9701Chemistry

AS Level Chemistry, one topic at a time

Detailed notes for all 22 AS topics of the Cambridge 9701 syllabus. Each topic has worked examples, diagrams, interactive tools and exam-style questions. Multiple-choice questions mark themselves. Structured questions come with mark schemes laid out the way examiners write them.

Paper 1: Multiple Choice

1 h 15 min40 marks31% of AS

40 four-option questions on the AS topics.

Paper 2: AS Structured Questions

1 h 15 min60 marks46% of AS

Structured questions on the AS topics: equations, calculations, explanations and mechanisms.

Paper 3: Advanced Practical Skills

2 h40 marks23% of AS

Lab work: titration, rates or enthalpy experiments, qualitative analysis, and handling results.

If you go on to the full A Level, these three papers make up 50% of it. Papers 4 and 5, in the A Level tab, make up the other 50%.

Topics

How the marks work

Chemistry mark schemes list the marking points separately. Each point (labelled M1, M2 and so on) is worth one mark. The same words come up again and again in them:

TermMeaning
M1, M2…Separate marking points, one mark each. Some depend on an earlier point (for example, a calculation mark needs the right method).
ecfError carried forward. If you get an early value wrong but use it correctly later, you can still earn the later marks.
owtte"Or words to that effect": the idea matters, not the exact wording.
allow / ignore / rejectAlternatives that also score, extra words that don't change the mark, and answers that score nothing even if part of them is right.
Three habits that save marks

Always give state symbols when an equation asks for them. Give answers to the number of significant figures the data supports (usually 3). Write units on every final answer.

Command words

WordWhat to do
StateGive a short answer with no explanation.
DefineGive the precise meaning. Learn the definitions on this site word for word.
DescribeSay what happens or what something is like, including observations such as colours.
ExplainGive reasons, usually in terms of particles, bonding or electrons.
DeduceReach a conclusion from the information given.
SuggestApply what you know to an unfamiliar situation. More than one answer may be accepted.
CalculateWork out a number, showing your working.
SketchDraw the general shape, with labelled axes and key features, not to scale.

The Data Booklet

In every theory paper you get a Data Booklet with a Periodic Table and tables of data. You don't need to memorise these values:

QuantityValue
Avogadro constantL = 6.02 × 1023 mol−1
molar gas constantR = 8.31 J K−1 mol−1
Faraday constantF = 9.65 × 104 C mol−1
charge on an electrone = −1.60 × 10−19 C
molar volume of a gasVm = 22.4 dm3 mol−1 at s.t.p. (273 K and 101 kPa); Vm = 24.0 dm3 mol−1 at room conditions
ionic product of waterKw = 1.00 × 10−14 mol2 dm−6 at 298 K
specific heat capacity of waterc = 4.18 kJ kg−1 K−1 (that is, 4.18 J g−1 K−1)

The booklet also lists ionisation energies, bond energies, standard electrode potentials, atomic and ionic radii, and the infrared and NMR absorption tables. Values in different books differ slightly, so in the exam always use the values printed in the booklet or in the question. The questions on this site give any data you need.

About the questions

The questions here are written in the style of 9701 papers and test the same skills. They are not copies of official papers. For genuine past papers and mark schemes, use the Cambridge School Support Hub or ask your teacher.

Almost everything in chemistry comes down to electrons: where they are, how strongly the nucleus holds them, and how easily they are lost or shared. This topic builds that picture, from the particles in the atom to orbitals and ionisation energies.

Inside the atom

An atom is mostly empty space. At its centre is a very small, dense nucleus that contains the protons and neutrons. The nucleus is about 10−15 m across, while the whole atom is about 10−10 m across, roughly 100 000 times larger. The electrons are found in shells (energy levels) in the space around the nucleus.

ParticleWhere it isRelative massRelative charge
protonnucleus1+1
neutronnucleus10
electronshells around the nucleus1/1836 (often written as negligible)−1

Distribution of mass and charge: nearly all the mass of an atom is in the nucleus, which is positively charged. The negative charge is spread through the much larger volume around it. An atom is neutral because it has equal numbers of protons and electrons.

Beams of particles in an electric field

If beams of protons, neutrons and electrons travel at the same speed between two charged plates:

  • Protons are deflected towards the negative plate.
  • Electrons are deflected towards the positive plate, and through a much larger angle because their mass is about 1/1836 of a proton's, but the size of their charge is the same.
  • Neutrons are not deflected because they have no charge.
+−positive platenegative platesourceneutronsprotonselectronssame speed
Not to scale. The amount of deflection depends on charge ÷ mass, so for the same charge, the lighter particle is deflected more.
Exam tip

For ions, the same rule applies: deflection ∝ charge/mass. A 2H+ ion is deflected half as much as a 1H+ ion, and a 4He2+ ion the same amount as 2H+.

Counting protons, neutrons and electrons

Proton number (atomic number), Z
The number of protons in the nucleus. It identifies the element.
Nucleon number (mass number), A
The total number of protons and neutrons in the nucleus.

A nuclide is written 3717Cl, with the nucleon number at the top and the proton number at the bottom. So:

  • number of protons = Z
  • number of neutrons = A − Z
  • number of electrons = Z − (charge on the ion). A 2+ ion has lost 2 electrons; a 2− ion has gained 2.
SpeciesProtonsNeutronsElectrons
3717Cl172017
4020Ca2+202018
3216S2−161618
2713Al3+131410

Atomic radius and ionic radius

The atomic radius is usually measured as half the distance between the nuclei of two bonded atoms of the same element. Two trends matter:

Across a period: radius decreases

The nuclear charge increases by one each time, but the electrons go into the same shell, so shielding hardly changes. The outer electrons are pulled closer to the nucleus.

Down a group: radius increases

Each element has one more shell of electrons. The outer electrons are further from the nucleus and more shielded by the inner shells, which outweighs the larger nuclear charge.

Ions compared with atoms

  • Positive ions are smaller than their atoms. A metal atom such as Na loses its whole outer shell, and the remaining electrons are pulled in more strongly because there are now more protons than electrons.
  • Negative ions are larger than their atoms. The extra electron(s) increase repulsion between electrons, while the nuclear charge stays the same, so the electron cloud spreads out.

An isoelectronic series (ions with the same number of electrons) shows the effect of nuclear charge clearly. All of these ions have the configuration 1s2 2s2 2p6:

IonN3−O2−F−Na+Mg2+Al3+
Protons789111213
Ionic radius / nm0.1710.1400.1360.0950.0650.050

Ten electrons are held by more and more protons, so the radius falls steadily.

Across Period 3

NaMgAlSiPSCl
Atomic radius / nm0.1860.1600.1430.1170.1100.1040.099
IonNa+Mg2+Al3+Si4+P3−S2−Cl−
Ionic radius / nm0.0950.0650.0500.0410.2120.1840.181

The cations Na+ to Si4+ all have 2 shells of electrons and rising nuclear charge, so they get smaller. The anions P3− to Cl− have 3 full shells, which is why there is a jump in size between Si4+ and P3−. They then get smaller too, as nuclear charge rises.

Isotopes

Isotopes
Atoms of the same element with the same number of protons but different numbers of neutrons.

For example, chlorine has two isotopes, 3517Cl (18 neutrons) and 3717Cl (20 neutrons). Carbon has 126C, 136C and the radioactive 146C.

Same chemical properties

Isotopes have the same number of electrons in the same arrangement. Chemical reactions involve the electrons, so isotopes react in the same way.

Different physical properties

Different numbers of neutrons give different masses, so isotopes (and their compounds) have different masses and densities. For example, "heavy water", D2O, is about 11% denser than H2O.

Shells, sub-shells and orbitals

Shell
A main energy level, numbered by the principal quantum number, n = 1, 2, 3, … The higher n is, the further the electrons are from the nucleus and the higher their energy.
Sub-shell
A group of orbitals of the same type within a shell: s, p, d or f.
Orbital
A region of space around the nucleus that can hold up to two electrons, with opposite spins.
Ground state
The lowest-energy arrangement of the electrons in an atom.
Sub-shellNumber of orbitalsMaximum electrons
s12
p36
d510
f714
Shell, nSub-shellsOrbitalsMaximum electrons (2n2)
11s12
22s, 2p48
33s, 3p, 3d918
44s, 4p, 4d, 4f1632

The order of filling

energy1s2s2p3s3p4s3d4p4s is slightly lower in energythan 3d, so it fills first
Each box is one orbital. The order of increasing energy is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p.

Three rules decide where the electrons go in the ground state:

  1. Lowest energy first: electrons fill the sub-shell of lowest energy first, so 4s fills before 3d.
  2. Two per orbital: an orbital holds at most two electrons, and they must have opposite spins (shown ↑↓).
  3. Single first (Hund's rule): in a sub-shell with several orbitals of equal energy, electrons go into separate orbitals with parallel spins before they pair up. This is because two electrons in the same orbital repel each other more than two electrons in different orbitals.

Writing electronic configurations

The full configuration lists every sub-shell with the number of electrons as a superscript. The shorthand form starts with the previous noble gas in square brackets.

AtomFull configurationShorthand
N (7)1s2 2s2 2p3[He] 2s2 2p3
Na (11)1s2 2s2 2p6 3s1[Ne] 3s1
Cl (17)1s2 2s2 2p6 3s2 3p5[Ne] 3s2 3p5
K (19)1s2 2s2 2p6 3s2 3p6 4s1[Ar] 4s1
Fe (26)1s2 2s2 2p6 3s2 3p6 3d6 4s2[Ar] 3d6 4s2
Cr (24)1s2 2s2 2p6 3s2 3p6 3d5 4s1[Ar] 3d5 4s1
Cu (29)1s2 2s2 2p6 3s2 3p6 3d10 4s1[Ar] 3d10 4s1
Br (35)1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5[Ar] 3d10 4s2 4p5

Cambridge writes 3d before 4s, grouping each shell together, even though 4s fills first. Either order is accepted.

The two exceptions: chromium and copper

Chromium is [Ar] 3d5 4s1, not [Ar] 3d4 4s2, and copper is [Ar] 3d10 4s1, not [Ar] 3d9 4s2. The 3d and 4s energies are very close. For Cr, putting one electron in each of the five 3d orbitals avoids pairing two electrons in 4s, which lowers the repulsion. For Cu, a full 3d sub-shell with one 4s electron is the lower-energy arrangement.

Configurations of ions

  • Negative ions: add electrons in the normal filling order. O2− is 1s2 2s2 2p6.
  • Positive ions of the d-block: remove the 4s electrons first, then 3d. Once the 3d orbitals contain electrons, 4s is the higher-energy, outer sub-shell.
IonConfigurationIonConfiguration
Fe2+[Ar] 3d6Cu2+[Ar] 3d9
Fe3+[Ar] 3d5Cr3+[Ar] 3d3
Zn2+[Ar] 3d10Cl−[Ne] 3s2 3p6

Electrons in boxes

Each box is an orbital and each arrow is an electron. Arrows in the same box point in opposite directions. Use the explorer below to see the boxes for any element up to krypton, and for its ions.

Shapes of s and p orbitals

xzyxzyxzyxzyspxpypz
An s orbital is spherical. A p orbital is dumb-bell shaped. The three p orbitals in a sub-shell point along the x, y and z axes, at 90° to each other.

Orbitals in higher shells are the same shape but larger: a 2s orbital is bigger than a 1s orbital.

Free radicals

A free radical is a species with one or more unpaired electrons. It is shown with a dot, for example the chlorine atom Cl• and the methyl radical CH3•. Nitrogen monoxide, NO, and nitrogen dioxide, NO2, are also free radicals because each molecule has an odd number of electrons. Radicals are very reactive, as you will see in Topic 14.

Ionisation energy

First ionisation energy
The energy needed to remove one electron from each atom in one mole of gaseous atoms, forming one mole of gaseous 1+ ions.
Mg(g) → Mg+(g) + e− first ionisation energy of magnesium, +738 kJ mol−1
Mg+(g) → Mg2+(g) + e− second ionisation energy of magnesium, +1451 kJ mol−1

Each successive ionisation removes one more electron from the ion formed in the step before. All ionisation energies are positive (endothermic), because energy is needed to pull an electron away from the attraction of the nucleus. Always include the (g) state symbols.

What decides the size of an ionisation energy?

Ionisation energy measures the attraction between the nucleus and the electron being removed. Four factors control it:

  1. Nuclear charge: more protons attract the electron more strongly.
  2. Distance from the nucleus (atomic or ionic radius): an electron further out is attracted less.
  3. Shielding: inner shells and sub-shells repel the outer electron and reduce the attraction it feels.
  4. Spin-pair repulsion: an electron sharing an orbital with another is repelled by it, so it is slightly easier to remove.

Trends in first ionisation energy

HHeLiBeBCNOFNeNaMgAlSiPSClArKCa05001000150020002500atomic number (element)first IE / kJ mol⁻¹2p startsfirst pair in 2p3p startsfirst pair in 3p
First ionisation energies of hydrogen to calcium. Each period starts low with a Group 1 metal and ends high with a noble gas.
Across a period: general increase

Nuclear charge increases, the electrons are in the same shell with similar shielding, and the atomic radius decreases. The outer electron is held more strongly.

Down a group: decrease

The outer electron is in a shell further from the nucleus and is shielded by more inner shells. This outweighs the increase in nuclear charge.

There are two small dips in each period, which give evidence for sub-shells:

DipExplanation
Be → B and Mg → AlThe electron removed from B is in a 2p orbital (from Al, 3p). This is higher in energy than the 2s (3s), slightly further from the nucleus, and shielded by the filled s sub-shell, so less energy is needed.
N → O and P → SIn O ([He] 2s2 2p4), one 2p orbital contains a pair of electrons. The repulsion between the paired electrons makes one of them easier to remove than any 2p electron in N, where all three are unpaired.

The big drop from a noble gas to the next Group 1 metal (Ne → Na) happens because the outer electron of Na is in a new shell, further away and shielded by two full inner shells.

Successive ionisation energies

For any element, each successive ionisation energy is larger than the one before. After each electron is removed, the remaining electrons are attracted by the same number of protons, there is less repulsion between them, and the ion is smaller.

A big jump appears whenever the next electron has to come from a shell closer to the nucleus. Counting the electrons removed before the first big jump gives the number of outer-shell electrons, and so the group.

123456789101133.544.555.5number of the electron removedlog₁₀(IE / kJ mol⁻¹)3s2s and 2p: 8 electrons1s
Sodium's 11 successive ionisation energies, on a log scale, show its configuration 2, 8, 1: one easily removed 3s electron, then eight from the second shell, then two from the first shell. Within the second shell there is a slightly bigger step between the 7th electron (the last from 2p) and the 8th (the first from 2s).

Worked example: finding the group

The first five ionisation energies of an element X, in kJ mol−1, are 578, 1817, 2745, 11 577 and 14 842. Which group is X in?

  1. Look at the ratio between each pair: 1817/578 ≈ 3.1, 2745/1817 ≈ 1.5, 11 577/2745 ≈ 4.2, 14 842/11 577 ≈ 1.3.
  2. The big jump is between the 3rd and 4th ionisation energies, so 3 electrons are in the outer shell.

X is in Group 13. (These are the values for aluminium, [Ne] 3s2 3p1.)

Questions

Questions 1 to 5 are multiple choice, like Paper 1. Questions 6 to 8 are structured, like Paper 2.

1Paper 1 style

How many protons, neutrons and electrons are in one 5626Fe3+ ion?

  1. 26 protons, 30 neutrons, 26 electrons
  2. 26 protons, 30 neutrons, 23 electrons
  3. 26 protons, 56 neutrons, 23 electrons
  4. 30 protons, 26 neutrons, 29 electrons

Protons = Z = 26. Neutrons = 56 − 26 = 30. A 3+ ion has lost 3 electrons, so 26 − 3 = 23 electrons.

2Paper 1 style

Which species has the electronic configuration 1s2 2s2 2p6 3s2 3p6 3d5?

  1. Cr
  2. Fe2+
  3. Fe3+
  4. Cu+

Fe is [Ar] 3d6 4s2. Fe3+ loses the two 4s electrons and one 3d electron, leaving [Ar] 3d5. Cr is [Ar] 3d5 4s1 (it still has a 4s electron), Fe2+ is 3d6 and Cu+ is 3d10.

3Paper 1 style

In which pair does the second element have a lower first ionisation energy than the first element?

  1. C, N
  2. N, O
  3. O, F
  4. Na, Mg

Oxygen's first ionisation energy is lower than nitrogen's because the electron removed from O is one of a pair in a 2p orbital, and spin-pair repulsion makes it easier to remove. The other pairs follow the normal increase across the period.

4Paper 1 style

A beam of particles passes between two charged plates. The beam is deflected towards the positive plate, through a larger angle than a beam of protons travelling at the same speed. What could the particles be?

  1. neutrons
  2. 2H+ ions
  3. 4He2+ ions
  4. electrons

Deflection towards the positive plate means the particles are negative. Electrons have a charge of −1 and a tiny mass, so their charge/mass ratio is far larger than a proton's.

5Paper 1 style

The first five ionisation energies of an element, in kJ mol−1, are 590, 1145, 4912, 6491 and 8153. In which group of the Periodic Table is the element?

  1. 1
  2. 2
  3. 13
  4. 14

The biggest jump is from 1145 to 4912, after the second electron. So there are 2 outer electrons and the element is in Group 2. (These are the values for calcium.)

6Paper 2 style
(a)
Define the term first ionisation energy.
[2]
(b)
Write an equation, with state symbols, for the second ionisation energy of aluminium.
[1]
(c)
Explain why the first ionisation energy generally increases across Period 3.
[3]
(d)
Explain why the first ionisation energy of aluminium is lower than that of magnesium.
[2]
  • Energy needed to remove one electron from each atom in one mole of atoms
  • …of gaseous atoms, to form one mole of gaseous 1+ ions
  • Al+(g) → Al2+(g) + e−
  • Nuclear charge (number of protons) increases
  • Electrons are added to the same shell, so shielding is similar
  • Atomic radius decreases, so the attraction between the nucleus and the outer electron increases
  • The electron removed from Al is in a 3p orbital; in Mg it is in 3s
  • 3p is higher in energy / further from the nucleus / shielded by the 3s electrons, so less energy is needed

M3 needs both state symbols and the correct charges.

7Paper 2 style
(a)
Give the full electronic configurations of (i) a chromium atom, (ii) a Cu2+ ion and (iii) a S2− ion.
[3]
(b)
Draw the electrons-in-boxes diagram for the 3d and 4s sub-shells of an Fe2+ ion, and state how many unpaired electrons it has.
[2]
(c)
Sketch the shape of a p orbital.
[1]
(d)
State what is meant by a free radical.
[1]
  • Cr: 1s2 2s2 2p6 3s2 3p6 3d5 4s1
  • Cu2+: 1s2 2s2 2p6 3s2 3p6 3d9
  • S2−: 1s2 2s2 2p6 3s2 3p6
  • 3d: ↑↓ ↑ ↑ ↑ ↑ (one pair and four single electrons, all singles parallel); 4s: empty box
  • 4 unpaired electrons
  • A dumb-bell shape (two lobes on opposite sides of the nucleus)
  • A species with one or more unpaired electrons
8Paper 2 style

Magnesium has three naturally occurring isotopes: 24Mg (78.99%), 25Mg (10.00%) and 26Mg (11.01%).

(a)
State what is meant by the term isotopes.
[1]
(b)
Calculate the relative atomic mass of magnesium to 3 significant figures.
[2]
(c)
Explain why 24Mg and 26Mg have the same chemical properties.
[1]
(d)
The radius of a Mg2+ ion is much smaller than the radius of a Mg atom. Explain why.
[2]
  • Atoms of the same element (same number of protons) with different numbers of neutrons
  • (24 × 78.99 + 25 × 10.00 + 26 × 11.01) ÷ 100
  • = 24.3
  • They have the same number of electrons / the same electronic configuration (and chemical reactions involve electrons)
  • Mg2+ has lost the whole third (outer) shell
  • The same nuclear charge now attracts fewer electrons, with less repulsion between them, so they are pulled closer

Stoichiometry is the arithmetic of chemistry: using formulas and balanced equations to work out how much of each substance reacts and forms. Almost every Paper 2 question and every titration in Paper 3 uses it, so practise until it is automatic.

Relative masses

Atoms are far too light to weigh in grams, so their masses are compared with a standard, the carbon-12 atom.

Unified atomic mass unit, u
One twelfth of the mass of one atom of carbon-12. (1 u = 1.66 × 10−27 kg.)
Relative isotopic mass
The mass of one atom of an isotope compared with one twelfth of the mass of one atom of carbon-12.
Relative atomic mass, Ar
The weighted average mass of the atoms of an element (taking account of the abundance of its isotopes), compared with one twelfth of the mass of one atom of carbon-12.
Relative molecular mass, Mr
The weighted average mass of a molecule, compared with one twelfth of the mass of one atom of carbon-12. It is the sum of the Ar values of the atoms in the molecule.
Relative formula mass
The same idea for ionic compounds, which do not contain molecules, for example NaCl. It is often still written Mr.

Relative masses have no units, because they are ratios.

Worked example: relative atomic mass from isotopic abundances

Chlorine is 75.77% 35Cl and 24.23% 37Cl. Calculate Ar(Cl).

  1. Multiply each relative isotopic mass by its percentage abundance and add: 35 × 75.77 + 37 × 24.23 = 2651.95 + 896.51 = 3548.46
  2. Divide by the total abundance, 100.

Ar(Cl) = 35.48, so 35.5 to 3 significant figures.

The abundances often come from a mass spectrum, where the height of each peak is proportional to the abundance of that isotope. See Topic 22.

The mole and the Avogadro constant

Mole
The amount of substance that contains 6.02 × 1023 particles (atoms, molecules, ions or electrons). This number is the Avogadro constant, L.
Molar mass, M
The mass of one mole of a substance, in g mol−1. It has the same numerical value as Ar or Mr.
Amount from massn = mMLearn it
Number of particlesN = n × LLearn it
Gas at room conditionsn = V (dm3)24.0Vm in Data Booklet
Solutionn = c × V (dm3)Learn it
amount, nin molmass, m, in gnumber of particles, Ngas volume, V, in dm³solution: c and V (dm³)÷ M× M÷ L× L÷ 24.0× 24.0c × Vn ÷ V
Every calculation goes through the amount in moles. Convert what you are given into moles, use the equation's ratio, then convert to what you are asked for.

Worked example: particles

How many molecules, and how many atoms, are in 9.0 g of water? (Mr of H2O = 18.0)

  1. n = 9.0/18.0 = 0.50 mol
  2. molecules = 0.50 × 6.02 × 1023 = 3.01 × 1023
  3. Each molecule has 3 atoms, so atoms = 3 × 3.01 × 1023 = 9.03 × 1023

3.0 × 1023 molecules and 9.0 × 1023 atoms (2 s.f., matching the data)

Formulas

Ions you must know

The charge on a simple ion can be predicted from the group: Group 1 forms 1+, Group 2 forms 2+, Group 13 forms 3+, Group 15 forms 3−, Group 16 forms 2− and Group 17 forms 1−. Learn these compound ions and two common d-block ions:

IonFormulaIonFormula
nitrateNO3−ammoniumNH4+
carbonateCO32−hydrogencarbonateHCO3−
sulfateSO42−phosphatePO43−
hydroxideOH−zincZn2+
silverAg+These nine are the ions the syllabus lists.

For elements with more than one possible charge, a Roman numeral gives the oxidation number: iron(II) is Fe2+, iron(III) is Fe3+, copper(II) is Cu2+, and manganese(IV) oxide is MnO2.

Writing formulas of ionic compounds

The total positive charge must equal the total negative charge. Use brackets when there is more than one compound ion.

CompoundIonsFormula
aluminium sulfateAl3+ and SO42−: 2 × (+3) = 3 × (−2)Al2(SO4)3
calcium phosphateCa2+ and PO43−: 3 × (+2) = 2 × (−3)Ca3(PO4)2
ammonium carbonateNH4+ and CO32−(NH4)2CO3
iron(III) hydroxideFe3+ and OH−Fe(OH)3
silver carbonateAg+ and CO32−Ag2CO3

Equations

A balanced equation has the same number of each type of atom, and the same total charge, on both sides. State symbols show the physical state: (s) solid, (l) liquid, (g) gas, (aq) dissolved in water.

An ionic equation shows only the species that change. Spectator ions, which are present but unchanged, are left out.

ReactionIonic equation
Any acid with any alkaliH+(aq) + OH−(aq) → H2O(l)
Silver nitrate solution with sodium chloride solutionAg+(aq) + Cl−(aq) → AgCl(s)
Zinc with copper(II) sulfate solutionZn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
A carbonate solution with an acidCO32−(aq) + 2H+(aq) → CO2(g) + H2O(l)
Barium chloride solution with a sulfate solutionBa2+(aq) + SO42−(aq) → BaSO4(s)

Empirical and molecular formulas

Empirical formula
The simplest whole-number ratio of the atoms of each element in a compound.
Molecular formula
The actual number of atoms of each element in one molecule of the compound.

For example, ethane C2H6 has the empirical formula CH3. Glucose C6H12O6 has the empirical formula CH2O.

Worked example: from percentage composition

A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its Mr is 56.0. Find its empirical and molecular formulas.

CH
mass in 100 g85.714.3
÷ Ar85.7 ÷ 12.0 = 7.1414.3 ÷ 1.0 = 14.3
÷ smallest12.00

Empirical formula CH2, with empirical formula mass 14.0. Then 56.0 ÷ 14.0 = 4.

The molecular formula is C4H8.

Worked example: from combustion data

Complete combustion of 0.230 g of a compound containing only C, H and O gives 0.440 g of CO2 and 0.270 g of H2O. Find its empirical formula.

  1. C: n(CO2) = 0.440/44.0 = 0.0100 mol, so 0.0100 mol C, mass 0.120 g
  2. H: n(H2O) = 0.270/18.0 = 0.0150 mol, so 0.0300 mol H, mass 0.0300 g
  3. O by difference: 0.230 − 0.120 − 0.0300 = 0.0800 g, so 0.0800 / 16.0 = 0.00500 mol O
  4. Ratio C : H : O = 0.0100 : 0.0300 : 0.00500 = 2 : 6 : 1

The empirical formula is C2H6O.

Hydrated salts

Hydrated
A crystalline compound that contains water molecules in its lattice, such as CuSO4·5H2O.
Water of crystallisation
The water molecules that are part of the crystal structure of a hydrated salt.
Anhydrous
A substance with no water of crystallisation, such as CuSO4. It forms when the hydrated salt is heated.

Worked example: finding the water of crystallisation

5.00 g of hydrated copper(II) sulfate, CuSO4.xH2O, is heated to constant mass. 3.20 g of anhydrous CuSO4 remains. Find x.

  1. n(CuSO4) = 3.20/159.6 = 0.0200 mol
  2. mass of water lost = 5.00 − 3.20 = 1.80 g, so n(H2O) = 1.80/18.0 = 0.100 mol
  3. ratio = 0.100 ÷ 0.0200 = 4.99, which rounds to 5

x = 5, so the formula is CuSO4·5H2O. Heating "to constant mass" makes sure all the water has gone.

Reacting masses and percentage yield

To find the mass of a product: (1) find the moles of the substance you know, (2) use the ratio from the balanced equation, (3) convert back to mass.

Worked example

Iron is extracted by Fe2O3 + 3CO → 2Fe + 3CO2. What mass of iron can be made from 800 g of Fe2O3? (Mr: Fe2O3 159.6; Ar: Fe 55.8)

  1. n(Fe2O3) = 800/159.6 = 5.01 mol
  2. 1 mol Fe2O3 gives 2 mol Fe, so n(Fe) = 10.0 mol
  3. mass = 10.03 × 55.8 = 559 g

559 g of iron. If only 500 g is actually obtained, the percentage yield = 500 / 559 × 100 = 89.4%.

Percentage yield% yield = actual yieldtheoretical yield × 100Learn it

Yields are below 100% because the reaction may be reversible or incomplete, side reactions form other products, and some product is lost when it is separated and purified.

Volumes of gases

One mole of any gas occupies the same volume at the same temperature and pressure: 24.0 dm3 at room conditions, or 22.4 dm3 at s.t.p. So for gases, the ratio of volumes equals the ratio of moles in the equation.

Worked example: finding a hydrocarbon's formula from gas volumes

10 cm3 of a gaseous hydrocarbon CxHy is burned in 80 cm3 of oxygen (an excess). After cooling to room temperature, 60 cm3 of gas remains. This decreases to 30 cm3 when the gas is passed through sodium hydroxide solution, which absorbs CO2. All volumes are at room conditions.

  1. After cooling, the water is a liquid, so the 60 cm3 is CO2 plus unreacted O2.
  2. NaOH removes CO2: volume of CO2 = 60 − 30 = 30 cm3. The 30 cm3 left is unreacted O2.
  3. O2 used = 80 − 30 = 50 cm3.
  4. Ratio CxHy : O2 : CO2 = 10 : 50 : 30 = 1 : 5 : 3, so x = 3.
  5. C3Hy + 5O2 → 3CO2 + y/2H2O. Balancing O: 10 = 6 + y/2, so y = 8.

The hydrocarbon is propane, C3H8.

Solutions and titrations

Concentration is the amount of solute in 1 dm3 of solution, in mol dm−3. Multiply by M to get g dm−3. Volumes measured in cm3 must be divided by 1000 to get dm3.

Worked example: a titration

25.0 cm3 of sodium carbonate solution is titrated with 0.100 mol dm−3 hydrochloric acid. The mean titre is 22.50 cm3. Find the concentration of the sodium carbonate.

Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)

  1. n(HCl) = 0.100 × 22.50/1000 = 2.250 × 10−3 mol
  2. ratio 1 : 2, so n(Na2CO3) = 1.125 × 10−3 mol
  3. c = 1.125 × 10−3/(25.0/1000) = 0.0450 mol dm−3

0.0450 mol dm−3, which is 0.0450 × 106.0 = 4.77 g dm−3

Limiting and excess reagents

The limiting reagent is the one that is completely used up. It decides how much product forms. The other reagent is in excess, and some of it is left over. To decide which is which, use the equation to find how much of one reactant is needed to react with all of the other.

Worked example

5.00 g of zinc is added to 50.0 cm3 of 2.00 mol dm−3 hydrochloric acid. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

  1. n(Zn) = 5.00 / 65.4 = 0.0765 mol. n(HCl) = 2.00 × 0.0500 = 0.100 mol.
  2. 0.100 mol HCl reacts with only 0.0500 mol Zn, so HCl is the limiting reagent and zinc is in excess.
  3. n(H2) = 0.100 ÷ 2 = 0.0500 mol, so volume = 0.0500 × 24.0 = 1.20 dm3.
  4. Zinc left = (0.0765 − 0.0500) × 65.4 = 1.73 g.
Significant figures

Give your answer to the same number of significant figures as the least precise data (usually 3). Keep at least one extra figure in intermediate steps, and only round at the end. Rounding too early is a common way to lose the final mark.

Questions

Use Ar values: H 1.0, C 12.0, O 16.0, Mg 24.3, S 32.1, Ca 40.1. Questions 1 to 5 are multiple choice. Questions 6 to 8 are structured.

1Paper 1 style

Which sample contains the greatest number of atoms?

  1. 1.0 g of H2
  2. 4.0 g of He
  3. 9.0 g of H2O
  4. 16 g of O2

Moles of atoms: A 0.50 mol H2 × 2 = 1.0 mol; B 1.0 mol; C 0.50 mol H2O × 3 = 1.5 mol; D 0.50 mol O2 × 2 = 1.0 mol. So C has the most atoms.

2Paper 1 style

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. What is its empirical formula?

  1. CHO
  2. CH2O
  3. C2H4O
  4. CH3O

C: 40.0 / 12.0 = 3.33; H: 6.7 / 1.0 = 6.7; O: 53.3 / 16.0 = 3.33. Dividing by 3.33 gives 1 : 2 : 1.

3Paper 1 style

5.00 g of calcium carbonate is heated until it has completely decomposed: CaCO3(s) → CaO(s) + CO2(g). What volume of carbon dioxide forms, measured at room conditions?

  1. 0.600 dm3
  2. 1.12 dm3
  3. 1.20 dm3
  4. 2.40 dm3

n(CaCO3) = 5.00 / 100.1 = 0.0500 mol, giving 0.0500 mol CO2. At room conditions, 0.0500 × 24.0 = 1.20 dm3. (B uses the s.t.p. value, 22.4.)

4Paper 1 style

25.0 cm3 of 0.100 mol dm−3 sodium hydroxide is exactly neutralised by 20.0 cm3 of sulfuric acid. What is the concentration of the sulfuric acid?

  1. 0.0625 mol dm−3
  2. 0.125 mol dm−3
  3. 0.0500 mol dm−3
  4. 0.250 mol dm−3

2NaOH + H2SO4 → Na2SO4 + 2H2O. n(NaOH) = 2.50 × 10−3 mol, so n(H2SO4) = 1.25 × 10−3 mol. c = 1.25 × 10−3 / 0.0200 = 0.0625 mol dm−3. Answer B forgets the 2 : 1 ratio.

5Paper 1 style

What is the formula of aluminium sulfate?

  1. AlSO4
  2. Al2SO4
  3. Al3(SO4)2
  4. Al2(SO4)3

Two Al3+ ions (+6) balance three SO42− ions (−6).

6Paper 2 style

A student heats 2.46 g of hydrated magnesium sulfate, MgSO4.xH2O, until all the water of crystallisation is removed. 1.20 g of anhydrous MgSO4 remains.

(a)
Explain what is meant by water of crystallisation.
[1]
(b)
Calculate the value of x.
[3]
(c)
Describe how the student could make sure that all of the water had been removed.
[1]
  • Water molecules that are part of / chemically bonded within the crystal lattice
  • n(MgSO4) = 1.20 / 120.4 = 9.97 × 10−3 mol
  • mass of water = 1.26 g, so n(H2O) = 1.26 / 18.0 = 0.0700 mol
  • ratio 0.0700 / 0.00997 = 7.02, so x = 7
  • Heat, cool and weigh repeatedly until the mass stays the same (heat to constant mass)
7Paper 2 style

4.00 g of magnesium ribbon is added to 100 cm3 of 0.800 mol dm−3 hydrochloric acid.

(a)
Write the equation for the reaction, including state symbols.
[1]
(b)
Show by calculation which reagent is in excess.
[2]
(c)
Calculate the volume of hydrogen formed at room conditions.
[1]
(d)
Calculate the mass of magnesium left at the end.
[1]
(e)
Write the ionic equation for the reaction.
[1]
  • Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
  • n(Mg) = 4.00 / 24.3 = 0.165 mol; n(HCl) = 0.800 × 0.100 = 0.0800 mol
  • 0.0800 mol HCl needs only 0.0400 mol Mg, so Mg is in excess
  • n(H2) = 0.0400 mol, so V = 0.0400 × 24.0 = 0.960 dm3
  • 4.00 − 0.0400 × 24.3 = 3.03 g
  • Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g)
8Paper 2 style
(a)
When 10.0 g of calcium carbonate is heated strongly, 4.80 g of calcium oxide is obtained. Calculate the percentage yield.
[2]
(b)
Suggest one reason why the yield is less than 100%.
[1]
(c)
20 cm3 of ethane, C2H6, is burned completely in 100 cm3 of oxygen. Calculate the total volume of gas left after cooling to room temperature, and state what it consists of. All volumes are measured at the same temperature and pressure.
[3]
  • Theoretical mass of CaO = (10.0 / 100.1) × 56.1 = 5.60 g
  • % yield = 4.80 / 5.60 × 100 = 85.6%
  • Decomposition was incomplete (not heated for long enough) / some solid was lost
  • 2C2H6 + 7O2 → 4CO2 + 6H2O, so 20 cm3 C2H6 uses 70 cm3 O2
  • 40 cm3 CO2 forms, and 30 cm3 O2 is left over (water is liquid after cooling)
  • Total 70 cm3: 40 cm3 CO2 and 30 cm3 O2

Bonds hold atoms together; intermolecular forces hold molecules together. This topic explains both, shows you how to draw dot-and-cross diagrams, and uses electron-pair repulsion to predict the shapes of molecules.

Electronegativity

Electronegativity
The power of an atom to attract electrons to itself (in a covalent bond).

Electronegativity depends on the same factors as ionisation energy: nuclear charge, atomic radius and shielding by inner shells and sub-shells. A small atom with a large nuclear charge and little shielding attracts bonding electrons strongly.

Across a period: increases

Nuclear charge increases while shielding stays about the same and the radius decreases, so bonding electrons are attracted more strongly.

Down a group: decreases

The atoms get bigger and have more shielding, so the bonding electrons are further from the nucleus and attracted less.

Fluorine is the most electronegative element. On the Pauling scale (values given in the exam when needed):

HCNOFNaMgAlSiPSClBrI
2.202.553.043.443.980.931.311.611.902.192.583.162.962.66

Predicting the type of bond

The difference in electronegativity between two atoms tells you how unevenly the bonding electrons are shared:

DifferenceBondExamples (difference)
zeronon-polar covalent: electrons shared equallyCl2 (0), H2 (0)
small to moderatepolar covalent: electrons shared unequally, giving partial charges δ+ and δ−HCl (0.96), O–H (1.24), AlCl3 (1.55)
large (roughly 2 or more)ionic: electrons transferredNaCl (2.23), MgO (2.13), NaF (3.05)

There is no sharp cut-off, so treat the boundary as a guide. Aluminium chloride is a useful case: its difference (1.55) is too small for ionic bonding, and it is indeed a covalent compound that forms Al2Cl6 molecules.

Ionic bonding

Ionic bonding
The electrostatic attraction between oppositely charged ions (positive cations and negative anions).

Ions form when a metal atom transfers its outer electrons to a non-metal atom, so that both end up with a noble gas configuration. In a dot-and-cross diagram, show only the outer shells, put each ion in square brackets, and write the charge at the top right.

Na+Cl−sodium chloride
Mg2+O2−magnesium oxide
F−Ca2+F−calcium fluoride
Outer shells only. Crosses are electrons that came from the metal; dots came from the non-metal. The metal ions are drawn with their new outer shell, the full shell below the one they lost (you may also draw them with an empty outer shell). Calcium fluoride has two fluoride ions for each calcium ion.

In the solid, each ion is surrounded by ions of the opposite charge in a giant ionic lattice (Topic 4). Ionic bonding is stronger when the ions have larger charges and smaller radii, which is why magnesium oxide (2+ and 2−) melts at 2852 °C and sodium chloride (1+ and 1−) at 801 °C.

Metallic bonding

Metallic bonding
The electrostatic attraction between positive metal ions and delocalised electrons.

The outer electrons of the metal atoms are not tied to any one atom: they move freely through a lattice of positive ions (a "sea" of delocalised electrons). Metallic bonding is stronger when there are more delocalised electrons per atom and the ions are smaller and more highly charged. This explains why the melting points rise from Na (98 °C) to Mg (650 °C) to Al (660 °C), and why metals conduct electricity.

Covalent bonding

Covalent bond
The electrostatic attraction between the nuclei of two atoms and a shared pair of electrons.

Each shared pair is one bond. Two shared pairs make a double bond and three make a triple bond. Pairs that are not shared are called lone pairs.

HHH2
ClClCl2
HClHCl
OOO2double bond
NNN2triple bond
OCOCO2
NHHHNH3
CHHHHCH4
OHHH2O
CCHHHHHHethane, C2H6
CCHHHHethene, C2H4
Outer shells only. Each overlap region holds the shared pairs. Notice the lone pairs: one on N in NH3, two on O in H2O, and three on each Cl.

Expanding the octet

Elements in Period 2 can hold at most 8 electrons in their outer shell, because the second shell has only 2s and 2p orbitals. Elements from Period 3 onwards can use empty d orbitals, so they can have more than 8 outer electrons. Examples: sulfur dioxide (10 electrons around S), phosphorus(V) chloride PCl5 (10 around P) and sulfur hexafluoride SF6 (12 around S). This is why nitrogen forms only NCl3 but phosphorus forms both PCl3 and PCl5.

SOOSO210 electrons around S
PClClClClClPCl510 electrons around P
Dot-and-cross diagrams don't show shape: SO2 is actually bent (see below).

Coordinate (dative covalent) bonding

Coordinate bond
A covalent bond in which both electrons in the shared pair come from the same atom.

The donor atom needs a lone pair, and the acceptor needs an empty orbital. Once formed, a coordinate bond is identical to any other covalent bond. It is often drawn as an arrow from the donor to the acceptor.

Ammonia and hydrogen chloride: the gases react to form white solid ammonium chloride. The lone pair on N forms a coordinate bond to H+: NH3(g) + HCl(g) → NH4Cl(s). All four N–H bonds in NH4+ are then identical, and the ion is tetrahedral.

Aluminium chloride: in AlCl3 the aluminium has only six outer electrons, so it has an empty orbital. In the vapour at lower temperatures, two AlCl3 molecules join to form Al2Cl6: a lone pair on a chlorine atom of each molecule forms a coordinate bond to the aluminium of the other.

+NHHHHNH4+the top N–H bond is the coordinate bond: both electrons (×) came from N
AlAlClClClClClClAl2Cl6arrows show the two coordinate bonds

σ and π bonds

A covalent bond forms when orbitals on two atoms overlap, so that the shared pair is attracted to both nuclei.

  • A σ (sigma) bond forms by direct, end-on overlap of orbitals between the two nuclei. Every single bond is a σ bond.
  • A π (pi) bond forms by sideways overlap of adjacent p orbitals, with electron density above and below the σ bond. A double bond is one σ + one π; a triple bond is one σ + two π.
σ bond: s + sfor example H₂σ bond: p + p end-ondirect overlap between the nucleiππ bond: p + p sidewaysabove and below the σ bond
The π bond has less overlap than a σ bond, so it is weaker and its electrons are more exposed to attack. This is why alkenes are much more reactive than alkanes.
Moleculeσ bondsπ bondsNotes
H210overlap of two 1s orbitals
ethane, C2H6706 C–H and 1 C–C
ethene, C2H4514 C–H, and C=C is 1 σ + 1 π
hydrogen cyanide, HC≡N22C–H, and C≡N is 1 σ + 2 π
nitrogen, N212the two π bonds are at right angles to each other

Hybridisation

Carbon's 2s and 2p orbitals mix (hybridise) to form a set of identical hybrid orbitals, which form the σ bonds. The number of hybrid orbitals equals the number of orbitals mixed.

HybridMade fromArrangementLeft overExample
sp3one s + three p4 orbitals, tetrahedral, 109.5°noneC in methane and ethane
sp2one s + two p3 orbitals, trigonal planar, 120°one p orbital, which forms the π bondC in ethene
spone s + one p2 orbitals, linear, 180°two p orbitals, which form two π bondsC in HCN and ethyne

In ethene, the π bond can only form if the p orbitals on the two carbons are parallel. So the molecule is planar and there is no rotation about C=C. This is the origin of cis/trans isomerism (Topic 13).

Bond energy and bond length

Bond energy
The energy required to break one mole of a particular covalent bond in the gaseous state.
Bond length
The internuclear distance between two covalently bonded atoms.

Some bond energies are exact: for a diatomic molecule such as H2 there is only one bond in one environment. Others are averages: a C–H bond in methane has a slightly different energy from a C–H bond in ethanol, so the table gives a mean value.

BondEnergy / kJ mol−1Length / nmBondEnergy / kJ mol−1Length / nm
C–C3500.154H–H4360.074
C=C6100.134O=O4960.121
C≡C8400.120N≡N9440.110
C–H4100.109Cl–Cl2420.199
C–O3600.143Br–Br1930.228
C=O7400.122I–I1510.267
O–H4600.096H–Cl4310.127
N–H3900.101H–I2990.161
  • Multiple bonds are shorter and stronger: C≡C > C=C > C–C. But C=C is less than twice C–C, because the π bond is weaker than the σ bond.
  • Longer bonds are generally weaker: the shared pair is further from the nuclei. H–Cl (0.127 nm, 431) is stronger than H–I (0.161 nm, 299), which is why hydrogen iodide decomposes much more easily (Topic 11).
  • Strong bonds mean low reactivity: N≡N (944 kJ mol−1) is one of the strongest bonds known, so nitrogen gas is very unreactive (Topic 12).

Shapes of molecules

Valence shell electron pair repulsion (VSEPR) theory: pairs of electrons around a central atom repel each other and move as far apart as possible. The shape depends on the number of bonding pairs and lone pairs.

  • A double or triple bond counts as one region of electron density, just like a single bond.
  • Lone pairs repel more strongly than bonding pairs, because they are closer to the central atom. The order is lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. Each lone pair squeezes the bond angles by about 2.5°.
  • The shape is named from the positions of the atoms, not the lone pairs.
Bonding pairsLone pairsShapeBond angleExample
20linear180°CO2, BeCl2
30trigonal planar120°BF3
40tetrahedral109.5°CH4, NH4+
31pyramidal107°NH3, H3O+
22non-linear (bent)104.5°H2O
50trigonal bipyramidal120° and 90°PF5, PCl5
60octahedral90°SF6
COO180°linear
BFFF120°trigonal planar
CHHHH109.5°tetrahedral
NHHH107°pyramidal
OHH104.5°non-linear
PFFFFFtrigonal bipyramidal
SFFFFFFoctahedral
A wedge points towards you; a hashed bond points away. Lone pairs are shown as pairs of dots.

Worked example: predicting a shape you haven't seen

Predict the shape of the oxonium ion, H3O+, and of sulfur dichloride, SCl2.

  1. H3O+: O has 6 outer electrons. Each H adds 1 (total 9), and the positive charge removes 1, leaving 8 electrons = 4 pairs. Three are bonding pairs (to H) and one is a lone pair, so the shape is pyramidal, about 107°, like NH3.
  2. SCl2: S has 6 outer electrons, and each Cl adds 1 for its bond (total 8) = 4 pairs. Two bonding pairs and two lone pairs give a non-linear shape with an angle of about 104.5°, like water.

More analogous examples: NH2− is non-linear (2 bonding, 2 lone pairs); CO32− and NO3− are trigonal planar (3 regions, no lone pairs on the central atom); SO2 is non-linear, about 120° (2 double bonds and 1 lone pair on S); PCl4+ is tetrahedral; PCl6− is octahedral.

Intermolecular forces

Van der Waals' forces is the general name for all the attractive forces between molecules, other than those due to bond formation. There are two types in this course:

Instantaneous dipole–induced dipole (id-id) forces

Also called London dispersion forces, these act between all molecules, polar or not. Electrons are always moving, so at any instant the electron cloud may be unevenly spread, giving a temporary (instantaneous) dipole. This induces a dipole in a neighbouring molecule, and the two attract. The forces are stronger when:

  • there are more electrons (larger molecules), so bigger dipoles form. Boiling points rise from F2 (−188 °C) to Cl2 (−34 °C), Br2 (59 °C) and I2 (184 °C).
  • there is more surface contact between molecules. Pentane (straight chain, bp 36 °C) boils higher than its isomer 2,2-dimethylpropane (nearly spherical, bp 10 °C).

Permanent dipole–permanent dipole (pd-pd) forces

A polar molecule has a permanent dipole: one end δ+ and the other δ−. The δ+ end of one molecule attracts the δ− end of the next. These forces act in addition to id-id forces. For example, propanone (CH3COCH3, Mr 58, bp 56 °C) boils much higher than butane (C4H10, Mr 58, bp −1 °C), which has a similar number of electrons but no permanent dipole.

Hydrogen bonding

Hydrogen bonding is a special, strong case of permanent dipole–permanent dipole attraction. It happens between molecules where hydrogen is bonded to a highly electronegative atom. In this syllabus, that means molecules with N–H or O–H groups, such as ammonia, water and alcohols. (H–F also forms hydrogen bonds.)

  1. The N–H or O–H bond is very polar, leaving the small H atom with a large δ+ charge.
  2. This H is attracted to a lone pair on an N or O atom of a neighbouring molecule.
  3. The three atoms O–H···O are in a straight line (180°), because this keeps the electron pairs as far apart as possible.
Oδ−Hδ+Hδ+Oδ−Hδ+Hδ+hydrogen bondO–H···O is close to 180°
A hydrogen bond (dashed) forms between a δ+ hydrogen of one water molecule and a lone pair on the oxygen of another.

The unusual properties of water

Each water molecule has two H atoms and two lone pairs, so it can form up to four hydrogen bonds. This explains why water:

  • has relatively high melting and boiling points. Hydrogen bonds are much stronger than the id-id forces in H2S, so more energy is needed to separate the molecules. Water boils at 100 °C; H2S, which has more electrons, boils at −60 °C.
  • has a high surface tension. Molecules at the surface are pulled strongly inwards and sideways by hydrogen bonds.
  • is less dense as ice than as liquid water. In ice, each molecule forms four hydrogen bonds in a regular tetrahedral arrangement, which holds the molecules further apart in an open lattice. When ice melts, some hydrogen bonds break and the molecules move closer together.
2345−180−120−60060120periodboiling point / °CCH₄NH₃H₂OHFGroup 14Group 15Group 16Group 17
From Period 3 downwards, the boiling points of the hydrides rise as the number of electrons (and so the id-id forces) increases. NH3, H2O and HF are far higher than the trend predicts because of hydrogen bonding. Group 14 hydrides, which cannot hydrogen bond, follow the trend all the way.

Bond polarity and polar molecules

A bond between atoms of different electronegativity is polar: the more electronegative atom gets a δ− charge. A molecule has an overall dipole moment (it is polar) only if its bond dipoles do not cancel out, which depends on its shape.

Polar molecules (dipoles don't cancel)Non-polar molecules with polar bonds (dipoles cancel)
H2O (non-linear), NH3 (pyramidal), HCl, CHCl3, CH3Cl, propanoneCO2 (linear), BF3 (trigonal planar), CH4 and CCl4 (tetrahedral), SF6 (octahedral)

How strong are the forces?

In general, ionic, covalent and metallic bonds are much stronger than intermolecular forces. Breaking a covalent bond takes hundreds of kJ mol−1. Hydrogen bonds are typically 5 to 40 kJ mol−1, and other van der Waals' forces are usually weaker still. When a simple molecular substance melts or boils, only the intermolecular forces are overcome. The covalent bonds inside the molecules stay intact.

Dot-and-cross diagram checklist

  • Show outer-shell electrons only, unless the question asks for all shells.
  • Use dots for one atom's electrons and crosses for the other's, so the reader can see where each shared pair came from.
  • Put shared pairs in the overlap between the two atoms, and include every lone pair.
  • For ions, use square brackets and show the charge. For a coordinate bond, both electrons of the pair have the donor's symbol.
  • Species with an odd number of electrons are allowed, such as the methyl radical CH3•: C has 3 shared pairs and 1 unpaired electron.

Questions

Questions 1 to 6 are multiple choice. Questions 7 to 9 are structured.

1Paper 1 style

Which molecule has a bond angle closest to 107°?

  1. H2O
  2. NH3
  3. CH4
  4. BF3

NH3 has 3 bonding pairs and 1 lone pair: pyramidal, 107°. Water is 104.5°, methane 109.5° and BF3 120°.

2Paper 1 style

Which molecule contains polar bonds but is not polar overall?

  1. NH3
  2. CHCl3
  3. CO2
  4. HCl

CO2 is linear, so its two C=O dipoles point in opposite directions and cancel. The others have dipoles that do not cancel.

3Paper 1 style

How many σ bonds and π bonds are there in one molecule of hydrogen cyanide, HCN?

  1. 2 σ and 2 π
  2. 1 σ and 3 π
  3. 3 σ and 1 π
  4. 2 σ and 1 π

H–C is one σ bond. C≡N is one σ and two π bonds. Total: 2 σ, 2 π.

4Paper 1 style

Hydrogen bonds form between molecules of which substance?

  1. methane, CH4
  2. hydrogen chloride, HCl
  3. methoxymethane, CH3OCH3
  4. methanol, CH3OH

Only methanol has an H attached to O. Methoxymethane has O with lone pairs, but no O–H, so its molecules can't hydrogen bond with each other.

5Paper 1 style

Which species contains a coordinate (dative covalent) bond?

  1. NH4+
  2. CH4
  3. H2O
  4. CO2

NH4+ forms when the lone pair on N bonds to H+, which has no electrons of its own.

6Paper 1 style

The boiling points of the halogens increase from fluorine to iodine. What is the main reason?

  1. The covalent bonds in the molecules get weaker.
  2. The molecules become more polar.
  3. The number of electrons increases, so the instantaneous dipole–induced dipole forces get stronger.
  4. Hydrogen bonding gets stronger.

Halogen molecules are non-polar, so only id-id forces act between them. These increase with the number of electrons. Boiling doesn't break the covalent bonds at all.

7Paper 2 style

Ammonia, NH3, reacts with boron trifluoride, BF3, to form the compound H3NBF3.

(a)
Draw dot-and-cross diagrams for NH3 and BF3, showing outer electrons only.
[2]
(b)
State the shape of each molecule and its bond angle.
[2]
(c)
Explain how a bond forms between N and B in H3NBF3.
[2]
(d)
Suggest the F–B–F bond angle in H3NBF3. Explain your answer.
[2]
  • NH3: three N–H shared pairs and one lone pair on N
  • BF3: three B–F shared pairs, three lone pairs on each F, and only 6 electrons around B
  • NH3 pyramidal, 107°
  • BF3 trigonal planar, 120°
  • The lone pair on N is donated to B…
  • …which has an empty orbital (electron deficient), forming a coordinate (dative covalent) bond
  • About 109.5° (allow 107° to 109.5°)
  • B now has four bonding pairs and no lone pairs, so the arrangement around B is tetrahedral
8Paper 2 style
(a)
Draw a diagram to show a hydrogen bond between two water molecules. Include partial charges and the relevant lone pair.
[3]
(b)
Explain why water has a much higher boiling point than hydrogen sulfide, H2S.
[2]
(c)
Explain why ice is less dense than liquid water.
[2]
(d)
Explain why the boiling points increase from H2S to H2Se to H2Te.
[2]
  • Two water molecules, with δ+ on H and δ− on O
  • Lone pair shown on the O of the second molecule
  • Hydrogen bond drawn between the H and that lone pair, with O–H···O roughly in a straight line
  • Water molecules are held by hydrogen bonds; H2S has only (weaker) van der Waals' forces (id-id and pd-pd)
  • Hydrogen bonds are stronger, so more energy is needed to separate the water molecules
  • In ice, each molecule forms (four) hydrogen bonds in a regular, open (tetrahedral) lattice
  • This holds the molecules further apart than in the liquid, where some hydrogen bonds are broken
  • The number of electrons in the molecules increases
  • So the instantaneous dipole–induced dipole forces get stronger and need more energy to overcome
9Paper 2 style
(a)
Describe how the σ bond and the π bond form in the C=C double bond of ethene.
[2]
(b)
State the hybridisation of the carbon atoms in ethene and the H–C–H bond angle.
[2]
(c)
The bond energy of C=C is 610 kJ mol−1 and of C–C is 350 kJ mol−1. Explain why the C=C bond energy is less than twice the C–C bond energy.
[1]
(d)
Nitrogen, N2, is much less reactive than ethene, even though both contain a multiple bond. Use the data N≡N 944 kJ mol−1 to suggest why.
[2]
  • σ: direct / end-on overlap of (sp2 hybrid) orbitals between the two carbon nuclei
  • π: sideways overlap of p orbitals, with electron density above and below the σ bond
  • sp2
  • 120°
  • The π bond is weaker than the σ bond, because there is less overlap of the p orbitals
  • N≡N has a very high bond energy, so a lot of energy is needed to break it
  • N2 is also non-polar, so it doesn't attract electrophiles or nucleophiles

The properties of a substance depend on its particles and on the forces between them. This topic covers how gases behave, how to use the ideal gas equation, and how the four types of structure explain melting points, conductivity and solubility.

The gaseous state

In a gas, the particles move rapidly and randomly in straight lines between collisions. Gas pressure comes from the particles colliding with the walls of the container: each collision exerts a force on the wall, and pressure is the total force per unit area.

Ideal gases

An ideal gas is a model that follows the gas laws exactly. Its key assumptions are:

  • the particles have zero volume (they are points)
  • there are no intermolecular forces of attraction between the particles
  • (also) the particles move randomly and their collisions are perfectly elastic

Real gases

Real gases behave most like an ideal gas at high temperature and low pressure. They deviate when:

The pressure is high

The particles are pushed close together, so the volume of the particles themselves is no longer negligible compared with the volume of the container.

The temperature is low

The particles move slowly, so intermolecular forces have a significant effect: the particles attract each other and hit the walls less hard.

Gases with small, non-polar particles, such as helium and hydrogen, are closest to ideal. Gases with strong intermolecular forces, such as ammonia (hydrogen bonding) and steam, deviate the most.

The ideal gas equation

Ideal gas equationpV = nRTLearn it
Using mass and molar masspV = mMRT, so M = mRTpVLearn it
SymbolQuantitySI unitConversion
ppressurePa1 kPa = 103 Pa
Vvolumem31 dm3 = 10−3 m3; 1 cm3 = 10−6 m3
namountmoln = mass ÷ M
Rmolar gas constant8.31 J K−1 mol−1given in the Data Booklet
TtemperatureKT / K = θ / °C + 273
The most common mistake

Forgetting to convert cm3 to m3 (× 10−6) and °C to K. If your Mr comes out a million times too big or small, check your units.

Worked example: relative molecular mass of a volatile liquid

0.150 g of a volatile liquid is injected into a gas syringe in an oven at 100 °C. It vaporises completely, and the vapour occupies 50.0 cm3 at a pressure of 101 kPa. Calculate Mr.

  1. Convert units: p = 101 000 Pa, V = 50.0 × 10−6 m3, T = 373 K
  2. n = pVRT = 101000 × 50.0 × 10−68.31 × 373 = 1.63 × 10−3 mol
  3. M = mn = 0.1501.63 × 10−3 = 92.1 g mol−1

Mr = 92.1 (this could be methylbenzene, C7H8, Mr 92.0)

Using the ideal gas equation shows where the molar volume comes from: at 20 °C and 101 kPa, V = nRT/p = 1 × 8.31 × 293 / 101 000 = 0.0241 m3 = 24.1 dm3, close to the 24.0 dm3 used for room conditions.

Bonding and structure

Solids are classified by the particles in their lattice and the forces between them. There are four types:

Na⁺Cl⁻giant ionic: NaCl
each C bonded to 4 othersgiant molecular: diamond
weak forces between layersgiant molecular: graphite
++++++++++++++e⁻e⁻e⁻e⁻e⁻e⁻e⁻e⁻e⁻e⁻e⁻e⁻delocalised electrons between the ionsgiant metallic: copper

Giant ionic: sodium chloride and magnesium oxide

A regular lattice of positive and negative ions held by strong ionic bonds in all directions. In sodium chloride, each Na+ ion is surrounded by six Cl− ions and each Cl− by six Na+ (6 : 6 coordination). Magnesium oxide has the same arrangement, with Mg2+ and O2− ions.

Simple molecular: iodine, buckminsterfullerene and ice

Small molecules held in a lattice by weak intermolecular forces. The covalent bonds inside each molecule are strong, but only the weak forces between molecules are broken on melting.

  • Iodine: I2 molecules held by id-id forces. It forms shiny grey-black crystals that sublime on gentle warming to a purple vapour.
  • Buckminsterfullerene, C60: 60 carbon atoms joined in a hollow ball of 12 pentagons and 20 hexagons. The C60 molecules are held by id-id forces, so it is soft, has a much lower sublimation point than diamond, does not conduct, and dissolves in non-polar solvents such as methylbenzene.
  • Ice: H2O molecules held in an open tetrahedral lattice by hydrogen bonds (Topic 3), which is why ice floats on water.

Giant molecular (giant covalent): diamond, graphite and silicon(IV) oxide

  • Diamond: each carbon atom is covalently bonded to four others in a tetrahedral arrangement (109.5°), forming a 3D network. Very hard, very high melting point, does not conduct (no delocalised electrons).
  • Graphite: each carbon is bonded to three others in flat layers of hexagons. The fourth outer electron of each carbon is delocalised over the layer, so graphite conducts electricity along the layers. The layers are held together by weak id-id forces, so they slide over each other: graphite is soft and slippery and is used as a lubricant and in pencils. Its melting point is still very high because the covalent bonds within the layers must break.
  • Silicon(IV) oxide, SiO2: each Si atom is bonded to four O atoms tetrahedrally, and each O atom to two Si atoms. Hard, high melting point (about 1700 °C), insoluble and non-conducting.

Giant metallic: copper

A regular lattice of positive copper ions in a sea of delocalised electrons. The metallic bonding is strong and non-directional, so copper has a high melting point (1085 °C). It conducts electricity because the delocalised electrons move through the structure. It is malleable and ductile because layers of ions can slide over each other without the metallic bonding breaking.

How structure explains properties

StructureMelting and boiling pointsElectrical conductivitySolubility
Giant ionicHigh: many strong ionic bonds must be brokenNot as a solid (ions can't move); conducts when molten or dissolved (ions free to move)Often soluble in water (ions attracted to polar water molecules); insoluble in non-polar solvents
Simple molecularLow: only weak intermolecular forces are overcomeDoes not conduct: no ions or delocalised electronsUsually soluble in non-polar solvents; water only if they can hydrogen bond with it (e.g. ethanol, sugar)
Giant molecularVery high: many strong covalent bonds must be brokenDoes not conduct, except graphite (delocalised electrons)Insoluble in all solvents
Giant metallicUsually high: strong metallic bondingConducts as solid and liquid: delocalised electronsInsoluble (some react with water)

Worked example: deducing the structure from data

SubstanceMelting point / °CConducts as a solid?Conducts when molten?Soluble in water?
W801noyesyes
X1085yesyesno
Y114nonovery slightly
Z1710nonono

W is giant ionic (conducts only when its ions are free: sodium chloride). X is giant metallic (conducts as a solid: copper). Y is simple molecular (low melting point: iodine). Z is giant molecular (very high melting point, never conducts: silicon(IV) oxide).

Questions

Use R = 8.31 J K−1 mol−1. Questions 1 to 4 are multiple choice. Questions 5 and 6 are structured.

1Paper 1 style

What volume does 0.50 mol of an ideal gas occupy at 300 K and 100 kPa?

  1. 12.5 dm3
  2. 1.25 dm3
  3. 0.0125 dm3
  4. 125 dm3

V = nRT/p = 0.50 × 8.31 × 300 / 100 000 = 0.0125 m3 = 12.5 dm3. Answer C is the value in m3 written with the wrong unit.

2Paper 1 style

Under which conditions does a gas behave most like an ideal gas?

  1. helium at high temperature and low pressure
  2. ammonia at low temperature and high pressure
  3. helium at low temperature and high pressure
  4. ammonia at high temperature and low pressure

Helium has small, non-polar atoms with very weak forces between them. High temperature and low pressure make intermolecular forces and particle volume least important.

3Paper 1 style

Which substance has a giant molecular (giant covalent) structure?

  1. carbon dioxide
  2. iodine
  3. sodium chloride
  4. silicon(IV) oxide

SiO2 is a 3D network of Si–O covalent bonds. CO2 and I2 are simple molecular, and NaCl is giant ionic.

4Paper 1 style

Why does graphite conduct electricity?

  1. It contains ions that move through the layers.
  2. Each carbon atom has one electron that is delocalised along its layer.
  3. The layers can slide over each other.
  4. The forces between the layers are weak.

Each carbon uses three electrons in σ bonds to its neighbours; the fourth is delocalised and moves along the layer when a voltage is applied.

5Paper 2 style

A sample of a gas has a mass of 0.355 g. At 27 °C and 98.0 kPa, it occupies a volume of 128 cm3.

(a)
Calculate the relative molecular mass of the gas.
[3]
(b)
State two assumptions of the kinetic theory for an ideal gas.
[2]
(c)
Explain why the gas would deviate more from ideal behaviour at a very high pressure.
[1]
  • Correct units: 98 000 Pa, 1.28 × 10−4 m3, 300 K
  • n = pV/RT = 98 000 × 1.28 × 10−4 / (8.31 × 300) = 5.03 × 10−3 mol
  • Mr = 0.355 / 5.03 × 10−3 = 70.6 (it could be chlorine, Cl2, 71.0)
  • The particles have zero / negligible volume
  • There are no intermolecular forces between the particles (allow: collisions are elastic; random motion)
  • At high pressure the particles are close together, so their own volume is significant compared with the volume of the container (and intermolecular forces become significant)
6Paper 2 style
(a)
Diamond and graphite are both forms of carbon. Explain why diamond is very hard but graphite is soft.
[3]
(b)
Explain why iodine has a much lower melting point than diamond, even though both contain only covalent bonds.
[2]
(c)
Explain why magnesium oxide conducts electricity when molten but not when solid.
[2]
(d)
Explain why copper is malleable.
[1]
  • Diamond: each C atom covalently bonded to four others in a rigid 3D (tetrahedral) network
  • Graphite: layers in which each C is bonded to three others
  • The layers are held by weak (id-id / van der Waals') forces, so they slide over each other
  • Iodine is simple molecular: only weak id-id forces between I2 molecules are overcome on melting
  • Diamond is giant molecular: many strong covalent bonds must be broken
  • In the solid, the ions are held in fixed positions in the lattice
  • When molten, the ions are free to move and carry charge
  • Layers of ions can slide over each other without breaking the metallic bonding (the delocalised electrons move with them)

Every reaction transfers energy to or from its surroundings, because bonds are broken and made. This topic covers how to define, measure and calculate enthalpy changes, including those you can't measure directly, using Hess's law.

Enthalpy changes

The enthalpy change, ΔH, is the heat energy transferred in a reaction at constant pressure, in kJ mol−1.

Exothermic: ΔH is negative

Energy is transferred to the surroundings, so the temperature of the surroundings rises. The products have less enthalpy than the reactants. Examples: combustion, neutralisation, most oxidation reactions.

Endothermic: ΔH is positive

Energy is taken in from the surroundings, so the temperature falls. The products have more enthalpy than the reactants. Examples: thermal decomposition of CaCO3, photosynthesis, dissolving ammonium nitrate.

Reaction pathway diagrams

A reaction pathway diagram (energy profile) shows the enthalpy of the reactants and products, and the energy barrier between them. The activation energy, EA, is the minimum energy needed for a reaction to happen (Topic 8). It is measured from the reactants up to the top of the barrier.

enthalpy, Hprogress of reactionreactantsproductsEAΔH negativeexothermic
enthalpy, Hprogress of reactionreactantsproductsEAΔH positiveendothermic
Label both axes, the reactant and product levels, ΔH (an arrow from reactants to products) and EA (an arrow from reactants to the peak).

Standard conditions and definitions

Enthalpy changes depend on conditions, so they are quoted under standard conditions, shown by the symbol : a temperature of 298 K and a pressure of 101 kPa, with solutions at 1 mol dm−3, and every substance in its standard state (its normal physical state under those conditions).

Enthalpy change of reaction, ΔHr
The enthalpy change when the amounts of reactants shown in the equation react under standard conditions.
Enthalpy change of formation, ΔHf
The enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions.
Enthalpy change of combustion, ΔHc
The enthalpy change when one mole of a substance is burned completely in oxygen under standard conditions.
Enthalpy change of neutralisation, ΔHneut
The enthalpy change when one mole of water is formed by the reaction of an acid with an alkali under standard conditions.
2C(s) + 3H2(g) + ½O2(g) → C2H5OH(l) ΔHf of ethanol: elements in their standard states, one mole of product
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ΔHc of ethanol: one mole of fuel, complete combustion
½H2SO4(aq) + NaOH(aq) → ½Na2SO4(aq) + H2O(l) ΔHneut: one mole of water. For any strong acid with a strong alkali it is about −57 kJ mol−1.
Watch the details

By definition, ΔHf of an element in its standard state is zero. Fractions such as ½O2 are fine in these equations, because the definition fixes one mole of the substance. Carbon's standard state is graphite, C(s).

Bond energies

Breaking a bond needs energy (endothermic). Making a bond releases energy (exothermic). The overall enthalpy change depends on the balance between the two:

From bond energiesΔHr = Σ(bonds broken) − Σ(bonds made)Learn it

Worked example: combustion of methane

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g). Bond energies / kJ mol−1: C–H 410, O=O 496, C=O (in CO2) 805, O–H 460.

  1. Bonds broken: 4 C–H + 2 O=O = 4(410) + 2(496) = 2632 kJ
  2. Bonds made: 2 C=O + 4 O–H = 2(805) + 4(460) = 3450 kJ
  3. ΔH = 2632 − 3450

ΔH = −818 kJ mol−1

The experimental ΔHc of methane is −890 kJ mol−1. The answers differ because bond energies are averages taken over many compounds, and because the bond energy calculation is for gases, while under standard conditions water is a liquid (condensing releases more energy).

Worked example: hydrogenation of ethene

C2H4(g) + H2(g) → C2H6(g). Only the bonds that change need counting: break C=C (610) and H–H (436); make C–C (350) and two new C–H (2 × 410).

ΔH = (610 + 436) − (350 + 820) = 1046 − 1170 = −124 kJ mol−1

Measuring enthalpy changes

In calorimetry, the reaction heats (or cools) a known mass of water or solution, and you measure the temperature change.

Energy transferredq = mcΔTLearn it
Enthalpy change per moleΔH = −mcΔTnLearn it

m is the mass of the liquid being heated (for dilute solutions, take 1 cm3 = 1 g), c is 4.18 J g−1 K−1, ΔT is the temperature change, and n is the amount of the limiting reagent (or of water formed, for neutralisation). The minus sign makes ΔH negative when the temperature rises.

Worked example: enthalpy change of neutralisation

25.0 cm3 of 2.00 mol dm−3 HCl is mixed with 25.0 cm3 of 2.00 mol dm−3 NaOH in a polystyrene cup. The temperature rises by 13.5 K.

  1. q = mcΔT = 50.0 × 4.18 × 13.5 = 2822 J
  2. n(H2O) = n(HCl) = 2.00 × 0.0250 = 0.0500 mol
  3. ΔH = −2822/0.0500 = −56 400 J mol−1

ΔHneut = −56.4 kJ mol−1

Worked example: enthalpy change of combustion with a spirit burner

Burning 0.580 g of ethanol (C2H5OH, Mr 46.0) raises the temperature of 200 g of water in a copper can by 18.0 K.

  1. q = 200 × 4.18 × 18.0 = 15 048 J = 15.05 kJ
  2. n(ethanol) = 0.580 / 46.0 = 0.01261 mol
  3. ΔHc = −15.05/0.01261

ΔHc = −1190 kJ mol−1 (3 s.f.). The data book value is −1367 kJ mol−1.

The experimental value is less exothermic mainly because of heat losses to the surroundings (heating the air and the can), and also incomplete combustion (soot forms) and evaporation of ethanol from the wick.

Correcting for heat loss

In Paper 3 you may record the temperature every minute before and after mixing. Extrapolating the cooling line back to the time of mixing gives the maximum temperature the mixture would have reached with no heat loss.

024681018212427303336time / mintemperature / °CΔT = 12.2 Kmixed at 3.5 minextrapolate the cooling line back
The highest reading (33.0 °C) underestimates the true rise. Extrapolating back gives 33.2 °C, so ΔT = 33.2 − 21.0 = 12.2 K.

Hess's law

Hess's law
The total enthalpy change for a reaction is independent of the route taken, provided the starting and finishing conditions are the same.

Hess's law lets you calculate enthalpy changes that can't be measured directly, such as the enthalpy of formation of ethanol (you can't make ethanol by mixing carbon, hydrogen and oxygen). Draw a cycle with the reaction you want along the top and an alternative route through known enthalpy changes.

Using enthalpy changes of formation

Both reactants and products can be made from the same elements, so the arrows go up from the elements:

From formation dataΔHr = ΣΔHf(products) − ΣΔHf(reactants)Learn it

Worked example

Calculate ΔH for CaCO3(s) → CaO(s) + CO2(g). ΔHf / kJ mol−1: CaCO3 −1207, CaO −635, CO2 −394.

ΔH = (−635 + (−394)) − (−1207) = +178 kJ mol−1

Using enthalpy changes of combustion

Both reactants and products burn to give the same combustion products, so the arrows go down to the combustion products:

From combustion dataΔHr = ΣΔHc(reactants) − ΣΔHc(products)Learn it
2C(s) + 3H2(g) + ½O2(g)C2H5OH(l)ΔHf2CO2(g) + 3H2O(l)2ΔHc(C) + 3ΔHc(H₂)ΔHc(C₂H₅OH)
The direct route (top) and the indirect route (down then up) must have the same total enthalpy change.

Worked example: enthalpy change of formation of ethanol

ΔHc / kJ mol−1: C(s) −394, H2(g) −286, C2H5OH(l) −1367.

  1. Route via combustion: ΔHf = 2ΔHc(C) + 3ΔHc(H2) − ΔHc(C2H5OH)
  2. = 2(−394) + 3(−286) − (−1367) = −788 − 858 + 1367

ΔHf(ethanol) = −279 kJ mol−1

Avoiding sign errors

Write each step of the indirect route along the direction of its arrow. If you go against an arrow, change the sign of that value. Multiply each value by the number of moles in the equation.

Questions

Use c = 4.18 J g−1 K−1. Questions 1 to 4 are multiple choice. Questions 5 to 7 are structured.

1Paper 1 style

Which equation represents the standard enthalpy change of formation of ethanol?

  1. 2C(g) + 6H(g) + O(g) → C2H5OH(l)
  2. C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
  3. 2C(s) + 3H2(g) + ½O2(g) → C2H5OH(l)
  4. 4C(s) + 6H2(g) + O2(g) → 2C2H5OH(l)

Formation uses the elements in their standard states (C(s), H2(g), O2(g)) and makes exactly one mole of the compound. D makes two moles; A uses gaseous atoms.

2Paper 1 style

Use the bond energies H–H 436, Cl–Cl 242 and H–Cl 431 kJ mol−1. What is ΔH for H2(g) + Cl2(g) → 2HCl(g)?

  1. −184 kJ mol−1
  2. +184 kJ mol−1
  3. −247 kJ mol−1
  4. +247 kJ mol−1

Broken: 436 + 242 = 678. Made: 2 × 431 = 862. ΔH = 678 − 862 = −184 kJ mol−1. C and D forget that two H–Cl bonds form.

3Paper 1 style

ΔHc values / kJ mol−1: C(s) −394, H2(g) −286, CH4(g) −890. What is ΔHf of methane?

  1. +76 kJ mol−1
  2. −76 kJ mol−1
  3. −210 kJ mol−1
  4. −1570 kJ mol−1

ΔHf = ΔHc(C) + 2ΔHc(H2) − ΔHc(CH4) = −394 + 2(−286) + 890 = −76 kJ mol−1.

4Paper 1 style

Which statement about the standard enthalpy change of formation of oxygen, O2(g), is correct?

  1. It is equal to the O=O bond energy.
  2. It is equal to half the O=O bond energy.
  3. It is negative because forming bonds releases energy.
  4. It is zero, because oxygen is an element in its standard state.

Forming an element from itself in its standard state involves no change, so ΔHf = 0.

5Paper 2 and 3 style

A student adds 2.00 g of zinc powder (an excess) to 50.0 cm3 of 0.400 mol dm−3 copper(II) sulfate solution in a polystyrene cup. The temperature rises from 20.5 °C to 40.0 °C.

Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)

(a)
Calculate the energy released.
[1]
(b)
Calculate the enthalpy change of reaction, in kJ mol−1.
[2]
(c)
The data book value is −217 kJ mol−1. Suggest two improvements to the method that would give a more accurate value.
[2]
  • q = 50.0 × 4.18 × 19.5 = 4076 J (4.08 kJ)
  • n(CuSO4) = 0.400 × 0.0500 = 0.0200 mol (zinc is in excess, so copper(II) sulfate is limiting)
  • ΔH = −4.076 / 0.0200 = −204 kJ mol−1 (negative sign needed)
  • Insulate the cup / use a lid, to reduce heat loss
  • Record the temperature over time and extrapolate the cooling curve back to the time of mixing (allow: use a more precise thermometer)
6Paper 2 style

The enthalpy change for CuSO4(s) + 5H2O(l) → CuSO4·5H2O(s) can't be measured directly. Instead, the enthalpy changes of solution are measured: anhydrous CuSO4(s) −66.5 kJ mol−1; hydrated CuSO4·5H2O(s) +11.5 kJ mol−1.

(a)
State Hess's law.
[1]
(b)
Draw an energy cycle and use it to calculate ΔH for the hydration of anhydrous copper(II) sulfate.
[2]
(c)
Suggest why this enthalpy change can't be measured directly.
[1]
  • The enthalpy change of a reaction is independent of the route, provided the initial and final conditions are the same
  • Cycle: CuSO4(s) → CuSO4(aq) directly (−66.5), and CuSO4(s) → CuSO4·5H2O(s) → CuSO4(aq) (ΔH then +11.5)
  • ΔH = −66.5 − (+11.5) = −78.0 kJ mol−1
  • You can't add exactly 5 mol of water so that it all becomes water of crystallisation; the solid would dissolve or react too slowly / incompletely
7Paper 2 style

Propene reacts with hydrogen: CH2=CHCH3(g) + H2(g) → CH3CH2CH3(g).

(a)
Define bond energy.
[1]
(b)
Use the bond energies C=C 610, C–C 350, C–H 410 and H–H 436 kJ mol−1 to calculate ΔH for this reaction.
[2]
(c)
Sketch a labelled reaction pathway diagram for this reaction.
[2]
  • The energy needed to break one mole of a particular covalent bond in the gaseous state
  • Broken: C=C + H–H = 610 + 436 = 1046; made: C–C + 2 C–H = 350 + 820 = 1170
  • ΔH = −124 kJ mol−1
  • Products drawn lower than reactants, with ΔH shown as a downward arrow labelled
  • Hump with EA shown from the reactant level to the peak; axes labelled enthalpy and progress of reaction

Redox reactions transfer electrons. Oxidation numbers are the bookkeeping that tracks where the electrons go, and they let you name compounds, identify oxidising and reducing agents, and balance difficult equations.

Oxidation numbers

An oxidation number is the charge an atom would have if all its bonds were ionic. Work it out with these rules, in order:

RuleExample
An uncombined element is 0.Na, O2, Cl2, S8: all 0
A simple ion: its charge.Fe3+ is +3; Cl− is −1
The oxidation numbers in a compound add up to 0; in an ion, to the ion's charge.In SO42−: S + 4(−2) = −2
Group 1 metals are +1, Group 2 are +2, aluminium is +3.In MgCl2, Mg is +2 and each Cl is −1
Fluorine is always −1.In OF2, F is −1, so O is +2
Hydrogen is +1, except in metal hydrides (e.g. NaH), where it is −1.In H2O, H is +1; in NaH, H is −1
Oxygen is −2, except in peroxides (e.g. H2O2, −1) and in OF2 (+2).In CO2, O is −2, so C is +4
Chlorine is −1, except when combined with O or F.In ClO3−, Cl is +5

Worked examples

SpeciesWorkingOxidation number
Mn in MnO4−Mn + 4(−2) = −1+7
Cr in Cr2O72−2Cr + 7(−2) = −2+6
S in S2O32−2S + 3(−2) = −2+2 (average)
N in NH4+N + 4(+1) = +1−3
N in NO3−N + 3(−2) = −1+5
V in VO2+V + 2(−2) = +1+5

Roman numerals in names give the oxidation number of the element before them: iron(III) chloride FeCl3, potassium manganate(VII) KMnO4, potassium dichromate(VI) K2Cr2O7, sodium chlorate(I) NaClO, sodium chlorate(V) NaClO3.

Redox

TermIn terms of electronsIn terms of oxidation number
Oxidationloss of electronsincrease in oxidation number
Reductiongain of electronsdecrease in oxidation number
Redoxa reaction in which oxidation and reduction happen together
Oxidising agenta species that oxidises another species, and is itself reduced (it gains electrons)
Reducing agenta species that reduces another species, and is itself oxidised (it loses electrons)
Disproportionationa reaction in which the same element is both oxidised and reduced

Remember "OIL RIG": Oxidation Is Loss, Reduction Is Gain (of electrons).

Worked example: identifying what is oxidised

2Fe3+(aq) + 2I−(aq) → 2Fe2+(aq) + I2(aq)

  • Fe: +3 → +2. Iron is reduced, so Fe3+ is the oxidising agent.
  • I: −1 → 0. Iodide is oxidised, so I− is the reducing agent.

Disproportionation

Cl2(aq) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H2O(l) Cl goes from 0 to −1 (reduced) and from 0 to +1 (oxidised)
2H2O2(aq) → 2H2O(l) + O2(g) O goes from −1 to −2 (reduced) and from −1 to 0 (oxidised)
2Cu+(aq) → Cu(s) + Cu2+(aq) Cu goes from +1 to 0 and from +1 to +2

Balancing redox equations

Method 1: half-equations

Balance each half separately, then add them so that the electrons cancel. In acidic solution, balance O atoms with H2O and H atoms with H+, then balance charge with electrons.

Worked example: manganate(VII) and iron(II)

  1. Reduction: MnO4− → Mn2+. Add 4 H2O to balance O, then 8 H+ to balance H: MnO4− + 8H+ → Mn2+ + 4H2O. Charge on the left is +7 and on the right +2, so add 5 electrons on the left: MnO4− + 8H+ + 5e− → Mn2+ + 4H2O
  2. Oxidation: Fe2+ → Fe3+ + e−. Multiply by 5 so the electrons match.
  3. Add and cancel the electrons:

MnO4−(aq) + 8H+(aq) + 5Fe2+(aq) → Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)

Method 2: changes in oxidation number

The total increase in oxidation number must equal the total decrease.

Worked example: copper and concentrated nitric acid

Unbalanced: Cu + HNO3 → Cu(NO3)2 + NO2 + H2O

  1. Cu: 0 → +2, an increase of 2. N: +5 → +4 (in NO2), a decrease of 1.
  2. So 2 N atoms must be reduced for every Cu oxidised: Cu + … → … + 2NO2.
  3. Another 2 HNO3 provide the nitrate ions in Cu(NO3)2, so 4 HNO3 in total. Then balance H and O with water.

Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

Topic 24 (A Level) builds on this with electrode potentials, which predict whether a redox reaction will happen.

Questions

1Paper 1 style

What is the oxidation number of chromium in K2Cr2O7?

  1. +3
  2. +5
  3. +6
  4. +7

2(+1) + 2Cr + 7(−2) = 0, so 2Cr = +12 and Cr = +6.

2Paper 1 style

Which reaction is a disproportionation?

  1. Zn + Cu2+ → Zn2+ + Cu
  2. 3Cl2 + 6OH− → 5Cl− + ClO3− + 3H2O
  3. 2Fe2+ + Cl2 → 2Fe3+ + 2Cl−
  4. CaCO3 → CaO + CO2

In B, chlorine goes from 0 to −1 (in Cl−) and from 0 to +5 (in ClO3−): the same element is both reduced and oxidised. D is not a redox reaction at all.

3Paper 1 style

In the reaction SO2 + 2H2S → 3S + 2H2O, which statement is correct?

  1. SO2 is the oxidising agent.
  2. SO2 is oxidised.
  3. H2S gains electrons.
  4. The oxidation number of hydrogen changes.

S in SO2 goes from +4 to 0 (reduced), so SO2 is the oxidising agent. S in H2S goes from −2 to 0 (oxidised, loses electrons). H stays at +1.

4Paper 2 style

Acidified potassium dichromate(VI) oxidises iron(II) ions to iron(III) ions. The dichromate(VI) ion is reduced to chromium(III) ions.

(a)
Construct the half-equation for the reduction of Cr2O72− to Cr3+ in acid.
[2]
(b)
Hence write the overall ionic equation for the reaction.
[1]
(c)
State the role of the Fe2+ ion in this reaction, and explain your answer in terms of electrons.
[2]
  • Correct species: Cr2O72−, H+, Cr3+, H2O with 6 electrons
  • Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O
  • Cr2O72− + 14H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+
  • Reducing agent
  • It loses an electron (it is oxidised from +2 to +3), giving electrons to the dichromate(VI)
5Paper 2 style

When chlorine reacts with hot aqueous sodium hydroxide, sodium chloride and sodium chlorate(V) form.

(a)
Give the formula of sodium chlorate(V).
[1]
(b)
Use oxidation numbers to construct a balanced equation for the reaction.
[2]
(c)
Explain why this is a disproportionation reaction.
[1]
  • NaClO3
  • Cl: 0 → −1 (decrease of 1) and 0 → +5 (increase of 5), so 5 Cl reduced for each 1 oxidised
  • 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O
  • Chlorine is both oxidised (to +5) and reduced (to −1) in the same reaction

Many reactions never go to completion: they reach a balance in which reactants and products are both present. This topic explains that balance, how to shift it, how to calculate equilibrium constants, and how acids and bases behave.

Reversible reactions and dynamic equilibrium

A reversible reaction can go in both directions: the products can react to re-form the reactants. It is shown with the symbol ⇌, for example N2(g) + 3H2(g) ⇌ 2NH3(g).

Dynamic equilibrium
The state in which the rate of the forward reaction equals the rate of the reverse reaction, so the concentrations of reactants and products stay constant.

"Dynamic" means both reactions are still happening; there is just no overall change. Equilibrium can only be reached in a closed system, where nothing can enter or leave. In an open system, a gaseous product would escape and the reverse reaction couldn't keep up.

Equilibrium is the same whichever side you start from: starting with only NH3 gives the same equilibrium mixture as starting with the equivalent amounts of N2 and H2 under the same conditions.

Le Chatelier's principle

Le Chatelier's principle
If a change is made to a system at dynamic equilibrium, the position of equilibrium moves to minimise this change.
ChangeEffect on the position of equilibriumEffect on K
Increase the concentration of a reactantMoves to the right, using up some of the added reactantnone
Remove a productMoves to the right, replacing some of the productnone
Increase the pressure (gases)Moves towards the side with fewer gas molecules, reducing the pressure. No effect if both sides have the same number.none
Increase the temperatureMoves in the endothermic direction, absorbing the extra heatchanges
Decrease the temperatureMoves in the exothermic directionchanges
Add a catalystNo change. It speeds up the forward and reverse reactions equally, so equilibrium is reached faster.none
Only temperature changes K

A change in concentration or pressure moves the position of equilibrium, but the system then readjusts so that K has the same value. For an exothermic forward reaction, raising the temperature decreases K. For an endothermic forward reaction, it increases K.

Equilibrium constants

Kc, in terms of concentrations

For a general reaction aA + bB ⇌ cC + dD at equilibrium:

Equilibrium constantKc = [C]c[D]d[A]a[B]bLearn it

Square brackets mean concentration at equilibrium in mol dm−3. Products go on top, and each concentration is raised to the power of its coefficient. Work out the units from the expression: for N2 + 3H2 ⇌ 2NH3, Kc = [NH3]2[N2][H2]3 has units (mol dm−3)2 / (mol dm−3)4 = mol−2 dm6.

A large Kc means the equilibrium lies to the right (mostly products); a small one means it lies to the left.

Worked example: calculating Kc

1.00 mol of H2 and 1.00 mol of I2 are heated in a sealed 1.00 dm3 flask. At equilibrium, 1.56 mol of HI is present. H2(g) + I2(g) ⇌ 2HI(g)

H2I2HI
Start / mol1.001.000
Change / mol−0.78−0.78+1.56
Equilibrium / mol0.220.221.56
Concentration / mol dm−30.220.221.56

Use the equation's ratio for the change row: forming 1.56 mol HI uses 0.78 mol each of H2 and I2.

Kc = 1.5620.22 × 0.22 = 50 (no units, because the powers cancel)

Worked example: an esterification

1.00 mol of ethanoic acid and 1.00 mol of ethanol reach equilibrium. 0.667 mol of ethyl ethanoate is then present. CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O

At equilibrium: acid 0.333 mol, ethanol 0.333 mol, ester 0.667 mol, water 0.667 mol. The volume V cancels because there are equal numbers of moles on each side.

Kc = 0.667 × 0.6670.333 × 0.333 = 4.0

Kp, in terms of partial pressures

Mole fraction
The amount of one gas divided by the total amount of gas: xA = nAntotal
Partial pressure
The pressure that one gas in a mixture would exert if it occupied the container on its own: pA = xA × ptotal. The partial pressures add up to the total pressure.

For gas-phase equilibria, Kp uses partial pressures in place of concentrations. For 2SO2(g) + O2(g) ⇌ 2SO3(g): Kp = pSO32pSO22 × pO2, with units Pa−1 (or kPa−1).

Worked example: calculating Kp

An equilibrium mixture contains 0.20 mol SO2, 0.10 mol O2 and 1.70 mol SO3 at a total pressure of 200 kPa.

  1. Total amount = 2.00 mol. Mole fractions: SO2 0.10, O2 0.050, SO3 0.85.
  2. Partial pressures: SO2 20 kPa, O2 10 kPa, SO3 170 kPa.
  3. Kp = 1702202 × 10 = 28 9004000

Kp = 7.2 kPa−1

In both kinds of calculation, set up a table of start, change and equilibrium amounts. At AS Level you won't need to solve a quadratic equation.

Equilibria in industry

Industrial conditions are a compromise between a good yield (position of equilibrium), a good rate, and cost and safety.

The Haber process

N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH = −92 kJ mol−1

ConditionValue usedReasoning
Temperatureabout 400 to 450 °CThe forward reaction is exothermic, so a low temperature gives a higher yield, but the rate would be too slow. This is a compromise.
Pressureabout 20 MPa (200 atm)4 gas molecules become 2, so high pressure increases the yield and the rate. Higher pressures would cost too much to generate and contain safely.
CatalystironIncreases the rate. It does not change the yield.
Removing productammonia is cooled and liquefied; unreacted gases are recycledThe yield per pass is only about 15%, but recycling gives an overall conversion of about 98%.

The Contact process

2SO2(g) + O2(g) ⇌ 2SO3(g) ΔH = −197 kJ mol−1. The SO3 is then used to make sulfuric acid.

ConditionValue usedReasoning
Temperatureabout 450 °CA compromise between yield (favoured by low temperature) and rate.
Pressurejust above atmospheric, 1 to 2 atm3 gas molecules become 2, so high pressure would increase the yield, but the conversion is already about 99% at low pressure, so the extra cost isn't worth it.
Catalystvanadium(V) oxide, V2O5Increases the rate.

Acids and bases

Common acids and alkalis

AcidFormulaAlkaliFormula
hydrochloric acidHClsodium hydroxideNaOH
sulfuric acidH2SO4potassium hydroxideKOH
nitric acidHNO3ammoniaNH3
ethanoic acidCH3COOHcalcium hydroxideCa(OH)2

The Brønsted–Lowry theory

Brønsted–Lowry acid
A proton (H+) donor.
Brønsted–Lowry base
A proton (H+) acceptor.

In HCl(g) + H2O(l) → H3O+(aq) + Cl−(aq), HCl donates a proton (acid) and water accepts it (base). In NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq), water donates a proton (acid) and ammonia accepts it (base). Water can act as either, depending on what it reacts with.

Strong and weak acids and bases

Strong

Fully dissociated (ionised) in aqueous solution. HCl, H2SO4, HNO3, NaOH, KOH.

HCl(aq) → H+(aq) + Cl−(aq)

Weak

Only partially dissociated in aqueous solution; most molecules stay intact. Ethanoic acid and ammonia.

CH3COOH(aq) ⇌ CH3COO−(aq) + H+(aq)

"Strong" and "weak" describe the extent of dissociation, not the concentration. A dilute solution of a strong acid can have a lower [H+] than a concentrated solution of a weak one.

pH

The pH scale measures acidity: pure water has a pH of 7, acidic solutions are below 7 and alkaline solutions are above 7. Each unit of pH is a factor of 10 in the concentration of H+ ions (the pH calculations are in Topic 25).

Telling strong and weak acids apart

Compare solutions of the same concentration, for example 0.1 mol dm−3 HCl and 0.1 mol dm−3 CH3COOH:

TestStrong acid (HCl)Weak acid (CH3COOH)
pH meter or universal indicatorlower pH (about 1)higher pH (about 3)
Electrical conductivityhigher: more ionslower: fewer ions
Reaction with magnesiumfaster fizzing: higher [H+]slower fizzing
Total volume of H2 with excess Mgthe same: as H+ is used up, more CH3COOH dissociates, so both eventually release the same amount of H+

Neutralisation and salts

In every neutralisation between an acid and an alkali, H+(aq) + OH−(aq) → H2O(l). The other ions form a salt, for example KOH(aq) + HNO3(aq) → KNO3(aq) + H2O(l).

pH titration curves

A titration curve plots pH against the volume of alkali added. The shape depends on whether the acid and base are strong or weak. All four below use 25.0 cm3 of 0.100 mol dm−3 acid and 0.100 mol dm−3 base, so the equivalence point is at 25.0 cm3.

025500714volume of base added / cm³pHstrong acid + strong base
025500714volume of base added / cm³pHweak acid + strong base
025500714volume of base added / cm³pHstrong acid + weak base
025500714volume of base added / cm³pHweak acid + weak base
The shaded bands show where two indicators change colour: methyl orange (orange, pH 3.1 to 4.4) and phenolphthalein (green, pH 8.3 to 10.0).
TitrationpH at equivalenceSteep sectionSuitable indicator
strong acid + strong base7about pH 3 to 11methyl orange or phenolphthalein
weak acid + strong baseabove 7 (about 8.7)about pH 7 to 11phenolphthalein
strong acid + weak basebelow 7 (about 5.3)about pH 3 to 7methyl orange
weak acid + weak baseabout 7no steep sectionnone is suitable

Choosing an indicator: its colour-change range must lie entirely within the steep (vertical) part of the curve. Then one drop of alkali at the end-point changes the pH right across the indicator's range.

Questions

1Paper 1 style

For N2(g) + O2(g) ⇌ 2NO(g), ΔH = +180 kJ mol−1. Which change increases the equilibrium yield of NO?

  1. increasing the pressure
  2. increasing the temperature
  3. adding a catalyst
  4. decreasing the temperature

The forward reaction is endothermic, so a higher temperature moves the equilibrium to the right. There are 2 gas molecules on each side, so pressure has no effect, and a catalyst never changes the yield.

2Paper 1 style

What are the units of Kc for 2SO2(g) + O2(g) ⇌ 2SO3(g)?

  1. no units
  2. mol dm−3
  3. mol−1 dm−3
  4. mol−1 dm3

Kc = [SO3]2[SO2]2[O2], so the units are (mol dm−3)2 / (mol dm−3)3 = (mol dm−3)−1 = mol−1 dm3.

3Paper 1 style

In which reaction is water acting as a Brønsted–Lowry base?

  1. NH3 + H2O ⇌ NH4+ + OH−
  2. CO32− + H2O ⇌ HCO3− + OH−
  3. HNO3 + H2O → H3O+ + NO3−
  4. CH3COO− + H2O ⇌ CH3COOH + OH−

In C, water accepts a proton from HNO3 to become H3O+. In A, B and D, water donates a proton, so it is acting as an acid.

4Paper 1 style

Which indicator is suitable for titrating ethanoic acid with sodium hydroxide?

  1. phenolphthalein, range pH 8.3 to 10.0
  2. methyl orange, range pH 3.1 to 4.4
  3. methyl red, range pH 4.4 to 6.2
  4. no indicator is suitable

For a weak acid and a strong base, the steep section is at about pH 7 to 11, so phenolphthalein changes colour right at the end-point.

5Paper 2 style

Hydrogen is made by the reaction CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g), ΔH = +206 kJ mol−1.

(a)
State and explain the effect on the equilibrium yield of hydrogen of (i) increasing the temperature and (ii) increasing the pressure.
[4]
(b)
Write the expression for Kp, and give its units.
[2]
(c)
At equilibrium at a total pressure of 300 kPa, the mixture contains 0.40 mol CH4, 0.40 mol H2O, 0.30 mol CO and 0.90 mol H2. Calculate Kp.
[3]
  • (i) Yield increases
  • The forward reaction is endothermic, so equilibrium moves right to absorb the added heat
  • (ii) Yield decreases
  • There are more gas molecules on the right (4) than the left (2), so equilibrium moves left to reduce the pressure
  • Kp = pCO × pH23pCH4 × pH2O
  • kPa2 (or Pa2)
  • Total = 2.00 mol; mole fractions CH4 0.20, H2O 0.20, CO 0.15, H2 0.45
  • Partial pressures: CH4 60, H2O 60, CO 45, H2 135 kPa
  • Kp = 45 × 1353 / (60 × 60) = 3.08 × 104 kPa2
6Paper 2 style

2.00 mol of PCl5 is heated in a sealed 4.00 dm3 container until it reaches equilibrium: PCl5(g) ⇌ PCl3(g) + Cl2(g). At equilibrium, 0.50 mol of Cl2 is present.

(a)
Explain what is meant by dynamic equilibrium.
[2]
(b)
Calculate Kc, giving its units.
[3]
(c)
State the effect on the value of Kc of adding more Cl2 at constant temperature.
[1]
  • The rates of the forward and reverse reactions are equal
  • The concentrations of reactants and products stay constant (in a closed system)
  • Equilibrium amounts: PCl5 1.50 mol, PCl3 0.50 mol, Cl2 0.50 mol; concentrations 0.375, 0.125, 0.125 mol dm−3
  • Kc = 0.125 × 0.125 / 0.375 = 0.0417
  • mol dm−3
  • No change (only temperature changes Kc)

Kinetics is about how fast reactions go. Collision theory and the Boltzmann distribution explain why concentration, pressure, temperature and catalysts change the rate.

Rate of reaction

Rate of reaction
The change in concentration of a reactant or product per unit time. Units: mol dm−3 s−1.

Rates are measured by following a property that changes during the reaction: the volume of gas given off (gas syringe), the loss in mass as a gas escapes, a colour change (colorimeter), or by taking samples and titrating them.

  • The average rate over a period = change in concentration ÷ time taken.
  • The rate at a particular moment is the gradient of the tangent to the concentration–time curve at that time. The initial rate is the gradient at t = 0.

Worked example

The concentration of a reactant falls from 0.050 mol dm−3 to 0.030 mol dm−3 in the first 40 s. What is the average rate over this time?

rate = (0.050 − 0.030) / 40 = 5.0 × 10−4 mol dm−3 s−1

Collision theory

For particles to react, they must collide, with at least the activation energy and in the right orientation.

Activation energy, EA
The minimum energy required for a collision to be effective.
Effective collision
A collision that results in a reaction.
Non-effective collision
A collision in which the particles bounce apart unchanged, because they have too little energy or the wrong orientation.

The rate depends on the frequency of effective collisions: the number per second.

Higher concentration

More particles in the same volume, so collisions are more frequent, and so are effective collisions.

Higher gas pressure

The same as higher concentration: the gas particles are closer together.

Larger surface area

For a solid, more particles are exposed, so collisions with it are more frequent.

Temperature and the Boltzmann distribution

In a gas (or solution), particles have a range of energies. The Boltzmann distribution shows how many particles have each energy.

energynumber of molecules with energy ET₁T₂ (higher)EAmore molecules with E ≥ EA at T₂
Increasing the temperature: the peak moves to a higher energy and gets lower, and the curve spreads out. The total area (the number of particles) stays the same, but the area beyond EA is much larger.

Key features to show when you sketch the curve: it starts at the origin, has no particles with zero energy, rises to a peak (the most likely energy) and then falls towards the energy axis without touching or crossing it, because there is no maximum energy.

Why a higher temperature increases the rate:

  1. Particles have more kinetic energy, so a much larger proportion of collisions have energy equal to or greater than EA. This is the main effect.
  2. Particles also move faster, so they collide more often. This is a much smaller effect.

Together, these increase the frequency of effective collisions. For many reactions near room temperature, a rise of 10 °C roughly doubles the rate.

Catalysts

Catalyst
A substance that increases the rate of a reaction without being used up overall. It provides an alternative reaction pathway (a different mechanism) with a lower activation energy.

With a lower EA, a greater proportion of particles have enough energy to react, so the frequency of effective collisions increases. A catalyst does not change ΔH, and it does not change the position of equilibrium.

energynumber of molecules with energy EEAEA with catalystextra molecules that can react
enthalpy, Hprogress of reactionno catalystwith catalystreactantsproductsEAΔH negative
Left: a catalyst lowers EA, so more particles have enough energy to react; the curve itself does not change. Right: the catalysed pathway has a lower peak, but the reactants and products (and so ΔH) are unchanged.

Homogeneous and heterogeneous catalysts

TypeMeaningExamples
HomogeneousThe catalyst is in the same phase as the reactants.H+ ions in esterification (all in solution); Fe2+ in the reaction of I− with S2O82−; enzymes in solution
HeterogeneousThe catalyst is in a different phase from the reactants, usually a solid with gaseous or dissolved reactants.iron in the Haber process; V2O5 in the Contact process; Pt, Pd and Rh in catalytic converters; nickel in hydrogenating alkenes

How these catalysts work in detail (adsorption, and catalysts that change oxidation state) is covered in Topic 26.

Questions

1Paper 1 style

Why does increasing the temperature increase the rate of a reaction?

  1. The activation energy decreases.
  2. The particles collide more often, and this is the main effect.
  3. A greater proportion of collisions have energy equal to or greater than the activation energy.
  4. The reaction becomes more exothermic.

The main effect is that many more particles have at least EA. Collisions do become more frequent, but that is a much smaller effect. EA and ΔH don't change.

2Paper 1 style

What happens to the Boltzmann distribution curve of a gas when the temperature is increased?

  1. The peak moves to higher energy and becomes lower.
  2. The peak moves to higher energy and becomes higher.
  3. The peak stays at the same energy and becomes lower.
  4. The area under the curve increases.

The particles spread out over a wider range of energies. The number of particles is unchanged, so the area stays the same and the peak must get lower.

3Paper 1 style

Which is an example of heterogeneous catalysis?

  1. H+(aq) catalysing the hydrolysis of an ester in solution
  2. Fe2+(aq) catalysing the reaction between I−(aq) and S2O82−(aq)
  3. an enzyme in solution catalysing a reaction in solution
  4. solid iron catalysing the reaction between gaseous nitrogen and hydrogen

In D the catalyst (solid) is in a different phase from the reactants (gases). The others are all in solution.

4Paper 2 style
(a)
Define activation energy.
[1]
(b)
Sketch the Boltzmann distribution for a gas at temperature T1. On the same axes, sketch the curve at a higher temperature T2. Label the axes and mark EA.
[3]
(c)
Use your sketch to explain why the rate of reaction is greater at T2.
[2]
  • The minimum energy required for a collision to be effective / for a reaction to occur
  • Axes: number of molecules (with a given energy) against energy
  • T1 curve starts at the origin and does not touch or cross the energy axis at high energy
  • T2 curve has a lower peak, moved to the right, and is above T1 at high energies
  • More molecules have energy ≥ EA at T2 (larger area beyond EA)
  • So there are more effective / successful collisions per unit time
5Paper 2 and 3 style

Marble chips (calcium carbonate) react with excess hydrochloric acid: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g).

(a)
Describe how you could follow the rate of this reaction.
[2]
(b)
The experiment is repeated with the same mass of marble as a powder. Explain the effect on the rate.
[2]
(c)
The experiment is repeated using hydrochloric acid of half the concentration (still in excess). Explain the effect on the rate, and state the effect on the total volume of gas produced.
[3]
  • Collect the CO2 in a gas syringe (or measure mass loss on a balance)
  • Record the volume (or mass) at regular time intervals
  • The rate increases: a powder has a larger surface area
  • More particles are exposed, so collisions with acid particles are more frequent
  • The rate decreases
  • Fewer acid particles per unit volume, so less frequent (effective) collisions
  • The total volume of gas is the same, because the amount of marble (the limiting reagent) is the same

Period 3 runs from sodium to argon. Across it, the structure changes from giant metallic to giant covalent to simple molecular, and the oxides and chlorides change from ionic and basic to covalent and acidic. Learning why lets you predict the chemistry of any element from its position.

Physical properties across Period 3

Atomic and ionic radius

Atomic radius decreases from Na to Cl: the nuclear charge increases but the extra electrons go into the same (third) shell with similar shielding, so the outer electrons are pulled closer. For the ions, Na+ to Si4+ have lost their third shell and get smaller as nuclear charge rises; P3− to Cl− keep three shells, so they are much larger, and they also get smaller across the period. The values are in the table in Topic 1.

Melting point

NaMgAlSiPSClAr−20020060010001400melting point / °C98650660141444115-102-189giant metallicgiant covalentsimple molecularatoms
Melting points of the Period 3 elements. The structure, not the period number, explains the pattern.
ElementsStructureExplanation of melting point
Na, Mg, Algiant metallicRises from Na to Al: the number of delocalised electrons per atom increases (1, 2, 3) and the ions get smaller and more highly charged, so the metallic bonding gets stronger.
Sigiant covalent (like diamond)Highest: every atom is held by four strong covalent bonds, and very many must be broken.
P4, S8, Cl2simple molecularLow: only weak id-id forces between molecules are overcome. S8 (128 electrons) > P4 (60) > Cl2 (34), because larger molecules have stronger id-id forces.
Arsingle atomsLowest: very weak id-id forces between separate atoms (18 electrons).

Electrical conductivity

Na, Mg and Al are good conductors because of their delocalised electrons. Conductivity increases from Na to Al, as the number of delocalised electrons per atom increases. Silicon is a semiconductor: a poor conductor whose conductivity rises with temperature. Phosphorus, sulfur, chlorine and argon do not conduct, because all their electrons are held in covalent bonds or in single atoms.

Reactions of the elements

With oxygen

ElementEquationObservations
Na4Na(s) + O2(g) → 2Na2O(s)burns with a bright yellow-orange flame; white solid
Mg2Mg(s) + O2(g) → 2MgO(s)burns with a brilliant white flame; white solid
Al4Al(s) + 3O2(g) → 2Al2O3(s)a thin, protective oxide layer forms at once; the powder burns with white sparks; white solid
SiSi(s) + O2(g) → SiO2(s)only on strong heating; white solid
PP4(s) + 5O2(g) → P4O10(s)white phosphorus catches fire in air with a bright flame and dense white smoke
SS(s) + O2(g) → SO2(g)burns with a blue flame; colourless, choking gas. SO2 is oxidised further to SO3 only with a catalyst (Topic 7).

With chlorine

ElementEquationProduct
Na2Na(s) + Cl2(g) → 2NaCl(s)white ionic solid; bright yellow-orange flame
MgMg(s) + Cl2(g) → MgCl2(s)white ionic solid
Al2Al(s) + 3Cl2(g) → 2AlCl3(s)pale yellow / white solid that sublimes; covalent, forms Al2Cl6
SiSi(s) + 2Cl2(g) → SiCl4(l)colourless covalent liquid
P2P(s) + 5Cl2(g) → 2PCl5(s)pale yellow covalent solid (with excess chlorine; PCl3, a liquid, with limited chlorine)

Sodium and magnesium with water

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) vigorous: sodium floats, melts into a ball, fizzes and moves about. The solution is strongly alkaline, pH 13 to 14.
Mg(s) + 2H2O(l) → Mg(OH)2(aq) + H2(g) very slow with cold water: a few bubbles over days. The solution is weakly alkaline, about pH 9 to 10, because Mg(OH)2 is only slightly soluble.
Mg(s) + H2O(g) → MgO(s) + H2(g) with steam: magnesium burns with a bright white glow.

Oxidation numbers of the oxides and chlorides

OxideNa2OMgOAl2O3SiO2P4O10SO2 / SO3
Oxidation number+1+2+3+4+5+4 / +6
ChlorideNaClMgCl2AlCl3SiCl4PCl5
Oxidation number+1+2+3+4+5

The maximum oxidation number equals the number of outer-shell electrons, because in the highest oxide or chloride the element uses all of its outer electrons in bonding. In the chlorides it stops at +5 for phosphorus, because there isn't room around a sulfur atom for six chlorine atoms.

The oxides: acid–base behaviour

OxideBondingWith waterpHNature
Na2OionicNa2O + H2O → 2NaOH13–14basic
MgOionicMgO + H2O → Mg(OH)2 (slightly soluble)about 9–10basic
Al2O3ionic, with some covalent characterinsoluble7amphoteric
SiO2giant covalentinsoluble7acidic
P4O10simple molecular covalentP4O10 + 6H2O → 4H3PO4about 2acidic
SO2simple molecular covalentSO2 + H2O → H2SO3about 2–3acidic
SO3simple molecular covalentSO3 + H2O → H2SO4about 1acidic

Reactions with acids and bases

Na2O(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l) basic oxide with acid
MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l) basic oxide with acid
Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(l) amphoteric oxide reacting as a base
Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2NaAl(OH)4(aq) amphoteric oxide reacting as an acid: sodium tetrahydroxoaluminate forms
SiO2(s) + 2NaOH(aq) → Na2SiO3(aq) + H2O(l) acidic oxide; needs hot, concentrated NaOH. Sodium silicate forms.
P4O10(s) + 12NaOH(aq) → 4Na3PO4(aq) + 6H2O(l) acidic oxide
SO2(g) + 2NaOH(aq) → Na2SO3(aq) + H2O(l) and SO3(g) + 2NaOH(aq) → Na2SO4(aq) + H2O(l) acidic oxides

The hydroxides follow the same pattern: NaOH and Mg(OH)2 are basic, and aluminium hydroxide is amphoteric:

Al(OH)3(s) + 3HCl(aq) → AlCl3(aq) + 3H2O(l) and Al(OH)3(s) + NaOH(aq) → NaAl(OH)4(aq)

Explaining the trend

Across the period, the electronegativity of the element increases, so the difference from oxygen decreases. The bonding in the oxides changes from ionic (Na, Mg) to covalent (Si to S), with aluminium in between.

  • Ionic oxides contain O2− ions, which accept protons: O2− + H2O → 2OH−. So they are basic.
  • Covalent oxides of non-metals react with water to form acids, or react with bases. So they are acidic.
  • Aluminium oxide sits between the two and shows both behaviours: it is amphoteric.

The chlorides with water

ChlorideBondingWhat happens in waterpH
NaClionicdissolves; the ions are surrounded by water molecules but don't react7
MgCl2ionicdissolves, with very slight hydrolysisabout 6.5
AlCl3covalentreacts (hydrolyses) to form an acidic solution; with a little water, steamy fumes of HClabout 3
SiCl4covalenthydrolyses violently: white solid SiO2 and steamy fumes of HClabout 1–2
PCl5covalenthydrolyses violently; steamy fumes of HClabout 1–2
AlCl3(s) + 6H2O(l) → [Al(H2O)6]3+(aq) + 3Cl−(aq), then [Al(H2O)6]3+(aq) ⇌ [Al(H2O)5(OH)]2+(aq) + H+(aq) the small, highly charged Al3+ ion pulls electron density from a water ligand, which releases H+
SiCl4(l) + 2H2O(l) → SiO2(s) + 4HCl(aq)
PCl5(s) + 4H2O(l) → H3PO4(aq) + 5HCl(aq)

Explaining the trend: ionic chlorides simply dissolve, giving neutral solutions. Covalent chlorides are hydrolysed by water to form HCl, so their solutions are strongly acidic. As with the oxides, the bonding changes from ionic to covalent across the period as electronegativity increases.

Inferring bonding from properties

A chloride or oxide with a high melting point that conducts when molten is ionic. One with a low melting point that reacts with water to give an acidic solution is covalent. Expect to be given data and asked to decide which.

Periodicity of other elements

The same patterns repeat in every period, so you can predict an element's properties from its position. For example, germanium (Group 14, Period 4) should be a semiconductor like silicon, form GeO2 and GeCl4, and its chloride should be hydrolysed by water. Use the explorer to compare electronegativity and first ionisation energy across the table.

Worked example: identifying an unknown element

Element X is in Period 3. Its oxide has a very high melting point and does not dissolve in water, but it does react with hot, concentrated sodium hydroxide. It does not react with acids. Its chloride is a liquid that fumes in moist air. Identify X.

X is silicon. The oxide is giant covalent and acidic (SiO2); the chloride SiCl4 is a covalent liquid that hydrolyses to give HCl fumes. (Aluminium oxide would also react with acids.)

Questions

1Paper 1 style

Which Period 3 element has the highest melting point?

  1. aluminium
  2. silicon
  3. phosphorus
  4. sulfur

Silicon has a giant covalent structure, so many strong covalent bonds must be broken.

2Paper 1 style

Which oxide reacts with both hydrochloric acid and aqueous sodium hydroxide?

  1. MgO
  2. SiO2
  3. Al2O3
  4. P4O10

Aluminium oxide is amphoteric. MgO is basic only; SiO2 and P4O10 are acidic only.

3Paper 1 style

A Period 3 chloride is added to water. The solution has a pH of 7. Which chloride was it?

  1. AlCl3
  2. SiCl4
  3. PCl5
  4. NaCl

Sodium chloride is ionic and dissolves without hydrolysis. The others are hydrolysed to give acidic solutions.

4Paper 2 style
(a)
Explain why the melting point of aluminium is higher than that of sodium.
[2]
(b)
Explain why the melting point of sulfur (S8) is higher than that of phosphorus (P4).
[2]
(c)
Write equations to show the amphoteric nature of aluminium oxide.
[2]
(d)
Write an equation for the reaction of silicon(IV) chloride with water, and state the approximate pH of the resulting mixture.
[2]
  • Al has more delocalised electrons per atom (3 compared with 1) and a smaller ion with a higher charge
  • So the metallic bonding (attraction between ions and delocalised electrons) is stronger
  • Both are simple molecular, so id-id forces are overcome
  • S8 has more electrons than P4, so its id-id forces are stronger
  • Al2O3 + 6HCl → 2AlCl3 + 3H2O
  • Al2O3 + 2NaOH + 3H2O → 2NaAl(OH)4
  • SiCl4 + 2H2O → SiO2 + 4HCl
  • pH 1 to 2
5Paper 2 style

Sodium oxide and sulfur trioxide both react with water.

(a)
Write equations for both reactions and state the approximate pH of each solution.
[3]
(b)
Explain, in terms of bonding, why the two oxides give solutions with such different pH values.
[2]
(c)
Write an equation for the reaction of sulfur trioxide with sodium hydroxide.
[1]
  • Na2O + H2O → 2NaOH
  • SO3 + H2O → H2SO4
  • Na2O: pH 13–14; SO3: pH 0–1
  • Na2O is ionic: its oxide ions react with water to form hydroxide ions
  • SO3 is covalent (simple molecular): it reacts with water to form an acid that releases H+
  • SO3 + 2NaOH → Na2SO4 + H2O

The Group 2 metals, magnesium to barium, all form 2+ ions and get more reactive down the group. Their compounds show clear trends in solubility and thermal stability, which come up often in Paper 1.

The elements

MgCaSrBa
Electronic configuration[Ne] 3s2[Ar] 4s2[Kr] 5s2[Xe] 6s2
Flame colournone (bright white when burning)brick redcrimson redapple green

Reactivity increases down the group. Each atom reacts by losing its two outer electrons to form an M2+ ion. Down the group, the atoms get larger and the outer electrons are more shielded, so the first and second ionisation energies decrease. The electrons are lost more easily.

Reactions of the metals

ReactantGeneral equationNotes
oxygen2M(s) + O2(g) → 2MO(s)burn with the flame colours above to give white oxides; faster down the group
cold waterM(s) + 2H2O(l) → M(OH)2(aq) + H2(g)Mg reacts extremely slowly; Ca fizzes steadily, giving a cloudy white suspension of Ca(OH)2; Sr and Ba react faster still
steamMg(s) + H2O(g) → MgO(s) + H2(g)magnesium burns with a bright white flame
dilute hydrochloric acidM(s) + 2HCl(aq) → MCl2(aq) + H2(g)all fizz; faster down the group; the chlorides are soluble
dilute sulfuric acidM(s) + H2SO4(aq) → MSO4 + H2(g)Mg reacts fully. With Ca, Sr and Ba the reaction quickly slows and stops, because an insoluble sulfate coating forms on the metal

Reactions of the compounds

CompoundReactionEquation
oxideswith waterMO(s) + H2O(l) → M(OH)2; CaO reacts vigorously and gives out heat (making "slaked lime")
oxideswith dilute acidMO(s) + 2HCl(aq) → MCl2(aq) + H2O(l)
hydroxideswith dilute acidM(OH)2 + 2HCl(aq) → MCl2(aq) + 2H2O(l)
carbonateswith dilute acidMCO3(s) + 2HCl(aq) → MCl2(aq) + H2O(l) + CO2(g); with sulfuric acid, the reaction of CaCO3, SrCO3 and BaCO3 soon stops (insoluble sulfate coating)
carbonateswith waterinsoluble; no reaction

The oxides and hydroxides are bases. Solutions of the hydroxides are alkaline, and the pH increases down the group because the hydroxides get more soluble.

Thermal decomposition

MCO3(s) → MO(s) + CO2(g) carbonates decompose on heating to the oxide and carbon dioxide
2M(NO3)2(s) → 2MO(s) + 4NO2(g) + O2(g) nitrates decompose to the oxide, brown nitrogen dioxide and oxygen

Thermal stability increases down the group: you need a higher temperature to decompose BaCO3 than MgCO3. (The explanation, in terms of how much the cation polarises the large carbonate or nitrate ion, is an A Level topic: see Topic 27.)

Solubility trends

Hydroxides: more soluble down the group

Mg(OH)2 is only sparingly soluble; Ba(OH)2 is quite soluble. So limewater, Ca(OH)2(aq), is only weakly alkaline, while barium hydroxide solution is strongly alkaline.

Sulfates: less soluble down the group

MgSO4 is soluble; BaSO4 is insoluble. Adding acidified barium chloride solution to a solution containing sulfate ions gives a white precipitate: Ba2+(aq) + SO42−(aq) → BaSO4(s).

Some uses follow from these properties: calcium hydroxide is used to neutralise acidic soil, magnesium hydroxide is used in indigestion remedies, and insoluble barium sulfate can be swallowed safely for X-ray images of the gut.

Making predictions

Use the trends to predict beryllium or radium chemistry, or unfamiliar compounds. For example, radium should react with water even more vigorously than barium, and radium sulfate should be even less soluble.

Questions

1Paper 1 style

Which statement about the Group 2 elements, from magnesium to barium, is correct?

  1. The solubility of the hydroxides increases.
  2. The solubility of the sulfates increases.
  3. The thermal stability of the carbonates decreases.
  4. The first ionisation energy increases.

Hydroxides become more soluble down the group. Sulfates become less soluble, carbonates become more stable and first ionisation energy decreases.

2Paper 1 style

What are the products when calcium nitrate is heated strongly?

  1. Ca(NO2)2 and O2
  2. CaO and NO2 only
  3. CaO, NO2 and O2
  4. Ca, N2 and O2

2Ca(NO3)2 → 2CaO + 4NO2 + O2. (Group 1 nitrates other than lithium form the nitrite and oxygen instead.)

3Paper 1 style

Barium reacts with dilute sulfuric acid, but the reaction stops very soon. Why?

  1. Barium is less reactive than magnesium.
  2. The acid is used up.
  3. Hydrogen gas forms a layer over the metal.
  4. Insoluble barium sulfate coats the metal.

BaSO4 is insoluble, so it forms a layer on the barium that stops the acid reaching the metal.

4Paper 2 style
(a)
Write an equation, with state symbols, for the reaction of strontium with water.
[2]
(b)
Explain why strontium reacts with water more vigorously than calcium.
[2]
(c)
The solution formed in (a) is tested with universal indicator. Suggest its pH, compared with the solution formed by calcium. Explain your answer.
[2]
(d)
Describe a test to show that a solution contains sulfate ions. Give the observation and an ionic equation.
[3]
  • Sr + 2H2O → Sr(OH)2 + H2
  • State symbols: Sr(s), H2O(l), Sr(OH)2(aq), H2(g)
  • Sr atoms are larger / outer electrons more shielded and further from the nucleus
  • So the (first and second) ionisation energies are lower and electrons are lost more easily
  • Higher pH (about 13 compared with about 12)
  • Sr(OH)2 is more soluble, so the concentration of OH− is higher
  • Add dilute HCl (or HNO3) then barium chloride (or barium nitrate) solution
  • White precipitate
  • Ba2+(aq) + SO42−(aq) → BaSO4(s)

The halogens are reactive non-metals that exist as diatomic molecules. Down the group they become less volatile, weaker oxidising agents and more strongly coloured, while their halide ions become stronger reducing agents. These trends explain a set of classic test-tube reactions.

Physical properties

Chlorine, Cl2Bromine, Br2Iodine, I2
State at room temperaturegasliquidsolid
Appearancepale yellow-green gasred-brown liquid, orange-brown vapourgrey-black shiny solid, purple vapour
In watervery pale green (almost colourless)orange to yellowbrown (dissolves only slightly, better in KI solution)
In a non-polar solvent such as hexanevery pale greenorange-redpurple / violet
Boiling point / °C−3459184
Bond energy, X–X / kJ mol−1242193151

Volatility decreases down the group

The halogens are simple molecular. Down the group, the molecules have more electrons, so the instantaneous dipole–induced dipole forces between them get stronger. More energy is needed to separate the molecules, so boiling points rise and volatility falls.

Bond strength decreases down the group

The atoms get larger, so the bonding pair of electrons is further from the two nuclei and is attracted less strongly. (Fluorine, F–F 158 kJ mol−1, is an exception: its atoms are so small that the lone pairs on the two atoms repel each other.)

The halogens as oxidising agents

A halogen reacts by gaining an electron: X2 + 2e− → 2X−. So halogens are oxidising agents. Oxidising power decreases down the group: chlorine is the strongest and iodine the weakest of the three.

Explanation: down the group the atoms get larger and the outer shell is more shielded, so there is less attraction between the nucleus and an incoming electron.

Displacement reactions

A halogen displaces a less reactive halogen from a solution of its halide ions.

Halogen added ↓ / halide ion →Cl−(aq)Br−(aq)I−(aq)
Cl2(aq)–yellow-orange solution: Br2 formsbrown solution: I2 forms
Br2(aq)no reaction–brown solution: I2 forms
I2(aq)no reactionno reaction–
Cl2(aq) + 2Br−(aq) → 2Cl−(aq) + Br2(aq)
Cl2(aq) + 2I−(aq) → 2Cl−(aq) + I2(aq)
Br2(aq) + 2I−(aq) → 2Br−(aq) + I2(aq)

Bromine and iodine solutions can look similar. Shaking with hexane makes it clear: bromine gives an orange-red upper layer, iodine a purple one.

Reactions with hydrogen

All the halogens react with hydrogen to form hydrogen halides, H2(g) + X2 → 2HX(g), but reactivity falls sharply down the group:

HalogenReaction with hydrogen
fluorineexplosive, even in the cold and dark
chlorineexplosive in sunlight (or a flame); slow in the dark
bromineneeds heating (about 300 °C) with a platinum catalyst; not explosive
iodineslow and incomplete even when heated with a catalyst; the reaction is reversible

Thermal stability of the hydrogen halides

Down the group the hydrogen halides become less stable to heat. Hydrogen chloride does not decompose at ordinary temperatures, hydrogen bromide decomposes slightly on strong heating, and a hot glass rod plunged into hydrogen iodide produces purple fumes of iodine: 2HI(g) → H2(g) + I2(g).

Explanation: the H–X bond energy decreases (H–Cl 431, H–Br 366, H–I 299 kJ mol−1), because the bond gets longer as the halogen atom gets larger. A weaker bond is broken more easily.

Reactions of the halide ions

Halide ions as reducing agents

A halide ion acts as a reducing agent by losing an electron: 2X− → X2 + 2e−. Reducing power increases down the group: iodide is the strongest reducing agent. Down the group, the ions get larger and the outer electrons are further from the nucleus and more shielded, so they are lost more easily.

Testing with silver nitrate and ammonia

Acidify the solution with dilute nitric acid (to remove carbonate ions, which would also give a precipitate), then add aqueous silver nitrate. Then test the precipitate with aqueous ammonia.

HalideWith AgNO3(aq)Add dilute NH3(aq)Add concentrated NH3(aq)
chloridewhite precipitate, AgCldissolvesdissolves
bromidecream precipitate, AgBrdoes not dissolvedissolves
iodideyellow precipitate, AgIdoes not dissolvedoes not dissolve
Ag+(aq) + X−(aq) → AgX(s) the ionic equation for all three precipitates
AgCl(s) + 2NH3(aq) → [Ag(NH3)2]+(aq) + Cl−(aq) the precipitate dissolves because a soluble complex ion forms

The precipitates get less soluble from AgCl to AgI, so it takes more concentrated ammonia to dissolve them, and AgI doesn't dissolve at all.

Reactions with concentrated sulfuric acid

Adding concentrated sulfuric acid to a solid halide first gives the hydrogen halide (an acid–base reaction). Bromide and iodide are strong enough reducing agents to go on and reduce the sulfuric acid; chloride is not.

HalideEquationsObservationsS oxidation number
NaClNaCl + H2SO4 → NaHSO4 + HClsteamy white fumes of HClstays +6 (no redox)
NaBrNaBr + H2SO4 → NaHSO4 + HBr
2HBr + H2SO4 → Br2 + SO2 + 2H2O
steamy fumes of HBr, and orange-brown vapour of Br2; colourless, choking SO2+6 → +4
NaINaI + H2SO4 → NaHSO4 + HI
2HI + H2SO4 → I2 + SO2 + 2H2O
6HI + H2SO4 → 3I2 + S + 4H2O
8HI + H2SO4 → 4I2 + H2S + 4H2O
some steamy fumes, purple vapour and black solid of I2, yellow solid sulfur, and a smell of rotten eggs (H2S)+6 → +4, 0 and −2

The further the sulfur is reduced, the stronger the reducing agent. Iodide reduces sulfur all the way from +6 to −2.

The reactions of chlorine

With sodium hydroxide: disproportionation

Cl2(aq) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H2O(l) cold, dilute NaOH. Cl: 0 → −1 (in NaCl) and 0 → +1 (in sodium chlorate(I), NaClO). This solution is household bleach.
3Cl2(aq) + 6NaOH(aq) → 5NaCl(aq) + NaClO3(aq) + 3H2O(l) hot, concentrated NaOH. Cl: 0 → −1 and 0 → +5 (in sodium chlorate(V), NaClO3).

Both are disproportionation reactions: chlorine is oxidised and reduced at the same time.

Chlorine in water purification

Chlorine is added to drinking water to kill bacteria. It reacts with the water (another disproportionation):

Cl2(aq) + H2O(l) ⇌ HCl(aq) + HOCl(aq) chloric(I) acid, HOCl, forms; Cl: 0 → −1 and +1
HOCl(aq) ⇌ H+(aq) + ClO−(aq) HOCl is a weak acid, so it partly dissociates to chlorate(I) ions

Both HOCl and ClO− are strong oxidising agents that kill bacteria. They are the active agents in the water.

Questions

1Paper 1 style

Why does the boiling point of the halogens increase from chlorine to iodine?

  1. The X–X bond energy increases.
  2. The molecules become more polar.
  3. Hydrogen bonding between molecules increases.
  4. Instantaneous dipole–induced dipole forces increase.

More electrons per molecule means stronger id-id forces. The X–X bond energy actually decreases, and boiling doesn't break these bonds anyway.

2Paper 1 style

A white precipitate forms when silver nitrate is added to an acidified solution of X. The precipitate dissolves in dilute aqueous ammonia. What is X?

  1. sodium bromide
  2. sodium chloride
  3. sodium iodide
  4. sodium nitrate

Silver chloride is white and dissolves in dilute ammonia. AgBr is cream and AgI is yellow; sodium nitrate gives no precipitate.

3Paper 1 style

Concentrated sulfuric acid is added to solid sodium iodide. Which product shows that sulfur has been reduced to an oxidation number of −2?

  1. SO2
  2. S
  3. H2S
  4. NaHSO4

In H2S, S is −2. In SO2 it is +4, in S it is 0, and in NaHSO4 it is still +6.

4Paper 1 style

What is the oxidation number of chlorine in the chlorine-containing product (other than NaCl) of the reaction between chlorine and hot, concentrated sodium hydroxide?

  1. +5
  2. +1
  3. +3
  4. +7

Hot, concentrated NaOH gives sodium chlorate(V), NaClO3: Na + Cl + 3(−2) = 0, so Cl = +5. Cold, dilute NaOH gives NaClO, with Cl at +1.

5Paper 2 style
(a)
Chlorine water is added to aqueous potassium bromide. Describe what you would see, and write an ionic equation.
[2]
(b)
Explain why chlorine is a stronger oxidising agent than bromine.
[2]
(c)
A hot glass rod is placed in a jar of hydrogen iodide. Purple fumes are seen. Explain this observation, using bond energies H–Cl 431 and H–I 299 kJ mol−1 to compare HI with HCl.
[2]
  • Colourless solution turns orange / yellow-orange
  • Cl2 + 2Br− → 2Cl− + Br2
  • Chlorine atoms are smaller / the outer shell is closer to the nucleus and less shielded
  • So there is a stronger attraction for an electron being gained
  • HI decomposes to hydrogen and (purple) iodine vapour: 2HI → H2 + I2
  • The H–I bond is much weaker than H–Cl (299 compared with 431), so HI is less thermally stable
6Paper 2 style
(a)
Concentrated sulfuric acid is added separately to solid sodium chloride and to solid sodium bromide. Describe what you would see in each case, and write equations for the reactions with sodium bromide.
[4]
(b)
Explain the difference between the two reactions in terms of the reducing power of the halide ions.
[2]
(c)
Chlorine is used to treat drinking water. Write an equation for the reaction of chlorine with water and name the species that kill bacteria.
[2]
  • NaCl: steamy (misty) white fumes only
  • NaBr: steamy fumes and orange-brown vapour (bromine)
  • NaBr + H2SO4 → NaHSO4 + HBr
  • 2HBr + H2SO4 → Br2 + SO2 + 2H2O
  • Bromide is a stronger reducing agent than chloride
  • Bromide (HBr) reduces S from +6 to +4; chloride is too weak to reduce the sulfuric acid
  • Cl2 + H2O ⇌ HCl + HOCl
  • HOCl and ClO−

Nitrogen makes up 78% of the air but is remarkably unreactive. Its compounds, though, matter a great deal: ammonia is a base and a fertiliser, and oxides of nitrogen from engines cause smog and acid rain, helped along by sulfur dioxide.

The lack of reactivity of nitrogen

Nitrogen, N2, is unreactive for two reasons:

  • The N≡N triple bond is very strong (944 kJ mol−1), so a lot of energy is needed to break it: reactions have a very high activation energy.
  • The molecule is non-polar: both atoms are identical, so there are no δ+ or δ− sites to attract electrophiles or nucleophiles.

Ammonia and the ammonium ion

Ammonia, NH3, is a weak Brønsted–Lowry base: the lone pair on nitrogen accepts a proton.

NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq) aqueous ammonia is weakly alkaline (about pH 11)
NH3(aq) + H+(aq) → NH4+(aq) with acids, forming ammonium salts, for example NH3 + HCl → NH4Cl

The ammonium ion, NH4+, is tetrahedral (109.5°). The new N–H bond is a coordinate bond, but once formed all four bonds are identical (Topic 3).

Displacing ammonia from ammonium salts

Warming any ammonium salt with a stronger base, such as aqueous sodium hydroxide or calcium hydroxide, releases ammonia gas. This is an acid–base reaction: the hydroxide ion takes a proton from the ammonium ion.

NH4Cl(s) + NaOH(aq) → NaCl(aq) + H2O(l) + NH3(g)
NH4+(aq) + OH−(aq) → NH3(g) + H2O(l) the ionic equation

This is the test for ammonium ions: the gas turns damp red litmus paper blue. It is also why ammonium fertilisers should not be spread at the same time as lime (calcium hydroxide): ammonia would be lost.

Oxides of nitrogen

Nitrogen and oxygen in the air react only at very high temperatures:

N2(g) + O2(g) → 2NO(g) nitrogen monoxide forms naturally in lightning, and in car engines and power stations, where fuel burns at high temperature
2NO(g) + O2(g) → 2NO2(g) in the air, NO is oxidised to brown nitrogen dioxide

Removing oxides of nitrogen from exhaust gases

A catalytic converter in a car exhaust contains platinum, palladium and rhodium on a honeycomb support with a large surface area. It converts oxides of nitrogen and carbon monoxide into harmless gases:

2CO(g) + 2NO(g) → 2CO2(g) + N2(g)

How the metal surface catalyses this (adsorption of the gases, bond weakening, desorption of the products) is covered in Topic 26.

Photochemical smog and PAN

In sunny, still weather, oxides of nitrogen react with unburnt hydrocarbons from vehicle exhausts to form peroxyacetyl nitrate, PAN, CH3CO–O–O–NO2. PAN is a key part of photochemical smog: it irritates the eyes and lungs and damages plants. Sunlight starts the process by splitting NO2 into NO and oxygen atoms; the radicals formed react with the hydrocarbons, and the products combine with NO2.

Acid rain

Fossil fuels contain sulfur, which burns to sulfur dioxide. Oxides of nitrogen act as a catalyst for oxidising atmospheric sulfur dioxide to sulfur trioxide:

SO2(g) + NO2(g) → SO3(g) + NO(g) NO2 oxidises SO2
NO(g) + ½O2(g) → NO2(g) NO2 is regenerated, so it acts as a catalyst
SO3(g) + H2O(l) → H2SO4(aq) sulfuric acid in rain

Nitrogen dioxide also forms nitric acid directly: 4NO2(g) + O2(g) + 2H2O(l) → 4HNO3(aq).

Acid rain damages buildings made of limestone and marble (CaCO3), corrodes metals, acidifies lakes so that fish die, and damages trees by leaching nutrients from the soil.

Questions

1Paper 1 style

What is the role of nitrogen dioxide in the formation of acid rain from sulfur dioxide?

  1. It is a reducing agent.
  2. It is a catalyst.
  3. It is a product.
  4. It absorbs sunlight.

NO2 oxidises SO2 to SO3 and is regenerated when NO reacts with oxygen, so it acts as a catalyst overall.

2Paper 1 style

Which statement explains why ammonia acts as a base?

  1. It has polar N–H bonds.
  2. It forms hydrogen bonds with water.
  3. Its nitrogen atom has a lone pair that can accept a proton.
  4. It dissolves to form NH4+ and H+ ions.

A Brønsted–Lowry base accepts a proton, using the lone pair on nitrogen.

3Paper 1 style

Why is nitrogen gas, N2, very unreactive?

  1. N2 molecules are polar
  2. the N≡N triple bond has a very high bond energy
  3. nitrogen has a very high electronegativity
  4. nitrogen has a giant covalent structure

The N≡N bond energy is 994 kJ mol−1, so a lot of energy is needed to break it. The molecule is also non-polar, so it doesn't attract electrophiles or nucleophiles.

4Paper 1 style

Nitrogen dioxide catalyses the oxidation of sulfur dioxide in the atmosphere. Which equation shows the step in which SO2 is oxidised?

  1. SO2 + NO2 → SO3 + NO
  2. 2NO + O2 → 2NO2
  3. SO3 + H2O → H2SO4
  4. 2SO2 + O2 → 2SO3

NO2 oxidises SO2 and is reduced to NO. The NO is then reoxidised by oxygen (B), so NO2 is regenerated, which is why it is a catalyst. D is the overall equation.

5Paper 1 style

Which statement about the ammonium ion, NH4+, is correct?

  1. it is pyramidal, with bond angles of 107°
  2. it has a lone pair of electrons on the nitrogen atom
  3. ammonia is released when an ammonium salt is warmed with dilute acid
  4. it forms when ammonia accepts a proton, acting as a Brønsted–Lowry base

NH3 uses its lone pair to form a dative bond to H+, so NH4+ is tetrahedral (109.5°) with no lone pair. Ammonia is released by warming with an alkali, not an acid.

6Paper 2 style
(a)
Give two reasons why nitrogen gas is unreactive.
[2]
(b)
Explain how nitrogen monoxide forms in a car engine, and write an equation for its removal in a catalytic converter.
[3]
(c)
Describe, with equations, how nitrogen dioxide helps to convert sulfur dioxide into sulfuric acid in the atmosphere.
[3]
(d)
Name the pollutant that forms when oxides of nitrogen react with unburnt hydrocarbons in sunlight.
[1]
  • Strong N≡N triple bond (high bond energy)
  • The molecule is non-polar
  • The high temperature in the engine provides enough energy for N2 and O2 from the air to react
  • N2 + O2 → 2NO
  • 2CO + 2NO → 2CO2 + N2
  • SO2 + NO2 → SO3 + NO
  • NO + ½O2 → NO2, so NO2 is regenerated / acts as a catalyst
  • SO3 + H2O → H2SO4
  • PAN / peroxyacetyl nitrate

Organic chemistry is the chemistry of carbon compounds. Millions exist, but they fall into a few families, each defined by its functional group. This topic sets up the language you need for every organic topic: formulas, names, reaction types, mechanisms and isomers.

Hydrocarbons and homologous series

Hydrocarbon
A compound containing carbon and hydrogen only.
Alkane
A saturated hydrocarbon, with only single bonds. General formula CnH2n+2 (for non-cyclic alkanes).
Homologous series
A family of compounds with the same functional group and general formula, in which each member differs from the next by CH2. Members have similar chemical properties and a gradual trend in physical properties such as boiling point.
Functional group
The atom or group of atoms that gives a family of compounds its characteristic reactions.
Saturated / unsaturated
Saturated compounds have only single C–C bonds. Unsaturated compounds contain at least one C=C (or C≡C) multiple bond.

Functional groups at AS Level

FamilyFunctional groupName ending or prefixExample
alkaneC–C, C–H only-anepropane, CH3CH2CH3
alkeneC=C-enepropene, CH3CH=CH2
halogenoalkaneC–X (X = F, Cl, Br, I)fluoro-, chloro-, bromo-, iodo-bromoethane, CH3CH2Br
alcohol–OH-ol (or hydroxy-)ethanol, CH3CH2OH
aldehyde–CHO-alethanal, CH3CHO
ketoneC–CO–C-onepropanone, CH3COCH3
carboxylic acid–COOH-oic acidethanoic acid, CH3COOH
ester–COO–-yl -oateethyl ethanoate, CH3COOCH2CH3
amine (primary)–NH2-amine (or amino-)ethylamine, CH3CH2NH2
nitrile–C≡N-nitrilepropanenitrile, CH3CH2CN

Types of formula

FormulaShowsButan-1-ol
molecularthe number of each type of atomC4H10O
empiricalthe simplest whole-number ratioC4H10O
generalthe formula for a whole seriesCnH2n+1OH
structuralthe arrangement of atoms, without drawing every bondCH3CH2CH2CH2OH
displayedevery atom and every bondsee below
skeletalthe carbon skeleton as zig-zag lines; C atoms at the corners and ends, H atoms on carbon not shown, functional groups drawn in fullsee below
CCCCOHHHHHHHHHHdisplayed formula
OHskeletal formula

Naming organic compounds

  1. Find the longest carbon chain containing the functional group. Its length gives the stem: meth (1), eth (2), prop (3), but (4), pent (5), hex (6).
  2. Add the suffix for the main functional group (-ane, -ene, -ol, -al, -one, -oic acid, -nitrile, -amine).
  3. Number the chain so that the functional group (or the first branch) gets the lowest number.
  4. Add prefixes for branches and halogens in alphabetical order, with position numbers: methyl, ethyl, bromo, chloro. Use di-, tri- for repeats (these don't count for alphabetical order).
StructureNameWhy
CH3CH(CH3)CH2CH32-methylbutanelongest chain 4 C; methyl on C2
CH2=CHCH2CH3but-1-eneC=C starts at C1
CH3CHBrCH32-bromopropaneBr on the middle carbon
CH3CH(OH)CH2CH3butan-2-olsecondary alcohol, OH on C2
CH3CH2CHOpropanalthe aldehyde C is always C1, so no number is needed
CH3COCH2CH3butanonethe C=O can only be at C2 in a 4-carbon ketone
CH2ClCH2Cl1,2-dichloroethanetwo Cl atoms, one on each carbon
(CH3)2C(OH)CH32-methylpropan-2-oltertiary alcohol
CH3CH2CNpropanenitrilethe carbon of C≡N is counted in the chain

Characteristic organic reactions

How bonds break

Homolytic fission
A covalent bond breaks so that each atom takes one electron of the shared pair, forming two free radicals: Cl–Cl → Cl• + Cl•.
Heterolytic fission
A covalent bond breaks so that one atom takes both electrons, forming a positive ion and a negative ion: H–Br → H+ + Br−.

Attacking species

Nucleophile
An electron-pair donor. It is attracted to a δ+ or positive carbon. Examples: OH−, CN−, NH3, H2O.
Electrophile
An electron-pair acceptor. It is attracted to an electron-rich region such as a C=C. Examples: H+, Br+ (from a polarised Br2), NO2+.
Free radical
A species with an unpaired electron, such as Cl• or CH3•.

Types of reaction

TypeWhat happensExample
additiontwo molecules join to make oneC2H4 + Br2 → C2H4Br2
substitutionone atom or group is replaced by anotherCH3CH2Br + OH− → CH3CH2OH + Br−
eliminationa small molecule is removed from a larger one, often forming a C=CCH3CH2OH → CH2=CH2 + H2O
hydrolysisa compound is broken down by reaction with water (or OH− or H+)ester + water → acid + alcohol
condensationtwo molecules join, losing a small molecule such as wateracid + alcohol → ester + water
oxidationgain of O or loss of H, shown by [O]CH3CH2OH + [O] → CH3CHO + H2O
reductiongain of H or loss of O, shown by [H]CH3CHO + 2[H] → CH3CH2OH

Mechanisms and curly arrows

A mechanism shows how a reaction happens, step by step. A curly arrow shows the movement of a pair of electrons. It must start from a lone pair or a bond, and point to where the electrons end up: an atom (forming a bond) or another atom (when a bond breaks). The four mechanisms at AS Level are:

  • free-radical substitution: alkanes with halogens (Topic 14)
  • electrophilic addition: alkenes (Topic 14)
  • nucleophilic substitution: halogenoalkanes (Topic 15)
  • nucleophilic addition: aldehydes and ketones (Topic 17)

Shapes of organic molecules

Carbon chains can be straight, branched or cyclic. The shape around each carbon depends on its bonds (Topic 3):

Carbon atomHybridisationShape and angleExample
four single bondssp3tetrahedral, 109.5°every C in an alkane
one double bond (3 σ + 1 π)sp2trigonal planar, 120°the C atoms of C=C and C=O
one triple bond (2 σ + 2 π)splinear, 180°the C of C≡N

Atoms attached directly to a C=C double bond lie in the same plane as the two carbons: ethene is a planar molecule. Because the π bond prevents rotation about C=C, groups attached to each end are locked in place.

Isomerism

Structural isomers
Compounds with the same molecular formula but different structural formulas.
Stereoisomers
Compounds with the same structural formula but a different arrangement of atoms in space.

Structural isomerism

butanechain isomers of C4H10
methylpropanechain isomers of C4H10
OHpropan-1-olpositional isomers
OHpropan-2-olpositional isomers
Opropanalfunctional group isomers of C3H6O
Opropanonefunctional group isomers of C3H6O
  • Chain isomers have different carbon skeletons (straight or branched).
  • Positional isomers have the same functional group at different positions on the chain.
  • Functional group isomers have different functional groups, for example an aldehyde and a ketone, or an alcohol and an ether.

Geometrical (cis/trans) isomerism

Geometrical isomerism happens when there is restricted rotation about a C=C double bond (because of the π bond) and each carbon of the C=C has two different groups attached. In the cis isomer the two similar groups are on the same side; in the trans isomer they are on opposite sides.

CCH3CCH3HHcis-but-2-eneboth CH₃ on the same side; bp 4 °C
CCH3CCH3HHtrans-but-2-eneCH₃ groups on opposite sides; bp 1 °C
But-1-ene, CH2=CHCH2CH3, has no cis/trans isomers, because one carbon of the C=C carries two identical H atoms.

Optical isomerism

A chiral centre is a carbon atom bonded to four different atoms or groups. A molecule with one chiral centre exists as two optical isomers (enantiomers): non-superimposable mirror images of each other, like a left and a right hand.

COHHC2H5CH3CHOHH5C2H3Cmirror
The two optical isomers of butan-2-ol. The chiral centre carries H, OH, CH3 and C2H5. A wedge comes towards you; a hashed bond goes away.

When you draw optical isomers, show the chiral carbon with two bonds in the plane, one wedge and one hashed bond, and draw the second isomer as a reflection of the first.

Worked example: how many isomers?

Deduce all the isomers of C4H9Br.

  1. Straight chain: 1-bromobutane, CH3CH2CH2CH2Br, and 2-bromobutane, CH3CHBrCH2CH3.
  2. Branched chain: 1-bromo-2-methylpropane, (CH3)2CHCH2Br, and 2-bromo-2-methylpropane, (CH3)3CBr.
  3. Check for chiral centres: in 2-bromobutane, C2 carries H, Br, CH3 and C2H5, so it has two optical isomers.

4 structural isomers, 5 isomers in total including the pair of enantiomers.

Questions

1Paper 1 style

What is the name of (CH3)2CHCH2OH?

  1. butan-1-ol
  2. 2-methylpropan-2-ol
  3. 2-methylpropan-1-ol
  4. 1-methylpropan-1-ol

The longest chain containing the OH has 3 carbons, with OH on C1 and a methyl branch on C2.

2Paper 1 style

Which compound shows cis/trans (geometrical) isomerism?

  1. CH2=CHCH3
  2. CH3CH=CHCl
  3. (CH3)2C=CH2
  4. CH2=CCl2

In B each carbon of the C=C has two different groups (H and CH3; H and Cl). Each of the others has one carbon carrying two identical groups.

3Paper 1 style

Which molecule contains a chiral centre?

  1. CH3CH2CH2OH
  2. (CH3)2CHOH
  3. CH3CH2Cl
  4. CH3CH(OH)COOH

In D (lactic acid) the second carbon carries H, OH, CH3 and COOH: four different groups.

4Paper 1 style

Which species is a nucleophile?

  1. CN−
  2. H+
  3. NO2+
  4. Cl•

A nucleophile donates an electron pair. CN− has a lone pair on carbon. H+ and NO2+ are electrophiles, and Cl• is a free radical.

5Paper 2 style

Compound P has the molecular formula C4H8.

(a)
Draw skeletal formulas of all the alkenes with molecular formula C4H8, including any geometrical isomers, and name each one.
[4]
(b)
Explain why but-2-ene has geometrical isomers but but-1-ene does not.
[2]
(c)
Draw a cyclic isomer of C4H8.
[1]
  • but-1-ene, CH2=CHCH2CH3
  • cis-but-2-ene and trans-but-2-ene drawn correctly
  • names cis- and trans-but-2-ene (or Z and E)
  • methylpropene, (CH3)2C=CH2
  • Rotation about C=C is restricted (by the π bond)
  • In but-2-ene each C of the C=C carries two different groups (H and CH3); in but-1-ene, C1 carries two H atoms
  • cyclobutane (a square) or methylcyclopropane
6Paper 2 style

Lactic acid is CH3CH(OH)COOH.

(a)
Identify the two functional groups in lactic acid.
[1]
(b)
Draw three-dimensional diagrams of the two optical isomers of lactic acid.
[2]
(c)
State the bond angles around the carbon of the COOH group and around the chiral carbon.
[2]
  • hydroxyl (alcohol) and carboxylic acid
  • One isomer drawn with the chiral C having H, OH, CH3 and COOH, using wedge and hashed bonds
  • The second drawn as a correct mirror image
  • COOH carbon: 120° (trigonal planar)
  • chiral carbon: 109.5° (tetrahedral)

Alkanes are fuels: unreactive apart from burning and free-radical substitution. Alkenes, with their electron-rich C=C bond, take part in a wide range of addition reactions that make them the starting point for plastics, alcohols and many other chemicals.

Alkanes

Making alkanes

  • Hydrogenation of alkenes: C2H4(g) + H2(g) → C2H6(g), with a Pt or Ni catalyst, heat.
  • Cracking of long-chain alkanes: breaking them into shorter alkanes and alkenes, using high temperature and a catalyst (such as aluminium oxide or zeolites), for example C10H22 → C8H18 + C2H4. Cracking matches the supply of crude oil fractions to the demand for petrol and for alkenes.

Why alkanes are unreactive

The C–C and C–H bonds are strong, and they are non-polar (carbon and hydrogen have similar electronegativities, 2.55 and 2.20). So there are no δ+ or δ− sites to attract nucleophiles or electrophiles. Alkanes don't react with acids, alkalis or most oxidising agents.

Combustion

C8H18(l) + 12½O2(g) → 8CO2(g) + 9H2O(l) complete combustion, in plenty of oxygen
C8H18(l) + 8½O2(g) → 8CO(g) + 9H2O(l) incomplete combustion, in limited oxygen, gives carbon monoxide (and soot, C)

Environmental consequences of burning hydrocarbons

PollutantSourceEffect
carbon dioxidecomplete combustiongreenhouse gas, contributes to climate change
carbon monoxideincomplete combustiontoxic: binds to haemoglobin and reduces oxygen transport
soot (carbon particles)incomplete combustionbreathing problems; blackens buildings
oxides of nitrogenN2 and O2 reacting in hot enginesacid rain, photochemical smog (Topic 12)
unburnt hydrocarbonsincomplete combustionphotochemical smog
sulfur dioxidesulfur impurities in fuelsacid rain

Catalytic converters remove CO, NO and unburnt hydrocarbons: 2CO + 2NO → 2CO2 + N2.

Free-radical substitution

In ultraviolet light, alkanes react with chlorine or bromine. For example, C2H6 + Cl2 → C2H5Cl + HCl. The mechanism has three stages:

StageEquationsWhat happens
InitiationCl2 → 2Cl•UV light breaks the Cl–Cl bond by homolytic fission, forming two chlorine radicals. (Cl–Cl is the weakest bond present.)
PropagationC2H6 + Cl• → C2H5• + HCl
C2H5• + Cl2 → C2H5Cl + Cl•
A radical reacts with a molecule to make a new radical. The second step regenerates Cl•, so this is a chain reaction: one radical can lead to thousands of product molecules.
TerminationCl• + Cl• → Cl2
C2H5• + Cl• → C2H5Cl
C2H5• + C2H5• → C4H10
Two radicals join and the chain stops. Notice that a little butane forms.

The reaction gives a mixture: Cl• can also attack chloroethane, giving C2H4Cl2, C2H3Cl3 and so on. Using an excess of the alkane favours the monosubstituted product.

Alkenes

Making alkenes

FromReagents and conditionsExample
halogenoalkane (elimination of HX)NaOH in ethanol, heatCH3CH2Br + NaOH → CH2=CH2 + NaBr + H2O
alcohol (dehydration)heated Al2O3 catalyst or concentrated H2SO4 or H3PO4, heatCH3CH2OH → CH2=CH2 + H2O
long-chain alkane (cracking)high temperature, catalystC10H22 → C8H18 + C2H4

Reactions of alkenes

The C=C bond is a region of high electron density (the π bond), and the π electrons are exposed above and below the plane of the molecule. So alkenes attract electrophiles and undergo addition reactions, in which the π bond breaks.

ReagentConditionsProductExample
hydrogenPt or Ni catalyst, heatalkaneCH2=CH2 + H2 → CH3CH3
steamH3PO4 catalyst, about 300 °C, 6 MPaalcoholCH2=CH2 + H2O → CH3CH2OH
hydrogen halide, HX(g)room temperaturehalogenoalkaneCH2=CH2 + HBr → CH3CH2Br
halogen, X2room temperaturedihalogenoalkaneCH2=CH2 + Br2 → CH2BrCH2Br
cold, dilute, acidified KMnO4room temperaturediol; purple → colourlessCH2=CH2 + [O] + H2O → HOCH2CH2OH
hot, concentrated, acidified KMnO4heatthe C=C is broken (oxidative cleavage)see below
itselfsee Topic 20addition polymernCH2=CH2 → (CH2CH2)n
The test for an alkene

Shake with aqueous bromine (bromine water). An alkene turns it from orange to colourless. Alkanes don't react, so the colour stays.

Oxidative cleavage with hot, concentrated manganate(VII)

Hot, concentrated, acidified KMnO4 breaks the C=C completely. What forms from each end depends on what is attached to that carbon:

End of the C=CProduct
=CH2CO2 and H2O
=CHRa carboxylic acid, RCOOH (via the aldehyde)
=CR2a ketone, R2C=O

For example, but-2-ene, CH3CH=CHCH3, gives two molecules of ethanoic acid; 2-methylpropene, (CH3)2C=CH2, gives propanone and carbon dioxide. Working backwards from the products tells you where the double bond was.

Mechanism: electrophilic addition

Bromine and ethene. As a Br2 molecule approaches the electron-rich C=C, its electrons are repelled, inducing a dipole (Brδ+–Brδ−). The π electrons attack the δ+ bromine, the Br–Br bond breaks heterolytically, and a carbocation forms. The bromide ion then attacks the positive carbon.

CCHHHHBrδ+Brδ−C+CHHHHBrBr−CCHHBrHHBrcarbocation1,2-dibromoethane
Electrophilic addition of bromine to ethene. Each curly arrow shows a pair of electrons moving.

Hydrogen bromide and propene. H–Br is already polar. The π bond attacks the δ+ hydrogen. Two carbocations could form, and the more stable one is formed in greater amount.

CCHHHCH3Hδ+Brδ−CC+HHHCH3HBr−CCHHHBrCH3Hsecondary carbocation2-bromopropane (major)
Electrophilic addition of HBr to propene, via the more stable secondary carbocation.

Markovnikov addition and carbocation stability

Alkyl groups have a positive inductive effect: they push electron density towards a neighbouring positive carbon, spreading out its charge and stabilising it. So carbocation stability is:

tertiary (R3C+) > secondary (R2CH+) > primary (RCH2+)

With an unsymmetrical alkene, the major product comes from the more stable carbocation. In practice: the hydrogen adds to the carbon of the C=C that already has more hydrogen atoms (Markovnikov's rule). Propene + HBr gives mainly 2-bromopropane, with a little 1-bromopropane (from the primary carbocation).

Questions

1Paper 1 style

Which equation is a propagation step in the reaction of methane with chlorine in UV light?

  1. Cl2 → 2Cl•
  2. CH4 + Cl• → CH3• + HCl
  3. CH3• + Cl• → CH3Cl
  4. CH4 + Cl• → CH3Cl + H•

A propagation step uses one radical and makes another. A is initiation, C is termination, and D doesn't happen (an H radical is not formed).

2Paper 1 style

What is the major product of the reaction between 2-methylpropene, (CH3)2C=CH2, and hydrogen bromide?

  1. 1-bromo-2-methylpropane
  2. 1-bromobutane
  3. 2-bromo-2-methylpropane
  4. 1,2-dibromo-2-methylpropane

H adds to the CH2 end, forming the tertiary carbocation (CH3)3C+, which is the most stable. Br then adds to give (CH3)3CBr.

3Paper 1 style

2-methylbut-2-ene, (CH3)2C=CHCH3, is heated with concentrated, acidified potassium manganate(VII). What are the organic products?

  1. propanone and ethanoic acid
  2. propanone and ethanal
  3. propanoic acid and ethanoic acid
  4. butanone and methanoic acid

The (CH3)2C= end becomes the ketone (CH3)2C=O (propanone). The =CHCH3 end becomes ethanoic acid, because aldehydes are oxidised further under these conditions.

4Paper 1 style

Which reagent and conditions convert ethene into ethane-1,2-diol?

  1. steam with a phosphoric acid catalyst
  2. aqueous sodium hydroxide, heat
  3. hot, concentrated, acidified KMnO4
  4. cold, dilute, acidified KMnO4

Cold, dilute manganate(VII) adds two OH groups across the C=C. The hot, concentrated reagent would break the molecule apart.

5Paper 2 style

Ethane reacts with bromine in ultraviolet light.

(a)
Name the type of reaction and the mechanism.
[1]
(b)
Write equations for the initiation, two propagation steps and one termination step.
[4]
(c)
Explain why a small amount of butane is found among the products.
[1]
(d)
Suggest why the product is a mixture of several bromo-compounds, and how the proportion of bromoethane could be increased.
[2]
  • Free-radical substitution
  • Initiation: Br2 → 2Br•
  • C2H6 + Br• → C2H5• + HBr
  • C2H5• + Br2 → C2H5Br + Br•
  • Termination, e.g. C2H5• + Br• → C2H5Br or 2Br• → Br2
  • Two ethyl radicals combine in a termination step: 2C2H5• → C4H10
  • Bromine radicals also substitute H atoms in bromoethane (further substitution), giving C2H4Br2 etc.
  • Use an excess of ethane
6Paper 2 style

Propene reacts with hydrogen bromide to form two products. The major product is 2-bromopropane.

(a)
Describe the mechanism of this reaction. Include curly arrows, relevant dipoles and the intermediate.
[4]
(b)
Explain why 2-bromopropane is the major product.
[2]
(c)
Give the reagents and conditions to convert propene into propan-2-ol in one step.
[1]
  • Dipole on H–Br (Hδ+, Brδ−) and a curly arrow from the C=C to the H
  • Curly arrow from the H–Br bond to Br
  • Secondary carbocation, CH3C+HCH3, and Br− with a lone pair
  • Curly arrow from the lone pair on Br− to the positive carbon
  • The secondary carbocation is more stable than the primary one
  • Because of the positive inductive effect of two alkyl groups, which push electrons towards the positive carbon
  • Steam, H3PO4 catalyst (high temperature and pressure)

The polar C–X bond makes halogenoalkanes useful starting materials: nucleophiles replace the halogen to make alcohols, nitriles and amines, while a base in ethanol removes HX to make alkenes. How fast they react depends on whether they are primary, secondary or tertiary, and on the halogen.

Making halogenoalkanes

FromReagents and conditionsExample
alkaneCl2 or Br2, UV light (free-radical substitution)C2H6 + Cl2 → C2H5Cl + HCl
alkeneHX(g) or X2, room temperature (electrophilic addition)CH2=CH2 + HBr → CH3CH2Br
alcohol (substitution of OH)HX(g)CH3CH2OH + HBr → CH3CH2Br + H2O
KCl (or KBr) and concentrated H2SO4 or H3PO4, which make HX in the flaskKBr + H2SO4 → KHSO4 + HBr, then as above
PCl3 and heat3CH3CH2OH + PCl3 → 3CH3CH2Cl + H3PO3
PCl5, room temperatureCH3CH2OH + PCl5 → CH3CH2Cl + POCl3 + HCl (steamy fumes of HCl: a test for OH)
SOCl2CH3CH2OH + SOCl2 → CH3CH2Cl + SO2 + HCl (both by-products are gases, so the product is easy to separate)

Primary, secondary and tertiary

A halogenoalkane is classified by the number of alkyl groups attached to the carbon that carries the halogen: primary (one, e.g. 1-bromobutane), secondary (two, e.g. 2-bromobutane) or tertiary (three, e.g. 2-bromo-2-methylpropane).

Nucleophilic substitution

Halogens are more electronegative than carbon, so the C–X bond is polar: Cδ+–Xδ−. A nucleophile attacks the δ+ carbon and replaces the halogen, which leaves as a halide ion.

NucleophileReagents and conditionsProductEquation
OH−NaOH(aq), heat under refluxalcoholCH3CH2Br + NaOH → CH3CH2OH + NaBr
CN−KCN in ethanol, heat under refluxnitrile (one more carbon in the chain)CH3CH2Br + KCN → CH3CH2CN + KBr
NH3NH3 in ethanol, heat in a sealed tube (under pressure)primary amineCH3CH2Br + 2NH3 → CH3CH2NH2 + NH4Br
H2OAgNO3(aq) in ethanol, warmalcohol; the halide forms a silver halide precipitateCH3CH2Br + H2O → CH3CH2OH + H+ + Br−, then Ag+ + Br− → AgBr
  • Heating with ammonia uses an excess of ammonia, because the amine product is also a nucleophile and could react further with the halogenoalkane.
  • The reaction with KCN is useful in synthesis because it lengthens the carbon chain by one. The nitrile can then be hydrolysed to a carboxylic acid (Topic 19).

Mechanisms: SN2 and SN1

SN2 (substitution, nucleophilic, bimolecular) happens mainly with primary halogenoalkanes. It is one step: the nucleophile attacks the δ+ carbon from the side opposite the halogen as the C–X bond breaks. In the middle there is a transition state in which the carbon is partly bonded to both.

HO−CHHH3CBrδ−HOCHHCH3BrHOCHHCH3Br−δ+[]−transition state
SN2 hydrolysis of bromoethane. Two species take part in the single step.

SN1 (substitution, nucleophilic, unimolecular) happens mainly with tertiary halogenoalkanes. It is two steps: first the C–X bond breaks heterolytically to form a carbocation (slow), then the nucleophile attacks it (fast).

CCH3H3CCH3Brδ−C+CH3H3CCH3Br−HO−CCH3H3CCH3OHδ+tertiary carbocationslowfast
SN1 hydrolysis of 2-bromo-2-methylpropane. Only one species takes part in the slow first step.
Why tertiary favours SN1

The tertiary carbocation is stabilised by the positive inductive effect of three alkyl groups. Also, the three bulky groups get in the way of a nucleophile approaching the carbon (steric hindrance), which blocks SN2.

Why primary favours SN2

A primary carbocation would be very unstable (only one alkyl group), so SN1 is unlikely. The carbon is easy to reach, so SN2 is fast.

Secondary halogenoalkanes react by a mixture of the two mechanisms.

Elimination

With NaOH in ethanol, heated, the hydroxide ion acts as a base rather than a nucleophile. It removes an H from the carbon next to the C–X, and HX is eliminated, forming an alkene:

CH3CH2Br + NaOH → CH2=CH2 + NaBr + H2O bromoethane → ethene. 2-Bromobutane gives a mixture of but-1-ene and cis- and trans-but-2-ene.
Same reagent, different product

NaOH in water (aqueous) gives substitution → alcohol. NaOH in ethanol (ethanolic) gives elimination → alkene. Always state the solvent.

Reactivity and the C–X bond

Warming each halogenoalkane with aqueous silver nitrate in ethanol shows how fast it hydrolyses, because the halide ions released form a precipitate.

HalogenoalkanePrecipitateTime for precipitateC–X bond energy / kJ mol−1
1-chlorobutanewhite AgClslowest (may need several minutes or more)C–Cl 340
1-bromobutanecream AgBrmoderateC–Br 280
1-iodobutaneyellow AgIfastestC–I 240

The C–Cl bond is the most polar, so you might expect it to react fastest. In fact the bond strength decides the rate: the C–I bond is the weakest, so it breaks most easily and iodoalkanes are the most reactive. (C–F is so strong that fluoroalkanes hardly react at all.)

The same test also identifies the halogen from the colour of the precipitate.

Questions

1Paper 1 style

Which reagent and conditions convert 1-bromopropane into butanenitrile?

  1. HCN, room temperature
  2. KCN in water, room temperature
  3. KCN in ethanol, heat under reflux
  4. NH3 in ethanol, heat under pressure

Cyanide ions in ethanol, heated, substitute for Br, adding one carbon: CH3CH2CH2Br + CN− → CH3CH2CH2CN + Br−, butanenitrile.

2Paper 1 style

Which halogenoalkane gives a precipitate most quickly when warmed with aqueous silver nitrate in ethanol?

  1. 2-iodo-2-methylpropane
  2. 2-chloro-2-methylpropane
  3. 1-chlorobutane
  4. 1-bromobutane

Iodo-compounds are fastest because C–I is the weakest bond, and tertiary halogenoalkanes react quickly through the stable tertiary carbocation.

3Paper 1 style

Which statement about the SN1 hydrolysis of 2-bromo-2-methylpropane is correct?

  1. It takes place in one step.
  2. The hydroxide ion attacks the carbon at the same time as the C–Br bond breaks.
  3. A transition state with five groups around carbon is formed.
  4. The intermediate is stabilised by the inductive effect of three methyl groups.

SN1 goes via a tertiary carbocation, which is stabilised by electron donation from the three methyl groups. A, B and C describe SN2.

4Paper 2 style

Bromoethane reacts with sodium hydroxide in two different ways, depending on the conditions.

(a)
Give the conditions and name the organic product for each reaction. Name the type of reaction in each case.
[4]
(b)
Draw the mechanism of the substitution reaction. Show the transition state.
[3]
(c)
Explain why 2-bromo-2-methylpropane reacts with aqueous NaOH by a different mechanism.
[2]
  • Aqueous NaOH, heat under reflux
  • Ethanol; nucleophilic substitution
  • NaOH in ethanol, heat
  • Ethene; elimination
  • Cδ+–Brδ− dipole and a curly arrow from a lone pair on OH− to the carbon
  • Curly arrow from the C–Br bond to Br
  • Transition state [HO···C···Br]− with partial bonds, then ethanol and Br−
  • It is tertiary: it forms a stable tertiary carbocation (SN1)
  • Three bulky methyl groups hinder attack by OH− on the carbon (so SN2 is blocked)

Alcohols are the hub of AS organic chemistry: they are made from alkenes, halogenoalkanes, carbonyl compounds and esters, and they turn into halogenoalkanes, alkenes, aldehydes, ketones, carboxylic acids and esters. Whether an alcohol is primary, secondary or tertiary decides how it oxidises.

Making alcohols

FromReagents and conditionsExample
alkenesteam, H3PO4 catalyst (electrophilic addition)CH2=CH2 + H2O → CH3CH2OH
alkenecold, dilute, acidified KMnO4CH2=CH2 + [O] + H2O → HOCH2CH2OH
halogenoalkaneNaOH(aq), heat (nucleophilic substitution)CH3CH2Br + OH− → CH3CH2OH + Br−
aldehyde or ketoneNaBH4 or LiAlH4 (reduction)CH3COCH3 + 2[H] → CH3CH(OH)CH3
carboxylic acidLiAlH4 (reduction)CH3COOH + 4[H] → CH3CH2OH + H2O
esterdilute acid or dilute alkali, heat (hydrolysis)CH3COOCH2CH3 + H2O ⇌ CH3COOH + CH3CH2OH

Classifying alcohols

An alcohol is primary if the carbon carrying the OH has one alkyl group (e.g. ethanol, CH3CH2OH), secondary if it has two (e.g. propan-2-ol, CH3CH(OH)CH3) and tertiary if it has three (e.g. 2-methylpropan-2-ol, (CH3)3COH). Molecules with more than one OH group are classified one OH at a time: propane-1,2-diol, CH3CH(OH)CH2OH, has one primary and one secondary OH.

Reactions of alcohols

ReactionReagents and conditionsExample
combustionburn in oxygenC2H5OH + 3O2 → 2CO2 + 3H2O
substitution → halogenoalkaneHX, KX + conc. H2SO4, PCl3 + heat, PCl5 or SOCl2 (see Topic 15)C2H5OH + PCl5 → C2H5Cl + POCl3 + HCl
with sodiumNa(s), room temperature2C2H5OH + 2Na → 2C2H5O−Na+ + H2; fizzing, less vigorous than sodium with water. Sodium ethoxide forms.
oxidationacidified K2Cr2O7 or acidified KMnO4, heatsee below
dehydration → alkeneheated Al2O3 or conc. H2SO4 or H3PO4, heatCH3CH2OH → CH2=CH2 + H2O
esterificationcarboxylic acid, conc. H2SO4 catalyst, warmCH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O

Oxidation

The usual oxidising agent is potassium dichromate(VI) acidified with dilute sulfuric acid. When it oxidises an alcohol, its colour changes from orange (Cr2O72−) to green (Cr3+). Acidified manganate(VII) also works (purple to colourless).

AlcoholProductConditionsColour change with Cr2O72−/H+
primaryaldehyde: CH3CH2OH + [O] → CH3CHO + H2Oheat and distil off the aldehyde as it forms (it has a lower boiling point than the alcohol), so it isn't oxidised furtherorange → green
primarycarboxylic acid: CH3CH2OH + 2[O] → CH3COOH + H2Oexcess oxidising agent, heat under reflux, so the aldehyde can't escape and is oxidised furtherorange → green
secondaryketone: CH3CH(OH)CH3 + [O] → CH3COCH3 + H2Oheat (under reflux or with distillation)orange → green
tertiaryno reactionheatstays orange

Tertiary alcohols are not oxidised because the carbon carrying the OH has no hydrogen atom to lose. This colour change is a test that distinguishes tertiary alcohols from primary and secondary ones.

Distillation

The vapour leaves the flask, is cooled in a condenser and collected. Used to remove a volatile product (an aldehyde) as soon as it forms.

Reflux

A vertical condenser returns the vapour to the flask, so the mixture can be heated for a long time without losing volatile substances. Used to make the reaction go to completion.

The tri-iodomethane (iodoform) test

Warm the compound with alkaline aqueous iodine (iodine and NaOH). A pale yellow precipitate of tri-iodomethane, CHI3 (with an antiseptic smell), shows that the compound contains the group CH3CH(OH)–. The rest of the molecule ends up as a carboxylate ion, RCO2−.

Positive: ethanol (the only primary alcohol to give it), propan-2-ol, butan-2-ol. Negative: methanol, propan-1-ol, 2-methylpropan-2-ol. Ethanal and methyl ketones (CH3CO–) also give a positive result (Topic 17), because the iodine first oxidises CH3CH(OH)– to CH3CO–.

Acidity: why alcohols are weaker acids than water

Alcohols can lose the H of the OH group, as their reaction with sodium shows, but they are even weaker acids than water. The alkyl group has a positive inductive effect: it pushes electron density onto the oxygen of the alkoxide ion, RO−. This intensifies the negative charge, making RO− less stable than OH− and more likely to take back a proton.

Questions

1Paper 1 style

Which alcohol gives a positive result with both acidified potassium dichromate(VI) and alkaline aqueous iodine?

  1. propan-1-ol
  2. butan-2-ol
  3. 2-methylpropan-2-ol
  4. methanol

Butan-2-ol is secondary (oxidised: orange to green) and contains CH3CH(OH)– (iodoform positive). Propan-1-ol and methanol lack the CH3CH(OH)– group; 2-methylpropan-2-ol is tertiary.

2Paper 1 style

Which conditions convert propan-1-ol into propanal?

  1. acidified K2Cr2O7, heat and distil the product off immediately
  2. acidified K2Cr2O7 in excess, heat under reflux
  3. NaBH4
  4. concentrated H2SO4, heat

Distilling the aldehyde off as it forms stops it being oxidised further to propanoic acid (which is what B gives). D would dehydrate the alcohol to propene.

3Paper 1 style

Which alcohol gives a yellow precipitate with alkaline aqueous iodine and is oxidised to a ketone by acidified potassium dichromate(VI)?

  1. ethanol
  2. propan-2-ol
  3. 2-methylpropan-2-ol
  4. propan-1-ol

The tri-iodomethane test needs a CH3CH(OH)– group. Propan-2-ol has one and is a secondary alcohol, so it oxidises to propanone. Ethanol gives the test but oxidises to ethanal and ethanoic acid.

4Paper 1 style

Butan-2-ol is heated with concentrated phosphoric acid. How many different alkenes, including stereoisomers, could form?

  1. 1
  2. 2
  3. 3
  4. 4

Water can be lost towards C1, giving but-1-ene, or towards C3, giving but-2-ene, which exists as cis and trans isomers: 3 alkenes.

5Quick check

Which reagent reacts with ethanol to give hydrogen gas?

  1. sodium
  2. sodium hydroxide
  3. sodium carbonate
  4. sodium chloride

2C2H5OH + 2Na → 2C2H5ONa + H2. Alcohols are too weakly acidic to react with NaOH or with carbonates.

6Paper 2 and 3 style

Three isomeric alcohols have the formula C4H10O: butan-1-ol (A), butan-2-ol (B) and 2-methylpropan-2-ol (C).

(a)
Classify each as primary, secondary or tertiary.
[1]
(b)
Each is warmed with acidified potassium dichromate(VI). Describe what you would see, and give the structural formula of any organic product formed under reflux.
[4]
(c)
Which of A, B and C gives a yellow precipitate with alkaline aqueous iodine? Explain.
[2]
(d)
Write an equation for the reaction of B with sodium.
[1]
  • A primary, B secondary, C tertiary
  • A and B: orange to green; C: stays orange
  • A gives butanoic acid, CH3CH2CH2COOH
  • B gives butanone, CH3COCH2CH3
  • C: no reaction, because there is no H on the carbon bearing the OH
  • B
  • It contains the CH3CH(OH)– group
  • 2CH3CH(OH)CH2CH3 + 2Na → 2CH3CH(O−Na+)CH2CH3 + H2

Aldehydes and ketones both contain the C=O carbonyl group. They are made by oxidising alcohols, reduced back to alcohols, and attacked by nucleophiles. A set of simple test-tube tests identifies them and tells them apart.

Structure and naming

In an aldehyde the C=O is at the end of the chain, with an H attached: –CHO (e.g. ethanal, CH3CHO; propanal, CH3CH2CHO). In a ketone the C=O is within the chain, with two carbon groups attached: C–CO–C (e.g. propanone, CH3COCH3; butanone, CH3COCH2CH3).

The C=O bond is polar: Cδ+=Oδ−. The δ+ carbon attracts nucleophiles.

Making aldehydes and ketones

ProductFromReagents and conditions
aldehydeprimary alcoholacidified K2Cr2O7 (or KMnO4), heat, distil the aldehyde off
ketonesecondary alcoholacidified K2Cr2O7 (or KMnO4), heat (with distillation)

Reactions

Reduction to alcohols

Reducing agents: NaBH4 (sodium tetrahydridoborate, in water or ethanol) or LiAlH4 (lithium tetrahydridoaluminate, in dry ether).

CH3CHO + 2[H] → CH3CH2OH aldehydes give primary alcohols
CH3COCH3 + 2[H] → CH3CH(OH)CH3 ketones give secondary alcohols

Addition of hydrogen cyanide

With HCN, a KCN catalyst, and heat, aldehydes and ketones form hydroxynitriles, adding a carbon to the chain:

CH3CHO + HCN → CH3CH(OH)CN ethanal → 2-hydroxypropanenitrile
CH3COCH3 + HCN → (CH3)2C(OH)CN propanone → 2-hydroxy-2-methylpropanenitrile

Mechanism: nucleophilic addition

The KCN provides cyanide ions, the nucleophile. CN− attacks the δ+ carbon of C=O, and the π electrons move onto the oxygen. The negatively charged oxygen in the intermediate then takes a proton from an HCN molecule, which regenerates CN−. That is why KCN acts as a catalyst.

NC−COδ−CH3HCO−CH3HCNHCNCOHCH3HCNδ++ CN⁻2-hydroxypropanenitrile
Nucleophilic addition of HCN to ethanal.

The product from ethanal has a chiral centre. The C=O group is planar, so CN− is equally likely to attack from above or below: the product is a 50 : 50 mixture of the two optical isomers (a racemic mixture, Topic 29).

Tests for carbonyl compounds

TestReagent and methodAldehydeKetone
any C=O2,4-dinitrophenylhydrazine (2,4-DNPH)orange (or yellow-orange) precipitate with both. The melting point of the purified precipitate can identify the compound.
Tollens'ammoniacal silver nitrate, [Ag(NH3)2]+, warmsilver mirror (or grey-black precipitate): Ag+ reduced to Agno change
Fehling'sFehling's solution (Cu2+ complex in alkali), warmblue solution → brick-red precipitate of copper(I) oxide, Cu2Ostays blue
dichromateacidified K2Cr2O7, warmorange → greenstays orange
tri-iodomethanealkaline I2(aq), warmpale yellow precipitate of CHI3 for ethanal and any methyl ketone, CH3CO– (e.g. propanone, butanone)

Aldehydes are distinguished from ketones because aldehydes are easily oxidised to carboxylic acids (they still have an H on the carbonyl carbon). Ketones are not, so they don't reduce Tollens', Fehling's or dichromate.

RCHO + [O] → RCOOH in Tollens' (alkaline) the acid forms as RCOO−

Questions

1Paper 1 style

A compound gives an orange precipitate with 2,4-DNPH and a yellow precipitate with alkaline aqueous iodine, but does not react with Tollens' reagent. What could it be?

  1. ethanal
  2. propanal
  3. butanone
  4. pentan-3-one

2,4-DNPH: a carbonyl. No silver mirror: a ketone. Iodoform positive: a methyl ketone, CH3CO–. Butanone fits; pentan-3-one is not a methyl ketone.

2Paper 1 style

What is the product when propanone reacts with HCN in the presence of KCN?

  1. CH3CH2CH2CN
  2. (CH3)2C(OH)CN
  3. CH3COCH2CN
  4. (CH3)2CHCN

CN adds to the carbonyl carbon and H adds to the oxygen, giving 2-hydroxy-2-methylpropanenitrile.

3Paper 1 style

Which reagent gives an orange precipitate with both propanal and propanone?

  1. Tollens' reagent
  2. Fehling's solution
  3. 2,4-dinitrophenylhydrazine
  4. acidified potassium dichromate(VI)

2,4-DNPH reacts with the C=O group of any aldehyde or ketone. Tollens', Fehling's and dichromate(VI) react only with the aldehyde.

4Paper 1 style

Hydrogen cyanide, with a little KCN, reacts with ethanal. Why is the product, 2-hydroxypropanenitrile, optically inactive?

  1. the product has no chiral centre
  2. the CN− ion attacks the planar C=O group equally from either side, giving a racemic mixture
  3. the reaction goes by an SN1 mechanism
  4. the product is rapidly hydrolysed

The product does have a chiral carbon. Because the carbonyl group is planar, attack from above and below is equally likely, so equal amounts of the two enantiomers form and their effects on plane-polarised light cancel.

5Quick check

What forms when butanone is reduced by NaBH4?

  1. butanal
  2. butan-1-ol
  3. butanoic acid
  4. butan-2-ol

Ketones are reduced to secondary alcohols: CH3COCH2CH3 + 2[H] → CH3CH(OH)CH2CH3.

6Paper 2 style

Butanal and butanone are isomers with formula C4H8O.

(a)
Describe a chemical test that gives a positive result for both compounds, and one that distinguishes them. Give the observations.
[3]
(b)
Name the product formed when butanone is reduced with NaBH4, and classify it.
[2]
(c)
Describe the mechanism of the reaction between butanal and HCN, with KCN as catalyst. Explain why KCN is described as a catalyst.
[4]
  • 2,4-DNPH: orange precipitate with both
  • Tollens' reagent, warm (or Fehling's, or acidified dichromate)
  • Butanal gives a silver mirror (brick-red precipitate / orange to green); butanone: no change
  • Butan-2-ol
  • Secondary alcohol
  • Dipole on C=O, and a curly arrow from the lone pair on C of CN− to the carbonyl C
  • Curly arrow from the C=O π bond to O, giving the alkoxide intermediate
  • Curly arrow from O− to H of HCN, and from the H–C bond to C, giving CH3CH2CH2CH(OH)CN
  • CN− is used in the first step and regenerated in the second

Carboxylic acids contain the –COOH group. They are weak acids, so they show all the typical reactions of acids, and they react with alcohols to make esters, the compounds behind many fruity smells and flavours.

Carboxylic acids

The functional group is the carboxyl group, –COOH. Names end in -oic acid: methanoic acid HCOOH, ethanoic acid CH3COOH, propanoic acid CH3CH2COOH. They are weak acids: CH3COOH(aq) ⇌ CH3COO−(aq) + H+(aq).

Making carboxylic acids

FromReagents and conditionsExample
primary alcohol or aldehydeacidified K2Cr2O7 or KMnO4, heat under refluxCH3CH2OH + 2[O] → CH3COOH + H2O
nitriledilute acid, heat under refluxCH3CH2CN + 2H2O + H+ → CH3CH2COOH + NH4+
nitriledilute alkali, heat under reflux, then acidifyCH3CH2CN + OH− + H2O → CH3CH2COO− + NH3, then CH3CH2COO− + H+ → CH3CH2COOH
esterdilute acid or alkali, heat (then acidify if alkali was used)CH3COOC2H5 + H2O ⇌ CH3COOH + C2H5OH

Reactions of carboxylic acids

ReactantProductsExample
reactive metalsalt + hydrogen2CH3COOH + Mg → (CH3COO)2Mg + H2
alkalisalt + water (neutralisation)CH3COOH + NaOH → CH3COONa + H2O
carbonatesalt + water + carbon dioxide2CH3COOH + Na2CO3 → 2CH3COONa + H2O + CO2
alcoholester + water (conc. H2SO4 catalyst, warm)CH3COOH + CH3OH ⇌ CH3COOCH3 + H2O
LiAlH4primary alcohol (reduction)CH3COOH + 4[H] → CH3CH2OH + H2O

Fizzing with a carbonate or hydrogencarbonate is the test for a carboxylic acid: alcohols and phenols don't do this. The salts are named as -oates: sodium ethanoate, magnesium ethanoate.

Esters

Naming esters

An ester is named from the alcohol part first (as an alkyl group) and then the acid part (as -oate). Ethanoic acid + methanol gives methyl ethanoate, CH3COOCH3. Propanoic acid + ethanol gives ethyl propanoate, CH3CH2COOCH2CH3.

OOfrom ethanoic acidfrom ethanolethyl ethanoate

Making esters (esterification)

Warm a carboxylic acid with an alcohol and a few drops of concentrated H2SO4 as a catalyst. This is a condensation reaction (water is lost) and it is reversible:

CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O ethanoic acid + ethanol ⇌ ethyl ethanoate + water

Hydrolysis of esters

With dilute acid, heat under reflux

The reverse of esterification. It is reversible, so it does not go to completion: CH3COOC2H5 + H2O ⇌ CH3COOH + C2H5OH

With dilute alkali (NaOH), heat under reflux

Goes to completion, because the acid is converted into its salt, which doesn't react with the alcohol: CH3COOC2H5 + NaOH → CH3COONa + C2H5OH. Add acid afterwards to obtain the carboxylic acid.

Esters have sweet, fruity smells, so they are used in flavourings and perfumes. They are also good solvents (ethyl ethanoate is used in nail varnish remover and glues).

Questions

1Paper 1 style

What is the name of CH3CH2COOCH3?

  1. ethyl ethanoate
  2. propyl methanoate
  3. ethyl methanoate
  4. methyl propanoate

The acid part, CH3CH2COO–, comes from propanoic acid; the alcohol part, –CH3, from methanol: methyl propanoate.

2Paper 1 style

Which reagent distinguishes ethanoic acid from ethanol?

  1. aqueous sodium carbonate
  2. sodium metal
  3. PCl5
  4. concentrated sulfuric acid

Only the acid fizzes with sodium carbonate (giving CO2). Both compounds react with Na and with PCl5, because both have an OH group.

3Paper 1 style

Which substances react with ethanoic acid to produce a gas?
1 sodium carbonate
2 magnesium
3 sodium hydroxide

  1. 1, 2 and 3
  2. 1 and 2 only
  3. 2 and 3 only
  4. 1 only

Sodium carbonate gives CO2 and magnesium gives H2. Sodium hydroxide neutralises the acid, but no gas forms.

4Paper 1 style

Ethyl propanoate is heated under reflux with excess NaOH(aq). Which pair of products forms?

  1. propanoic acid and ethanol
  2. sodium ethanoate and propan-1-ol
  3. sodium propanoate and ethanol
  4. ethanoic acid and propan-1-ol

Alkaline hydrolysis gives the salt of the acid part (propanoate) and the alcohol part (ethanol): CH3CH2COOC2H5 + NaOH → CH3CH2COONa + C2H5OH.

5Paper 2 style

Propyl ethanoate is used as a flavouring.

(a)
Give the structural formulas of the acid and the alcohol needed to make it, and the conditions.
[3]
(b)
Write an equation for the hydrolysis of propyl ethanoate with aqueous sodium hydroxide, and explain why alkaline hydrolysis goes to completion but acid hydrolysis does not.
[3]
(c)
Ethanoic acid can be made from ethanenitrile. Give the reagents and conditions, and write an equation.
[2]
  • CH3COOH
  • CH3CH2CH2OH
  • Concentrated H2SO4 catalyst, warm
  • CH3COOCH2CH2CH3 + NaOH → CH3COONa + CH3CH2CH2OH
  • Acid hydrolysis is reversible / an equilibrium
  • In alkali the carboxylic acid is removed as its salt (carboxylate ion), which can't react with the alcohol
  • Dilute HCl (or H2SO4), heat under reflux
  • CH3CN + 2H2O + H+ → CH3COOH + NH4+

At AS Level you need to know how to make primary amines, nitriles and hydroxynitriles, and how to turn nitriles into carboxylic acids. Nitriles are especially useful because making them adds a carbon atom to the chain.

Primary amines

A primary amine has an –NH2 group attached to a carbon chain, for example ethylamine, CH3CH2NH2. It forms when a halogenoalkane is heated with ammonia in ethanol, in a sealed tube (under pressure):

CH3CH2Br + 2NH3 → CH3CH2NH2 + NH4Br

An excess of ammonia is used. The amine product has a lone pair on N, so it is itself a nucleophile and could react with more halogenoalkane to form secondary and tertiary amines. (Classifying amines and their basicity come at A Level, Topic 34.)

Nitriles

Nitriles contain the –C≡N group. They are named from the total number of carbon atoms, including the carbon of C≡N: CH3CN is ethanenitrile and CH3CH2CN is propanenitrile.

MakingReagents and conditionsExample
nitrile from a halogenoalkaneKCN in ethanol, heat under refluxCH3CH2Br + KCN → CH3CH2CN + KBr
hydroxynitrile from an aldehyde or ketoneHCN, KCN catalyst, heatCH3CHO + HCN → CH3CH(OH)CN

Hydrolysis of nitriles

Heating a nitrile under reflux with a dilute acid or a dilute alkali hydrolyses the –C≡N to –COOH:

CH3CH2CN + 2H2O + H+ → CH3CH2COOH + NH4+ with dilute acid, e.g. HCl(aq): the carboxylic acid forms directly
CH3CH2CN + H2O + OH− → CH3CH2COO− + NH3 with dilute alkali, e.g. NaOH(aq): the carboxylate salt forms, and ammonia is given off. Then acidify to obtain the acid.

Worked example: a two-step synthesis

How can propanoic acid be made from bromoethane?

  1. Bromoethane → propanenitrile: KCN in ethanol, heat under reflux. The chain grows from 2 to 3 carbons.
  2. Propanenitrile → propanoic acid: dilute HCl, heat under reflux.

Questions

1Paper 1 style

Which compound is formed when 2-hydroxypropanenitrile is heated under reflux with dilute hydrochloric acid?

  1. propanoic acid
  2. 2-hydroxypropanoic acid
  3. propan-2-ol
  4. 2-aminopropanoic acid

Only the C≡N is hydrolysed, to COOH. The OH group is unchanged: CH3CH(OH)COOH, 2-hydroxypropanoic acid (lactic acid).

2Paper 1 style

Ethylamine is made by heating bromoethane with ammonia in ethanol in a sealed tube. Why is excess ammonia used?

  1. ammonia is a catalyst
  2. to reduce further substitution, which would give secondary and tertiary amines
  3. to make the reaction go by an SN1 mechanism
  4. to dissolve the bromoethane

Ethylamine is itself a nucleophile and can attack more bromoethane. A large excess of NH3 makes it much more likely that bromoethane meets an ammonia molecule instead.

3Paper 1 style

1-bromopropane is heated under reflux with KCN in ethanol. The organic product is then heated with dilute hydrochloric acid. What is the final organic product?

  1. propanoic acid
  2. propan-1-amine
  3. propanenitrile
  4. butanoic acid

CN− replaces Br, giving butanenitrile, CH3CH2CH2CN (4 carbons). Acid hydrolysis turns –CN into –COOH: butanoic acid.

4Paper 2 style
(a)
Give the reagents and conditions to convert 1-bromopropane into propylamine, and explain why one reagent is used in excess.
[3]
(b)
Suggest a two-step synthesis of butanoic acid from 1-bromopropane. Give the reagents, conditions and intermediate.
[3]
  • Ammonia in ethanol
  • Heat in a sealed tube / under pressure
  • Excess ammonia, to stop the amine product reacting further (forming secondary / tertiary amines)
  • Step 1: KCN in ethanol, heat under reflux
  • Intermediate: butanenitrile, CH3CH2CH2CN
  • Step 2: dilute acid (HCl), heat under reflux

Alkenes join together in long chains to form addition polymers: poly(ethene) for bags and bottles, poly(chloroethene) for window frames and pipes. You need to draw repeat units, work back to the monomer, and explain why these plastics are hard to dispose of.

Addition polymerisation

In addition polymerisation, many small unsaturated molecules (monomers) join together to form one very long molecule (a polymer). The π bond of each C=C opens up and the monomers link through new C–C single bonds. No other product is formed.

CCHHHHnn ethene
nCCHHHHpoly(ethene)
nCCHHHClpoly(chloroethene), PVCfrom chloroethene, CH2=CHCl
nCCHHHCH3poly(propene)from propene, CH2=CHCH3
The repeat unit has the two carbons of the original C=C, now joined by a single bond, with bonds extending through the brackets on each side.
nCH2=CHCl → (CH2–CHCl)n chloroethene → poly(chloroethene)

Finding the monomer from a polymer

Find the repeating pattern of two carbons along the main chain, draw one repeat unit, and put a C=C between those two carbons.

CCCCCCHHHHHHHCOOCH3HCOOCH3HCOOCH3
Three repeat units of a polymer. Each repeat unit is –CH2–CH(COOCH3)–, so the monomer is methyl propenoate, CH2=CHCOOCH3.

Disposal of poly(alkene)s

  • Non-biodegradable: poly(alkene)s are chemically inert, like alkanes. Their C–C and C–H bonds are strong and non-polar, so microorganisms can't break them down, and they last for centuries in landfill or the sea.
  • Harmful combustion products: burning them releases energy but can produce carbon monoxide (incomplete combustion), and burning PVC releases hydrogen chloride, which is toxic and causes acid rain. Other toxic chlorinated compounds can also form.
  • Recycling and the development of degradable polymers (Topic 35) help reduce these problems.

Questions

1Paper 1 style

What is the repeat unit of the polymer formed from but-2-ene, CH3CH=CHCH3?

  1. –CH2–CH2–CH2–CH2–
  2. –CH2–CH(C2H5)–
  3. –CH(CH3)–CH(CH3)–
  4. –CH2–CH(CH3)–CH2–

The two carbons of the C=C form the main chain; each still carries its methyl group.

2Paper 1 style

Which is the repeat unit of poly(propene)?

  1. –CH2–CH2–CH2–
  2. –CH2–CH(CH3)–
  3. –CH(CH3)–CH(CH3)–
  4. –CH2=C(CH3)–

Propene is CH2=CHCH3. The C=C opens and the CH3 becomes a side group on every other carbon. No C=C bonds remain in the polymer.

3Paper 1 style

A polymer has the repeat unit –CH2–CHCl–. What is its monomer?

  1. chloroethane
  2. 1,2-dichloroethene
  3. chloroethene
  4. ethene

The monomer has the same atoms with a C=C bond: CH2=CHCl, chloroethene. The polymer is PVC.

4Quick check

Why are poly(alkene)s difficult to dispose of?

  1. they are chemically inert, so they are non-biodegradable, and burning some of them releases toxic gases
  2. they contain reactive C=C bonds
  3. they are soluble in water
  4. they are ionic, so they have very high melting points

Poly(alkene)s are saturated, with strong, non-polar C–C and C–H bonds, so microorganisms can't break them down. Burning PVC releases HCl.

5Paper 2 style

Poly(tetrafluoroethene), PTFE, is made from tetrafluoroethene, CF2=CF2.

(a)
Draw the repeat unit of PTFE.
[1]
(b)
Explain why poly(alkene)s such as poly(ethene) are not biodegradable.
[2]
(c)
Suggest why burning waste PVC is a particular problem.
[1]
  • –CF2–CF2– drawn with C–C single bond, four F atoms and continuation bonds through brackets
  • They are saturated / have strong C–C and C–H bonds
  • The bonds are non-polar, so they are unreactive and can't be broken down by microorganisms (enzymes)
  • It produces toxic hydrogen chloride (which also causes acid rain)

Synthesis questions ask you to plan a route from one compound to another using the reactions in Topics 13 to 20, or to identify the functional groups in an unfamiliar molecule and predict how it reacts. The key is knowing which reagent turns each functional group into which other.

The AS reaction map

Choose a functional group to see every way to make it and everything it can be turned into, with reagents and conditions.

Planning a route

  1. Compare the start and target: has the number of carbons changed? Which functional groups are lost and which are gained?
  2. Need an extra carbon? Use a cyanide: halogenoalkane + KCN, or aldehyde/ketone + HCN. Then hydrolyse or reduce the nitrile.
  3. Work backwards from the target: what can make it? Can that be made from the starting material?
  4. Give full conditions for each step: reagent, solvent, catalyst, temperature (heat, reflux, distil, UV light).

Worked example 1

Suggest how to make ethanoic acid from ethene in two steps.

  1. Ethene → ethanol: steam, H3PO4 catalyst
  2. Ethanol → ethanoic acid: acidified K2Cr2O7, heat under reflux

Worked example 2

Suggest a three-step route from propene to 2-hydroxy-2-methylpropanenitrile, (CH3)2C(OH)CN.

  1. Propene → propan-2-ol: steam, H3PO4 catalyst (OH goes on the middle carbon, Markovnikov)
  2. Propan-2-ol → propanone: acidified K2Cr2O7, heat
  3. Propanone → product: HCN with KCN catalyst

Worked example 3: a molecule with several groups

Predict what happens when HOCH2CH2CHO (3-hydroxypropanal) reacts with (a) excess acidified dichromate under reflux, (b) NaBH4, (c) Tollens' reagent.

  1. Both the primary OH and the CHO are oxidised to COOH: HOOCCH2COOH (propanedioic acid).
  2. Only the CHO is reduced: HOCH2CH2CH2OH (propane-1,3-diol).
  3. Silver mirror; the CHO is oxidised (to –COO−); the OH is unchanged.

Questions

1Paper 1 style

Which two-step route converts ethene into ethanoic acid?

  1. steam with H3PO4 catalyst; then acidified K2Cr2O7, heat under reflux
  2. HBr; then KCN in ethanol
  3. H2 with Ni; then Cl2 in UV light
  4. cold, dilute, acidified KMnO4; then NaOH(aq)

Hydration gives ethanol, and refluxing with excess acidified dichromate(VI) oxidises it fully to ethanoic acid. Route B gives propanenitrile; route D gives a diol.

2Paper 1 style

Which is the best way to convert bromoethane into propanoic acid?

  1. NaOH(aq); then acidified K2Cr2O7 under reflux
  2. KCN in ethanol; then heat with dilute HCl
  3. excess NH3 in ethanol; then dilute HCl
  4. NaOH in ethanol; then steam with H3PO4

Propanoic acid has 3 carbons and bromoethane 2, so a carbon must be added: substitution by CN−, then hydrolysis of the nitrile. Route A gives ethanoic acid.

3Paper 1 style

A compound decolourises bromine water and fizzes with sodium carbonate solution. Which could it be?

  1. CH3CH2COOH
  2. CH2=CHCH2OH
  3. CH2=CHCOOH
  4. CH3COOCH=CH2

Decolourising bromine water shows a C=C bond; fizzing with a carbonate shows a carboxylic acid. Only propenoic acid has both.

4Paper 2 style
(a)
Suggest a three-step synthesis of propanoic acid from ethene. Give reagents and conditions for each step, and name the intermediates.
[4]
(b)
The compound CH2=CHCH2OH is treated with (i) bromine dissolved in hexane and (ii) acidified potassium dichromate(VI), heated under reflux. Give the structural formula of each product.
[2]
  • Ethene + HBr (room temperature) → bromoethane
  • Bromoethane + KCN in ethanol, heat under reflux → propanenitrile
  • Propanenitrile + dilute HCl, heat under reflux → propanoic acid
  • Intermediates named: bromoethane and propanenitrile
  • (i) CH2BrCHBrCH2OH
  • (ii) CH2=CHCOOH

Going through ethanol doesn't help here: the chain must gain a carbon, and only the cyanide step does that.

Infrared spectroscopy tells you which functional groups a molecule contains. Mass spectrometry tells you its relative molecular mass, how many carbon atoms it has, whether it contains chlorine or bromine, and which fragments it breaks into. Together they can identify an unknown compound.

Infrared spectroscopy

Covalent bonds vibrate (stretch and bend). Each type of bond absorbs infrared radiation at a characteristic frequency, quoted as a wavenumber in cm−1. An IR spectrum plots transmittance against wavenumber; each absorption shows as a dip (often called a peak).

BondFound inWavenumber / cm−1Appearance
C–Oalcohols, carboxylic acids, esters1040–1300strong
C=Calkenes1500–1680weak to medium
C=Oaldehydes, ketones, carboxylic acids, esters, amides1640–1750strong, sharp
C≡Nnitriles2200–2250medium, sharp
C–Halkanes, alkenes, arenes2850–3100strong
O–Hcarboxylic acids (hydrogen bonded)2500–3000very broad
O–Halcohols, phenols (hydrogen bonded)3200–3600broad
N–Hamines, amides3300–3500medium

In the exam, use the ranges printed in the Data Booklet; they are close to these. The region below about 1500 cm−1 is the fingerprint region: it has many overlapping absorptions that are unique to each compound, so it can be matched against a database.

40003000200015001000500050100wavenumber / cm⁻¹transmittance / %ethanol, CH3CH2OHO–H (alcohol)C–HC–O40003000200015001000500050100wavenumber / cm⁻¹transmittance / %propanone, CH3COCH3C=OC–Hno O–H40003000200015001000500050100wavenumber / cm⁻¹transmittance / %ethanoic acid, CH3COOHvery broad O–H (acid)C=OC–O
Simplified spectra. Ethanol: broad O–H at about 3350 and C–O near 1050. Propanone: strong C=O near 1715 and no O–H. Ethanoic acid: both C=O and a very broad O–H that overlaps the C–H absorptions.
Reading a spectrum

Start above 1500 cm−1. Is there a strong C=O near 1700? A broad O–H? If both, it's a carboxylic acid; C=O without O–H suggests an aldehyde, ketone or ester. Then use other evidence (tests, mass spectrum) to decide.

Mass spectrometry

In a mass spectrometer, molecules are ionised and the positive ions are separated by their mass-to-charge ratio, m/e. (You don't need to know how the instrument works.) The spectrum plots relative abundance against m/e. Most ions have a charge of 1+, so m/e equals the mass of the ion.

Elements: isotopes and relative atomic mass

For an element, each peak is one isotope, and its height gives the isotope's relative abundance.

18192021222324050100m/erelative abundance / %90.48%0.27%9.25%neon

Worked example

Use the spectrum of neon to calculate its relative atomic mass.

Ar = (20 × 90.48 + 21 × 0.27 + 22 × 9.25) ÷ 100 = 20.19, so 20.2

Compounds: the molecular ion and fragments

  • The molecular ion, M+, is the whole molecule with one electron removed. Its m/e (the last major peak) gives the relative molecular mass.
  • Molecular ions can break apart (fragment). Only the charged fragment is detected: M+ → X+ + Y•.
  • The tallest peak is the base peak, the most abundant ion, given 100%.
05101520253035404550556065050100m/erelative abundance / %CH₃⁺CH₃CO⁺M⁺M+1propanone, CH₃COCH₃
Propanone, Mr 58. The base peak at 43 is CH3CO+, formed when the molecular ion loses a methyl radical. The peak at 15 is CH3+.
m/e15172931434577
Possible fragmentCH3+OH+C2H5+ or CHO+CH2OH+C3H7+ or CH3CO+COOH+C6H5+

The [M+1] peak: counting carbon atoms

About 1.1% of carbon atoms are 13C. So a small peak appears one unit above the molecular ion, and its height depends on the number of carbon atoms, n:

Number of carbon atomsn = 100 × abundance of [M + 1]+1.1 × abundance of M+Learn it

Worked example

A compound has its M+ peak at m/e 72 with a relative abundance of 40.0, and an [M+1]+ peak with an abundance of 1.76.

  1. n = 100 × 1.761.1 × 40.0 = 4.0, so 4 carbon atoms.
  2. 4 C account for 48 of the 72 mass units. The remaining 24 could be H8O (8 + 16 = 24), giving C4H8O: for example, butanone or butanal.

The [M+2] peak: chlorine and bromine

HalogenIsotopesPattern
one Cl atom35Cl : 37Cl ≈ 3 : 1M+ and [M+2]+ peaks in the ratio 3 : 1
one Br atom79Br : 81Br ≈ 1 : 1M+ and [M+2]+ peaks of about equal height
2025303540455055606570050100m/erelative abundance / %C₂H₅⁺M⁺ (³⁵Cl)M+2 (³⁷Cl)chloroethane, C₂H₅Cl2030405060708090100110050100m/erelative abundance / %C₂H₅⁺M⁺ (⁷⁹Br)M+2 (⁸¹Br)bromoethane, C₂H₅Br
Two peaks separated by 2 units in the ratio 3 : 1 show one chlorine atom; 1 : 1 shows one bromine atom. With two Cl atoms the pattern is M : M+2 : M+4 = 9 : 6 : 1; with two Br atoms it is 1 : 2 : 1.

Questions

1Paper 1 style

The IR spectrum of a compound has a strong absorption at 1715 cm−1 and no broad absorption above 2500 cm−1. Which compound could it be?

  1. propan-1-ol
  2. butanone
  3. propanoic acid
  4. but-1-ene

1715 cm−1 is C=O. No O–H rules out the alcohol and the acid. Butanone has C=O and no O–H.

2Paper 1 style

The mass spectrum of a compound has a molecular ion peak at m/e 78 and a peak at m/e 80 of one third of its height. What does the compound contain?

  1. one bromine atom
  2. two chlorine atoms
  3. one chlorine atom
  4. one bromine and one chlorine atom

M : M+2 = 3 : 1 is the pattern for one chlorine atom. (The compound could be 2-chloropropane, C3H7Cl, Mr 78 with 35Cl.)

3Paper 1 style

The infrared spectrum of a compound has a very broad absorption from 2500 to 3000 cm−1 and a strong, sharp absorption at 1710 cm−1. Which compound could it be?

  1. ethanol
  2. propanone
  3. propanoic acid
  4. ethyl ethanoate

The very broad band at 2500–3000 cm−1 is the O–H of a carboxylic acid, and 1710 cm−1 is C=O. Ethanol has no C=O; propanone and the ester have no O–H.

4Paper 1 style

The mass spectrum of a compound has two molecular-ion peaks, at m/e 94 and 96, of almost equal height. What is the compound?

  1. CH3Cl
  2. CH3Br
  3. CH2Cl2
  4. C2H5Br

Equal M and M+2 peaks mean one bromine atom. CH379Br has M = 12 + 3 + 79 = 94; C2H5Br would be at 108 and 110.

5Paper 2 style

Compound Q contains carbon, hydrogen and oxygen only. Its mass spectrum has the M+ peak at m/e 60 (abundance 24.0) and an [M+1]+ peak (abundance 0.53). Its IR spectrum has a very broad absorption from 2500 to 3000 cm−1 and a strong absorption at 1710 cm−1.

(a)
Calculate the number of carbon atoms in Q.
[1]
(b)
Identify the bonds responsible for the two IR absorptions.
[2]
(c)
Deduce the structural formula of Q.
[1]
(d)
Suggest the formula of the fragment responsible for a peak at m/e 45, and write an equation for its formation from the molecular ion.
[2]
  • n = 100 × 0.53 / (1.1 × 24.0) = 2.0, so 2 carbon atoms
  • 2500–3000 broad: O–H in a carboxylic acid
  • 1710: C=O
  • CH3COOH (ethanoic acid, Mr 60)
  • COOH+
  • CH3COOH+ → COOH+ + CH3•

A Level Chemistry, built on AS

Notes for all 15 A Level topics of the Cambridge 9701 syllabus, from Born–Haber cycles to NMR spectroscopy. Each topic has worked examples, diagrams, interactive tools and Paper 4 style questions with mark schemes. A Level questions often expect AS knowledge too, so revise both.

Paper 4: A Level Structured Questions

2 h100 marks38.5% of A Level

Structured questions on the A Level topics; knowledge of the AS topics is assumed.

Paper 5: Planning, Analysis and Evaluation

1 h 15 min30 marks11.5% of A Level

Planning an experiment, processing data, drawing graphs, and evaluating results and errors. May use unfamiliar contexts.

Papers 1, 2 and 3 (the AS papers) make up the other 50% of the A Level.

Topics

Paper 5 skills

Paper 5 has no practical work, but it tests the thinking behind practical work. Expect to:

  • Plan: identify the independent, dependent and control variables; describe a method with specific apparatus and quantities; explain how to make it fair and safe (with a hazard and a precaution).
  • Process data: calculate means, rates, ln values or reciprocals; give answers to a sensible number of significant figures.
  • Draw and read graphs: choose scales that use at least half the grid, plot accurately, draw a best-fit line, calculate a gradient using a large triangle, and identify anomalous points.
  • Evaluate: calculate percentage uncertainties (uncertainty ÷ reading × 100), compare them, decide whether the data support a conclusion, and suggest improvements.
About the questions

The questions here are written in the style of 9701 papers and test the same skills. They are not copies of official papers. For genuine past papers and mark schemes, use the Cambridge School Support Hub or ask your teacher.

A Level energetics answers two big questions. Why do ionic compounds form, and why do some dissolve? (Lattice and hydration energies.) And will a reaction happen at all? (Entropy and Gibbs free energy.)

New enthalpy terms

Enthalpy change of atomisation, ΔHat
The enthalpy change when one mole of gaseous atoms is formed from the element in its standard state under standard conditions. Always positive. For example Na(s) → Na(g) and ½Cl2(g) → Cl(g).
Lattice energy, ΔHlatt
The enthalpy change when one mole of an ionic compound is formed from its gaseous ions under standard conditions. Always negative (exothermic). For example Na+(g) + Cl−(g) → NaCl(s).
First electron affinity, EA1
The enthalpy change when one electron is added to each atom in one mole of gaseous atoms, to form one mole of gaseous 1− ions. For example Cl(g) + e− → Cl−(g), −349 kJ mol−1.
Second electron affinity, EA2
The enthalpy change when one electron is added to each ion in one mole of gaseous 1− ions, to form one mole of gaseous 2− ions. For example O−(g) + e− → O2−(g), +798 kJ mol−1.

Electron affinity

First electron affinities are usually exothermic: the incoming electron is attracted by the nucleus. The same factors as for ionisation energy apply: nuclear charge, distance from the nucleus and shielding. Second electron affinities are always endothermic, because the electron is being added to a negative ion, which repels it.

Group 17FClBrIGroup 16OS
EA1 / kJ mol−1−328−349−325−295EA1 / kJ mol−1−141−200
  • Down a group, EA1 becomes less exothermic: the atoms are larger and the outer shell more shielded, so the incoming electron is attracted less.
  • Fluorine and oxygen are exceptions: they are less exothermic than chlorine and sulfur. Their atoms are so small that the outer electrons are crowded together, and they repel the incoming electron strongly.
  • Group 17 vs Group 16: the halogens have more exothermic EA1 values, because they have a larger nuclear charge for the same shell.

Born–Haber cycles

Lattice energy can't be measured directly, so it is calculated with a Hess's law cycle, the Born–Haber cycle. One route forms the ionic solid directly from its elements (ΔHf). The other goes through gaseous atoms and then gaseous ions.

Na(s) + ½Cl₂(g)Na(g) + ½Cl₂(g)ΔHat(Na) = +107Na⁺(g) + e⁻ + ½Cl₂(g)IE₁(Na) = +496Na⁺(g) + e⁻ + Cl(g)ΔHat(Cl) = +122Na⁺(g) + Cl⁻(g)EA₁(Cl) = −349NaCl(s)ΔHf = −411lattice energy = ?enthalpy (kJ mol⁻¹)
Born–Haber cycle for sodium chloride, with values in kJ mol−1. Upward arrows are endothermic; downward arrows are exothermic.
From the cycleΔHf = ΔHat(metal) + ΣIE + ΔHat(non-metal) + ΣEA + ΔHlattLearn it

Worked example: sodium chloride

−411 = 107 + 496 + 122 + (−349) + ΔHlatt

ΔHlatt = −411 − 376 = −787 kJ mol−1

Worked example: magnesium oxide

Data / kJ mol−1: ΔHf(MgO) −602; ΔHat(Mg) +148; IE1(Mg) +736; IE2(Mg) +1450; ΔHat(O) +249; EA1(O) −141; EA2(O) +798.

  1. Sum of the steps to the gaseous ions: 148 + 736 + 1450 + 249 − 141 + 798 = +3240
  2. ΔHlatt = −602 − 3240

ΔHlatt(MgO) = −3842 kJ mol−1

Note that both ionisation energies and both electron affinities are needed for 2+ and 2− ions. For MgCl2 or CaF2, the atomisation and electron affinity of the halogen must each be multiplied by 2.

What makes a lattice energy large?

CompoundNaClKClNaFMgCl2CaOMgO
Lattice energy / kJ mol−1 (typical values)−787−711−918−2526−3401−3791
  • Larger ionic charge → much more exothermic lattice energy: the attraction between ions is stronger. MgO (2+, 2−) is about five times NaCl (1+, 1−).
  • Smaller ionic radius → more exothermic: the ions get closer together. NaCl > KCl (K+ is larger) and NaF > NaCl (F− is smaller); MgO > CaO.

Experimental lattice energies (from Born–Haber cycles) can differ slightly from values calculated for a perfectly ionic model, but that comparison isn't assessed.

Enthalpies of solution and hydration

Enthalpy change of hydration, ΔHhyd
The enthalpy change when one mole of a specified gaseous ion dissolves in enough water to form an infinitely dilute solution. Always exothermic. For example Na+(g) → Na+(aq).
Enthalpy change of solution, ΔHsol
The enthalpy change when one mole of an ionic solid dissolves in enough water to form an infinitely dilute solution. Can be exothermic or endothermic. For example NaCl(s) → Na+(aq) + Cl−(aq).
Na⁺(g) + Cl⁻(g)NaCl(s)Na⁺(aq) + Cl⁻(aq)lattice energyΔHhyd(Na⁺) + ΔHhyd(Cl⁻)ΔHsol
Dissolving can be imagined as breaking the lattice into gaseous ions (the reverse of the lattice energy), then hydrating each ion.
Solution cycleΔHsol = −ΔHlatt + ΣΔHhydLearn it

Worked example

ΔHlatt(NaCl) = −787; ΔHhyd(Na+) = −406; ΔHhyd(Cl−) = −378 kJ mol−1.

ΔHsol = +787 + (−406) + (−378) = +3 kJ mol−1: very slightly endothermic, which is why salt dissolving barely changes the temperature.

Size of hydration enthalpies: smaller ions with larger charges attract the polar water molecules more strongly (δ− oxygen towards cations, δ+ hydrogen towards anions), so their hydration enthalpies are more exothermic. For example ΔHhyd: Li+ −519, Na+ −406, K+ −322 kJ mol−1, and 2+ ions are much more exothermic still.

Entropy

Entropy, S
The number of possible arrangements of the particles and their energy in a given system. Units: J K−1 mol−1.

The more ways the particles and their energy can be arranged, the more disordered the system and the higher its entropy. Systems tend to move towards higher entropy. Every substance above 0 K has a positive entropy.

Predicting the sign of ΔS

ChangeΔSReason
melting, boiling, dissolving a solidpositiveparticles become freer to move, with more ways to arrange them: S(solid) < S(liquid) ≪ S(gas)
freezing, condensing, precipitationnegativethe reverse
increasing temperaturepositiveparticles have more energy, spread over more energy levels
more moles of gas formed, e.g. CaCO3(s) → CaO(s) + CO2(g)positivegases have much higher entropy than solids or liquids
fewer moles of gas, e.g. N2(g) + 3H2(g) → 2NH3(g)negative4 mol of gas become 2
Entropy changeΔS = ΣS(products) − ΣS(reactants)Learn it

Worked example

S / J K−1 mol−1: CaCO3(s) 92.9, CaO(s) 39.7, CO2(g) 213.6.

ΔS = (39.7 + 213.6) − 92.9 = +160.4 J K−1 mol−1. Positive, as expected, because a gas forms.

Gibbs free energy

Gibbs free energy changeΔG = ΔH − TΔSLearn it

A reaction is feasible (it can happen spontaneously) when ΔG is negative. When ΔG = 0, the system is at the borderline (equilibrium).

Watch the units

ΔH is in kJ mol−1 but ΔS is in J K−1 mol−1. Divide ΔS by 1000 before using it, and use T in kelvin.

Worked example: when does calcium carbonate decompose?

For CaCO3(s) → CaO(s) + CO2(g), ΔH = +178 kJ mol−1 and ΔS = +160.4 J K−1 mol−1.

  1. At 298 K: ΔG = 178 − 298 × 0.1604 = +130 kJ mol−1. Positive, so not feasible at room temperature.
  2. It becomes feasible when ΔG ≤ 0, that is when T ≥ ΔH / ΔS = 178 / 0.1604

T = 1110 K (about 837 °C). Limestone kilns run above this temperature.

The effect of temperature on feasibility

ΔHΔSΔG = ΔH − TΔSFeasible?
negativepositivealways negativeat all temperatures
positivenegativealways positivenever
negativenegativenegative at low Tbelow T = ΔH/ΔS
positivepositivenegative at high Tabove T = ΔH/ΔS
030060090012001500−120−60060120temperature / KΔG / kJ mol⁻¹ΔH −, ΔS +ΔH +, ΔS −ΔH −, ΔS −ΔH +, ΔS +ΔG = 0
ΔG plotted against temperature is a straight line with intercept ΔH and gradient −ΔS. Where a line crosses ΔG = 0 (dashed lines at 667 K and 900 K for these examples), feasibility changes.

Feasible doesn't mean fast. A negative ΔG says a reaction can happen, not how quickly. Diamond turning into graphite has a negative ΔG at room temperature, but the activation energy is so high that it never happens noticeably. Such systems are kinetically stable.

Questions

1Quick check

Which ionic compound has the most exothermic lattice energy?

  1. NaCl
  2. MgO
  3. KBr
  4. CaO

MgO has 2+ and 2− ions (high charge) and the small Mg2+ ion (smaller than Ca2+).

2Quick check

For a reaction, ΔH = −50 kJ mol−1 and ΔS = −100 J K−1 mol−1. Above which temperature does it stop being feasible?

  1. 0.5 K
  2. 200 K
  3. 500 K
  4. 5000 K

T = ΔH/ΔS = −50 / −0.100 = 500 K. Below this, ΔG is negative.

3Quick check

Which change has the most negative entropy change, ΔS?

  1. H2O(l) → H2O(g)
  2. CaCO3(s) → CaO(s) + CO2(g)
  3. N2(g) + 3H2(g) → 2NH3(g)
  4. 2H2(g) + O2(g) → 2H2O(l)

In D three moles of gas become liquid (ΔS about −327 J K−1 mol−1), a bigger loss of disorder than four moles of gas becoming two in C (−199 J K−1 mol−1). A and B increase entropy.

4Quick check

The ions in magnesium oxide and in sodium fluoride all have the electron configuration of neon. Why is the lattice energy of MgO much more exothermic than that of NaF?

  1. Mg2+ and O2− have twice the charges of Na+ and F−
  2. Mg2+ and O2− are larger ions
  3. magnesium has a higher first ionisation energy than sodium
  4. oxygen is more electronegative than fluorine

Lattice energy depends on the charges and radii of the ions. The radii are similar, but doubling both charges makes the attraction about four times stronger.

5Quick check

Which ion has the most exothermic enthalpy change of hydration?

  1. K+
  2. Na+
  3. Li+
  4. Mg2+

Mg2+ has the highest charge and a small radius, so it attracts the δ− oxygen atoms of water molecules most strongly.

6Paper 4 style

Data / kJ mol−1: ΔHf(CaCl2(s)) −796; ΔHat(Ca) +178; IE1(Ca) +590; IE2(Ca) +1145; ΔHat(Cl) +122; EA1(Cl) −349.

(a)
Define lattice energy.
[2]
(b)
Construct a Born–Haber cycle and calculate the lattice energy of calcium chloride.
[3]
(c)
Suggest, with a reason, whether the lattice energy of CaBr2 is more or less exothermic than that of CaCl2.
[2]
  • Enthalpy change when one mole of an ionic compound is formed
  • …from its gaseous ions (under standard conditions)
  • Correct cycle, with 2 × atomisation of Cl and 2 × EA of Cl
  • Sum to gaseous ions: 178 + 590 + 1145 + 2(122) + 2(−349) = +1459
  • ΔHlatt = −796 − 1459 = −2255 kJ mol−1
  • Less exothermic
  • Br− has a larger ionic radius than Cl−, so the ions are further apart and attract less strongly
7Paper 4 style

Ammonia is made by N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ mol−1. Standard entropies / J K−1 mol−1: N2 192, H2 131, NH3 193.

(a)
Explain why the entropy change for this reaction is negative, without calculation.
[1]
(b)
Calculate ΔS.
[1]
(c)
Calculate ΔG at 298 K and state whether the reaction is feasible.
[2]
(d)
Calculate the temperature above which the reaction is no longer feasible. Suggest why the Haber process still runs at about 700 K.
[3]
  • 4 mol of gas become 2 mol of gas, so there are fewer ways of arranging the particles
  • ΔS = 2(193) − [192 + 3(131)] = 386 − 585 = −199 J K−1 mol−1
  • ΔG = −92 − 298 × (−0.199) = −32.7 kJ mol−1
  • Negative, so feasible
  • T = 92 / 0.199 = 462 K
  • Above this, the equilibrium yield is low (ΔG positive means K < 1) but the rate is much faster
  • The high pressure and removal of ammonia push the equilibrium to the right, so a useful yield is still obtained

Electrolysis uses electricity to drive a redox reaction; an electrochemical cell uses a redox reaction to produce electricity. Standard electrode potentials link the two, and predict which redox reactions can happen.

Electrolysis

In electrolysis, a direct current passes through an electrolyte (a molten or dissolved ionic compound). Positive ions move to the cathode (negative electrode), where they gain electrons (reduction). Negative ions move to the anode (positive electrode), where they lose electrons (oxidation).

Predicting the products

Molten compounds give the metal at the cathode and the non-metal at the anode: molten lead(II) bromide gives lead and bromine.

Aqueous solutions also contain water, which can be reduced or oxidised instead. The product depends on the electrode potentials and on concentration:

At the cathode

If the metal ion has a more positive E⦵ than hydrogen (such as Cu2+, Ag+), the metal forms: Cu2+ + 2e− → Cu. Otherwise (Na, K, Mg, Al…) hydrogen forms instead: 2H2O + 2e− → H2 + 2OH− (or 2H+ + 2e− → H2 in acid).

At an inert anode

Iodide, bromide and concentrated chloride ions give the halogen: 2Cl− → Cl2 + 2e−. Otherwise (sulfate, nitrate, dilute chloride) oxygen forms: 4OH− → O2 + 2H2O + 4e− or 2H2O → O2 + 4H+ + 4e−.

ElectrolyteCathode productAnode product (inert electrodes)
concentrated NaCl(aq)hydrogenchlorine
dilute NaCl(aq)hydrogenmainly oxygen (some chlorine)
CuSO4(aq)copperoxygen
dilute H2SO4(aq)hydrogenoxygen (in a 2 : 1 volume ratio)
KI(aq)hydrogeniodine
CuSO4(aq) with copper electrodescopperthe copper anode dissolves: Cu → Cu2+ + 2e−

Why concentration matters: E⦵(Cl2/Cl−) = +1.36 V and E⦵(O2/H2O) = +1.23 V are close. A high concentration of Cl− makes the Cl2/Cl− electrode potential less positive (see the Nernst equation below), so chloride is oxidised more easily than water.

Quantitative electrolysis

ChargeQ = ItLearn it
Faraday constantF = LeLearn it
Moles of electronsn(e−) = QFLearn it

The Faraday constant is the charge on one mole of electrons: 6.02 × 1023 × 1.60 × 10−19 = 9.63 × 104 C mol−1 (the Data Booklet gives 9.65 × 104). Use the half-equation to link moles of electrons to moles of product.

Worked example

Copper(II) sulfate solution is electrolysed with a current of 0.500 A for 30.0 minutes. Calculate the mass of copper deposited and the volume of oxygen released (room conditions).

  1. Q = It = 0.500 × 30.0 × 60 = 900 C
  2. n(e−) = 900/96 500 = 9.33 × 10−3 mol
  3. Cu2+ + 2e− → Cu: n(Cu) = 4.66 × 10−3 mol, mass = 4.66 × 10−3 × 63.5 = 0.296 g
  4. 2H2O → O2 + 4H+ + 4e−: n(O2) = 9.33 × 10−3 ÷ 4 = 2.33 × 10−3 mol, volume = 2.33 × 10−3 × 24.0 = 0.0560 dm3

0.296 g of copper and 56.0 cm3 of oxygen

Determining the Avogadro constant by electrolysis

  1. Weigh a clean, dry copper cathode. Set up a circuit with copper electrodes in copper(II) sulfate solution, an ammeter and a variable resistor to keep the current constant.
  2. Pass a measured current for a measured time.
  3. Remove the cathode, rinse it with water then propanone, dry and reweigh.
  4. Calculate Q = It, and moles of electrons = 2 × moles of copper. Then F = Q ÷ moles of electrons, and L = F ÷ e.

Worked example

A current of 0.200 A for 40.0 min increases the mass of the cathode by 0.158 g.

  1. Q = 0.200 × 2400 = 480 C
  2. n(Cu) = 0.158 / 63.5 = 2.49 × 10−3 mol, so n(e−) = 4.98 × 10−3 mol
  3. charge per mole of electrons = 480 / 4.98 × 10−3 = 9.65 × 104 C mol−1

L = 9.65 × 104 / 1.60 × 10−19 = 6.03 × 1023 mol−1

Standard electrode potentials

Standard electrode potential, E⦵
The electrode potential of a half-cell measured relative to the standard hydrogen electrode, under standard conditions: 298 K, 1.00 mol dm−3 solutions, gases at 101 kPa.
Standard cell potential, E⦵cell
The potential difference between two half-cells under standard conditions.

Electrode potentials are always written for the reduction half-equation, for example Cu2+(aq) + 2e− ⇌ Cu(s), E⦵ = +0.34 V.

The standard hydrogen electrode

The reference electrode is hydrogen gas at 101 kPa bubbling over a platinum electrode coated in platinum black, dipping into a solution with [H+] = 1.00 mol dm−3 (such as 1.00 mol dm−3 HCl), at 298 K. Its potential is defined as exactly 0.00 V.

1.00 mol dm⁻³ H⁺(aq)platinised PtH₂(g), 101 kPaV1.00 mol dm⁻³ Mⁿ⁺(aq)Msalt bridgeall at 298 K; the voltmeter reading is E⦵ of the Mⁿ⁺/M half-cell

Measuring electrode potentials

  • A metal in contact with its ions (such as Cu in 1.00 mol dm−3 Cu2+): connect to the standard hydrogen electrode with a salt bridge and a high-resistance voltmeter.
  • Ions of the same element in two oxidation states (such as Fe3+/Fe2+): use a solution containing both ions at 1.00 mol dm−3, with an inert platinum electrode to carry the electrons.
  • A non-metal (such as Cl2/Cl−): bubble the gas at 101 kPa over a platinum electrode in a 1.00 mol dm−3 solution of its ions.

The salt bridge (filter paper soaked in aqueous KNO3) completes the circuit by letting ions move between the half-cells, without the solutions mixing. A high-resistance voltmeter draws almost no current, so it measures the maximum potential difference.

Half-equationE⦵ / VHalf-equationE⦵ / V
Li+ + e− ⇌ Li−3.04Cu2+ + 2e− ⇌ Cu+0.34
Mg2+ + 2e− ⇌ Mg−2.38I2 + 2e− ⇌ 2I−+0.54
Zn2+ + 2e− ⇌ Zn−0.76Fe3+ + e− ⇌ Fe2++0.77
Fe2+ + 2e− ⇌ Fe−0.44Ag+ + e− ⇌ Ag+0.80
Pb2+ + 2e− ⇌ Pb−0.13Br2 + 2e− ⇌ 2Br−+1.07
2H+ + 2e− ⇌ H20.00Cr2O72− + 14H+ + 6e− ⇌ 2Cr3+ + 7H2O+1.33
Sn4+ + 2e− ⇌ Sn2++0.15Cl2 + 2e− ⇌ 2Cl−+1.36
O2 + 4H+ + 4e− ⇌ 2H2O+1.23MnO4− + 8H+ + 5e− ⇌ Mn2+ + 4H2O+1.52
  • The more positive the E⦵, the more easily the species on the left is reduced, so the stronger the oxidising agent on the left. MnO4−/H+ and Cl2 are strong oxidising agents.
  • The more negative the E⦵, the more easily the species on the right is oxidised, so the stronger the reducing agent on the right. Li, Mg and Zn are strong reducing agents.

Electrochemical cells

1.00 mol dm⁻³ Zn²⁺(aq)1.00 mol dm⁻³ Cu²⁺(aq)ZnCusalt bridge (KNO₃(aq) on filter paper)Vhigh-resistance voltmeter: +1.10 Velectronsnegative electrode: Zn → Zn²⁺ + 2e⁻positive electrode: Cu²⁺ + 2e⁻ → Cu
The zinc–copper cell. Electrons flow in the external wire from the more negative electrode (Zn) to the more positive one (Cu).
Cell potentialEcell = E(positive electrode) − E(negative electrode)Learn it
  • The half-cell with the more positive E⦵ is the positive electrode; reduction happens there.
  • The half-cell with the more negative E⦵ is the negative electrode; oxidation happens there.
  • Electrons flow through the external circuit from negative to positive.

Predicting feasibility

A redox reaction is feasible if E⦵cell for it is positive. To check a reaction, find the half-equation for the species being reduced and the one for the species being oxidised; E⦵cell = E⦵(reduced) − E⦵(oxidised).

Worked example: will iron(III) ions oxidise iodide ions?

  1. Reduction: Fe3+ + e− → Fe2+, +0.77 V. Oxidation: 2I− → I2 + 2e−, which is the reverse of +0.54 V.
  2. E⦵cell = 0.77 − 0.54 = +0.23 V, positive, so the reaction is feasible.
  3. Combine the half-equations so the electrons cancel (double the iron one):

2Fe3+(aq) + 2I−(aq) → 2Fe2+(aq) + I2(aq), E⦵cell = +0.23 V

Limitations: a positive E⦵cell means a reaction can happen, but it may be too slow to notice (high activation energy), and the conditions may not be standard (different concentrations or temperature change E).

The effect of concentration: the Nernst equation

By Le Chatelier's principle, increasing the concentration of the oxidised species (on the left of the half-equation) pushes the equilibrium to the right, making E more positive. Decreasing it makes E less positive.

Nernst equation (298 K)E = E + 0.059zlog [oxidised species][reduced species]Learn it

z is the number of electrons in the half-equation. Solid metals are left out of the expression (their "concentration" is constant).

Worked examples

  1. Cu2+/Cu with [Cu2+] = 0.010 mol dm−3: E = 0.34 + 0.0592log (0.010) = 0.34 − 0.059 = +0.28 V
  2. Fe3+/Fe2+ with [Fe3+] = 0.10 and [Fe2+] = 1.0 mol dm−3: E = 0.77 + 0.059log 0.101.0 = 0.77 − 0.059 = +0.71 V

Linking Ecell to Gibbs free energy

Free energy from a cellΔG = −nFEcellLearn it

n is the number of moles of electrons transferred in the equation. A positive E⦵cell gives a negative ΔG, consistent with a feasible reaction. For the zinc–copper cell: ΔG = −2 × 96 500 × 1.10 = −212 000 J mol−1 = −212 kJ mol−1.

Questions

1Quick check

Which species is the strongest reducing agent? Use the table above.

  1. Cl−
  2. Fe2+
  3. Cu
  4. Zn

The strongest reducing agent is on the right of the half-equation with the most negative E⦵: Zn (−0.76 V).

2Quick check

A silver half-cell is connected to a magnesium half-cell under standard conditions. What is E⦵cell?

  1. 1.58 V
  2. 3.18 V
  3. −3.18 V
  4. 0.80 V

E⦵cell = 0.80 − (−2.38) = 3.18 V. Silver is the positive electrode.

3Quick check

What mass of aluminium is produced when 0.30 mol of electrons passes through molten aluminium oxide? (Ar: Al, 27.0)

  1. 0.90 g
  2. 2.7 g
  3. 8.1 g
  4. 27 g

Al3+ + 3e− → Al: 0.30 mol of electrons gives 0.10 mol of Al, which is 0.10 × 27.0 = 2.7 g.

4Quick check

E⦵ values: Fe3+/Fe2+ +0.77 V; I2/I− +0.54 V; Cu2+/Cu +0.34 V; Br2/Br− +1.07 V; Zn2+/Zn −0.76 V. Which species can oxidise Fe2+ to Fe3+ under standard conditions?

  1. I2
  2. Cu2+
  3. Br2
  4. Zn2+

The oxidising agent needs a more positive E⦵ than +0.77 V. Only Br2/Br− qualifies: E⦵cell = 1.07 − 0.77 = +0.30 V.

5Quick check

The concentration of Cu2+(aq) in a Cu2+/Cu half-cell is lowered from 1.00 to 0.010 mol dm−3. What happens to the electrode potential?

  1. it becomes less positive, by about 0.06 V
  2. it becomes more positive, by about 0.06 V
  3. it becomes less positive, by about 0.12 V
  4. it doesn't change

E = E⦵ + (0.059/2) log[Cu2+] = 0.34 + 0.0295 × (−2) = 0.28 V, a fall of 0.059 V.

6Paper 4 style
(a)
Describe the standard hydrogen electrode.
[3]
(b)
A cell is made from a Br2/Br− half-cell and an Fe3+/Fe2+ half-cell. Calculate E⦵cell, state which electrode is positive, and write the overall equation for the cell reaction.
[3]
(c)
Use the Nernst equation to calculate the electrode potential of the Fe3+/Fe2+ half-cell when [Fe3+] = 0.050 mol dm−3 and [Fe2+] = 0.50 mol dm−3.
[2]
(d)
Calculate ΔG⦵ for the cell reaction in (b).
[2]
  • Hydrogen gas at 101 kPa (1 atm)
  • Platinum (platinised / coated with platinum black) electrode
  • 1.00 mol dm−3 H+(aq), at 298 K
  • E⦵cell = 1.07 − 0.77 = +0.30 V
  • The Br2/Br− electrode is positive
  • Br2 + 2Fe2+ → 2Br− + 2Fe3+
  • E = 0.77 + 0.059 log(0.050/0.50)
  • = 0.77 − 0.059 = +0.71 V
  • ΔG = −2 × 96 500 × 0.30 = −57 900 J mol−1
  • = −57.9 kJ mol−1
7Paper 4 style

Concentrated aqueous sodium chloride is electrolysed using inert electrodes.

(a)
Write half-equations for the reactions at the anode and cathode.
[2]
(b)
Explain why chlorine, not oxygen, is the main product at the anode.
[2]
(c)
A current of 2.50 A is passed for 1.00 hour. Calculate the volume of chlorine produced at room conditions.
[3]
  • Anode: 2Cl− → Cl2 + 2e−
  • Cathode: 2H2O + 2e− → H2 + 2OH−
  • E⦵ values for Cl2/Cl− (+1.36) and O2/H2O (+1.23) are close
  • The high concentration of Cl− makes its electrode potential less positive, so Cl− is oxidised in preference
  • Q = 2.50 × 3600 = 9000 C
  • n(e−) = 9000 / 96 500 = 0.0933 mol; n(Cl2) = 0.0466 mol
  • V = 0.0466 × 24.0 = 1.12 dm3

This topic makes acid–base chemistry quantitative: calculating pH for strong and weak acids, designing buffers that resist pH change, and using solubility products and partition coefficients to describe how solutes distribute themselves.

Conjugate acids and bases

When a Brønsted–Lowry acid donates a proton, what is left can accept a proton back: it is the conjugate base of that acid. Every acid–base reaction has two conjugate pairs.

CH3COOH + H2O ⇌ CH3COO− + H3O+ acid 1 (CH3COOH) and its conjugate base 1 (CH3COO−); base 2 (H2O) and its conjugate acid 2 (H3O+)
NH3 + H2O ⇌ NH4+ + OH− base 1 (NH3) and its conjugate acid (NH4+); acid 2 (H2O) and its conjugate base (OH−)

A conjugate pair differs by exactly one H+. The stronger an acid, the weaker its conjugate base.

pH, Kw and strong acids and bases

pHpH = −log [H+], so [H+] = 10−pHLearn it
Ionic product of waterKw = [H+][OH−] = 1.00 × 10−14 mol2 dm−6 (298 K)Kw in Data Booklet

Water ionises slightly: H2O ⇌ H+ + OH−. In pure water at 298 K, [H+] = [OH−] = 1.00 × 10−7 mol dm−3, so pH = 7.00. (The ionisation is endothermic, so at higher temperatures Kw is larger and neutral water has a pH below 7.)

SolutionWorkingpH
0.0200 mol dm−3 HCl (strong, monoprotic)[H+] = 0.0200; pH = −log 0.02001.70
0.0500 mol dm−3 NaOH (strong base)[OH−] = 0.0500; [H+] = 1.00 × 10−14 / 0.0500 = 2.00 × 10−1312.70
0.0500 mol dm−3 Ba(OH)2two OH− per formula: [OH−] = 0.100; [H+] = 1.00 × 10−1313.00

Weak acids: Ka and pKa

For a weak acid HA ⇌ H+ + A−:

Acid dissociation constantKa = [H+][A−][HA]Learn it
pKapKa = −log KaLearn it

The larger Ka (the smaller pKa), the stronger the acid. Ethanoic acid has Ka = 1.74 × 10−5 mol dm−3 (pKa 4.76).

pH of a weak acid

Two approximations make the calculation simple: (1) [H+] = [A−], because water's own ionisation is negligible; (2) [HA] at equilibrium ≈ the initial concentration c, because so little dissociates. Then Ka = [H+]2c, so:

Weak acid[H+] = √Ka × cLearn it

Worked example

Calculate the pH of 0.100 mol dm−3 ethanoic acid. Ka = 1.74 × 10−5 mol dm−3.

[H+] = √(1.74 × 10−5 × 0.100) = 1.32 × 10−3 mol dm−3; pH = 2.88

Worked example: finding Ka from pH

A 0.0500 mol dm−3 solution of a weak acid has pH 3.10. Find Ka.

[H+] = 10−3.10 = 7.94 × 10−4; Ka = (7.94 × 10−4)2 / 0.0500 = 1.26 × 10−5 mol dm−3

Buffer solutions

Buffer solution
A solution that resists changes in pH when small amounts of acid or alkali are added.

An acidic buffer contains a weak acid and its conjugate base (a salt of the acid), both in large amounts, for example ethanoic acid and sodium ethanoate. It can also be made by partly neutralising a weak acid with a strong base. (A basic buffer uses a weak base and its salt, such as ammonia and ammonium chloride.)

In the ethanoic acid / sodium ethanoate buffer there is a large reservoir of both CH3COOH and CH3COO−:

CH3COOH ⇌ CH3COO− + H+
  • Adding a little acid (H+): the added H+ reacts with the large amount of CH3COO−: CH3COO− + H+ → CH3COOH. The ratio [CH3COOH]/[CH3COO−] hardly changes, so [H+] and the pH hardly change.
  • Adding a little alkali (OH−): the OH− reacts with the large amount of CH3COOH: CH3COOH + OH− → CH3COO− + H2O. Again the ratio barely changes.
Buffer pH[H+] = Ka × [HA][A−]Learn it

Worked example 1

A buffer contains 0.100 mol dm−3 ethanoic acid and 0.200 mol dm−3 sodium ethanoate. Ka = 1.74 × 10−5.

[H+] = 1.74 × 10−5 × 0.100 / 0.200 = 8.70 × 10−6; pH = 5.06

Worked example 2: by partial neutralisation

25.0 cm3 of 0.200 mol dm−3 NaOH is added to 50.0 cm3 of 0.200 mol dm−3 ethanoic acid.

  1. n(acid) = 0.0100 mol; n(NaOH) = 0.00500 mol, which converts 0.00500 mol of acid into ethanoate.
  2. Left: 0.00500 mol CH3COOH and 0.00500 mol CH3COO−, in the same volume, so the ratio is 1.

[H+] = Ka, so pH = pKa = 4.76. (This is the half-neutralisation point on a titration curve.)

Buffers in the body

Blood must stay at pH 7.35 to 7.45. The main buffer is carbonic acid / hydrogencarbonate: H2CO3 ⇌ H+ + HCO3−, where H2CO3 comes from dissolved carbon dioxide (CO2 + H2O ⇌ H2CO3). Extra acid is removed by HCO3− + H+ → H2CO3; the H2CO3 breaks down to CO2, which is breathed out. Extra alkali is removed by H2CO3 + OH− → HCO3− + H2O. Buffers are also used in shampoos, food, fermentation and to calibrate pH meters.

The pH titration curves from Topic 7 are now fully explained: the flat region for a weak acid is a buffer region, with pH = pKa at half-neutralisation. Try changing the acid and base in the titration simulator below.

Solubility product, Ksp

For a sparingly soluble salt in equilibrium with its saturated solution, such as AgCl(s) ⇌ Ag+(aq) + Cl−(aq):

Solubility product, Ksp
The product of the concentrations of the ions in a saturated solution of a sparingly soluble salt, each raised to the power of its coefficient in the equation, at a given temperature. The solid is not included.
SaltEquilibriumKsp expressionUnits
AgClAgCl(s) ⇌ Ag+ + Cl−[Ag+][Cl−]mol2 dm−6
CaF2CaF2(s) ⇌ Ca2+ + 2F−[Ca2+][F−]2mol3 dm−9
Ag2CrO4Ag2CrO4(s) ⇌ 2Ag+ + CrO42−[Ag+]2[CrO42−]mol3 dm−9

Worked example 1: Ksp from solubility

The solubility of silver chloride is 1.30 × 10−5 mol dm−3.

Ksp = (1.30 × 10−5)2 = 1.69 × 10−10 mol2 dm−6

Worked example 2: solubility from Ksp

Ksp(CaF2) = 3.90 × 10−11 mol3 dm−9. If the solubility is s, then [Ca2+] = s and [F−] = 2s, so Ksp = s(2s)2 = 4s3.

s = ∛(3.90 × 10−11 / 4) = 2.14 × 10−4 mol dm−3

The common ion effect

A sparingly soluble salt is less soluble in a solution that already contains one of its ions. Adding the common ion pushes the equilibrium to the left, so more solid forms, until the ion product again equals Ksp.

Worked example

What is the solubility of AgCl (Ksp 1.69 × 10−10) in 0.100 mol dm−3 NaCl?

[Cl−] ≈ 0.100 (the AgCl adds very little), so [Ag+] = 1.69 × 10−10 / 0.100 = 1.69 × 10−9 mol dm−3. That's nearly 10 000 times less than in pure water.

A precipitate forms whenever mixing two solutions makes the ion product exceed Ksp.

Partition coefficients

Partition coefficient, Kpc
The ratio of the concentrations of a solute in two immiscible solvents when it has reached equilibrium, at a given temperature (the solute is in the same physical state in both solvents).
Partition coefficientKpc = [X in solvent 1][X in solvent 2]Learn it

The value depends on the relative polarities: a non-polar solute (such as iodine) is much more soluble in a non-polar organic solvent than in water, so Kpc(organic/water) is large. A polar solute that can hydrogen bond with water (such as ammonia or ethanoic acid) has a smaller value.

Worked example: solvent extraction

Kpc for a compound Y between ether and water is 4.0. 100 cm3 of an aqueous solution contains 2.00 g of Y. How much Y is extracted by shaking with 50 cm3 of ether?

  1. Let x g go into the ether: x/50(2.00 − x)/100 = 4.0
  2. So 2x = 4.0(2.00 − x), giving x = 1.33 g (67% extracted).

Two separate extractions with 25 cm3 of ether each remove 1.00 g then 0.50 g: 1.50 g in total (75%). Several small extractions are more efficient than one large one.

Questions

1Quick check

What is the pH of 0.0010 mol dm−3 potassium hydroxide at 298 K?

  1. 3.0
  2. 10.0
  3. 11.0
  4. 14.0

[OH−] = 1.0 × 10−3, so [H+] = 1.0 × 10−14 / 1.0 × 10−3 = 1.0 × 10−11, pH 11.0.

2Quick check

Which pair is a conjugate acid–base pair?

  1. H2SO4 and HSO4−
  2. H3O+ and OH−
  3. NH4+ and NH2−
  4. HCl and Cl2

A conjugate pair differs by one H+. H3O+ and OH− differ by two; so do NH4+ and NH2−.

3Quick check

Which mixture is a buffer solution?

  1. HCl(aq) and NaCl(aq)
  2. NaOH(aq) and NaCl(aq)
  3. CH3COOH(aq) and CH3COONa(aq)
  4. NH3(aq) and excess HCl(aq)

A buffer needs a weak acid and its conjugate base (or a weak base and its conjugate acid), both in reasonable amounts. With excess HCl, all the NH3 becomes NH4+, leaving no weak base.

4Quick check

What are the expression for the solubility product of silver chromate(VI), Ag2CrO4, and its units?

  1. [Ag+][CrO42−]; mol2 dm−6
  2. [Ag+]2[CrO42−]; mol3 dm−9
  3. [Ag+]2[CrO42−] / [Ag2CrO4]; no units
  4. 2[Ag+][CrO42−]; mol2 dm−6

Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO42−(aq). The solid isn't included, and three concentration terms give mol3 dm−9.

5Quick check

Solute X has a partition coefficient Kpc = [X]hexane / [X]water = 4.0. 100 cm3 of water containing 0.50 g of X is shaken with 100 cm3 of hexane. What mass of X stays in the water?

  1. 0.10 g
  2. 0.125 g
  3. 0.25 g
  4. 0.40 g

Let x g stay in the water: ((0.50 − x)/100) ÷ (x/100) = 4.0, so 0.50 − x = 4.0x and x = 0.10 g.

6Paper 4 style

Propanoic acid, CH3CH2COOH, is a weak acid with Ka = 1.35 × 10−5 mol dm−3.

(a)
Write the expression for Ka of propanoic acid.
[1]
(b)
Calculate the pH of 0.150 mol dm−3 propanoic acid.
[2]
(c)
A buffer is made by adding 0.0300 mol of sodium propanoate to 200 cm3 of the 0.150 mol dm−3 acid. Calculate its pH.
[2]
(d)
Explain, with an equation, how this buffer resists a decrease in pH when a small amount of acid is added.
[2]
  • Ka = [CH3CH2COO−][H+][CH3CH2COOH]
  • [H+] = √(1.35 × 10−5 × 0.150) = 1.42 × 10−3
  • pH = 2.85
  • n(acid) = 0.0300 mol, n(salt) = 0.0300 mol, so the ratio is 1 and [H+] = Ka
  • pH = −log(1.35 × 10−5) = 4.87
  • CH3CH2COO− + H+ → CH3CH2COOH
  • There is a large reservoir of the conjugate base, so the ratio [HA]/[A−] (and so [H+]) changes very little
7Paper 4 style

Lead(II) iodide, PbI2, is sparingly soluble. Its solubility at 298 K is 1.52 × 10−3 mol dm−3.

(a)
Write the expression for Ksp of PbI2 and calculate its value, with units.
[3]
(b)
Calculate the solubility of PbI2 in 0.100 mol dm−3 potassium iodide. Explain the difference from (a).
[3]
  • Ksp = [Pb2+][I−]2
  • = (1.52 × 10−3)(3.04 × 10−3)2 = 1.40 × 10−8
  • mol3 dm−9
  • [Pb2+] = 1.40 × 10−8 / (0.100)2
  • = 1.40 × 10−6 mol dm−3
  • Common ion effect: the I− from KI shifts the equilibrium to the left, so less PbI2 dissolves

A rate equation tells you exactly how the rate depends on each concentration. Finding it from experimental data, then working backwards to a mechanism, is one of the main Paper 4 and Paper 5 skills.

Rate equations

For a reaction A + B → products, experiments give a rate equation of the form:

Rate equationrate = k[A]m[B]nLearn it
Order of reaction with respect to a reactant
The power to which the concentration of that reactant is raised in the rate equation (here m for A, n for B). In this course it is 0, 1 or 2.
Overall order
The sum of the individual orders, m + n.
Rate constant, k
The constant of proportionality in the rate equation. It is constant at a given temperature, and increases as the temperature increases.
Half-life, t½
The time taken for the concentration of a reactant to fall to half its initial value.
Rate-determining step
The slowest step in a multi-step reaction mechanism; it limits the overall rate.
Intermediate
A species formed in one step of a mechanism and used up in a later step. It does not appear in the overall equation.

Orders can't be found from the balanced equation; they must be measured. The units of k depend on the overall order:

Overall order0123
Units of kmol dm−3 s−1s−1dm3 mol−1 s−1dm6 mol−2 s−1

Finding orders from graphs

02040608010000.250.50.751time / s[A] / mol dm⁻³zero orderfirst ordersecond ordert½2t½3t½
[A]ratezerofirstsecond
Left: concentration–time graphs. Right: rate–concentration graphs. The dashed lines on the first-order curve show successive equal half-lives.
Order in AConcentration–time graphRate–concentration graphHalf-life
0straight line sloping downhorizontal line: rate doesn't depend on [A]decreases as [A] falls
1curve with a constant half-lifestraight line through the originconstant, whatever the starting concentration
2steeper curve that levels out more; half-life gets longerupward curve (rate ∝ [A]2)increases as [A] falls

To get rates from a concentration–time graph, draw tangents at several points and find their gradients. The initial rate is the gradient at t = 0.

First-order reactions and half-life

For a first-order reaction the half-life is independent of concentration: going from 1.00 to 0.50, 0.50 to 0.25 or 0.25 to 0.125 mol dm−3 all take the same time. This is a quick test for first order. Radioactive decay is also first order.

First-order half-lifek = 0.693t1/2Learn it

Worked example

The decomposition of a compound is first order with a half-life of 120 s. Find k, and the fraction left after 360 s.

k = 0.693 / 120 = 5.78 × 10−3 s−1. 360 s is 3 half-lives, so (½)3 = 1/8 remains.

Finding orders from initial rates

Run the reaction several times, changing the concentration of one reactant at a time, and measure the initial rate each time. Compare pairs of experiments in which only one concentration changes.

Worked example

Experiment[A] / mol dm−3[B] / mol dm−3Initial rate / mol dm−3 s−1
10.100.102.0 × 10−4
20.200.108.0 × 10−4
30.100.306.0 × 10−4
  1. Experiments 1 and 2: [A] doubles, rate × 4 = 22, so second order in A.
  2. Experiments 1 and 3: [B] × 3, rate × 3, so first order in B.
  3. Rate = k[A]2[B]; overall third order.
  4. From experiment 1: k = 2.0 × 10−4 / (0.102 × 0.10) = 0.20

k = 0.20 dm6 mol−2 s−1

Rate equations and mechanisms

The rate equation contains only the species that take part in (or before) the rate-determining step, in the numbers that appear in that step. Species that react only after the slow step are zero order.

Example 1: nucleophilic substitution

  • 2-bromo-2-methylpropane + OH−: rate = k[(CH3)3CBr]. Only the halogenoalkane is in the slow step (forming the carbocation): SN1.
  • Bromoethane + OH−: rate = k[CH3CH2Br][OH−]. Both are in the single step: SN2.

Example 2: proposing a mechanism

For NO2(g) + CO(g) → NO(g) + CO2(g), rate = k[NO2]2. CO is zero order, so it isn't in the slow step, and two NO2 molecules are.

  1. Step 1 (slow): NO2 + NO2 → NO3 + NO
  2. Step 2 (fast): NO3 + CO → NO2 + CO2

The steps add up to the overall equation, and NO3 is an intermediate.

Example 3: a catalyst in the rate equation

For the acid-catalysed iodination of propanone, CH3COCH3 + I2 → CH3COCH2I + HI, rate = k[CH3COCH3][H+]. Iodine is zero order, so it reacts after the slow step, and H+ is a catalyst: it appears in the rate equation but not in the overall equation.

Temperature and the rate constant

Increasing the temperature increases k, and so the rate, because a greater proportion of molecules have energy at least equal to EA (Topic 8). The orders don't change.

Catalysts in more detail

Heterogeneous catalysis

The catalyst is a solid; the reactants are gases or in solution. It works in three stages:

  1. Adsorption: reactant molecules form weak bonds to active sites on the catalyst surface. This weakens bonds within the reactants and holds them close together in a favourable orientation.
  2. Reaction on the surface, with a lower activation energy.
  3. Desorption: the product molecules break away from the surface, freeing the sites.

Examples: iron in the Haber process (N2 and H2 are adsorbed and their bonds weakened); platinum, palladium and rhodium in catalytic converters (CO and NO are adsorbed and react to form CO2 and N2). A good catalyst adsorbs strongly enough to weaken bonds, but not so strongly that the products can't desorb.

Homogeneous catalysis

The catalyst is in the same phase as the reactants. It is used up in one step and regenerated in a later step, usually by changing oxidation state.

Iron ions and the iodide–peroxodisulfate reaction

S2O82−(aq) + 2I−(aq) → 2SO42−(aq) + I2(aq) is slow, because both reactants are negative ions that repel each other, giving a high activation energy. Fe2+ catalyses it:

S2O82− + 2Fe2+ → 2SO42− + 2Fe3+
2Fe3+ + 2I− → 2Fe2+ + I2

Each step is between oppositely charged ions, so the activation energy is lower. Fe3+ works equally well, starting with the second step. Both steps are feasible: E⦵(S2O82−/SO42−) = +2.01 V > +0.77 V > +0.54 V (Topic 24).

Oxides of nitrogen and atmospheric sulfur dioxide

SO2 + NO2 → SO3 + NO
NO + ½O2 → NO2

NO2 is regenerated, so it catalyses the oxidation of SO2 to SO3, which forms acid rain (Topic 12).

Questions

1Quick check

The rate equation for a reaction is rate = k[X][Y]2. What are the units of k?

  1. dm3 mol−1 s−1
  2. dm6 mol−2 s−1
  3. mol2 dm−6 s−1
  4. s−1

Third order overall: k = rate / concentration3 = (mol dm−3 s−1) / (mol dm−3)3 = dm6 mol−2 s−1.

2Quick check

A reaction is first order. Its concentration falls from 0.80 to 0.20 mol dm−3 in 50 s. What is its half-life?

  1. 25 s
  2. 50 s
  3. 12.5 s
  4. 100 s

0.80 → 0.40 → 0.20 is two half-lives in 50 s, so t½ = 25 s.

3Quick check

For the reaction A + B → C, rate = k[A][B]2. What are the units of k?

  1. s−1
  2. dm3 mol−1 s−1
  3. mol dm−3 s−1
  4. dm6 mol−2 s−1

k = rate / ([A][B]2) = mol dm−3 s−1 ÷ (mol dm−3)3 = dm6 mol−2 s−1.

4Quick check

Which statement about a heterogeneous catalyst, such as iron in the Haber process, is correct?

  1. it is in the same phase as the reactants
  2. reactants are adsorbed onto its surface, which weakens their bonds
  3. it is used up and must be replaced
  4. it increases the equilibrium yield of ammonia

Adsorption onto active sites weakens bonds and holds molecules close together; the products then desorb. A catalyst speeds up the forward and reverse reactions equally, so the equilibrium yield doesn't change.

5Quick check

The half-life of a first-order reaction is 20 s. What fraction of the reactant remains after 60 s?

  1. 1/3
  2. 1/6
  3. 1/8
  4. 1/4

60 s is three half-lives: ½ × ½ × ½ = 1/8.

6Paper 4 style

Hydrogen peroxide reacts with iodide ions in acid: H2O2 + 2I− + 2H+ → I2 + 2H2O. Initial rates were measured:

Experiment[H2O2][I−][H+]Initial rate / mol dm−3 s−1
10.0100.0100.101.75 × 10−6
20.0300.0100.105.25 × 10−6
30.0300.0200.101.05 × 10−5
40.0300.0200.201.05 × 10−5
(a)
Deduce the order with respect to each reactant. Explain your reasoning.
[3]
(b)
Write the rate equation and calculate the rate constant, with units.
[3]
(c)
Suggest the species involved in the rate-determining step.
[1]
  • H2O2: first order: experiments 1 and 2, concentration × 3, rate × 3
  • I−: first order: experiments 2 and 3, concentration × 2, rate × 2
  • H+: zero order: experiments 3 and 4, rate unchanged
  • rate = k[H2O2][I−]
  • k = 1.75 × 10−6 / (0.010 × 0.010) = 0.0175
  • dm3 mol−1 s−1
  • One H2O2 and one I− (the acid reacts in a later, fast step)
7Paper 4 style
(a)
Explain the difference between homogeneous and heterogeneous catalysis.
[1]
(b)
Describe how iron catalyses the reaction between nitrogen and hydrogen in the Haber process.
[3]
(c)
Explain, with equations, how Fe3+ ions catalyse the reaction between peroxodisulfate and iodide ions, and why the uncatalysed reaction is slow.
[3]
  • Homogeneous: catalyst in the same phase as the reactants; heterogeneous: different phase
  • N2 and H2 are adsorbed onto the iron surface (active sites)
  • Bonds in the reactants are weakened and the molecules are held close together, so they react
  • The ammonia molecules desorb from the surface
  • 2Fe3+ + 2I− → 2Fe2+ + I2
  • 2Fe2+ + S2O82− → 2Fe3+ + 2SO42−
  • Uncatalysed: both reactant ions are negative and repel each other, so the activation energy is high

At AS Level you learned the trends in Group 2. At A Level you explain two of them: why the nitrates and carbonates get more stable down the group, and why the hydroxides get more soluble while the sulfates get less soluble.

Thermal stability of the nitrates and carbonates

On heating, MCO3 → MO + CO2 and 2M(NO3)2 → 2MO + 4NO2 + O2. The temperature needed increases down the group.

  1. The cation M2+ attracts the electron cloud of the large carbonate (or nitrate) anion, polarising it: electron density is pulled towards the cation.
  2. This distortion weakens the C–O bonds in the carbonate ion (N–O in nitrate), making it easier to break them and release CO2 (or NO2 and O2).
  3. Down the group, the cation has the same charge but a larger radius, so its charge density is lower. It polarises the anion less, so the compound is more stable and needs a higher temperature to decompose.

Mg2+ is the smallest, most polarising ion, so magnesium carbonate decomposes most easily; barium carbonate needs the highest temperature.

Solubility of the hydroxides and sulfates

Whether a salt dissolves depends on the balance in the solution cycle (Topic 23): ΔHsol = −ΔHlatt + ΔHhyd(cation) + ΔHhyd(anion). The more exothermic (or less endothermic) ΔHsol, the more soluble the salt tends to be.

Down the group, the cation gets larger, so both the lattice energy and the hydration enthalpy of the cation become less exothermic. Which change wins depends on the size of the anion:

Sulfates: less soluble down the group

The sulfate ion is large, so the inter-ionic distance (r+ + r−) is dominated by the anion and changes only a little as the cation grows. The lattice energy decreases only slightly. But the hydration enthalpy of the cation decreases a lot. So ΔHsol becomes more endothermic, and solubility falls.

Hydroxides: more soluble down the group

The hydroxide ion is small, so a larger cation increases the inter-ionic distance significantly. The lattice energy decreases more than the cation's hydration enthalpy. So ΔHsol becomes more exothermic (less endothermic), and solubility rises.

(Entropy also plays a part in solubility, but the enthalpy argument above is what the syllabus asks for.)

Questions

1Quick check

Which carbonate decomposes at the lowest temperature?

  1. MgCO3
  2. CaCO3
  3. SrCO3
  4. BaCO3

Mg2+ is the smallest Group 2 ion, with the highest charge density, so it polarises (distorts) the carbonate ion's electron cloud the most and weakens a C–O bond.

2Quick check

How do the solubilities of the Group 2 sulfates and hydroxides change down the group?

  1. both increase
  2. both decrease
  3. sulfates decrease, hydroxides increase
  4. sulfates increase, hydroxides decrease

Down the group both the lattice energy and the hydration enthalpy become less exothermic. For the sulfates the hydration enthalpy falls faster, so solubility decreases; for the hydroxides the lattice energy falls faster, so solubility increases.

3Paper 4 style
(a)
Write an equation for the thermal decomposition of strontium nitrate.
[1]
(b)
Explain why magnesium carbonate decomposes at a lower temperature than barium carbonate.
[3]
(c)
Explain, in terms of enthalpy changes, why barium sulfate is less soluble than magnesium sulfate.
[3]
  • 2Sr(NO3)2 → 2SrO + 4NO2 + O2
  • Mg2+ is smaller than Ba2+ (same charge), so it has a higher charge density
  • It polarises (distorts) the carbonate ion's electron cloud more
  • This weakens the C–O bonds, so less energy is needed to decompose it
  • The hydration enthalpy of Ba2+ is much less exothermic than that of Mg2+
  • The lattice energy changes only slightly, because the large sulfate ion dominates the inter-ionic distance
  • So ΔHsol is more endothermic for BaSO4, making it less soluble
4Paper 4 style
(a)
Write an equation for the thermal decomposition of magnesium nitrate.
[2]
(b)
Describe and explain the trend in the thermal stability of the Group 2 nitrates.
[3]
  • Correct formulas: Mg(NO3)2, MgO, NO2, O2
  • Balanced: 2Mg(NO3)2 → 2MgO + 4NO2 + O2
  • Thermal stability increases down the group
  • The cation radius increases, so its charge density decreases
  • So the cation polarises (distorts) the nitrate ion's electron cloud less, and the N–O bonds are weakened less

The first-row transition elements, titanium to copper, have partly filled 3d orbitals. That one feature explains their variable oxidation states, their catalytic activity, their complex ions and their colours.

What is a transition element?

Transition element
A d-block element that forms one or more stable ions with incompletely filled d orbitals.
ElementTiVCrMnFeCoNiCu
Configuration (after [Ar])3d2 4s23d3 4s23d5 4s13d5 4s23d6 4s23d7 4s23d8 4s23d10 4s1
Common oxidation states+2, +3, +4+2, +3, +4, +5+2, +3, +6+2, +4, +6, +7+2, +3+2, +3+2+1, +2

Scandium and zinc are d-block elements but not transition elements: scandium's only ion, Sc3+, has an empty 3d sub-shell ([Ar]), and zinc's only ion, Zn2+, has a full one ([Ar] 3d10).

Shapes of d orbitals

xy3dxyxz3dz²
The 3dxy orbital has four lobes between the x and y axes (3dxz and 3dyz are similar in other planes). The 3dz² orbital has two lobes along the z axis and a ring around the middle. (The fifth, 3dx²−y², has lobes along the x and y axes.)

Characteristic properties

Compared with s-block metals, transition elements are dense, hard and have high melting points. Chemically, they have four characteristic properties:

Variable oxidation states

The 3d and 4s sub-shells are close in energy, so different numbers of electrons can be lost (or used in bonding) for similar amounts of energy. The highest state rises to +7 at manganese (all 4s and 3d electrons), then falls.

Catalytic activity

They can change oxidation state easily (passing electrons to and from reactants), and they have vacant, energetically accessible d orbitals to form dative bonds with reactant molecules (adsorption).

Complex ion formation

Vacant d orbitals that are energetically accessible accept lone pairs from ligands.

Coloured compounds

Partly filled d orbitals that are split in energy by ligands absorb visible light (section below).

Vanadium shows variable oxidation states clearly: reducing an acidified solution of ammonium vanadate(V) with zinc changes the colour from yellow (VO2+, +5) to blue (VO2+, +4) to green (V3+, +3) to violet (V2+, +2).

Complexes and ligands

Ligand
A molecule or ion with one or more lone pairs available to donate to a transition element ion, forming a dative covalent bond.
Complex
A molecule or ion formed by a central metal atom or ion surrounded by one or more ligands.
Coordination number
The number of dative bonds formed between the ligands and the central metal ion.
TypeBonds to the metalExamples
monodentate1H2O, NH3, Cl−, CN−
bidentate21,2-diaminoethane, H2NCH2CH2NH2 ("en", through both N); ethanedioate, C2O42− (through two O)
polydentateseveralEDTA4−, which bonds through six atoms (four O and two N), wrapping around the metal ion

Shapes of complexes

CuOH2OH2H2OOH2OH2H2O2+octahedral, 90°[Cu(H2O)6]2+, coordination number 6
CuClClClCl2−tetrahedral, 109.5°[CuCl4]2−, coordination number 4
PtClClNH3NH3square planar, 90°[Pt(NH3)2Cl2], coordination number 4
AgH3NNH3+linear, 180°[Ag(NH3)2]+ (Tollens' reagent)
Each line from a ligand to the metal is a dative bond from a lone pair on the ligand's donor atom (O, N or Cl).

Working out the formula and charge: overall charge = charge on the metal ion + total charge of the ligands. For example: Fe3+ with six CN− gives [Fe(CN)6]3−; Co3+ with six NH3 gives [Co(NH3)6]3+; Ni2+ with three en gives [Ni(en)3]2+ (coordination number 6); Cr3+ with four H2O and two Cl− gives [Cr(H2O)4Cl2]+.

Ligand exchange

In ligand exchange, one ligand replaces another. If the new ligands are larger (like Cl−), fewer fit around the metal, so the coordination number and shape can change.

Copper(II)

Add to [Cu(H2O)6]2+ (blue)ObservationEquation
NaOH(aq)pale blue precipitate[Cu(H2O)6]2+ + 2OH− → [Cu(OH)2(H2O)4] + 2H2O
NH3(aq), a littlepale blue precipitate (ammonia acts as a base)as above
NH3(aq), excessprecipitate dissolves to a deep blue solution[Cu(H2O)6]2+ + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 4H2O
concentrated HClyellow-green (then yellow) solution; octahedral → tetrahedral[Cu(H2O)6]2+ + 4Cl− ⇌ [CuCl4]2− + 6H2O

Cobalt(II)

Add to [Co(H2O)6]2+ (pink)ObservationEquation
NaOH(aq)blue precipitate, which turns pink on standing and darkens (brown) in air[Co(H2O)6]2+ + 2OH− → [Co(OH)2(H2O)4] + 2H2O
NH3(aq), a littleblue precipitateas above
NH3(aq), excessprecipitate dissolves to a pale brown (straw) solution, which darkens in air as Co(II) is oxidised to Co(III)[Co(H2O)6]2+ + 6NH3 → [Co(NH3)6]2+ + 6H2O
concentrated HClblue solution; octahedral → tetrahedral[Co(H2O)6]2+ + 4Cl− ⇌ [CoCl4]2− + 6H2O

The reactions with HCl are reversible: adding water moves the equilibrium back to the aqua complex.

Redox chemistry

Use E⦵ values (Topic 24) to predict whether a transition-metal redox reaction is feasible. For example:

  • Acidified manganate(VII) can oxidise chloride ions (E⦵cell = 1.52 − 1.36 = +0.16 V), so manganate(VII) titrations are acidified with sulfuric acid, not hydrochloric acid.
  • Dichromate(VI) cannot oxidise chloride ions under standard conditions (1.33 − 1.36 = −0.03 V), so it can be used with HCl.

Redox titrations

(a) Manganate(VII) and iron(II)

MnO4− + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+ no indicator needed: the end point is the first permanent pale pink colour (excess MnO4−)

25.0 cm3 of an iron(II) solution needs 21.60 cm3 of 0.0200 mol dm−3 KMnO4.

  1. n(MnO4−) = 0.0200 × 0.02160 = 4.32 × 10−4 mol
  2. n(Fe2+) = 5 × 4.32 × 10−4 = 2.16 × 10−3 mol

[Fe2+] = 2.16 × 10−3 / 0.0250 = 0.0864 mol dm−3

(b) Manganate(VII) and ethanedioate

2MnO4− + 16H+ + 5C2O42− → 2Mn2+ + 8H2O + 10CO2 warm the flask to about 60 °C; the reaction speeds up as Mn2+ forms, because Mn2+ catalyses it (autocatalysis)

What volume of 0.0200 mol dm−3 KMnO4 reacts with 0.150 g of sodium ethanedioate, Na2C2O4 (Mr 134.0)?

n(C2O42−) = 1.12 × 10−3 mol; n(MnO4−) = 2/5 × that = 4.48 × 10−4 mol; volume = 4.48 × 10−4 / 0.0200 = 0.0224 dm3 = 22.4 cm3

(c) Copper(II) and iodide

2Cu2+ + 4I− → 2CuI + I2 white precipitate of copper(I) iodide, coloured brown by the iodine
I2 + 2S2O32− → 2I− + S4O62− the iodine is titrated with sodium thiosulfate; starch is added near the end point (blue-black → colourless)

Overall, 2 mol Cu2+ give 1 mol I2, which reacts with 2 mol S2O32−: so n(Cu2+) = n(S2O32−). If 25.0 cm3 of copper(II) solution needs 20.40 cm3 of 0.100 mol dm−3 thiosulfate:

n(Cu2+) = 2.04 × 10−3 mol, so [Cu2+] = 0.0816 mol dm−3

Why complexes are coloured

Degenerate orbitals
Orbitals of the same energy. In an isolated ion, all five 3d orbitals are degenerate.
Non-degenerate orbitals
Orbitals of different energy. Ligands split the d orbitals into two non-degenerate sets.
free ion:five degenerate d orbitalsΔEoctahedral: 2 higher, 3 lowerΔE (smaller)tetrahedral: 3 higher, 2 lower
The ligands' lone pairs repel electrons in the d orbitals. In an octahedral complex, the two orbitals pointing at the ligands (along the axes) are raised more. In a tetrahedral complex the pattern is reversed and the splitting is smaller.
  1. When white light passes through the solution, an electron in a lower d orbital absorbs a photon and is promoted to a higher d orbital.
  2. The frequency of light absorbed depends on the energy gap: ΔE = hf. This is in the visible region.
  3. The colour we see is the complementary colour: the light that is transmitted. [Cu(H2O)6]2+ absorbs mainly red/orange light, so it looks blue.

If the d sub-shell is empty (d0, such as Sc3+ and Ti4+) or full (d10, such as Zn2+ and Cu+), no d–d transition is possible, so the compounds are colourless (white solids).

Different ligands, different colours: ligands split the d orbitals by different amounts. NH3 causes a larger splitting than H2O, so [Cu(NH3)4(H2O)2]2+ absorbs a higher frequency (shorter wavelength) than [Cu(H2O)6]2+, and looks a deeper blue. Changing the oxidation state or the coordination number (shape) also changes ΔE and the colour.

Stereoisomerism in complexes

Geometrical (cis/trans) isomerism

PtClClNH3NH3cis-[Pt(NH3)2Cl2]"cisplatin", an anticancer drug; polar
PtClNH3ClNH3trans-[Pt(NH3)2Cl2]non-polar: the dipoles cancel
CoOH2NH3H3NOH2NH3H3N2+cis-[Co(NH3)4(H2O)2]2+H₂O ligands at 90°
CoOH2OH2H3NNH3NH3H3N2+trans-[Co(NH3)4(H2O)2]2+H₂O ligands at 180°

Cisplatin works by binding to DNA in cancer cells (the two Cl ligands are replaced by N atoms of the DNA bases), which stops the cells dividing. The trans isomer cannot do this.

Optical isomerism

Complexes with two or three bidentate ligands can be chiral. [Ni(en)3]2+ exists as two non-superimposable mirror images. [Ni(en)2(H2O)2]2+ has a trans isomer (not chiral) and a cis isomer that exists as a pair of enantiomers: three stereoisomers in total.

NNNNNNNiNNNNNNNimirroreach green arc is one en ligand

Polarity: a complex is non-polar if its shape makes the bond dipoles cancel, as in trans-[Pt(NH3)2Cl2]. The cis isomer is polar.

Stability constants

Stability constant, Kstab
The equilibrium constant for the formation of a complex ion in a solvent from its constituent ions or molecules.

For [Cu(H2O)6]2+ + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + 4H2O, the concentration of water is effectively constant, so it is left out:

Stability constantKstab = [[Cu(NH3)4(H2O)2]2+][Cu2+][NH3]4Learn it

(Here [Cu2+] means the hexaaqua ion.) The units are mol−4 dm12. Values are often quoted as lg Kstab:

Complex[CuCl4]2−[Cu(NH3)4(H2O)2]2+[Cu(EDTA)]2−
lg Kstab5.613.118.8

A larger Kstab means a more stable complex. Ligand exchange tends to go towards the complex with the larger Kstab: adding ammonia to [CuCl4]2− forms the ammine complex, and adding EDTA to either forms [Cu(EDTA)]2−. Polydentate ligands form very stable complexes partly because one ligand replaces several water molecules, increasing the number of particles and so the entropy.

Worked example

In a solution, [NH3] = 0.100 mol dm−3 and lg Kstab = 13.1. Find the ratio [complex] : [Cu2+].

ratio = Kstab × [NH3]4 = 1013.1 × 10−4 = 1.3 × 109: essentially all the copper is in the ammine complex.

Questions

1Quick check

Which ion is colourless in aqueous solution?

  1. Cu2+
  2. Fe3+
  3. Zn2+
  4. Ni2+

Zn2+ is 3d10: with a full d sub-shell, no d–d transitions can happen.

2Quick check

What is the coordination number of nickel in [Ni(en)2(H2O)2]2+?

  1. 2
  2. 3
  3. 4
  4. 6

Each en forms two dative bonds (4 in total) and each water forms one (2 more): 6.

3Quick check

What is the oxidation state of cobalt in [CoCl4]2−, and what is the shape of the complex?

  1. +2, square planar
  2. +2, tetrahedral
  3. +4, tetrahedral
  4. −2, octahedral

x + 4(−1) = −2, so x = +2. The large chloride ligands give four-coordinate, tetrahedral complexes such as [CoCl4]2− and [CuCl4]2−.

4Quick check

Why is [Cu(H2O)6]2+ coloured?

  1. Cu2+ has a full 3d sub-shell
  2. the water ligands absorb visible light
  3. electrons move from the 4s to the 3d orbitals
  4. the ligands split the 3d orbitals, and an electron absorbs visible light to move to a higher d orbital

Cu2+ is 3d9, so there is space in the higher-energy d orbitals. The gap ΔE matches visible light; red-orange light is absorbed and the complementary colour, blue, is seen.

5Quick check

Which ligand is bidentate?

  1. ethane-1,2-diamine, H2NCH2CH2NH2
  2. ammonia, NH3
  3. cyanide, CN−
  4. EDTA4−

Ethane-1,2-diamine has two N atoms, each with a lone pair, so it forms two dative bonds to the metal ion. EDTA4− is hexadentate; NH3 and CN− are monodentate.

6Paper 4 style
(a)
Define the terms ligand and complex.
[2]
(b)
Aqueous ammonia is added dropwise, until in excess, to aqueous copper(II) sulfate. Describe what you would see and write equations for the reactions.
[4]
(c)
Explain why [Cu(H2O)6]2+ is coloured.
[3]
(d)
Suggest why the colour changes when the water ligands are replaced by ammonia.
[1]
  • Ligand: a species with a lone pair that forms a dative bond to a (central) metal atom or ion
  • Complex: a central metal atom or ion surrounded by (dative-bonded) ligands
  • Pale blue precipitate
  • [Cu(H2O)6]2+ + 2OH− → [Cu(OH)2(H2O)4] + 2H2O (or with NH3 as the base)
  • Dissolves in excess to give a deep blue solution
  • [Cu(H2O)6]2+ + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 4H2O
  • Ligands split the 3d orbitals into two non-degenerate sets
  • An electron is promoted from a lower to a higher d orbital, absorbing light of a frequency corresponding to ΔE (in the visible region)
  • The complementary colour (the light not absorbed) is seen
  • Ammonia splits the d orbitals by a different (larger) amount, so a different frequency is absorbed
7Paper 4 style

An iron tablet weighing 0.650 g is dissolved in dilute sulfuric acid, and all the iron is present as Fe2+. The solution needs 18.50 cm3 of 0.0200 mol dm−3 KMnO4 to reach the end point.

(a)
Write the ionic equation for the reaction.
[1]
(b)
Calculate the percentage by mass of iron in the tablet.
[3]
(c)
Explain why hydrochloric acid is not used to acidify the solution. Use E⦵ values: MnO4−/Mn2+ +1.52 V, Cl2/Cl− +1.36 V.
[2]
  • MnO4− + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
  • n(MnO4−) = 0.0200 × 0.01850 = 3.70 × 10−4 mol
  • n(Fe) = 5 × 3.70 × 10−4 = 1.85 × 10−3 mol; mass = 1.85 × 10−3 × 55.8 = 0.103 g
  • % Fe = 0.103 / 0.650 × 100 = 15.9%
  • E⦵cell = 1.52 − 1.36 = +0.16 V, so manganate(VII) would oxidise chloride ions (to chlorine)
  • So more KMnO4 would be used and the titre would be too high

A Level organic chemistry adds aromatic compounds (built on the benzene ring), acyl chlorides, amides and amino acids. This topic covers their names, the special bonding in benzene, and why optical isomers matter so much for medicines.

New functional groups

FamilyFunctional groupExample
arenebenzene ring, C6H5– (phenyl)methylbenzene, C6H5CH3
halogenoarenehalogen on a benzene ringchlorobenzene, C6H5Cl
phenol–OH directly on a benzene ringphenol, C6H5OH
acyl chloride–COClethanoyl chloride, CH3COCl
amide–CONH2, –CONHRethanamide, CH3CONH2; N-methylethanamide, CH3CONHCH3
secondary amine–NH–dimethylamine, (CH3)2NH
amino acid–NH2 and –COOH on the same molecule2-aminopropanoic acid (alanine), CH3CH(NH2)COOH
aromatic amine–NH2 on a benzene ringphenylamine, C6H5NH2

Naming aromatic compounds

Number the ring carbons from the carbon carrying the main functional group (C1), going the way that gives the lowest numbers to the other substituents.

OHphenol
COOHNO23-nitrobenzoic acid
OHBrBrBr2,4,6-tribromophenol
CH3Cl2-chloromethylbenzene

Esters and amides are named from both parts and can have up to six carbons in each: for example propyl butanoate, or N-ethylpropanamide (CH3CH2CONHCH2CH3), where "N-ethyl" means the ethyl group is on the nitrogen.

New reaction types

Electrophilic substitution
An electrophile replaces an atom (usually H) on an aromatic ring. This is how arenes react (Topic 30).
Addition–elimination
A nucleophile adds to a C=O group, then a small molecule (such as HCl) is eliminated. This is how acyl chlorides react (Topic 33).

The structure of benzene

  • Benzene, C6H6, is a planar regular hexagon. Each carbon is sp2 hybridised, forming three σ bonds (to two carbons and one hydrogen) at 120°.
  • Each carbon has one remaining p orbital, at right angles to the plane of the ring. The six p orbitals overlap sideways all the way round, forming a delocalised π system above and below the plane, containing six electrons.
  • All six C–C bonds are the same length, 0.139 nm: between a C–C single bond (0.154 nm) and a C=C double bond (0.134 nm).
Kekulé structurealternating single and double bonds (not correct)
delocalised structurethe circle shows the delocalised π system

Evidence for delocalisation: hydrogenating cyclohexene releases 120 kJ mol−1, so a structure with three C=C bonds would release about 3 × 120 = 360 kJ mol−1. Benzene releases only 208 kJ mol−1, so it is about 152 kJ mol−1 more stable than expected. This extra stability is the delocalisation energy. It also explains why benzene does not decolourise bromine water: addition would destroy the delocalised system.

Optical isomerism in more depth

A pair of enantiomers (optical isomers) has identical physical and chemical properties, with two exceptions:

  • Plane-polarised light: each enantiomer rotates the plane of polarisation by the same angle, but in opposite directions. A compound that rotates plane-polarised light is optically active.
  • Biological activity: enzymes and receptors in the body are themselves chiral, so the two enantiomers can fit them differently and have different effects.
Racemic mixture
A 50 : 50 mixture of two enantiomers. It is not optically active, because the rotations cancel out.

Reactions that make a new chiral centre from a planar starting material (such as HCN adding to an aldehyde, or an SN1 reaction through a planar carbocation) give a racemic mixture, because attack from either side is equally likely.

Chirality and drugs

  • Often only one enantiomer of a drug is active. The other may be inactive, or even harmful. The best-known example is thalidomide in the 1950s and 1960s: one enantiomer is a sedative, while the other caused severe birth defects.
  • Synthesis in the lab usually gives a racemic mixture. Separating the enantiomers is difficult and expensive, and half the product is wasted.
  • Using chiral catalysts (or enzymes) makes only the required enantiomer. This means smaller doses, fewer side effects and less waste.

Questions

1Quick check

Which statement about the bonding in benzene is correct?

  1. It contains three C=C and three C–C bonds.
  2. Each carbon atom uses three sp2 orbitals in σ bonds and one p orbital in the delocalised π system.
  3. The bond angles are 109.5°.
  4. The π electrons are located between alternate pairs of carbon atoms.

The six p electrons are delocalised around the whole ring; all bonds are equal and the angles are 120°.

2Quick check

How many chiral centres are there in 2,3-dibromobutane, CH3CHBrCHBrCH3?

  1. 0
  2. 1
  3. 2
  4. 3

C2 and C3 each have four different groups attached: H, Br, CH3 and the rest of the molecule.

3Quick check

What is the bond angle around each carbon atom in benzene, and how is each carbon hybridised?

  1. 109.5°, sp3
  2. 120°, sp2
  3. 180°, sp
  4. 90°, sp2

Each carbon forms three σ bonds (two C–C, one C–H) in a plane at 120°. The remaining p orbital overlaps sideways with its neighbours to form the delocalised π system.

4Paper 4 style

The enthalpy change of hydrogenation of cyclohexene is −120 kJ mol−1, and of benzene is −208 kJ mol−1.

(a)
Use these data to explain what they show about the stability of benzene.
[2]
(b)
Describe the bonding in benzene in terms of orbital overlap.
[3]
(c)
A drug is made as a racemic mixture. Explain what is meant by a racemic mixture, and give two reasons why manufacturers try to produce a single enantiomer.
[3]
  • Expected for three C=C: 3 × −120 = −360 kJ mol−1
  • Benzene releases 152 kJ mol−1 less, so it is more stable (because of delocalisation)
  • Each C is sp2 hybridised, forming σ bonds to two C and one H
  • The remaining p orbital on each C overlaps sideways with p orbitals on both neighbours
  • Forming a delocalised π system (ring of electron density) above and below the plane
  • A 50 : 50 mixture of two enantiomers (optically inactive)
  • Only one enantiomer may be active, so the dose can be halved / less waste
  • The other enantiomer may cause harmful side effects

Benzene and methylbenzene are electron-rich but very stable. They react with electrophiles, but by substitution rather than addition, so that the delocalised ring survives. Groups already on the ring decide where the next substituent goes.

Reactions of benzene and methylbenzene

ReactionReagents and conditionsProduct from benzene
halogenation (ring)Cl2 or Br2 with AlCl3 or AlBr3 (or Fe) catalyst, room temperature, no lightchlorobenzene or bromobenzene + HCl/HBr
nitrationconcentrated HNO3 and concentrated H2SO4, 25–60 °Cnitrobenzene, C6H5NO2 + H2O
Friedel–Crafts alkylationCH3Cl (a chloroalkane) and AlCl3, heatmethylbenzene, C6H5CH3 + HCl
Friedel–Crafts acylationCH3COCl (an acyl chloride) and AlCl3, heatphenylethanone, C6H5COCH3 + HCl
hydrogenationH2, Pt or Ni catalyst, heat (and pressure)cyclohexane, C6H12
side-chain oxidation (alkylbenzenes only)hot alkaline KMnO4, then dilute acidmethylbenzene → benzoic acid, C6H5COOH

Above 60 °C, nitration goes further and gives 1,3-dinitrobenzene. Any alkyl side chain, however long, is oxidised to –COOH by hot alkaline manganate(VII); the benzene ring itself is not attacked.

Mechanism: electrophilic substitution

Making the electrophile

HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4− nitration: the sulfuric acid generates the nitronium ion, NO2+
Br2 + AlBr3 → Br+ + AlBr4− bromination: the "halogen carrier" polarises Br2 and generates Br+
CH3Cl + AlCl3 → CH3+ + AlCl4− and CH3COCl + AlCl3 → CH3CO+ + AlCl4− Friedel–Crafts

The mechanism

ON+OHNO2NO2H⁺++intermediate
Nitration of benzene. (1) Two π electrons from the delocalised ring attack the electrophile, forming a positively charged intermediate in which the delocalisation is partly broken. (2) The C–H bond breaks and its electrons return to the ring, restoring the stable delocalised system and releasing H+.

The H+ is picked up to regenerate the catalyst: H+ + HSO4− → H2SO4 in nitration, or H+ + AlBr4− → AlBr3 + HBr in bromination.

Why substitution, not addition?

Addition would permanently destroy the delocalised π system, losing about 150 kJ mol−1 of stability. Substitution keeps the ring intact, so it is energetically favoured.

Side chain or ring?

Methylbenzene can be chlorinated in two different places, depending on the conditions:

In the side chain

Cl2, UV light (or boiling): free-radical substitution, like an alkane, giving (chloromethyl)benzene, C6H5CH2Cl, and further substitution products.

On the ring

Cl2, AlCl3 catalyst, room temperature, dark: electrophilic substitution, giving a mixture of 2-chloromethylbenzene and 4-chloromethylbenzene.

Directing effects

A group already on the ring decides where a second substituent goes:

Group on the ringDirects toEffect on reactivity
–NH2, –OH, –R (alkyl)2- and 4- positionsactivate the ring (donate electron density); –OH and –NH2 strongly
–NO2, –COOH, –COR3- positiondeactivate the ring (withdraw electron density)

For example, nitrating methylbenzene gives mainly 2-nitromethylbenzene and 4-nitromethylbenzene, and faster than benzene. Nitrating nitrobenzene gives 1,3-dinitrobenzene, more slowly.

Questions

1Quick check

Methylbenzene reacts with chlorine in bright sunlight. What is the main organic product?

  1. 2-chloromethylbenzene
  2. 4-chloromethylbenzene
  3. (chloromethyl)benzene, C6H5CH2Cl
  4. 1,2,3,4,5,6-hexachlorocyclohexane

UV light means free-radical substitution in the methyl side chain. Ring substitution needs a halogen carrier such as AlCl3 and no light.

2Quick check

What is the electrophile in the nitration of benzene?

  1. NO2−
  2. HNO3
  3. NO3−
  4. NO2+

HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4−. The nitronium ion is attacked by the delocalised π electrons of the ring.

3Quick check

Why does benzene react with bromine by substitution rather than addition?

  1. substitution keeps the stable delocalised π system
  2. benzene has no π electrons
  3. bromine is not an electrophile
  4. addition would need a catalyst

Addition would destroy the delocalisation, losing about 150 kJ mol−1 of stabilisation. Substitution restores the delocalised ring.

4Paper 4 style
(a)
Write an equation for the formation of the electrophile in the nitration of benzene.
[1]
(b)
Describe the mechanism of the nitration of benzene, using curly arrows.
[3]
(c)
Explain why benzene undergoes substitution rather than addition.
[1]
(d)
Suggest the reagents and conditions to convert benzene into phenylethanone, C6H5COCH3, and name the reaction.
[2]
  • HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4−
  • Curly arrow from the delocalised π system (ring) to N of NO2+
  • Intermediate with a broken ring of delocalisation, a positive charge, and H and NO2 on the same carbon
  • Curly arrow from the C–H bond back into the ring, releasing H+
  • Substitution keeps the stable delocalised π system; addition would destroy it
  • Ethanoyl chloride, CH3COCl, with an AlCl3 catalyst, heat
  • Friedel–Crafts acylation (electrophilic substitution)

Halogenoarenes, with the halogen joined directly to a benzene ring, are far less reactive than halogenoalkanes. The reason lies in how the halogen's lone pair interacts with the ring.

Making halogenoarenes

React the arene with Cl2 or Br2 and an AlCl3 or AlBr3 catalyst (electrophilic substitution, Topic 30):

C6H6 + Cl2 → C6H5Cl + HCl benzene → chlorobenzene
C6H5CH3 + Cl2 → ClC6H4CH3 + HCl methylbenzene → 2-chloromethylbenzene and 4-chloromethylbenzene (the methyl group directs to 2 and 4)

Why halogenoarenes are unreactive

Chloroethane is hydrolysed by heating with aqueous sodium hydroxide; chlorobenzene is not hydrolysed under these conditions at all.

  1. A lone pair on the chlorine atom overlaps with the delocalised π system of the ring. This gives the C–Cl bond partial double bond character, so it is stronger (and shorter) than in a halogenoalkane, and harder to break.
  2. The electron-rich π system repels approaching nucleophiles such as OH−.
  3. A nucleophile can't approach from the side opposite the halogen (as in SN2), because the ring is in the way.

The ease of hydrolysis is: acyl chlorides (fast at room temperature) > halogenoalkanes (need heating with NaOH(aq)) > halogenoarenes (no reaction under normal conditions). See Topic 33.

Questions

1Quick check

Why doesn't chlorobenzene undergo nucleophilic substitution with NaOH(aq) under normal conditions?

  1. the C–Cl bond is very polar
  2. a lone pair on Cl overlaps with the delocalised π system, strengthening the C–Cl bond
  3. chlorine is too electronegative
  4. chlorobenzene doesn't dissolve in water

The overlap gives the C–Cl bond partial double-bond character, so it is stronger than in a halogenoalkane. The π electrons also repel incoming nucleophiles.

2Quick check

Which reagent and conditions convert methylbenzene into (chloromethyl)benzene, C6H5CH2Cl?

  1. Cl2 with AlCl3 catalyst, room temperature
  2. HCl(g), room temperature
  3. Cl2 in UV light
  4. NaCl(aq), heat

UV light gives free-radical substitution in the side chain. With AlCl3, chlorine substitutes into the ring instead, at the 2- and 4-positions.

3Paper 4 style
(a)
Give the reagents and conditions to make chlorobenzene from benzene.
[1]
(b)
Chloroethane and chlorobenzene are each warmed with aqueous silver nitrate in ethanol. Describe and explain the difference in observations.
[3]
  • Cl2 with AlCl3 (or Fe / FeCl3) catalyst, room temperature
  • Chloroethane: white precipitate (slowly); chlorobenzene: no precipitate
  • In chlorobenzene a lone pair on Cl overlaps with the π system, strengthening the C–Cl bond
  • The π electron cloud also repels the (water) nucleophile
4Paper 4 style
(a)
Describe a test to distinguish 1-chlorobutane from chlorobenzene. Give the observations.
[3]
(b)
Explain the difference in reactivity.
[2]
  • Warm each with NaOH(aq) (or aqueous ethanolic AgNO3)
  • Acidify with dilute HNO3, then add AgNO3(aq)
  • 1-chlorobutane gives a white precipitate (AgCl); chlorobenzene gives no precipitate
  • In chlorobenzene a lone pair on Cl overlaps with the delocalised π system of the ring
  • So the C–Cl bond is stronger (partial double-bond character) and is not broken / not hydrolysed

Phenol has an –OH group joined directly to a benzene ring. The ring and the oxygen affect each other: the ring makes phenol much more acidic than an alcohol, and the oxygen makes the ring much more reactive towards electrophiles than benzene.

Alcohols and acyl chlorides

Alcohols react with acyl chlorides at room temperature to form esters. Unlike esterification with a carboxylic acid, this is fast and goes to completion:

CH3COCl + CH3CH2OH → CH3COOCH2CH3 + HCl ethanoyl chloride + ethanol → ethyl ethanoate + steamy fumes of HCl

Making phenol

In the lab, phenol is made from phenylamine in two steps:

  1. Diazotisation: phenylamine with HNO2 (made from NaNO2 and dilute HCl), below 10 °C gives benzenediazonium chloride: C6H5NH2 + HNO2 + HCl → C6H5N2+Cl− + 2H2O
  2. Warming with water (above 10 °C) decomposes the diazonium salt, giving off nitrogen: C6H5N2+Cl− + H2O → C6H5OH + N2 + HCl

The temperature must be kept below 10 °C in step 1, because the diazonium ion decomposes on warming.

Reactions of phenol

ReagentConditionsProduct and observationEquation
NaOH(aq)room temperaturesodium phenoxide; phenol dissolves (it is only slightly soluble in water)C6H5OH + NaOH → C6H5O−Na+ + H2O
Na(s)room temperature (molten phenol)sodium phenoxide and hydrogen; fizzing2C6H5OH + 2Na → 2C6H5O−Na+ + H2
diazonium saltin NaOH(aq), below 10 °Can orange-yellow azo dye (precipitate)see Topic 34
dilute HNO3room temperaturemixture of 2-nitrophenol and 4-nitrophenolC6H5OH + HNO3 → HOC6H4NO2 + H2O
Br2(aq)room temperatureorange colour disappears, white precipitate of 2,4,6-tribromophenolC6H5OH + 3Br2 → C6H2Br3OH + 3HBr

Phenol does not react with carbonates: it is too weak an acid to release CO2. This distinguishes phenols from carboxylic acids.

The acidity of phenol

Phenol is a weak acid: C6H5OH ⇌ C6H5O− + H+. It is more acidic than water, which is more acidic than ethanol:

acidity: phenol > water > ethanol

  • Phenol: in the phenoxide ion, a lone pair on the oxygen overlaps with the π system of the ring, so the negative charge is delocalised over the ring. This stabilises the ion, so phenol loses H+ more readily.
  • Water: the charge on OH− is on the oxygen alone.
  • Ethanol: the ethyl group has a positive inductive effect, pushing extra electron density onto the oxygen of C2H5O−. This intensifies the negative charge and destabilises the ion, so ethanol is the weakest acid.

Why phenol reacts so easily with electrophiles

Benzene needs concentrated acids or a halogen carrier to react, but phenol reacts with dilute nitric acid and with bromine water at room temperature, without a catalyst. A lone pair on the oxygen is delocalised into the ring, increasing the electron density of the π system (especially at positions 2, 4 and 6). The ring attracts electrophiles more strongly and can polarise a bromine molecule without a catalyst. This is why the –OH group activates the ring and directs to the 2-, 4- and 6- positions, so bromine substitutes three times.

The same ideas apply to other phenols, such as naphthol (an –OH on a two-ring naphthalene system): they are weakly acidic, form salts with NaOH, react with bromine water and couple with diazonium salts to form dyes.

Questions

1Quick check

Which reagent reacts with phenol but not with ethanol?

  1. aqueous sodium hydroxide
  2. sodium metal
  3. ethanoyl chloride
  4. acidified potassium dichromate(VI)

Phenol is acidic enough to react with NaOH; ethanol isn't. Both react with Na and with ethanoyl chloride. Dichromate oxidises ethanol, not phenol.

2Quick check

Which list shows the compounds in order of increasing acid strength?

  1. ethanol, water, phenol, ethanoic acid
  2. water, ethanol, phenol, ethanoic acid
  3. ethanol, phenol, water, ethanoic acid
  4. phenol, ethanol, water, ethanoic acid

In the phenoxide ion the negative charge is delocalised into the ring, which stabilises it; in the ethoxide ion the ethyl group pushes electrons onto O, which destabilises it. Ethanoate is stabilised more, by delocalisation over two O atoms.

3Quick check

Phenol is added to bromine water at room temperature. What is seen, and what forms?

  1. no change: a halogen carrier is needed
  2. the orange colour disappears and a white precipitate of 2,4,6-tribromophenol forms
  3. the orange colour disappears and colourless bromobenzene forms
  4. a violet colour forms

A lone pair on the O atom overlaps with the ring, raising its electron density, so phenol reacts with bromine water without a catalyst, at the 2-, 4- and 6-positions.

4Paper 4 style
(a)
Describe what you would see when aqueous bromine is added to phenol, and write an equation.
[3]
(b)
Explain why phenol reacts with bromine under much milder conditions than benzene.
[2]
(c)
Place phenol, ethanol and water in order of increasing acidity and explain the order.
[3]
  • Orange (bromine) colour disappears
  • White precipitate
  • C6H5OH + 3Br2 → C6H2Br3OH + 3HBr
  • A lone pair on O is delocalised into the ring
  • Increasing the electron density of the ring, so it polarises / attracts the electrophile (Br2) more strongly
  • ethanol < water < phenol
  • Phenoxide: the negative charge is delocalised over the ring, stabilising the anion
  • Ethoxide: the alkyl group pushes electrons onto O (positive inductive effect), destabilising the anion

Acyl chlorides are the most reactive carboxylic acid derivatives: they react at room temperature with water, alcohols, phenols, ammonia and amines. This topic also explains the relative acidity of carboxylic acids, phenols and alcohols, and two oxidisable acids.

More about carboxylic acids

Making benzoic acid

An alkylbenzene is oxidised by hot alkaline KMnO4, then acidified with dilute acid: methylbenzene gives benzoic acid, C6H5COOH (Topic 30).

Two acids that can be oxidised further

  • Methanoic acid, HCOOH, still has a C–H on its carbonyl carbon (it is also an aldehyde). It is oxidised to CO2 and H2O by Fehling's solution (red precipitate), Tollens' reagent (silver mirror), and acidified KMnO4 or K2Cr2O7: HCOOH + [O] → CO2 + H2O.
  • Ethanedioic acid, HOOCCOOH, is oxidised by warm acidified KMnO4 to CO2 (the basis of the titration in Topic 28).

Making acyl chlorides

ReagentConditionsEquation
PCl5room temperatureCH3COOH + PCl5 → CH3COCl + POCl3 + HCl
PCl3heat3CH3COOH + PCl3 → 3CH3COCl + H3PO3
SOCl2(warm)CH3COOH + SOCl2 → CH3COCl + SO2 + HCl

Comparing acidity

acidity: carboxylic acid > phenol > water > alcohol

In the carboxylate ion, RCOO−, the negative charge is delocalised over two oxygen atoms (both C–O bonds become equal). This stabilises the ion more than in phenoxide, where the charge spreads into the less electronegative carbon ring.

Chlorine-substituted carboxylic acids

AcidCH3COOHClCH2COOHCl2CHCOOHCl3CCOOH
pKa4.762.861.290.65

Chlorine is electronegative, so it withdraws electron density from the carboxylate group (a negative inductive effect). This spreads out and stabilises the negative charge of the anion, so the acid dissociates more. The effect is larger with more Cl atoms and when the Cl is closer to the COOH group: 2-chlorobutanoic acid is stronger than 4-chlorobutanoic acid.

Acyl chlorides

Acyl chlorides contain –COCl, for example ethanoyl chloride, CH3COCl, and benzoyl chloride, C6H5COCl. They react with nucleophiles at room temperature, always producing HCl (steamy white fumes):

NucleophileProductEquation
watercarboxylic acidCH3COCl + H2O → CH3COOH + HCl
alcoholesterCH3COCl + CH3CH2OH → CH3COOCH2CH3 + HCl
phenolester (phenyl ester)C6H5COCl + C6H5OH → C6H5COOC6H5 + HCl (phenyl benzoate)
ammoniaprimary amideCH3COCl + 2NH3 → CH3CONH2 + NH4Cl
primary aminesecondary amide (N-substituted)CH3COCl + CH3NH2 → CH3CONHCH3 + HCl

Phenols don't form esters with carboxylic acids easily, so acyl chlorides are the usual way to make phenyl esters.

Mechanism: addition–elimination

The carbonyl carbon in an acyl chloride is very δ+, because it is bonded to two electronegative atoms (O and Cl). A nucleophile adds to it, the C=O re-forms, and HCl is eliminated.

Cδ+Oδ−H3CClOH2CO−H3CClOH2+COH3COH+ HCladditionelimination
Hydrolysis of ethanoyl chloride. The same mechanism applies with alcohols, phenols, ammonia and amines as the nucleophile.

Ease of hydrolysis

CompoundHydrolysisWhy
acyl chloride, CH3COClfast, with water at room temperaturethe carbonyl carbon is strongly δ+ (bonded to O and Cl) and planar, so easily attacked
chloroalkane, CH3CH2Clslow, needs heating with NaOH(aq)the carbon is only δ+ from Cl; attack is via SN1 or SN2
chloroarene, C6H5Clno reaction under normal conditionsthe C–Cl bond is strengthened by overlap with the ring, and the π system repels nucleophiles

Questions

1Quick check

Which acid is the strongest?

  1. ethanoic acid
  2. 3-chloropropanoic acid
  3. chloroethanoic acid
  4. dichloroethanoic acid

Two Cl atoms close to the COOH group withdraw the most electron density, stabilising the anion most.

2Quick check

Which is the strongest acid?

  1. CH3COOH
  2. ClCH2COOH
  3. Cl2CHCOOH
  4. Cl3CCOOH

Each electronegative Cl atom draws electron density away from the –COO− group, spreading out the negative charge and stabilising the anion. Trichloroethanoic acid has three.

3Quick check

Which compound is hydrolysed most rapidly by water at room temperature?

  1. chloroethane
  2. chlorobenzene
  3. ethanoyl chloride
  4. ethyl ethanoate

The carbonyl carbon of an acyl chloride is bonded to both O and Cl, so it is strongly δ+ and is attacked rapidly by water, giving ethanoic acid and steamy fumes of HCl.

4Quick check

Ethanoyl chloride reacts with methylamine. What is the organic product?

  1. N-methylethanamide, CH3CONHCH3
  2. ethanamide, CH3CONH2
  3. methyl ethanoate
  4. ethylamine

Nucleophilic addition–elimination: the lone pair on N attacks the carbonyl carbon, and HCl is eliminated (it then reacts with more methylamine).

5Paper 4 style
(a)
Give a reagent to convert benzoic acid into benzoyl chloride, and write the equation.
[2]
(b)
Benzoyl chloride reacts with ethylamine. Name the type of reaction and give the structural formula of the organic product.
[2]
(c)
Describe the mechanism of the reaction between ethanoyl chloride and methanol.
[4]
(d)
Explain why ethanoyl chloride is hydrolysed much more readily than chloroethane.
[2]
  • PCl5 (or SOCl2, or PCl3 with heat)
  • C6H5COOH + PCl5 → C6H5COCl + POCl3 + HCl
  • Addition–elimination (condensation)
  • C6H5CONHCH2CH3
  • Dipole on C=O (C δ+); curly arrow from a lone pair on the methanol O to the carbonyl C
  • Curly arrow from the C=O π bond to O, forming a tetrahedral intermediate (O− and O+H)
  • Curly arrow re-forming C=O and from the C–Cl bond to Cl
  • Loss of H+ (as HCl); product methyl ethanoate
  • In ethanoyl chloride the C is bonded to two electronegative atoms (O and Cl), so it is more δ+
  • and the planar C=O carbon is easier for a nucleophile to attack (addition to C=O is easy)

Amines are organic bases, thanks to the lone pair on nitrogen. Phenylamine leads to diazonium salts and azo dyes. Amides and amino acids join together to form the peptide links in proteins.

Amines

Amines are classified by the number of carbon groups on the nitrogen: primary (RNH2, e.g. ethylamine), secondary (R2NH, e.g. diethylamine), tertiary (R3N).

Making amines

MethodReagents and conditionsExample
halogenoalkane + ammoniaNH3 in ethanol, heat under pressure (sealed tube)CH3CH2Br + 2NH3 → CH3CH2NH2 + NH4Br
halogenoalkane + primary amine (gives a secondary amine)amine in ethanol, heat in a sealed tubeCH3CH2Br + CH3CH2NH2 → (CH3CH2)2NH + HBr
reduction of an amideLiAlH4 (in dry ether)CH3CONH2 + 4[H] → CH3CH2NH2 + H2O
reduction of a nitrileLiAlH4 or H2 with Ni catalystCH3CN + 4[H] → CH3CH2NH2

Reducing a nitrile is a useful way to make a primary amine with one more carbon than the halogenoalkane you started from, without the mixture of products that ammonia gives.

Reaction with acyl chlorides

Amines (and ammonia) react with acyl chlorides at room temperature to form amides: CH3COCl + CH3CH2NH2 → CH3CONHCH2CH3 + HCl (Topic 33).

Basicity

Amines are weak bases: the lone pair on nitrogen accepts a proton. CH3CH2NH2 + H2O ⇌ CH3CH2NH3+ + OH−. With acids they form salts, such as ethylammonium chloride, CH3CH2NH3+Cl−.

base strength: ethylamine > ammonia > phenylamine

  • Ethylamine is stronger than ammonia: the ethyl group has a positive inductive effect, pushing electron density onto the nitrogen, so its lone pair is more available to accept a proton.
  • Phenylamine is much weaker: the lone pair on nitrogen overlaps with (is delocalised into) the π system of the benzene ring, so it is less available to accept a proton.

Phenylamine

Making phenylamine

  1. Nitrate benzene: conc. HNO3 and conc. H2SO4, below 60 °C → nitrobenzene.
  2. Reduce: tin and concentrated HCl, heat under reflux: C6H5NO2 + 6[H] → C6H5NH2 + 2H2O. In the acid, the product is present as the salt C6H5NH3+.
  3. Add NaOH(aq) to release the free amine, which is then separated by steam distillation.

Reactions of phenylamine

  • With bromine water at room temperature: the orange colour disappears and a white precipitate of 2,4,6-tribromophenylamine forms: C6H5NH2 + 3Br2 → C6H2Br3NH2 + 3HBr. Like –OH in phenol, the –NH2 lone pair is delocalised into the ring, activating it at positions 2, 4 and 6.
  • Diazotisation: with HNO2 (NaNO2 + dilute HCl), below 10 °C, phenylamine forms benzenediazonium chloride, C6H5N2+Cl−. Warming this with water gives phenol and nitrogen gas (Topic 32).

Azo dyes

Benzenediazonium chloride reacts with phenol dissolved in NaOH(aq), below 10 °C (a coupling reaction) to give an orange-yellow precipitate of an azo compound. The diazonium ion acts as an electrophile and attacks the 4-position of the phenol (or phenoxide) ring.

C6H5N2+ + C6H5OH → C6H5N=NC6H4OH + H+ 4-hydroxyazobenzene (in the alkaline solution it forms as its sodium salt)
NNOHazo group, –N=N–
The –N=N– azo group links two benzene rings. The extended delocalised system absorbs visible light, so azo compounds are coloured and used as dyes (such as methyl orange). Other azo dyes are made the same way from other aromatic amines and phenols (or naphthols).

Amides

Amides contain –CONH2 (primary) or –CONHR (secondary, N-substituted). They are made from an acyl chloride and ammonia or a primary amine, at room temperature.

ReactionConditionsProducts
hydrolysis with aciddilute HCl, heat under refluxCH3CONH2 + H2O + HCl → CH3COOH + NH4Cl
hydrolysis with alkaliNaOH(aq), heat under refluxCH3CONH2 + NaOH → CH3COONa + NH3 (ammonia gas given off)
reduction of C=OLiAlH4CH3CONH2 + 4[H] → CH3CH2NH2 + H2O

Amides are much weaker bases than amines (effectively neutral): the lone pair on nitrogen is delocalised over the N–C=O group, because the electronegative oxygen of the carbonyl withdraws electron density. So the lone pair is not available to accept a proton.

Amino acids

Amino acids have an amine group and a carboxylic acid group on the same carbon (2-amino acids): H2NCH(R)COOH. In all except glycine (R = H), that carbon is chiral.

Zwitterions

In the solid, and in solution near neutral pH, the acidic –COOH donates a proton to the basic –NH2 on the same molecule, forming a zwitterion: an ion with both a positive and a negative charge but no overall charge. This explains why amino acids are crystalline solids with high melting points and are soluble in water: they are held by strong ionic attractions.

H3N+CRHCOO−zwitterionat the isoelectric point
H3N+CRHCOOHin acid (low pH)positive ion
H2NCRHCOO−in alkali (high pH)negative ion
Isoelectric point
The pH at which an amino acid exists as the zwitterion, with no overall charge.

Amino acids are amphoteric: the –COO− group accepts H+ from acids, and the –NH3+ group gives H+ to bases.

Peptides

The –COOH of one amino acid can react with the –NH2 of another in a condensation reaction, forming an amide (peptide) link, –CONH–, and water. Two amino acids give a dipeptide; three give a tripeptide; many give a polypeptide or protein.

H2NCH2CONHCHCH3COOHpeptide link
The dipeptide formed from glycine (left) and alanine (right). Because either amino acid can provide the –COOH, two different dipeptides (Gly-Ala and Ala-Gly) are possible. Hydrolysis (heating with acid or alkali) breaks the peptide links again.

Electrophoresis

A mixture of amino acids (or peptides) is placed on a gel or paper soaked in a buffer at a fixed pH, and a voltage is applied. At that pH, each amino acid has a charge that depends on its isoelectric point:

  • if pH < isoelectric point, the amino acid is positive and moves to the negative electrode;
  • if pH > isoelectric point, it is negative and moves to the positive electrode;
  • at its isoelectric point, it doesn't move.

Smaller ions and ions with a larger charge move faster. After separation, the spots are made visible (for example with ninhydrin) and identified.

−+start line (buffer at pH 6.0)lysine (pI 9.7): positive at pH 6glycine (pI 6.0): no overall chargeaspartic acid (pI 2.8): negative at pH 6

Questions

1Quick check

Which list shows the compounds in order of increasing base strength?

  1. ammonia, phenylamine, ethylamine
  2. phenylamine, ammonia, ethylamine
  3. ethylamine, ammonia, phenylamine
  4. phenylamine, ethylamine, ammonia

Phenylamine's lone pair is delocalised into the ring (weakest); ethylamine's alkyl group pushes electrons onto N (strongest).

2Quick check

An amino acid is placed in a buffer with a pH below its isoelectric point, and a voltage is applied. What happens?

  1. it doesn't move
  2. it moves towards the positive electrode
  3. it moves towards the negative electrode
  4. it is hydrolysed

Below the isoelectric point the –COO− group is protonated, so the amino acid is a cation, H3N+CHRCOOH, and moves towards the negative electrode.

3Quick check

Why is ethanamide, CH3CONH2, not basic?

  1. the nitrogen atom has no lone pair
  2. the lone pair on nitrogen is delocalised over the C=O group
  3. it doesn't dissolve in water
  4. the C=O group accepts the proton instead

Overlap of the N lone pair with the π system of the C=O group makes it much less available to accept a proton.

4Paper 4 style
(a)
Describe the two-step synthesis of phenylamine from nitrobenzene, then its conversion into benzenediazonium chloride. Give reagents and conditions.
[4]
(b)
Benzenediazonium chloride is added to phenol in aqueous NaOH. Name the type of product, draw its structure, and give one use.
[3]
(c)
Explain why the temperature is kept below 10 °C.
[1]
  • Tin and concentrated HCl, heat (under reflux)
  • then NaOH(aq) to free the amine from its salt
  • NaNO2 and dilute HCl (to form HNO2)
  • below 10 °C (0–10 °C)
  • An azo compound / azo dye
  • C6H5–N=N–C6H4–OH with the OH in the 4-position
  • Dye (or indicator)
  • The diazonium salt decomposes (to phenol and nitrogen) at higher temperatures
5Paper 4 style

Alanine is 2-aminopropanoic acid, CH3CH(NH2)COOH. Its isoelectric point is 6.0.

(a)
Draw the structure of alanine at pH 6.0, at pH 2 and at pH 11.
[3]
(b)
Draw the two dipeptides that can form from one glycine (H2NCH2COOH) and one alanine, and circle the peptide link in one of them.
[3]
(c)
Explain why ethanamide is a much weaker base than ethylamine.
[2]
  • pH 6: CH3CH(NH3+)COO−
  • pH 2: CH3CH(NH3+)COOH
  • pH 11: CH3CH(NH2)COO−
  • H2NCH2CONHCH(CH3)COOH
  • H2NCH(CH3)CONHCH2COOH
  • –CONH– circled
  • In ethanamide the N lone pair is delocalised over the C=O group (by the electronegative oxygen)
  • so it is not available to accept a proton

Condensation polymers form when monomers with two functional groups join, losing a small molecule each time. Polyesters and polyamides, including proteins, are made this way, and unlike poly(alkene)s they can be broken down by hydrolysis.

Condensation polymerisation

Each monomer has two functional groups that can react. Each link formed releases a small molecule: water (from –COOH) or HCl (from –COCl).

Polyesters

Made from (a) a diol and a dicarboxylic acid (or a dioyl dichloride), or (b) a single monomer with both an –OH and a –COOH group (a hydroxycarboxylic acid).

The best-known polyester is PET, poly(ethylene terephthalate), from ethane-1,2-diol, HOCH2CH2OH, and benzene-1,4-dicarboxylic acid, HOOCC6H4COOH. It is used for drinks bottles and polyester fibres.

Polyamides

Made from (a) a diamine and a dicarboxylic acid (or a dioyl dichloride), (b) an aminocarboxylic acid, or (c) amino acids (giving polypeptides and proteins).

The best-known synthetic polyamide is nylon-6,6, from 1,6-diaminohexane, H2N(CH2)6NH2, and hexanedioic acid, HOOC(CH2)4COOH (or hexanedioyl dichloride, ClOC(CH2)4COCl). Kevlar is an aromatic polyamide used in body armour.

nH2N(CH2)6NH2 + nHOOC(CH2)4COOH → (NH(CH2)6NHCO(CH2)4CO)n + 2nH2O

Repeat units and monomers

nOCH2CH2OCOCOPET (a polyester)ester link: –COO–
nNH(CH2)6NHCO(CH2)4COnylon-6,6 (a polyamide)amide link: –CONH–
To find the monomers, break the ester or amide link in the repeat unit and add back –OH to the C=O and –H to the O or N.

Predicting the type of polymerisation

Monomer(s)PolymerisationPolymer
one monomer with a C=Cadditionpoly(alkene); the main chain is all carbon; no small molecule lost
diol + dicarboxylic acid (or dioyl chloride); or a hydroxy acidcondensationpolyester (–COO– links); water or HCl lost
diamine + dicarboxylic acid (or dioyl chloride); or an amino acidcondensationpolyamide (–CONH– links); water or HCl lost

Looking at a section of polymer: if the main chain contains only carbon atoms, it is an addition polymer. If it contains –COO– or –CONH– links, it is a condensation polymer.

Degradable polymers

  • Poly(alkene)s are chemically inert: their non-polar C–C and C–H bonds aren't attacked by acids, bases or enzymes, so they are non-biodegradable (Topic 20).
  • Photodegradable polymers contain groups (such as C=O) that absorb ultraviolet light, breaking bonds in the chain so the plastic crumbles in sunlight.
  • Polyesters and polyamides can be broken down by hydrolysis of their ester and amide links, by acids or alkalis (and by enzymes in the environment), so they are biodegradable. Poly(lactic acid), made from lactic acid, CH3CH(OH)COOH, is a biodegradable polyester used for packaging and dissolvable stitches.

Questions

1Quick check

Which pair of monomers forms a polyamide?

  1. hexane-1,6-diamine and hexanedioic acid
  2. ethane-1,2-diol and benzene-1,4-dicarboxylic acid
  3. ethene and chloroethene
  4. ethane-1,2-diol and hexane-1,6-diamine

An amide link forms between –NH2 and –COOH (or –COCl), losing water (or HCl). B makes a polyester, C an addition polymer, and D has no acid group.

2Quick check

Which polymer can be broken down by hydrolysis?

  1. poly(ethene)
  2. poly(chloroethene)
  3. poly(propene)
  4. a polyester such as PET

Ester and amide links can be hydrolysed by acids, alkalis or enzymes, so condensation polymers can be biodegradable. Poly(alkene)s have only C–C bonds in their chains.

3Quick check

A polymer has the repeat unit –O–CH(CH3)–CO–. What type of polymer is it, and what is its monomer?

  1. addition polymer; propene
  2. polyester; 2-hydroxypropanoic acid
  3. polyamide; 2-aminopropanoic acid
  4. polyester; propane-1,2-diol

The repeat unit contains an ester link, –CO–O–. A single monomer with both –OH and –COOH groups, 2-hydroxypropanoic acid (lactic acid), condenses with itself to give poly(lactic acid).

4Paper 4 style

Poly(lactic acid) is made from 2-hydroxypropanoic acid, CH3CH(OH)COOH.

(a)
Name the type of polymerisation and draw the repeat unit of the polymer.
[2]
(b)
Explain why poly(lactic acid) is biodegradable but poly(propene) is not.
[2]
(c)
Nylon-6,6 can be made from hexanedioyl dichloride instead of hexanedioic acid. Name the other product of each condensation step, and give one advantage of using the acyl chloride.
[2]
  • Condensation polymerisation
  • Repeat unit –O–CH(CH3)–CO– with continuation bonds
  • Poly(lactic acid) has ester links that can be hydrolysed (by acid, alkali or enzymes)
  • Poly(propene) has only non-polar C–C and C–H bonds, which are unreactive
  • Hydrogen chloride
  • The reaction is faster / happens at room temperature (acyl chlorides are more reactive)

Synthesis questions at A Level combine aliphatic and aromatic chemistry. You may be asked to plan routes of several steps, to spot every functional group in a complex molecule such as a drug, and to predict how it reacts with a given reagent.

The full reaction map

This map includes every AS and A Level reaction. A Level reactions are tagged.

Multi-step routes

Worked example 1: an azo dye from benzene

  1. Benzene → nitrobenzene: conc. HNO3, conc. H2SO4, below 60 °C
  2. Nitrobenzene → phenylamine: Sn and conc. HCl, heat under reflux; then NaOH(aq)
  3. Phenylamine → benzenediazonium chloride: NaNO2 and dilute HCl, below 10 °C
  4. Diazonium salt → azo dye: phenol in NaOH(aq), below 10 °C

Worked example 2: phenyl benzoate from methylbenzene and phenol

  1. Methylbenzene → benzoic acid: hot alkaline KMnO4, then dilute acid
  2. Benzoic acid → benzoyl chloride: PCl5 (or SOCl2)
  3. Benzoyl chloride + phenol → phenyl benzoate: room temperature (an ester can't easily be made from phenol and a carboxylic acid)

Worked example 3: a longer amine

Make propylamine, CH3CH2CH2NH2, from bromoethane without a mixture of amines:

  1. Bromoethane → propanenitrile: KCN in ethanol, heat under reflux (adds a carbon)
  2. Propanenitrile → propylamine: LiAlH4 (or H2/Ni)

Molecules with several functional groups

Treat each group separately, then check whether a reagent affects more than one.

Worked example: aspirin and paracetamol

MoleculeFunctional groupsPredictions
aspirin, 2-ethanoyloxybenzoic acidcarboxylic acid; ester (phenyl ester); benzene ringfizzes with Na2CO3 (CO2); hydrolysed by NaOH(aq) to a carboxylate, a phenoxide and ethanoate; no colour with neutral iron(III) chloride until hydrolysed (no free phenol)
paracetamol, N-(4-hydroxyphenyl)ethanamidephenol; secondary amide; benzene ringdissolves in NaOH(aq) (phenol); decolourises bromine water with a precipitate (activated ring); hydrolysed by acid or alkali at the amide link to 4-aminophenol and ethanoic acid
Checklist for routes

Give the reagent and the conditions for every step (temperature, solvent, catalyst, reflux or distillation). Watch for reagents that would also attack another group: hot acidified manganate(VII) oxidises alcohols and alkene C=C bonds; LiAlH4 reduces acids, esters, amides, aldehydes, ketones and nitriles; NaBH4 reduces only aldehydes and ketones.

Questions

1Quick check

Which reagent reacts with both functional groups in 4-hydroxybenzoic acid, HOC6H4COOH?

  1. Na2CO3(aq)
  2. NaOH(aq)
  3. ethanol with concentrated H2SO4
  4. 2,4-DNPH

NaOH neutralises the carboxylic acid and the phenol, giving NaOC6H4COONa. Phenols are too weakly acidic to react with carbonates.

2Quick check

Which sequence converts benzene into phenylamine?

  1. Cl2 with AlCl3; then NH3
  2. NaNO2 and HCl; then Sn and HCl
  3. concentrated HNO3 and H2SO4; then Sn and concentrated HCl, followed by NaOH(aq)
  4. CH3Cl with AlCl3; then hot alkaline KMnO4

Nitration gives nitrobenzene, which tin and hydrochloric acid reduce to phenylamine. Chlorobenzene doesn't undergo nucleophilic substitution with ammonia.

3Quick check

Which reagent distinguishes phenylmethanol, C6H5CH2OH, from phenol, C6H5OH?

  1. sodium metal
  2. dilute hydrochloric acid
  3. 2,4-DNPH
  4. neutral iron(III) chloride solution

Phenol gives a violet colour with neutral FeCl3(aq); phenylmethanol is an alcohol, not a phenol, so it doesn't. Both fizz with sodium.

4Paper 4 style

Compound X is HOCH2C6H4CHO (4-(hydroxymethyl)benzaldehyde): a benzene ring with a –CH2OH group and a –CHO group in the 1,4-positions.

(a)
Give the structural formula of the organic product when X reacts with (i) NaBH4, (ii) excess acidified K2Cr2O7 under reflux, (iii) HCN with KCN.
[3]
(b)
Describe a test to distinguish X from HOCH2C6H4COCH3. Give the observations.
[2]
(c)
Suggest a two-step synthesis of 4-methylbenzoyl chloride from 4-methylbenzaldehyde, CH3C6H4CHO.
[2]
  • (i) HOCH2C6H4CH2OH
  • (ii) HOOCC6H4COOH
  • (iii) HOCH2C6H4CH(OH)CN
  • Tollens' reagent (or Fehling's), warm
  • X gives a silver mirror (or red precipitate); the ketone gives no change
  • Oxidise: acidified K2Cr2O7, heat under reflux, giving 4-methylbenzoic acid (side chain CH3 unaffected)
  • Then PCl5 (or SOCl2)

Chromatography separates mixtures and identifies their components. NMR spectroscopy shows how the carbon and hydrogen atoms in a molecule are arranged, which, with IR and mass spectra (Topic 22), lets you deduce a structure with confidence.

Chromatography

Every chromatography method has a stationary phase (which stays still) and a mobile phase (which moves through it). Components of a mixture separate because they distribute themselves differently between the two phases: those that interact more with the stationary phase move more slowly.

Thin-layer chromatography (TLC)

  • Stationary phase: a thin layer of a polar solid such as aluminium oxide (or silica) on a glass, metal or plastic sheet.
  • Mobile phase: a solvent (polar or non-polar) that rises up the plate.
  • Spots are placed on a pencil baseline; the plate is stood in a little solvent (below the baseline) in a covered tank; it is removed before the solvent reaches the top and the solvent front is marked. Colourless spots are shown with UV light or iodine vapour.
baseline (pencil)solvent frontmixtureABspot: 3.6 cm from baselinesolvent: 8.0 cm
The mixture contains A and B: each of its spots matches a reference spot at the same height.
Retention factorRf = distance moved by the spotdistance moved by the solvent frontLearn it

For A: Rf = 3.6 / 8.0 = 0.45. For a given stationary phase, solvent and temperature, each compound has a characteristic Rf. A compound has a low Rf if it is strongly adsorbed by (interacts strongly with) the polar stationary phase, for example through hydrogen bonding; it has a high Rf if it is more soluble in the mobile phase.

Gas/liquid chromatography (GLC)

  • Stationary phase: a high-boiling, non-polar liquid coated on a solid support inside a long, coiled column in an oven.
  • Mobile phase: an unreactive carrier gas, such as helium or nitrogen.
  • The retention time is the time between injecting the sample and a component reaching the detector. Components that are more soluble in (interact more with) the stationary liquid, or that are less volatile, have longer retention times.
024681012retention time / mindetector responseABC

The area under each peak is proportional to the amount of that component, so the percentage composition = area of one peak ÷ total area of all peaks × 100. Peaks are identified by comparing retention times with known samples run under the same conditions (or by passing them into a mass spectrometer).

Worked example

The peak areas in a chromatogram are A 1.2 cm2, B 3.6 cm2 and C 3.2 cm2. What is the percentage of B?

3.6 / (1.2 + 3.6 + 3.2) × 100 = 45%

Carbon-13 NMR spectroscopy

Each peak in a 13C NMR spectrum corresponds to one carbon environment: carbon atoms in identical positions in the molecule give a single peak. The chemical shift, δ (in ppm, relative to TMS), shows what type of environment it is.

Carbon environmentδ / ppm (typical)
C–C (alkyl)5–55
C–Cl, C–Br10–70
C–N (amine)35–60
C–O (alcohol, ether, ester)50–70
C=C (alkene) and aromatic C110–160
C≡N110–125
C=O in carboxylic acids, esters, amides160–185
C=O in aldehydes, ketones190–220
220200180160140120100806040200chemical shift, δ / ppmC–O, 642610C=O, 206CH₃, 30tall lines: propan-1-ol (3 peaks) · short lines: propanone (2 peaks)
Propan-1-ol, CH3CH2CH2OH, has three carbon environments, so three peaks. Propanone, CH3COCH3, has only two: the two CH3 groups are equivalent. The heights are not proportional to the number of carbons.

Counting environments: look for symmetry. Propan-2-ol has 2 environments; butan-1-ol 4; butan-2-ol 4; 2-methylpropan-2-ol 2; benzene 1; methylbenzene 5 (the CH3 and four ring environments).

Proton (¹H) NMR spectroscopy

A 1H NMR spectrum gives four pieces of information:

  1. Number of peaks: the number of different hydrogen environments.
  2. Chemical shift: the type of environment (table below).
  3. Relative peak areas (integration): the relative number of hydrogens in each environment.
  4. Splitting pattern: the number of hydrogens on the adjacent carbon atoms.
Type of protonδ / ppm (typical)
alkyl: R–CH3, R–CH2–R0.9–1.7
next to C=O: CH3–CO–2.0–2.5
next to an aromatic ring: CH3–Ar2.2–3.0
next to N: R–CH2–NH22.3–3.0
next to a halogen: CH–Cl, CH–Br3.0–4.2
next to O: CH–O (alcohol, ether, ester)3.3–4.8
alkene: C=CH4.5–6.0
aromatic: Ar–H6.0–9.0
aldehyde: –CHO9.3–10.5
alcohol O–H0.5–6.0 (variable)
phenol O–H4.5–7.0
carboxylic acid O–H9.0–13.0
amine N–H / amide N–H1.0–5.0 / 5.0–12.0

Use the Data Booklet's table in the exam; ranges differ slightly between sources.

Splitting: the n + 1 rule

The signal from a set of equivalent protons is split into n + 1 lines by n equivalent protons on the adjacent carbon atoms. Protons in the same environment don't split each other. O–H and N–H protons usually appear as singlets and don't split their neighbours, because they exchange rapidly between molecules.

H on adjacent C, n0123more
Patternsingletdoublet (1 : 1)triplet (1 : 2 : 1)quartet (1 : 3 : 3 : 1)multiplet
543210chemical shift, δ / ppmCH₂: quartet, 2HOH: singlet, 1HCH₃: triplet, 3Hethanol, CH₃CH₂OH543210chemical shift, δ / ppmOCH₂: quartet, 2HCH₃CO: singlet, 3HCH₃: triplet, 3Hethyl ethanoate, CH₃COOCH₂CH₃
In ethanol, the CH3 protons (next to a CH2) give a triplet, and the CH2 protons (next to a CH3) give a quartet; the OH is a singlet. In ethyl ethanoate, the CH3CO has no neighbouring H, so it is a singlet; the OCH2 is shifted to about 4.1 ppm by the oxygen.

Practical points

  • TMS, tetramethylsilane, Si(CH3)4, is the reference standard at δ = 0. Its 12 equivalent protons give one strong singlet away from almost all other peaks; it is also inert, non-toxic and volatile (easily removed afterwards).
  • Samples are dissolved in a deuterated solvent, such as CDCl3, which has no 1H atoms, so the solvent doesn't give a signal.
  • Identifying O–H and N–H: run a second spectrum after shaking with D2O. The O–H and N–H protons exchange with deuterium (–OH → –OD), and their peaks disappear.

Worked example: deducing a structure

A compound C3H6O has an IR absorption at 1715 cm−1. Its 1H NMR spectrum has one singlet at δ 2.1 ppm. Its 13C spectrum has two peaks, at 30 and 206 ppm. Identify it.

  1. 1715 cm−1 and 206 ppm: a C=O in an aldehyde or ketone.
  2. One 1H environment at 2.1 ppm (H next to C=O), and no aldehyde peak at 9–10 ppm: a ketone with equivalent methyl groups.

Propanone, CH3COCH3. (Its isomer propanal, CH3CH2CHO, would show three ¹H environments, including an aldehyde H near 9.8 ppm, and three ¹³C peaks.)

Questions

1Quick check

How many peaks are in the 13C NMR spectrum of 2-methylpropan-1-ol, (CH3)2CHCH2OH?

  1. 2
  2. 4
  3. 3
  4. 5

The two CH3 carbons are equivalent; then CH and CH2OH: 3 environments.

2Quick check

In the ¹H NMR spectrum of 1-chloropropane, CH3CH2CH2Cl, what is the splitting pattern of the CH2Cl protons?

  1. singlet
  2. triplet
  3. quartet
  4. sextet

The CH2Cl protons have 2 H on the adjacent carbon: 2 + 1 = 3, a triplet. (The middle CH2 has 5 neighbours and gives a multiplet.)

3Paper 4 style

Compound Y, C4H8O2, is an ester. Its ¹H NMR spectrum (in CDCl3) shows: δ 1.2 (triplet, 3H), δ 2.3 (quartet, 2H), δ 3.7 (singlet, 3H).

(a)
Explain why CDCl3 is used as the solvent and what TMS is used for.
[2]
(b)
Use the data to deduce the structure of Y. Explain each peak.
[4]
(c)
Predict the number of peaks in the ¹³C NMR spectrum of Y.
[1]
  • CDCl3 contains no ¹H, so it gives no signal / doesn't interfere
  • TMS is the reference (standard) for chemical shift, δ = 0
  • δ 3.7, singlet, 3H: an OCH3 group with no neighbouring H
  • δ 2.3, quartet, 2H: a CH2 next to C=O with 3 neighbouring H
  • δ 1.2, triplet, 3H: a CH3 next to a CH2
  • Y is methyl propanoate, CH3CH2COOCH3
  • 4 peaks
4Paper 4 and 5 style

A student analyses a painkiller tablet by TLC on aluminium oxide, using a non-polar solvent.

(a)
A spot moves 2.7 cm when the solvent front has moved 7.5 cm. Calculate Rf.
[1]
(b)
Explain why a compound containing –OH and –COOH groups has a lower Rf than a hydrocarbon on this plate.
[2]
(c)
Explain why the baseline must be drawn in pencil and must be above the solvent level.
[2]
  • Rf = 2.7 / 7.5 = 0.36
  • The polar groups interact strongly (hydrogen bond) with the polar stationary phase (aluminium oxide)
  • and are less soluble in the non-polar mobile phase, so they move more slowly
  • Pencil (graphite) does not dissolve in the solvent / move up the plate (ink would)
  • If below the solvent level, the spots would dissolve into the solvent in the tank instead of moving up the plate