Pearson Edexcel · International A-Level · 化学

An atom 原子 is one particle of an element 元素. An element contains only one type of atom, defined by proton number. An ion 离子 has an electrical charge because its electron count differs from its proton count. A molecule 分子 is a discrete group of covalently bonded atoms. A compound 化合物 contains different elements chemically combined. An empirical formula gives the simplest whole-number atom ratio; a molecular formula 分子式 gives actual atom numbers in one molecule. Ionic lattices have formula units, not separate molecules.

The mole 摩尔 is the unit of amount of substance. The particle count is $N=nL$, where $L=6.02\times10^{23}\ \mathrm{mol^{-1}}$. Thus one mole contains $6.02\times10^{23}$ of the specified entities. Always say whether you count atoms, molecules, ions or formula units. A mole of methane molecules contains five moles of atoms.

Relative atomic mass 相对原子质量, $A_r$, compares the abundance-weighted mean atom mass with one-twelfth of a carbon-12 atom's mass. Relative molecular mass 相对分子质量, $M_r$, adds the relative atomic masses in a molecule. Relative formula mass 相对式量 applies the same sum to a formula unit, including giant structures. Both are ratios without units. Molar mass 摩尔质量, $M$, is mass per mole, in $\mathrm{g\,mol^{-1}}$.

Use $n=m/M$ for a mass, $n=cV$ for a solution with $V$ in $\mathrm{dm^3}$, and $pV=nRT$ for a gas with pressure in Pa, volume in $\mathrm{m^3}$ and temperature in K. A stated molar gas volume can replace the gas equation only under its stated conditions. At the usual classroom RTP approximation, use $24\ \mathrm{dm^3\,mol^{-1}}$. Divide $\mathrm{cm^3}$ by 1000 to obtain $\mathrm{dm^3}$.

Worked reacting-mass method. The equation is $\mathrm{Mg+2HCl\rightarrow MgCl_2+H_2}$. There is excess acid and $0.480\ \mathrm g$ magnesium, with $M(\mathrm{Mg})=24.0\ \mathrm{g\,mol^{-1}}$.

The coefficient ratio is Mg:H₂ = 1:1, so $n(\mathrm{H_2})=0.0200\ \mathrm{mol}$. At the stated RTP approximation:

If both reactant amounts are given, compare amount divided by coefficient. The smaller value identifies the limiting reagent 限量试剂. Do not compare masses alone.

Mass concentration 质量浓度 is $m/V$ in $\mathrm{g\,dm^{-3}}$; amount concentration 物质的量浓度 is $n/V$ in $\mathrm{mol\,dm^{-3}}$. Divide mass concentration by molar mass to convert between them. Parts per million 百万分率, ppm, is a fraction multiplied by $10^6$; state whether it is a mass fraction or, for gases at the same temperature and pressure, a volume fraction.

To find an empirical formula 实验式, divide each element's mass by its $A_r$ and divide the resulting amounts by the smallest amount. Multiply the whole ratio if necessary to obtain integers. Do not round 1.5 to 2: multiply all ratios by 2. To find a molecular formula, divide measured $M_r$ by empirical-formula mass and multiply every subscript by this integer.

Worked check. A compound contains 2.40 g carbon and 0.600 g hydrogen. Carbon amount is $n=m/M=2.40/12.0=0.200\ \mathrm{mol}$; hydrogen atom amount is $n=m/M=0.600/1.00=0.600\ \mathrm{mol}$. C:H = 1:3, so the empirical formula is CH₃. If $M_r=30.0$, the factor is $30.0/15.0=2$, giving C₂H₆. The element masses add to the sample mass, 3.00 g.

Balance full equations without changing substance formulae, and include states: (s), (l), (g), (aq). For a precipitation, $\mathrm{Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)}$ omits spectator ions. For an acid and carbonate, $\mathrm{CO_3^{2-}(aq)+2H^+(aq)\rightarrow CO_2(g)+H_2O(l)}$ conserves both atoms and charge. A precipitate or gas is an observation; an equation explains the change. Zinc in copper(II) sulfate gives a reddish copper deposit while the blue solution fades: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). Magnesium in dilute acid dissolves with hydrogen bubbles: Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g). An acid with a soluble base can warm without gas or precipitate: H⁺(aq) + OH⁻(aq) → H₂O(l). Match each equation to the actual observation rather than treating every acid reaction as gas formation.

Percentage yield 产率 compares actual with theoretical product amount. For a theoretical 4.00 g and actual isolated dry 3.20 g, yield = 3.20/4.00 × 100% = 80.0%. Atom economy 原子经济 compares the balanced desired-product mass with all balanced product masses. Include equation coefficients. A high atom economy does not guarantee a high experimental yield.

In the molar-gas-volume practical, use a known limiting amount and collect the gas under measured conditions. Account for leaks, gas dissolving, incomplete reaction and dead space. A leak lowers measured gas volume and therefore the calculated molar volume. To confirm a formula by reaction with oxygen, heat to constant mass, cool before weighing and explain how loss of solid or incomplete oxidation changes the inferred ratio. Practical reasoning remains part of this theory unit.

Protons and neutrons are in the nucleus; electrons occupy orbitals around it. Relative particle masses are approximately 1, 1 and $1/1836$; charges are +1, 0 and −1. Atomic number 原子序数 $Z$ counts protons; mass number 质量数 $A$ counts protons plus neutrons. Thus neutrons = $A-Z$. A positive ion has fewer electrons than protons. Isotopes 同位素 have the same proton number but different neutron numbers.

A mass spectrometer forms gaseous positive ions, separates them by mass-to-charge ratio 质荷比, $m/z$, and detects relative abundance 相对丰度, the proportion of the sample contributing each ion type. Vaporisation, ionisation, acceleration, separation and detection have distinct roles. In magnetic separation at a given accelerating potential, a lower $m/z$ ion bends more; a higher charge lowers $m/z$, rather than making bending smaller. A doubly charged mass-40 ion appears at $m/z=20$.

Worked isotope mean. For 60% mass-69 and 40% mass-71 gallium, $A_r=(69\times60+71\times40)/100=69.8$. The mean lies between the isotope masses, closer to the more abundant isotope. Molecular-ion peaks can identify $M_r$ when charge is known. With chlorine isotope proportions 3:1, Cl₂ molecular-ion peaks at 70, 72 and 74 have probabilities $9/16$, $6/16$ and $1/16$: the middle combination occurs in two orders.

An orbital 轨道 holds at most two electrons with opposite spins. An s orbital is spherical; a p orbital has two lobes. There are 1, 3 and 5 orbitals in s, p and d subshells, holding at most 2, 6 and 10 electrons. Fill equal-energy orbitals singly before pairing. The first four quantum shells have maximum capacities 2, 8, 18 and 32 electrons. Their available subshells are 1s; 2s/2p; 3s/3p/3d; and 4s/4p/4d/4f. An f subshell has seven orbitals and holds fourteen electrons. The periodic table has s, p and d blocks according to the subshell receiving the differentiating electron. Write configurations through krypton, including the chromium and copper exceptions: Cr is [Ar]3d⁵4s¹ and Cu is [Ar]3d¹⁰4s¹. For their positive ions, remove 4s electrons before 3d electrons. Electron-box notation must show both occupancy and spin.

First ionisation energy 电离能 is the energy per mole for $\mathrm{X(g)\rightarrow X^+(g)+e^-}$. Second ionisation removes an electron from X⁺(g); third ionisation is X²⁺(g) → X³⁺(g) + e⁻. All ionisation steps are endothermic. Greater nuclear charge strengthens attraction; more shielding and greater electron distance weaken it. A large successive-energy jump reveals that all outer-shell electrons have been removed. This identifies the main-group outer electron count but does not alone count every occupied shell.

Across Period 3, nuclear charge rises while inner-shell shielding changes little, giving a broad first-ionisation increase. Al loses a higher-energy 3p electron more easily than Mg loses a 3s electron. S has a paired 3p electron with greater repulsion than the singly occupied 3p orbitals in P. Down a group, increased distance and shielding outweigh increased nuclear charge. Recurring trends across periods are periodicity 周期性; use labelled axes, and interpret a logarithmic axis as ratios rather than equal energy increments.

An ionic bond 离子键 is the net electrostatic attraction between oppositely charged ions throughout a giant lattice. Electron transfer forms ions; the attraction is the bond. Conductivity when molten or dissolved, ion migration and electron-density evidence support the ionic model. Coloured cations move toward the negative electrode in solution. Electron-density maps show electron distribution around nuclei; compare the distribution with the ion model, rather than assuming a contour directly records the history of electron transfer. Ionic solids lack mobile ions. Smaller ions and higher charges usually strengthen lattice attraction. In an isoelectronic series, greater proton number pulls the same electron population closer: the radius decreases from N³⁻ toward Al³⁺. Down a group, extra occupied shells increase radius.

A covalent bond 共价键 is attraction between two nuclei and a shared electron pair. Dot-and-cross diagrams identify electron origin. Include lone pairs and brackets/charge for ions. A coordinate bond 配位键 uses a shared pair initially supplied by one atom, as when NH₃ donates a pair to H⁺ to form NH₄⁺. Aluminium chloride can form Al₂Cl₆ with two chloride bridges and coordinate bonding; electron diagrams must account for every outer electron.

Electron pairs around a central atom repel and arrange apart. Lone pairs repel more strongly than bonding pairs, but there is no universal fixed angle reduction per lone pair. Count regions around the central atom; a multiple bond occupies one region. Bond length 键长 is the distance between nuclei; bond angle 键角 is the angle between adjacent bonds.

Each carbon in ethene has three regions and a planar arrangement, with angles about 120°. Electronegativity 电负性 is attraction for a bonding electron pair. Different electronegativities produce polar bonds. A molecule is polar only if its bond dipoles do not cancel: CO₂ is non-polar; H₂O is polar. Ionic and covalent bonding are ends of a continuum. Polarising power 极化能力 describes a cation’s ability to distort another ion’s electron cloud; polarisability 极化性 describes how readily that cloud is distorted. A small, highly charged cation strongly distorts an anion's electron cloud; large anions are easier to polarise. Do not use one melting point alone as proof of structure.

Small molecular substances usually melt or boil by overcoming intermolecular attractions rather than breaking bonds inside molecules. Diamond has four covalent bonds per atom in a giant network; graphite has three per atom in layers and mobile delocalised electrons; graphene is a single layer. Diamond is hard and insulating; graphite layers can slide and conduct along them; graphene combines strength and conductivity. Metals contain positive ions attracted to delocalised electrons, which carry current in solid and molten metal.

The Period 2/3 melting trend follows changing structures: metallic elements, then giant covalent carbon/silicon, then small molecular non-metals and monatomic noble gases. In Period 2, lithium and beryllium are metallic; boron and carbon form giant covalent structures, then nitrogen, oxygen and fluorine are molecular and neon is monatomic. Molecular electron count and attractions help explain differences among the non-metals. A smooth nuclear-charge explanation alone cannot explain the whole melting trend. Stronger metallic attraction, involving ionic charge, ion size and the number of delocalised electrons, usually raises melting temperature; the detailed metal structure also matters. Heating a metal to melt it does not remove all its electrons.

A functional group 官能团 gives characteristic reactions. A homologous series 同系列 shares a functional group and general formula, with successive members differing by CH₂. Structural isomers 结构异构体 have the same molecular formula but different atom connectivity. In displayed formulae show every atom and bond; in skeletal formulae each unlabelled end or corner is carbon, with enough implied H atoms for four bonds. Show heteroatoms and their attached hydrogen explicitly.

Use meth-, eth-, prop-, but-, pent-, hex-, hept-, oct-, non-, dec- for one to ten carbon atoms. Choose a parent containing the relevant functional group and number to give its required lowest locant, then locate substituents. Acyclic alkanes have formula CₙH₂ₙ₊₂; single-ring cycloalkanes have CₙH₂ₙ. Both are saturated: they contain only carbon–carbon single bonds. CₙH₂ₙ alone does not prove an alkene because a cycloalkane can share it.

Generate structural isomers systematically: keep the formula fixed, change the longest chain or ring size, distribute the remaining carbon atoms, then remove rotated or renumbered duplicates. Pentane has three chain isomers; hexane has five. A different drawing direction alone does not make a new structure.

For saturated single rings, begin with the largest ring and shorten it while moving carbons into substituents. There is one C₃H₆ ring structure, two C₄H₈ and five C₅H₁₀ structures. The C₆H₁₂ set has twelve constitutional structures; distinct cis/trans forms are not counted again as structural isomers.

Fractional distillation separates crude oil into fractions with different boiling ranges. Cracking converts larger hydrocarbons to smaller products, including alkenes; reforming can change straight chains to branched or cyclic structures. Balance atoms in every proposed equation. Complete combustion of a hydrocarbon forms CO₂ and water. Incomplete combustion can form toxic CO or carbon particulates; CO reduces blood oxygen transport by binding haemoglobin. Fuel sulfur and high-temperature nitrogen/oxygen reactions can produce acidic oxides. Carbon particulates harm air quality; unburned hydrocarbons can contribute to photochemical pollution. Removing fuel sulfur does not prevent nitrogen oxides formed from hot air. Carbon neutrality 碳中和 means no net carbon dioxide increase over the stated life-cycle boundary. Bioethanol's growth can take up CO₂, but processing and transport emissions prevent automatic carbon-neutral claims. Hydrogen makes water at use; its production method still matters.

A hazard 危害 is a potential source of harm; risk 风险 considers its likelihood and severity under actual conditions. Use the school's risk assessment, small quantities, ventilation and controls specific to flammability or toxicity. A safer alternative can reduce risk without pretending the original substance has no hazard.

A free radical 自由基 has an unpaired electron. Homolytic fission 均裂 gives one electron to each fragment; heterolytic fission 异裂 gives both to one fragment, forming ions. UV light starts chlorine substitution by breaking Cl₂ homolytically: Cl₂ → 2Cl·. Use single-headed curly arrows for individual electrons.

Propagation: Cl· + CH₄ → HCl + CH₃·, then CH₃· + Cl₂ → CH₃Cl + Cl·. One radical is regenerated, so the chain continues. Termination joins two radicals, for example 2CH₃· → C₂H₆. Further substitution and multiple possible positions produce mixtures, limiting the method's value for making one pure product.

An acyclic alkene with one C=C has formula CₙH₂ₙ. A cycloalkene with one ring and one double bond has CₙH₂ₙ₋₂: the ring removes another two hydrogens. A double bond consists of a sigma bond σ键 from head-on orbital overlap and a pi bond π键 from sideways p-orbital overlap. Rotating one carbon would break pi overlap, so rotation is restricted. E/Z isomerism 顺反异构 requires two different groups on each double-bond carbon. Rank the directly attached atoms by atomic number; if tied, compare the next atoms. Compare priorities separately on the two carbons. Higher-priority groups together give Z, opposite give E. Cis/trans is insufficient where no suitable identical substituents can be compared.

An electrophile 亲电试剂 accepts an electron pair. In HBr addition, a curly arrow begins at the pi bond and ends at H, while another begins at the H–Br bond and ends at Br. This forms a carbocation 碳正离子 and Br⁻; a bromide lone pair then forms the new carbon–bromine bond. Propene mainly gives 2-bromopropane because its route forms a secondary rather than primary carbocation. Tertiary carbocations are generally more stable than secondary, which are more stable than primary, under these conditions.

Bromine is polarised by the electron-rich double bond and adds across it; ethene forms 1,2-dibromoethane. Bromine decolourisation supports unsaturation in an appropriate organic test but does not identify one unique alkene. Avoid unrequired claims about stereochemical product mixtures.

Other additions: hydrogen with nickel gives an alkane; steam with an acid catalyst gives an alcohol; cold, dilute acidified manganate(VII) oxidises C=C to a diol. State reagents and conditions, not only product names. In addition polymerisation 加成聚合, open each C=C into the polymer backbone, retain its side groups, and show brackets with bonds extending through them and $n$ outside. A repeat unit 重复单元 is the smallest backbone segment whose repetition describes the chain, including its side groups. No small molecule is eliminated. Polymer disposal needs evidence: sorting/recycling, persistence, biodegradable 可生物降解的 alternatives and removal of toxic incineration gases have different advantages and costs.

A flask can get warmer while the reacting chemicals lose energy. A clear solution can form a white precipitate 沉淀 without any change in oxidation number. This unit connects observations to models: energy transfer, attractions, electron transfer, collisions and reaction pathways. Start by identifying which model explains the evidence.

Prerequisites: Unit 1 mole ratios, electronic configuration, bond polarity, structural formulae and electron-pair arrows. The scope is Pearson Topics 6–10. Core practicals 2–8 are taught where they support these topics; a reference cannot replace supervised practical experience. Follow the school risk assessment. Hazardous halogen, concentrated-acid and organic preparations require approved laboratory supervision.

An enthalpy change 焓变, $\Delta H$, is the heat energy change at constant pressure. Standard conditions mean a pressure of 100 kPa and a stated temperature, usually 298 K. State symbols matter: making liquid water releases a different amount of energy from making water vapour. An exothermic 放热的 reaction has $\Delta H<0$; an endothermic 吸热的 reaction has $\Delta H>0$. In an enthalpy-level diagram, put reactants and products on horizontal levels. The arrow runs from reactants to products; its vertical change gives $\Delta H$.

These definitions specify different reactions, each under standard conditions with substances in their standard states:

For example, $\frac12\mathrm{Cl_2(g)\rightarrow Cl(g)}$ is an atomisation equation, whereas $\mathrm{Cl_2(g)\rightarrow2Cl(g)}$ describes twice that amount. The standard formation enthalpy of an element in its standard state is zero by convention, not because the element has no energy.

A complete calorimetry calculation

Mix 50.0 cm³ of 1.00 mol dm⁻³ HCl with 50.0 cm³ of 1.00 mol dm⁻³ NaOH. Both start at the same temperature; the corrected rise is 6.8 °C. Assume density 1.00 g cm⁻³, specific heat capacity 4.18 J g⁻¹ °C⁻¹ and negligible cup heat capacity.

The solution gains energy, so the reaction loses it. Use the total heated mass, but the limiting reacting amount. For unequal acid and alkali amounts, calculate both before finding the water amount. In a combustion experiment, water heated is not the fuel amount: obtain the latter from the burner's mass loss and the fuel molar mass.

Use a lid, insulation and stirring, and measure temperatures before and after mixing at regular times. Extrapolate the post-reaction cooling line to the mixing time when the method requires a cooling correction. Heat lost to the surroundings or absorbed by the cup makes an uncorrected exothermic result less negative. Incomplete combustion and fuel evaporation also make a combustion result less negative. Repeats help random variation; they do not remove these systematic effects. A large temperature rise reduces percentage thermometer uncertainty but must remain safe.

Hess cycles and bond enthalpies

Hess's law 赫斯定律 says that the enthalpy change depends on the initial and final states, not the route. Reverse an equation: reverse the sign. Multiply its coefficients: multiply its enthalpy. Add equations and cancel species, including their states, before adding the enthalpies.

For $\mathrm{C(s)+\frac12O_2(g)\rightarrow CO(g)}$, use $\mathrm{C+O_2\rightarrow CO_2}$, $\Delta H=-394$ kJ mol⁻¹, and the reverse of $\mathrm{CO+\frac12O_2\rightarrow CO_2}$, whose forward $\Delta H=-283$ kJ mol⁻¹. The target is the sum, so $\Delta H=-394-(-283)=-111$ kJ mol⁻¹. A cycle using formation data gives $\sum\Delta H_f(\text{products})-\sum\Delta H_f(\text{reactants})$, with coefficients included. A combustion cycle gives the reactant combustion sum minus the product combustion sum.

Core practical 2 compares dissolution/reaction routes to the same final solution. Match amounts and final states, account for each measured temperature change, and test whether experimental error explains disagreement. A cycle with unmatched final concentrations is not exactly the same thermodynamic route.

A mean bond enthalpy 平均键焓 is the average energy needed to break one mole of a specified covalent bond in gaseous molecules. Bond breaking takes energy; bond formation releases it.

For $\mathrm{H_2(g)+Cl_2(g)\rightarrow2HCl(g)}$, let H–H = 436, Cl–Cl = 243 and H–Cl = 431 kJ mol⁻¹. Then $\Delta H\approx436+243-2(431)=-183$ kJ mol⁻¹. Conversely, if $\Delta H=-184$ kJ mol⁻¹, $E(\mathrm{H-Cl})=[436+243-(-184)]/2=431.5$ kJ mol⁻¹. Mean values average different molecular environments and assume gas-phase bonds, so the estimate may differ from measured reaction enthalpy. A weak bond may break more readily, but reaction speed also depends on the whole activation pathway; $\Delta H$ alone does not predict rate.

All atoms and molecules have London forces 伦敦力, caused by instantaneous dipoles 偶极 inducing dipoles in neighbouring particles. More electrons and a more polarisable electron cloud usually strengthen these forces. Permanent dipole–dipole attractions also occur between polar molecules 极性分子. A hydrogen bond 氢键 is an attraction between H bonded to N, O or F and a lone pair on N, O or F of another molecule. An O–H covalent bond within ethanol is not itself a hydrogen bond.

Water, ammonia and hydrogen fluoride hydrogen-bond. Ethanol also has an O–H donor and oxygen lone pairs. An ether has oxygen lone pairs and can accept hydrogen bonds from water, but cannot hydrogen-bond to another ether molecule using an O–H donor. Use the actual structure rather than just finding oxygen in the formula.

Explain boiling by the attractions between molecules, not by breaking every covalent bond. Longer alkane chains usually have more electrons and stronger London forces, so boiling temperatures rise. For isomeric alkanes, branching makes molecules more compact and reduces effective contact, so boiling temperature usually falls. An alcohol has a higher boiling temperature than a similar-size alkane because additional hydrogen bonding requires more energy to overcome. From HCl to HBr to HI, larger electron clouds strengthen London forces; HF is unusually high because it hydrogen-bonds.

Water has unusually high melting and boiling temperatures for its molecular size. Ice contains an open hydrogen-bonded arrangement. On melting, some of this arrangement collapses, bringing molecules closer; liquid water is denser than ice near the melting point. It is incorrect to explain floating ice by saying its molecules become larger.

Solubility depends on the balance between attractions broken and attractions formed. Water hydrates ions: the partially negative oxygen faces cations and partially positive hydrogens face anions. Small alcohols mix well with water because their O–H groups hydrogen-bond; a growing non-polar carbon chain reduces this advantage. Many halogenoalkanes have polar C–X bonds yet dissolve poorly in water: they cannot replace the water–water hydrogen-bond network effectively. A non-aqueous solvent with similar intermolecular attractions can be more suitable. “Polar dissolves polar” is a useful starting comparison, not a universal rule.

An oxidation number 氧化数 is a bookkeeping value, written with a sign; the total equals the charge on the species. Elements have zero. Group 1 is usually +1 and Group 2 +2. Oxygen is usually −2, but −1 in peroxides; hydrogen is usually +1, but −1 in metal hydrides. Use the sum rule when a familiar rule has an exception. For example, in $\mathrm{Cr_2O_7^{2-}}$, $2x+7(-2)=-2$, so chromium is +6. Iron(III) oxide is $\mathrm{Fe_2O_3}$ because two +3 and three −2 sum to zero.

Oxidation is electron loss and an increase in oxidation number. Reduction is electron gain and a decrease. An oxidising agent 氧化剂 gains electrons and is reduced; a reducing agent 还原剂 loses electrons and is oxidised. To combine half-equations 半反应方程式, multiply them until electrons lost equal electrons gained, then cancel electrons. For example, $\mathrm{Mg\rightarrow Mg^{2+}+2e^-}$ and $\mathrm{Cl_2+2e^-\rightarrow2Cl^-}$ give $\mathrm{Mg+Cl_2\rightarrow Mg^{2+}+2Cl^-}$. Check both atoms and charge.

Down Groups 1 and 2, increasing radius and shielding 屏蔽 reduce attraction to the outer electron despite increasing nuclear charge. First ionisation energy decreases and reaction by electron loss becomes easier. Group 1 metals Li to K and Group 2 metals Mg to Ba therefore become more reactive down their groups.

Group 2 hydroxide solubility increases down the group; sulfate solubility decreases. Thus BaSO₄ is a useful insoluble precipitate, whereas Ba(OH)₂ is more soluble than Mg(OH)₂. Keep solubility distinct from strength: the dissolved Group 2 hydroxide supplies hydroxide ions, but a small dissolved amount limits concentration.

Small cations with high charge density strongly polarise large anions. Lithium and the smaller Group 2 cations therefore make their carbonates and nitrates less thermally stable. Stability increases down each group as cation size increases. Most Group 1 carbonates resist ordinary laboratory heating; Li₂CO₃ decomposes to Li₂O + CO₂. Group 2 carbonates give MO + CO₂. Group 1 nitrates except lithium give nitrite + oxygen, $2\mathrm{MNO_3}\rightarrow2\mathrm{MNO_2}+\mathrm{O_2}$. Lithium and Group 2 nitrates give oxide, brown NO₂ and oxygen; e.g. $2\mathrm{Mg(NO_3)_2}\rightarrow2\mathrm{MgO}+4\mathrm{NO_2}+\mathrm{O_2}$. Compare equal amounts under controlled heating; do not confuse a faster gas-production rate with a directly measured decomposition temperature. NO₂ is toxic and acidic; use approved ventilation. CO₂ turns limewater milky; oxygen relights a glowing splint.

Flame colours arise when excited electrons fall to lower energy levels and emit photons of characteristic energies. Li is crimson, Na yellow, K lilac, Ca brick-red, Sr crimson-red and Ba apple-green; Mg has no characteristic flame-test colour. Use a clean wire and avoid sodium contamination. White burning magnesium is not a diagnostic magnesium-ion flame colour.

Use separate aliquots 等分试样 so one test's reagent cannot contaminate the next.

AgCl is white and dissolves in dilute ammonia. AgBr is cream, insoluble in dilute ammonia but dissolves in concentrated ammonia. AgI is yellow and insoluble in both. These tests identify halide ions already in solution; a covalently bonded halogen in a halogenoalkane first needs hydrolysis 水解.

A standard solution 标准溶液 has an accurately known concentration. For core practical 4, accurately weigh a suitable solid acid, dissolve it in deionised water, transfer quantitatively to a volumetric flask 容量瓶, rinse the vessel/funnel into the flask, make to the mark at eye level and mix by repeated inversion. Choose its concentration using mass, molar mass and flask volume. Do not make to the mark before the solid has dissolved and reached room temperature.

For core practicals 3 and 4, rinse the burette with its titrant and the pipette with the solution it will transfer; rinse the conical flask with deionised water. Remove the filling funnel, ensure the jet is filled, read the meniscus at eye level and record final minus initial readings. Swirl and add drops near the endpoint over a white tile. Use a rough titre to locate the endpoint, then obtain concordant 一致的 accurate titres. Do not average the rough titre with accurate ones. Methyl orange is yellow in alkali and red in acid; phenolphthalein is pink in alkali and colourless in acid. Select an indicator whose colour change lies in the sharp endpoint region.

Amount ratio before concentration

25.00 cm³ of 0.0500 mol dm⁻³ Na₂CO₃ needs 24.60 cm³ of HCl. The equation is $\mathrm{Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O}$.

Mass concentration is $cM$ in g dm⁻³. If each burette reading has uncertainty ±0.05 cm³, the maximum uncertainty in a titre is ±0.10 cm³; here $100(0.10/24.60)=0.41\%$. Add relevant percentage uncertainties for a product/quotient as a maximum estimate. A larger appropriate titre lowers relative reading uncertainty. Repeating an endpoint does not correct a wrongly prepared standard.

Chlorine is a pale green gas, bromine a red-brown liquid and iodine a grey-black solid at room temperature; iodine vapour is purple. Down Cl₂, Br₂, I₂, larger electron clouds strengthen London forces, raising melting and boiling temperatures. Electronegativity 电负性 and oxidising power decrease because increasing radius and shielding weaken attraction for an incoming electron. This electron-gain argument differs from the Group 1 electron-loss argument.

Chlorine displaces bromide and iodide; bromine displaces iodide; iodine displaces neither. For example, $\mathrm{Cl_2+2Br^-\rightarrow2Cl^-+Br_2}$. Bromide loses electrons and is the reducing agent. In water, chlorine is pale green, bromine orange/brown and iodine brown; in a suitable non-polar organic solvent, bromine is orange and iodine violet. State the solvent when using colour as evidence. Predict fluorine to be a stronger oxidant and astatine a weaker oxidant from the trend; do not invent a safe school test for either.

In disproportionation 歧化, the same element in one species is both oxidised and reduced. Chlorine with water gives $\mathrm{Cl_2+H_2O\rightleftharpoons HCl+HClO}$: Cl goes from 0 to −1 and +1. Chlorine water disinfects because chlorine-containing oxidants destroy microorganisms. Balance the benefit against toxicity and harmful by-products rather than claiming it is risk-free.

Cold dilute alkali gives $\mathrm{Cl_2+2OH^-\rightarrow Cl^-+ClO^-+H_2O}$, the basis of hypochlorite bleach. Hot concentrated alkali gives $3\mathrm{Cl_2}+6\mathrm{OH^-}\rightarrow5\mathrm{Cl^-}+\mathrm{ClO_3^-}+3\mathrm{H_2O}$; chlorine becomes −1 and +5. Analogous bromine and iodine equations follow the same atom/charge accounting under the specified conditions.

With concentrated H₂SO₄, solid chloride salts undergo acid–base reaction to release HCl; chloride is not a sufficiently strong reducing agent to reduce the acid under these conditions. Bromide first gives HBr, which can reduce sulfur from +6 to +4 in SO₂ while forming Br₂: $2\mathrm{HBr}+\mathrm{H_2SO_4}\rightarrow\mathrm{Br_2}+\mathrm{SO_2}+2\mathrm{H_2O}$. Iodide is a stronger reducing agent and can give sulfur and H₂S as well as SO₂, with iodine formed. For H₂S: $8\mathrm{HI}+\mathrm{H_2SO_4}\rightarrow4\mathrm{I_2}+\mathrm{H_2S}+4\mathrm{H_2O}$. These are hazardous supervised demonstrations, not instructions for independent gas generation. Hydrogen halides dissolve in water to form acids and react with ammonia to form ammonium halides: $\mathrm{HX+NH_3\rightarrow NH_4X}$.

Reaction needs collisions with sufficient energy to overcome the activation energy 活化能 and a suitable orientation. Greater concentration, gas pressure or exposed solid surface increases collision frequency. Higher temperature increases collision frequency and, more importantly, the fraction with energy at least $E_a$.

On a Maxwell–Boltzmann 麦克斯韦–玻尔兹曼 energy distribution, temperature increase makes the peak lower and shifts it to higher energy; the area stays the same for the same number of molecules. The high-energy area beyond $E_a$ increases. A catalyst 催化剂 provides another pathway with lower activation energy. At the same temperature it does not change the distribution, but moves the energy threshold left so more collisions can react. A catalysed reaction profile may have two peaks separated by an intermediate; neither the reactant/product levels nor $\Delta H$ changes.

For a fixed visible endpoint, $1/t$ is a comparative rate measure in s⁻¹, provided each trial reaches the same extent of reaction. It is not automatically a rate in mol dm⁻³ s⁻¹. On a gas-volume–time graph, average rate is $\Delta V/\Delta t$; instantaneous rate is a tangent gradient. If a tangent passes through (10 s, 12 cm³) and (50 s, 36 cm³), its gradient is $(36-12)/(50-10)=0.60$ cm³ s⁻¹. Use points on the tangent, not two nearby curve points. A faster reaction need not make more final product if the limiting amount is unchanged.

Dynamic equilibrium 动态平衡 requires a closed system. Forward and reverse reactions continue at equal rates; concentrations remain constant but need not be equal. Increasing a reactant concentration shifts equilibrium toward its consumption. Increasing pressure favours the side with fewer gas moles; if gas coefficients are equal, there is no composition shift. Increasing temperature favours the endothermic direction. A catalyst speeds both directions and reaches the same equilibrium sooner.

For $\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}$, with exothermic forward reaction, high pressure and low temperature favour ammonia yield. Low temperature slows reaction; very high pressure raises energy, equipment and safety costs. Industrial conditions balance rate, yield, separation/recycling and cost. A catalyst permits an adequate rate at lower temperature; it does not itself increase the equilibrium yield. More selective catalysts can also improve atom economy by reducing unwanted pathways, but this is a different claim from shifting equilibrium.

A nucleophile 亲核试剂 donates an electron pair to form a covalent bond. The polar C–X bond has a partially positive carbon. A curly arrow starts at a lone pair or bond, not at a positive charge. Heterolytic 异裂的 breaking gives both bond electrons to one atom, producing ions. A mechanism shows electron movement, not just an overall equation.

Classify a halogenoalkane by the number of carbon groups attached to the carbon bearing X: one primary, two secondary, three tertiary. This is not the position number in its name. 1-bromobutane is primary; 2-bromobutane is secondary; 2-bromo-2-methylpropane is tertiary.

For primary bromoethane and OH⁻, draw one arrow from an oxygen lone pair to the carbon bearing Br and another from the C–Br bond to Br. Products are ethanol and Br⁻. For ammonia, the same electron-pair attack gives CH₃CH₂NH₃⁺ and Br⁻; a second ammonia removes H⁺ to give CH₃CH₂NH₂ and NH₄⁺. Include this proton-transfer step and charges. Detailed S_N1/S_N2 kinetics are Unit 4 material.

In core practical 5, compare equal concentrations/volumes of halogenoalkanes with the same silver nitrate/ethanol mixture at controlled temperature. Time formation of a comparable precipitate. Iodoalkanes hydrolyse faster than bromoalkanes, which hydrolyse faster than chloroalkanes, because C–I is weaker than C–Br and C–Cl. C–F is strongest and least easily broken. Do not use C–X polarity alone to predict this order. In the specified aqueous/ethanolic hydrolysis comparison, tertiary halogenoalkanes generally react faster than secondary, then primary. Solvent, concentration and detection conditions must be matched; precipitate appearance is an indirect rate measure.

Core practical 6 converts 2-methylpropan-2-ol to 2-chloro-2-methylpropane using concentrated HCl. Explain substitution of OH by Cl, separation of organic/aqueous layers, removal of acidic impurities, drying and purification. A separating funnel must be vented as specified by the approved method; identify layers using density/evidence, not an assumption that all organic liquids float. No naked flame near flammable organics.

Classify alcohols by the carbon attached to OH, not by counting all OH groups. Primary propan-1-ol oxidises with potassium dichromate(VI) in dilute sulfuric acid. Distil the aldehyde 醛 as it forms to limit further oxidation: $\mathrm{CH_3CH_2CH_2OH+[O]\rightarrow CH_3CH_2CHO+H_2O}$. Heat under reflux with excess oxidant to make the acid: $\mathrm{CH_3CH_2CH_2OH+2[O]\rightarrow CH_3CH_2COOH+H_2O}$. Secondary propan-2-ol gives propanone: $\mathrm{CH_3CH(OH)CH_3+[O]\rightarrow CH_3COCH_3+H_2O}$. Tertiary alcohols resist oxidation under these conditions because the OH-bearing carbon lacks H. Orange dichromate turns green on reduction.

Aldehydes give a brick-red precipitate with warm Fehling's or Benedict's solution; ordinary ketones do not. Carboxylic acids react with carbonate/hydrogencarbonate to release CO₂. These distinguish products of core practical 7; do not identify an unknown from the dichromate colour alone.

Alcohol combustion produces CO₂ and H₂O when oxygen is sufficient. PCl₅ converts an alcohol to a chloroalkane, producing POCl₃ and HCl fumes: $\mathrm{ROH+PCl_5\rightarrow RCl+POCl_3+HCl}$. The fumes support an OH-group test but are not unique proof of an alcohol. KBr with 50% concentrated sulfuric acid supplies HBr for conversion to a bromoalkane; red phosphorus and iodine supply a reagent for an iodoalkane. Heating with concentrated phosphoric acid eliminates water to make an alkene. Distinguish dehydration from oxidation and substitution; state conditions on the arrow.

Reflux heats the mixture while a condenser returns vapour; distillation collects vapour after condensation and separates by volatility. Water enters the condenser at the bottom and leaves at the top. Solvent extraction partitions a product between immiscible layers. Use a suitable anhydrous 无水的 drying agent to remove residual water, then decant/filter it before final distillation. A narrow boiling range near a reference value supports purity but is not conclusive identification; pressure and thermometer position affect the measurement.

The molecular-ion peak 分子离子峰 in a mass spectrum gives molecular mass for a singly charged ion. It need not be the largest peak. Fragment ions 碎片离子 help suggest structures: loss of CH₃ corresponds to 15 mass units, but a single fragment rarely proves one isomer. Distinguish isotope peaks from fragments. In an IR spectrum, compare absorption positions with the provided data: C=O, O–H, N–H, C=C, C–H and C–X can identify possible functional groups. Alcohol O–H is broad; acid O–H is very broad and appears with C=O. Both aldehydes and ketones have C=O, so combine IR with formula, mass spectrum and chemical tests. An absent strong expected absorption can rule out a proposal, subject to spectrum quality. Core practical 8 combines ion tests and organic tests using clean separate aliquots and a recorded chain of evidence.

The rate equation 速率方程 relates rate to concentrations at a specified temperature:

The order 反应级数 with respect to A is $m$; overall order is $m+n$. These powers are found experimentally. They need not match the balanced equation's coefficients. In this course, individual orders are 0, 1 or 2. The rate constant 速率常数 $k$ depends on temperature and pathway, not the reactant concentrations in the stated model. Its units follow by rearranging the equation.

Compare trials 1 and 2: only [A] doubles and rate doubles, so order in A is one. Compare 1 and 3: only [B] doubles and rate quadruples, so order in B is two. Therefore $r=k[\mathrm A][\mathrm B]^2$ and overall order is three.

Using trial 2 or 3 gives the same $k$. That is an internal check, not independent proof that the model works outside the measured range. With both concentrations doubled, the model predicts a factor $2\times2^2=8$ increase in initial rate.

Graphs, half-life and method choice

A zero-order rate–concentration graph is horizontal; a first-order graph is a straight line through the origin; a second-order graph curves upward as concentration squared. A zero-order concentration–time graph falls linearly while the model remains valid. For first order, concentration falls exponentially and the half-life 半衰期 is constant: the time from 0.080 to 0.040 mol dm⁻³ equals that from 0.040 to 0.020. For second order, successive half-lives grow as concentration falls. Read half-life from the graph and show the concentration pair used; one interval alone cannot establish constant half-life.

Initial-rate experiments vary one concentration between separate trials and hold others and temperature constant. Continuous monitoring follows one trial over time. Select a measurement that changes with reaction progress: gas volume for a gas-forming reaction; mass loss for escaping gas; colorimetry for a suitable coloured species; or timed samples quenched and titrated. Mass loss is unsuitable if no material leaves the apparatus. Calibrate a colorimeter and control path length/wavelength. Account for sampling delay, gas leaks and temperature drift.

A clock endpoint can approximate initial rate if the same small amount reacts before the endpoint in each trial. Then $1/t$ is proportional to the initial rate. Keep total volume, endpoint chemistry and temperature fixed. If a large changing fraction reacts before the endpoint, inverse time is not a valid initial-rate comparison.

Mechanisms must fit both kinetics and stoichiometry

The rate-determining step 决速步骤 is the slow step controlling the overall rate under the stated conditions. A proposed mechanism must add to the overall equation and be consistent with the observed rate law. Intermediates are made in one step and consumed in another; a catalyst is consumed then regenerated. A rate equation supports a mechanism but rarely proves it uniquely.

For acid-catalysed iodination of propanone, rate is first order in propanone and H⁺ and zero order in iodine under the usual measured conditions. Changing iodine concentration alone does not change the initial rate. This supports iodine reacting after the slow sequence; it does not mean iodine is absent from the overall reaction. In core practical 9a, follow iodine amount through appropriately quenched samples and titration, or use a calibrated colour method. Core practical 9b uses an iodine clock: justify the fixed-extent approximation before treating $1/t$ as rate.

For primary halogenoalkane substitution by OH⁻, an S_N2 model has one concerted step and rate depending on both reactants. Backside attack produces inversion at a suitable chiral carbon. For tertiary hydrolysis, an S_N1 model has slow C–X heterolysis forming a carbocation, then faster nucleophile attack. Its rate depends on halogenoalkane concentration but not nucleophile concentration under the stated conditions. A freely accessible planar carbocation permits attack from both sides, giving a racemic product in the ideal model. Solvent and substrate matter; do not apply either rate law to every substitution without evidence.

Temperature and catalysis

When supplied, use the Arrhenius relation $k=Ae^{-E_a/(RT)}$, or $\ln k=\ln A-E_a/(RT)$. On a graph of $\ln k$ against $1/T$, gradient $=-E_a/R$. Use kelvin and match energy units to $R=8.31$ J mol⁻¹ K⁻¹. A measured gradient of −6200 K gives:

Core practical 10 measures comparable rates across controlled temperatures. Let the mixture reach the chosen temperature, measure the reacting mixture rather than only the water bath, and justify the same rate proxy at every temperature. A graph against $1000/T$ has a gradient smaller by a factor of 1000 than one against $1/T$; label the actual axis before using the equation.

A homogeneous catalyst is in the same phase as the reacting mixture; a heterogeneous catalyst is in another phase. A solid industrial catalyst can adsorb gaseous reactants onto its surface, weaken bonds/provide another pathway, then release products by desorption. Greater accessible surface gives more sites. A poison blocks sites. The catalyst lowers activation energy and changes rate, not the equilibrium constant 平衡常数 at a fixed temperature.

Entropy 熵 describes the random dispersal of particles and energy. For a substance, entropy generally rises with temperature and from solid to liquid to gas. A perfect crystal at zero kelvin has zero entropy. Gas spreading through a room is favoured because many more arrangements distribute its molecules through the larger volume.

The natural direction has positive total entropy change 总熵变:

Calculate system entropy using products minus reactants, with balanced coefficients. Standard molar entropy units are J mol⁻¹ K⁻¹, so convert an enthalpy in kJ mol⁻¹ to J mol⁻¹ before dividing by kelvin.

Suppose a reaction has $\Delta S_{\mathrm{system}}=+120$ J mol⁻¹ K⁻¹ and $\Delta H=+30.0$ kJ mol⁻¹. At 298 K:

The process is feasible despite being endothermic. Assuming the given enthalpy/entropy values stay approximately constant, the boundary is $T=\Delta H/\Delta S_{\mathrm{system}}=30000/120=250$ K; above 250 K total entropy is positive. At the boundary it is zero, not positive.

An exothermic reaction increases surroundings entropy. An endothermic one decreases it, but a sufficiently positive system change can outweigh that decrease. More gas moles often suggest a positive system entropy change; use data for a numerical conclusion. Dissolving an ionic solid disperses ions, but strong hydration can order surrounding water, so solution entropy is not automatically positive. Neither sign of system entropy alone decides feasibility.

Thermodynamic stability 热力学稳定性 concerns the energetically/entropically favoured state under the stated conditions. Kinetic stability 动力学稳定性 concerns a high activation barrier and slow reaction. A feasible reaction can remain imperceptibly slow. Increasing temperature changes both kinetics and total entropy; these are different explanations.

Pearson defines lattice energy 晶格能 here as the exothermic formation of one mole of ionic solid from its gaseous ions. For NaCl, the equation is Na⁺(g) + Cl⁻(g) → NaCl(s), so this lattice formation value is negative. Separating the lattice is the reverse and positive.

First electron affinity 电子亲和能 is the enthalpy change when one mole of gaseous atoms each gains one electron to form gaseous 1− ions. Write the gas states and electron: Cl(g) + e⁻ → Cl⁻(g). Second electron affinity adds an electron to an already negative gaseous ion and is usually endothermic because of repulsion.

A Born–Haber cycle for NaCl separates formation from the elements into atomisation, ionisation, electron affinity and lattice formation. Using illustrative values in kJ mol⁻¹, $\Delta H_f=-411$, Na atomisation +108, Cl atomisation +121, Na ionisation +496 and Cl electron affinity −349:

For MgCl₂, include two chlorine atomisations, two electron affinities and both Mg ionisation energies. The cycle must form the ions in the actual solid formula. A theoretical lattice value from a purely ionic model may differ from a Born–Haber value derived from experiment. A substantial extra stabilisation supports covalent character: a small highly charged cation polarises a large anion's electron cloud. Agreement supports the model but does not prove a bond is perfectly ionic.

Hydration enthalpy 水合焓 forms one mole of aqueous ions from gaseous ions, normally exothermically. Solution enthalpy dissolves one mole of solute in water under the stated conditions. With the lattice formation convention:

If lattice formation is −780 and the total hydration enthalpy is −760 kJ mol⁻¹, solution enthalpy is $+780-760=+20$ kJ mol⁻¹. Smaller ions and greater ionic charge usually make both lattice formation and hydration more negative. Because these effects compete, one trend alone cannot establish solubility. Combine solution enthalpy with entropy to assess dissolution. In Group 2 sulfate trends, changes in hydration and lattice terms differ in size; do not say every less exothermic hydration term necessarily makes every salt insoluble.

Down Group 2, larger cations have less exothermic hydration. For the large sulfate anion, the lattice-dissociation term changes comparatively little, so the loss of hydration stabilisation helps explain decreasing sulfate solubility. For hydroxides, the decrease in lattice-dissociation enthalpy is more important and can outweigh the less exothermic hydration, supporting increasing solubility. These explanations compare both terms within each family; a final numerical prediction also needs solution entropy at the stated temperature.

For $a\mathrm A+b\mathrm B\rightleftharpoons c\mathrm C+d\mathrm D$, an appropriate concentration expression is $K_c=[\mathrm C]^c[\mathrm D]^d/([\mathrm A]^a[\mathrm B]^b)$. Use equilibrium concentrations, not the amounts initially mixed. Omit pure solids and pure liquids in heterogeneous expressions; do not omit an aqueous solute simply because its formula contains no gas state.

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g):

With equilibrium concentrations 0.200, 0.100 and 0.300 mol dm⁻³ for NH₃, N₂ and H₂ respectively, $K_c=(0.200)^2/[0.100(0.300)^3]=14.8$ dm⁶ mol⁻². Derive the units from the expression rather than memorising one unit for all constants.

Use an amount table before the expression

For A(g) + B(g) ⇌ C(g), initially 0.500 mol each of A and B and no C are in 2.00 dm³. If 0.200 mol C forms, equilibrium amounts are 0.300, 0.300 and 0.200 mol. Concentrations are 0.150, 0.150 and 0.100 mol dm⁻³, so:

Using the mole amounts directly would give a different wrong result because the concentration powers do not cancel all volume factors.

For gases, partial pressure 分压 is mole fraction times total pressure: $p_i=(n_i/n_{\mathrm{total}})P$. Pearson's $K_p$ data use pressures in atm. For N₂O₄(g) ⇌ 2NO₂(g), $K_p=p(\mathrm{NO_2})^2/p(\mathrm{N_2O_4})$ with units atm. At total pressure 3.00 atm and equilibrium amounts 1.00 mol N₂O₄ and 2.00 mol NO₂, the partial pressures are 1.00 and 2.00 atm. Thus $K_p=4.00$ atm. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), $K_p=p(\mathrm{CO_2})$ while both solids are present; their amounts do not enter the expression.

Changing concentration or pressure changes composition as the system returns to the same constant at fixed temperature. A catalyst does not change the constant. For an exothermic forward reaction, raising temperature lowers K; for an endothermic forward reaction, it raises K. The change in K explains the new equilibrium composition. A very large K favours products in the stated expression but does not guarantee a fast reaction or literally zero reactant.

Where supplied, $\Delta S_{\mathrm{total}}=R\ln K$ relates the entropy change and an appropriately standardised dimensionless equilibrium constant. Positive total entropy corresponds to $K>1$; zero to $K=1$; negative to $K<1$. Use the defined constant and standard-state conventions. Do not take a logarithm of an unconverted dimensional pressure value or confuse this with a claim that all classroom Kc/Kp expressions have identical units.

A Brønsted–Lowry acid 酸 donates a proton; a base 碱 accepts one. Conjugate 共轭的 acid–base partners differ by one H⁺. In NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, the pairs are NH₄⁺/NH₃ and H₂O/OH⁻. A strong acid dissociates essentially completely in the specified dilute aqueous model; a weak acid only partly dissociates. Strength describes dissociation 解离, whereas concentration describes amount per volume.

Use numerical hydrogen-ion concentration in mol dm⁻³ in this course expression. A 0.0200 mol dm⁻³ monoprotic strong acid has pH $=-\log_{10}(0.0200)=1.70$. A pH of 3.40 gives $[\mathrm{H^+}]=3.98\times10^{-4}$ mol dm⁻³. Do not automatically double a diprotic acid concentration unless the stated dissociation model justifies both protons being fully released.

For water, $K_w=[\mathrm{H^+}][\mathrm{OH^-}]$. At 298 K use $K_w=1.00\times10^{-14}$ mol² dm⁻⁶, when given. A strong base giving [OH⁻] = 0.0100 mol dm⁻³ gives [H⁺] = $K_w/[\mathrm{OH^-}]=1.00\times10^{-12}$ mol dm⁻³ and pH = 12.00. In general neutral means [H⁺] = [OH⁻]; pH 7 is neutral at this temperature, not a universal temperature-independent rule. $\mathrm{p}K_a=-\log_{10}K_a$ and $\mathrm{p}K_w=-\log_{10}K_w$ using the course's concentration conventions.

Weak acids and measured Ka

For HA ⇌ H⁺ + A⁻, $K_a=[\mathrm{H^+}][\mathrm{A^-}]/[\mathrm{HA}]$. If water's H⁺ contribution is too small to matter and dissociation $x$ is small compared with initial concentration $c$, then [H⁺] ≈ [A⁻] = $x$ and [HA] ≈ $c$:

For $K_a=1.80\times10^{-5}$ mol dm⁻³ and $c=0.100$ mol dm⁻³, [H⁺] ≈ $1.34\times10^{-3}$ mol dm⁻³, so pH ≈ 2.87. The fraction dissociated is $0.00134/0.100=1.34\%$, supporting the approximation. The specification does not require quadratic solutions; recognise when supplied data do not justify the simple approximation rather than hiding the issue.

Core practical 11 finds $K_a$ from calibrated pH and known acid concentration. If pH = 2.90 for 0.100 mol dm⁻³ acid, $x=10^{-2.90}=1.26\times10^{-3}$ mol dm⁻³. Then $K_a=x^2/(c-x)=1.61\times10^{-5}$ mol dm⁻³. If concentration is obtained from a weighed acid, first use $c=m/(MV)$ with the final solution volume. Calibrate the pH meter with suitable buffers, rinse the electrode between samples and control temperature.

For a tenfold dilution of an ideal strong acid in the appropriate concentration range, pH increases by one. For a weak acid obeying the small-dissociation approximation, [H⁺] falls by $\sqrt{10}$ and pH rises by about 0.5. Compare equimolar acid/base/salt measurements using dissociation and reaction of ions with water; a salt solution is not automatically neutral. Very dilute solutions need care because water's own ions matter.

A buffer solution 缓冲溶液 resists small pH changes when small amounts of acid or alkali are added. A weak-acid buffer contains substantial HA and its conjugate base A⁻. Added H⁺ is consumed by A⁻; added OH⁻ is consumed by HA. Both components must remain after any initial neutralisation. Buffer capacity is finite.

With $K_a=1.80\times10^{-5}$, [HA] = 0.200 and [A⁻] = 0.100 mol dm⁻³, [H⁺] ≈ $3.60\times10^{-5}$ mol dm⁻³ and pH ≈ 4.44. To make pH 5.00, the required ratio is $[\mathrm{A^-}]/[\mathrm{HA}]=K_a/[\mathrm{H^+}]=1.80$. If [HA] is 0.100 mol dm⁻³, [A⁻] should be 0.180 mol dm⁻³ under the stated approximation.

When a strong base is added to a weak acid, do the mole reaction first: HA + OH⁻ → A⁻ + H₂O. Suppose 0.0100 mol HA initially meets 0.00400 mol OH⁻. Afterwards HA = 0.00600 mol and A⁻ = 0.00400 mol, so use their 2:3 ratio; common final volume cancels in the ratio. Using the initial acid amount would be wrong. Beyond equivalence 等当点 the excess strong base controls pH and the acid-buffer expression no longer applies.

On a titration curve, distinguish initial pH, buffer region, steep change, equivalence and excess titrant. Strong acid–strong base has equivalence near pH 7 at 298 K. Weak acid–strong base has equivalence above 7; strong acid–weak base below 7. Weak acid–weak base usually lacks a sharp enough vertical region for a simple colour indicator 指示剂. Choose an indicator whose transition interval lies within the steep region shown by the actual curve; do not choose just because its midpoint matches the starting pH.

At half-neutralisation of a weak acid by strong base, [HA] = [A⁻], so pH = pKa. The analogous strong-acid/weak-base curve supports analysis of the conjugate weak-acid equilibrium when the species and direction are stated. For a diprotic acid, equivalence amounts reflect two proton equivalents, but two separate steep jumps are visible only when the dissociations are sufficiently distinct. Track each neutralisation stage before applying a buffer or excess-reagent formula.

Buffers in cells and blood help limit changes in enzyme conditions; the carbonic acid/hydrogencarbonate pair is one example. Food buffers can reduce pH changes associated with microbial activity and deterioration. A buffer does not sterilise food or make unlimited acid addition harmless.

A chiral centre 手性中心 in these examples is a tetrahedral carbon attached to four different groups. Its two enantiomers 对映异构体 are non-superimposable mirror images. Flat formulae can hide this difference; use wedge/dash bonds to show the three-dimensional arrangement. A pure enantiomer rotates plane-polarised monochromatic light; its partner gives the equal opposite rotation under identical conditions. A racemic mixture 外消旋混合物 contains equal amounts of both, so their rotations cancel. Zero rotation alone does not prove a substance is racemic: an achiral substance is also inactive.

S_N2 attack at a chiral carbon gives inversion, whereas ideal S_N1 reaction through a planar carbocation allows attack from both sides. Use the stated starting material and observed optical activity to assess the mechanism. Do not infer the sign of optical rotation just by inspecting a wedge/dash drawing.

Aldehydes have a terminal –CHO group; ketones have C=O between two carbon groups. Propanal is CH₃CH₂CHO and propanone CH₃COCH₃. Carbonyl 羰基 oxygen accepts hydrogen bonds from water. Simple aldehydes/ketones lack O–H/N–H/F–H donors, so their pure molecules do not form the alcohol-like hydrogen-bond network. Small ones are water-soluble; increasing the non-polar chain reduces solubility.

For example, CH₃CHO + 2[H] → CH₃CH₂OH. Prepare derivatives with 2,4-DNPH by the approved method and compare their melting temperatures with supplied reference data. A sharp matching melting range supports identification; mixture contamination and reference overlap are limits. No derivative equation is required here. Tollens' reagent and cyanide chemistry require controlled laboratory handling and disposal; do not store or improvise these reagents.

HCN, with KCN providing CN⁻, adds to a carbonyl by nucleophilic addition 亲核加成. A carbon lone pair in CN⁻ attacks the partially positive carbonyl carbon; the C=O π pair moves to oxygen. The O⁻ intermediate then gains H⁺ from HCN, regenerating CN⁻. Propanal gives CH₃CH₂CH(OH)CN. The planar carbonyl can be attacked from either face; with achiral conditions, the new chiral centre forms a racemic mixture. Propanone's product has two identical methyl groups, so it is not chiral for that reason.

The iodoform test uses iodine in alkali: compounds with CH₃CO– give a yellow CHI₃ precipitate. Ethanal also fits this pattern with H as its other carbonyl substituent. Ethanol and secondary alcohols of type CH₃CH(OH)R can first oxidise to a suitable carbonyl and also give a positive result. A positive test does not prove every unknown is propanone.

A carboxylic acid has –COOH; name its carbonyl carbon as carbon 1. Hydrogen-bonded associations raise boiling temperatures. Small acids hydrogen-bond with water; a growing hydrocarbon part reduces solubility. Make acids by oxidation of primary alcohols/aldehydes or by nitrile hydrolysis with the stated aqueous acidic or alkaline conditions. The nitrile carbon becomes the acid/carboxylate carbon, so count it in the product chain.

Carboxylic acids form salts with bases and release CO₂ with carbonate/hydrogencarbonate. LiAlH₄ in dry ether reduces RCOOH to RCH₂OH after the required work-up. PCl₅ converts RCOOH to the acyl chloride 酰氯 RCOCl, with POCl₃ and HCl also formed. Heating an acid with an alcohol and an acid catalyst establishes an esterification equilibrium: CH₃COOH + CH₃CH₂OH ⇌ CH₃COOCH₂CH₃ + H₂O. The product is ethyl ethanoate; the alcohol supplies the alkyl name and the acid supplies the carboxylate name.

Acyl chlorides react readily with nucleophiles. RCOCl + H₂O → RCOOH + HCl; with R′OH they give RCOOR′ + HCl. With excess NH₃ they give RCONH₂ and NH₄Cl overall: RCOCl + 2NH₃ → RCONH₂ + NH₄Cl. With a primary amine R′NH₂ they give an N-substituted amide; a second amine molecule accepts HCl. Keep carbonyl carbon in the product and count the two nitrogen-reactant molecules in the overall equation.

Acid hydrolysis of an ester reversibly gives carboxylic acid and alcohol. Alkaline hydrolysis gives carboxylate salt and alcohol and is effectively driven to products by salt formation. Ethyl ethanoate + NaOH → sodium ethanoate + ethanol. Do not write a free carboxylic acid as the immediate product in alkaline solution.

A polyester 聚酯 forms by condensation 缩合 between suitable bifunctional monomers, producing ester links and a small molecule. Terylene can use benzene-1,4-dicarboxylic acid and ethane-1,2-diol. Trace –O–CH₂–CH₂–O–CO–C₆H₄–CO– through the repeat unit, with open bonds crossing brackets. The para benzene arrangement matters. Do not use an alkene addition-polymer template: condensation links functional groups and releases water when a diacid and diol react.

Accurate mass can distinguish formulae with the same nominal integer mass. With C = 12.0000, H = 1.0078 and O = 15.9949, C₂H₆O has mass $2(12.0000)+6(1.0078)+15.9949=46.0417$. Add the given precise atomic values before rounding; a nominal 46 alone does not establish this formula.

Carbon-13 NMR gives one signal per distinct carbon environment in the simple spectra considered here. Symmetry can make different carbon atoms equivalent: propanone has two carbon environments, not three; propanal has three. Use supplied chemical-shift data to distinguish carbonyl, C–O and hydrocarbon environments. Signal count is not simply total carbon count.

Proton NMR uses chemical shifts for environments, relative integrated areas 积分面积 for proton numbers, and splitting by neighbouring non-equivalent protons. For a simple set coupled to $n$ equivalent neighbouring H atoms, the $n+1$ rule predicts splitting. A CH₃CH₂– group often gives a 3H triplet and 2H quartet. Exchangeable OH/NH signals may be broad and need not follow simple neighbouring-H splitting; use the stated spectrum. Ethyl ethanoate has a 3H singlet at the acyl methyl, a 2H quartet at OCH₂ and a 3H triplet at the terminal methyl, with three proton environments. Combine these with IR carbonyl evidence and molecular mass rather than treating one quartet as unique proof.

Chromatography 色谱法 separates by different interactions with a stationary 固定的 and a mobile 流动的 phase. For paper/TLC, $R_f=\text{distance of spot}/\text{distance of solvent front}$, both from the baseline. A spot 3.6 cm from the baseline with a 6.0 cm solvent front has $R_f=0.60$, without units. Solvent and stationary-phase conditions affect the result; compare references under matched conditions. One spot can hide co-eluting substances.

HPLC and gas chromatography separate compounds through a column, giving different retention times under specified conditions. A matching retention time 保留时间 supports a candidate but is not unique proof. Coupling chromatography to mass spectrometry separates mixture components and gives mass/fragment evidence for each, useful in forensic work and drug testing in sport. Keep sampling provenance, standards, contamination controls and uncertainty in the interpretation.

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.