HCl or HNO₃). Mohr’s salt, FeSO₄·(NH₄)₂SO₄·6H₂O, is preferred for Fe²⁺ because it resists air oxidation.
Qualitative salt analysis identifies ions by systematic tests. Anions: carbonate (effervescence, CO₂ turns lime water milky), chloride (white AgCl, soluble in NH₄OH), sulphate (white BaSO₄, insoluble in acids), nitrate (brown ring test), acetate. Cation groups: I (Pb²⁺, with dilute HCl, white ppt), II (Cu²⁺, Pb²⁺ with H₂S in acid, black), III (Fe³⁺, Al³⁺ with NH₄OH in presence of NH₄Cl), IV (Zn²⁺, Mn²⁺, Ni²⁺, Co²⁺ with H₂S in alkaline medium), V (Ca²⁺, Ba²⁺, Sr²⁺ with (NH₄)₂CO₃), VI (Mg²⁺), and NH₄⁺ detected with Nessler’s reagent (brown precipitate). Chromatography separates mixtures by differential movement; Rf = distance moved by solute / distance moved by solvent.
n₂M₁V₁ = n₁M₂V₂ | Normality-free approach: compare moles using the balanced equation
Rf = dsolute/dsolvent (always < 1) | Percentage error = (|observed − true|/true) × 100
NaOH: M = (0.100 × 25.0)/20.0 = 0.125 M. Oxalate: 25.0 mL × 0.050 M = 1.25 mmol; the ratio MnO₄⁻:C₂O₄²⁻ = 2:5, so MnO₄⁻ = 0.50 mmol and V = 0.50/0.020 = 25.0 mL.
- Explain why concordant titration readings improve confidence in a result.
- Why is KMnO₄ titration performed in hot, acidified solution?
- A chromatogram shows a spot 3.2 cm from the baseline while the solvent front is 8.0 cm. Find Rf (answer: 0.4).
📋 Constants, Named Reactions and Exam Tips
Important Constants
| Quantity | Value |
|---|---|
| Gas constant R | 8.314 J K⁻¹ mol⁻¹ = 0.0821 L atm K⁻¹ mol⁻¹ |
| Faraday constant F | 96500 C mol⁻¹ |
| 2.303RT/F at 298 K | 0.059 V |
| Avogadro constant | 6.022 × 10²³ mol⁻¹ |
| Kf, Kb of water | 1.86 and 0.52 K kg mol⁻¹ |
| Molar volume of gas at STP (273 K, 1 bar) | 22.7 L mol⁻¹ (22.4 L at 1 atm) |
Named reactions at a glance
| Reaction | Change |
|---|---|
| Finkelstein / Swarts | R–Cl or R–Br → R–I (NaI, acetone) / R–F (AgF) |
| Wurtz / Fittig | 2R–X + 2Na → R–R / aryl halides couple to biaryls |
| Williamson | RONa + R′X → R–O–R′ |
| Kolbe / Reimer–Tiemann | Phenol → salicylic acid / salicylaldehyde |
| Rosenmund / Stephen / Etard | Acid chloride, nitrile, toluene → aldehyde |
| Clemmensen / Wolff–Kishner | C=O → CH₂ (acidic / basic conditions) |
| Aldol / Cannizzaro | With α-H: β-hydroxy carbonyl / without α-H: acid + alcohol |
| Hell–Volhard–Zelinsky | α-halogenation of carboxylic acids |
| Hofmann bromamide / Gabriel | Amide → amine (one C less) / pure 1° aliphatic amines |
| Sandmeyer / Gattermann / Balz–Schiemann | ArN₂⁺ → Ar–Cl, Br, CN / Ar–X (Cu) / Ar–F |
| Carbylamine / Hinsberg | Tests for primary amines / for classifying amines |
Exam strategy
- Learn the exceptions. Many questions test exceptions such as Cr and Cu configurations, benzaldehyde not reducing Fehling’s solution, and fructose being a reducing ketose.
- Write balanced equations with conditions. Reagent, temperature and solvent often carry marks.
- Use the right sign conventions. Cathode and anode, E° as reduction potential, ΔG° = −nFE° are frequent sources of lost marks.
- Explain “why” using structure. For acidity and basicity, think inductive effect, resonance and solvation.
- Practise conversions. Chain conversions between organic compounds appear every year, so memorise key reagent chains such as nitrobenzene → aniline → diazonium → product.
- Show calculations in steps. Write formula, substitution, result and unit, and check units for rate constants and conductivity.