01Stphysical
Physical · Ch. 01 / 20
Stoichiometry & the Mole
counting atoms by weighing matter
Moles from mass
n = m ⁄ M
m in g, M in g·mol⁻¹
Moles from particles
n = N ⁄ NA
NA = 6.022×10²³ mol⁻¹
Gas at STP (273 K, 1 atm)
n = V ⁄ 22.4 L
22.7 L at SATP (298 K, 1 bar)
Molarity
M = n ⁄ V(L) = (w ⁄ M) · 1000 ⁄ V(mL)
temperature-dependent
Molality
m = nsolute ⁄ kgsolvent
temperature-independent; preferred for colligatives
Mole fraction
xi = ni ⁄ Σn · Σx = 1
dimensionless
Normality
N = M · n-factor · N₁V₁ = N₂V₂
equivalence-based
Dilution
M₁V₁ = M₂V₂
moles conserved
Mass %
%w = (mass of element × 100) ⁄ molar mass
empirical formula route
Empirical ↔ molecular
M = k · MEF · k = M ⁄ MEF
k is whole number
Equivalent mass
E = M ⁄ n-factor
n-factor = valency, acidity, basicity, or e⁻ transferred
Volume strength of H₂O₂
V = 5.6 × M
"10 volume" ⇒ 1 L gives 10 L O₂ at STP
⚗ Lab Notes
- Limiting reagent alone sets theoretical yield. Find by dividing moles by stoichiometric coefficient — the smallest wins.
- % yield = actual ⁄ theoretical × 100. Atom economy = (M of desired product ⁄ ΣM) × 100.
- Always balance before calculating. Atoms & charge must be conserved.
- Mole ratio = coefficient ratio: aA + bB → cC ⇒ nA⁄a = nB⁄b at stoichiometric mix.
- Convert cm³ → L and mg → g before plugging in.
∫ Key Derivations
- Molarity from density & %: M = (% · ρ · 10) ⁄ Msolute. Derives from 1 L of solution containing %·ρ·10 g of solute.
- M–m relation: M = m · ρsoln ⁄ (1 + m·Msolute⁄1000). Used when only molality is known.
- Mixing two solutions: Mmix = (M₁V₁ + M₂V₂) ⁄ (V₁ + V₂) for non-reacting solutes of same type.
▶ Worked Example
10.0 g of CaCO₃ reacts with excess HCl. What volume of CO₂ is produced at STP?
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
n(CaCO₃) = 10.0 ⁄ 100 = 0.10 mol
n(CO₂) = 0.10 mol (1:1 ratio)
V = 0.10 × 22.4 = 2.24 L at STP
✗ Common Exam Mistakes
- Forgetting to convert °C to K in gas-volume problems.
- Using mass of solution instead of mass of solvent in molality — the two differ by the solute mass.
- Assuming the largest number of moles is the limiting reagent — it isn't; it's the smallest ratio n/coeff.
- Mixing up equivalent mass and molar mass when using N₁V₁ = N₂V₂ for titrations.
02Atphysical
Physical · Ch. 02 / 20
Atomic Structure
quanta, orbits & the hydrogen spectrum
Photon energy
E = hν = hc ⁄ λ
E(eV) = 1240 ⁄ λ(nm)
Photoelectric
hν = φ + ½mv²max
φ = hν₀ (work function); threshold frequency ν₀
Bohr radius
rn = 0.529 · n² ⁄ Z Å
hydrogen-like ions only
Bohr velocity
vn = (2.19 × 10⁶) · Z ⁄ n m·s⁻¹
electron speed in nth orbit
Bohr energy
En = −13.6 · Z² ⁄ n² eV
IE from level n = 13.6Z²⁄n² eV
Bohr kinetic ⁄ potential
KE = −En · PE = 2En
PE = 2·KE (virial theorem)
Rydberg formula
1 ⁄ λ = RH Z² (1 ⁄ n₁² − 1 ⁄ n₂²)
RH = 1.097×10⁷ m⁻¹
Number of spectral lines
N = n(n − 1) ⁄ 2
transitions from level n to ground
de Broglie
λ = h ⁄ mv = h ⁄ √(2mK)
K = kinetic energy of particle
Bohr–de Broglie link
2πr = nλ
standing wave around orbit
Uncertainty
Δx · Δp ≥ h ⁄ 4π
also Δx·mΔv ≥ h ⁄ 4π
Effective nuclear charge
Zeff = Z − S
Slater's rules give S; shielding s > p > d > f
⚗ Lab Notes
- Quantum numbers: n=1,2,…; l=0…n−1; ml=−l…+l; ms=±½. Total orbitals in shell n = n²; max electrons = 2n².
- Aufbau via (n + l) rule: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d…
- Hund: maximise spin before pairing. Pauli: no two electrons share all four quantum numbers.
- Exceptions from half/full-shell stability: Cr = [Ar]3d⁵4s¹, Cu = [Ar]3d¹⁰4s¹, Mo, Ag, Au likewise.
- Series: Lyman n₁=1 (UV), Balmer n₁=2 (visible), Paschen n₁=3 (IR), Brackett n₁=4, Pfund n₁=5.
- Wave-function sign: nodes = n−l−1 (radial); l (angular). Total nodes = n−1.
∫ Key Derivations
- Bohr radius from force balance: equate Coulomb force to centripetal: m v² ⁄ r = kZe² ⁄ r² and quantise mvr = nh ⁄ 2π. Eliminate v ⇒ rn = n²h² ⁄ 4π²mkZe².
- Bohr energy: E = KE + PE = ½mv² − kZe² ⁄ r = −kZe² ⁄ 2r. Substitute rn ⇒ En ∝ −Z² ⁄ n².
- de Broglie from Bohr: quantisation 2πr = nλ ⇒ mv = nh ⁄ 2πr ⇒ λ = h ⁄ mv.
- Photoelectric stopping potential: eV0 = hν − hν₀, so V0 vs ν is linear with slope h ⁄ e.
▶ Worked Example
Calculate the shortest wavelength in the Balmer series of hydrogen.
Balmer: n₁ = 2. Shortest λ ⇒ maximum energy ⇒ n₂ = ∞.
1⁄λ = RH(1⁄2² − 1⁄∞) = 1.097×10⁷ × 1⁄4
λ = 4 ⁄ 1.097×10⁷ = 364.6 nm (Balmer limit, near UV)
✗ Common Exam Mistakes
- Writing λ in metres but plugging c = 3×10⁸ and E in eV — always match SI or use 1240 ⁄ λ(nm) directly.
- Forgetting that the Rydberg formula applies only to hydrogen-like (1-electron) species; use Z² for He⁺, Li²⁺ etc.
- Confusing radial nodes (n−l−1) with angular nodes (l). Total nodes = n−1, always.
- Stating that all half-filled and fully-filled configurations are exceptions — only d and f orbitals produce them.
03Gsphysical
Physical · Ch. 03 / 20
States of Matter: Gases & Liquids
PV = nRT and its rebellions
Boyle's law
P₁V₁ = P₂V₂
constant T, n
Charles's law
V₁ ⁄ T₁ = V₂ ⁄ T₂
constant P, n; T in K
Gay-Lussac
P₁ ⁄ T₁ = P₂ ⁄ T₂
constant V, n
Avogadro's law
V ∝ n
constant T, P
Ideal gas
PV = nRT
R = 0.0821 L·atm·mol⁻¹K⁻¹ = 8.314 J·mol⁻¹K⁻¹
Gas density
d = PM ⁄ RT · M = dRT ⁄ P
d in g·L⁻¹ with R = 0.0821
Combined gas law
P₁V₁ ⁄ T₁ = P₂V₂ ⁄ T₂
closed system
Dalton's law
Ptot = Σpi · pi = xiPtot
non-reacting mixture
Amagat's law
Vtot = ΣVi
partial volumes; ideal gas only
Graham's law
r₁ ⁄ r₂ = √(M₂ ⁄ M₁) = √(d₂ ⁄ d₁)
diffusion ⁄ effusion
Kinetic gas equation
PV = ⅓ N m u²rms
microscopic form of ideal-gas law
Molecular speeds
ump = √(2RT ⁄ M) · ū = √(8RT ⁄ πM)
urms = √(3RT ⁄ M); ratio 1 : 1.128 : 1.224
KE of gas
KE = 3 ⁄ 2 nRT
3 ⁄ 2 kBT per molecule
van der Waals
(P + an² ⁄ V²)(V − nb) = nRT
a = attraction, b = volume
Compressibility
Z = PV ⁄ nRT
ideal Z = 1 always
Critical constants
Tc = 8a ⁄ 27Rb · Pc = a ⁄ 27b² · Vc = 3b
van der Waals gas
Clausius–Clapeyron
ln(P₂ ⁄ P₁) = (ΔHvap ⁄ R)(1 ⁄ T₁ − 1 ⁄ T₂)
vapour pressure vs T
⚗ Lab Notes
- Kinetic-theory assumptions: point particles, no intermolecular forces, elastic collisions, random motion, Newtonian mechanics.
- Compressibility Z: H₂ and He have Z > 1 at all pressures (repulsion dominates); CO₂, NH₃ show Z < 1 at moderate P (attraction).
- Above critical temperature Tc a gas cannot be liquefied however high the pressure.
- Boyle temperature TB = a ⁄ Rb: at TB the gas behaves ideally over a wide pressure range.
- Surface tension and viscosity fall with T; vapour pressure rises (Clausius–Clapeyron).
- Payman's gas collected over water: Pgas = Patm − PH₂O (aqueous tension).
∫ Key Derivations
- Gas density from ideal gas: PV = (m ⁄ M)RT ⇒ P = (m ⁄ V)(RT ⁄ M) ⇒ P = dRT ⁄ M ⇒ d = PM ⁄ RT.
- Speed ratios: ump = √(2RT ⁄ M), ū = √(8RT ⁄ πM), urms = √(3RT ⁄ M) ⇒ 1 : 1.128 : 1.224.
- KE derivation: PV = ⅓Nmū² and PV = NkBT ⇒ ⅓mū² = kBT ⇒ KE = ½mū² = (3⁄2)kBT per molecule.
- Critical constants: at critical point ∂P ⁄ ∂V = 0 and ∂²P ⁄ ∂V² = 0 applied to van der Waals ⇒ Tc, Pc, Vc.
▶ Worked Example
A gas diffuses 4 times as fast as SO₂. Find its molar mass.
r₁ ⁄ r₂ = √(M₂ ⁄ M₁) = 4 ⇒ M₂ ⁄ M₁ = 16
M(SO₂) = 32 + 32 = 64 g·mol⁻¹
M₁ = 64 ⁄ 16 = 4 g·mol⁻¹ (helium)
✗ Common Exam Mistakes
- Plugging °C into any gas equation — always convert to kelvin first.
- Using R = 0.0821 when pressure is in Pa — R must match the unit system.
- Forgetting that Graham's law uses molar masses, not densities directly (though d ∝ M).
- Claiming van der Waals constants a, b are universal — they differ per gas.
04Thphysical
Physical · Ch. 04 / 20
Chemical Thermodynamics
energy, heat & the direction of change
First law
ΔU = q + w
expansion work w = −PextΔV
Enthalpy link
H = U + PV · ΔH = ΔU + ΔngRT
qp = ΔH, qv = ΔU
Calorimetry
q = mcΔT · q = CΔT
cwater = 4.184 J·g⁻¹K⁻¹
Hess's law
ΔH°rxn = ΣΔH°f(prod) − ΣΔH°f(react)
elements in standard state: ΔH°f = 0
Bond enthalpies
ΔH = Σ(broken) − Σ(formed)
average values; gas-phase only
Kirchhoff's equation
ΔHT₂ = ΔHT₁ + ΔCp(T₂ − T₁)
temperature dependence of ΔH
Entropy
ΔS = qrev ⁄ T
ΔS°rxn = ΣS°(prod) − ΣS°(react)
ΔS of gas on expansion
ΔS = nR ln(V₂ ⁄ V₁) = nR ln(P₁ ⁄ P₂)
isothermal, ideal gas
Gibbs energy
ΔG = ΔH − TΔS
ΔG < 0 → spontaneous
Equilibrium link
ΔG° = −RT ln K · ΔG = ΔG° + RT ln Q
ties to Ch. 05 & 08
Reversible isothermal work
w = −nRT ln(V₂ ⁄ V₁)
maximum work extracted
Adiabatic relation
TVγ−1 = const · PVγ = const
γ = Cp ⁄ Cv
Cp − Cv
Cp − Cv = R (per mole)
ideal gas only; Mayer's relation
⚗ Lab Notes
- State functions: U, H, S, G, T, P, V. Path functions: q, w.
- IUPAC sign convention: q > 0 heat into system; w > 0 work done on system (compression).
- Crossover temperature where ΔG flips sign: T = ΔH ⁄ ΔS (when ΔG = 0).
- Melting & vapourisation always raise entropy; S° is never zero for any substance at T > 0 (3rd law exception: perfect crystal at 0 K has S = 0).
- Spontaneity checklist: (ΔH−, ΔS−) spontaneous at low T; (+,+) never; (−,+) always; (+,−) only at high T.
- Bond enthalpies are averages over many molecules — don't use them for small molecules where ΔHf is available.
∫ Key Derivations
- Cp − Cv = R: at constant P, qp = ΔU + PΔV = nCvΔT + nRΔT, so Cp = Cv + R.
- ΔG° = −RT ln K: at equilibrium ΔG = 0, so 0 = ΔG° + RT ln K ⇒ ΔG° = −RT ln K.
- Maximum (reversible) work: w = −∫P dV with P = nRT ⁄ V ⇒ w = −nRT ln(V₂ ⁄ V₁).
- Adiabatic PVγ = const: from dU = −P dV and dU = nCv dT ⇒ dT ⁄ T = (γ−1)(−dV ⁄ V), integrate.
▶ Worked Example
ΔH = +178 kJ, ΔS = +161 J·K⁻¹ for CaCO₃ → CaO + CO₂. Find T above which the reaction is spontaneous.
ΔG = ΔH − TΔS < 0 ⇒ T > ΔH ⁄ ΔS
T > 178000 ⁄ 161 = 1106 K ≈ 833 °C
✗ Common Exam Mistakes
- Mixing ΔH in kJ with ΔS in J — always convert one before T = ΔH ⁄ ΔS.
- Forgetting that bond enthalpy method gives an approximate ΔH (average values); Hess with ΔHf is exact.
- Using ΔS°(elements) = 0 — that's only true for ΔH°f. S°(O₂) ≠ 0.
- Treating q and w as state functions — they depend on path, not endpoints.
05Eqphysical
Physical · Ch. 05 / 20
Chemical Equilibrium
the dynamic balance of forward & back
Law of mass action
Kc = [C]ᶜ[D]ᵈ ⁄ [A]ᵃ[B]ᵇ
aA + bB ⇌ cC + dD
Kp vs Kc
Kp = Kc(RT)Δn
Δn = Σngas,prod − Σngas,react
Kx (mole fraction)
Kx = Kp ⁄ PΔn
pressure-dependent when Δn ≠ 0
Reaction quotient
Q < K → fwd · Q > K → rev · Q = K ⇌
Q has K's form but at any moment
Free energy link
ΔG° = −RT ln K · ΔG = RT ln(Q ⁄ K)
K large ⇒ ΔG° very negative
Degree of dissociation
Kc = Cα² ⁄ (1 − α)
A ⇌ B + C; small α ⇒ K ≈ Cα²
Vapour density method
α = (D − d) ⁄ (d − D ⁄ n)
D = initial vapour density, d = at equilibrium, n = product moles
van 't Hoff
ln(K₂ ⁄ K₁) = (ΔH ⁄ R)(1 ⁄ T₁ − 1 ⁄ T₂)
K's temperature dependence
Reversing & scaling K
K′ = 1 ⁄ K · nK: Kⁿ
add reactions: multiply K's
Heterogeneous eq.
solids & liquids omitted
activity = 1 for pure phases
⚗ Lab Notes
- Pure solids & liquids are omitted from K (activity = 1).
- K changes only with temperature; a catalyst speeds both directions and leaves K untouched.
- Reverse reaction: K′ = 1⁄K. Multiply equation by n: Kⁿ. Add equations: multiply K values.
- Le Chatelier: add reactant → forward; compress (raise P) → fewer gas moles; heat → endothermic side; inert gas at constant V does nothing.
- K ≫ 1: product-favoured. K ≪ 1: reactant-favoured. Use approximations when K is small.
- For 2SO₂ + O₂ ⇌ 2SO₃: high pressure and ~450 °C with V₂O₅ catalyst is the contact process compromise.
∫ Key Derivations
- Kp = Kc(RT)Δn: pi = (ni ⁄ V)RT = [i]RT. Substitute into Kp = Π(pi)νi.
- α from vapour density: Mobs = Minitial(1 + α(n−1)) and d ∝ 1⁄M ⇒ α = (D−d) ⁄ ((n−1)d).
- van 't Hoff: from ΔG° = ΔH° − TΔS° and ΔG° = −RT ln K ⇒ ln K = −ΔH° ⁄ RT + ΔS° ⁄ R; subtract two temperatures.
▶ Worked Example
For N₂ + 3H₂ ⇌ 2NH₃, Kc = 41 at 400 °C. Find Kp.
Δn = 2 − (1 + 3) = −2
T = 400 + 273 = 673 K
Kp = Kc(RT)−2 = 41 ⁄ (0.0821 × 673)² = 1.35 × 10⁻²
✗ Common Exam Mistakes
- Forgetting that K is dimensionless (activities), so units should be dropped in the final answer.
- Adding an inert gas at constant V doesn't shift equilibrium — only changes in partial pressures or T matter.
- Believing catalysts change K — they don't; they only help reach equilibrium faster.
- Using Δn = (moles of products − moles of reactants) including solids — it's gaseous moles only.
06Iophysical
Physical · Ch. 06 / 20
Ionic Equilibrium
acids, bases, buffers & pH
Water ion product
Kw = [H⁺][OH⁻] = 1.0×10⁻¹⁴
25 °C ⇒ pH + pOH = 14
pH scale
pH = −log[H⁺] · pK = −log K
lower pKa ⇒ stronger acid
Weak acid
Ka = Cα² ⁄ (1 − α) ≈ Cα²
pH = ½(pKa − log C) for weak acid
Weak base
Kb ≈ Cα² · pOH = ½(pKb − log C)
pKa + pKb = pKw = 14
Henderson–Hasselbalch
pH = pKa + log([A⁻] ⁄ [HA])
acid buffer; best near pH = pKa
Basic buffer
pOH = pKb + log([BH⁺] ⁄ [B])
e.g. NH₃ ⁄ NH₄Cl
Solubility product
Ksp = [A⁺]ᵃ[B⁻]ᵇ
precipitation when Qsp > Ksp
Solubility ↔ Ksp
AxBy: Ksp = (xs)ˣ(ys)ʸ
s = solubility in mol·L⁻¹
Salt hydrolysis
Kh = Kw ⁄ Ka · h = √(Kh ⁄ C)
weak acid + strong base salt
pH of weak-acid salt
pH = 7 + ½(pKa + log C)
salt of weak acid + strong base
pH of weak-base salt
pH = 7 − ½(pKb + log C)
salt of strong acid + weak base
Common ion effect
α = Ka ⁄ [A⁻]added
suppresses ionisation; basis of buffers
⚗ Lab Notes
- Definitions: Arrhenius (H⁺/OH⁻ in water), Brønsted–Lowry (proton transfer), Lewis (e⁻-pair acceptor/donor).
- Strong acids to memorise: HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄ — assume full dissociation.
- Indicators switch near pKa ± 1: methyl orange 3.1–4.4, bromothymol blue 6.0–7.6, phenolphthalein 8.2–10.0.
- Common-ion effect suppresses ionisation — the engine of every buffer.
- Watch: dilution raises α but lowers [H⁺]; pH of a weak acid creeps toward 7, never past it.
- Conjugate base of a strong acid (Cl⁻, NO₃⁻, ClO₄⁻) is too weak to hydrolyse.
∫ Key Derivations
- pH of weak acid: Ka = [H⁺]² ⁄ (C − [H⁺]) ≈ [H⁺]² ⁄ C ⇒ [H⁺] = √(KaC) ⇒ pH = ½(pKa − log C).
- H–H equation: Ka = [H⁺][A⁻] ⁄ [HA] ⇒ [H⁺] = Ka[HA] ⁄ [A⁻] ⇒ pH = pKa + log([A⁻] ⁄ [HA]).
- pH of weak-acid salt: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. Kh = Kw ⁄ Ka = [CH₃COOH][OH⁻] ⁄ [CH₃COO⁻] = x² ⁄ C ⇒ [OH⁻] = √(KhC), etc.
- Buffer capacity β = dC ⁄ dpH = 2.303 × C · Ka[H⁺] ⁄ (Ka + [H⁺])²; maximum at pH = pKa.
▶ Worked Example
Find pH of 0.1 M acetic acid (Ka = 1.8×10⁻⁵).
pH = ½(pKa − log C) = ½(4.74 − log 0.1)
= ½(4.74 + 1) = 2.87
✗ Common Exam Mistakes
- Applying pH = ½(pKa − log C) to strong acids — for strong acids, pH = −log C directly.
- Forgetting that [H⁺] + [Na⁺] = [OH⁻] + [A⁻] (charge balance) — always use it to verify approximations.
- Thinking Ksp alone compares solubilities — only valid when stoichiometry is the same; otherwise solve for s.
- Ignoring water autoionisation when [H⁺] < 10⁻⁶ M (very dilute acids).
07Sophysical
Physical · Ch. 07 / 20
Solutions & Colligative Properties
what particle number alone can do
Henry's law
p = KH · x
gas solubility ∝ pressure; KH rises with T ⇒ solubility falls
Raoult's law
pi = xi · p°i
for volatile ideal solutions
Relative lowering
Δp ⁄ p° = xsolute = n ⁄ (n + N)
for non-volatile solute
Boiling elevation
ΔTb = i · Kb · m
Kb(water) = 0.52 K·kg·mol⁻¹
Freezing depression
ΔTf = i · Kf · m
Kf(water) = 1.86 K·kg·mol⁻¹
Osmotic pressure
π = i · CRT
reverse osmosis when applied P > π
Molar mass from ΔTf
M = Kf · w · 1000 ⁄ (ΔTf · W)
w = solute g, W = solvent g
Molar mass from π
M = wRT ⁄ πV
preferred for polymers
van 't Hoff factor
i = Mnormal ⁄ Mobs
i > 1 dissociation, i < 1 association
Degree of dissociation
α = (i − 1) ⁄ (n − 1)
n = particles per molecule
Degree of association
α = (1 − i) ⁄ (1 − 1⁄n)
e.g. benzoic acid dimer (n = 2)
Isotonic solutions
π₁ = π₂ ⇔ C₁ = C₂ (non-electrolyte)
no net water flow across membrane
⚗ Lab Notes
- Colligative properties depend on particle count only: NaCl ⇒ i ≈ 2, CaCl₂ ≈ 3; acetic acid dimerising in benzene ⇒ i ≈ 0.5.
- Ideal solution: obeys Raoult at all compositions, ΔHmix = 0, ΔVmix = 0.
- Positive deviation → min-boiling azeotrope (EtOH–water, 95.4%); negative → max-boiling (HNO₃–water).
- For macromolecules, π is the most sensitive colligative route to molar mass.
- Watch: colligatives use molality (π uses molarity) — never mass percent.
- Azeotropes cannot be separated by simple distillation — that's why 95% ethanol is the commercial "absolute" grade.
∫ Key Derivations
- ΔTb = Kbm: from Clausius–Clapeyron applied to dilute solution: ΔTb = (RT² ⁄ ΔHvap) · xsolute. Group the solvent constants into Kb.
- Molar mass from Δp: Δp ⁄ p° ≈ n ⁄ N = (w ⁄ M) ⁄ (W ⁄ Msolv) ⇒ M = (w · p° · Msolv) ⁄ (Δp · W).
- i and α: If each molecule splits into n particles, initial 1 → 1−α+nα total, so i = 1 + α(n−1).
▶ Worked Example
What is the freezing point of a 0.1 m NaCl solution (assume i = 1.9)?
ΔTf = i · Kf · m = 1.9 × 1.86 × 0.1
ΔTf = 0.353 K
Tf = 0 − 0.353 = −0.353 °C
✗ Common Exam Mistakes
- Forgetting the van 't Hoff factor i for electrolytes — NaCl isn't 1 particle.
- Using molarity in ΔTb or ΔTf — only molality works (π is the exception).
- Treating azeotropes as compounds — they are mixtures with fixed vapour composition.
- Assuming i = 2 for NaCl always — in reality ion pairing reduces i slightly below 2.
08Elphysical
Physical · Ch. 08 / 20
Electrochemistry
where chemistry meets the circuit
Cell emf
E°cell = E°cathode − E°anode
both as reduction potentials
Nernst equation
E = E° − (0.0591 ⁄ n) log Q
298 K; Q from cell reaction
Equilibrium condition
E = 0 ⇒ E° = (0.0591 ⁄ n) log K
dead battery
Energy links
ΔG° = −nFE°
F = 96 485 C·mol⁻¹
Resistance & conductance
R = ρl ⁄ A · G = 1 ⁄ R = κA ⁄ l
κ = conductivity
Cell constant
G* = l ⁄ A = κ ⁄ G = κR
determined with KCl standard
Molar conductivity
Λm = 1000κ ⁄ C
in S·cm²·mol⁻¹
Kohlrausch's law
Λ°m = ν₊λ°₊ + ν₋λ°₋
independent migration at infinite dilution
Weak electrolyte Λ°
Λ°CH₃COOH = Λ°CH₃COONa + Λ°HCl − Λ°NaCl
algebraic combination
Degree of dissociation
α = Λm ⁄ Λ°m
weak electrolyte at concentration C
Faraday's first law
m = (M · I · t) ⁄ (n · F) = Z · Q
Z = electrochemical equivalent
Faraday's second law
m₁ ⁄ m₂ = E₁ ⁄ E₂
same charge, different cells in series
⚗ Lab Notes
- SHE: Pt | H₂(1 bar) | H⁺(1 M) — defined exactly 0.000 V.
- AN-OX: anode = oxidation (− in galvanic, + in electrolytic). Electrons flow anode → cathode externally; salt bridge maintains charge balance.
- Concentration cell: E° = 0 but E >0 when concentrations differ (Nernst only).
- Batteries: lead-acid ≈ 2 V per cell; Li-ion shuttles Li⁺ between intercalation hosts; fuel cells run continuously on H₂/O₂.
- Corrosion is electrochemical: Fe → Fe²⁺ at anodic sites. Defence: galvanising, sacrificial Zn ⁄ Mg anodes, cathodic protection.
- Λm rises with dilution; for strong electrolytes linearly with √C (Debye–Hückel–Onsager), for weak electrolytes sharply.
∫ Key Derivations
- Nernst: ΔG = ΔG° + RT ln Q and ΔG = −nFE, ΔG° = −nFE° ⇒ −nFE = −nFE° + RT ln Q ⇒ E = E° − (RT ⁄ nF) ln Q.
- At 298 K: (RT ⁄ F) ln 10 = 0.0591 V, so E = E° − (0.0591 ⁄ n) log Q.
- α = Λm ⁄ Λ°m: at infinite dilution all molecules dissociate, so α = 1 and Λ = Λ°. At finite C, fewer ions contribute.
- Ka from Λ: Ka = Cα² ⁄ (1 − α) = C(Λ ⁄ Λ°)² ⁄ (1 − Λ ⁄ Λ°).
▶ Worked Example
For Zn | Zn²⁺(0.1 M) || Cu²⁺(1 M) | Cu with E° = 1.10 V, find E.
n = 2, Q = [Zn²⁺] ⁄ [Cu²⁺] = 0.1 ⁄ 1 = 0.1
E = 1.10 − (0.0591 ⁄ 2) log(0.1) = 1.10 − 0.02955 × (−1)
E = 1.13 V (concentrating reactants boosts E)
✗ Common Exam Mistakes
- Flipping the sign in E°cell = E°cathode − E°anode. Both are tabulated as reduction potentials — don't reverse one.
- Forgetting that n in Nernst is the number of electrons transferred per mole of cell reaction, not per half-cell.
- Confusing molar conductivity (Λm, per mole) with specific conductivity (κ, per length).
- In Faraday problems, mixing up valency (n in Mⁿ⁺) with the metal symbol — always balance the electrode reaction first.
09Kiphysical
Physical · Ch. 09 / 20
Chemical Kinetics
rates, orders & the half-life clock
Rate of reaction
r = −(1 ⁄ a) d[A] ⁄ dt = (1 ⁄ p) d[P] ⁄ dt
divide by stoichiometric coeff.
Rate law
r = k[A]ᵐ[B]ⁿ
m + n = order; experimental
Zero order
[A] = [A]₀ − kt · t½ = [A]₀ ⁄ 2k
surface-catalysed, photochemical
First order
k = (2.303 ⁄ t) log([A]₀ ⁄ [A])
t½ = 0.693 ⁄ k (independent of [A]₀)
nth-order time to fraction
t1⁄2 ∝ 1 ⁄ [A]₀n−1
diagnostic for n
Second order
1 ⁄ [A] = 1 ⁄ [A]₀ + kt
t½ = 1 ⁄ k[A]₀
Arrhenius equation
k = A e−Ea ⁄ RT
A = frequency factor
Two-temperature form
log(k₂ ⁄ k₁) = (Ea ⁄ 2.303R)(1 ⁄ T₁ − 1 ⁄ T₂)
find Ea from two rate measurements
Linear Arrhenius
ln k = ln A − Ea ⁄ RT
slope of ln k vs 1⁄T = −Ea ⁄ R
Collision theory
r = P · Z · e−Ea ⁄ RT
P = steric factor
Pseudo-first order
r = k'[A] with k' = k[B]excess
hydrolysis of ester in dilute water
Orders at a glance
| order | integrated | t½ | units of k |
|---|---|---|---|
| 0 | [A] = [A]₀ − kt | [A]₀ ⁄ 2k | mol·L⁻¹·s⁻¹ |
| 1 | ln([A]₀⁄[A]) = kt | 0.693 ⁄ k | s⁻¹ |
| 2 | 1⁄[A] − 1⁄[A]₀ = kt | 1 ⁄ k[A]₀ | L·mol⁻¹·s⁻¹ |
| n | — | ∝ 1⁄[A]₀n−1 | mol1−n·Ln−1·s⁻¹ |
⚗ Lab Notes
- Units of k for order n: mol1−n·Ln−1·s⁻¹.
- Molecularity is a whole number ≥ 1 (mechanism step); order can be 0 or fractional and is experimental.
- A catalyst lowers Ea for both directions; a rule of thumb: rate roughly doubles per +10 °C (temperature coefficient ≈ 2).
- Collision theory: r = P · Z · e−Ea ⁄ RT; P is the orientation factor.
- Constant half-life ⇒ first order — radioactive decay is the classic example.
- For A → products, t75% ⁄ t50% = 2 for first order, 3 for zero order.
∫ Key Derivations
- First-order integrated: d[A] ⁄ [A] = −k dt ⇒ ∫ = −k∫dt ⇒ ln([A] ⁄ [A]₀) = −kt ⇒ [A] = [A]₀ e−kt.
- t½ for first order: [A] = [A]₀ ⁄ 2 ⇒ ln(1⁄2) = −k t½ ⇒ t½ = ln 2 ⁄ k = 0.693 ⁄ k.
- Arrhenius from rate theory: k = (kBT ⁄ h) e−ΔG‡ ⁄ RT; combine with ΔG‡ = ΔH‡ − TΔS‡ to link A and Ea.
- Graphical order test: plot [A] vs t (0), ln[A] vs t (1), 1⁄[A] vs t (2) — the linear one identifies the order.
▶ Worked Example
A first-order reaction is 75% complete in 60 min. Find k and t½.
[A] ⁄ [A]₀ = 0.25, t = 60 min
k = (2.303 ⁄ 60) log(1 ⁄ 0.25) = 2.303 ⁄ 60 × log 4 = 0.0231 min⁻¹
t½ = 0.693 ⁄ 0.0231 = 30 min (note: t₇₅% = 2 × t½)
✗ Common Exam Mistakes
- Writing the order from stoichiometry — order is always experimental; stoichiometry only matches for elementary reactions.
- Using the Arrhenius slope as Ea — slope = −Ea ⁄ R, so Ea = −slope × R.
- For pseudo-first order, forgetting that k' = k[B]excess; to get true k you must divide by [B].
- Confusing average rate (Δ[A] ⁄ Δt) with instantaneous rate (tangent slope at a point).
10Peinorganic
Inorganic · Ch. 10 / 20
Periodicity & the Periodic Table
the trends behind every element
Atomic radius trend
across period ↓ · down group ↑
Zeff ↑, shells ↑
Ionic radii
cation < atom < anion
isoelectronic: more p⁺ = smaller
Ionization energy
M(g) → M⁺(g) + e⁻
always endothermic; IE₂ > IE₁
IE anomaly
Be > B · N > O
half-filled & full subshell stability
Electron affinity
X(g) + e⁻ → X⁻(g)
EA(Cl) > EA(F): 2p too compact
Electronegativity
F = 4.0 (Pauling max)
across ↑, down ↓
Effective nuclear charge
Zeff = Z − S
Slater rules; shielding s > p > d > f
Inert-pair effect
heavier p-block prefers lower ox. state
Pb²⁺ more stable than Pb⁴⁺; Bi³⁺ more than Bi⁵⁺
Diagonal relationship
Li – Mg · Be – Al · B – Si
similar Zeff ⁄ radius
Periodic Table · Trends at a glance
1H1.008
2He4.00
4Be9.01
5B10.81
6C12.01
7N14.01
8O16.00
9F19.00
10Ne20.18
13Al26.98
14Si28.09
15P30.97
16S32.06
17Cl35.45
18Ar39.95
21Sc44.96
22Ti47.87
23V50.94
24Cr52.00
25Mn54.94
26Fe55.85
27Co58.93
28Ni58.69
29Cu63.55
30Zn65.38
31Ga69.72
32Ge72.63
33As74.92
34Se78.97
35Br79.90
36Kr83.80
shown: periods 1–4 · metal non-metal metalloid transition
⚗ Lab Notes
- Radius: down a group ↑ (new shell), across a period ↓ (Zeff rises). Cation < atom < anion.
- Isoelectronic series: more protons = smaller — Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻.
- IE anomalies: Be > B (full 2s vs 2p¹); N > O (half-filled 2p³ vs 2p⁴).
- Oxide character runs basic → amphoteric → acidic across a period (Na₂O → Al₂O₃ → SO₃).
- Diagonal relationships from similar Zeff⁄radius: Li–Mg, Be–Al, B–Si.
- Second-period anomalies: F has lower EA than Cl; N has no pentahalide (no d-orbitals); Be and Al are amphoteric.
∫ Key Derivations
- Slater's rules (quick form): group electrons as (1s)(2s,2p)(3s,3p)(3d)(4s,4p)…; contributions to S: same group 0.35 each (0.30 for 1s); one shell lower 0.85 for s,p, 1.00 for d,f; two or more shells lower 1.00.
- IE vs Z graph: sharp drops at group 1 after each noble gas; peaks at group 18 and at Be, N (full/half subshells).
▶ Worked Example
Arrange in order of increasing size: O²⁻, F⁻, Na⁺, Mg²⁺, Al³⁺.
All are isoelectronic (10 electrons).
More protons = smaller radius.
Proton count: Al(13) > Mg(12) > Na(11) > F(9) > O(8)
Order: Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻
✗ Common Exam Mistakes
- Assuming noble gases have the largest atomic radii in their period — their van der Waals radii do, but their covalent radii are not directly comparable.
- Claiming electron affinity is always exothermic — second EA is always endothermic (adding an electron to an anion).
- Thinking transition metals show smooth trends — they're irregular because inner d electrons shield poorly.
11Boinorganic
Inorganic · Ch. 11 / 20
Chemical Bonding & Molecular Structure
shapes, orbitals & why water bends
Formal charge
FC = V − L − ½B
V valence, L lone e⁻, B bonding e⁻
Octet exceptions
incomplete (BF₃) · expanded (SF₆) · odd (NO)
period 3+ can expand via d orbitals
Steric number
SN = σ bonds + lone pairs
2 sp · 3 sp² · 4 sp³ · 5 sp³d · 6 sp³d²
% s-character
sp 50% · sp² 33% · sp³ 25%
more s = shorter, stronger, more acidic
Dipole moment
μ = q · d · % ionic = μobs ⁄ μionic × 100
1 D = 3.34×10⁻³⁰ C·m
Bond order (MO)
BO = (Nb − Na) ⁄ 2
higher BO ⇒ shorter, stronger
Magnetic moment
μ = √(n(n + 2)) BM
n = unpaired e⁻
Hydrogen bond strength
F–H···F > O–H···O > N–H···N
5–40 kJ·mol⁻¹, directional
Lattice energy (Born–Landé)
U ∝ Z⁺Z⁻ ⁄ (r⁺ + r⁻)
Born–Mayer adds the Born exponent
Fajans' rules
small cation + large anion + high charge ⇒ more covalent
LiI is most covalent of lithium halides
VSEPR quick shapes
| SN | LP | shape | angle | example |
|---|---|---|---|---|
| 2 | 0 | linear | 180° | CO₂ |
| 3 | 0 | trig. planar | 120° | BF₃ |
| 3 | 1 | bent | ~120° | SO₂ |
| 4 | 0 | tetrahedral | 109.5° | CH₄ |
| 4 | 1 | trig. pyramidal | 107° | NH₃ |
| 4 | 2 | bent | 104.5° | H₂O |
| 5 | 0 | trig. bipyramidal | 90°, 120° | PCl₅ |
| 5 | 1 | see-saw | — | SF₄ |
| 6 | 0 | octahedral | 90° | SF₆ |
| 6 | 1 | square pyramidal | — | BrF₅ |
| 6 | 2 | square planar | 90° | XeF₄ |
MO configurations (homonuclear diatomics)
| molecule | configuration | BO | magnetic |
|---|---|---|---|
| H₂ | σ1s² | 1 | dia |
| He₂ | σ1s² σ*1s² | 0 | doesn't exist |
| N₂ | (σ2s)²(σ*2s)²(π2p)⁴(σ2p)² | 3 | dia |
| O₂ | …(π2p)⁴(σ2p)²(π*2p)² | 2 | para (2 unpaired) |
| F₂ | …(π*2p)⁴ | 1 | dia |
⚗ Lab Notes
- Repulsion order LP–LP > LP–BP > BP–BP ⇒ angles CH₄ 109.5° > NH₃ 107° > H₂O 104.5°.
- MO theory: O₂ has two unpaired π* electrons ⇒ paramagnetic; N₂ diamagnetic.
- Hydrogen bonding needs H attached to N, O or F; ice's open lattice makes it less dense than water.
- Fajans' rules: small cation + large anion + high charge ⇒ more covalent character (LiI > LiF covalency).
- In trigonal bipyramids, lone pairs sit equatorial; in octahedra all sites are equivalent.
- Resonance structures aren't flipping — the real molecule is a hybrid always. More covalent bonds & full octets dominate.
∫ Key Derivations
- % ionic character: μionic = (1.602×10⁻¹⁹ C)(bond length in m); % = μobs ⁄ μionic × 100. Hannay–Smith: % = 16|Δχ| + 3.5(Δχ)².
- Born–Haber cycle: ΔHf = ΔHsub + IE + ½D + EA + U ⇒ solve for lattice energy U.
- MO energy levels for O₂ vs N₂: below N₂ the σ2p is higher than π2p; above N₂ it flips (s–p mixing).
▶ Worked Example
Predict the shape and magnetic behaviour of O₂²⁻ (peroxide ion).
Total electrons = 16 + 2 = 18 (same as F₂)
MO: (π*2p)⁴ — all paired
BO = (10 − 8) ⁄ 2 = 1 (single bond)
Geometry: linear; diamagnetic
✗ Common Exam Mistakes
- Counting double bonds as two electron domains in VSEPR — they count as one domain.
- Assuming all AB₂ molecules are linear — they are only linear with SN = 2 (no lone pairs).
- Saying O₂ is diamagnetic because it has an even number of electrons — MO shows 2 unpaired.
- Forgetting that resonance structures differ only in electron placement, never in atom positions.
12Crinorganic
Inorganic · Ch. 12 / 20
Solid State & Crystal Chemistry
unit cells, packing & defects
Unit-cell density
ρ = Z · M ⁄ (a³ · NA)
a in cm ⇒ ρ in g·cm⁻³
Edge–radius contact
sc: a = 2r · bcc: a = 4r ⁄ √3 · fcc: a = 2√2 r
touch along edge ⁄ body ⁄ face diagonal
Atoms per unit cell
sc: 1 · bcc: 2 · fcc: 4
corner 1⁄8, face 1⁄2, body 1
Packing efficiency
sc 52.4% · bcc 68% · fcc ⁄ hcp 74%
void fraction is the complement
Coordination number
sc 6 · bcc 8 · fcc ⁄ hcp 12
nearest neighbours
Bragg's law
nλ = 2d · sin θ
X-ray diffraction
Radius-ratio rule
r⁺ ⁄ r⁻ → CN 3 (<0.225) · 4 (≤0.414) · 6 (≤0.732) · 8 (>0.732)
triangular · tetrahedral · octahedral · cubic
Interstitial radii
tetrahedral 0.225R · octahedral 0.414R
in fcc packing
Void count in fcc
tetrahedral = 2N · octahedral = N
N = atoms per cell
⚗ Lab Notes
- Schottky: equal cation–anion vacancies ⇒ density drops (NaCl, KCl). Frenkel: small ion dislodged to interstitial (AgCl, ZnS) — density unchanged.
- Structures: NaCl 6:6 rock-salt; CsCl 8:8; ZnS 4:4 zinc blende; CaF₂ 8:4 fluorite; Na₂O 4:8 antifluorite.
- Doping Si with P ⇒ n-type; with B ⇒ p-type — together they make the diode.
- 7 crystal systems, 14 Bravais lattices; ionic solids conduct only when molten ⁄ dissolved.
- fcc = cubic close-packed; hcp = ABAB… stacking; fcc = ABCABC…
- F-centres (colour centres): trapped electrons in anion vacancies give NaCl its yellow, KCl its violet tinge.
∫ Key Derivations
- Packing efficiency of fcc: atoms touch along face diagonal: 4r = √2 a ⇒ a = 2√2 r. Volume of 4 spheres = 4 × (4⁄3)πr³ = 16πr³ ⁄ 3. Cell volume = a³ = 16√2 r³. Ratio = π ⁄ 3√2 ≈ 0.74.
- bcc packing: 4r = √3 a ⇒ a = 4r ⁄ √3. 2 spheres per cell. PE = 2 × (4⁄3)πr³ ⁄ a³ = √3π ⁄ 8 ≈ 0.68.
- Density formula: ρ = mass ⁄ volume = (Z × M ⁄ NA) ⁄ a³.
▶ Worked Example
Copper (fcc, M = 63.5) has a = 361 pm. Find its density.
Z = 4, a = 3.61 × 10⁻⁸ cm
ρ = 4 × 63.5 ⁄ (3.61 × 10⁻⁸)³ × (6.022 × 10²³)
ρ = 254 ⁄ (4.70 × 10⁻²³ × 6.022 × 10²³) = 8.97 g·cm⁻³ (matches literature ≈ 8.96)
✗ Common Exam Mistakes
- Forgetting that a in pm must be converted to cm (1 pm = 10⁻¹⁰ cm) when computing density.
- Thinking Frenkel defect changes density — it doesn't; only Schottky does.
- Using CN 12 for bcc — that's fcc/hcp; bcc has CN 8.
13Coinorganic
Inorganic · Ch. 13 / 20
Coordination Compounds
complexes, colours & crystal fields
Coordination number
CN = Σ (ligand denticity × count)
most common: 4 (tet/sq.pl) & 6 (oct)
Oxidation state
M ox + Σ(charge of ligands) = overall charge
neutral ligands (NH₃, H₂O, CO): 0
Spin-only μ
μ = √(n(n + 2)) BM
n = unpaired electrons
CFSE octahedral
t₂g: −0.4Δₒ each · eg: +0.6Δₒ each
subtract P for each paired pair
CFSE tetrahedral
e: −0.6Δt · t₂: +0.4Δt · Δt = (4⁄9)Δₒ
small splitting ⇒ high-spin always
EAN rule
EAN = Z − ox + 2 × CN
= 36 (Kr) for stable carbonyls
Spectrochemical series
I⁻ < Br⁻ < Cl⁻ < F⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO
weak → strong field
d-electron count
dⁿ where n = (group of M) − ox
Cr³⁺ is d³, Fe²⁺ is d⁶
⚗ Lab Notes
- Werner: primary valence = oxidation state (ionisable); secondary = coordination number (fixed geometry).
- Strong field ⇒ low spin (pair before promoting); weak field ⇒ high spin. Crossover around d⁴–d⁷.
- Colour arises from d–d transitions absorbing the complementary colour — [Ti(H₂O)₆]³⁺ is violet (absorbs green-yellow).
- Isomerism: ionisation ([Co(NH₃)₅Br]SO₄ vs [Co(NH₃)₅SO₄]Br); linkage (NO₂ vs ONO); coordination; cis–trans (e.g. [Pt(NH₃)₂Cl₂]); optical (chelates).
- Real-world: EDTA titrates water hardness; cisplatin is anticancer; haemoglobin carries O₂ via Fe²⁺; vitamin B₁₂ is Co³⁺.
- Chelate effect: multidentate ligands form more stable complexes (entropy-driven).
∫ Key Derivations
- CFSE calculation: sum each electron's contribution: t₂g −0.4Δₒ, eg +0.6Δₒ; subtract pairing energy P for every extra paired pair beyond the free ion.
- Why tetrahedral complexes are high-spin: Δt ≈ 4Δₒ ⁄ 9 is always smaller than pairing energy P.
- Δₒ from absorption: Δₒ = hc ⁄ λmax (in J per photon) × NA for per mole.
▶ Worked Example
For [Fe(CN)₆]³⁻ (d⁵, CN⁻ = strong field), find μ and CFSE.
Strong field ⇒ low spin ⇒ t₂g⁵ eg⁰
Unpaired electrons n = 1
μ = √(1 × 3) = 1.73 BM
CFSE = 5(−0.4Δₒ) − 2P (2 extra pairs vs free d⁵) = −2.0Δₒ − 2P
✗ Common Exam Mistakes
- Forgetting that CO, NH₃, H₂O, en are neutral ligands — they don't contribute to oxidation state.
- Counting d electrons using the atomic number of the metal instead of the metal ion's group position minus oxidation state.
- Assuming all octahedral d⁶ complexes are diamagnetic — only low-spin (strong field) are.
- Confusing linkage isomers (same formula, different donor atom) with ionisation isomers (different counterions).
14Ocorganic
Organic · Ch. 14 / 20
Organic Basics — GOC & Isomerism
the grammar every reaction obeys
Degree of unsaturation
DBE = (2C + 2 + N − H − X) ⁄ 2
double-bond equivalents
Carbocation stability
3° > 2° > 1° > CH₃⁺
+I & hyperconjugation; benzylic/allylic boosted by resonance
Carbanion stability
CH₃⁻ > 1° > 2° > 3°
opposite order; sp > sp² > sp³
Radical stability
3° > 2° > 1° > CH₃•
follows carbocations
Acidity (pKa)
lower pKa ⇒ stronger acid
EWGs stabilise conjugate base
Aromaticity — Hückel
planar + cyclic + conjugated + (4n + 2) π e⁻
benzene: n = 1 (6 π e⁻)
Antiaromaticity
planar + cyclic + conjugated + 4n π e⁻
cyclobutadiene (4 e⁻) — highly unstable
Inductive effect
−I withdraw · +I donate
fades beyond ~2 C atoms
Resonance (mesomeric)
−M: NO₂, CN, COOH · +M: OH, NH₂, OR
dominates over induction in conjugated systems
Hyperconjugation
more α-H ⇒ more stable carbocation & alkene
Baker–Nathan effect
⚗ Lab Notes
- −I withdrawers: NO₂, CN, halogens, COOH, CF₃. +I donors: alkyl groups. Inductive effect fades beyond ~2 carbons.
- Resonance outranks induction inside a conjugated chain; more covalent structures & full octets dominate.
- Hyperconjugation needs α-H — more α-H ⇒ more stable carbocation and more stable alkene.
- Nucleophiles donate pairs (OH⁻, CN⁻, NH₃, H₂O); electrophiles accept them (H⁺, NO₂⁺, R₃C⁺, BF₃).
- SN1: two steps, racemisation, polar protic solvent, 3° favoured. SN2: one step, Walden inversion, polar aprotic, CH₃⁄1° favoured.
- Zaitsev: the more substituted alkene dominates elimination unless the base is bulky (Hofmann product).
∫ Key Derivations
- DBE derivation: an alkane CnH2n+2 has zero DBE. Each ring or double bond removes 2 H. Each N adds one H. Halogens count like H.
- Hammett equation: log(K ⁄ K₀) = ρσ — linear free-energy relationship for substituted benzoic acids. ρ is reaction sensitivity, σ is substituent constant.
▶ Worked Example
Rank acidity: CH₃COOH, ClCH₂COOH, Cl₂CHCOOH, Cl₃CCOOH.
All are carboxylic acids; −I effect of Cl stabilises the carboxylate.
More Cl = stronger −I = stronger acid.
Order: CH₃COOH < ClCH₂COOH < Cl₂CHCOOH < Cl₃CCOOH
pKa: 4.76 → 2.86 → 1.29 → 0.65
✗ Common Exam Mistakes
- Claiming halogens are +M — they are −M weakly but −I strongly; overall they deactivate benzene rings.
- Forgetting that NH₂ is +M only when lone pair can delocalise; aniline is less basic than alkyl amines.
- Treating carbocation rearrangements as optional — if a more stable cation can form by H⁻ or R⁻ shift, it will.
- Mixing up antiaromatic (unstable) with non-aromatic (doesn't meet criteria, e.g. cyclooctatetraene is tub-shaped).
15Hyorganic
Organic · Ch. 15 / 20
Hydrocarbons
alkanes, alkenes, alkynes & benzene
General formulas
CnH2n+2 (alkanes) · CnH2n (alkenes) · CnH2n−2 (alkynes)
alkadienes share alkynes formula
Markovnikov
H → C with more H already
peroxides flip for HBr only
Anti-Markovnikov
HBr + ROOR · or hydroboration
radical mechanism · syn addition
Ozonolysis
C=C + O₃ ⁄ Zn → carbonyls
reductive: aldehydes ⁄ ketones
Oxidative ozonolysis
O₃ ⁄ H₂O₂ → acids ⁄ ketones
aldehydes over-oxidised
Combustion
CnH2n+2 + (3n+1)⁄2 O₂ → nCO₂ + (n+1)H₂O
highly exothermic
Wurtz
2RX + 2Na → R–R + 2NaX
best for even-carbon alkanes
Friedel–Crafts
ArH + RX ⁄ AlCl₃ → ArR
or ArH + RCOCl ⁄ AlCl₃ → ArCOR
Directing effects
o,p: −OH, −OR, −R, −NH₂ · meta: −NO₂, −COOH, −CN
halogens: o,p but deactivating
Partial reductions
Lindlar: cis · Na/NH₃(liq): trans
alkyne → alkene
Mechanism · Electrophilic Aromatic Substitution (nitration)
⚗ Lab Notes
- Radical halogenation: initiation (hν), propagation, termination. Selectivity Br₂ ≫ Cl₂; reactivity F₂ > Cl₂ > Br₂.
- Terminal alkynes are weakly acidic (sp C–H, pKa ≈ 25) — NaNH₂ forms acetylides; Ag⁺ ⁄ Cu⁺ give precipitates (identification test).
- Friedel–Crafts needs AlCl₃; fails on strongly deactivated rings and on amines (Lewis base ties up the catalyst).
- Partial reductions: Lindlar catalyst (Pd/BaSO₄ + quinoline) ⇒ cis-alkene; Na ⁄ liquid NH₃ ⇒ trans-alkene.
- Benzene prefers substitution over addition to preserve aromaticity; UV + Cl₂ forces hexachlorocyclohexane (BHC).
- Conjugated dienes undergo 1,4-addition under thermodynamic control (high T, long time).
∫ Key Derivations
- Markovnikov by carbocation stability: HX adds to form the more stable carbocation; 2° or 3° intermediates are lower in energy.
- Anti-Markovnikov by radical chain: Br• adds to give the more stable radical (2° or 3°); H atom ends up on the more substituted carbon.
- Ozonolysis location test: cleaving the C=C gives two carbonyls; identifying them pinpoints the double bond in the original alkene.
▶ Worked Example
An unknown alkene on ozonolysis gives acetone and formaldehyde. What is it?
Acetone: (CH₃)₂C=O · Formaldehyde: H₂C=O
Join the two carbonyl carbons with a C=C, remove the O's.
Structure: (CH₃)₂C=CH₂ · 2-methylpropene
✗ Common Exam Mistakes
- Applying anti-Markovnikov to HCl or HI with peroxides — only HBr works (the other HX bonds are too strong ⁄ too weak).
- Forgetting that Friedel–Crafts alkylation can polyalkylate (the product is more activated than the starting material).
- Claiming cyclooctatetraene is antiaromatic — it's tub-shaped and non-planar, so it's non-aromatic.
16Hxorganic
Organic · Ch. 16 / 20
Haloalkanes & Haloarenes
the versatile C–X bond
Bond strength
C–F > C–Cl > C–Br > C–I
reactivity reversed
SN2 reactivity
CH₃X > 1° > 2° ≫ 3°
backside attack ⇒ inversion
SN1 reactivity
3° > 2° ≫ 1°
carbocation intermediate ⇒ racemic
SN2 rate law
rate = k[RX][Nu⁻]
bimolecular
SN1 rate law
rate = k[RX]
unimolecular (RDS = ionisation)
E2 elimination
strong base, anti-periplanar H & X
Zaitsev product dominant
E1 elimination
carbocation, then base takes β-H
competes with SN1 in protic solvents
Grignard formation
RX + Mg →(dry ether) RMgX
destroyed by H₂O, ROH, CO₂
Wurtz–Fittig
ArX + RX + 2Na → ArR + 2NaX
alkylation of arenes
Finkelstein
RCl + NaI →(acetone) RI + NaCl↓
halide exchange
Swarts
RCl + AgF → RF + AgCl
makes fluorides (CFC synthesis)
Mechanism · SN2 (Walden inversion)
⚗ Lab Notes
- Polar aprotic solvents (DMSO, acetone) accelerate SN2; polar protic (H₂O, ROH) stabilise SN1 ions.
- Allylic & benzylic halides are highly reactive in both SN1 and SN2 (resonance-stabilised intermediates).
- Aryl halides are inert: C–X has partial double-bond character (resonance) and sp² carbon resists backside attack.
- Hazards: chloroform oxidises to phosgene in air (stored with ~1% ethanol); CFCs deplete ozone; AgNO₃ test ranks halide lability.
- 2° halides can go either SN1 or SN2 — solvent and nucleophile strength decide.
∫ Key Derivations
- SN2 activation energy: backside attack avoids the filled orbital lobe of the leaving group — lowest-energy path.
- SN1 rate law: RDS is R–X → R⁺ + X⁻ (slow); then Nu⁻ + R⁺ → R–Nu (fast). Overall rate = k₁[RX].
- E1cB mechanism: base removes proton first giving carbanion, then leaving group departs — seen when leaving group is poor and α-H is acidic.
▶ Worked Example
Predict the major product of 2-bromo-2-methylpropane + aq. NaOH.
Substrate is 3° — SN1 favoured in aqueous (protic) medium.
RDS: (CH₃)₃C–Br → (CH₃)₃C⁺ + Br⁻
Fast: (CH₃)₃C⁺ + OH⁻ → (CH₃)₃COH (tert-butyl alcohol)
Some E1 also gives isobutylene as minor product.
✗ Common Exam Mistakes
- Assuming all 2° substrates are SN1 — strong nucleophile + polar aprotic solvent flips to SN2.
- Forgetting that Grignard reagents are destroyed by any acidic hydrogen (H₂O, ROH, NH₃, RCOOH) — use dry glassware.
- Thinking aryl halides undergo SN reactions easily — they need activating groups (NO₂) at ortho/para for SNAr.
17Ohorganic
Organic · Ch. 17 / 20
Alcohols, Phenols & Ethers
the −OH family and its reactions
Oxidation ladder
1° → RCHO → RCOOH · 2° → R₂C=O · 3° inert (mild)
PCC stops 1° at aldehyde
Lucas test
3° instant · 2° ~5 min · 1° none at RT
ZnCl₂ ⁄ conc. HCl
Iodoform test
CH₃CH(OH)– or CH₃CO– → CHI₃↓
yellow ppt · I₂ ⁄ NaOH
Acidity order
H₂O > ROH > RC≡CH > NH₃ > RH
PhOH > H₂O > ROH
Phenol acidity
PhOH + NaOH → PhO⁻Na⁺ + H₂O
phenoxide is resonance-stabilised
Williamson synthesis
R′O⁻ + RX → R′–O–R + X⁻
best with 1° halide; 3° halide eliminates
Dehydration of ROH
443 K H₂SO₄: alkene · 413 K: ether
ease: 3° > 2° > 1°
Kolbe's reaction
PhO⁻Na⁺ + CO₂ →(Δ) salicylic acid
ortho to OH
Reimer–Tiemann
PhOH + CHCl₃ ⁄ NaOH → salicylaldehyde
:CCl₂ (dichlorocarbene) is the electrophile
Ether cleavage
R–O–R′ + excess HI → RI + R′I + H₂O
ethers are otherwise inert
Hydroboration–oxidation
RCH=CH₂ + (BH₃)₂ →(H₂O₂⁄OH⁻) RCH₂CH₂OH
anti-Markovnikov 1° alcohol
Mechanism · Dehydration of ethanol to ethene
⚗ Lab Notes
- H-bonding ⇒ b.p. far above alkanes of equal mass; water solubility fades beyond ~3 carbons.
- Dehydration with conc. H₂SO₄: 443 K → alkene, 413 K → ether; 3° alcohols dehydrate easiest (carbocation).
- Iodoform test: yellow CHI₃ from CH₃CH(OH)– or CH₃CO– groups; distinguishes ethanol from methanol.
- Phenol + neutral FeCl₃ → violet complex (characteristic test); glycerol + Cu(OH)₂ → deep blue.
- Ethers as solvents: they resist most reagents; diethyl ether forms explosive peroxides on long storage with air.
- Anisole (PhOCH₃) undergoes EAS ortho/para; ring is activated by the methoxy group.
∫ Key Derivations
- E1 mechanism for dehydration: protonation makes −OH₂⁺ a good leaving group; loss of H₂O gives carbocation; base (HSO₄⁻) removes β-H.
- Kolbe and Reimer–Tiemann: both involve electrophilic aromatic substitution on phenoxide (more reactive than phenol). The former uses CO₂ (CO₂H electrophile); the latter uses :CCl₂.
- Williamson limits: if the halide is tertiary, E2 dominates — you get alkene, not ether. Use 1° halides with alkoxides.
▶ Worked Example
Distinguish 1-propanol from 2-propanol chemically.
Iodoform test: 2-propanol (CH₃CHOHCH₃) has CH₃CH(OH)– group → yellow ppt of CHI₃.
1-propanol does NOT give the test (no CH₃CH(OH)– group).
Iodoform distinguishes them clearly.
Also: Lucas test — 2-propanol (2°) turbid in ~5 min; 1-propanol (1°) stays clear at room temperature.
✗ Common Exam Mistakes
- Writing "Na" as the product of ROH + Na — the products are sodium alkoxide and H₂ gas.
- Confusing PCC (mild, stops at aldehyde) with KMnO₄ or CrO₃ (strong, goes to acid).
- Claiming phenol is a stronger acid than HCl — it isn't; phenol is only stronger than water and alcohols.
- Assuming all ethers are cleaved by HI at the same rate — benzyl and allyl ethers cleave very fast via SN1.
18Akorganic
Organic · Ch. 18 / 20
Aldehydes & Ketones
the carbonyl at the centre of it all
Nucleophilic addition
C=O + Nu⁻ → alkoxide → product
defining carbonyl reaction
Reactivity order
HCHO > RCHO > R₂CO
sterics + only one +I donor in aldehydes
Aldol condensation
2 RCH₂CHO →(OH⁻) β-OH-ald →(Δ) enal
needs α-H
Crossed aldol
one partner without α-H
else you get a statistical mixture
Cannizzaro
2 HCHO →(conc. OH⁻) CH₃OH + HCOO⁻
no α-H ⇒ disproportionation
Crossed Cannizzaro
HCHO + RCHO → HCOO⁻ + RCH₂OH
HCHO is always oxidised
Haloform
CH₃CO–R + 3X₂ ⁄ OH⁻ → CHX₃ + RCOO⁻
methyl ketones
Reductions to CH₂
Clemmensen (Zn–Hg ⁄ HCl) · Wolff–Kishner (NH₂NH₂ ⁄ KOH)
acidic vs basic routes
Reductions to alcohol
NaBH₄, LiAlH₄, H₂ ⁄ Ni
1° from aldehyde · 2° from ketone
Classic tests
Tollens → Ag mirror · Fehling → Cu₂O · 2,4-DNP → orange ppt
Tollens ⁄ Fehling: aldehydes
Acetal formation
C=O + 2ROH →(dry HCl) R₂C(OR)₂ + H₂O
protects carbonyl through basic steps
Mechanism · Nucleophilic addition (HCN to a ketone)
⚗ Lab Notes
- Aldehydes out-react ketones — less steric bulk and only one +I donor.
- Grignard outcomes: HCHO → 1° alcohol; RCHO → 2°; R₂CO → 3°. Carbon chain grows by the Grignard skeleton.
- Acetals protect carbonyls through basic steps; form in dry ROH ⁄ HCl, hydrolyse back in aqueous acid.
- Aldehydes oxidise easily (KMnO₄, CrO₃, even air); ketones resist mild oxidants but are cleaved by hot KMnO₄ or HNO₃.
- Schiff's reagent turns magenta with aldehydes only — a clean split from ketones.
- MP-derivatives (2,4-DNP) are crystalline solids with characteristic melting points — used to identify unknown carbonyls.
∫ Key Derivations
- Aldol mechanism: OH⁻ abstracts α-H giving enolate; enolate attacks another carbonyl carbon; protonation gives β-hydroxyaldehyde; heat drives elimination of water.
- Cannizzaro: OH⁻ attacks one aldehyde giving a tetrahedral intermediate; hydride transfer to a second aldehyde yields acid + alcohol (disproportionation).
- Haloform: three successive α-halogenations; the resulting CX₃ group is expelled as CX₃⁻ which picks up H⁺ to give CHX₃.
▶ Worked Example
Convert acetone to 2-methylpropan-2-ol using a Grignard reagent.
Product is (CH₃)₃COH — a 3° alcohol with an extra methyl vs acetone.
Use methylmagnesium bromide: (CH₃)₂C=O + CH₃MgBr → (CH₃)₃CO⁻MgBr⁺
Acid work-up (H₃O⁺) gives (CH₃)₃COH (tert-butyl alcohol)
✗ Common Exam Mistakes
- Saying Tollens' test is for all carbonyls — it's specific to aldehydes (and some α-hydroxy ketones).
- Forgetting that Clemmensen fails on acid-sensitive substrates — use Wolff–Kishner (basic conditions) instead.
- Writing aldol as always giving the dehydrated enal — dehydration needs heat; at room T you get the β-hydroxy aldehyde.
19Caorganic
Organic · Ch. 19 / 20
Carboxylic Acids & Derivatives
−COOH chemistry and the reactivity ladder
Acid dissociation
RCOOH ⇌ RCOO⁻ + H⁺ · pKa ≈ 4–5
carboxylate resonance-stabilised
Substituent effect
Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH > CH₃COOH
−I strengthens; fades with distance
Fischer esterification
RCOOH + R′OH ⇌(H⁺) RCOOR′ + H₂O
equilibrium; drive with Le Chatelier
Saponification
RCOOR′ + NaOH → RCOONa + R′OH
base-driven; irreversible
HVZ reaction
RCH₂COOH + X₂ ⁄ red P → RCHXCOOH
α-halogenation of the acid
Decarboxylation
RCOONa + NaOH ⁄ CaO →(Δ) RH + Na₂CO₃
soda lime; β-keto acids decarboxylate on gentle heating
Hunsdiecker
RCOOAg + Br₂ →(Δ) RBr + CO₂ + AgBr
chain shortens by 1 C
Derivative reactivity
RCOCl > (RCO)₂O > RCOOR′ > RCONH₂
better leaving group = more reactive
Amide hydrolysis
acid: RCOOH + NH₄⁺ · base: RCOO⁻ + NH₃
peptide bonds in proteins
Hell–Volhard–Zelinsky
via RCOBr enolisation
catalysed by PBr₃ formed in situ
Reduction
LiAlH₄: RCOOH → RCH₂OH · NaBH₄: no reaction
acids resist mild reducing agents
⚗ Lab Notes
- NaHCO₃ test: brisk CO₂ effervescence confirms −COOH (phenols are too weak to react).
- Carboxylic acids dimerise through twin H-bonds ⇒ anomalously high b.p.; C₁–C₄ are water-soluble.
- Formic acid is the odd one out: it reduces Tollens' & Fehling's (its H acts aldehydic).
- Order of acidity across families: RCOOH > PhOH > H₂O > ROH > HC≡CH.
- Amides are the least reactive derivatives — that's why peptide bonds survive in biology.
- Aspirin is acetylsalicylic acid — made by esterifying the phenolic −OH of salicylic acid with acetic anhydride.
∫ Key Derivations
- Fischer esterification mechanism: acid protonates the carbonyl O; ROH attacks; proton transfer; loss of H₂O; deprotonation gives ester. Each step reversible.
- Saponification: OH⁻ attacks ester carbonyl giving tetrahedral intermediate; alkoxide leaves; final proton transfer drives it to completion (acid + base → salt).
- HVZ mechanism: PBr₃ converts the acid to the more easily enolised acyl bromide; Br₂ halogenates the α-position; hydrolysis restores the acid.
▶ Worked Example
Convert acetic acid to ethylamine (shorter route).
CH₃COOH → CH₃CONH₂ (NH₃, heat)
Hofmann bromamide: CH₃CONH₂ + Br₂ + 4KOH → CH₃NH₂ + K₂CO₃ + 2KBr + 2H₂O
Chain shortens by one carbon — but wait, we lost one! For ethylamine from acetic:
Alternative: CH₃COOH →(LiAlH₄) CH₃CH₂OH →(PBr₃) CH₃CH₂Br →(NH₃ excess) CH₃CH₂NH₂
✗ Common Exam Mistakes
- Assuming all derivatives convert to each other freely — uphill conversions (amide → ester) don't work; downhill (ester → amide) does.
- Forgetting that NaBH₄ does not reduce carboxylic acids — only LiAlH₄ does among common hydrides.
- Believing decarboxylation of simple acids happens easily — it requires soda lime and very high temperature, except for β-keto acids.
- Confusing saponification (base hydrolysis of esters) with acidic hydrolysis (Fischer esterification in reverse).
20Amorganic
Organic · Ch. 20 / 20
Amines & Biomolecules
nitrogen's lone pair — and life's polymers
Basicity in water
2° > 3° > 1° > NH₃ (Et) · 2° > 1° > 3° > NH₃ (Me)
+I vs solvation trade-off
Aromatic weakness
ArNH₂ ≪ RNH₂
lone pair delocalised into ring
Aniline + Br₂ water
→ 2,4,6-tribromoaniline ↓
white ppt, no catalyst needed
Diazotization
ArNH₂ + NaNO₂ ⁄ HCl →(0–5 °C) ArN₂⁺Cl⁻
keep ice-cold
Sandmeyer
ArN₂⁺ + CuX → ArX · CuCN → ArCN · H₂O → ArOH
radical chain via Cu⁺
Carbylamine test
RNH₂ + CHCl₃ + 3KOH → RNC↑ + 3KCl + 3H₂O
foul · primary amines only
Hinsberg test
PhSO₂Cl + RNH₂ → PhSO₂NHR (soluble in NaOH)
1° soluble · 2° insoluble · 3° unreactive
Gabriel synthesis
K⁺phthalimide + RX → RNH₂ (after hydrazinolysis)
pure 1° amine
Hofmann bromamide
RCONH₂ + Br₂ ⁄ NaOH → RNH₂
chain shortens by 1 C
Zwitterion & pI
⁺H₃N–CHR–COO⁻ · pI = ½(pK₁ + pK₂)
no net charge at pI
DNA pairs
A=T (2 H-bonds) · G≡C (3 H-bonds)
Chargaff's rules: A=T, G=C
⚗ Lab Notes
- Hinsberg test (PhSO₂Cl): 1° product soluble in alkali; 2° insoluble; 3° unreactive.
- From diazonium: Sandmeyer CuX → ArX; CuCN → ArCN; H₂O ⁄ Δ → ArOH; KI → ArI. Coupling with phenols ⁄ anilines gives azo dyes (−N=N−).
- Gabriel phthalimide synthesis delivers pure primary amines; Hofmann bromamide degradation shortens the chain by one carbon.
- Biomolecules: glycosidic linkage joins sugars; peptide bond (−CO−NH−) joins amino acids; reducing sugars retain a hemiacetal.
- DNA denaturation unfolds 2° ⁄ 3° structure but leaves the primary chain intact; renaturation ("annealing") is the basis of PCR.
- Proteins: primary (sequence) → secondary (α-helix, β-sheet via H-bonds) → tertiary (3D) → quaternary (subunits).
∫ Key Derivations
- Basicity order explained: in gas phase only +I matters ⇒ 3° > 2° > 1° > NH₃. In water, the cation must be solvated; the bulkier the alkyl groups the poorer the solvation. The 2° cation hits the sweet spot.
- Sandmeyer mechanism: Cu⁺ reduces ArN₂⁺ to Ar• + N₂ + Cu²⁺; Cu²⁺ then re-oxidises with X⁻ to give ArX and Cu⁺ (catalytic cycle).
- Peptide bond geometry: resonance with C=O gives partial double-bond character; the six atoms Cα–C–N–Cα are planar, and trans dominates.
▶ Worked Example
Convert aniline to iodobenzene.
Step 1: diazotize — C₆H₅NH₂ + NaNO₂ + HCl (0–5 °C) → C₆H₅N₂⁺Cl⁻
Step 2: treat with KI (no catalyst needed):
C₆H₅N₂⁺Cl⁻ + KI → C₆H₅I + N₂↑ + KCl
✗ Common Exam Mistakes
- Assuming aniline is as basic as methylamine — it's roughly 10⁻⁴ as basic because the lone pair is delocalised.
- Heating a diazonium salt — it decomposes violently to N₂ and a phenol (if water present) or tar (dry).
- Confusing Sandmeyer (CuX catalyst) with Gattermann (Cu powder + HX) — similar but different reagents.
- Saying amino acids exist as neutral molecules in solution — they exist as zwitterions; only at pI do they have no net charge.
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