Transition Metals and Organic Nitrogen Chemistry

A battery transfers electrons through a wire. A metal catalyst changes a reaction pathway. An organic synthesis joins chosen carbon groups. In each case, success depends on identifying the particles and the conditions. This unit connects redox measurements, metal complexes and organic routes.

Prerequisites: oxidation numbers, half-equations, equilibrium, entropy, organic functional groups and Unit 4 analytical methods. This reference covers Pearson Topics 16–20 and the theory of core practicals 12–16. Separate practical Unit 6 requires practical-source preparation. Laboratory work needs the approved supervised procedure and its risk assessment.

A half-cell 半电池 contains an oxidised/reduced pair in contact with a conducting electrode. A metal electrode can be the reacting solid. An inert platinum electrode carries electrons when the pair contains only solutions or a gas and solution. A salt bridge 盐桥 allows ions to move and completes the circuit without directly mixing the main solutions. Electrons flow through the wire, not the bridge.

An individual electrode potential cannot be measured alone. Connect it to a reference. The standard hydrogen electrode 标准氢电极 has hydrogen gas at 100 kPa, aqueous H⁺ at 1.00 mol dm⁻³ and platinum, at 298 K. Its standard potential is defined as zero. The standard electrode potential 标准电极电势 of another couple is its potential relative to that reference under standard conditions. Other solutes have concentration 1.00 mol dm⁻³; pure solids and liquids are in their standard states.

Write table entries as reductions. For example:

The more positive reduction is favoured at the positive electrode of the spontaneous cell. Zinc is oxidised at the negative electrode; Cu²⁺ is reduced at the positive electrode. The net equation is Zn + Cu²⁺ → Zn²⁺ + Cu. Cancel equal electron amounts after multiplying half-equations, but never multiply an electrode potential by a coefficient.

The cell diagram is Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s). A single line represents a phase boundary; the double line represents the salt bridge. State concentrations and conditions alongside the diagram. For an Fe³⁺/Fe²⁺ solution half-cell, include Pt(s) as the inert contact. Species in one aqueous phase are separated by a comma, not an extra phase boundary.

In core practical 12, clean the metal surfaces and use a high-resistance voltmeter. Check electrode identity, solution concentration, temperature and good electrical contacts. Do not assume a smaller measured voltage disproves the table: actual conditions, oxide films and internal resistance affect measurements.

Feasibility has limits

A positive standard cell potential predicts thermodynamic feasibility for the stated overall reaction under standard conditions. It does not predict a visible reaction rate. A high activation barrier can prevent an observable change. Non-standard concentrations can change the electrode potentials and even the predicted direction. Compare the same conditions; do not treat a standard table as a concentration-independent rule.

For a defined cell reaction, $E^\circ_{cell}$ is proportional to its standard total entropy change. It is also related to $\ln K$, with electron number and temperature included in the proportionality. Thus positive E° corresponds to positive total entropy and K > 1. Reversing the reaction changes the sign of E° and inverts K. Changing the balanced equation's scale changes the entropy change and electron number, but not the voltage.

A reducing agent 还原剂 donates electrons and is oxidised. The reduced form of a very negative-potential couple is usually a strong reducing agent. An oxidising agent 氧化剂 accepts electrons and is reduced. The oxidised form of a very positive-potential couple is usually a strong oxidising agent. Name the actual species, not just “the half-cell”.

A disproportionation 歧化 reaction oxidises and reduces the same element from one starting oxidation state. To test it, use the reductions leading to the starting state and away from it. If the reduction of the starting species is more positive than reduction into that state, the proposed disproportionation has positive E°. For illustrative A³⁺/A²⁺ = +0.80 V and A²⁺/A = −0.20 V, disproportionation of A²⁺ to A³⁺ and A gives −0.20 − 0.80 = −1.00 V. It is not feasible under those standard conditions. This example concerns hypothetical A, not measured values for a named metal.

Acidified manganate(VII) is reduced from purple MnO₄⁻ to nearly colourless Mn²⁺:

Iron(II) loses one electron: Fe²⁺ → Fe³⁺ + e⁻. Therefore one mole MnO₄⁻ reacts with five moles Fe²⁺. Use dilute sulfuric acid. Hydrochloric acid can be oxidised; nitric acid can oxidise Fe²⁺ before titration. Near the endpoint add manganate(VII) dropwise while swirling. The first persistent faint pink shows a slight excess. No separate indicator is needed.

A 25.00 cm³ Fe²⁺ sample requires 20.00 cm³ of 0.0200 mol dm⁻³ MnO₄⁻:

For iodine and thiosulfate:

Add starch when the iodine solution becomes pale yellow. Adding it to concentrated iodine can make a strongly bound complex and an indistinct endpoint. The final blue-black colour disappears. If another oxidant liberates iodine from excess iodide, first use its equation to connect oxidant to iodine, then use the 1:2 iodine/thiosulfate ratio. These are different stoichiometric steps.

Core practicals 13a and 13b use redox titration to determine concentrations. Rinse a burette with its titrant and a pipette with the sampled solution. Water remaining in a conical flask does not change the moles transferred by the pipette. A burette reading uncertainty of ±0.05 cm³ at each end gives a worst-case titre uncertainty of ±0.10 cm³. For 20.00 cm³, percentage uncertainty is $(0.10/20.00)100=0.50\%$. Include dilution and other apparatus uncertainties when the question requires the final concentration uncertainty. Repeat concordant titres reduce random variation; they do not remove a systematic concentration error.

Fuel cells require a continuing supply

A fuel cell 燃料电池 produces electricity while fuel and oxidant enter from outside. For hydrogen and oxygen, the overall reaction is 2H₂ + O₂ → 2H₂O.

Cancel H⁺ or OH⁻ only after matching electrons. Both media give the same overall equation. Hydrogen-cell operation produces water, but hydrogen production, storage and transport have environmental costs. Methanol is easier to store as a liquid but produces CO₂. Its overall oxygen-cell equation is 2CH₃OH + 3O₂ → 2CO₂ + 4H₂O. Compare total production pathways and energy efficiency, not just the cell's exhaust.

A transition element 过渡元素 forms at least one stable ion with an incomplete d subshell. Scandium forms Sc³⁺, which is d⁰, and zinc forms Zn²⁺, which is d¹⁰. They do not meet this definition. Copper qualifies because Cu²⁺ is d⁹, even though Cu⁺ is d¹⁰.

For Sc to Zn, the neutral atom configurations after [Ar] are: Sc 3d¹4s²; Ti 3d²4s²; V 3d³4s²; Cr 3d⁵4s¹; Mn 3d⁵4s²; Fe 3d⁶4s²; Co 3d⁷4s²; Ni 3d⁸4s²; Cu 3d¹⁰4s¹; Zn 3d¹⁰4s². Chromium and copper are exceptions to the simple filling pattern. When these atoms form positive ions, remove 4s electrons before 3d electrons. Fe²⁺ is [Ar]3d⁶, not [Ar]3d⁴4s². Similar 3d and 4s energies help explain variable oxidation states.

A ligand 配体 donates a lone pair to a metal ion, forming a coordinate bond. A complex ion 配离子 contains a central metal ion surrounded by ligands. The coordination number 配位数 counts donor atoms directly bonded to the metal, not the number of ligand molecules. Six water ligands give [Cu(H₂O)₆]²⁺, coordination number six. Three bidentate ethanedioate ligands also give coordination number six.

Water, OH⁻, NH₃ and Cl⁻ are monodentate 单齿的. Ethane-1,2-diamine, often written en, is bidentate 双齿的. EDTA⁴⁻ can be hexadentate 六齿的. Neutral ligands do not change the ion's charge; each anionic ligand contributes its charge. In [Fe(CN)₆]³⁻, six CN⁻ contribute −6, so iron is +3.

Six ligand electron-pair regions repel and arrange around the metal with the greatest available separation. This gives an octahedral 八面体的 arrangement for the six-coordinate complexes here. Larger chloride ligands favour four-coordinate complexes such as tetrahedral [CoCl₄]²⁻ 四面体的. Some four-coordinate complexes are square planar 平面正方形的. Cisplatin has two NH₃ and two Cl ligands around platinum in a square plane. The two chloride ligands are adjacent in the cis form; opposite positions define the trans form. Cisplatin is used in cancer treatment as the single cis isomer, not a mixture with the trans form. A four-coordinate formula alone does not prove tetrahedral geometry.

Colour and ligand exchange

Ligands split the d orbitals into different energy levels. Absorbing visible light can promote a d electron between these levels. The remaining transmitted/reflected light produces the observed colour. Changing the metal oxidation state, ligand or coordination number can change the energy gap and colour. A d⁰ or d¹⁰ ion has no suitable d–d transition; this explains many colourless complexes, but does not claim every colour in chemistry is a d–d transition.

A ligand exchange 配体交换 replaces one ligand with another. Excess ammonia changes the pale blue copper(II) aqueous complex into a deep blue solution:

Initially, a small amount of NH₃ acts as a base and makes a pale blue hydroxide precipitate. In excess NH₃, that precipitate dissolves as the ammine complex forms. Do not describe both stages as one precipitate colour change. Excess concentrated chloride gives yellow [CuCl₄]²⁻, often through an observed green mixture:

For cobalt, pink [Co(H₂O)₆]²⁺ exchanges with chloride to form blue [CoCl₄]²⁻. Adding water shifts towards pink; increasing chloride concentration favours blue. State whether water or chloride concentration changed before explaining the equilibrium.

Replacing six water ligands with three bidentate ligands releases six free water molecules while using three free ligand molecules. The increase in freely moving particles often favours the chelate 螯合 complex through entropy. It is not simply “more coordinate bonds”: both complexes can have six metal–donor bonds. In haemoglobin, Fe²⁺ is held by a polydentate 多齿的 ligand and binds oxygen reversibly. Carbon monoxide competes strongly for binding and prevents normal oxygen transport. The detailed haem structure is not required here.

Hydroxide removes protons from coordinated water, producing a hydrated hydroxide precipitate. A simplified equation is Fe³⁺ + 3OH⁻ → Fe(OH)₃(s). A hydrated-complex equation is:

Ammonia first accepts protons from coordinated water; write NH₄⁺ as a product. For copper:

The following table distinguishes initial precipitates from the solution in excess reagent. Observations refer to fresh aqueous tests. Air oxidation and slow ligand substitution can change colours during standing; record those conditions rather than silently combining stages.

The initial precipitate column applies to small additions of either hydroxide or ammonia in these tests. For a divalent metal M, the simplified equations are M²⁺ + 2OH⁻ → M(OH)₂(s) and M²⁺ + 2NH₃ + 2H₂O → M(OH)₂(s) + 2NH₄⁺. Here M can be Mn, Fe, Co, Ni, Cu or Zn. For Cr³⁺ and Fe³⁺, replace the coefficients 2 with 3 and use M(OH)₃. These formulae describe the same proton-removal chemistry as the hydrated-complex equations.

For the excess-ammonia stage, Co(OH)₂ or Ni(OH)₂ can form [M(NH₃)₆]²⁺ with six NH₃, releasing two OH⁻. Zinc forms [Zn(NH₃)₄]²⁺ with four NH₃, releasing two OH⁻. For copper, Cu(OH)₂ + 4NH₃ + 2H₂O ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺ + 2OH⁻. These equilibria account for dissolution in sufficient ammonia; the observed extent and subsequent cobalt oxidation depend on conditions. Chromium, manganese and iron precipitates do not dissolve in the usual excess-ammonia test. Do not use the deep-blue copper colour for every ammine complex.

The amphoteric 两性的 Cr(OH)₃ and Zn(OH)₂ react with excess hydroxide. For example Zn(OH)₂(s) + 2OH⁻ → [Zn(OH)₄]²⁻. Zinc also forms [Zn(NH₃)₄]²⁺ in excess ammonia. “Dissolves in both” does not mean both ligands produce the same complex. Chromium(III) may be represented as [Cr(OH)₆]³⁻ in excess hydroxide; follow the species convention supplied in the question.

Vanadium shows several oxidation states in acidic solution: V(V), yellow VO₂⁺; V(IV), blue VO²⁺; V(III), green V³⁺; V(II), violet V²⁺. Zinc in acid reduces through these stages. Yellow and blue mixed during a change can look green; this alone is not proof of pure V³⁺. Combine colour with the reaction conditions and electrode data.

Acidified orange dichromate(VI) is reduced to green Cr³⁺. Zinc in acid can reduce it further to blue Cr²⁺; air readily oxidises Cr²⁺ back. In alkaline solution, hydrogen peroxide can oxidise Cr(III) to yellow chromate(VI). Subsequent acidification favours orange dichromate:

Chromium stays +6 in this equilibrium. It is an acid–base equilibrium, not a redox change. Check oxidation numbers before assuming every colour change transfers electrons.

Catalysts can change oxidation state

In the Contact process, V₂O₅ catalyses SO₂ oxidation. A useful cycle is V₂O₅ + SO₂ → V₂O₄ + SO₃, then 2V₂O₄ + O₂ → 2V₂O₅. Add the steps to cancel the catalyst and recover 2SO₂ + O₂ → 2SO₃. On catalytic-converter surfaces, reactants adsorb, react through a lower-energy pathway and desorb. For example 2CO + 2NO → 2CO₂ + N₂. Surface blockage reduces the available sites.

For the aqueous iodide/peroxodisulfate reaction, Fe²⁺/Fe³⁺ provides a homogeneous cycle:

Adding cancels iron ions. For acidified manganate(VII) and ethanedioate, Mn²⁺ product acts as an autocatalyst 自催化剂. The reaction can start slowly, accelerate as Mn²⁺ builds up, then slow as reactants are used. Core practical 14 prepares a transition-metal complex: calculate limiting amount and theoretical yield, isolate and dry crystals, and distinguish product loss from impurities that falsely raise mass.

Benzene is an arene 芳烃. Its six p orbitals overlap into a delocalised 离域的 electron system above and below the carbon plane. The carbon–carbon bonds have equal lengths, between typical single and double bonds. X-ray evidence supports equal bond lengths. Its infrared spectrum has aromatic ring vibrations and aromatic C–H absorption above 3000 cm⁻¹. Use supplied reference data to compare its bands with a localised-alkene prediction; IR alone does not count three ordinary isolated double bonds. Together with its lower-than-expected hydrogenation enthalpy, the evidence disagrees with the three-isolated-bond model.

If one isolated double bond gives hydrogenation enthalpy −120 kJ mol⁻¹, three would predict −360. An illustrative measured benzene value of −208 is 152 kJ mol⁻¹ less exothermic. Benzene starts at lower enthalpy than that localised model. This comparison uses a hypothetical reference; benzene does not rapidly switch between three fixed double bonds.

Benzene usually resists addition because it would destroy ring delocalisation. It does not decolourise bromine water like an alkene under the usual test conditions. It burns with a smoky flame when combustion is incomplete because of its high carbon proportion. Substitution replaces H while restoring the delocalised ring.

An electrophilic substitution 亲电取代 reaction has three essential stages: generate the electrophile, use a ring electron pair to form a C–E bond, then lose H⁺ to restore delocalisation. The intermediate has a positive charge and a ring carbon bonded to both H and E. Draw a curly arrow from the ring electron system to E⁺, then from the C–H bond back into the ring. An arrow starting at H⁺ is not electron flow.

For nitration, HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O generates the nitronium 硝基正离子 ion. Ring substitution gives nitrobenzene and releases H⁺, regenerating the acid catalyst. For Friedel–Crafts acylation, RCOCl + AlCl₃ → RCO⁺ + AlCl₄⁻; substitution then forms HCl and regenerates AlCl₃. For bromination, the simple electrophile model is Br₂ + FeBr₃ → Br⁺ + FeBr₄⁻. FeBr₄⁻ accepts the released H⁺ to form HBr and regenerate FeBr₃. For alkylation, RCl + AlCl₃ → R⁺ + AlCl₄⁻; AlCl₄⁻ accepts H⁺ to form HCl and regenerate AlCl₃. In each case the ring-attack and ring-restoration electron-pair arrows follow the nitration pattern. These are simplified mechanism models of catalyst-assisted electrophile generation; free ions are not assumed to exist in the original reagent bottle. Sulfonation conditions and products are required here; its detailed mechanism is outside this specification statement.

Phenol reacts much more easily with bromine water, giving a white precipitate of 2,4,6-tribromophenol and decolourisation without a halogen carrier. Oxygen's lone pair increases ring electron density by interaction with the ring. This makes attack by an electrophile easier. Do not transfer benzene's catalyst requirement to phenol.

An amine 胺 derives from ammonia by replacing H with carbon groups. Butan-1-amine is CH₃CH₂CH₂CH₂NH₂; phenylamine is C₆H₅NH₂. Nitrogen's lone pair accepts H⁺, making amines bases and nucleophiles. RNH₂ + H₂O ⇌ RNH₃⁺ + OH⁻, and RNH₂ + HCl → RNH₃⁺Cl⁻. Small amines dissolve in water through hydrogen bonding. Solubility generally falls as the non-polar carbon group grows.

Aliphatic 脂肪族的 amines are generally stronger bases than ammonia in the comparisons required here. Alkyl groups increase electron density at nitrogen. Phenylamine is a weaker base because its lone pair interacts with the aromatic ring and is less available to bind H⁺. Basicity is not decided by counting N–H bonds. Hydrogen bonding also raises boiling points compared with similar-sized hydrocarbons.

Excess ethanolic ammonia with a halogenoalkane, heated in a sealed apparatus under the approved procedure, forms a primary amine. Excess ammonia reduces further substitution but does not make it impossible. The amine product can attack more halogenoalkane, giving secondary/tertiary amines and eventually a quaternary ammonium salt. Butylamine can also act as a ligand towards Cu²⁺ through its lone pair.

Reducing a nitrile with LiAlH₄ in dry ether followed by suitable aqueous work-up gives a primary amine. R–C≡N becomes R–CH₂NH₂. The nitrile carbon remains in the chain. Making a nitrile from a halogenoalkane using ethanolic cyanide then reducing it increases the carbon chain by one overall.

Reduce nitrobenzene using tin and concentrated HCl under reflux. The acidic mixture initially contains the phenylammonium salt. Add alkali during work-up to liberate phenylamine. Writing free phenylamine as the only species in the strongly acidic reaction mixture misses this stage.

An amide 酰胺 has nitrogen bonded to a carbonyl carbon. Ethanoyl chloride reacts with a primary amine to form a substituted amide. CH₃COCl + 2C₄H₉NH₂ → CH₃CONHC₄H₉ + C₄H₉NH₃⁺Cl⁻. One amine molecule forms the amide; another neutralises HCl. The amide lone pair interacts with the carbonyl system, so amides are much less basic than amines. Do not name CH₃CONH₂ as an amine or confuse its C=O with a ketone.

Names and complete structures

Use –amine for the amine, –amide for the amide and an amino– prefix with the carboxylic-acid parent for an amino acid. Ethanamine is CH₃CH₂NH₂; ethanamide is CH₃CONH₂; alanine is 2-aminopropanoic acid. A displayed formula includes every atom and bond. A skeletal formula omits carbon labels and their attached hydrogens, but retains nitrogen, oxygen and their attached hydrogens.

Diazotisation and colour

Cold nitrous acid, made from NaNO₂ and dilute acid, reacts with phenylamine below about 10 °C to form a diazonium 重氮 salt. Keep the mixture cold because these salts are unstable when warmed. Coupling with phenol in alkaline solution produces an azo 偶氮 compound containing –N=N– between aromatic systems. The extended delocalised system absorbs visible light, producing colour. Distinguish making the diazonium ion from the separate coupling step.

An amino acid 氨基酸 contains both amine and carboxylic acid groups. A typical α-amino acid is H₂N–CH(R)–COOH. Most have a chiral central carbon, but glycine has two H atoms and is achiral. In the solid and much aqueous solution, proton transfer gives a zwitterion 两性离子, ⁺H₃N–CH(R)–COO⁻. Its net charge is zero, but it contains two charged groups. It is not an uncharged H₂N/COOH molecule.

In acid the principal form is ⁺H₃N–CH(R)–COOH. In alkali it is H₂N–CH(R)–COO⁻. Explain changes by adding/removing protons at the correct group. To investigate acid/base behaviour, add acid and alkali separately to measured amino-acid samples and follow pH with a calibrated meter. To investigate optical activity, zero a polarimeter with the solvent, then measure a clear solution with plane-polarised monochromatic light. Control concentration, path length, temperature and wavelength when comparing rotations. Pure opposite enantiomers rotate equally in opposite directions under matched conditions. A racemic mixture has zero net rotation; glycine is also inactive, so zero alone does not prove a racemate.

Linking an amino group and carboxyl group produces an amide or peptide bond 肽键, –C(=O)–NH–, with loss of water. Two different amino acids can give two ordered dipeptides; reversing their order changes the molecule.

A polyamide 聚酰胺 forms by condensation between suitable diamines and dicarboxylic acids or their acyl chlorides. Nylon-6,6 uses H₂N(CH₂)₆NH₂ and HOOC(CH₂)₄COOH. Its repeat is [–NH(CH₂)₆NHCO(CH₂)₄CO–] with continuing bonds at both ends. Protein chains are polyamides formed from amino acids. Their N–H and C=O groups form hydrogen bonds between chains, affecting strength and melting behaviour.

Poly(propenamide), [–CH₂–CH(CONH₂)–], is an addition polymer. The side group contains an amide, but the backbone forms by opening the C=C bond without elimination of a small molecule. Poly(ethenol), [–CH₂–CH(OH)–], has many hydroxyl groups that hydrogen-bond to water. Its water solubility allows suitable laundry bags or liquid-detergent capsules to dissolve during washing. Distinguish water-soluble polymers from polymers that merely absorb water; formulation and conditions matter.

A Grignard reagent 格氏试剂 is made by reacting a halogenoalkane with magnesium in dry ether: RX + Mg → RMgX. Water destroys it, forming RH. Dry glassware and solvent matter because this reaction consumes the intended reagent before it can build the carbon chain.

With CO₂ followed by acidic work-up, RMgX forms RCOOH, adding one carbon. With methanal, it forms a primary alcohol; with another aldehyde, a secondary alcohol; with a ketone, a tertiary alcohol. The carbon group R bonds to the former carbonyl carbon. Oxygen becomes OH during the separate acidic work-up. Count every carbon from both starting fragments.

For example ethylmagnesium bromide plus ethanal gives CH₃CH(OH)CH₂CH₃ after work-up: butan-2-ol. Using methanal instead gives propan-1-ol. Adding acid before the carbonyl consumes the Grignard reagent, so the order cannot be reversed.

When a task specifies four synthetic conversions, one valid route is bromoethane → ethanol (aqueous OH⁻, heat) → ethanal (controlled oxidation and distillation) → butan-2-ol (ethylmagnesium bromide, dry ether, then acidic work-up) → butan-2-one (oxidation). Each arrow changes the main organic compound. The Grignard reagent is prepared separately. Check functional group, carbon count and conditions at every arrow.

Evidence can constrain an unfamiliar product

Use combustion data to determine the empirical formula; combine it with molar mass for the molecular formula. Element tests identify elements, not a unique structure. IR constrains functional groups; mass spectra constrain molecular mass and fragments. Carbon-13 NMR counts distinct carbon environments; proton NMR adds chemical shifts, integration and splitting. Several structures may fit one spectrum. Seek one structure fitting all the observations.

An unknown has formula C₃H₆O, a strong C=O absorption, a proton signal near 9–10 ppm, and a positive Tollens' test. Propanal fits the aldehyde proton and oxidation test. Propanone fits the formula and C=O but lacks an aldehyde H and normally gives a negative Tollens' test. Predict propanal's three carbon environments and its aldehyde/CH₂/CH₃ proton environments as a further check. Core practical 15 identifies both inorganic and organic unknowns through a planned sequence of small-scale tests, with positive controls and separate samples. Use metal-ion precipitation/excess-reagent tests for appropriate inorganic samples. Avoid adding every reagent to one tube: later changes may come from previous reagents rather than the original unknown.

Isolation is not the same as reaction

Reflux 回流 heats a mixture without losing volatile reactants: vapour condenses and returns. Distillation collects a volatile component separately. Never seal a heated reflux apparatus. During extraction 萃取, a separating funnel divides immiscible layers. Identify which layer contains the product from solvent density and solubility; the organic layer is not always uppermost. Vent the funnel away from people.

Water washes remove water-soluble impurities. Sodium carbonate washes remove acidic impurities but release CO₂, so vent carefully. A suitable anhydrous drying agent removes residual water from an organic liquid; filter or decant before distillation. Steam distillation carries a suitable water-insoluble volatile compound over with steam at a lower temperature than its normal boiling point.

Recrystallisation 重结晶 dissolves an impure solid in a minimum amount of hot solvent. Cool to crystallise; filter, wash with a little cold solvent and dry. Too much solvent lowers recovery. Insoluble impurities can be removed by hot filtration, while soluble impurities remain mainly in the mother liquor. A narrow melting range near the expected value supports purity; a lower, broader range suggests impurities. For melting temperature, use a dry finely powdered sample in a capillary and heat slowly near the expected range. Record the first melting and completion temperatures. For a liquid boiling temperature, place the thermometer bulb at the side-arm entrance so vapour reaching the condenser passes it. Record the pressure and a steady boiling temperature; compare reference values at matching pressure. Mixtures can boil over a range. Neither melting point nor percentage yield alone proves identity.

Core practical 16 prepares aspirin by acylating salicylic acid with ethanoic anhydride under the approved method, then purifying the solid. Calculate theoretical yield from the limiting reagent using the 1:1 salicylic acid/aspirin ratio. If 1.38 g salicylic acid (M = 138 g mol⁻¹) is limiting, $n=1.38/138=0.0100$ mol and theoretical aspirin mass is $0.0100\times180=1.80$ g. A dry yield of 1.26 g gives $(1.26/1.80)100=70.0\%$. Wet crystals can give an artificially high yield and a misleading melting range.

1. A cell uses reduction couples +0.54 V and −0.25 V. Give the positive electrode, E°cell, and one reason a predicted reaction may not be seen.
2. A 25.00 cm³ aliquot liberates iodine requiring 18.00 cm³ of 0.100 mol dm⁻³ thiosulfate. Calculate iodine moles. State what extra equation is needed to find the original oxidant concentration.
3. A four-coordinate complex is blue. Explain why these facts alone do not establish its geometry or the metal oxidation state.
4. Methanal reacts with propylmagnesium bromide, then acid. Identify the alcohol and explain the effect of adding water too early.
5. A polymer has –CONH₂ side groups and a carbon-only backbone. Decide whether the presence of amide groups proves condensation polymerisation.

Answers: 1. The +0.54 V reduction is at the positive electrode; E°cell = 0.79 V. A high activation barrier can make the reaction too slow. Non-standard conditions also limit the standard prediction. 2. Thiosulfate amount = 0.100 × 0.01800 = 0.00180 mol; iodine = 0.000900 mol. The oxidant/iodide equation gives the separate mole ratio. 3. Four-coordinate complexes may be tetrahedral or square planar; colour also depends on ligand and coordination, so further evidence is needed. 4. Butan-1-ol; water consumes the Grignard reagent and forms propane. 5. No: poly(propenamide) forms by addition across C=C while retaining amide side groups.