Organic chemistry I: A-level Chemistry AQA revision notes

3.3.1

Introduction to organic chemistry

Types of formula 3.3.1.1

Definitions

Molecular formula: the actual number of atoms of each element in a molecule. Empirical formula: the simplest whole-number ratio.
General formula: the algebraic formula for a homologous series, e.g. $C_{n} H_{2 n}$ for alkenes.
Structural formula: the minimal detail showing the arrangement of atoms, e.g. $CH_{2} = CHCH_{2} CH_{3}$ . Displayed formula: every atom and bond shown. Skeletal formula: carbon skeleton as lines, with hydrogens on carbon omitted.

Key results

But-1-ene: molecular $C_{4} H_{8}$ ; empirical $CH_{2}$ ; general $C_{n} H_{2 n}$ ; structural $CH_{2} = CHCH_{2} CH_{3}$ .

Notes

In skeletal formulae every line end and corner is a carbon carrying enough hydrogens to make four bonds; atoms other than C and H (and H atoms bonded to them) are always drawn.
A molecular or general formula does not identify a single compound: cycloalkanes share $C_{n} H_{2 n}$ with alkenes.
Worked example: ethanoic acid has molecular formula $C_{2} H_{4} O_{2}$ , empirical formula $CH_{2} O$ , structural formula $CH_{3} COOH$ , and in skeletal form a line ending in a carbon bearing =O and –OH.
Questions specify which formula they want, and each earns marks only in that form: a displayed formula must show every bond, including O–H; a skeletal formula must not show C or H atoms on carbon.
Benzene rings in skeletal formulae are drawn as a hexagon with a circle, representing the delocalised electrons (see 3.3.10).

Functional groups and homologous series 3.3.1.1

Definitions

Functional group: an atom or group of atoms responsible for a compound's characteristic reactions.
Homologous series: compounds with the same functional group and general formula, in which successive members differ by $CH_{2}$ .

Key results

Name endings and prefixes: alkene −ene; halogenoalkane halo−; alcohol −ol; aldehyde −al; ketone −one; carboxylic acid −oic acid; ester alkyl alkanoate; amine −amine; nitrile −nitrile.
General formulae: alkanes $C_{n} H_{2 n + 2}$ ; alkenes and cycloalkanes $C_{n} H_{2 n}$ ; alcohols $C_{n} H_{2 n + 1} OH$ ; aldehydes and ketones $C_{n} H_{2 n} O$ ; carboxylic acids and esters $C_{n} H_{2 n} O_{2}$ .

Notes

Members of a series have similar chemical properties (same functional group) and a trend in physical properties: boiling point rises with chain length as London forces increase.
The alcohols ethanol, propan-1-ol and butan-1-ol fit $C_{n} H_{2 n + 1} OH$ .
Successive members differ by $CH_{2}$ , a mass difference of 14, so their relative molecular masses increase in steps of 14.
Within a series, physical properties change gradually: boiling point and viscosity rise, and solubility in water falls as the non-polar hydrocarbon chain lengthens (methanol and ethanol mix with water in all proportions; hexan-1-ol is only slightly soluble).
Recognising functional groups is the first step in every organic question, because the functional group decides which reagents react and by which mechanism.

IUPAC nomenclature 3.3.1.1

Key results

3-methylpentan-2-ol: five-carbon chain, OH on C2, methyl branch on C3.
2,3-dimethylbutane; cyclohexane ( $C_{6} H_{12}$ , a ring indicated by cyclo-).

Method

Find the longest carbon chain containing the principal functional group; its length gives the stem (meth-, eth-, prop-, but-, pent-, hex-).
Number the chain to give the functional group the lowest possible locant.
Name and number side chains and substituents as prefixes in alphabetical order, using di-, tri- for repeats and commas between locants.

In practiceName $CH_{3} CH(OH)CH(CH_{3})CH_{2} CH_{3}$ .

The longest chain containing the OH group has 5 carbons: pentan- … -ol.
Number from the end nearer the OH: it is on C2 (not C4).
The methyl group is then on C3: 3-methylpentan-2-ol.

Notes

The principal functional group sets the suffix; other groups (halogens, alkyl branches) become prefixes.
Numbers are separated by commas; numbers and letters by hyphens.
Worked examples: $CH_{3} CH(Cl)CH_{2} CH_{3}$ is 2-chlorobutane; $CH_{2} = C(CH_{3}) CH_{3}$ is 2-methylpropene; $CH_{3} CH_{2} CHO$ is propanal; $CH_{3} COCH_{2} CH_{3}$ is butanone.
Priority of suffixes: carboxylic acid > ester > aldehyde > ketone > alcohol > amine; alkenes are indicated in the stem (-ene) alongside the suffix, e.g. but-3-en-2-ol.
Some positions need no number because there is no alternative: propanal and propanone, butanoic acid, and ethanol cannot be numbered any other way.
Check a name by drawing the structure from it. If the drawing has a longer chain or allows lower numbers than the name gives, the name is wrong.

Reaction mechanisms 3.3.1.2

Definitions

Curly arrow: shows the movement of a pair of electrons, starting from a lone pair or a bond and ending where the pair forms a new bond or lone pair.
Heterolytic fission: both electrons of a bond go to one atom, forming ions. Homolytic fission: one electron goes to each atom, forming radicals.
Radical: a species with an unpaired electron, shown with a dot. Nucleophile: an electron-pair donor. Electrophile: an electron-pair acceptor.

Notes

In nucleophilic substitution, hydroxide's lone pair attacks the $δ +$ carbon while the C–Br pair moves onto bromine (heterolytic).
Radical mechanisms have three stages: initiation (radicals formed, e.g. by UV), propagation (a radical is used and another formed) and termination (two radicals combine).
Arrows must start exactly at a lone pair or bond; this precision is often what earns or loses mechanism marks.
Mechanism types at A-level: free-radical substitution (alkanes), nucleophilic substitution and elimination (halogenoalkanes), electrophilic addition (alkenes), nucleophilic addition (carbonyls), nucleophilic addition–elimination (acyl chlorides) and electrophilic substitution (arenes). Each is named by the attacking species and what happens overall.
Partial charges explain where attack happens: a carbon bonded to a more electronegative atom is $δ +$ and attracts nucleophiles; an electron-rich C=C or benzene ring attracts electrophiles.
A half-headed arrow (fish-hook) represents the movement of a single electron in radical reactions; AQA does not require these, but you must not use full curly arrows for radical steps.
Common nucleophiles: $OH^{-}$ , $CN^{-}$ , $NH_{3}$ , $H_{2} O$ , alcohols and amines. Common electrophiles: $H^{+}$ , $NO_{2}^{+}$ , $Br_{2}$ (polarised), carbocations and acylium ions.

Structural isomerism 3.3.1.3

Definitions

Structural isomers: compounds with the same molecular formula but different structural formulae.

Key results

Chain isomers: butane and 2-methylpropane ( $C_{4} H_{10}$ ).
Position isomers: but-1-ene and but-2-ene ( $C_{4} H_{8}$ ).
Functional group isomers: propanal (aldehyde) and propanone (ketone), both $C_{3} H_{6} O$ .

Notes

Functional group isomers have different chemical properties; chain and position isomers differ mainly in physical properties.
Worked example: $C_{4} H_{9} Br$ has four structural isomers: 1-bromobutane, 2-bromobutane, 1-bromo-2-methylpropane and 2-bromo-2-methylpropane. Systematically vary the chain first, then the position of the functional group.
Functional group isomers occur in predictable pairs: alkenes and cycloalkanes ( $C_{n} H_{2 n}$ ), alcohols and ethers, aldehydes and ketones, carboxylic acids and esters.
Drawing the 'same' molecule bent or rotated does not make a new isomer. Check by naming each structure: two drawings with the same name are the same compound.
Isomers can be distinguished by their reactions (e.g. Tollens' reagent for an aldehyde but not a ketone) or by spectroscopy (different numbers of NMR environments).

E–Z stereoisomerism 3.3.1.3

Definitions

Stereoisomers: compounds with the same structural formula but a different arrangement of atoms in space.
E–Z isomerism: arises because rotation about a C=C bond is restricted, when each carbon of the double bond carries two different groups.

Key results

Cahn–Ingold–Prelog priority: on each carbon, the attached atom with the higher atomic number has priority (Br > Cl > O > N > C > H). If tied, compare the next atoms along.
Z (zusammen): the higher-priority groups are on the same side. E (entgegen): they are on opposite sides.

Notes

(E)-but-2-ene has the two methyl groups on opposite sides; (Z)-but-2-ene on the same side.
If either carbon of the double bond carries two identical groups (as in propene's $= CH_{2}$ ), E–Z isomerism is impossible.
Multiple bonds count as duplicated atoms when ranking: –CHO (O, O, H) outranks –CH₂OH (O, H, H).
Worked example: 1-bromo-2-chloroethene, CHBr=CHCl. On C1, Br outranks H; on C2, Cl outranks H. If Br and Cl are on the same side the isomer is Z, otherwise E.
E–Z isomers have different physical properties: (Z)-but-2-ene boils at about 4 °C and (E)-but-2-ene at about 1 °C. They often react similarly because they have the same functional group.
The cis–trans names are only safe when each carbon carries one H; E–Z works for every case, and E is not always trans (e.g. when a higher-priority group is not the one you expect).
Interconversion requires breaking the π bond, which needs a lot of energy (heat or UV light), which is why the two isomers can be separated and stored at room temperature.

Worked examples

Worked example

But-2-ene shows E/Z isomerism, but but-1-ene does not.

Explain why.

Show worked solution

E/Z isomerism requires restricted rotation about a bond (present in any C=C double bond, since rotating would require breaking the π bond) and each carbon of that double bond carrying two different attached groups.

In but-2-ene, CH₃CH=CHCH₃, both double-bond carbons each carry one methyl group and one hydrogen atom, which are different from each other, so distinct E (methyl groups on opposite sides) and Z (methyl groups on the same side) isomers exist.

In but-1-ene, CH₂=CHCH₂CH₃, one of the double-bond carbons carries two hydrogen atoms, which are identical to one another; swapping two identical groups on the same carbon gives back the same structure rather than a genuinely different isomer, so but-1-ene fails the second condition and shows no E/Z isomerism.

Worked example

A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass and has $M_{r} = 42.0$ .

Determine its molecular formula.

Show worked solution

Moles of C (per 100 g):

= \frac{85.7}{12.0} = 7.14

moles of H:

= \frac{14.3}{1.0} = 14.3

Dividing both by the smaller value: C, $1.00$ ; H, $2.00$ , giving empirical formula CH₂, with empirical formula mass:

12.0 + 2 (1.0) = 14.0

Since:

\frac{42.0}{14.0} = 3

the molecular formula is three times the empirical formula: C₃H₆. (This formula is consistent with either propene or cyclopropane; a further test, such as bromine water, would be needed to distinguish between them.)

3.3.2

Alkanes

Alkanes: structure and boiling points 3.3.2

Definitions

Alkanes: saturated hydrocarbons (only single bonds) with general formula $C_{n} H_{2 n + 2}$ .

Notes

Each carbon is tetrahedral (bond angles about 109.5°); all bonds are σ bonds.
Boiling point rises along the unbranched series because larger molecules have stronger London forces.
Branched isomers have lower boiling points than their straight-chain isomers: a more compact shape gives less surface contact between molecules, so weaker London forces.
Alkanes are saturated: every carbon forms four single bonds, so no more atoms can be added without breaking a C–C or C–H bond.
Alkanes are unreactive because C–C and C–H bonds are strong and almost non-polar, so they do not attract nucleophiles or electrophiles. They react with oxygen (combustion) and halogens (under UV light) only.
Boiling points: methane −162 °C, ethane −89 °C, propane −42 °C, butane −1 °C, pentane 36 °C. The first four are gases at room temperature, which is why they are used as fuel gases.
Cycloalkanes such as cyclohexane have the general formula $C_{n} H_{2 n}$ and are isomers of alkenes, but they are saturated.

Fractional distillation of crude oil 3.3.2.1

Notes

Crude oil is heated and vaporised and enters a column that is hottest at the bottom and coolest at the top.
Vapour rises; each component condenses where the temperature falls below its boiling point. Larger molecules with higher boiling points condense lower down.
Each fraction is a mixture of hydrocarbons with similar boiling points: refinery gases, petrol, naphtha, kerosene, diesel, heavy fuel oil and bitumen.
Fractional distillation is a physical separation: no bonds within the molecules are broken.
Crude oil is a mixture of mainly alkanes of many chain lengths; individual compounds are separated further only where needed. Fractions are named by use: refinery gases (fuel, bottled gas), petrol (cars), naphtha (petrochemical feedstock), kerosene (jet fuel), diesel (lorries, trains), fuel oil (ships, power stations) and bitumen (roads, roofing).
The bubble caps on each tray let rising vapour pass through the condensed liquid, so vapour and liquid repeatedly exchange heat: lower-boiling components re-evaporate and higher-boiling ones condense.
Residues with very high boiling points are distilled again under reduced pressure (vacuum distillation), which lowers boiling points so they can be separated without cracking.
Larger molecules in lower fractions are more viscous, darker and harder to ignite, and burn with smokier flames.

Cracking 3.3.2.2

Definitions

Cracking: breaking C–C bonds in large alkane molecules to form smaller alkanes and alkenes.

Key results

Example: $C_{10} H_{22} \to C_{6} H_{14} + C_{4} H_{8}$ .
Thermal cracking: high temperature and high pressure; gives a high proportion of alkenes.
Catalytic cracking: high temperature, slight pressure, zeolite catalyst; gives motor fuels with branched alkanes and aromatic hydrocarbons.

Notes

Cracking matches supply to demand: heavy fractions are abundant but less useful, while smaller molecules make better fuels and alkenes are feedstocks for polymers.
Thermal cracking proceeds through free radicals formed by homolytic fission of C–C bonds; conditions are typically 700–1200 K and up to 7000 kPa, with short reaction times to stop the products decomposing completely.
Catalytic cracking uses a zeolite catalyst at about 720 K and slightly above atmospheric pressure; it is cheaper to run because it needs a lower temperature, and it proceeds through carbocations.
Worked example of balancing: dodecane can crack as $C_{12} H_{26} \to C_{8} H_{18} + 2 C_{2} H_{4}$ . Check that carbon and hydrogen balance, and that at least one product is an alkene (there are not enough H atoms for all products to be alkanes).
Branched and cyclic hydrocarbons burn more smoothly in engines than straight-chain ones, which is why the motor-fuel products of catalytic cracking are valuable.

Combustion 3.3.2.3

Key results

Complete: $CH_{4} + 2 O_{2} \to CO_{2} + 2 H_{2} O$ .
Incomplete: $2 CH_{4} + 3 O_{2} \to 2 CO + 4 H_{2} O$ , or with less oxygen still $CH_{4} + O_{2} \to C + 2 H_{2} O$ .

Notes

Alkanes are used as fuels because combustion is highly exothermic.
Carbon monoxide is toxic (it binds to haemoglobin); soot (carbon particulates) causes respiratory problems and global dimming.
Worked example: complete combustion of octane, $C_{8} H_{18} + 12 \frac{1}{2} O_{2} \to 8 CO_{2} + 9 H_{2} O$ . Balance C first, then H, then O, using a half if necessary (or doubling everything).
Incomplete combustion occurs when the oxygen supply is limited, as in engines and faulty boilers. Carbon monoxide is colourless and odourless, so detectors are needed.
Carbon dioxide is a greenhouse gas: burning fossil fuels releases carbon that has been locked away for millions of years, increasing atmospheric $CO_{2}$ and contributing to climate change (see infrared absorption, 3.3.6).
Larger alkanes need more oxygen per molecule and so burn incompletely more readily, producing smoky flames.

Pollutants and their removal 3.3.2.3

Key results

Catalytic converter (Pt, Pd, Rh): $2 CO + 2 NO \to 2 CO_{2} + N_{2}$ ; unburned hydrocarbons are also oxidised.
Flue-gas desulfurisation: $CaO + SO_{2} \to CaSO_{3}$ ; $CaCO_{3} + SO_{2} \to CaSO_{3} + CO_{2}$ .

Notes

Nitrogen oxides form when nitrogen and oxygen from the air react at the high temperature in an engine; they contribute to acid rain and photochemical smog.
Sulfur impurities in fuel burn to $SO_{2}$ , an acidic gas that causes acid rain; basic calcium compounds remove it.
Converters do not remove $CO_{2}$ , a greenhouse gas.
Nitrogen monoxide forms in the engine, $N_{2} + O_{2} \to 2 NO$ , and in air is oxidised to $NO_{2}$ . The oxides of nitrogen form nitric acid in rain and contribute to photochemical smog and ground-level ozone.
Unburned hydrocarbons also contribute to photochemical smog; the catalytic converter oxidises them, e.g. $C_{8} H_{18} + 25 NO \to 8 CO_{2} + 12 \frac{1}{2} N_{2} + 9 H_{2} O$ .
A catalytic converter works only when warm and is poisoned by lead, which is one reason leaded petrol was phased out. The ceramic honeycomb gives a large surface with a thin, cheap coating of the metals.
Sulfur is now largely removed from fuels before combustion, so less $SO_{2}$ forms in the first place; flue-gas desulfurisation treats emissions from power stations.

Chlorination: free-radical substitution 3.3.2.4

Key results

Initiation: $Cl_{2} \to 2 Cl ∙$ (UV light).
Propagation: $Cl ∙ + CH_{4} \to HCl + ∙ CH_{3}$ ; $∙ CH_{3} + Cl_{2} \to CH_{3} Cl + Cl ∙$ .
Termination: $2 Cl ∙ \to Cl_{2}$ ; $∙ CH_{3} + Cl ∙ \to CH_{3} Cl$ ; $2 ∙ CH_{3} \to C_{2} H_{6}$ .

Notes

The propagation steps regenerate the chlorine radical, so one initiation can produce many product molecules: a chain reaction.
Further substitution gives a mixture ( $CH_{2} Cl_{2}$ , $CHCl_{3}$ , $CCl_{4}$ ). An excess of methane makes chloromethane the main product.
UV light supplies the energy for homolytic fission of the relatively weak Cl–Cl bond.
The same mechanism works with bromine, and with other alkanes. Ethane gives chloroethane via $Cl ∙ + C_{2} H_{6} \to HCl + ∙ C_{2} H_{5}$ then $∙ C_{2} H_{5} + Cl_{2} \to C_{2} H_{5} Cl + Cl ∙$ .
Further substitution example: $Cl ∙ + CH_{3} Cl \to HCl + ∙ CH_{2} Cl$ , then $∙ CH_{2} Cl + Cl_{2} \to CH_{2} Cl_{2} + Cl ∙$ . Write these to show where dichloromethane comes from.
Longer alkanes give mixtures of structural isomers, because a hydrogen on any carbon can be replaced; propane gives both 1-chloropropane and 2-chloropropane. Free-radical substitution is therefore a poor way to make a single pure product.
In the propagation steps, the dot goes on the atom with the unpaired electron: write $∙ CH_{3}$ , not $CH_{3} ∙$ attached ambiguously.

3.3.3

Halogenoalkanes

Halogenoalkanes and C–X bond enthalpy 3.3.3.1

Key results

Primary, secondary and tertiary: the carbon bonded to the halogen has one, two or three carbon neighbours.
Bond enthalpy: C–F > C–Cl > C–Br > C–I; reactivity in substitution increases from fluoro to iodo compounds.
Bond enthalpies (kJ mol⁻¹): C–F 467, C–Cl 346, C–Br 290, C–I 228.

Notes

The C–X bond is polar ( $C^{δ +}$ – $X^{δ -}$ ), making the carbon open to attack by nucleophiles.
Bond enthalpy, not polarity, explains the reactivity order: C–F is the most polar but the strongest, so fluoroalkanes are the least reactive; the weak C–I bond breaks most easily.
Experimental comparison: warm each halogenoalkane with aqueous silver nitrate in ethanol. Water acts as the nucleophile, and the halide released precipitates as AgX. Iodoalkanes give a yellow precipitate fastest and chloroalkanes a white one slowest, confirming that bond enthalpy controls the rate.
Ethanol is used as a co-solvent because halogenoalkanes do not mix with water; it lets both reagents come into contact.
Halogenoalkanes are useful synthetic intermediates precisely because the halogen is easily replaced; they are made from alkanes (radical substitution), alkenes (addition of HX) or alcohols.

Nucleophilic substitution: hydroxide and cyanide 3.3.3.1

Key results

With aqueous NaOH or KOH, heated under reflux: $CH_{3} CH_{2} Br + OH^{-} \to CH_{3} CH_{2} OH + Br^{-}$ .
With KCN in ethanol and water, heated under reflux: $CH_{3} CH_{2} Br + CN^{-} \to CH_{3} CH_{2} CN + Br^{-}$ (propanenitrile).

Notes

Mechanism: the nucleophile's lone pair attacks the $δ +$ carbon; the C–Br bonding pair moves onto bromine, which leaves as bromide.
Cyanide attacks through carbon, so the chain gains one carbon atom: a useful way to lengthen a carbon chain.
In the mechanism draw: the lone pair on $OH^{-}$ (or the carbon of $CN^{-}$ ), an arrow from it to the $δ +$ carbon, the $δ + / δ -$ dipole on C–Br, and an arrow from the C–Br bond to the Br atom. Then show the products with $Br^{-}$ .
Reflux means heating with a vertical condenser so that volatile reactants and solvent condense and return to the flask, allowing prolonged heating without loss.
Nitriles are named by counting the CN carbon as part of the chain: $CH_{3} CN$ is ethanenitrile and $CH_{3} CH_{2} CN$ propanenitrile. They can be hydrolysed to carboxylic acids or reduced to amines (3.3.14).
With aqueous hydroxide some elimination also occurs, especially for secondary and tertiary halogenoalkanes, so yields of alcohol are rarely 100%.

Nucleophilic substitution with ammonia 3.3.3.1

Key results

Excess ethanolic ammonia, heated in a sealed vessel: $CH_{3} CH_{2} Br + 2 NH_{3} \to CH_{3} CH_{2} NH_{2} + NH_{4} Br$ .

Notes

Step 1: ammonia's lone pair attacks the carbon, forming an ethylammonium ion. Step 2: a second ammonia molecule removes a proton, giving ethylamine.
Excess ammonia is used because the amine product is itself a nucleophile and would otherwise react further to give secondary and tertiary amines and quaternary ammonium salts.
A sealed vessel keeps the volatile ammonia in the reaction mixture.
Mechanism details: in step 1, show the lone pair on N and the arrow to carbon, with the C–Br bond breaking; the product is $CH_{3} CH_{2} NH_{3}^{+}$ . In step 2, a lone pair on a second $NH_{3}$ attacks an H on the $N^{+}$ , and the N–H bonding pair moves back onto nitrogen.
Amines are named by the alkyl group plus -amine (ethylamine, or ethanamine in IUPAC style); the $NH_{2}$ group is called amino as a prefix, as in 2-aminopropanoic acid.
This reaction links to 3.3.11, where amines react further with halogenoalkanes to give secondary and tertiary amines and quaternary ammonium salts.

Elimination 3.3.3.2

Key results

Hot ethanolic KOH: $CH_{3} CHBrCH_{3} + OH^{-} \to CH_{2} = CHCH_{3} + H_{2} O + Br^{-}$ .

Notes

Here hydroxide acts as a base: it removes a proton from a carbon next to the C–Br carbon; that C–H pair forms the C=C, and the C–Br pair leaves with bromine.
Substitution and elimination compete. Aqueous conditions and lower temperature favour substitution (alcohol); ethanolic conditions and higher temperature favour elimination (alkene).
Tertiary halogenoalkanes favour elimination more than primary ones.
Mechanism: draw the lone pair on $OH^{-}$ attacking an H on the carbon next to C–Br; an arrow from that C–H bond to between the two carbons (forming C=C); and an arrow from the C–Br bond to Br.
Unsymmetrical halogenoalkanes can give more than one alkene: 2-bromobutane gives but-1-ene and both E and Z but-2-ene, depending on which neighbouring carbon loses a hydrogen.
Elimination is useful for making alkenes, which can then be polymerised or converted into other products by addition.
Hydroxide is the same species in substitution and elimination: as a nucleophile it attacks carbon; as a base it removes a proton. Which role dominates depends on the conditions and the structure of the halogenoalkane.

CFCs and ozone depletion 3.3.3.3

Key results

UV in the stratosphere breaks C–Cl bonds homolytically: $CF_{2} Cl_{2} \to ∙ CF_{2} Cl + Cl ∙$ .
$Cl ∙ + O_{3} \to ClO ∙ + O_{2}$ ; $ClO ∙ + O_{3} \to 2 O_{2} + Cl ∙$ . Overall $2 O_{3} \to 3 O_{2}$ .

Notes

The chlorine radical is regenerated, so it acts as a catalyst and one radical can destroy thousands of ozone molecules.
The ozone layer absorbs harmful UV that causes skin cancer and cataracts. Scientific evidence of ozone loss led to international restrictions on CFCs (the Montreal Protocol) and chlorine-free alternatives such as HFCs.
Ozone forms naturally in the stratosphere, $O_{2} \to 2 O$ (UV) then $O + O_{2} \to O_{3}$ , and absorbs UV-B and UV-C, so a balance of formation and destruction normally keeps its concentration roughly constant.
CFCs were chosen as refrigerants, aerosol propellants and solvents because they are unreactive, non-toxic and non-flammable; that same stability lets them survive long enough to reach the stratosphere.
C–F bonds are too strong to be broken by the UV that reaches the stratosphere; it is the weaker C–Cl bond that breaks, which is why HFCs (no chlorine) do not deplete ozone.
Nitrogen monoxide from aircraft engines can catalyse ozone breakdown in the same way: $NO + O_{3} \to NO_{2} + O_{2}$ , then $NO_{2} + O \to NO + O_{2}$ .

3.3.4

Alkenes

Structure and reactivity of alkenes 3.3.4.1

Definitions

Alkenes: unsaturated hydrocarbons containing a C=C double bond; general formula $C_{n} H_{2 n}$ .
Electrophile: an electron-pair acceptor.

Notes

The double bond is one σ bond plus one π bond, formed by sideways overlap of p orbitals above and below the plane of the molecule.
Each carbon is trigonal planar (about 120°); in ethene all six atoms are in one plane.
The π bond is a region of high electron density that attracts electrophiles, and it prevents rotation about the C=C bond.
The π bond is weaker than the σ bond (C=C 612 kJ mol⁻¹ compared with C–C 348), so it breaks first in addition reactions, while the σ bond remains.
Alkenes are much more reactive than alkanes: the exposed π electrons are both a concentrated negative charge and relatively easily polarised.
Addition reactions of alkenes have 100% atom economy, which makes them valuable in industry (e.g. making ethanol, poly(ethene) and halogenoalkanes).
Restricted rotation about the C=C bond, caused by the π bond's position above and below the plane, is the origin of E–Z isomerism (3.3.1).

Electrophilic addition of hydrogen bromide 3.3.4.2

Key results

Propene + HBr gives 2-bromopropane (major) and 1-bromopropane (minor).

Notes

Step 1: the π electrons attack the $δ +$ H of HBr; the H–Br pair moves onto Br, forming $Br^{-}$ and a carbocation. Step 2: $Br^{-}$ donates a lone pair to the carbocation.
Carbocation stability: tertiary > secondary > primary, because alkyl groups push electron density towards the positive carbon. The more stable secondary carbocation forms more readily, so 2-bromopropane is the major product.
With a symmetrical alkene such as ethene only one product is possible.
Mechanism details: show the $δ +$ H and $δ -$ Br on H–Br; an arrow from the C=C to the H; an arrow from the H–Br bond to Br; the carbocation with its + on the correct carbon; and an arrow from a lone pair on $Br^{-}$ to the positive carbon.
A tertiary carbocation is the most stable because three alkyl groups each release electron density (a positive inductive effect), spreading the positive charge. 2-methylpropene with HBr gives mainly 2-bromo-2-methylpropane.
The reaction with HBr works at room temperature with gaseous hydrogen bromide; HCl and HI react similarly. The rate increases from HCl to HI because the H–X bond becomes weaker.
Markovnikov's rule (H adds to the carbon that already has more hydrogens) is a description of this outcome; the explanation is carbocation stability.

Addition of sulfuric acid 3.3.4.2

Key results

Cold concentrated $H_{2} SO_{4}$ with ethene gives ethyl hydrogensulfate, $CH_{3} CH_{2} OSO_{3} H$ ; warming with water then hydrolyses it to ethanol, regenerating the acid.

Notes

The mechanism mirrors HBr addition: protonation by the acid gives a carbocation, and the hydrogensulfate ion attacks through an oxygen lone pair.
Overall this is a route from an alkene to an alcohol, with sulfuric acid acting as a catalyst.
Mechanism: the C=C attacks an H of $H_{2} SO_{4}$ (the O–H bond breaking towards O), forming a carbocation and $HSO_{4}^{-}$ ; a lone pair on an oxygen of $HSO_{4}^{-}$ then bonds to the carbocation.
With propene, the more stable secondary carbocation forms, so propan-2-ol is the main alcohol produced after hydrolysis.
This is the laboratory-scale route to alcohols from alkenes; the industrial route, hydration with steam and a phosphoric acid catalyst, follows the same idea of protonation followed by attack on a carbocation (3.3.5).
Overall equations: $C_{2} H_{4} + H_{2} SO_{4} \to CH_{3} CH_{2} OSO_{2} OH$ , then $CH_{3} CH_{2} OSO_{2} OH + H_{2} O \to CH_{3} CH_{2} OH + H_{2} SO_{4}$ .

Addition of bromine and the test for alkenes 3.3.4.2

Key results

$CH_{2} = CH_{2} + Br_{2} \to CH_{2} BrCH_{2} Br$ (1,2-dibromoethane).
Test: shaking an alkene with bromine water turns it from orange to colourless.

Notes

Bromine is non-polar, but the π electrons repel the electrons in Br–Br, inducing a dipole; the $δ +$ bromine is attacked and a bromide ion leaves.
AQA mechanisms show a carbocation intermediate; the sheet's bromonium ion is the more detailed description and leads to the same product.
Decolourisation of bromine water shows unsaturation (a C=C bond); saturated alkanes give no change.
Mechanism details: draw the induced dipole $Br^{δ +} - Br^{δ -}$ ; an arrow from the C=C to the $δ +$ Br; an arrow from the Br–Br bond to the $δ -$ Br; then $Br^{-}$ attacking the carbocation.
With bromine water, water molecules compete with bromide ions for the carbocation, so some 2-bromoethanol forms as well as 1,2-dibromoethane. Either way, the orange colour disappears.
The test also works with bromine dissolved in an organic solvent; its decolourisation is used to compare the degree of unsaturation of fats and oils.
The induced dipole is the key idea: a non-polar molecule becomes an electrophile when it approaches the π electrons.

Addition polymerisation 3.3.4.3

Definitions

Addition polymer: a long-chain molecule formed when many alkene monomers join by opening their C=C bonds, with no other product.

Key results

Ethene → poly(ethene); propene → poly(propene); chloroethene → poly(chloroethene), PVC.
The repeating unit is two carbons from the monomer's C=C, with its side groups, drawn with extension bonds through the brackets.

Notes

All atoms of the monomers end up in the polymer, so atom economy is 100%.
To find the monomer from a polymer, take one repeating unit and put the C=C back between its two backbone carbons.
Worked example: but-2-ene gives a polymer whose repeating unit is $- CH(CH_{3}) - CH(CH_{3}) -$ ; propene gives $- CH_{2} - CH(CH_{3}) -$ , with the methyl groups as side chains, not in the main chain.
Polymerisation of ethene can be carried out at high pressure with an initiator (giving branched, low-density poly(ethene)) or with a Ziegler–Natta catalyst at low pressure (giving linear, high-density poly(ethene)). Different chain packing gives different properties.
Poly(tetrafluoroethene), PTFE, from $CF_{2} = CF_{2}$ , is used as a non-stick coating because its C–F bonds are very strong and unreactive.
The name of the polymer is poly(monomer): poly(chloroethene) is PVC.

Properties of addition polymers 3.3.4.3

Notes

Chains are held together only by intermolecular forces: London forces in poly(ethene) and poly(propene), plus permanent dipole–dipole forces in PVC from its polar C–Cl bonds.
Unplasticised PVC is rigid (pipes, window frames). A plasticiser pushes the chains apart, weakening the forces between them so they slide more easily, giving flexible PVC (cable insulation, flooring).
Polyalkenes are unreactive because their backbone has only strong, non-polar C–C and C–H bonds; this makes them durable but non-biodegradable.
Longer chains and less branching give stronger London forces between chains, so higher melting points and greater strength; this is why high-density poly(ethene) is stiffer than low-density poly(ethene).
Disposal options include landfill (persists for centuries), incineration (energy recovery, but toxic gases such as HCl from PVC must be removed), and recycling (mechanical recycling needs sorting by polymer type).
Biodegradable and photodegradable polymers are being developed, but most polyalkenes are not broken down by microorganisms, which have no enzymes for their non-polar backbones.
Compare with condensation polymers (3.3.12), whose polar ester and amide links can be hydrolysed.

3.3.5

Alcohols

Primary, secondary and tertiary alcohols 3.3.5

Key results

Primary: the C–OH carbon is bonded to one other carbon (ethanol). Secondary: two (propan-2-ol). Tertiary: three (2-methylpropan-2-ol).

Notes

The classification predicts oxidation behaviour, because oxidation needs a hydrogen on the C–OH carbon.
Methanol has no carbon neighbours but behaves as a primary alcohol.
Worked examples for $C_{4} H_{10} O$ alcohols: butan-1-ol (primary), butan-2-ol (secondary), 2-methylpropan-1-ol (primary) and 2-methylpropan-2-ol (tertiary).
Alcohols have much higher boiling points than alkanes of similar $M_{r}$ because their O–H groups form hydrogen bonds between molecules: ethanol boils at 78 °C, propane (similar $M_{r}$ ) at −42 °C.
Short-chain alcohols mix with water because they form hydrogen bonds with water molecules; solubility falls as the non-polar chain lengthens.
Uses: ethanol as a fuel, solvent and in drinks; methanol as a feedstock and fuel.

Producing ethanol 3.3.5.1

Key results

Hydration of ethene: $CH_{2} = CH_{2} + H_{2} O \to CH_{3} CH_{2} OH$ ; steam, phosphoric acid catalyst, about 300 °C and 60–70 atm.
Fermentation: $C_{6} H_{12} O_{6} \to 2 C_{2} H_{5} OH + 2 CO_{2}$ ; yeast, aqueous, anaerobic, about 35 °C.

Notes

Hydration is fast, continuous and gives pure ethanol, but uses a non-renewable feedstock (ethene from crude oil). Fermentation uses renewable sugar but is slow, a batch process, and gives dilute ethanol needing distillation.
Bioethanol is called carbon-neutral because the 6 $CO_{2}$ absorbed in photosynthesis equals the 2 released in fermentation plus 4 in combustion.
In practice it is not fully carbon-neutral: farming, processing and transport use energy, and growing fuel crops can compete with food and habitats.
Fermentation must be kept free of air: in the presence of oxygen, microorganisms oxidise ethanol to ethanoic acid (vinegar). Above about 15% ethanol the yeast is killed, so stronger solutions need distillation.
The optimum temperature balances enzyme activity against denaturation: below about 25 °C fermentation is slow; above about 40 °C the yeast enzymes denature.
Write the carbon-neutrality equations in full: photosynthesis $6 CO_{2} + 6 H_{2} O \to C_{6} H_{12} O_{6} + 6 O_{2}$ ; fermentation; and combustion $2 C_{2} H_{5} OH + 6 O_{2} \to 4 CO_{2} + 6 H_{2} O$ .

Mechanism of acid-catalysed hydration 3.3.5.1

Notes

Step 1: the alkene's π electrons attack $H^{+}$ , forming a carbocation (secondary for propene, so propan-2-ol is the major product).
Step 2: a water molecule attacks the carbocation through an oxygen lone pair, giving a protonated alcohol.
Step 3: the protonated alcohol loses $H^{+}$ , regenerating the acid catalyst.
The catalyst appears in the first step and is released in the last, so it is not used up; this is shown by $H^{+}$ appearing on both sides of the overall mechanism.
Industrially the catalyst is phosphoric acid adsorbed on silica, at about 300 °C and 60–70 atm with excess steam. Conversion per pass is only about 5%, so unreacted ethene is recycled, giving an overall conversion of about 95%.
The reaction is the reverse of dehydration: high pressure and excess water favour the alcohol; heating with concentrated acid and removing water favours the alkene.
For unsymmetrical alkenes, the major alcohol comes from the more stable carbocation, exactly as with HBr.

Oxidation of alcohols 3.3.5.2

Key results

Primary alcohol → aldehyde (warm, distil the product off as it forms): $CH_{3} CH_{2} OH + [O] \to CH_{3} CHO + H_{2} O$ .
Primary alcohol → carboxylic acid (excess oxidant, heat under reflux): $CH_{3} CH_{2} OH + 2 [O] \to CH_{3} COOH + H_{2} O$ .
Secondary alcohol → ketone: $CH_{3} CH(OH)CH_{3} + [O] \to CH_{3} COCH_{3} + H_{2} O$ . Tertiary alcohols are not oxidised.

Notes

The oxidising agent is acidified potassium dichromate(VI); it turns from orange ( $Cr_{2} O_{7}^{2 -}$ ) to green ( $Cr^{3 +}$ ) when an alcohol is oxidised.
Distillation removes the aldehyde before it can be oxidised further; reflux keeps it in the flask so it becomes the acid.
Tertiary alcohols have no hydrogen on the C–OH carbon, so the dichromate stays orange.
Worked equation with dichromate: $3 CH_{3} CH_{2} OH + Cr_{2} O_{7}^{2 -} + 8 H^{+} \to 3 CH_{3} CHO + 2 Cr^{3 +} + 7 H_{2} O$ . In exams [O] notation is accepted unless the full equation is asked for.
Aldehydes have lower boiling points than their alcohols (ethanal 21 °C, ethanol 78 °C) because they cannot hydrogen bond to each other, which is why distilling as it forms separates the aldehyde from unreacted alcohol.
The mixture is heated in a flask with anti-bumping granules; for distillation, a condenser sloping downwards leads to a receiver cooled in ice.

Distinguishing aldehydes and ketones 3.3.5.2

Key results

Tollens' reagent (warm): an aldehyde gives a silver mirror; a ketone gives no change.
Fehling's solution (warm): an aldehyde gives a brick-red precipitate of $Cu_{2} O$ ; a ketone leaves it blue.

Notes

Aldehydes can be oxidised further (to carboxylic acids), so they reduce these mild oxidising agents; ketones cannot be oxidised easily.
Tollens' reagent is made by adding a drop of NaOH to silver nitrate solution (brown precipitate of $Ag_{2} O$ ) and then dilute ammonia until the precipitate just dissolves, forming $[Ag(NH_{3})_{2}]^{+}$ . It is warmed with the sample in a water bath, never over a flame.
Half-equations: $Ag^{+} + e^{-} \to Ag$ (silver mirror) and $Cu^{2 +} + e^{-} \to Cu^{+}$ (in Fehling's, giving red $Cu_{2} O$ ). The aldehyde is oxidised to the carboxylic acid (as its salt in the alkaline reagents).
Acidified dichromate also distinguishes them: aldehydes turn it green, ketones do not. Primary and secondary alcohols also give a positive dichromate test, so it cannot identify an aldehyde on its own.

Dehydration of alcohols 3.3.5.3

Key results

Heat with concentrated $H_{2} SO_{4}$ or $H_{3} PO_{4}$ : $CH_{3} CH(OH)CH_{3} \to CH_{2} = CHCH_{3} + H_{2} O$ .

Notes

Mechanism: the OH is protonated; water leaves, forming a carbocation; a proton is lost from a neighbouring carbon to form the C=C, regenerating the acid.
Some alcohols give more than one alkene, including E–Z isomers, depending on which neighbouring carbon loses a proton.
Dehydrating bioethanol gives ethene from a renewable source, which can be polymerised without using crude oil.
Worked example: butan-2-ol gives three alkenes on dehydration: but-1-ene (H lost from C1) and both (E)- and (Z)-but-2-ene (H lost from C3).
Laboratory method: heat the alcohol with concentrated phosphoric acid and collect the alkene, which distils out, then purify it by washing with sodium carbonate solution (to remove acid), drying with anhydrous calcium chloride and redistilling.
Concentrated phosphoric acid is often preferred to sulfuric acid because sulfuric acid is a strong oxidising agent and gives side products such as $CO_{2}$ and $SO_{2}$ .
Elimination from alcohols and from halogenoalkanes both form C=C; the reagents and mechanisms differ but the outcome is the same.

3.3.6

Organic analysis

Tests for functional groups 3.3.6.1

Key results

Alkene: bromine water decolourised (orange → colourless).
Primary or secondary alcohol, or aldehyde: warm acidified potassium dichromate(VI) turns from orange to green.
Aldehyde: silver mirror with Tollens' reagent; brick-red precipitate with Fehling's solution.
Carboxylic acid: effervescence with sodium hydrogencarbonate; the $CO_{2}$ turns limewater milky.

Notes

Use a fresh sample for each test.
No single test identifies a compound: aldehydes also turn dichromate green, so combine observations.
Tertiary alcohols give no change with dichromate, which distinguishes them from primary and secondary alcohols.
Halogenoalkanes: warm with NaOH(aq), acidify with dilute nitric acid, then add silver nitrate; the precipitate colour identifies the halogen. The halogen must first be released as a halide ion.
Ammonium ions and amines are not on this list but can be detected by their smell and by turning damp red litmus blue.
Order the tests sensibly: use non-destructive or quick tests (bromine water, sodium hydrogencarbonate) before oxidation tests, and always describe the observation for both a positive and a negative result where asked.
A good answer names the reagent precisely (e.g. 'acidified potassium dichromate(VI)', not 'dichromate'), the conditions (warm) and the observation.

Mass spectrometry 3.3.6.2

Key results

The molecular ion peak gives $M_{r}$ : propanone's $M^{+}$ is at $m / z$ 58.
High-resolution mass spectrometry measures precise masses: $C_{3} H_{6} O$ is 58.0419 while $C_{4} H_{10}$ is 58.0783, so only one formula fits.

Notes

The tallest (base) peak is set at 100%; it need not be the molecular ion. For propanone it is $CH_{3} CO^{+}$ at 43, from loss of a methyl radical.
Precise masses work because isotopic masses are not whole numbers ( $^{1} H$ = 1.007825, $^{16} O$ = 15.994915, $^{12} C$ = 12.000000 exactly).
Fragmentation: the molecular ion splits into a fragment ion and a radical, $M^{+∙} \to X^{+} + Y^{∙}$ ; only the charged fragment is detected. For propanone, loss of $CH_{3}^{∙}$ (mass 15) gives $CH_{3} CO^{+}$ at 43.
Common fragments: 15 ( $CH_{3}^{+}$ ), 29 ( $C_{2} H_{5}^{+}$ or $CHO^{+}$ ), 43 ( $C_{3} H_{7}^{+}$ or $CH_{3} CO^{+}$ ), 45 ( $COOH^{+}$ ). Fragment patterns distinguish isomers that have the same molecular ion.
A small M+1 peak arises from molecules containing one $^{13} C$ atom (1.1% of carbon); compounds with Cl or Br show characteristic M+2 peaks.

Infrared spectroscopy 3.3.6.3

Key results

O–H (alcohol): 3230–3550 cm⁻¹, broad. O–H (acid): 2500–3000 cm⁻¹, very broad.
C=O: 1680–1750 cm⁻¹, strong. C–O: 1000–1300 cm⁻¹. C=C: 1620–1680 cm⁻¹. C≡N: 2220–2260 cm⁻¹.

Notes

Bonds absorb IR at characteristic wavenumbers, which appear as dips in transmittance.
Below about 1500 cm⁻¹ is the fingerprint region: the pattern is unique to a compound and is matched against a reference spectrum.
Hydrogen bonding broadens O–H absorptions; the very broad acid O–H band overlapping the C–H region, together with C=O, is characteristic of a carboxylic acid.
Spectra plot transmittance (%) against wavenumber (cm⁻¹, from 4000 down to about 500). An absorption appears as a downward trough; strong, broad and sharp are part of how each band is described.
Worked reading: a broad band at 3300 cm⁻¹ with a strong band at 1050 cm⁻¹ and no C=O suggests an alcohol; a very broad band from 3300 to 2500 cm⁻¹ plus a strong 1710 cm⁻¹ band is a carboxylic acid; a strong 1720 cm⁻¹ band with no O–H is an aldehyde, ketone or ester.
IR is used to monitor reactions: oxidising an alcohol to a ketone removes the O–H band and adds a C=O band. Breathalysers detect ethanol by its C–H absorption.
Use the data booklet values in exams rather than memorised ones; they are given as ranges because the exact wavenumber depends on the rest of the molecule.

Combining the evidence 3.3.6

Notes

Worked identification: precise mass 58.0419 gives $C_{3} H_{6} O$ ; IR shows C=O but no O–H; Tollens' gives no silver mirror. So the compound is propanone, not propanal.
Each technique answers a different question: mass spectrometry gives the formula, IR the bonds present, chemical tests the functional group's behaviour.
Another worked identification: a compound has $M_{r}$ 60 and percentage composition 60.0% C, 13.3% H, 26.7% O, giving $C_{3} H_{8} O$ . IR shows a broad band at 3350 cm⁻¹ (O–H, alcohol). It is oxidised by acidified dichromate to a product that gives no silver mirror, so it is a secondary alcohol: propan-2-ol.
Work systematically: molecular formula first (from composition and $M_{r}$ ), then degree of unsaturation (compare with the saturated formula), then functional groups (IR and chemical tests), then the carbon skeleton (fragments, NMR in 3.3.15).
State the evidence for each conclusion explicitly. Examiners award marks for linking each observation to an inference, not just for the final structure.
If two structures remain possible, suggest a test that would distinguish them and say what each would show.

Infrared absorption and the greenhouse effect 3.3.6.3

Notes

The Earth emits infrared radiation; $CO_{2}$ , $CH_{4}$ and $H_{2} O$ absorb some of it because their bonds vibrate at the same frequencies.
Only vibrations that change the molecule's dipole moment absorb IR, which is why $N_{2}$ and $O_{2}$ do not contribute.
Absorbed energy is re-emitted in all directions, some back towards the surface, warming it: the greenhouse effect.
The Sun's radiation reaches the surface largely as visible light; the warmed surface re-emits energy at longer, infrared wavelengths, which greenhouse gases absorb. That asymmetry is why the gases warm the lower atmosphere.
Methane is a more potent greenhouse gas than carbon dioxide molecule for molecule, but carbon dioxide's much larger emissions make it the main contributor to human-caused warming. Water vapour is the most abundant greenhouse gas, but its concentration is controlled by temperature rather than directly by emissions.
The evidence that $CO_{2}$ concentration has risen comes from direct atmospheric measurement since 1958 and from air trapped in ice cores; the rise correlates with fossil-fuel use.
Reducing the greenhouse effect involves cutting emissions (renewable energy, efficiency, carbon-neutral fuels) and capturing $CO_{2}$ , for example by storing it underground.