Flashcards · IB Diploma Programme Chemistry · Topic 6

Reactivity 3 – What are the mechanisms of chemical change? flashcards

41 cards on Topic 6, subtopics 3.1 to 3.4. Pick a subtopic or study them all.

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3.1 Proton transfer reactions

Brønsted–Lowry acid and base, and a conjugate pair: Acid: proton (H⁺) donor. Base: proton acceptor. Conjugate pair: two species whose formulas differ by a single H⁺, e.g. H₂PO₄⁻ / HPO₄²⁻.
Amphiprotic vs amphoteric: Amphiprotic: can donate or accept a proton (H₂O, HSO₄⁻). Amphoteric: a substance, such as Al₂O₃, that reacts with both acids and bases.
pH of 0.0040 mol dm⁻³ NaOH at 298 K: [H⁺] = Kw ÷ [OH⁻] = 1.0 × 10⁻¹⁴ ÷ 0.0040 = 2.5 × 10⁻¹² mol dm⁻³. pH = 11.60
Strong vs weak acids, with examples: Strong: completely ionized (HCl, HBr, HI, HNO₃, H₂SO₄). Weak: partially ionized, an equilibrium (carboxylic acids, H₂CO₃, HF). Strong is not the same as concentrated.
Products of an acid with a carbonate, and with ammonia: Carbonate (or hydrogencarbonate): salt + water + CO₂. Ammonia: an ammonium salt only, e.g. HNO₃ + NH₃ → NH₄NO₃ (no water).
Why is pure water above 298 K not pH 7, and is it still neutral?: Water’s ionization is endothermic, so Kw rises with temperature and the pH falls below 7. It is still neutral because [H⁺] = [OH⁻].
Strong acid titrated with strong base: shape of the pH curve: Starts low, rises slowly while acid is in excess, then rises almost vertically (about pH 3 to 11) around the equivalence point at pH 7 (298 K), then levels off.
Relationship between Ka and Kb for a conjugate pair: Ka × Kb = Kw, so pKa + pKb = 14.00 at 298 K. A weaker acid has a stronger conjugate base. (HL only)
pH of 0.0800 mol dm⁻³ weak acid with pKa 4.87: [H⁺] = √(Ka × c) = √(10⁻⁴·⁸⁷ × 0.0800) = 1.04 × 10⁻³ mol dm⁻³. pH = 2.98 (HL only)
Is NH₄Br solution acidic, neutral or basic? Why?: Acidic: it comes from a strong acid and a weak base, and NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺. (HL only)
pH of an acidic buffer, and its pH when [HA] = [A⁻]: pH = pKa + log₁₀([A⁻]/[HA]). Equal concentrations give pH = pKa. Dilution leaves the pH nearly unchanged but lowers the capacity. (HL only)
How is an acid–base indicator chosen?: It changes colour near pH = pKa(HInd). Its range must sit inside the vertical section of the pH curve around the equivalence pH. Weak acid with weak base: no indicator is suitable. (HL only)

3.2 Electron transfer reactions

Oxidation and reduction (four definitions): Oxidation: electrons lost, oxidation state rises, oxygen gained, hydrogen lost. Reduction: electrons gained, oxidation state falls, oxygen lost, hydrogen gained.
Cu²⁺ + Fe → Cu + Fe²⁺: oxidizing agent and reducing agent: Fe 0 → +2 is oxidized, so Fe is the reducing agent. Cu +2 → 0 is reduced, so Cu²⁺ is the oxidizing agent.
Oxidation state of sulfur in HSO₃⁻: +1 + x + 3(−2) = −1, so x = +4.
Half-equation for manganate(VII) being reduced in acid: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Charge: −1 + 8 − 5 = +2 on each side.
Electrode signs in voltaic and electrolytic cells: Oxidation at the anode and reduction at the cathode in both. Voltaic: anode negative, cathode positive. Electrolytic: anode positive, cathode negative.
What does the salt bridge do in a voltaic cell?: Anions move towards the anode half-cell and cations towards the cathode half-cell, keeping each solution electrically neutral and completing the circuit. Electrons travel only through the wire.
Oxidizing a primary alcohol: aldehyde vs carboxylic acid: Aldehyde: distil it off as it forms (lower boiling point, no hydrogen bonding between its molecules). Carboxylic acid: reflux with excess oxidizing agent. A tertiary alcohol does not react.
Standard hydrogen electrode: H⁺(aq) + e⁻ ⇌ ½H₂(g): platinum electrode, H₂ at 100 kPa, [H⁺] = 1.00 mol dm⁻³, 298 K. E⦵ = 0.00 V by convention. (HL only)
Mg²⁺/Mg −2.37 V and Sn²⁺/Sn −0.14 V. E⦵cell and ΔG⦵ (F = 96 500 C mol⁻¹)?: E⦵cell = E⦵(cathode) − E⦵(anode) = −0.14 − (−2.37) = +2.23 V. ΔG⦵ = −nFE⦵cell = −2 × 96 500 × 2.23 = −430 kJ mol⁻¹. (HL only)
Electrolysis of concentrated vs dilute NaCl(aq) with inert electrodes: Both: hydrogen at the cathode. Concentrated: chlorine at the anode (NaOH forms). Dilute: mainly oxygen at the anode. (HL only)
Electroplating: what is the anode, the cathode and the electrolyte?: Cathode: the object to be plated (Mⁿ⁺ + ne⁻ → M). Anode: the plating metal (M → Mⁿ⁺ + ne⁻). Electrolyte: a solution of the plating metal’s ions. (HL only)

3.3 Electron sharing reactions

Radical, and homolytic fission: Radical: a species with an unpaired electron (Cl•, CH₃•). Homolytic fission: a bond splits evenly, one electron to each atom, giving two radicals (X₂ → 2X• with UV light or heat).
How is homolytic fission shown with curly arrows?: Two single-barbed (fish-hook) arrows, each from the centre of the bond to one atom. Each shows one electron moving.
Why do alkanes react only by radical substitution?: C–C and C–H bonds are strong (high Ea) and essentially non-polar, so there is no δ+ or δ− site for nucleophiles or electrophiles.
Mechanism for an alkane RH with halogen X₂: Initiation: X₂ → 2X• (UV). Propagation: X• + RH → R• + HX; R• + X₂ → RX + X•. Termination: X• + X• → X₂; R• + X• → RX; R• + R• → R–R.
Why does radical substitution give a mixture of products?: Further substitution (RX still has C–H bonds), hydrogens in different environments can all be replaced, and termination R• + R• gives a longer alkane.
How many monochloro structural isomers does heptane give?: Four: heptane has four hydrogen environments (C1/C7, C2/C6, C3/C5, C4), giving 1-, 2-, 3- and 4-chloroheptane.
Why does an alkane not decolourize bromine water in the dark?: Without UV light (or heat) Br₂ does not undergo homolytic fission, so no radicals form, and the non-polar alkane offers nothing for bromine to attack.

3.4 Electron-pair sharing reactions

Nucleophile: A species that supplies both electrons of a new covalent bond. It can be neutral (H₂O, NH₃) or negative (OH⁻, CN⁻).
Electrophile: A species that takes in both electrons of a new covalent bond. It can be neutral (HBr, Br₂) or positive (H⁺, NO₂⁺).
Heterolytic fission: Bond breaking in which one fragment keeps the whole bonding pair, giving an anion and a cation; one double-barbed arrow. Example: (CH₃)₂CHBr → (CH₃)₂CH⁺ + Br⁻.
1-bromobutane + CN⁻: product: CH₃CH₂CH₂CH₂CN + Br⁻: pentanenitrile, because the CN carbon counts in the chain. Nucleophilic substitution; Br⁻ is the leaving group.
Electrophilic addition to alkenes: reagents and products: X₂ → dihalogenoalkane. HX → halogenoalkane. Steam with an acid catalyst → alcohol. Example: but-2-ene + Br₂ → 2,3-dibromobutane.
Lewis acid and Lewis base: Lewis acid: electron-pair acceptor (an electrophile). Lewis base: electron-pair donor (a nucleophile). In SnCl₄ + 2Cl⁻ → SnCl₆²⁻, Cl⁻ is the base. (HL only)
SN1 vs SN2: SN2 (primary): one step, rate = k[RX][Nu], attack opposite the leaving group with inversion. SN1 (tertiary): two steps via a carbocation, rate = k[RX], not stereospecific. Secondary can react by both. (HL only)
[RX] is halved and [Nu] tripled. How does the rate change for SN2 and for SN1?: SN2: × 0.5 × 3 = × 1.5. SN1: × 0.5, because only [RX] is in its rate equation. (HL only)
Why does rate follow RI > RBr > RCl in nucleophilic substitution?: C–I is the longest and weakest bond, and I⁻ is the best leaving group (largest, most stable ion). C–Cl is the most polar but the slowest. (HL only)
Major product of HBr with an unsymmetrical alkene: The product from the more stable carbocation (tertiary > secondary > primary, from the positive inductive effect of alkyl groups). 2-methylhex-1-ene gives mainly 2-bromo-2-methylhexane. (HL only)
Oxidation state of iron in [Fe(H₂O)₅(OH)]²⁺: x + 5(0) + (−1) = +2, so x = +3. Neutral ligands: H₂O, NH₃; anionic ligands: Cl⁻, OH⁻, CN⁻ (−1 each). (HL only)